Palindrome detection: Difference between revisions
| Line 3,881: | Line 3,881: | ||
} |
} |
||
?></lang> |
?></lang> |
||
=={{header|Picat}}== |
|||
<lang Picat>go => |
|||
Tests = ["In girum imus nocte et consumimur igni", |
|||
"this is a non palindrome string", |
|||
"anna ABcdcBA anna", |
|||
"anna ABcdcBA annax", |
|||
"A man, a plan, a canoe, pasta, heros, rajahs" ++ |
|||
"a coloratura, maps, snipe, percale, macaroni, " ++ |
|||
"a gag, a banana bag, a tan, a tag, " ++ |
|||
"a banana bag again (or a camel), a crepe, pins, " ++ |
|||
"Spam, a rut, a Rolo, cash, a jar, sore hats, " ++ |
|||
"a peon, a canal - Panama!", |
|||
10, |
|||
111111, |
|||
12221, |
|||
9384212, |
|||
10.01 |
|||
], |
|||
foreach(Test in Tests) |
|||
if is_palindrome(Test) then |
|||
println([Test, "exact palindrome"]) |
|||
elseif is_palindrome_inexact(Test) then |
|||
println([Test, "inexact palindrome"]) |
|||
else |
|||
println([Test, "no"]) |
|||
end |
|||
end, |
|||
nl. |
|||
% Detect palindromes for strings (and numbers). |
|||
is_palindrome(N), number(N) => is_palindrome(N.to_string()). |
|||
is_palindrome(S) => S == S.reverse(). |
|||
% Detect inexact palindromes. |
|||
is_palindrome_inexact(N), number(N) => is_palindrome_inexact(N.to_string()). |
|||
is_palindrome_inexact(S) => |
|||
is_palindrome(strip(S)).</lang> |
|||
{{out}} |
|||
<pre>[In girum imus nocte et consumimur igni,inexact palindrome] |
|||
[this is a non palindrome string,no] |
|||
[anna ABcdcBA anna,exact palindrome] |
|||
[anna ABcdcBA annax,no] |
|||
[A man, a plan, a canoe, pasta, heros, rajahsa coloratura, maps, snipe, percale, macaroni, a gag, a banana bag, a tan, a tag, a banana bag again (or a camel), a crepe, pins, Spam, a rut, a Rolo, cash, a jar, sore hats, a peon, a canal - Panama!,inexact palindrome] |
|||
[10,no] |
|||
[11,exact palindrome] |
|||
[111111,exact palindrome] |
|||
[12221,exact palindrome] |
|||
[9384212,no] |
|||
[10.01,exact palindrome]</pre> |
|||
=={{header|PicoLisp}}== |
=={{header|PicoLisp}}== |
||
Revision as of 07:09, 8 May 2022
You are encouraged to solve this task according to the task description, using any language you may know.
A palindrome is a phrase which reads the same backward and forward.
Write a function or program that checks whether a given sequence of characters (or, if you prefer, bytes) is a palindrome.
For extra credit:
- Support Unicode characters.
- Write a second function (possibly as a wrapper to the first) which detects inexact palindromes, i.e. phrases that are palindromes if white-space and punctuation is ignored and case-insensitive comparison is used.
- It might be useful for this task to know how to reverse a string.
- This task's entries might also form the subjects of the task Test a function.
- Metrics
- Counting
- Word frequency
- Letter frequency
- Jewels and stones
- I before E except after C
- Bioinformatics/base count
- Count occurrences of a substring
- Count how many vowels and consonants occur in a string
- Remove/replace
- XXXX redacted
- Conjugate a Latin verb
- Remove vowels from a string
- String interpolation (included)
- Strip block comments
- Strip comments from a string
- Strip a set of characters from a string
- Strip whitespace from a string -- top and tail
- Strip control codes and extended characters from a string
- Anagrams/Derangements/shuffling
- Word wheel
- ABC problem
- Sattolo cycle
- Knuth shuffle
- Ordered words
- Superpermutation minimisation
- Textonyms (using a phone text pad)
- Anagrams
- Anagrams/Deranged anagrams
- Permutations/Derangements
- Find/Search/Determine
- ABC words
- Odd words
- Word ladder
- Semordnilap
- Word search
- Wordiff (game)
- String matching
- Tea cup rim text
- Alternade words
- Changeable words
- State name puzzle
- String comparison
- Unique characters
- Unique characters in each string
- Extract file extension
- Levenshtein distance
- Palindrome detection
- Common list elements
- Longest common suffix
- Longest common prefix
- Compare a list of strings
- Longest common substring
- Find common directory path
- Words from neighbour ones
- Change e letters to i in words
- Non-continuous subsequences
- Longest common subsequence
- Longest palindromic substrings
- Longest increasing subsequence
- Words containing "the" substring
- Sum of the digits of n is substring of n
- Determine if a string is numeric
- Determine if a string is collapsible
- Determine if a string is squeezable
- Determine if a string has all unique characters
- Determine if a string has all the same characters
- Longest substrings without repeating characters
- Find words which contains all the vowels
- Find words which contains most consonants
- Find words which contains more than 3 vowels
- Find words which first and last three letters are equals
- Find words which odd letters are consonants and even letters are vowels or vice_versa
- Formatting
- Substring
- Rep-string
- Word wrap
- String case
- Align columns
- Literals/String
- Repeat a string
- Brace expansion
- Brace expansion using ranges
- Reverse a string
- Phrase reversals
- Comma quibbling
- Special characters
- String concatenation
- Substring/Top and tail
- Commatizing numbers
- Reverse words in a string
- Suffixation of decimal numbers
- Long literals, with continuations
- Numerical and alphabetical suffixes
- Abbreviations, easy
- Abbreviations, simple
- Abbreviations, automatic
- Song lyrics/poems/Mad Libs/phrases
- Mad Libs
- Magic 8-ball
- 99 Bottles of Beer
- The Name Game (a song)
- The Old lady swallowed a fly
- The Twelve Days of Christmas
- Tokenize
- Text between
- Tokenize a string
- Word break problem
- Tokenize a string with escaping
- Split a character string based on change of character
- Sequences
11l
<lang 11l>F is_palindrome(s)
R s == reversed(s)</lang>
360 Assembly
<lang 360asm>* Reverse b string 25/06/2018 PALINDRO CSECT
USING PALINDRO,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) prolog
ST R13,4(R15) "
ST R15,8(R13) "
LR R13,R15 "
LA R8,BB @b[1]
LA R9,AA+L'AA-1 @a[n-1]
LA R6,1 i=1
LOOPI C R6,=A(L'AA) do i=1 to length(a)
BH ELOOPI leave i
MVC 0(1,R8),0(R9) substr(b,i,1)=substr(a,n-i+1,1)
LA R8,1(R8) @b=@b+1
BCTR R9,0 @a=@a-1
LA R6,1(R6) i=i+1
B LOOPI end do
ELOOPI XPRNT AA,L'AA print a
CLC BB,AA if b=a
BNE SKIP
XPRNT MSG,L'MSG then print msg
SKIP L R13,4(0,R13) epilog
LM R14,R12,12(R13) "
XR R15,R15 "
BR R14 exit
AA DC CL32'INGIRUMIMUSNOCTEETCONSUMIMURIGNI' a BB DS CL(L'AA) b MSG DC CL23'IT IS A TRUE PALINDROME'
YREGS
END PALINDRO</lang>
- Output:
INGIRUMIMUSNOCTEETCONSUMIMURIGNI IT IS A TRUE PALINDROME
8080 Assembly
<lang 8080asm> org 100h jmp demo ;;; Is the $-terminated string at DE a palindrome? ;;; Returns: zero flag set if palindrome palin: mov h,d ; Find end of string mov l,e mvi a,'$' cmp m ; The empty string is a palindrome rz pend: inx h ; Scan until terminator found cmp m jnz pend dcx h ; Move to last byte of text ptest: ldax d ; Load char at left pointer cmp m ; Compare to char at right pointer rnz ; If not equal, not a palindrome inx d ; Move pointers dcx h mov a,d ; Check if left pointer is before right pointer cmp h ; High byte jc ptest mov a,e ; Low byte cmp l jc ptest xra a ; Made it to the end - set zero flag ret ; Return ;;; Test the routine on a few examples demo: lxi h,words ; Word list pointer loop: mov e,m ; Load word pointer inx h mov d,m inx h mov a,e ; Stop when zero reached ora d rz push h ; Keep word list pointer call pstr ; Print word call palin ; Check if palindrome lxi d,no jnz print ; Print "no" if not a palindrome lxi d,yes ; Print "yes" otherwise print: call pstr pop h jmp loop ;;; Print strint using CP/M keeping DEHL registers pstr: push d push h mvi c,9 call 5 pop h pop d ret yes: db ': yes',13,10,'$' no: db ': no',13,10,'$' words: dw w1,w2,w3,w4,0 w1: db 'rotor$' w2: db 'racecar$' w3: db 'level$' w4: db 'rosetta$'</lang>
- Output:
rotor: yes racecar: yes level: yes rosetta: no
8086 Assembly
<lang asm> cpu 8086 org 100h section .text jmp demo ;;; Check if the $-terminated string in [DS:SI] is a palindrome. ;;; Returns with zero flag set if so. ;;; Destroyed: AL, CX, SI, DI, ES. palin: push es ; Set ES=DS. pop ds mov al,'$' ; Find end of string mov cx,-1 mov di,si repne scasb dec di ; Move back to last actual character .loop: cmp si,di ja .ok ; If SI > DI, it is a palindrome lodsb dec di ; Compare left character to right character cmp al,[di] jne .no ; If not equal, not a palindrome jmp .loop ; Otherwise, try next pair of characters .ok: cmp al,al ; Set zero flag .no: ret ; Return ;;; Try the routine on a couple of strings demo: mov si,words .loop: lodsw ; Grab word pointer test ax,ax ; Zero? jz .done ; Then we are done mov dx,ax ; Otherwise, print word mov ah,9 int 21h xchg bp,si ; Keep array pointer in BP xchg si,dx ; Put word pointer in SI call palin ; Check if it is a palindrome mov dx,yes ; Print 'yes'... jz .print ; ...if it is a palindrome mov dx,no ; Otherwise, print 'no' .print: int 21h xchg si,bp ; Restore array pointer jmp .loop ; Get next word. .done: ret yes: db ': yes',13,10,'$' ; Yes and no no: db ': no',13,10,'$' words: dw .w1,.w2,.w3,.w4,.w5,0 .w1: db 'rotor$' ; Words to check .w2: db 'racecar$' .w3: db 'level$' .w4: db 'redder$' .w5: db 'rosetta$'</lang>
- Output:
rotor: yes racecar: yes level: yes redder: yes rosetta: no
ACL2
<lang Lisp>(defun reverse-split-at-r (xs i ys)
(if (zp i)
(mv xs ys)
(reverse-split-at-r (rest xs) (1- i)
(cons (first xs) ys))))
(defun reverse-split-at (xs i)
(reverse-split-at-r xs i nil))
(defun is-palindrome (str)
(let* ((lngth (length str))
(idx (floor lngth 2)))
(mv-let (xs ys)
(reverse-split-at (coerce str 'list) idx)
(if (= (mod lngth 2) 1)
(equal (rest xs) ys)
(equal xs ys)))))</lang>
Action!
<lang Action!>BYTE FUNC Palindrome(CHAR ARRAY s)
BYTE l,r
l=1 r=s(0) WHILE l<r DO IF s(l)#s(r) THEN RETURN (0) FI l==+1 r==-1 OD
RETURN (1)
BYTE FUNC IsIgnored(BYTE c)
IF (c>=' AND c<='/) OR
(c>=': AND c<='@) OR
(c>='[ AND c<='_) THEN
RETURN (1)
FI
RETURN (0)
BYTE FUNC ToUpper(BYTE c)
IF c>='a AND c<='z THEN RETURN (c-'a+'A) FI
RETURN (c)
BYTE FUNC InexactPalindrome(CHAR ARRAY s)
BYTE l,r,lc,rc
l=1 r=s(0)
WHILE l<r
DO
WHILE IsIgnored(s(l))
DO
l==+1
IF l>=r THEN RETURN (1) FI
OD
WHILE IsIgnored(s(r))
DO
r==-1
IF l>=r THEN RETURN (1) FI
OD
lc=ToUpper(s(l)) rc=ToUpper(s(r))
IF lc#rc THEN RETURN (0) FI l==+1 r==-1 OD
RETURN (1)
PROC Test(CHAR ARRAY s)
IF Palindrome(s) THEN
PrintF("'%S' is a palindrome%E%E",s)
ELSEIF InexactPalindrome(s) THEN
PrintF("'%S' is an inexact palindrome%E%E",s)
ELSE
PrintF("'%S' is not a palindrome%E%E",s)
FI
RETURN
PROC Main()
Test("rotavator")
Test("13231+464+989=989+464+13231")
Test("Was it a car or a cat I saw?")
Test("Did Hannah see bees? Hannah did.")
Test("This sentence is not a palindrome.")
Test("123 456 789 897 654 321")
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
'rotavator' is a palindrome '13231+464+989=989+464+13231' is a palindrome 'Was it a car or a cat I saw?' is an inexact palindrome 'Did Hannah see bees? Hannah did.' is an inexact palindrome 'This sentence is not a palindrome.' is not a palindrome '123 456 789 897 654 321' is not a palindrome
ActionScript
The following function handles non-ASCII characters properly, since charAt() returns a single Unicode character. <lang ActionScript>function isPalindrome(str:String):Boolean { for(var first:uint = 0, second:uint = str.length - 1; first < second; first++, second--) if(str.charAt(first) != str.charAt(second)) return false; return true; }</lang>
Ada
<lang ada>function Palindrome (Text : String) return Boolean is begin
for Offset in 0..Text'Length / 2 - 1 loop
if Text (Text'First + Offset) /= Text (Text'Last - Offset) then
return False;
end if;
end loop;
return True;
end Palindrome;</lang>
Ada 2012 version: <lang ada> function Palindrome (Text : String) return Boolean is (for all i in Text'Range => Text(i)= Text(Text'Last-i+Text'First)); </lang>
ALGOL 68
<lang algol68># Iterative # PROC palindrome = (STRING s)BOOL:(
FOR i TO UPB s OVER 2 DO
IF s[i] /= s[UPB s-i+1] THEN GO TO return false FI
OD;Power
else: TRUE EXIT
return false: FALSE
);
- Recursive #
PROC palindrome r = (STRING s)BOOL:
IF LWB s >= UPB s THEN TRUE ELIF s[LWB s] /= s[UPB s] THEN FALSE ELSE palindrome r(s[LWB s+1:UPB s-1]) FI
- Test #
main: (
STRING t = "ingirumimusnocteetconsumimurigni";
FORMAT template = $"sequence """g""" "b("is","isnt")" a palindrome"l$;
printf((template, t, palindrome(t)));
printf((template, t, palindrome r(t)))
)</lang>
- Output:
sequence "ingirumimusnocteetconsumimurigni" is a palindrome sequence "ingirumimusnocteetconsumimurigni" is a palindrome
APL
NARS2000 APL, dynamic function "if the argument matches the reverse of the argument", with Unicode character support: <lang APL> {⍵≡⌽⍵} 'abc' 0
{⍵≡⌽⍵} '⍋racecar⍋'
1</lang> Or in tacit function form, a combination of three functions, right tack (echo), reverse, then the result of each compared with the middle one, match (equals): <lang APL> (⊢≡⌽) 'abc' 0
(⊢≡⌽) 'nun'
1</lang> An inexact version is harder, because uppercase and lowercase with Unicode awareness depends on APL interpreter; NARS2000 has no support for it. Classic case conversion means lookup up the letters in an alphabet of UppercaseLowercase, then mapping those positions into an UppercaseUppercase or LowercaseLowercase array. Remove non-A-Za-z first to get rid of punctuation, and get an inexact dynamic function with just English letter support: <lang APL>inexact←{Aa←(⎕A,⎕a) ⋄ (⊢≡⌽)(⎕a,⎕a)[Aa⍳⍵/⍨⍵∊Aa]}
inexact 'abc,-cbA2z'
0
inexact 'abc,-cbA2'
1</lang> Dyalog APL has a Unicode-aware uppercase/lowercase function (819 I-beam), AFAIK no support for looking up Unicode character classes.
AppleScript
Using post-Yosemite AppleScript (to pull in lowercaseStringWithLocale from Foundation classes) <lang AppleScript>use framework "Foundation"
CASE-INSENSITIVE PALINDROME, IGNORING SPACES ? ----
-- isPalindrome :: String -> Bool on isPalindrome(s)
s = intercalate("", reverse of characters of s)
end isPalindrome
-- toSpaceFreeLower :: String -> String on spaceFreeToLower(s)
script notSpace
on |λ|(s)
s is not space
end |λ|
end script
intercalate("", filter(notSpace, characters of toLower(s)))
end spaceFreeToLower
TEST -------------------------
on run
isPalindrome(spaceFreeToLower("In girum imus nocte et consumimur igni"))
--> true
end run
GENERIC FUNCTIONS -------------------
-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter
-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- toLower :: String -> String
on toLower(str)
set ca to current application
((ca's NSString's stringWithString:(str))'s ¬
lowercaseStringWithLocale:(ca's NSLocale's currentLocale())) as text
end toLower</lang>
- Output:
true
Core language only
It's not clear if "sequence of characters" means an array thereof or a single piece of text. But the basic method in AppleScript would be: <lang applescript>on isPalindrome(txt)
set txt to join(txt, "") -- In case the input's a list (array). return (txt = join(reverse of txt's characters, ""))
end isPalindrome
on join(lst, delim)
set astid to AppleScript's text item delimiters set AppleScript's text item delimiters to delim set txt to lst as text set AppleScript's text item delimiters to astid return txt
end join
return isPalindrome("Radar")</lang>
Text comparisons in AppleScript are case-insensitive by default, so:
- Output:
<lang applescript>true</lang>
If case is to be taken into account, the call to the handler can be enclosed in a 'considering case' control statement. <lang applescript>considering case
return isPalindrome("Radar")
end considering</lang>
- Output:
<lang applescript>false</lang>
It's also possible to "ignore" white space, hyphens, and punctuation, which are considered by default. And of course case can be ignored explicitly, if desired, to ensure that this condition's in force during the call to the handler. The attributes can be combined in one statement. So to check for inexact palindromicity (if that's a word):
<lang applescript>ignoring case, white space, hyphens and punctuation
return isPalindrome("Was it a 😀car, or a c😀at-I-saw?")
end ignoring</lang>
- Output:
<lang applescript>true</lang>
Applesoft BASIC
<lang ApplesoftBasic>100 DATA"MY DOG HAS FLEAS" 110 DATA"MADAM, I'M ADAM." 120 DATA"1 ON 1" 130 DATA"IN GIRUM IMUS NOCTE ET CONSUMIMUR IGNI" 140 DATA"A man, a plan, a canal: Panama!" 150 DATA"KAYAK" 160 DATA"REDDER" 170 DATA"H" 180 DATA""
200 FOR L1 = 1 TO 9 210 READ W$ : GOSUB 300" IS PALINDROME? 220 PRINT CHR$(34); W$; CHR$(34); " IS "; 230 IF NOT PALINDROME THEN PRINT "NOT "; 240 PRINT "A PALINDROME" 250 NEXT 260 END
300 REMIS PALINDROME? 310 PA = 1 320 L = LEN(W$) 330 IF L = 0 THEN RETURN 340 FOR L0 = 1 TO L / 2 + .5 350 PA = MID$(W$, L0, 1) = MID$(W$, L - L0 + 1, 1) 360 IF PALINDROME THEN NEXT L0 370 RETURN</lang>
ARM Assembly
<lang>@ Check whether the ASCII string in [r0] is a palindrome @ Returns with zero flag set if palindrome. palin: mov r1,r0 @ Find end of string 1: ldrb r2,[r1],#1 @ Grab character and increment pointer tst r2,r2 @ Zero yet? bne 1b @ If not try next byte sub r1,r1,#2 @ Move R1 to last actual character. 2: cmp r0,r1 @ When R0 >= R1, cmpgt r2,r2 @ make sure zero is set, bxeq lr @ and stop (the string is a palindrome). ldrb r2,[r1],#-1 @ Grab [R1] (end) and decrement. ldrb r3,[r0],#1 @ Grab [R0] (begin) and increment cmp r2,r3 @ Are they equal? bxne lr @ If not, it's not a palindrome. b 2b @ Otherwise, try next pair.
@ Try the function on a couple of strings .global _start _start: ldr r8,=words @ Word pointer 1: ldr r9,[r8],#4 @ Grab word and move pointer tst r9,r9 @ Null? moveq r7,#1 @ Then we're done; syscall 1 = exit swieq #0 mov r1,r9 @ Print the word bl print mov r0,r9 @ Test if the word is a palindrome bl palin ldreq r1,=yes @ "Yes" if it is a palindrome ldrne r1,=no @ "No" if it's not a palindrome bl print b 1b @ Next word
@ Print zero-terminated string [r1] using Linux syscall print: push {r7,lr} @ Keep R7 and link register mov r2,r1 @ Find end of string 1: ldrb r0,[r2],#1 @ Grab character and increment pointer tst r0,r0 @ Zero yet? bne 1b @ If not, keep going sub r2,r2,r1 @ Calculate length of string (bytes to write) mov r0,#1 @ Stdout = 1 mov r7,#4 @ Syscall 4 = write swi #0 @ Make the syscall pop {r7,lr} @ Restore R7 and link register bx lr
@ Strings yes: .asciz ": yes\n" @ Output yes or no no: .asciz ": no\n" w1: .asciz "rotor" @ Words to test w2: .asciz "racecar" w3: .asciz "level" w4: .asciz "redder" w5: .asciz "rosetta" words: .word w1,w2,w3,w4,w5,0</lang>
- Output:
rotor: yes racecar: yes level: yes redder: yes rosetta: no
Arturo
<lang rebol>palindrome?: $[seq] -> seq = reverse seq
loop ["abba" "boom" "radar" "civic" "great"] 'wrd [ print [wrd ": palindrome?" palindrome? wrd] ]</lang>
- Output:
abba : palindrome? true boom : palindrome? false radar : palindrome? true civic : palindrome? true great : palindrome? false
AutoHotkey
Reversing the string: <lang AutoHotkey>IsPalindrome(Str){ Loop, Parse, Str ReversedStr := A_LoopField . ReversedStr return, (ReversedStr == Str)?"Exact":(RegExReplace(ReversedStr,"\W")=RegExReplace(Str,"\W"))?"Inexact":"False" }</lang>
AutoIt
<lang AutoIt>;== AutoIt Version: 3.3.8.1
Global $aString[7] = [ _ "In girum imus nocte, et consumimur igni", _ ; inexact palindrome "Madam, I'm Adam.", _ ; inexact palindrome "salàlas", _ ; exact palindrome "radar", _ ; exact palindrome "Lagerregal", _ ; exact palindrome "Ein Neger mit Gazelle zagt im Regen nie.", _ ; inexact palindrome "something wrong"] ; no palindrome Global $sSpace42 = " "
For $i = 0 To 6 If _IsPalindrome($aString[$i]) Then ConsoleWrite('"' & $aString[$i] & '"' & StringLeft($sSpace42, 42-StringLen($aString[$i])) & 'is an exact palindrome.' & @LF) Else If _IsPalindrome( StringRegExpReplace($aString[$i], '\W', ) ) Then ConsoleWrite('"' & $aString[$i] & '"' & StringLeft($sSpace42, 42-StringLen($aString[$i])) & 'is an inexact palindrome.' & @LF) Else ConsoleWrite('"' & $aString[$i] & '"' & StringLeft($sSpace42, 42-StringLen($aString[$i])) & 'is not a palindrome.' & @LF) EndIf EndIf Next
Func _IsPalindrome($_string) Local $iLen = StringLen($_string) For $i = 1 To Int($iLen/2) If StringMid($_string, $i, 1) <> StringMid($_string, $iLen-($i-1), 1) Then Return False Next Return True EndFunc </lang>
- Output:
<lang Text> "In girum imus nocte, et consumimur igni" is an inexact palindrome. "Madam, I'm Adam." is an inexact palindrome. "salàlas" is an exact palindrome. "radar" is an exact palindrome. "Lagerregal" is an exact palindrome. "Ein Neger mit Gazelle zagt im Regen nie." is an inexact palindrome. "something wrong" is not a palindrome. </lang>
--BugFix (talk) 14:26, 13 November 2013 (UTC)
AWK
Non-recursive
See Reversing a string.
<lang awk>function is_palindro(s) {
if ( s == reverse(s) ) return 1 return 0
}</lang>
Recursive
<lang awk>function is_palindro_r(s) {
if ( length(s) < 2 ) return 1 if ( substr(s, 1, 1) != substr(s, length(s), 1) ) return 0 return is_palindro_r(substr(s, 2, length(s)-2))
}</lang>
Testing <lang awk>BEGIN {
pal = "ingirumimusnocteetconsumimurigni" print is_palindro(pal) print is_palindro_r(pal)
}</lang>
BaCon
<lang freebasic> OPTION COMPARE TRUE
INPUT "Enter your line... ", word$
IF word$ = REVERSE$(word$) THEN
PRINT "This is an exact palindrome!"
ELIF EXTRACT$(word$, "punct:|blank:", TRUE) = REVERSE$(EXTRACT$(word$, "punct:|blank:", TRUE)) THEN
PRINT "This is an inexact palindrome!"
ELSE
PRINT "Not a palindrome."
ENDIF </lang>
- Output:
Enter your line... In girum imus nocte, et consumimur igni This is an inexact palindrome! Enter your line... Madam, I'm Adam. This is an inexact palindrome! Enter your line... radar This is an exact palindrome! Enter your line... Something else Not a palindrome.
Bash
<lang bash>
- ! /bin/bash
- very simple way to detect a palindrome in Bash
- output of bash --version -> GNU bash, version 4.4.7(1)-release x86_64 ...
echo "enter a string" read input
size=${#input} count=0
while (($count < $size)) do
array[$count]=${input:$count:1}
(( count+=1 ))
done
count=0
for ((i=0 ; i < $size; i+=1)) do
if [ "${array[$i]}" == "${array[$size - $i - 1]}" ]
then
(( count += 1 ))
fi
done
if (( $count == $size )) then
echo "$input is a palindrome"
fi </lang>
BASIC
<lang qbasic>' OPTION _EXPLICIT ' For QB64. In VB-DOS remove the underscore.
DIM txt$
' Palindrome CLS PRINT "This is a palindrome detector program." PRINT INPUT "Please, type a word or phrase: ", txt$
IF IsPalindrome(txt$) THEN
PRINT "Is a palindrome."
ELSE
PRINT "Is Not a palindrome."
END IF
END
FUNCTION IsPalindrome (AText$)
' Var DIM CleanTXT$, RvrsTXT$
CleanTXT$ = CleanText$(AText$) RvrsTXT$ = RvrsText$(CleanTXT$)
IsPalindrome = (CleanTXT$ = RvrsTXT$)
END FUNCTION
FUNCTION CleanText$ (WhichText$)
' Var DIM i%, j%, c$, NewText$, CpyTxt$, AddIt%, SubsTXT$ CONST False = 0, True = NOT False
SubsTXT$ = "AIOUE" CpyTxt$ = UCASE$(WhichText$) j% = LEN(CpyTxt$)
FOR i% = 1 TO j% c$ = MID$(CpyTxt$, i%, 1)
' See if it is a letter. Includes Spanish letters.
SELECT CASE c$
CASE "A" TO "Z"
AddIt% = True
CASE " ", "¡", "¢", "£"
c$ = MID$(SubsTXT$, ASC(c$) - 159, 1)
AddIt% = True
CASE "‚"
c$ = "E"
AddIt% = True
CASE "¤"
c$ = "¥"
AddIt% = True
CASE ELSE
AddIt% = False
END SELECT
IF AddIt% THEN
NewText$ = NewText$ + c$
END IF
NEXT i%
CleanText$ = NewText$
END FUNCTION
FUNCTION RvrsText$ (WhichText$)
' Var DIM i%, c$, NewText$, j%
j% = LEN(WhichText$) FOR i% = 1 TO j% NewText$ = MID$(WhichText$, i%, 1) + NewText$ NEXT i%
RvrsText$ = NewText$
END FUNCTION</lang>
- Output:
This is a palindrome detector program.
Please, type a word or phrase: Madam, I'm Adam. Is a palindrome.
This is a palindrome detector program.
Please, type a word or phrase: This is just a test. Is not a palindrome.
IS-BASIC
<lang IS-BASIC> 100 PROGRAM "Palindr.bas" 110 LINE INPUT PROMPT "Text: ":TX$ 120 PRINT """";TX$;""" is "; 130 IF PALIND(TX$) THEN 140 PRINT "a palindrome." 150 ELSE 160 PRINT "not a palindrome." 170 END IF 180 DEF TRIM$(TX$) 190 LET T$="" 200 FOR I=1 TO LEN(TX$) 210 IF TX$(I)>="A" AND TX$(I)<="Z" THEN LET T$=T$&TX$(I) 220 NEXT 230 LET TRIM$=T$ 240 END DEF 250 DEF PALIND(TX$) 260 LET PALIND=-1:LET TX$=TRIM$(UCASE$(TX$)) 270 FOR I=1 TO LEN(TX$)/2 280 IF TX$(I)<>TX$(LEN(TX$)-I+1) THEN LET PALIND=0:EXIT FOR 290 NEXT 300 END DEF</lang>
Sinclair ZX81 BASIC
Exact palindrome
The specification suggests, but does not insist, that we reverse the input string and then test for equality; this algorithm is more efficient. <lang basic> 10 INPUT S$
20 FOR I=1 TO LEN S$/2 30 IF S$(I)<>S$(LEN S$-I+1) THEN GOTO 60 40 NEXT I 50 GOTO 70 60 PRINT "NOT A "; 70 PRINT "PALINDROME"</lang>
Inexact palindrome
Add the following lines to convert the program into an inexact-palindrome checker (i.e. one that ignores non-alphabetic characters). The resulting program still works with only 1k of RAM. The ZX81 only supports its own character set, which does not include lower case, so that case-insensitive comparison and a fortiori Unicode are not possible. <lang basic> 15 GOSUB 90
80 STOP 90 LET T$=""
100 FOR I=1 TO LEN S$ 110 IF S$(I)>="A" AND S$(I)<="Z" THEN LET T$=T$+S$(I) 120 NEXT I 130 LET S$=T$ 140 RETURN</lang>
BBC BASIC
<lang bbcbasic> test$ = "A man, a plan, a canal: Panama!"
PRINT """" test$ """" ;
IF FNpalindrome(FNletters(test$)) THEN
PRINT " is a palindrome"
ELSE
PRINT " is not a palindrome"
ENDIF
END
DEF FNpalindrome(A$) = (A$ = FNreverse(A$))
DEF FNreverse(A$)
LOCAL B$, P%
FOR P% = LEN(A$) TO 1 STEP -1
B$ += MID$(A$,P%,1)
NEXT
= B$
DEF FNletters(A$)
LOCAL B$, C%, P%
FOR P% = 1 TO LEN(A$)
C% = ASC(MID$(A$,P%))
IF C% > 64 AND C% < 91 OR C% > 96 AND C% < 123 THEN
B$ += CHR$(C% AND &5F)
ENDIF
NEXT
= B$</lang>
- Output:
"A man, a plan, a canal: Panama!" is a palindrome
Batch File
<lang dos>@echo off setlocal enabledelayedexpansion set /p string=Your string : set count=0
- loop
if "!%string%:~%count%,1!" neq "" ( set reverse=!%string%:~%count%,1!!reverse! set /a count+=1 goto loop ) set palindrome=isn't if "%string%"=="%reverse%" set palindrome=is echo %string% %palindrome% a palindrome. pause exit</lang>
Or, recursive (and without setlocal enabledelayedexpansion) (compatible with ReactOS cmd.exe) <lang dos>@echo off set /p testString=Your string (all same case please) : call :isPalindrome result %testString: =% if %result%==1 echo %testString% is a palindrome if %result%==0 echo %testString% isn't a palindrome pause goto :eof
- isPalindrome
set %1=0 set string=%2 if "%string:~2,1%"=="" ( set %1=1 goto :eof ) if "%string:~0,1%"=="%string:~-1%" ( call :isPalindrome %1 %string:~1,-1% ) goto :eof</lang>
BCPL
<lang bcpl>get "libhdr"
let palindrome(s) = valof $( let l = s%0
for i = 1 to l/2
unless s%i = s%(l+1-i)
resultis false
resultis true
$)
let inexact(s) = valof $( let temp = vec 1+256/BYTESPERWORD
temp%0 := 0
for i = 1 to s%0 do
$( let ch = s%i | 32
if '0'<=ch & ch<='9' | 'a'<=ch & ch<='z' then
$( temp%0 := temp%0 + 1
temp%(temp%0) := ch
$)
$)
resultis palindrome(temp)
$)
let check(s) =
palindrome(s) -> "exact palindrome", inexact(s) -> "inexact palindrome", "not a palindrome"
let start() be $( let tests = vec 8
tests!0 := "rotor"
tests!1 := "racecar"
tests!2 := "RACEcar"
tests!3 := "level"
tests!4 := "redder"
tests!5 := "rosetta"
tests!6 := "A man, a plan, a canal: Panama"
tests!7 := "Egad, a base tone denotes a bad age"
tests!8 := "This is not a palindrome"
for i = 0 to 8 do
writef("'%S': %S*N", tests!i, check(tests!i))
$)</lang>
- Output:
'rotor': exact palindrome 'racecar': exact palindrome 'RACEcar': inexact palindrome 'level': exact palindrome 'redder': exact palindrome 'rosetta': not a palindrome 'A man, a plan, a canal: Panama': inexact palindrome 'Egad, a base tone denotes a bad age': inexact palindrome 'This is not a palindrome': not a palindrome
Befunge
The following code reads a line from stdin and prints "True" if it is a palindrome, or False" otherwise. <lang befunge>v_$0:8p>:#v_:18p08g1-08p >:08g`!v ~->p5p ^ 0v1p80-1g80vj!-g5g80g5_0'ev
- a^80+1:g8<>8g1+:18pv>0"eslaF">:#,_@
relet-2010------>003-x -^"Tru"<</lang>
To check a string, replace "dennis sinned" with your own string.
Note that this has some limits.:
- There must be a quotation mark immediately after the string, and then nothing but spaces for the rest of that line.
- The v at the end of that same line must remain immediately above the 2. (Very important.) The closing quotation mark can be against the v, but can't replace it.
- The potential palindrome can be no longer than 76 characters (which beats the previous version's 11), and everything (spaces, punctuation, capitalization, etc.) is considered part of the palindrome. (Best to just use lower case letters and nothing else.)
<lang befunge>v> "emordnilap a toN",,,,,,,,,,,,,,,,@,,,,,,,,,,,,,,,"Is a palindrome" < 2^ < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < 4 ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v 8 ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v
- ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v
+ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""
>"dennis sinned" v
" 2 """""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""" 0
> ^- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 9
_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ p
v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^
v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^
^< < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < <
>09g8p09g1+09pv
|: < <
^<</lang>
BQN
3 functions in three different styles to check if a string is a palindrome. All three forms return 1 for palindrome, and 0 for non-palindrome.
BQN considers characters as single units, and hence the functions support unicode by default.
<lang bqn>Pal ← ≡⊸⌽ Pal1 ← ⊢≡⌽ Pal2 ← {𝕩≡⌽𝕩}</lang>
Bracmat
<lang bracmat>( ( palindrome
= a
. @(!arg:(%?a&utf$!a) ?arg !a)
& palindrome$!arg
| utf$!arg
)
& ( desep
= x
. @(!arg:?x (" "|"-"|",") ?arg)
& !x desep$!arg
| !arg
)
& "In girum imus nocte et consumimur igni"
"Я иду с мечем, судия"
"The quick brown fox"
"tregða, gón, reiði - er nóg að gert"
"人人為我,我為人人"
"가련하시다 사장집 아들딸들아 집장사 다시 하련가"
: ?candidates
& whl
' ( !candidates:%?candidate ?candidates
& out
$ ( !candidate
is
( palindrome$(low$(str$(desep$!candidate)))
& indeed
| not
)
a
palindrome
)
)
& );</lang> Output:
In girum imus nocte et consumimur igni is indeed a palindrome Я иду с мечем, судия is indeed a palindrome The quick brown fox is not a palindrome tregða, gón, reiði - er nóg að gert is indeed a palindrome 人人為我,我為人人 is indeed a palindrome 가련하시다 사장집 아들딸들아 집장사 다시 하련가 is indeed a palindrome
Burlesque
<lang burlesque> zz{ri}f[^^<-== </lang>
C
Non-recursive
This function compares the first char with the last, the second with the one previous the last, and so on. The first different pair it finds, return 0 (false); if all the pairs were equal, then return 1 (true). You only need to go up to (the length) / 2 because the second half just re-checks the same stuff as the first half; and if the length is odd, the middle doesn't need to be checked (so it's okay to do integer division by 2, which rounds down).
<lang c>#include <string.h>
int palindrome(const char *s) {
int i,l;
l = strlen(s);
for(i=0; i<l/2; i++)
{
if ( s[i] != s[l-i-1] ) return 0;
}
return 1;
}</lang> More idiomatic version: <lang c>int palindrome(const char *s) {
const char *t; /* t is a pointer that traverses backwards from the end */
for (t = s; *t != '\0'; t++) ; t--; /* set t to point to last character */
while (s < t)
{
if ( *s++ != *t-- ) return 0;
}
return 1;
}</lang>
Recursive
A single char is surely a palindrome; a string is a palindrome if first and last char are the same and the remaining string (the string starting from the second char and ending to the char preceding the last one) is itself a palindrome.
<lang c>int palindrome_r(const char *s, int b, int e) {
if ( (e - 1) <= b ) return 1; if ( s[b] != s[e-1] ) return 0; return palindrome_r(s, b+1, e-1);
}</lang>
Testing
<lang c>#include <stdio.h>
- include <string.h>
/* testing */ int main() {
const char *t = "ingirumimusnocteetconsumimurigni";
const char *template = "sequence \"%s\" is%s palindrome\n";
int l = strlen(t);
printf(template,
t, palindrome(t) ? "" : "n't");
printf(template,
t, palindrome_r(t, 0, l) ? "" : "n't");
return 0;
}</lang>
C#
Non-recursive
<lang csharp>using System;
class Program {
static string Reverse(string value)
{
char[] chars = value.ToCharArray();
Array.Reverse(chars);
return new string(chars);
}
static bool IsPalindrome(string value)
{
return value == Reverse(value);
}
static void Main(string[] args)
{
Console.WriteLine(IsPalindrome("ingirumimusnocteetconsumimurigni"));
}
}</lang>
Using LINQ operators <lang csharp>using System; using System.Linq;
class Program { static bool IsPalindrome(string text) { return text == new String(text.Reverse().ToArray()); }
static void Main(string[] args) { Console.WriteLine(IsPalindrome("ingirumimusnocteetconsumimurigni")); } } </lang>
No string reversal
Reversing a string is very slow. A much faster way is to simply compare characters. <lang csharp>using System;
static class Program {
//As an extension method (must be declared in a static class)
static bool IsPalindrome(this string sentence)
{
for (int l = 0, r = sentence.Length - 1; l < r; l++, r--)
if (sentence[l] != sentence[r]) return false;
return true;
}
static void Main(string[] args)
{
Console.WriteLine("ingirumimusnocteetconsumimurigni".IsPalindrome());
}
}</lang>
C++
The C solutions also work in C++, but C++ allows a simpler one: <lang cpp>#include <string>
- include <algorithm>
bool is_palindrome(std::string const& s) {
return std::equal(s.begin(), s.end(), s.rbegin());
}</lang>
Or, checking half is sufficient (on odd-length strings, this will ignore the middle element): <lang cpp>#include <string>
- include <algorithm>
bool is_palindrome(std::string const& s) {
return std::equal(s.begin(), s.begin()+s.length()/2, s.rbegin());
}</lang>
Clojure
<lang clojure>(defn palindrome? [s]
(= s (clojure.string/reverse s)))</lang>
lower-level, but somewhat faster <lang clojure>(defn palindrome? [^String s]
(loop [front 0 back (dec (.length s))]
(or (>= front back)
(and (= (.charAt s front) (.charAt s back))
(recur (inc front) (dec back)))))</lang>
Test
user=> (palindrome? "amanaplanacanalpanama") true user=> (palindrome? "Test 1, 2, 3") false
CLU
<lang clu>% Reverse a string str_reverse = proc (s: string) returns (string)
chs: array[char] := array[char]$predict(0, string$size(s))
for c: char in string$chars(s) do
array[char]$addl(chs, c)
end
return (string$ac2s(chs))
end str_reverse
% 'Normalize' a string (remove everything but letters and make uppercase) normalize = proc (s: string) returns (string)
chs: array[char] := array[char]$predict(0, string$size(s))
for c: char in string$chars(s) do
if c>='a' cand c<='z' then
c := char$i2c(char$c2i(c) - 32)
end
if c>='A' cand c<='Z' then
array[char]$addh(chs, c)
end
end
return (string$ac2s(chs))
end normalize
% Check if a string is an exact palindrome palindrome = proc (s: string) returns (bool)
return (s = str_reverse(s))
end palindrome
% Check if a string is an inexact palindrome inexact_palindrome = proc (s: string) returns (bool)
return (palindrome(normalize(s)))
end inexact_palindrome
% Test cases start_up = proc ()
po: stream := stream$primary_output()
tests: array[string] := array[string]$[
"rotor", "racecar", "RACEcar", "level", "rosetta",
"A man, a plan, a canal: Panama",
"Egad, a base tone denotes a bad age",
"This is not a palindrome"
]
for test: string in array[string]$elements(tests) do
stream$puts(po, "\"" || test || "\": ")
if palindrome(test) then
stream$putl(po, "exact palindrome")
elseif inexact_palindrome(test) then
stream$putl(po, "inexact palindrome")
else
stream$putl(po, "not a palindrome")
end
end
end start_up</lang>
- Output:
"rotor": exact palindrome "racecar": exact palindrome "RACEcar": inexact palindrome "level": exact palindrome "rosetta": not a palindrome "A man, a plan, a canal: Panama": inexact palindrome "Egad, a base tone denotes a bad age": inexact palindrome "This is not a palindrome": not a palindrome
COBOL
<lang COBOL> identification division.
function-id. palindromic-test.
data division.
linkage section.
01 test-text pic x any length.
01 result pic x.
88 palindromic value high-value
when set to false low-value.
procedure division using test-text returning result.
set palindromic to false
if test-text equal function reverse(test-text) then
set palindromic to true
end-if
goback.
end function palindromic-test.
</lang>
CoffeeScript
<lang coffeescript>
String::isPalindrome = ->
for i in [0...@length / 2] when @[i] isnt @[@length - (i + 1)]
return no
yes
String::stripped = -> @toLowerCase().replace /\W/gi,
console.log "'#{ str }' : #{ str.stripped().isPalindrome() }" for str in [
'In girum imus nocte et consumimur igni'
'A man, a plan, a canal: Panama!'
'There is no spoon.'
]
</lang>
- Output:
'In girum imus nocte et consumimur igni' : true 'A man, a plan, a canal: Panama!' : true 'There is no spoon.' : false
Common Lisp
<lang lisp>(defun palindrome-p (s)
(string= s (reverse s)))</lang>
Alternate solution
I use Allegro CL 10.1
<lang lisp>
- Project
- Palindrome detection
(defun palindrome(x)
(if (string= x (reverse x))
(format t "~d" ": palindrome" (format t x))
(format t "~d" ": not palindrome" (format t x))))
(terpri) (setq x "radar") (palindrome x) (terpri) (setq x "books") (palindrome x) (terpri) </lang> Output:
radar: palindrome books: not palindrome
Component Pascal
BlackBox Component Builder <lang oberon2> MODULE BbtPalindrome; IMPORT StdLog;
PROCEDURE ReverseStr(str: ARRAY OF CHAR): POINTER TO ARRAY OF CHAR; VAR top,middle,i: INTEGER; c: CHAR; rStr: POINTER TO ARRAY OF CHAR; BEGIN NEW(rStr,LEN(str$) + 1); top := LEN(str$) - 1; middle := (top - 1) DIV 2; FOR i := 0 TO middle DO rStr[i] := str[top - i]; rStr[top - i] := str[i]; END; IF ODD(LEN(str$)) THEN rStr[middle + 1] := str[middle + 1] END; RETURN rStr; END ReverseStr;
PROCEDURE IsPalindrome(str: ARRAY OF CHAR): BOOLEAN; BEGIN RETURN str = ReverseStr(str)$; END IsPalindrome;
PROCEDURE Do*;
VAR
x: CHAR;
BEGIN
StdLog.String("'salalas' is palindrome?:> ");
StdLog.Bool(IsPalindrome("salalas"));StdLog.Ln;
StdLog.String("'madamimadam' is palindrome?:> ");
StdLog.Bool(IsPalindrome("madamimadam"));StdLog.Ln;
StdLog.String("'abcbda' is palindrome?:> ");
StdLog.Bool(IsPalindrome("abcbda"));StdLog.Ln;
END Do;
END BbtPalindrome.
</lang>
Execute: ^Q BbtPalindrome.Do
- Output:
'salalas' is palindrome?:> $TRUE 'madamimadam' is palindrome?:> $TRUE 'abcbda' is palindrome?:> $FALSE
Cowgol
<lang cowgol>include "cowgol.coh";
- Check if a string is a palindrome
sub palindrome(word: [uint8]): (r: uint8) is
r := 1;
# empty string is a palindrome
if [word] == 0 then
return;
end if;
# find the end of the word
var end_ := word;
while [@next end_] != 0 loop
end_ := @next end_;
end loop;
# check if bytes match in both directions
while word < end_ loop
if [word] != [end_] then
r := 0;
return;
end if;
word := @next word;
end_ := @prev end_;
end loop;
end sub;
- Check if a string is an inexact palindrome
sub inexact(word: [uint8]): (r: uint8) is
var buf: uint8[256];
var ptr := &buf[0];
# filter non-letters and non-numbers
while [word] != 0 loop
var c := [word];
if (c >= 'a' and c <= 'z') or (c >= '0' and c <= '9') then
# copy lowercase letters and numbers over verbatim
[ptr] := c;
ptr := @next ptr;
elseif c >= 'A' and c <= 'Z' then
# make uppercase letters lowercase
[ptr] := c | 32;
ptr := @next ptr;
end if;
word := @next word;
end loop;
[ptr] := 0;
r := palindrome(&buf[0]);
end sub;
var tests: [uint8][] := {
"civic", "level", "racecar", "A man, a plan, a canal: Panama", "Egad, a base tone denotes a bad age", "There is no spoon."
};
var i: @indexof tests := 0; while i < @sizeof tests loop
print(tests[i]);
print(": ");
if palindrome(tests[i]) == 1 then
print("exact palindrome\n");
elseif inexact(tests[i]) == 1 then
print("inexact palindrome\n");
else
print("not a palindrome\n");
end if;
i := i + 1;
end loop;</lang>
- Output:
civic: exact palindrome level: exact palindrome racecar: exact palindrome A man, a plan, a canal: Panama: inexact palindrome Egad, a base tone denotes a bad age: inexact palindrome There is no spoon.: not a palindrome
Crystal
Declarative
<lang Ruby> def palindrome(s)
s == s.reverse
end </lang>
Imperative
<lang Ruby> def palindrome_imperative(s) : Bool
mid = s.size // 2
last = s.size - 1
(0...mid).each do |i|
if s[i] != s[last - i]
return false
end
end
true
end </lang>
Also
<lang Ruby>def palindrome_2(s)
mid = s.size // 2
mid.times { |j| return false if s[j] != s[-1 - j] }
true
end</lang>
Performance comparison <lang Ruby> require "benchmark" Benchmark.ips do |x|
x.report("declarative") { palindrome("hannah") }
x.report("imperative1") { palindrome_imperative("hannah")}
x.report("imperative2") { palindrome_2("hannah")}
end </lang>
declarative 45.45M ( 22.00ns) (±11.16%) 32.0B/op fastest imperative1 35.49M ( 28.18ns) (± 2.82%) 0.0B/op 1.28× slower imperative2 40.73M ( 24.55ns) (± 3.82%) 0.0B/op 1.12× slower
D
High-level 32-bit Unicode Version
<lang d>import std.traits, std.algorithm;
bool isPalindrome1(C)(in C[] s) pure /*nothrow*/ if (isSomeChar!C) {
auto s2 = s.dup; s2.reverse(); // works on Unicode too, not nothrow. return s == s2;
}
void main() {
alias pali = isPalindrome1;
assert(pali(""));
assert(pali("z"));
assert(pali("aha"));
assert(pali("sees"));
assert(!pali("oofoe"));
assert(pali("deified"));
assert(!pali("Deified"));
assert(pali("amanaplanacanalpanama"));
assert(pali("ingirumimusnocteetconsumimurigni"));
assert(pali("salà las"));
}</lang>
Mid-level 32-bit Unicode Version
<lang d>import std.traits;
bool isPalindrome2(C)(in C[] s) pure if (isSomeChar!C) {
dchar[] dstr;
foreach (dchar c; s) // not nothrow
dstr ~= c;
for (int i; i < dstr.length / 2; i++)
if (dstr[i] != dstr[$ - i - 1])
return false;
return true;
}
void main() {
alias isPalindrome2 pali;
assert(pali(""));
assert(pali("z"));
assert(pali("aha"));
assert(pali("sees"));
assert(!pali("oofoe"));
assert(pali("deified"));
assert(!pali("Deified"));
assert(pali("amanaplanacanalpanama"));
assert(pali("ingirumimusnocteetconsumimurigni"));
assert(pali("salà las"));
}</lang>
Low-level 32-bit Unicode Version
<lang d>import std.stdio, core.exception, std.traits;
// assume alloca() to be pure for this program extern(C) pure nothrow void* alloca(in size_t size);
bool isPalindrome3(C)(in C[] s) pure if (isSomeChar!C) {
auto p = cast(dchar*)alloca(s.length * 4);
if (p == null)
// no fallback heap allocation used
throw new OutOfMemoryError();
dchar[] dstr = p[0 .. s.length];
// use std.utf.stride for an even lower level version
int i = 0;
foreach (dchar c; s) { // not nothrow
dstr[i] = c;
i++;
}
dstr = dstr[0 .. i];
foreach (j; 0 .. dstr.length / 2)
if (dstr[j] != dstr[$ - j - 1])
return false;
return true;
}
void main() {
alias isPalindrome3 pali;
assert(pali(""));
assert(pali("z"));
assert(pali("aha"));
assert(pali("sees"));
assert(!pali("oofoe"));
assert(pali("deified"));
assert(!pali("Deified"));
assert(pali("amanaplanacanalpanama"));
assert(pali("ingirumimusnocteetconsumimurigni"));
assert(pali("salà las"));
}</lang>
Low-level ASCII Version
<lang d>bool isPalindrome4(in string str) pure nothrow {
if (str.length == 0) return true;
immutable(char)* s = str.ptr;
immutable(char)* t = &(str[$ - 1]);
while (s < t)
if (*s++ != *t--) // ugly
return false;
return true;
}
void main() {
alias isPalindrome4 pali;
assert(pali(""));
assert(pali("z"));
assert(pali("aha"));
assert(pali("sees"));
assert(!pali("oofoe"));
assert(pali("deified"));
assert(!pali("Deified"));
assert(pali("amanaplanacanalpanama"));
assert(pali("ingirumimusnocteetconsumimurigni"));
//assert(pali("salà las"));
}</lang>
Dart
<lang dartlang> bool isPalindrome(String s){
for(int i = 0; i < s.length/2;i++){
if(s[i] != s[(s.length-1) -i])
return false;
}
return true;
} </lang>
Delphi
<lang Delphi>uses
SysUtils, StrUtils;
function IsPalindrome(const aSrcString: string): Boolean; begin
Result := SameText(aSrcString, ReverseString(aSrcString));
end;</lang>
Dyalect
<lang dyalect>func isPalindrom(str) {
str == str.Reverse()
}
print(isPalindrom("ingirumimusnocteetconsumimurigni"))</lang>
Déjà Vu
<lang dejavu>palindrome?: local :seq chars local :len-seq -- len seq
for i range 0 / len-seq 2: if /= seq! i seq! - len-seq i: return false true
!. palindrome? "ingirumimusnocteetconsumimurigni" !. palindrome? "nope"</lang>
- Output:
true false
E
It is only necessarily to scan the first half of the string, upper(0, upper.size() // 2), and compare each character to the corresponding character from the other end, upper[last - i].
The for loop syntax is for key pattern => value pattern in collection { ... }, ? imposes an additional boolean condition on a pattern (it may be read “such that”), and if the pattern does not match in a for loop then the iteration is skipped, so false is returned only if upper[last - i] != c.
<lang e>def isPalindrome(string :String) {
def upper := string.toUpperCase()
def last := upper.size() - 1
for i => c ? (upper[last - i] != c) in upper(0, upper.size() // 2) {
return false
}
return true
}</lang>
EchoLisp
<lang lisp>
- returns #t or #f
(define (palindrome? string) (equal? (string->list string) (reverse (string->list string))))
- to strip spaces, use the following
- (define (palindrome? string)
- (let ((string (string-replace string "/\ /" "" "g")))
- (equal? (string->list string) (reverse (string->list string)))))
</lang>
Eiffel
<lang Eiffel> is_palindrome (a_string: STRING): BOOLEAN -- Is `a_string' a palindrome? require string_attached: a_string /= Void local l_index, l_count: INTEGER do from Result := True l_index := 1 l_count := a_string.count until l_index >= l_count - l_index + 1 or not Result loop Result := (Result and a_string [l_index] = a_string [l_count - l_index + 1]) l_index := l_index + 1 end end </lang>
Ela
<lang ela>open list string
isPalindrome xs = xs == reverse xs isPalindrome <| toList "ingirumimusnocteetconsumimurigni" </lang>
Function reverse is taken from list module and is defined as:
<lang ela>reverse = foldl (flip (::)) (nil xs)
foldl f z (x::xs) = foldl f (f z x) xs foldl _ z [] = z </lang>
Elixir
<lang elixir> defmodule PalindromeDetection do
def is_palindrome(str), do: str == String.reverse(str)
end </lang> Note: Because of Elixir's strong Unicode support, this even supports graphemes:
iex(1)> PalindromeDetection.is_palindrome("salàlas")
true
iex(2)> PalindromeDetection.is_palindrome("as⃝df̅")
false
iex(3)> PalindromeDetection.is_palindrome("as⃝df̅f̅ds⃝a")
true
Elm
<lang elm>import String exposing (reverse, length) import Html exposing (Html, Attribute, text, div, input) import Html.Attributes exposing (placeholder, value, style) import Html.Events exposing (on, targetValue) import Html.App exposing (beginnerProgram)
-- The following function (copied from Haskell) satisfies the -- rosettacode task description. is_palindrome x = x == reverse x
-- The remainder of the code demonstrates the use of the function -- in a complete Elm program. main = beginnerProgram { model = "" , view = view , update = update }
update newStr oldStr = newStr
view : String -> Html String view candidate =
div []
([ input
[ placeholder "Enter a string to check."
, value candidate
, on "input" targetValue
, myStyle
]
[]
] ++
[ let testResult =
is_palindrome candidate
statement =
if testResult then "PALINDROME!" else "not a palindrome"
in div [ myStyle] [text statement]
])
myStyle : Attribute msg myStyle =
style
[ ("width", "100%")
, ("height", "20px")
, ("padding", "5px 0 0 5px")
, ("font-size", "1em")
, ("text-align", "left")
]</lang>
Link to live demo: http://dc25.github.io/palindromeDetectionElm/
Emacs Lisp
<lang lisp>(defun palindrome (s)
(string= s (reverse s)))</lang>
Erlang
<lang Erlang> -module( palindrome ).
-export( [is_palindrome/1, task/0] ).
is_palindrome( String ) -> String =:= lists:reverse(String).
task() -> display( "abcba" ), display( "abcdef" ), Latin = "In girum imus nocte et consumimur igni", No_spaces_same_case = lists:append( string:tokens(string:to_lower(Latin), " ") ), display( Latin, No_spaces_same_case ).
display( String ) -> io:fwrite( "Is ~p a palindrom? ~p~n", [String, is_palindrome(String)] ).
display( String1, String2 ) -> io:fwrite( "Is ~p a palindrom? ~p~n", [String1, is_palindrome(String2)] ). </lang>
- Output:
22> palindrome:task(). Is "abcba" a palindrom? true Is "abcdef" a palindrom? false Is "In girum imus nocte et consumimur igni" a Latin palindrom? true
Euphoria
<lang euphoria>function isPalindrome(sequence s)
for i = 1 to length(s)/2 do
if s[i] != s[$-i+1] then
return 0
end if
end for
return 1
end function</lang>
<lang euphoria> include std/sequence.e -- reverse include std/console.e -- display include std/text.e -- upper include std/utils.e -- iif
IsPalindrome("abcba") IsPalindrome("abcdef") IsPalindrome("In girum imus nocte et consumimur igni")
procedure IsPalindrome(object s)
display("Is '[]' a palindrome? ",{s},0)
s = remove_all(' ',upper(s))
display(iif(equal(s,reverse(s)),"true","false"))
end procedure</lang>
- Output:
Is 'abcba' a palindrome? true Is 'abcdef' a palindrome? false Is 'In girum imus nocte et consumimur igni' a palindrome? true
Excel
LAMBDA
Binding the following lambda expression to the name ISPALINDROME in the Name Manager for the Excel WorkBook:
(See LAMBDA: The ultimate Excel worksheet function)
<lang lisp>ISPALINDROME =LAMBDA(s,
LET(
lcs, FILTERP(
LAMBDA(c, " " <> c)
)(
CHARS(LOWER(s))
),
CONCAT(lcs) = CONCAT(REVERSE(lcs))
)
)</lang>
and assuming that the following generic lambdas are also bound to the names CHARS, FILTERP, and REVERSE in the Name Manager for the active WorkBook:
<lang lisp>CHARS =LAMBDA(s,
MID(s, ROW(INDIRECT("1:" & LEN(s))), 1)
)
FILTERP
=LAMBDA(p,
LAMBDA(xs,
FILTER(xs, p(xs))
)
)
REVERSE
=LAMBDA(xs,
LET(
n, ROWS(xs),
SORTBY(
xs,
SEQUENCE(n, 1, n, -1)
)
)
)</lang>
- Output:
| fx | =ISPALINDROME(A2) | ||
|---|---|---|---|
| A | B | ||
| 1 | Test string | Is palindrome ? | |
| 2 | In girum imus nocte et consumimur igni | TRUE | |
| 3 | abban | FALSE | |
| 4 | abba | TRUE | |
| 5 | aba | TRUE | |
| 6 | ab | FALSE | |
| 7 | a | TRUE | |
F#
<lang fsharp>let isPalindrome (s: string) =
let arr = s.ToCharArray() arr = Array.rev arr</lang>
Examples: <lang fsharp>isPalindrome "abcba" val it : bool = true isPalindrome ("In girum imus nocte et consumimur igni".Replace(" ", "").ToLower());; val it : bool = true isPalindrome "abcdef" val it : bool = false</lang>
Factor
<lang factor>USING: kernel sequences ;
- palindrome? ( str -- ? ) dup reverse = ;</lang>
Falcon
VBA/Python programmer's approach not sure if it's the most falconic way <lang falcon> /* created by Aykayayciti Earl Lamont Montgomery April 9th, 2018 */
function is_palindrome(a) a = strUpper(a).replace(" ", "") b = a[-1:0] return b == a end
a = "mom" > is_palindrome(a) </lang>
- Output:
true [Finished in 1.7s]
more falconic <lang falcon> /* created by Aykayayciti Earl Lamont Montgomery April 9th, 2018 */
b = "mom" > strUpper(b).replace(" ", "") == strUpper(b[-1:0]) ? "Is a palindrome" : "Is not a palindrome" </lang>
- Output:
Is a palindrome [Finished in 1.5s]
Fantom
<lang fantom> class Palindrome {
// Function to test if given string is a palindrome
public static Bool isPalindrome (Str str)
{
str == str.reverse
}
// Give it a test run
public static Void main ()
{
echo (isPalindrome(""))
echo (isPalindrome("a"))
echo (isPalindrome("aa"))
echo (isPalindrome("aba"))
echo (isPalindrome("abb"))
echo (isPalindrome("salàlas"))
echo (isPalindrome("In girum imus nocte et consumimur igni".lower.replace(" ","")))
}
} </lang>
FBSL
<lang qbasic>#APPTYPE CONSOLE
FUNCTION stripNonAlpha(BYVAL s AS STRING) AS STRING DIM sTemp AS STRING = "" DIM c AS STRING FOR DIM i = 1 TO LEN(s) c = MID(s, i, 1) IF INSTR("ABCDEFGHIJKLMNOPQRSTUVWXYZ", c, 0, 1) THEN sTemp = stemp & c END IF NEXT RETURN sTemp END FUNCTION
FUNCTION IsPalindrome(BYVAL s AS STRING) AS INTEGER FOR DIM i = 1 TO STRLEN(s) \ 2 ' only check half of the string, as scanning from both ends IF s{i} <> s{STRLEN - (i - 1)} THEN RETURN FALSE 'comparison is not case sensitive NEXT
RETURN TRUE END FUNCTION
PRINT IsPalindrome(stripNonAlpha("A Toyota")) PRINT IsPalindrome(stripNonAlpha("Madam, I'm Adam")) PRINT IsPalindrome(stripNonAlpha("the rain in Spain falls mainly on the rooftops"))
PAUSE </lang>
- Output:
1 1 0
Forth
<lang forth>: first over c@ ;
- last >r 2dup + 1- c@ r> swap ;
- palindrome? ( c-addr u -- f )
begin dup 1 <= if 2drop true exit then first last <> if 2drop false exit then 1 /string 1- again ;
</lang>
FIRST and LAST are once-off words that could be beheaded immediately afterwards. The version taking advantage of Tail Call Optimization or a properly tail-recursive variant of RECURSE (easily added to any Forth) is very similar. The horizontal formatting highlights the parallel code - and potential factor; a library of many string tests like this could have ?SUCCESS and ?FAIL .
Below is a separate Forth program that detects palindrome phrases as well as single word palindromes. It was programmed using gforth.
<lang forth> variable temp-addr
- valid-char? ( addr1 u -- f ) ( check for valid character )
+ dup C@ 48 58 within over C@ 65 91 within or swap C@ 97 123 within or ;
- >upper ( c1 -- c2 )
dup 97 123 within if 32 - then ;
- strip-input ( addr1 u -- addr2 u ) ( Strip characters, then copy stripped string to temp-addr )
pad temp-addr !
temp-addr @ rot rot 0 do dup I 2dup valid-char? if
+ C@ >upper temp-addr @ C! 1 temp-addr +!
else 2drop
then loop drop temp-addr @ pad - ;
- get-phrase ( -- addr1 u )
." Type a phrase: " here 1024 accept here swap -trailing cr ;
- position-phrase ( addr1 u -- addr1 u addr2 u addr1 addr2 u )
temp-addr @ over 2over 2over drop swap ;
- reverse-copy ( addr1 addr2 u -- addr1 addr2 )
0 do over I' 1- I - + over I + 1 cmove loop 2drop ;
- palindrome? ( -- )
get-phrase strip-input position-phrase reverse-copy compare 0= if ." << Valid >> Palindrome." else ." << Not >> a Palindrome." then cr ;
</lang>
Example:
palindrome?
Type a phrase: A man, a plan, a cat, a ham, a yak, a yam, a hat, a canal-Panama!
<< Valid >> Palindrome.
Fortran
<lang fortran>program palindro
implicit none
character(len=*), parameter :: p = "ingirumimusnocteetconsumimurigni"
print *, is_palindro_r(p)
print *, is_palindro_r("anothertest")
print *, is_palindro2(p)
print *, is_palindro2("test")
print *, is_palindro(p)
print *, is_palindro("last test")
contains</lang>
Non-recursive
<lang fortran>! non-recursive function is_palindro(t)
logical :: is_palindro character(len=*), intent(in) :: t
integer :: i, l
l = len(t)
is_palindro = .false.
do i=1, l/2
if ( t(i:i) /= t(l-i+1:l-i+1) ) return
end do
is_palindro = .true.
end function is_palindro
! non-recursive 2 function is_palindro2(t) result(isp)
logical :: isp character(len=*), intent(in) :: t
character(len=len(t)) :: s integer :: i
forall(i=1:len(t)) s(len(t)-i+1:len(t)-i+1) = t(i:i) isp = ( s == t )
end function is_palindro2</lang>
Recursive <lang fortran> recursive function is_palindro_r (t) result (isp)
implicit none character (*), intent (in) :: t logical :: isp
isp = len (t) == 0 .or. t (: 1) == t (len (t) :) .and. is_palindro_r (t (2 : len (t) - 1))
end function is_palindro_r</lang>
<lang fortran>end program palindro</lang>
FreeBASIC
<lang FreeBASIC>' version 20-06-2015 ' compile with: fbc -s console "filename".bas
- Ifndef TRUE ' define true and false for older freebasic versions
#Define FALSE 0 #Define TRUE Not FALSE
- EndIf
Function reverse(norm As String) As Integer
Dim As String rev Dim As Integer i, l = Len(norm) -1
rev = norm
For i = 0 To l
rev[l-i] = norm[i]
Next
If norm = rev Then
Return TRUE
Else
Return FALSE
End If
End Function
Function cleanup(in As String, action As String = "") As String
' action = "" do nothing, [l|L] = convert to lowercase, ' [s|S] = strip spaces, [p|P] = strip punctuation. If action = "" Then Return in
Dim As Integer i, p_, s_ Dim As String ch
action = LCase(action)
For i = 1 To Len(action)
ch = Mid(action, i, 1)
If ch = "l" Then in = LCase(in)
If ch = "p" Then
p_ = 1
ElseIf ch = "s" Then
s_ = 1
End If
Next
If p_ = 0 And s_ = 0 Then Return in
Dim As String unwanted, clean
If s_ = 1 Then unwanted = " "
If p_ = 1 Then unwanted = unwanted + "`~!@#$%^&*()-=_+[]{}\|;:',.<>/?"
For i = 1 To Len(in)
ch = Mid(in, i, 1)
If InStr(unwanted, ch) = 0 Then clean = clean + ch
Next
Return clean
End Function
' ------=< MAIN >=------
Dim As String test = "In girum imus nocte et consumimur igni" 'IIf ( cond, true, false ), true and false must be of the same type (num, string, UDT) Print Print " reverse(test) = "; IIf(reverse(test) = FALSE, "FALSE", "TRUE") Print " reverse(cleanup(test,""l"")) = "; IIf(reverse(cleanup(test,"l")) = FALSE, "FALSE", "TRUE") Print " reverse(cleanup(test,""ls"")) = "; IIf(reverse(cleanup(test,"ls")) = FALSE, "FALSE", "TRUE") Print "reverse(cleanup(test,""PLS"")) = "; IIf(reverse(cleanup(test,"PLS")) = FALSE, "FALSE", "TRUE")
' empty keyboard buffer While InKey <> "" : Wend Print : Print : Print "Hit any key to end program" Sleep End</lang>
- Output:
reverse(test) = FALSE reverse(cleanup(test,"l")) = FALSE reverse(cleanup(test,"ls")) = TRUE reverse(cleanup(test,"PLS")) = TRUE
Frink
This version will even work with upper-plane Unicode characters. Many languages will not work correctly with upper-plane Unicode characters because they are represented as Unicode "surrogate pairs" which are represented as two characters in a UTF-16 stream. In addition, Frink uses a grapheme-based reverse, which allows the algorithm below to operate on combined sequences of Unicode characters.
For example, the string "og\u0308o" represents an o, a g with combining diaeresis, followed by the letter o. Or, in other words, "og̈o". Note that while there are four Unicode codepoints, only three "graphemes" are displayed. Using Frink's smart "reverse" function preserves these combined graphemes and detects them correctly as palindromes.
<lang frink>isPalindrome[x] := x == reverse[x] </lang>
Test in Frink with upper-plane Unicode: <lang frink>isPalindrome["x\u{1f638}x"]</lang>
true
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.
In this page you can see the program(s) related to this task and their results.
GAP
<lang gap>ZapGremlins := function(s)
local upper, lower, c, i, n, t;
upper := "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
lower := "abcdefghijklmnopqrstuvwxyz";
t := [ ];
i := 1;
for c in s do
n := Position(upper, c);
if n <> fail then
t[i] := lower[n];
i := i + 1;
else
n := Position(lower, c);
if n <> fail then
t[i] := c;
i := i + 1;
fi;
fi;
od;
return t;
end;
IsPalindrome := function(s)
local t; t := ZapGremlins(s); return t = Reversed(t);
end;</lang>
GML
<lang go> //Setting a var from an argument passed to the script var str; str = argument0 //Takes out all spaces/anything that is not a letter or a number and turns uppercase letters to lowercase str = string_lettersdigits(string_lower(string_replace(str,' ',))); var inv; inv = ; //for loop that reverses the sequence var i; for (i = 0; i < string_length(str); i += 1;)
{
inv += string_copy(str,string_length(str)-i,1);
}
//returns true if the sequence is a palindrome else returns false return (str == inv); </lang>
Palindrome detection using a Downward For-Loop
<lang go>
//Remove everything except for letters and digits and convert the string to lowercase. source is what will be compared to str. var str = string_lower(string_lettersdigits(string_replace(argument0," ",""))), source = "";
//Loop through and store each character of str in source. for (var i = string_length(str); i > 0; i--) {
source += string_char_at(str,i);
}
//Return if it is a palindrome. return source == str; </lang>
Go
<lang go>package pal
func IsPal(s string) bool {
mid := len(s) / 2
last := len(s) - 1
for i := 0; i < mid; i++ {
if s[i] != s[last-i] {
return false
}
}
return true
}</lang>
This version works with Unicode,
<lang go> func isPalindrome(s string) bool { runes := []rune(s) numRunes := len(runes) - 1 for i := 0; i < len(runes)/2; i++ { if runes[i] != runes[numRunes-i] { return false } } return true }</lang>
Or using more slicing,
<lang go> func isPalindrome(s string) bool { runes := []rune(s) for len(runes) > 1 { if runes[0] != runes[len(runes)-1] { return false } runes = runes[1 : len(runes)-1] } return true }</lang>
Groovy
Trivial
Solution: <lang groovy>def isPalindrome = { String s ->
s == s?.reverse()
}</lang>
Test program: <lang groovy>println isPalindrome("") println isPalindrome("a") println isPalindrome("abcdefgfedcba") println isPalindrome("abcdeffedcba") println isPalindrome("abcedfgfedcb")</lang>
- Output:
true true true true false
This solution assumes nulls are palindromes.
Non-recursive
Solution: <lang groovy>def isPalindrome = { String s ->
def n = s.size() n < 2 || s[0..<n/2] == s[-1..(-n/2)]
}</lang>
Test program and output are the same. This solution does not handle nulls.
Recursive
Solution follows the C palindrome_r recursive solution: <lang groovy>def isPalindrome isPalindrome = { String s ->
def n = s.size() n < 2 || (s[0] == s[n-1] && isPalindrome(s[1..<(n-1)]))
}</lang>
Test program and output are the same. This solution does not handle nulls.
Haskell
Non-recursive
A string is a palindrome if reversing it we obtain the same string.
<lang haskell>is_palindrome x = x == reverse x</lang>
Or, applicative and point-free, with some pre-processing of data (shedding white space and upper case):
<lang haskell>import Data.Bifunctor (second) import Data.Char (toLower)
PALINDROME DETECTION -----------------
isPalindrome :: Eq a => [a] -> Bool isPalindrome = (==) <*> reverse
-- Or, comparing just the leftward characters with -- with a reflection of just the rightward characters.
isPal :: String -> Bool isPal s =
let (q, r) = quotRem (length s) 2
in uncurry (==) $
second (reverse . drop r) $ splitAt q s
TEST -------------------------
main :: IO () main =
mapM_ putStrLn $
(showResult <$> [isPalindrome, isPal])
<*> fmap
prepared
[ "",
"a",
"ab",
"aba",
"abba",
"In girum imus nocte et consumimur igni"
]
prepared :: String -> String prepared cs = [toLower c | c <- cs, ' ' /= c]
showResult f s = (show s) <> " -> " <> show (f s)</lang>
- Output:
"" -> True "a" -> True "ab" -> False "aba" -> True "abba" -> True "ingirumimusnocteetconsumimurigni" -> True "" -> True "a" -> True "ab" -> False "aba" -> True "abba" -> True "ingirumimusnocteetconsumimurigni" -> True
Recursive
See the C palindrome_r code for an explanation of the concept used in this solution.
<lang haskell>is_palindrome_r x | length x <= 1 = True
| head x == last x = is_palindrome_r . tail. init $ x
| otherwise = False</lang>
HicEst
<lang hicest> result = Palindrome( "In girum imus nocte et consumimur igni" ) ! returns 1 END
FUNCTION Palindrome(string)
CHARACTER string, CopyOfString
L = LEN(string)
ALLOCATE(CopyOfString, L)
CopyOfString = string
EDIT(Text=CopyOfString, UpperCase=L)
L = L - EDIT(Text=CopyOfString, End, Left=' ', Delete, DO=L) ! EDIT returns number of deleted spaces
DO i = 1, L/2
Palindrome = CopyOfString(i) == CopyOfString(L - i + 1)
IF( Palindrome == 0 ) RETURN
ENDDO
END</lang>
Icon and Unicon
<lang Icon>procedure main(arglist) every writes(s := !arglist) do write( if palindrome(s) then " is " else " is not", " a palindrome.") end</lang>
The following simple procedure uses the built-in reverse. Reverse creates a transient string which will get garbage collected. <lang Icon>procedure palindrome(s) #: return s if s is a palindrome return s == reverse(s) end</lang>
Note: The IPL procedure strings contains a palindrome tester called ispal that uses reverse and is equivalent to the version of palindrome above.
This version uses positive and negative sub-scripting and works not only on strings but lists of strings, such as ["ab","ab"] or ["ab","x"] the first list would pass the test but the second wouldn't. <lang Icon>procedure palindrome(x) #: return x if s is x palindrome local i every if x[i := 1 to (*x+ 1)/2] ~== x[-i] then fail return x end</lang>
Ioke
<lang ioke>Text isPalindrome? = method(self chars == self chars reverse)</lang>
J
Non-recursive
Reverse and match method <lang j>isPalin0=: -: |.</lang> Example usage <lang j> isPalin0 'ABBA' 1
isPalin0 -.&' ' tolower 'In girum imus nocte et consumimur igni'
1</lang>
Recursive
Tacit and explicit verbs: <lang j>isPalin1=: 0:`($:@(}.@}:))@.({.={:)`1:@.(1>:#)
isPalin2=: monad define
if. 1>:#y do. 1 return. end.
if. ({.={:)y do. isPalin2 }.}:y else. 0 end.
)</lang>
Note that while these recursive verbs are bulkier and more complicated, they are also several thousand times more inefficient than isPalin0.
<lang j> foo=: foo,|.foo=:2000$a.
ts=:6!:2,7!:2 NB. time and space required to execute sentence ts 'isPalin0 foo'
2.73778e_5 5184
ts 'isPalin1 foo'
0.0306667 6.0368e6
ts 'isPalin2 foo'
0.104391 1.37965e7
'isPalin1 foo' %&ts 'isPalin0 foo'
1599.09 1164.23
'isPalin2 foo' %&ts 'isPalin0 foo'
3967.53 2627.04</lang>
Java
Non-Recursive
<lang java>public static boolean pali(String testMe){ StringBuilder sb = new StringBuilder(testMe); return testMe.equals(sb.reverse().toString()); }</lang>
Non-Recursive using indexes (supports upper-plane Unicode) <lang java>public static boolean isPalindrome(String input) { for (int i = 0, j = input.length() - 1; i < j; i++, j--) { char startChar = input.charAt(i); char endChar = input.charAt(j);
// Handle surrogate pairs in UTF-16 if (Character.isLowSurrogate(endChar)) { if (startChar != input.charAt(--j)) { return false; } if (input.charAt(++i) != endChar) { return false; } } else if (startChar != endChar) { return false; } } return true; }</lang>
Recursive (this version does not work correctly with upper-plane Unicode)
<lang java>public static boolean rPali(String testMe){ if(testMe.length()<=1){ return true; } if(!(testMe.charAt(0)+"").equals(testMe.charAt(testMe.length()-1)+"")){ return false; } return rPali(testMe.substring(1, testMe.length()-1)); }</lang>
Recursive using indexes (this version does not work correctly with upper-plane Unicode)
<lang java>public static boolean rPali(String testMe){ int strLen = testMe.length(); return rPaliHelp(testMe, strLen-1, strLen/2, 0); }
public static boolean rPaliHelp(String testMe, int strLen, int testLen, int index){ if(index > testLen){ return true; } if(testMe.charAt(index) != testMe.charAt(strLen-index)){ return false; } return rPaliHelp(testMe, strLen, testLen, index + 1); } </lang>
Regular Expression (source) <lang java>public static boolean pali(String testMe){ return testMe.matches("|(?:(.)(?<=(?=^.*?(\\1\\2?)$).*))+(?<=(?=^\\2$).*)"); }</lang>
JavaScript
<lang javascript>function isPalindrome(str) {
return str === str.split("").reverse().join("");
}
console.log(isPalindrome("ingirumimusnocteetconsumimurigni"));</lang>
ES6 implementation
<lang javascript>var isPal = str => str === str.split("").reverse().join("");</lang>
Or, ignoring spaces and variations in case:
<lang JavaScript>(() => {
// isPalindrome :: String -> Bool
const isPalindrome = s => {
const cs = filter(c => ' ' !== c, s.toLocaleLowerCase());
return cs.join() === reverse(cs).join();
};
// TEST -----------------------------------------------
const main = () =>
isPalindrome(
'In girum imus nocte et consumimur igni'
)
// GENERIC FUNCTIONS ----------------------------------
// filter :: (a -> Bool) -> [a] -> [a]
const filter = (f, xs) => (
'string' !== typeof xs ? (
xs
) : xs.split()
).filter(f);
// reverse :: [a] -> [a]
const reverse = xs =>
'string' !== typeof xs ? (
xs.slice(0).reverse()
) : xs.split().reverse().join();
// MAIN --- return main();
})();</lang>
- Output:
true
jq
<lang jq>def palindrome: explode as $in | ($in|reverse) == $in;</lang> Example:
"salàlas" | palindrome
- Output:
true
Jsish
<lang javascript>/* Palindrome detection, in Jsish */ function isPalindrome(str:string, exact:boolean=true) {
if (!exact) {
str = str.toLowerCase();
str = str.replace(/[ \t,;:!?.]/g, );
}
return str === str.match(/./g).reverse().join();
}
- isPalindrome('BUB');
- isPalindrome('CUB');
- isPalindrome('Bub');
- isPalindrome('Bub', false);
- isPalindrome('In girum imus nocte et consumimur igni', false);
- isPalindrome('A man, a plan, a canal; Panama!', false);
- isPalindrome('Never odd or even', false);
/*
!EXPECTSTART!
isPalindrome('BUB') ==> true isPalindrome('CUB') ==> false isPalindrome('Bub') ==> false isPalindrome('Bub', false) ==> true isPalindrome('In girum imus nocte et consumimur igni', false) ==> true isPalindrome('A man, a plan, a canal; Panama!', false) ==> true isPalindrome('Never odd or even', false) ==> true
!EXPECTEND!
- /</lang>
Most of that code is for testing, using echo mode lines (semicolon in column 1)
- Output:
prompt$ jsish --U palindrome.jsi
isPalindrome('BUB') ==> true
isPalindrome('CUB') ==> false
isPalindrome('Bub') ==> false
isPalindrome('Bub', false) ==> true
isPalindrome('In girum imus nocte et consumimur igni', false) ==> true
isPalindrome('A man, a plan, a canal; Panama!', false) ==> true
isPalindrome('Never odd or even', false) ==> true
prompt$ jsish -u palindrome.jsi
[PASS] palindrome.jsi
Julia
<lang julia>palindrome(s) = s == reverse(s)</lang> Non-Recursive <lang julia> function palindrome(s)
len = length(s)
for i = 1:(len/2)
if(s[len-i+1]!=s[i])
return false
end
end
return true
end </lang> Recursive <lang julia> function palindrome(s)
len = length(s)
if(len==0 || len==1)
return true
end
if(s[1] == s[len])
return palindrome(SubString(s,2,len-1))
end
return false
end</lang>
k
<lang k>is_palindrome:{x~|x}</lang>
Kotlin
<lang scala>// version 1.1.2
/* These functions deal automatically with Unicode as all strings are UTF-16 encoded in Kotlin */
fun isExactPalindrome(s: String) = (s == s.reversed())
fun isInexactPalindrome(s: String): Boolean {
var t = "" for (c in s) if (c.isLetterOrDigit()) t += c t = t.toLowerCase() return t == t.reversed()
}
fun main(args: Array<String>) {
val candidates = arrayOf("rotor", "rosetta", "step on no pets", "été")
for (candidate in candidates) {
println("'$candidate' is ${if (isExactPalindrome(candidate)) "an" else "not an"} exact palindrome")
}
println()
val candidates2 = arrayOf(
"In girum imus nocte et consumimur igni",
"Rise to vote, sir",
"A man, a plan, a canal - Panama!",
"Ce repère, Perec" // note: 'è' considered a distinct character from 'e'
)
for (candidate in candidates2) {
println("'$candidate' is ${if (isInexactPalindrome(candidate)) "an" else "not an"} inexact palindrome")
}
}</lang>
- Output:
'rotor' is an exact palindrome 'rosetta' is not an exact palindrome 'step on no pets' is an exact palindrome 'été' is an exact palindrome 'In girum imus nocte et consumimur igni' is an inexact palindrome 'Rise to vote, sir' is an inexact palindrome 'A man, a plan, a canal - Panama!' is an inexact palindrome 'Ce repère, Perec' is not an inexact palindrome
LabVIEW
This image is a VI Snippet, an executable image of LabVIEW code. The LabVIEW version is shown on the top-right hand corner. You can download it, then drag-and-drop it onto the LabVIEW block diagram from a file browser, and it will appear as runnable, editable code.
langur
<lang langur>val .ispal = f len(.s) > 0 and .s == s2s .s, len(.s)..1
val .tests = h{
"": false, "z": true, "aha": true, "αηα": true, "αννα": true, "αννασ": false, "sees": true, "seas": false, "deified": true, "solo": false, "solos": true, "amanaplanacanalpanama": true, "a man a plan a canal panama": false, # true if we remove spaces "ingirumimusnocteetconsumimurigni": true,
}
for .word in sort(keys .tests) {
val .foundpal = .ispal(.word) writeln .word, ": ", .foundpal, if(.foundpal == .tests[.word]: ""; " (FAILED TEST)")
}</lang>
- Output:
: false a man a plan a canal panama: false aha: true amanaplanacanalpanama: true deified: true ingirumimusnocteetconsumimurigni: true seas: false sees: true solo: false solos: true z: true αηα: true αννα: true αννασ: false
Lasso
<lang Lasso>define ispalindrome(text::string) => {
local(_text = string(#text)) // need to make copy to get rid of reference issues
#_text -> replace(regexp(`(?:$|\W)+`), -ignorecase)
local(reversed = string(#_text)) #reversed -> reverse
return #_text == #reversed }
ispalindrome('Tätatät') // works with high ascii ispalindrome('Hello World')
ispalindrome('A man, a plan, a canoe, pasta, heros, rajahs, a coloratura, maps, snipe, percale, macaroni, a gag, a banana bag, a tan, a tag, a banana bag again (or a camel), a crepe, pins, Spam, a rut, a Rolo, cash, a jar, sore hats, a peon, a canal – Panama!')</lang>
- Output:
true false true
Liberty BASIC
<lang lb>print isPalindrome("In girum imus nocte et consumimur igni") print isPalindrome(removePunctuation$("In girum imus nocte et consumimur igni", "S")) print isPalindrome(removePunctuation$("In girum imus nocte et consumimur igni", "SC"))
function isPalindrome(string$)
isPalindrome = 1
for i = 1 to int(len(string$)/2)
if mid$(string$, i, 1) <> mid$(string$, len(string$)-i+1, 1) then isPalindrome = 0 : exit function
next i
end function
function removePunctuation$(string$, remove$)
'P = remove puctuation. S = remove spaces C = remove case
If instr(upper$(remove$), "C") then string$ = lower$(string$)
If instr(upper$(remove$), "P") then removeCharacters$ = ",.!'()-&*?<>:;~[]{}"
If instr(upper$(remove$), "S") then removeCharacters$ = removeCharacters$;" "
for i = 1 to len(string$)
if instr(removeCharacters$, mid$(string$, i, 1)) then string$ = left$(string$, i-1);right$(string$, len(string$)-i) : i = i - 1
next i
removePunctuation$ = string$
end function</lang>
- Output:
0 0 1
LiveCode
This implementation defaults to exact match, but has an optional parameter to do inexact.<lang LiveCode>function palindrome txt exact
if exact is empty or exact is not false then
set caseSensitive to true --default is false
else
replace space with empty in txt
put lower(txt) into txt
end if
return txt is reverse(txt)
end palindrome
function reverse str
repeat with i = the length of str down to 1
put byte i of str after revstr
end repeat
return revstr
end reverse</lang>
Logo
<lang logo>to palindrome? :w
output equal? :w reverse :w
end</lang>
Lua
<lang lua>function ispalindrome(s) return s == string.reverse(s) end</lang>
M4
Non-recursive
This uses the invert from Reversing a string.
<lang m4>define(`palindrorev',`ifelse(`$1',invert(`$1'),`yes',`no')')dnl
palindrorev(`ingirumimusnocteetconsumimurigni')
palindrorev(`this is not palindrome')</lang>
Recursive <lang m4>define(`striptwo',`substr(`$1',1,eval(len(`$1')-2))')dnl define(`cmplast',`ifelse(`striptwo(`$1')',,`yes',dnl substr(`$1',0,1),substr(`$1',eval(len(`$1')-1),1),`yes',`no')')dnl define(`palindro',`dnl ifelse(eval(len(`$1')<1),1,`yes',cmplast(`$1'),`yes',`palindro(striptwo(`$1'))',`no')')dnl palindro(`ingirumimusnocteetconsumimurigni') palindro(`this is not palindrome')</lang>
Maple
This uses functions from Maple's built-in StringTools package.
<lang Maple> with(StringTools):
IsPalindrome("ingirumimusnocteetconsumimurigni");
IsPalindrome("In girum imus nocte et consumimur igni");
IsPalindrome(LowerCase(DeleteSpace("In girum imus nocte et consumimur igni"))); </lang>
- Output:
true
false
true
Mathematica/Wolfram Language
Built-in function handling lists, numbers, and strings: <lang Mathematica>PalindromeQ</lang>
- Examples:
PalindromeQ["TNT"] PalindromeQ["test"] PalindromeQ["deified"] PalindromeQ["salálas"] PalindromeQ["ingirumimusnocteetconsumimurigni"]
- Output:
True False True True True
MATLAB
<lang MATLAB>function trueFalse = isPalindrome(string)
trueFalse = all(string == fliplr(string)); %See if flipping the string produces the original string
if not(trueFalse) %If not a palindrome
string = lower(string); %Lower case everything
trueFalse = all(string == fliplr(string)); %Test again
end
if not(trueFalse) %If still not a palindrome
string(isspace(string)) = []; %Strip all space characters out
trueFalse = all(string == fliplr(string)); %Test one last time
end
end</lang>
- Sample Usage:
<lang MATLAB>>> isPalindrome('In girum imus nocte et consumimur igni')
ans =
1
</lang>
Maxima
<lang maxima>palindromep(s) := block([t], t: sremove(" ", sdowncase(s)), sequal(t, sreverse(t)))$
palindromep("Sator arepo tenet opera rotas"); /* true */</lang>
MAXScript
Non-recursive
<lang maxscript>fn isPalindrome s = (
local reversed = "" for i in s.count to 1 by -1 do reversed += s[i] return reversed == s
)</lang>
Recursive
<lang maxscript>fn isPalindrome_r s = (
if s.count <= 1 then
(
true
)
else
(
if s[1] != s[s.count] then
(
return false
)
isPalindrome_r (substring s 2 (s.count-2))
)
)</lang>
Testing
<lang maxscript>local p = "ingirumimusnocteetconsumimurigni" format ("'%' is a palindrome? %\n") p (isPalindrome p) format ("'%' is a palindrome? %\n") p (isPalindrome_r p)</lang>
min
<lang min>(dup reverse ==) :palindrome? (dup "" split reverse "" join ==) :str-palindrome?
"apple" str-palindrome? puts "racecar" str-palindrome? puts (a b c) palindrome? puts (a b b a) palindrome? puts</lang>
- Output:
false true false true
MiniScript
<lang MiniScript>isPalindrome = function(s)
// convert to lowercase, and strip non-letters stripped = "" for c in s.lower if c >= "a" and c <= "z" then stripped = stripped + c end for
// check palindromity mid = floor(stripped.len/2) for i in range(0, mid) if stripped[i] != stripped[-i - 1] then return false end for return true
end function
testStr = "Madam, I'm Adam" answer = [testStr, "is"] if not isPalindrome(testStr) then answer.push "NOT" answer.push "a palindrome" print answer.join</lang>
- Output:
Madam, I'm Adam is a palindrome
Mirah
<lang mirah>def reverse(s:string)
StringBuilder.new(s).reverse.toString()
end
def palindrome?(s:string)
s.equals(reverse(s))
end
puts palindrome?("anna") # ==> true puts palindrome?("Erik") # ==> false puts palindrome?("palindroom-moordnilap") # ==> true puts nil # ==> null</lang>
ML
mLite
<lang ocaml>fun to_locase s = implode ` map (c_downcase) ` explode s
fun only_alpha s = implode ` filter (fn x = c_alphabetic x) ` explode s
fun is_palin ( h1 :: t1, h2 :: t2, n = 0 ) = true | ( h1 :: t1, h2 :: t2, n ) where ( h1 eql h2 ) = is_palin( t1, t2, n - 1) | ( h1 :: t1, h2 :: t2, n ) = false | (str s) = let val es = explode ` to_locase ` only_alpha s; val res = rev es; val k = (len es) div 2 in is_palin (es, res, k) end
fun test_is_palin s = (print "\""; print s; print "\" is a palindrome: "; print ` is_palin s; println "")
fun test (f, arg, res, ok, notok) = if (f arg eql res) then ("'" @ arg @ "' " @ ok) else ("'" @ arg @ "' " @ notok)
println ` test (is_palin, "In girum imus nocte, et consumimur igni", true, "is a palindrome", "is NOT a palindrome"); println ` test (is_palin, "Madam, I'm Adam.", true, "is a palindrome", "is NOT a palindrome"); println ` test (is_palin, "salàlas", true, "is a palindrome", "is NOT a palindrome"); println ` test (is_palin, "radar", true, "is a palindrome", "is NOT a palindrome"); println ` test (is_palin, "Lagerregal", true, "is a palindrome", "is NOT a palindrome"); println ` test (is_palin, "Ein Neger mit Gazelle zagt im Regen nie.", true, "is a palindrome", "is NOT a palindrome"); println ` test (is_palin, "something wrong", true, "is a palindrome", "is NOT a palindrome");</lang> Output:
'In girum imus nocte, et consumimur igni' is a palindrome 'Madam, I'm Adam.' is a palindrome 'salàlas' is a palindrome 'radar' is a palindrome 'Lagerregal' is a palindrome 'Ein Neger mit Gazelle zagt im Regen nie.' is a palindrome 'something wrong' is NOT a palindrome
Standard ML
<lang sml> fun palindrome s =
let val cs = explode s in cs = rev cs end
</lang>
MMIX
<lang mmix>argc IS $0 argv IS $1
LOC Data_Segment
DataSeg GREG @
LOC @+1000
ItsPalStr IS @-Data_Segment
BYTE "It's palindrome",10,0
LOC @+(8-@)&7
NoPalStr IS @-Data_Segment
BYTE "It is not palindrome",10,0
LOC #100
GREG @
% input: $255 points to where the string to be checked is % returns $255 0 if not palindrome, not zero otherwise % trashs: $0,$1,$2,$3 % return address $4 DetectPalindrome LOC @
ADDU $1,$255,0 % $1 = $255
2H LDB $0,$1,0 % get byte at $1
BZ $0,1F % if zero, end (length)
INCL $1,1 % $1++
JMP 2B % loop
1H SUBU $1,$1,1 % ptr last char of string
ADDU $0,DataSeg,0 % $0 to data seg.
3H CMP $3,$1,$255 % is $0 == $255?
BZ $3,4F % then jump
LDB $3,$1,0 % otherwise get the byte
STB $3,$0,0 % and copy it
INCL $0,1 % $0++
SUB $1,$1,1 % $1--
JMP 3B
4H LDB $3,$1,0
STB $3,$0,0 % copy the last byte
% now let us compare reversed string and straight string
XOR $0,$0,$0 % index
ADDU $1,DataSeg,0
6H LDB $2,$1,$0 % pick char from rev str
LDB $3,$255,$0 % pick char from straight str
BZ $3,PaliOk % finished as palindrome
CMP $2,$2,$3 % == ?
BNZ $2,5F % if not, exit
INCL $0,1 % $0++
JMP 6B
5H XOR $255,$255,$255
GO $4,$4,0 % return false
PaliOk NEG $255,0,1
GO $4,$4,0 % return true
% The Main for testing the function % run from the command line % $ mmix ./palindrome.mmo ingirumimusnocteetconsumimurigni Main CMP argc,argc,2 % argc > 2?
BN argc,3F % no -> not enough arg
ADDU $1,$1,8 % argv+1
LDOU $255,$1,0 % argv[1]
GO $4,DetectPalindrome
BZ $255,2F % if not palindrome, jmp
SETL $0,ItsPalStr % pal string
ADDU $255,DataSeg,$0
JMP 1F
2H SETL $0,NoPalStr % no pal string
ADDU $255,DataSeg,$0
1H TRAP 0,Fputs,StdOut % print 3H XOR $255,$255,$255
TRAP 0,Halt,0 % exit(0)</lang>
Modula-2
<lang modula2>MODULE Palindrome; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,ReadChar;
PROCEDURE IsPalindrome(str : ARRAY OF CHAR) : BOOLEAN; VAR i,m : INTEGER; VAR buf : ARRAY[0..63] OF CHAR; BEGIN
i := 0;
m := HIGH(str) - 1;
WHILE i<m DO
IF str[i] # str[m-i] THEN
RETURN FALSE
END;
INC(i)
END;
RETURN TRUE
END IsPalindrome;
PROCEDURE Print(str : ARRAY OF CHAR); VAR buf : ARRAY[0..63] OF CHAR; BEGIN
FormatString("%s: %b\n", buf, str, IsPalindrome(str));
WriteString(buf)
END Print;
BEGIN
Print("");
Print("z");
Print("aha");
Print("sees");
Print("oofoe");
Print("deified");
Print("Deified");
Print("amanaplanacanalpanama");
Print("ingirumimusnocteetconsumimurigni");
ReadChar
END Palindrome.</lang>
Modula-3
<lang modula3>MODULE Palindrome;
IMPORT Text;
PROCEDURE isPalindrome(string: TEXT): BOOLEAN =
VAR len := Text.Length(string);
BEGIN
FOR i := 0 TO len DIV 2 - 1 DO
IF Text.GetChar(string, i) # Text.GetChar(string, (len - i - 1)) THEN
RETURN FALSE;
END;
END;
RETURN TRUE;
END isPalindrome;
END Palindrome.</lang>
Nanoquery
<lang Nanoquery>def is_palindrome(s)
temp = ""
for char in s
if "abcdefghikjklmnopqrstuvwxyz" .contains. lower(char)
temp += lower(char)
end
end
return list(temp) = list(temp).reverse()
end</lang>
Nemerle
<lang Nemerle>using System; using System.Console; using Nemerle.Utility.NString; //contains methods Explode() and Implode() which convert string -> list[char] and back
module Palindrome {
IsPalindrome( text : string) : bool
{
Implode(Explode(text).Reverse()) == text;
}
Main() : void
{
WriteLine("radar is a palindrome: {0}", IsPalindrome("radar"));
}
}</lang> And a function to remove spaces and punctuation and convert to lowercase <lang Nemerle>Clean( text : string ) : string {
def sepchars = Explode(",.;:-?!()' ");
Concat( "", Split(text, sepchars)).ToLower()
}</lang>
NetRexx
<lang netrexx> y='In girum imus nocte et consumimur igni'
-- translation: We walk around in the night and -- we are burnt by the fire (of love) say say 'string = 'y say
pal=isPal(y)
if pal==0 then say "The string isn't palindromic."
else say 'The string is palindromic.'
method isPal(x) static
x=x.upper().space(0) /* removes all blanks (spaces) */
/* and translate to uppercase. */
return x==x.reverse() /* returns 1 if exactly equal */
</lang>
NewLISP
Works likewise for strings and for lists <lang lisp> (define (palindrome? s)
(setq r s) (reverse r) ; Reverse is destructive. (= s r))
- Make ‘reverse’ non-destructive and avoid a global variable
(define (palindrome? s)
(= s (reverse (copy s))))
</lang>
Nim
The following program detects if UTF-8 strings are exact palindromes. If "exact" is set to "false", it ignores the white spaces and the differences of letter case to detect inexact palindromes. Differences in punctuation are still relevant. <lang nim>import unicode
func isPalindrome(rseq: seq[Rune]): bool =
## Return true if a sequence of runes is a palindrome.
for i in 1..(rseq.len shr 1):
if rseq[i - 1] != rseq[^i]:
return false
result = true
func isPalindrome(str: string; exact = true): bool {.inline.} =
## Return true if a UTF-8 string is a palindrome. ## If "exact" is false, ignore white spaces and ignore case.
if exact:
result = str.toRunes.isPalindrome()
else:
var rseq: seq[Rune]
for rune in str.runes:
if not rune.isWhiteSpace:
rseq.add rune.toLower
result = rseq.isPalindrome()
when isMainModule:
proc check(s: string) =
var exact, inexact: bool
exact = s.isPalindrome()
if not exact:
inexact = s.isPalindrome(exact = false)
let txt = if exact: " is an exact palindrome."
elif inexact: " is an inexact palindrome."
else: " is not a palindrome."
echo '"', s, '"', txt
check "rotor" check "été" check "αννα" check "salà las" check "In girum imus nocte et consumimur igni" check "Esope reste ici et se repose" check "This is a palindrom"</lang>
- Output:
"rotor" is an exact palindrome. "été" is an exact palindrome. "αννα" is an exact palindrome. "salà las" is an inexact palindrome. "In girum imus nocte et consumimur igni" is an inexact palindrome. "Esope reste ici et se repose" is an inexact palindrome. "This is a palindrom" is not a palindrome.
Objeck
<lang objeck> bundle Default {
class Test {
function : Main(args : String[]) ~ Nil {
IsPalindrome("aasa")->PrintLine();
IsPalindrome("acbca")->PrintLine();
IsPalindrome("xx")->PrintLine();
}
function : native : IsPalindrome(s : String) ~ Bool {
l := s->Size();
for(i := 0; i < l / 2; i += 1;) {
if(s->Get(i) <> s->Get(l - i - 1)) {
return false;
};
};
return true;
}
}
} </lang>
OCaml
<lang ocaml>let is_palindrome s =
let l = String.length s in
let rec comp n =
n = 0 || (s.[l-n] = s.[n-1] && comp (n-1)) in
comp (l / 2)</lang>
and here a function to remove the white spaces in the string:
<lang ocaml>let rem_space str =
let len = String.length str in
let res = Bytes.create len in
let rec aux i j =
if i >= len
then (Bytes.sub_string res 0 j)
else match str.[i] with
| ' ' | '\n' | '\t' | '\r' ->
aux (i+1) (j)
| _ ->
Bytes.set res j str.[i];
aux (i+1) (j+1)
in
aux 0 0
</lang>
and to make the test case insensitive, just use the function String.lowercase_ascii.
Octave
Recursive <lang octave>function v = palindro_r(s)
if ( length(s) == 1 ) v = true; return; elseif ( length(s) == 2 ) v = s(1) == s(2); return; endif if ( s(1) == s(length(s)) ) v = palindro_r(s(2:length(s)-1)); else v = false; endif
endfunction</lang>
Non-recursive <lang octave>function v = palindro(s)
v = all( (s == s(length(s):-1:1)) == 1);
endfunction</lang>
Testing <lang octave>palindro_r("ingirumimusnocteetconsumimurigni") palindro("satorarepotenetoperarotas")</lang>
Oforth
<lang Oforth>String method: isPalindrome self reverse self == ;</lang>
Ol
<lang scheme>
- simple case - only lowercase letters
(define (palindrome? str)
(let ((l (string->runes str)))
(equal? l (reverse l))))
(print (palindrome? "ingirumimusnocteetconsumimurigni"))
- ==> #true
(print (palindrome? "thisisnotapalindrome"))
- ==> #false
- complex case - with ignoring letter case and punctuation
(define (alpha? x)
(<= #\a x #\z))
(define (lowercase x)
(if (<= #\A x #\Z)
(- x (- #\A #\a))
x))
(define (palindrome? str)
(let ((l (filter alpha? (map lowercase (string->runes str)))))
(equal? l (reverse l))))
(print (palindrome? "A man, a plan, a cat, a ham, a yak, a yam, a hat, a canal-Panama!"))
- ==> #true
(print (palindrome? "This is not a palindrome"))
- ==> #false
</lang>
Oz
<lang oz>fun {IsPalindrome S}
{Reverse S} == S
end</lang>
PARI/GP
<lang parigp>ispal(s)={
s=Vec(s); for(i=1,#v\2, if(v[i]!=v[#v-i+1],return(0)) ); 1
};</lang>
A version for numbers:
<lang parigp>ispal(s)={
my(d=digits(n)); for(i=1,#d\2, if(d[i]!=d[n+1=i],return(0)) ); 1
};</lang>
Pascal
<lang pascal>program Palindro;
{ RECURSIVE } function is_palindro_r(s : String) : Boolean; begin
if length(s) <= 1 then
is_palindro_r := true
else begin
if s[1] = s[length(s)] then
is_palindro_r := is_palindro_r(copy(s, 2, length(s)-2))
else
is_palindro_r := false
end
end; { is_palindro_r }
{ NON RECURSIVE; see Reversing a string for "reverse" } function is_palindro(s : String) : Boolean; begin
if s = reverse(s) then
is_palindro := true
else
is_palindro := false
end;</lang>
<lang pascal>procedure test_r(s : String; r : Boolean); begin
write('"', s, '" is ');
if ( not r ) then
write('not ');
writeln('palindrome')
end;
var
s1, s2 : String;
begin
s1 := 'ingirumimusnocteetconsumimurigni'; s2 := 'in girum imus nocte'; test_r(s1, is_palindro_r(s1)); test_r(s2, is_palindro_r(s2)); test_r(s1, is_palindro(s1)); test_r(s2, is_palindro(s2))
end.</lang>
<lang pascal>program PalindromeDetection; var
input, output: string; s: char; i: integer;
begin
writeln('write down your input:');
readln(input);
output:=;
for i:=1 to length(input) do
begin
s:=input[i];
output:=s+output;
end;
writeln();
if(input=output)then
writeln('input was palindrome')
else
writeln('input was not palindrome');
end.</lang>
Perl
There is more than one way to do this.
- palindrome uses the built-in function reverse().
- palindrome_c uses iteration; it is a translation of the C solution.
- palindrome_r uses recursion.
- palindrome_e uses a recursive regular expression.
All of these functions take a parameter, or default to $_ if there is no parameter. None of these functions ignore case or strip characters; if you want do that, you can use ($s = lc $s) =~ s/[\W_]//g before you call these functions.
<lang perl># Palindrome.pm package Palindrome;
use strict; use warnings;
use Exporter 'import'; our @EXPORT = qw(palindrome palindrome_c palindrome_r palindrome_e);
sub palindrome {
my $s = (@_ ? shift : $_); return $s eq reverse $s;
}
sub palindrome_c {
my $s = (@_ ? shift : $_);
for my $i (0 .. length($s) >> 1)
{
return 0 unless substr($s, $i, 1) eq substr($s, -1 - $i, 1);
}
return 1;
}
sub palindrome_r {
my $s = (@_ ? shift : $_);
if (length $s <= 1) { return 1; }
elsif (substr($s, 0, 1) ne substr($s, -1, 1)) { return 0; }
else { return palindrome_r(substr($s, 1, -1)); }
}
sub palindrome_e {
(@_ ? shift : $_) =~ /^(.?|(.)(?1)\2)$/ + 0
}</lang>
This example shows how to use the functions:
<lang perl># pbench.pl use strict; use warnings;
use Benchmark qw(cmpthese); use Palindrome;
printf("%d, %d, %d, %d: %s\n",
palindrome, palindrome_c, palindrome_r, palindrome_e, $_)
for
qw/a aa ab abba aBbA abca abba1 1abba ingirumimusnocteetconsumimurigni/, 'ab cc ba', 'ab ccb a';
printf "\n";
my $latin = "ingirumimusnocteetconsumimurigni"; cmpthese(100_000, {
palindrome => sub { palindrome $latin },
palindrome_c => sub { palindrome_c $latin },
palindrome_r => sub { palindrome_r $latin },
palindrome_e => sub { palindrome_e $latin },
});</lang>
- Output:
on a machine running Perl 5.10.1 on amd64-openbsd
$ perl pbench.pl
1, 1, 1, 1: a
1, 1, 1, 1: aa
0, 0, 0, 0: ab
1, 1, 1, 1: abba
0, 0, 0, 0: aBbA
0, 0, 0, 0: abca
0, 0, 0, 0: abba1
0, 0, 0, 0: 1abba
1, 1, 1, 1: ingirumimusnocteetconsumimurigni
1, 1, 1, 1: ab cc ba
0, 0, 0, 0: ab ccb a
(warning: too few iterations for a reliable count)
Rate palindrome_r palindrome_e palindrome_c palindrome
palindrome_r 51020/s -- -50% -70% -97%
palindrome_e 102041/s 100% -- -41% -94%
palindrome_c 172414/s 238% 69% -- -90%
palindrome 1666667/s 3167% 1533% 867% --
With this machine, palindrome() ran far faster than the alternatives (and too fast for a reliable count). The Perl regular expression engine recursed twice as fast as the Perl interpreter.
Phix
function is_palindrome(sequence s) return s==reverse(s) end function ?is_palindrome("rotator") -- prints 1 ?is_palindrome("tractor") -- prints 0 constant punctuation = " `~!@#$%^&*()-=_+[]{}\\|;:',.<>/?", nulls = repeat("",length(punctuation)) function extra_credit(sequence s) s = utf8_to_utf32(lower(substitute_all(s,punctuation,nulls))) return s==reverse(s) end function -- these all print 1 (true) ?extra_credit("Madam, I'm Adam.") ?extra_credit("A man, a plan, a canal: Panama!") ?extra_credit("In girum imus nocte et consumimur igni") ?extra_credit("人人為我,我為人人") ?extra_credit("Я иду с мечем, судия") ?extra_credit("아들딸들아") ?extra_credit("가련하시다 사장집 아들딸들아 집장사 다시 하련가") ?extra_credit("tregða, gón, reiði - er nóg að gert")
PHP
<lang php><?php function is_palindrome($string) {
return $string == strrev($string);
} ?></lang>
Regular expression-based solution (source) <lang php><?php function is_palindrome($string) {
return preg_match('/^(?:(.)(?=.*(\1(?(2)\2|))$))*.?\2?$/', $string);
} ?></lang>
Picat
<lang Picat>go =>
Tests = ["In girum imus nocte et consumimur igni",
"this is a non palindrome string",
"anna ABcdcBA anna",
"anna ABcdcBA annax",
"A man, a plan, a canoe, pasta, heros, rajahs" ++
"a coloratura, maps, snipe, percale, macaroni, " ++
"a gag, a banana bag, a tan, a tag, " ++
"a banana bag again (or a camel), a crepe, pins, " ++
"Spam, a rut, a Rolo, cash, a jar, sore hats, " ++
"a peon, a canal - Panama!",
10,
111111,
12221,
9384212,
10.01
],
foreach(Test in Tests)
if is_palindrome(Test) then
println([Test, "exact palindrome"])
elseif is_palindrome_inexact(Test) then
println([Test, "inexact palindrome"])
else
println([Test, "no"])
end
end,
nl.
% Detect palindromes for strings (and numbers). is_palindrome(N), number(N) => is_palindrome(N.to_string()). is_palindrome(S) => S == S.reverse().
% Detect inexact palindromes. is_palindrome_inexact(N), number(N) => is_palindrome_inexact(N.to_string()). is_palindrome_inexact(S) =>
is_palindrome(strip(S)).</lang>
- Output:
[In girum imus nocte et consumimur igni,inexact palindrome] [this is a non palindrome string,no] [anna ABcdcBA anna,exact palindrome] [anna ABcdcBA annax,no] [A man, a plan, a canoe, pasta, heros, rajahsa coloratura, maps, snipe, percale, macaroni, a gag, a banana bag, a tan, a tag, a banana bag again (or a camel), a crepe, pins, Spam, a rut, a Rolo, cash, a jar, sore hats, a peon, a canal - Panama!,inexact palindrome] [10,no] [11,exact palindrome] [111111,exact palindrome] [12221,exact palindrome] [9384212,no] [10.01,exact palindrome]
PicoLisp
<lang PicoLisp>(de palindrome? (S)
(= (setq S (chop S)) (reverse S)) )</lang>
- Output:
: (palindrome? "ingirumimusnocteetconsumimurigni") -> T
Pike
<lang pike>int main(){
if(pal("rotator")){
write("palindrome!\n");
}
if(!pal("asdf")){
write("asdf isn't a palindrome.\n");
}
}
int pal(string input){
if( reverse(input) == input ){
return 1;
} else {
return 0;
}
}</lang>
PL/I
To satisfy the revised specification (which contradicts the preceding explanation) the following trivially solves the problem in PL/I: <lang PL/I>is_palindrome = (text = reverse(text));</lang>
The following solution strips spaces: <lang>is_palindrome: procedure (text) returns (bit(1));
declare text character (*) varying;
text = remove_blanks(text); text = lowercase(text); return (text = reverse(text));
remove_blanks: procedure (text);
declare text character (*) varying;
declare (i, j) fixed binary (31);
j = 0;
do i = 1 to length(text);
if substr(text, i, 1) = ' ' then
do; j = j + 1; substr(text, j, 1) = substr(text, i, 1); end;
end;
return (substr(text, 1, j));
end remove_blanks; end is_palindrome;</lang>
PL/M
<lang plm>100H:
/* CHECK EXACT PALINDROME ASSUMING $-TERMINATED STRING */ PALINDROME: PROCEDURE(PTR) BYTE;
DECLARE (PTR, FRONT, BACK) ADDRESS, STR BASED PTR BYTE;
/* FIND END */
FRONT, BACK = 0;
DO WHILE STR(BACK) <> '$';
BACK = BACK + 1;
END;
BACK = BACK - 1;
/* CHECK MATCH */
DO WHILE (FRONT < BACK) AND (STR(FRONT) = STR(BACK));
FRONT = FRONT + 1;
BACK = BACK - 1;
END;
RETURN FRONT >= BACK;
END PALINDROME;
/* CHECK INEXACT PALINDROME: FILTER OUT NON-LETTERS AND NUMBERS */ INEXACT$PALINDROME: PROCEDURE(PTR) BYTE;
/* 256 BYTES OUGHT TO BE ENOUGH FOR EVERYONE */ DECLARE (PTR, OPTR) ADDRESS; DECLARE FILTER (256) BYTE; DECLARE (IN BASED PTR, OUT BASED OPTR) BYTE; OPTR = .FILTER;
DO WHILE IN <> '$';
OUT = IN OR 32;
/* LOWERCASE CHARACTERS ARE NOT IN THE PL/M CHARSET,
BUT WE CAN JUST WRITE THE ASCII VALUES AS NUMBERS */
IF (OUT >= '0' AND OUT <= '9')
OR (OUT >= 97 AND OUT <= 122) THEN
OPTR = OPTR + 1;
PTR = PTR + 1;
END;
OUT = '$';
RETURN PALINDROME(.FILTER);
END INEXACT$PALINDROME;
/* CP/M BDOS CALLS */ BDOS: PROCEDURE(FUNC, ARG);
DECLARE FUNC BYTE, ARG ADDRESS; GO TO 5;
END BDOS;
PRINT: PROCEDURE(STRING);
DECLARE STRING ADDRESS; CALL BDOS(9, STRING);
END PRINT;
/* TEST SOME STRINGS */ DECLARE STRINGS (8) ADDRESS; STRINGS(0) = .'ROTOR$'; STRINGS(1) = .'RACECAR$'; STRINGS(2) = .'LEVEL$'; STRINGS(3) = .'REDDER$'; STRINGS(4) = .'RACECAR$'; STRINGS(5) = .'A MAN, A PLAN, A CANAL: PANAMA$'; STRINGS(6) = .'EGAD, A BASE TONE DENOTES A BAD AGE.$'; STRINGS(7) = .'ROSETTA$';
DECLARE N BYTE; DO N = 0 TO LAST(STRINGS);
CALL PRINT(STRINGS(N));
CALL PRINT(.': $');
IF PALINDROME(STRINGS(N)) THEN
CALL PRINT(.'EXACT$');
ELSE IF INEXACT$PALINDROME(STRINGS(N)) THEN
CALL PRINT(.'INEXACT$');
ELSE
CALL PRINT(.'NOT A PALINDROME$');
CALL PRINT(.(13,10,'$'));
END;
CALL BDOS(0,0); EOF</lang>
- Output:
ROTOR: EXACT RACECAR: EXACT LEVEL: EXACT REDDER: EXACT RACECAR: EXACT A MAN, A PLAN, A CANAL: PANAMA: INEXACT EGAD, A BASE TONE DENOTES A BAD AGE.: INEXACT ROSETTA: NOT A PALINDROME
Plain English
Strings and substrings all come with two byte pointers by default:
first, which points to the first byte in the string.last, which points to the last byte in the string.
first is an address, while first's target is the byte at that address.
No need to actually reverse the string; just compare the first's target with the last's target until they meet in the middle.
<lang plainenglish>To decide if a string is palindromic:
Slap a substring on the string.
Loop.
If the substring's first is greater than the substring's last, say yes.
If the substring's first's target is not the substring's last's target, say no.
Add 1 to the substring's first.
Subtract 1 from the substring's last.
Repeat.</lang>
Pointless
Basic Function <lang pointless>isPalindrome(chars) =
chars == reverse(chars)</lang>
With Pre-processing <lang pointless>output =
"A man, a plan, a canal -- Panama" |> toList |> filter(inFunc(alNums)) |> map(toLower) |> isPalindrome |> println</lang>
- Output:
true
Potion
<lang Potion># The readable recursive version palindrome_i = (s, b, e):
if (e <= b): true. elsif (s ord(b) != s ord(e)): false. else: palindrome_i(s, b+1, e-1).
.
palindrome = (s):
palindrome_i(s, 0, s length - 1).
palindrome(argv(1))</lang>
PowerBASIC
The output is identical to the QBasic version, above.
<lang powerbasic>FUNCTION isPalindrome (what AS STRING) AS LONG
DIM whatcopy AS STRING, chk AS STRING, tmp AS STRING * 1, L0 AS LONG
FOR L0 = 1 TO LEN(what)
tmp = UCASE$(MID$(what, L0, 1))
SELECT CASE tmp
CASE "A" TO "Z"
whatcopy = whatcopy & tmp
chk = tmp & chk
CASE "0" TO "9"
MSGBOX "Numbers are cheating! (""" & what & """)"
FUNCTION = 0
EXIT FUNCTION
END SELECT
NEXT
FUNCTION = ISTRUE((whatcopy) = chk)
END FUNCTION
FUNCTION PBMAIN () AS LONG
DATA "My dog has fleas", "Madam, I'm Adam.", "1 on 1", "In girum imus nocte et consumimur igni"
DIM L1 AS LONG, w AS STRING
FOR L1 = 1 TO DATACOUNT
w = READ$(L1)
IF ISTRUE(isPalindrome(w)) THEN
MSGBOX $DQ & w & """ is a palindrome"
ELSE
MSGBOX $DQ & w & """ is not a palindrome"
END IF
NEXT
END FUNCTION</lang>
PowerShell
An exact version based on reversing the string: <lang PowerShell> Function Test-Palindrome( [String] $Text ){
$CharArray = $Text.ToCharArray() [Array]::Reverse($CharArray) $Text -eq [string]::join(, $CharArray)
} </lang>
PowerShell (Regex Version)
This version is much faster because it does not manipulate arrays. [This is not clear; the above version was slowed down by using -join instead of [string]::join, and -like instead of -eq. After changing those it is similar, if not faster, than this version]. <lang PowerShell> function Test-Palindrome {
<#
.SYNOPSIS
Tests if a string is a palindrome.
.DESCRIPTION
Tests if a string is a true palindrome or, optionally, an inexact palindrome.
.EXAMPLE
Test-Palindrome -Text "racecar"
.EXAMPLE
Test-Palindrome -Text '"Deliver desserts," demanded Nemesis, "emended, named, stressed, reviled."' -Inexact
#>
[CmdletBinding()]
[OutputType([bool])]
Param
(
# The string to test for palindrominity.
[Parameter(Mandatory=$true)]
[string]
$Text,
# When specified, detects an inexact palindrome.
[switch]
$Inexact
)
if ($Inexact)
{
# Strip all punctuation and spaces
$Text = [Regex]::Replace("$Text($7&","[^1-9a-zA-Z]","")
}
$Text -match "^(?'char'[a-z])+[a-z]?(?:\k'char'(?'-char'))+(?(char)(?!))$"
} </lang> <lang PowerShell> Test-Palindrome -Text 'radar' </lang>
- Output:
True
<lang PowerShell> Test-Palindrome -Text "In girum imus nocte et consumimur igni." </lang>
- Output:
False
<lang PowerShell> Test-Palindrome -Text "In girum imus nocte et consumimur igni." -Inexact </lang>
- Output:
True
PowerShell (Unicode category aware, no string reverse)
An inexact version can remove punctuation by looking at Unicode categories for each character, either using .Net methods or a regex. <lang PowerShell>Function Test-Palindrome { [CmdletBinding()] Param(
[Parameter(ValueFromPipeline)] [string[]]$Text
)
process {
:stringLoop foreach ($T in $Text)
{
# Normalize Unicode combining characters,
# so character á compares the same as (a+combining accent)
$T = $T.Normalize([Text.NormalizationForm]::FormC)
# Remove anything from outside the Unicode category
# "Letter from any language"
$T = $T -replace '\P{L}',
# Walk from each end of the string inwards,
# comparing a char at a time.
# Avoids string copy / reverse overheads.
$Left, $Right = 0, [math]::Max(0, ($T.Length - 1))
while ($Left -lt $Right)
{
if ($T[$Left] -ne $T[$Right])
{
# return early if string is not a palindrome
[PSCustomObject]@{
Text = $T
IsPalindrome = $False
}
continue stringLoop
}
else
{
$Left++
$Right--
}
}
# made it to here, then string is a palindrome
[PSCustomObject]@{
Text = $T
IsPalindrome = $True
}
}
} } 'ánu-ná', 'nowt' | Test-Palindrome</lang>
- Output:
PS C:\> 'ánu-ná', 'nowt' | Test-Palindrome Text IsPalindrome ---- ------------ ánuná True now False
Processing
<lang processing> void setup(){ println(isPalindrome(InsertPalindromeHere)); }
boolean isPalindrome(string check){ char[] letters = new char[check.length]; string invert = " "; string modCheck = " " + check; for(int i = 0; i < letters.length; i++){ letters[i] = check.charAt(i); } for(int i = letters.length-1; i >= 0; i--){ invert = invert + letters[i]; }
if(invert == modCheck){ return true; } else { return false; } } </lang>
- Output:
"true" or "false" depending
Alternative Implementation: using StringBuilder, implementing exact and inexact check
<lang processing> void setup(){
println("PalindromeDetection");
String[] tests = {
"abcba",
"aa",
"a",
"",
" ",
"ab",
"abcdba",
"A man, a plan, a canal: Panama!",
"Dammit, I’m Mad!",
"Never odd or even",
"ingirumimusnocteetconsumimurigni"
};
for (int i = 0; i < tests.length; i++){
println((i + 1) + ". '" + tests[i] + "' isExactPalindrome: " + isExactPalindrome(tests[i]) + " isInexactPalindrome: " + isInexactPalindrome(tests[i]));
}
}
/*
- Check for exact palindrome using StringBuilder and String since String in Java does not provide any reverse functionality because Strings are immutable.
- /
boolean isExactPalindrome(String s){ StringBuilder sb = new StringBuilder(s); return s.equals(sb.reverse().toString()); }
/*
- Check for inexact palindrome using the check for exact palindromeabove.
- /
boolean isInexactPalindrome(String s){
// removes all whitespaces and non-visible characters,
// remove anything besides alphabet characters
// ignore case
return isExactPalindrome(s.replaceAll("\\s+","").replaceAll("[^A-Za-z]+", "").toLowerCase());
} </lang>
- Output:
PalindromeDetection 1. 'abcba' isExactPalindrome: true isInexactPalindrome: true 2. 'aa' isExactPalindrome: true isInexactPalindrome: true 3. 'a' isExactPalindrome: true isInexactPalindrome: true 4. '' isExactPalindrome: true isInexactPalindrome: true 5. ' ' isExactPalindrome: true isInexactPalindrome: true 6. 'ab' isExactPalindrome: false isInexactPalindrome: false 7. 'abcdba' isExactPalindrome: false isInexactPalindrome: false 8. 'A man, a plan, a canal: Panama!' isExactPalindrome: false isInexactPalindrome: true 9. 'Dammit, I’m Mad!' isExactPalindrome: false isInexactPalindrome: true 10. 'Never odd or even' isExactPalindrome: false isInexactPalindrome: true 11. 'ingirumimusnocteetconsumimurigni' isExactPalindrome: true isInexactPalindrome: true
Prolog
Non-recursive
From this tutorial.
<lang prolog>palindrome(Word) :- name(Word,List), reverse(List,List).</lang>
Recursive
<lang prolog>pali(Str) :- sub_string(Str, 0, 1, _, X), string_concat(Str2, X, Str), string_concat(X, Mid, Str2), pali(Mid). pali(Str) :- string_length(Str, Len), Len < 2.</lang>
Changing string into atom makes the program run also on GNU Prolog. I.e.
<lang prolog>pali(Str) :- sub_atom(Str, 0, 1, _, X), atom_concat(Str2, X, Str), atom_concat(X, Mid, Str2), pali(Mid). pali(Str) :- atom_length(Str, Len), Len < 2.</lang>
PureBasic
<lang PureBasic>Procedure IsPalindrome(StringToTest.s)
If StringToTest=ReverseString(StringToTest) ProcedureReturn 1 Else ProcedureReturn 0 EndIf
EndProcedure</lang>
Python
Now that Python 2.7 and Python 3.4 are quite different, We should include the version IMHO.
Non-recursive
This one uses the reversing the string technique (to reverse a string Python can use the odd but right syntax string[::-1])
<lang python>def is_palindrome(s):
return s == s[::-1]</lang>
Non-recursive, Ignoring Punctuation/Case/Spaces
A word is a palindrome if the letters are the same forwards as backwards, but the other methods given here will return False for, e.g., an input of "Go hang a salami, I'm a lasagna hog" or "A man, a plan, a canal: Panama." An implementation that traverses the string and ignores case differences, spaces, and non-alpha characters is pretty trivial.
<lang python>def is_palindrome(s):
low = 0
high = len(s) - 1
while low < high:
if not s[low].isalpha():
low += 1
elif not s[high].isalpha():
high -= 1
else:
if s[low].lower() != s[high].lower():
return False
else:
low += 1
high -= 1
return True</lang>
Recursive
<lang python>def is_palindrome_r(s):
if len(s) <= 1: return True elif s[0] != s[-1]: return False else: return is_palindrome_r(s[1:-1])</lang>
Python has short-circuit evaluation of Boolean operations so a shorter and still easy to understand recursive function is
<lang python>def is_palindrome_r2(s):
return not s or s[0] == s[-1] and is_palindrome_r2(s[1:-1])</lang>
Testing
<lang python>def test(f, good, bad):
assert all(f(x) for x in good) assert not any(f(x) for x in bad) print '%s passed all %d tests' % (f.__name__, len(good)+len(bad))
pals = (, 'a', 'aa', 'aba', 'abba') notpals = ('aA', 'abA', 'abxBa', 'abxxBa') for ispal in is_palindrome, is_palindrome_r, is_palindrome_r2:
test(ispal, pals, notpals)</lang>
Palindrome Using Regular Expressions Python 2.7
<lang python>def p_loop():
import re, string
re1="" # Beginning of Regex
re2="" # End of Regex
pal=raw_input("Please Enter a word or phrase: ")
pd = pal.replace(' ',)
for c in string.punctuation:
pd = pd.replace(c,"")
if pal == "" :
return -1
c=len(pd) # Count of chars.
loops = (c+1)/2
for x in range(loops):
re1 = re1 + "(\w)"
if (c%2 == 1 and x == 0):
continue
p = loops - x
re2 = re2 + "\\" + str(p)
regex= re1+re2+"$" # regex is like "(\w)(\w)(\w)\2\1$"
#print(regex) # To test regex before re.search
m = re.search(r'^'+regex,pd,re.IGNORECASE)
if (m):
print("\n "+'"'+pal+'"')
print(" is a Palindrome\n")
return 1
else:
print("Nope!")
return 0</lang>
Checking the left half against a reflection of the right half
<lang python>Palindrome detection
- isPalindrome :: String -> Bool
def isPalindrome(s):
True if the string is unchanged under reversal.
(The left half is a reflection of the right half)
d, m = divmod(len(s), 2)
return s[0:d] == s[d + m:][::-1]
- ------------------------- TEST -------------------------
- main :: IO ()
def main():
Test
print('\n'.join(
f'{repr(s)} -> {isPalindrome(cleaned(s))}' for s in [
"",
"a",
"ab",
"aba",
"abba",
"In girum imus nocte et consumimur igni"
]
))
- cleaned :: String -> String
def cleaned(s):
A lower-case copy of s, with spaces pruned. return [c.lower() for c in s if ' ' != c]
- MAIN ---
if __name__ == '__main__':
main()
</lang>
- Output:
'' -> True 'a' -> True 'ab' -> False 'aba' -> True 'abba' -> True 'In girum imus nocte et consumimur igni' -> True
Twiddle Indexing
I have no idea what this technique is called, so I'm going with "Twiddle Indexing".
Twiddle Indexing v. Negative Indexing
0 1 2 3 4 <-- index
[ a, b, c, d, e ]
~4 ~3 ~2 ~1 ~0 <-- twiddle index
0 1 2 3 4 <-- index
[ a, b, c, d, e ]
-5 -4 -3 -2 -1 <-- negative index
<lang Python>def palindromic(str):
for i in range(len(str)//2):
if str[i] != str[~i]:
return(False)
return(True)</lang>
Quackery
<lang Quackery> [ dup reverse = ] is palindromic ( [ --> b )
[ [] swap witheach
[ upper dup
dup lower = iff
drop else join ]
palindromic ] is inexactpalindrome ( $ --> b )</lang>
Twiddle Indexing
<lang Quackery> [ true swap
dup size 2 / times
[ dup i peek
over i ~ peek != if
[ dip not conclude ] ]
drop ] is palindromic ( [ --> b )</lang>
R
Recursive
Note that the recursive method will fail if the string length is too long. R will assume an infinite recursion if a recursion nests deeper than 5,000. Options may be set in the environment to increase this to 500,000. <lang rsplus>palindro <- function(p) {
if ( nchar(p) == 1 ) {
return(TRUE)
} else if ( nchar(p) == 2 ) {
return(substr(p,1,1) == substr(p,2,2))
} else {
if ( substr(p,1,1) == substr(p, nchar(p), nchar(p)) ) {
return(palindro(substr(p, 2, nchar(p)-1)))
} else {
return(FALSE)
}
}
}</lang>
Iterative <lang rsplus>palindroi <- function(p) {
for(i in 1:floor(nchar(p)/2) ) {
r <- nchar(p) - i + 1
if ( substr(p, i, i) != substr(p, r, r) ) return(FALSE)
}
TRUE
}</lang>
Comparative
This method is somewhat faster than the other two.
Note that this method incorrectly regards an empty string as not a palindrome. Please leave this bug in the code, and take a look a the Testing_a_Function page. <lang rsplus>revstring <- function(stringtorev) {
return(
paste(
strsplit(stringtorev,"")1[nchar(stringtorev):1]
,collapse="")
)
} palindroc <- function(p) {return(revstring(p)==p)}</lang>
Rev
R has a built-in function for reversing vectors, so we only have to coerce our input in to the proper form.
Unicode is supported, but this ignores the "inexact palindromes" extra credit requirement because, without some sort of regex, supporting Unicode while stripping punctuation and white space is hard in R. <lang rsplus>is.Palindrome <- function(string) {
characters <- unlist(strsplit(string, "")) all(characters == rev(characters))
}</lang>
- Output:
The rev solution is not benchmarked.
test <- "ingirumimusnocteetconsumimurigni"
tester <- paste(rep(test,38),collapse="")
> test <- "ingirumimusnocteetconsumimurigni"
> tester <- paste(rep(test,38),collapse="")
> system.time(palindro(tester))
user system elapsed
0.04 0.00 0.04
> system.time(palindroi(tester))
user system elapsed
0.01 0.00 0.02
> system.time(palindroc(tester))
user system elapsed
0 0 0
Racket
<lang Racket> (define (palindromb str)
(let* ([lst (string->list (string-downcase str))]
[slst (remove* '(#\space) lst)])
(string=? (list->string (reverse slst)) (list->string slst))))
- example output
> (palindromb "able was i ere i saw elba")
- t
> (palindromb "waht the hey")
- f
> (palindromb "In girum imus nocte et consumimur igni")
- t
> </lang>
Raku
(formerly Perl 6) <lang perl6>subset Palindrom of Str where {
.flip eq $_ given .comb(/\w+/).join.lc
}
my @tests = q:to/END/.lines;
A man, a plan, a canal: Panama. My dog has fleas Madam, I'm Adam. 1 on 1 In girum imus nocte et consumimur igni END
for @tests { say $_ ~~ Palindrom, "\t", $_ }</lang>
- Output:
True A man, a plan, a canal: Panama. False My dog has fleas True Madam, I'm Adam. False 1 on 1 True In girum imus nocte et consumimur igni
Rascal
The most simple solution: <lang rascal>import String;
public bool palindrome(str text) = toLowerCase(text) == reverse(text);</lang>
A solution that handles sentences with spaces and capitals:
<lang rascal>import String;
public bool palindrome(str text){ text = replaceAll(toLowerCase(text), " ", ""); return text == reverse(text); } </lang>
Example: <lang rascal>rascal>palindrome("In girum imus nocte et consumimur igni") bool: true</lang>
REBOL
<lang REBOL>REBOL [
Title: "Palindrome Recognizer" URL: http://rosettacode.org/wiki/Palindrome
]
- In order to compete with all the one-liners, the operation is
- compressed
- parens force left hand side to evaluate first, where I
- copy the phrase, then uppercase it and assign it to 'p'. Now the
- right hand side is evaluated
- p is copied, then reversed in place;
- the comparison is made and implicitely returned.
palindrome?: func [ phrase [string!] "Potentially palindromatic prose." /local p ][(p: uppercase copy phrase) = reverse copy p]
- Teeny Tiny Test Suite
assert: func [code][print [either do code [" ok"]["FAIL"] mold code]]
print "Simple palindromes, with an exception for variety:" repeat phrase ["z" "aha" "sees" "oofoe" "Deified"][ assert compose [palindrome? (phrase)]]
print [crlf "According to the problem statement, these should fail:"] assert [palindrome? "A man, a plan, a canal, Panama."] ; Punctuation not ignored. assert [palindrome? "In girum imus nocte et consumimur igni"] ; Spaces not removed.
- I know we're doing palindromes, not alliteration, but who could resist...?</lang>
- Output:
Simple palindromes, with an exception for variety: ok [palindrome? "z"] ok [palindrome? "aha"] ok [palindrome? "sees"] FAIL [palindrome? "oofoe"] ok [palindrome? "Deified"] According to the problem statement, these should fail: FAIL [palindrome? "A man, a plan, a canal, Panama."] FAIL [palindrome? "In girum imus nocte et consumimur igni"]
Retro
<lang Retro>
- palindrome? (s-f) dup s:hash [ s:reverse s:hash ] dip eq? ;
'ingirumimusnocteetconsumimurigni palindrome? n:put </lang>
REXX
version 1
<lang REXX>/*REXX pgm checks if phrase is palindromic; ignores the case of the letters. */ parse arg y /*get (optional) phrase from the C.L. */ if y= then y='In girum imus nocte et consumimur igni' /*[↓] translation.*/
/*We walk around in the night and we are burnt by the fire (of love).*/
say 'string = ' y if isTpal(y) then say 'The string is a true palindrome.'
else if isPal(y) then say 'The string is an inexact palindrome.'
else say "The string isn't palindromic."
exit /*stick a fork in it, we're all done. */ /*────────────────────────────────────────────────────────────────────────────*/ isTpal: return reverse(arg(1))==arg(1) isPal: return isTpal(translate(space(x,0)))</lang>
- output :
string = In girum imus nocte et consumimur igni The string is an inexact palindrome.
version 2
(Works with Regina 3.8 and later, with options: AREXX_BIFS and AREXX_SEMANTICS)
It should be noted that the COMPRESS function is not a Classic REXX BIF and isn't present in many REXXes.
The SPACE(string,0) BIF can be used instead.
It should also be noted that UPPER BIF is not present in some REXXes.
Use the PARSE UPPER statement or TRANSLATE() BIF instead.
<lang REXX>
/* REXX */
/*Check whether a string is a palindrome */ parse pull string select when palindrome(string) then say string 'is an exact palindrome.' when palindrome(compress(upper(string))) then say string 'is an inexact palindrome.' otherwise say string 'is not palindromic.' end exit 0
palindrome: procedure parse arg string return string==reverse(string) </lang>
- Output:
ABBA is an exact palindrome. In girum imus nocte et consumimur igni is an inexact palindrome. djdjdj is not palindromic.
Ring
<lang ring> aString = "radar" bString = "" for i=len(aString) to 1 step -1
bString = bString + aString[i]
next see aString if aString = bString see " is a palindrome." + nl else see " is not a palindrome" + nl ok </lang>
Ruby
Non-recursive
<lang ruby>def palindrome?(s)
s == s.reverse
end</lang>
Recursive
<lang ruby>def r_palindrome?(s)
if s.length <= 1 true elsif s[0] != s[-1] false else r_palindrome?(s[1..-2]) end
end</lang>
Testing Note that the recursive method is much slower -- using the 2151 character palindrome by Dan Hoey here, we have: <lang ruby>str = "A man, a plan, a caret, [...2110 chars deleted...] a canal--Panama.".downcase.delete('^a-z') puts palindrome?(str) # => true puts r_palindrome?(str) # => true
require 'benchmark' Benchmark.bm do |b|
b.report('iterative') {10000.times {palindrome?(str)}}
b.report('recursive') {10000.times {r_palindrome?(str)}}
end</lang>
- Output:
true
true
user system total real
iterative 0.062000 0.000000 0.062000 ( 0.055000)
recursive 16.516000 0.000000 16.516000 ( 16.562000)
Run BASIC
<lang runbasic>data "My dog has fleas", "Madam, I'm Adam.", "1 on 1", "In girum imus nocte et consumimur igni"
for i = 1 to 4
read w$ print w$;" is ";isPalindrome$(w$);" Palindrome"
next
FUNCTION isPalindrome$(str$) for i = 1 to len(str$)
a$ = upper$(mid$(str$,i,1)) if (a$ >= "A" and a$ <= "Z") or (a$ >= "0" and a$ <= "9") then b$ = b$ + a$: c$ = a$ + c$
next i if b$ <> c$ then isPalindrome$ = "not"</lang>
- Output:
My dog has fleas is not Palindrome Madam, I'm Adam. is Palindrome 1 on 1 is not Palindrome In girum imus nocte et consumimur igni is Palindrome
Rust
<lang rust>fn is_palindrome(string: &str) -> bool {
let half_len = string.len() / 2;
string
.chars()
.take(half_len)
.eq(string.chars().rev().take(half_len))
}
macro_rules! test {
( $( $x:tt ),* ) => { $( println!("'{}': {}", $x, is_palindrome($x)); )* };
}
fn main() {
test!(
"",
"a",
"ada",
"adad",
"ingirumimusnocteetconsumimurigni",
"人人為我,我為人人",
"Я иду с мечем, судия",
"아들딸들아",
"The quick brown fox"
);
}</lang>
- Output:
'': true 'a': true 'ada': true 'adad': false 'ingirumimusnocteetconsumimurigni': true '人人為我,我為人人': true 'Я иду с мечем, судия': false '아들딸들아': true 'The quick brown fox': false
The above soluion checks if the codepoints form a pallindrome, but it is perhaps more correct to consider if the graphemes form a pallindrome. This can be accomplished with an external library and a slight modification to is_palindrome.
<lang rust>extern crate unicode_segmentation;
use unicode_segmentation::UnicodeSegmentation;
fn is_palindrome(string: &str) -> bool {
string.graphemes(true).eq(string.graphemes(true).rev())
}</lang>
SAS
Description <lang SAS> The macro "palindro" has two parameters: string and ignorewhitespace.
string is the expression to be checked. ignorewhitespace, (Y/N), determines whether or not to ignore blanks and punctuation.
This macro was written in SAS 9.2. If you use a version before SAS 9.1.3, the compress function options will not work. </lang> Code <lang SAS> %MACRO palindro(string, ignorewhitespace);
DATA _NULL_; %IF %UPCASE(&ignorewhitespace)=Y %THEN %DO;
/* The arguments of COMPRESS (sp) ignore blanks and puncutation */ /* We take the string and record it in reverse order using the REVERSE function. */
%LET rev=%SYSFUNC(REVERSE(%SYSFUNC(COMPRESS(&string,,sp))));
%LET string=%SYSFUNC(COMPRESS(&string.,,sp));
%END;
%ELSE %DO;
%LET rev=%SYSFUNC(REVERSE(&string));
%END;
/*%PUT rev=&rev.;*/
/*%PUT string=&string.;*/
/* Here we determine if the string and its reverse are the same. */
%IF %UPCASE(&string)=%UPCASE(&rev.) %THEN %DO;
%PUT TRUE;
%END;
%ELSE %DO;
%PUT FALSE;
%END;
RUN;
%MEND; </lang> Example macro call and output <lang SAS> %palindro("a man, a plan, a canal: panama",y);
TRUE
NOTE: DATA statement used (Total process time):
real time 0.00 seconds
cpu time 0.00 seconds
%palindro("a man, a plan, a canal: panama",n);
FALSE
NOTE: DATA statement used (Total process time):
real time 0.00 seconds
cpu time 0.00 seconds
</lang>
Scala
Non-recursive, robustified
<lang Scala> def isPalindrome(s: String): Boolean = (s.size >= 2) && s == s.reverse</lang>
Bonus: Detect and account for odd space and punctuation
<lang scala> def isPalindromeSentence(s: String): Boolean =
(s.size >= 2) && {
val p = s.replaceAll("[^\\p{L}]", "").toLowerCase
p == p.reverse
}
</lang>
Recursive
<lang Scala>import scala.annotation.tailrec
def isPalindromeRec(s: String) = {
@tailrec
def inner(s: String): Boolean =
(s.length <= 1) || (s.head == s.last) && inner(s.tail.init)
(s.size >= 2) && inner(s) }</lang>
Testing <lang Scala> // Testing
assert(!isPalindrome(""))
assert(!isPalindrome("z"))
assert(isPalindrome("amanaplanacanalpanama"))
assert(!isPalindrome("Test 1,2,3"))
assert(isPalindrome("1 2 1"))
assert(!isPalindrome("A man a plan a canal Panama."))
assert(!isPalindromeSentence(""))
assert(!isPalindromeSentence("z"))
assert(isPalindromeSentence("amanaplanacanalpanama"))
assert(!isPalindromeSentence("Test 1,2,3"))
assert(isPalindromeSentence("1 2 1"))
assert(isPalindromeSentence("A man a plan a canal Panama."))
assert(!isPalindromeRec(""))
assert(!isPalindromeRec("z"))
assert(isPalindromeRec("amanaplanacanalpanama"))
assert(!isPalindromeRec("Test 1,2,3"))
assert(isPalindromeRec("1 2 1"))
assert(!isPalindromeRec("A man a plan a canal Panama."))
println("Successfully completed without errors.")</lang>
Scheme
Non-recursive
<lang scheme>(define (palindrome? s)
(let ((chars (string->list s))) (equal? chars (reverse chars))))</lang>
Recursive <lang scheme>(define (palindrome? s)
(let loop ((i 0)
(j (- (string-length s) 1)))
(or (>= i j)
(and (char=? (string-ref s i) (string-ref s j))
(loop (+ i 1) (- j 1))))))
- Or
(define (palindrome? s)
(let loop ((s (string->list s))
(r (reverse (string->list s))))
(or (null? s)
(and (char=? (car s) (car r))
(loop (cdr s) (cdr r))))))
> (palindrome? "ingirumimusnocteetconsumimurigni")
- t
> (palindrome? "This is not a palindrome")
- f
></lang>
Seed7
<lang seed7>const func boolean: palindrome (in string: stri) is func
result
var boolean: isPalindrome is TRUE;
local
var integer: index is 0;
var integer: length is 0;
begin
length := length(stri);
for index range 1 to length div 2 do
if stri[index] <> stri[length - index + 1] then
isPalindrome := FALSE;
end if;
end for;
end func;</lang>
For palindromes where spaces shuld be ignore use: <lang seed7>palindrome(replace("in girum imus nocte et consumimur igni", " ", ""))</lang>
SequenceL
Using the Reverse Library Function <lang sequencel>import <Utilities/Sequence.sl>;
isPalindrome(string(1)) := equalList(string, reverse(string));</lang>
Version Using an Indexed Function <lang sequencel>isPalindrome(string(1)) := let compares[i] := string[i] = string[size(string) - (i - 1)] foreach i within 1 ... (size(string) / 2); in all(compares);</lang>
Sidef
Built-in <lang ruby>say "noon".is_palindrome; # true</lang>
Non-recursive
<lang ruby>func palindrome(s) {
s == s.reverse
}</lang>
Recursive
<lang ruby>func palindrome(s) {
if (s.len <= 1) {
true
}
elsif (s.first != s.last) {
false
}
else {
__FUNC__(s.ft(1, -2))
}
}</lang>
Simula
<lang simula>BEGIN
BOOLEAN PROCEDURE ISPALINDROME(T); TEXT T;
BEGIN
BOOLEAN RESULT;
INTEGER I, J;
I := 1;
J := T.LENGTH;
RESULT := TRUE;
WHILE RESULT AND I < J DO
BEGIN
CHARACTER L, R;
T.SETPOS(I); L := T.GETCHAR; I := I + 1;
T.SETPOS(J); R := T.GETCHAR; J := J - 1;
RESULT := L = R;
END;
ISPALINDROME := RESULT;
END ISPALINDROME;
TEXT T;
FOR T :- "", "A", "AA", "ABA", "SALALAS", "MADAMIMADAM",
"AB", "AAB", "ABCBDA"
DO
BEGIN
OUTTEXT(IF ISPALINDROME(T) THEN "IS " ELSE "ISN'T");
OUTTEXT(" PALINDROME: ");
OUTCHAR('"');
OUTTEXT(T);
OUTCHAR('"');
OUTIMAGE;
END;
END.</lang>
- Output:
IS PALINDROME: "" IS PALINDROME: "A" IS PALINDROME: "AA" IS PALINDROME: "ABA" IS PALINDROME: "SALALAS" IS PALINDROME: "MADAMIMADAM" ISN'T PALINDROME: "AB" ISN'T PALINDROME: "AAB" ISN'T PALINDROME: "ABCBDA"
Slate
Non-Recursive <lang slate>s@(String traits) isPalindrome [
(s lexicographicallyCompare: s reversed) isZero
].</lang>
Recursive Defined on Sequence since we are not using String-specific methods: <lang slate>s@(Sequence traits) isPalindrome [
s isEmpty ifTrue: [True] ifFalse: [(s first = s last) /\ [(s sliceFrom: 1 to: s indexLast - 1) isPalindrome]]
].</lang>
Testing <lang slate>define: #p -> 'ingirumimusnocteetconsumimurigni'. inform: 'sequence ' ; p ; ' is ' ; (p isPalindrome ifTrue: [] ifFalse: ['not ']) ; 'a palindrome.'.</lang>
Smalltalk
<lang smalltalk>isPalindrome := [:aString | str := (aString select: [:chr| chr isAlphaNumeric]) collect: [:chr | chr asLowercase]. str = str reversed. ]. </lang>
<lang smalltalk>String extend [
palindro [ "Non-recursive"
^ self = (self reverse)
]
palindroR [ "Recursive"
(self size) <= 1 ifTrue: [ ^true ]
ifFalse: [ |o i f| o := self asOrderedCollection.
i := o removeFirst.
f := o removeLast.
i = f ifTrue: [ ^ (o asString) palindroR ]
ifFalse: [ ^false ]
]
]
].</lang>
Testing
<lang smalltalk>('hello' palindro) printNl. ('hello' palindroR) printNl. ('ingirumimusnocteetconsumimurigni' palindro) printNl. ('ingirumimusnocteetconsumimurigni' palindroR) printNl.</lang>
<lang smalltalk>SequenceableCollection>>isPalindrome ^self reverse = self </lang>
SNOBOL4
<lang SNOBOL4> define('pal(str)') :(pal_end) pal str notany(&ucase &lcase) = :s(pal)
str = replace(str,&ucase,&lcase)
leq(str,reverse(str)) :s(return)f(freturn)
pal_end
define('palchk(str)tf') :(palchk_end)
palchk output = str;
tf = 'False'; tf = pal(str) 'True'
output = 'Palindrome: ' tf :(return)
palchk_end
- # Test and display
palchk('Able was I ere I saw Elba')
palchk('In girum imus nocte et consumimur igni')
palchk('The quick brown fox jumped over the lazy dogs')
end</lang>
- Output:
Able was I ere I saw Elba Palindrome: True In girum imus nocte et consumimur igni Palindrome: True The quick brown fox jumped over the lazy dogs Palindrome: False
SQL
<lang sql>SET @txt = REPLACE('In girum imus nocte et consumimur igni', ' ', ); SELECT REVERSE(@txt) = @txt;</lang>
Swift
<lang Swift>import Foundation
// Allow for easy character checking extension String {
subscript (i: Int) -> String {
return String(Array(self)[i])
}
}
func isPalindrome(str:String) -> Bool {
if (count(str) == 0 || count(str) == 1) {
return true
}
let removeRange = Range<String.Index>(start: advance(str.startIndex, 1), end: advance(str.endIndex, -1))
if (str[0] == str[count(str) - 1]) {
return isPalindrome(str.substringWithRange(removeRange))
}
return false
}</lang>
<lang swift>func isPal(str: String) -> Bool {
let c = str.characters return lazy(c).reverse() .startsWith(c[c.startIndex...advance(c.startIndex, c.count / 2)])
}</lang>
Tcl
Non-recursive
<lang tcl>package require Tcl 8.5 proc palindrome {s} {
return [expr {$s eq [string reverse $s]}]
}</lang>
Recursive
<lang tcl>proc palindrome_r {s} {
if {[string length $s] <= 1} {
return true
} elseif {[string index $s 0] ne [string index $s end]} {
return false
} else {
return [palindrome_r [string range $s 1 end-1]]
}
}</lang>
Testing
<lang tcl>set p ingirumimusnocteetconsumimurigni puts "'$p' is palindrome? [palindrome $p]" puts "'$p' is palindrome? [palindrome_r $p]"</lang>
TUSCRIPT
<lang tuscript> $$ MODE TUSCRIPT pal ="ingirumimusnocteetconsumimurigni" pal_r=TURN(pal) SELECT pal CASE $pal_r PRINT "true" DEFAULT PRINT/ERROR "untrue" ENDSELECT </lang>
- Output:
true
TypeScript
<lang javascript>const detectNonLetterRegexp=/[^A-ZÀ-ÞЀ-Я]/g;
function stripDiacritics(phrase:string){
return phrase.normalize('NFD').replace(/[\u0300-\u036f]/g, "")
}
function isPalindrome(phrase:string){
const TheLetters = stripDiacritics(phrase.toLocaleUpperCase().replace(detectNonLetterRegexp, )); const middlePosition = TheLetters.length/2; const leftHalf = TheLetters.substr(0, middlePosition); const rightReverseHalf = TheLetters.substr(-middlePosition).split().reverse().join(); return leftHalf == rightReverseHalf;
}
console.log(isPalindrome('Sueño que esto no es un palíndromo')) console.log(isPalindrome('Dábale arroz a la zorra el abad!')) console.log(isPalindrome('Я иду с мечем судия')) </lang>
UNIX Shell
<lang bash>if [[ "${text}" == "$(rev <<< "${text}")" ]]; then
echo "Palindrome"
else
echo "Not a palindrome"
fi</lang>
Ursala
The algorithm is to convert to lower case, and then compare the intersection of the argument and the set of letters (declared in the standard library) with its reversal. This is done using the built in operator suffixes for intersection (c), identity (i), reversal (x) and equality (E). <lang Ursala>#import std
palindrome = ~&cixE\letters+ * -:~& ~=`A-~rlp letters</lang> This test programs applies the function to each member of a list of three strings, of which only the first two are palindromes. <lang Ursala>#cast %bL
examples = palindrome* <'abccba','foo ba rra bo of','notone'></lang>
- Output:
<true,true,false>
Vala
Checks if a word is a palindrome ignoring the case and spaces. <lang vala>bool is_palindrome (string str) {
var tmp = str.casefold ().replace (" ", "");
return tmp == tmp.reverse ();
}
int main (string[] args) {
print (is_palindrome (args[1]).to_string () + "\n"); return 0;
}</lang>
VBA
This function uses function Reverse() (or Rreverse()) from Reverse a string, after first stripping spaces from the string using the built-in function Replace and converting it to lower case. It can't handle punctuation (yet). Just like the VBScript version it could also work using StrReverse.
<lang VBA> Public Function isPalindrome(aString as string) as Boolean dim tempstring as string
tempstring = Lcase(Replace(aString, " ", "")) isPalindrome = (tempstring = Reverse(tempstring))
End Function </lang>
- Example:
print isPalindrome("In girum imus nocte et consumimur igni")
True
VBScript
Implementation
<lang vb>function Squish( s1 ) dim sRes sRes = vbNullString dim i, c for i = 1 to len( s1 ) c = lcase( mid( s1, i, 1 )) if instr( "abcdefghijklmnopqrstuvwxyz0123456789", c ) then sRes = sRes & c end if next Squish = sRes end function
function isPalindrome( s1 ) dim squished squished = Squish( s1 ) isPalindrome = ( squished = StrReverse( squished ) ) end function</lang>
Invocation
<lang vb>wscript.echo isPalindrome( "My dog has fleas") wscript.echo isPalindrome( "Madam, I'm Adam.") wscript.echo isPalindrome( "1 on 1") wscript.echo isPalindrome( "In girum imus nocte et consumimur igni")</lang>
- Output:
0 -1 0 -1
Vedit macro language
This routine checks if current line is a palindrome:
<lang vedit>:PALINDROME: EOL #2 = Cur_Col-2 BOL for (#1 = 0; #1 <= #2/2; #1++) {
if (CC(#1) != CC(#2-#1)) { Return(0) }
} Return(1)</lang>
Testing:
<lang vedit>Call("PALINDROME") if (Return_Value) {
Statline_Message("Yes")
} else {
Statline_Message("No")
} Return</lang>
Visual Basic .NET
<lang vbnet>Module Module1
Function IsPalindrome(p As String) As Boolean
Dim temp = p.ToLower().Replace(" ", "")
Return StrReverse(temp) = temp
End Function
Sub Main()
Console.WriteLine(IsPalindrome("In girum imus nocte et consumimur igni"))
End Sub
End Module</lang>
- Output:
True
Wortel
<lang wortel>@let {
; Using a hook pal1 @(= @rev) ; Function with argument pal2 &s = s @rev s ; for inexact palindromes pal3 ^(@(= @rev) .toLowerCase. &\@replace[&"\s+"g ""]) [[ !pal1 "abcba" !pal2 "abcbac" !pal3 "In girum imus nocte et consumimur igni" ]]
}</lang>
Returns:
[true false true]
Wren
<lang javascript>var isPal = Fn.new { |word| word == ((word.count > 0) ? word[-1..0] : "") }
System.print("Are the following palindromes?") for (word in ["rotor", "rosetta", "step on no pets", "été", "wren", "🦊😀🦊"]) {
System.print(" %(word) => %(isPal.call(word))")
}</lang>
- Output:
Are the following palindromes? rotor => true rosetta => false step on no pets => true été => true wren => false 🦊😀🦊 => true
X86 Assembly
<lang x86asm>
- x86_84 Linux nasm
section .text
isPalindrome:
mov rsi, rax mov rdi, rax
get_end: cmp byte [rsi], 0 je get_result inc rsi jmp get_end
get_result: mov rax, 0 dec rsi
compare:
mov cl, byte [rdi]
cmp byte [rsi], cl
jne not_palindrome
cmp rsi, rdi
je palindrome
inc rdi
cmp rdi, rsi
je palindrome
dec rsi
jmp compare
not_palindrome: mov rax, 0 ret palindrome: mov rax, 1 ret
</lang>
XPL0
<lang XPL0>include c:\cxpl\codes; \intrinsic 'code' declarations string 0; \use zero-terminated strings
func StrLen(Str); \Return number of characters in an ASCIIZ string char Str; int I; for I:= 0 to -1>>1-1 do
if Str(I) = 0 then return I;
func Palindrome(S); \Return 'true' if S is a palindrome char S; int L, I; [L:= StrLen(S); for I:= 0 to L/2-1 do
if S(I) # S(L-1-I) then return false;
return true; ]; \Palindrome
int Word, I; [Word:=
["otto", "mary", "ablewasiereisawelba", "ingirumimusnocteetconsumimurigni"];
for I:= 0 to 4-1 do
[Text(0, if Palindrome(Word(I)) then "yes" else "no"); CrLf(0); ];
]</lang>
- Output:
yes no yes yes
Yorick
Function is_palindrome meets the task description. Function prep_palindrome demonstrates how to convert an English sentence into a form that can be tested with is_palindrome (by changing case and stripping non-alphabetical characters).
<lang yorick>func is_palindrome(str) {
s = strchar(str)(:-1); return allof(s == s(::-1));
}
func prep_palindrome(str) {
s = strchar(strlower(str)); w = where(s >= 'a' & s <= 'z'); return strchar(s(w));
}</lang>
zkl
<lang zkl>fcn pali(text){
if (text.len()<2) return(False); text==text.reverse();
} fcn pali2(text){ pali((text - " \t\n.,").toLower()) } // or whatever punctuation is</lang>
- Output:
pali("red rum sir is murder") //--> False
pali("red rum sir is murder" - " ") //-->True, remove spaces
pali2("In girum imus nocte et consumimur igni") //-->True
Zoea
<lang Zoea> program: palindrome
case: 1
input: abcdcba
output: true
case: 2
input: dog
output: false
case: 3
input: x
output: true
case: 4
input: abc
output: false
</lang>
Zoea Visual
- Programming Tasks
- Text processing
- Recursion
- String manipulation
- Classic CS problems and programs
- Palindromes
- 11l
- 360 Assembly
- 8080 Assembly
- 8086 Assembly
- ACL2
- Action!
- ActionScript
- Ada
- ALGOL 68
- APL
- AppleScript
- Applesoft BASIC
- ARM Assembly
- Arturo
- AutoHotkey
- AutoIt
- AWK
- BaCon
- Bash
- BASIC
- IS-BASIC
- Sinclair ZX81 BASIC
- BBC BASIC
- Batch File
- BCPL
- Befunge
- BQN
- Bracmat
- Burlesque
- C
- C sharp
- C++
- Clojure
- CLU
- COBOL
- CoffeeScript
- Common Lisp
- Component Pascal
- Cowgol
- Crystal
- D
- Dart
- Delphi
- Dyalect
- Déjà Vu
- E
- EchoLisp
- Eiffel
- Ela
- Elixir
- Elm
- Emacs Lisp
- Erlang
- Euphoria
- Excel
- F Sharp
- Factor
- Falcon
- Fantom
- FBSL
- Forth
- Fortran
- FreeBASIC
- Frink
- Fōrmulæ
- GAP
- GML
- Go
- Groovy
- Haskell
- HicEst
- HicEst examples needing attention
- Examples needing attention
- Icon
- Unicon
- Icon Programming Library
- Ioke
- J
- Java
- JavaScript
- Jq
- Jsish
- Julia
- K
- Kotlin
- LabVIEW
- Langur
- Lasso
- Liberty BASIC
- LiveCode
- Logo
- Lua
- M4
- Maple
- Mathematica
- Wolfram Language
- MATLAB
- Maxima
- MAXScript
- Min
- MiniScript
- Mirah
- ML
- MLite
- Standard ML
- MMIX
- Modula-2
- Modula-3
- Nanoquery
- Nemerle
- NetRexx
- NewLISP
- Nim
- Objeck
- OCaml
- Octave
- Oforth
- Ol
- Oz
- PARI/GP
- Pascal
- Perl
- Phix
- PHP
- Picat
- PicoLisp
- Pike
- PL/I
- PL/M
- Plain English
- Pointless
- Potion
- PowerBASIC
- PowerShell
- Processing
- Prolog
- PureBasic
- Python
- Quackery
- R
- Racket
- Raku
- Rascal
- REBOL
- Retro
- REXX
- Ring
- Ruby
- Run BASIC
- Rust
- SAS
- Scala
- Scheme
- Seed7
- SequenceL
- Sidef
- Simula
- Slate
- Smalltalk
- SNOBOL4
- SQL
- Swift
- Tcl
- TUSCRIPT
- TypeScript
- UNIX Shell
- Ursala
- Vala
- VBA
- VBScript
- Vedit macro language
- Visual Basic .NET
- Wortel
- Wren
- X86 Assembly
- XPL0
- Yorick
- Zkl
- Zoea
- Zoea Visual