Pangram checker

Pangram checker
You are encouraged to solve this task according to the task description, using any language you may know.

Write a function or method to check a sentence to see if it is a pangram or not and show its use.

A pangram is a sentence that contains all the letters of the English alphabet at least once, for example: The quick brown fox jumps over the lazy dog.

ActionScript 2.0

<lang ActionScript> //i know this method is very barbaric, but it does the job with just basic commands. function pangram(k:String) { var hasA:Boolean = false; var hasB:Boolean = false; var hasC:Boolean = false; var hasD:Boolean = false; var hasE:Boolean = false; var hasF:Boolean = false; var hasG:Boolean = false; var hasH:Boolean = false; var hasI:Boolean = false; var hasJ:Boolean = false; var hasK:Boolean = false; var hasL:Boolean = false; var hasM:Boolean = false; var hasN:Boolean = false; var hasO:Boolean = false; var hasP:Boolean = false; var hasQ:Boolean = false; var hasR:Boolean = false; var hasS:Boolean = false; var hasT:Boolean = false; var hasU:Boolean = false; var hasV:Boolean = false; var hasW:Boolean = false; var hasX:Boolean = false; var hasY:Boolean = false; var hasZ:Boolean = false; for (i=0; i<=k.length-1; i++) { if (k.charAt(i) == "A" or k.charAt(i) == "a") { hasA = true; } if (k.charAt(i) == "B" or k.charAt(i) == "b") { hasB = true; } if (k.charAt(i) == "C" or k.charAt(i) == "c") { hasC = true; } if (k.charAt(i) == "D" or k.charAt(i) == "d") { hasD = true; } if (k.charAt(i) == "E" or k.charAt(i) == "e") { hasE = true; } if (k.charAt(i) == "F" or k.charAt(i) == "f") { hasF = true; } if (k.charAt(i) == "G" or k.charAt(i) == "g") { hasG = true; } if (k.charAt(i) == "H" or k.charAt(i) == "h") { hasH = true; } if (k.charAt(i) == "I" or k.charAt(i) == "i") { hasI = true; } if (k.charAt(i) == "J" or k.charAt(i) == "j") { hasJ = true; } if (k.charAt(i) == "K" or k.charAt(i) == "k") { hasK = true; } if (k.charAt(i) == "L" or k.charAt(i) == "l") { hasL = true; } if (k.charAt(i) == "M" or k.charAt(i) == "m") { hasM = true; } if (k.charAt(i) == "N" or k.charAt(i) == "n") { hasN = true; } if (k.charAt(i) == "O" or k.charAt(i) == "o") { hasO = true; } if (k.charAt(i) == "P" or k.charAt(i) == "p") { hasP = true; } if (k.charAt(i) == "Q" or k.charAt(i) == "q") { hasQ = true; } if (k.charAt(i) == "R" or k.charAt(i) == "r") { hasR = true; } if (k.charAt(i) == "S" or k.charAt(i) == "s") { hasS = true; } if (k.charAt(i) == "T" or k.charAt(i) == "t") { hasT = true; } if (k.charAt(i) == "U" or k.charAt(i) == "u") { hasU = true; } if (k.charAt(i) == "V" or k.charAt(i) == "v") { hasV = true; } if (k.charAt(i) == "W" or k.charAt(i) == "w") { hasW = true; } if (k.charAt(i) == "X" or k.charAt(i) == "x") { hasX = true; } if (k.charAt(i) == "Y" or k.charAt(i) == "y") { hasY = true; } if (k.charAt(i) == "Z" or k.charAt(i) == "z") { hasZ = true; } } if (hasA == true and hasB == true and hasC == true and hasD == true and hasE == true and hasF == true and hasG == true and hasH == true and hasI == true and hasJ == true and hasK == true and hasL == true and hasM == true and hasN == true and hasO == true and hasP == true and hasQ == true and hasR == true and hasS == true and hasT == true and hasU == true and hasV == true and hasW == true and hasX == true and hasY == true and hasZ == true) { return true; } else { return false; } }

trace(pangram("the QuiCk Brown Fox jumps over the lazy dog")); //true </lang>

Ada

Using Sets <lang Ada> with Ada.Text_IO; use Ada.Text_IO; with Ada.Strings.Maps; use Ada.Strings.Maps; with Ada.Characters.Handling; use Ada.Characters.Handling; procedure pangram is

function ispangram(txt: String) return Boolean is lowtxt : String := To_Lower(txt); letset,txtset : Character_Set; begin letset := To_Set("abcdefghijklmnopqrstuvwxyz"); txtset := To_Set(lowtxt); return (letset-txtset)=Null_Set; end ispangram;

begin put_line(Boolean'Image(ispangram("This is a test"))); put_line(Boolean'Image(ispangram("The quick brown fox jumps over the lazy dog"))); put_line(Boolean'Image(ispangram("NOPQRSTUVWXYZ abcdefghijklm"))); put_line(Boolean'Image(ispangram("abcdefghijklopqrstuvwxyz"))); --Missing m, n end pangram; </lang> Output:

FALSE
TRUE
TRUE
FALSE

ALGOL 68

Works with: ALGOL 68 version Standard - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

Works with: ELLA ALGOL 68 version Any (with appropriate job cards)

<lang algol68># init pangram: # INT la = ABS "a", lz = ABS "z"; INT ua = ABS "A", uz = ABS "Z"; IF lz-la+1 > bits width THEN

 put(stand error, "Exception: insufficient bits in word for task");
 stop

FI;

PROC is a pangram = (STRING test)BOOL: (

 BITS a2z := BIN(ABS(2r1 SHL (lz-la))-1); # assume: ASCII & Binary #
 FOR i TO UPB test WHILE
   INT c = ABS test[i];
   IF la <= c AND c <= lz THEN
     a2z := a2z AND NOT(2r1 SHL (c-la))
   ELIF ua <= c AND c <= uz THEN
     a2z := a2z AND NOT(2r1 SHL (c-ua))
   FI;

1. WHILE # a2z /= 2r0 DO

   SKIP
 OD;
 a2z = 2r0

);

main:(

 []STRING test list = (
   "Big fjiords vex quick waltz nymph",
   "The quick brown fox jumps over a lazy dog",
   "A quick brown fox jumps over a lazy dog"
 );
 FOR key TO UPB test list DO
   STRING test = test list[key];
   IF is a pangram(test) THEN
     print(("""",test,""" is a pangram!", new line))
   FI
 OD

)</lang> Output:

"Big fjiords vex quick waltz nymph" is a pangram!
"The quick brown fox jumps over a lazy dog" is a pangram!

AutoHotkey

<lang autohotkey>Gui, -MinimizeBox Gui, Add, Edit, w300 r5 vText Gui, Add, Button, x105 w100 Default, Check Pangram Gui, Show,, Pangram Checker Return

GuiClose:

   ExitApp

Return

ButtonCheckPangram:

   Gui, Submit, NoHide
   Loop, 26
       If Not InStr(Text, Char := Chr(64 + A_Index)) {
           MsgBox,, Pangram, Character %Char% is missing!
           Return
       }
   MsgBox,, Pangram, OK`, this is a Pangram!

Return</lang>

BASIC

Works with: QBasic

<lang qbasic>DECLARE FUNCTION IsPangram! (sentence AS STRING)

DIM x AS STRING

x = "My dog has fleas." GOSUB doIt x = "The lazy dog jumps over the quick brown fox." GOSUB doIt x = "Jackdaws love my big sphinx of quartz." GOSUB doIt x = "What's a jackdaw?" GOSUB doIt

END

doIt:

   PRINT IsPangram!(x), x
   RETURN

FUNCTION IsPangram! (sentence AS STRING)

   'returns -1 (true) if sentence is a pangram, 0 (false) otherwise
   DIM l AS INTEGER, s AS STRING, t AS INTEGER
   DIM letters(25) AS INTEGER

   FOR l = 1 TO LEN(sentence)
       s = UCASE$(MID$(sentence, l, 1))
       SELECT CASE s
           CASE "A" TO "Z"
               t = ASC(s) - 65
               letters(t) = 1
       END SELECT
   NEXT

   FOR l = 0 TO 25
       IF letters(l) < 1 THEN
           IsPangram! = 0
           EXIT FUNCTION
       END IF
   NEXT

   IsPangram! = -1

END FUNCTION</lang>

Output:

 0            My dog has fleas.
-1            The quick brown fox jumps over the lazy dog.
-1            Jackdaws love my big sphinx of quartz.
 0            What's a jackdaw?

BBC BASIC

<lang bbcbasic> FOR test% = 1 TO 2

       READ test$
       PRINT """" test$ """ " ;
       IF FNpangram(test$) THEN
         PRINT "is a pangram"
       ELSE
         PRINT "is not a pangram"
       ENDIF
     NEXT test%
     END
     
     DATA "The quick brown fox jumped over the lazy dog"
     DATA "The five boxing wizards jump quickly"
     
     DEF FNpangram(A$)
     LOCAL C%
     A$ = FNlower(A$)
     FOR C% = ASC("a") TO ASC("z")
       IF INSTR(A$, CHR$(C%)) = 0 THEN = FALSE
     NEXT
     = TRUE
     
     DEF FNlower(A$)
     LOCAL A%, C%
     FOR A% = 1 TO LEN(A$)
       C% = ASCMID$(A$,A%)
       IF C% >= 65 IF C% <= 90 MID$(A$,A%,1) = CHR$(C%+32)
     NEXT
     = A$</lang>

Output:

"The quick brown fox jumped over the lazy dog" is not a pangram
"The five boxing wizards jump quickly" is a pangram

Brat

<lang brat>pangram? = { sentence |

 letters = [:a :b :c :d :e :f :g :h :i :j :k :l :m
   :n :o :p :q :r :s :t :u :v :w :x :y :z]

   sentence.downcase!

   letters.reject! { l |
     sentence.include? l
   }

 letters.empty?

}

p pangram? 'The quick brown fox jumps over the lazy dog.' #Prints true p pangram? 'Probably not a pangram.' #Prints false</lang>

Alternative version:

<lang brat>pangram? = { sentence |

 sentence.downcase.dice.unique.select(:alpha?).length == 26

}</lang>

C

<lang C>#include <stdio.h>

1. include <stdlib.h>
2. include <string.h>

int isPangram( char *string ) {

   static char *alphabet="abcdefghijklmnopqrstuvwxyz";
   char wasused[26], *sp, *ap;
   int j;

   for (j=0; j<26; j++) wasused[j]=0;
   for (sp=string;*sp; sp++) {
       ap = strchr(alphabet, tolower(*sp));
       if (ap != NULL) {
          wasused[ap-alphabet] = 1;
       }
   }
   for (j=0; (j<26) && wasused[j]; j++);

// if (j<26) printf("Missing character %c\n", alphabet[j]);

   return (j==26);

}

main( int argc, char *argv[]) {

  char *pangramTxt;
  if (argc < 2) {
      printf("usage %s (text to check for pangram)\n", argv[0]);
      exit(1);
   }
   pangramTxt = argv[1];

printf("%s\n",pangramTxt);

   printf("Is pangram? %s\n", (isPangram(pangramTxt)?"Yes":"No"));
   return 0;

}</lang> Usage:

>pangram "The quick brown fox jumps over lazy dogs."
Is Pangram? Yes

C#

<lang csharp>using System; using System.Linq;

static class Program {

   static bool IsPangram(this string text, string alphabet = "abcdefghijklmnopqrstuvwxyz")
   {
       return alphabet.All(text.ToLower().Contains);
   }

   static void Main(string[] arguments)
   {
       Console.WriteLine(arguments.Any() && arguments.First().IsPangram());
   }

}</lang>

C++

<lang cpp>#include <algorithm>

1. include <cctype>
2. include <string>

using namespace std;

const string alphabet("abcdefghijklmnopqrstuvwxyz"); // sorted, no duplicates

bool is_pangram(string s) {

 // Convert to lower case.
 transform(s.begin(), s.end(), s.begin(), ::tolower);
 // Convert to a sorted sequence of unique characters.
 sort(s.begin(), s.end());
 // Is the second sequence a subset of the first sequence?
 return includes(s.begin(), s.end(), alphabet.begin(), alphabet.end());

}</lang>

Common Lisp

<lang lisp>(defun pangramp (s)

 (null (set-difference
         (loop for c from (char-code #\A) upto (char-code #\Z) collect (code-char c))
         (coerce (string-upcase s) 'list))))</lang>

D

A low-level solution: <lang d>pure nothrow bool isPangram(string text) {

   uint bitset;
   foreach (c; text) {
       if (c >= 'a' && c <= 'z')
           bitset |= (1u << (c - 'a'));
       else if (c >= 'A' && c <= 'Z')
           bitset |= (1u << (c - 'A'));
   }
   return bitset == 0b11_11111111_11111111_11111111;

}

void main() {

   assert(isPangram("the quick brown fox jumps over the lazy dog"));
   assert(!isPangram("ABCDEFGHIJKLMNOPQSTUVWXYZ"));
   assert(!isPangram("ABCDEFGHIJKL.NOPQRSTUVWXYZ"));
   assert(isPangram("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ"));

}</lang>

Internationalized version:

<lang d>import std.string, std.traits, std.uni;

enum Alphabet : dstring {

   DE = "abcdefghijklmnopqrstuvwxyzßäöü",
   EN = "abcdefghijklmnopqrstuvwxyz",
   SV = "abcdefghijklmnopqrstuvwxyzåäö"

}

bool isPangram(T)(T[] s, dstring alpha = Alphabet.EN) if (isSomeChar!T) {

   foreach (c; alpha) 
       if (indexOf(s, c) == -1 && indexOf(s, toUniUpper(c)) == -1)
           return false;
   return true;

}</lang>

<lang d>unittest {

   assert(isPangram("the quick brown fox jumps over the lazy dog".dup, Alphabet.EN));
   assert(isPangram("Falsches Üben von Xylophonmusik quält jeden größeren Zwerg", Alphabet.DE));
   assert(isPangram("Yxskaftbud, ge vår wczonmö iqhjälp"w, Alphabet.SV));

}</lang>

Clojure

<lang lisp>(defn pangram? [s]

 (let [letters (into #{} "abcdefghijklmnopqrstuvwxyz")]
   (= (->> s .toLowerCase (filter letters) (into #{})) letters)))</lang>

Delphi

<lang Delphi>program PangramChecker;

{$APPTYPE CONSOLE}

uses StrUtils;

function IsPangram(const aString: string): Boolean; var

 c: char;

begin

 for c := 'a' to 'z' do
   if not ContainsText(aString, c) then
     Exit(False);

 Result := True;

end;

begin

 Writeln(IsPangram('The quick brown fox jumps over the lazy dog')); // true
 Writeln(IsPangram('Not a panagram')); // false

end.</lang>

E

<lang e>def isPangram(sentence :String) {

   return ("abcdefghijklmnopqrstuvwxyz".asSet() &! sentence.toLowerCase().asSet()).size() == 0

}</lang>

&! is the “but-not” or set difference operator.

F#

If the difference between the set of letters in the alphabet and the set of letters in the given string (after conversion to lower case) is the empty set then every letter appears somewhere in the given string: <lang fsharp>let isPangram (str: string) = (set['a'..'z'] - set(str.ToLower())).IsEmpty</lang>

Factor

Translation of: E

<lang factor>: pangram? ( str -- ? )

   [ "abcdefghijklmnopqrstuvwxyz" ] dip >lower diff length 0 = ;

"How razorback-jumping frogs can level six piqued gymnasts!" pangram? .</lang>

Forth

<lang forth>: pangram? ( addr len -- ? )

 0 -rot bounds do
   i c@ 32 or [char] a -
   dup 0 26 within if
     1 swap lshift or
   else drop then
 loop
 1 26 lshift 1- = ;

s" The five boxing wizards jump quickly." pangram? . \ -1</lang>

Fortran

Works with: Fortran version 90 and later

<lang fortran>module pangram

 implicit none
 private
 public :: is_pangram
 character (*), parameter :: lower_case = 'abcdefghijklmnopqrstuvwxyz'
 character (*), parameter :: upper_case = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

contains

 function to_lower_case (input) result (output)

   implicit none
   character (*), intent (in) :: input
   character (len (input)) :: output
   integer :: i
   integer :: j

   output = input
   do i = 1, len (output)
     j = index (upper_case, output (i : i))
     if (j /= 0) then
       output (i : i) = lower_case (j : j)
     end if
   end do

 end function to_lower_case

 function is_pangram (input) result (output)

   implicit none
   character (*), intent (in) :: input
   character (len (input)) :: lower_case_input
   logical :: output
   integer :: i

   lower_case_input = to_lower_case (input)
   output = .true.
   do i = 1, len (lower_case)
     if (index (lower_case_input, lower_case (i : i)) == 0) then
       output = .false.
       exit
     end if
   end do

 end function is_pangram

end module pangram</lang> Example: <lang fortran>program test

 use pangram, only: is_pangram

 implicit none
 character (256) :: string

 string = 'This is a sentence.'
 write (*, '(a)') trim (string)
 write (*, '(l1)') is_pangram (string)
 string = 'The five boxing wizards jumped quickly.'
 write (*, '(a)') trim (string)
 write (*, '(l1)') is_pangram (string)

end program test</lang> Output: <lang>This is a sentence. F The five boxing wizards jumped quickly. T</lang>

Go

<lang go>package main

import "fmt"

func main() {

   for _, s := range []string{
       "The quick brown fox jumps over the lazy dog.",
       `Watch "Jeopardy!", Alex Trebek's fun TV quiz game.`,
       "Not a pangram.",
   } {
       if pangram(s) {
           fmt.Println("Yes:", s)
       } else {
           fmt.Println("No: ", s)
       }
   }

}

func pangram(s string) bool {

   var rep [26]bool
   var count int
   for _, c := range s {
       if c >= 'a' {
           if c > 'z' {
               continue
           }
           c -= 'a'
       } else {
           if c < 'A' || c > 'Z' {
               continue
           }
           c -= 'A'
       }
       if !rep[c] {
           if count == 25 {
               return true
           }
           rep[c] = true
           count++
       }
   }
   return false

}</lang> Output:

Yes: The quick brown fox jumps over the lazy dog.
Yes: Watch "Jeopardy!", Alex Trebek's fun TV quiz game.
No:  Not a pangram.

Haskell

<lang haskell>import Data.Char (toLower) import Data.List ((\\))

pangram :: String -> Bool pangram = null . (['a' .. 'z'] \\) . map toLower

main = print $ pangram "How razorback-jumping frogs can level six piqued gymnasts!"</lang>

HicEst

<lang HicEst>PangramBrokenAt("This is a Pangram.") ! => 2 (b is missing) PangramBrokenAt("The quick Brown Fox jumps over the Lazy Dog") ! => 0 (OK)

FUNCTION PangramBrokenAt(string)

  CHARACTER string, Alfabet="abcdefghijklmnopqrstuvwxyz"
  PangramBrokenAt = INDEX(Alfabet, string, 64)
  ! option 64: verify = 1st letter of string not in Alfabet

END</lang>

Icon and Unicon

A panagram procedure: <lang Icon>procedure panagram(s) #: return s if s is a panagram and fail otherwise if (map(s) ** &lcase) === &lcase then return s end</lang>

And a main to drive it: <lang Icon>procedure main(arglist)

if *arglist > 0 then

  every ( s := "" ) ||:= !arglist || " " 

else

  s := "The quick brown fox jumps over the lazy dog."

writes(image(s), " -- is") writes(if not panagram(s) then "n't") write(" a panagram.") end</lang>

Ioke

<lang ioke>Text isPangram? = method(

 letters = "abcdefghijklmnopqrstuvwxyz" chars
 text = self lower chars
 letters map(x, text include?(x)) reduce(&&)

)</lang>

Here is an example of it's use in the Ioke REPL:

<lang ioke> iik> "The quick brown fox jumps over the lazy dog" isPangram? "The quick brown fox jumps over the lazy dog" isPangram? +> true

iik> "The quick brown fox jumps over the" isPangram? "The quick brown fox jumps over the" isPangram? +> false</lang>

J

Solution: <lang j>require 'strings' isPangram=: (a. {~ 97+i.26) */@e. tolower</lang>

Example use: <lang j> isPangram 'The quick brown fox jumps over the lazy dog.' 1

  isPangram 'The quick brown fox falls over the lazy dog.'

0</lang>

Java

Works with: Java version 1.5+

<lang java5>public class Pangram {

   public static boolean isPangram(String test){
       for (char a = 'A'; a <= 'Z'; a++)
           if (!test.contains(a) && !test.contains(Character.toLowerCase(a)))
               return false;
       return true;
   }

   public static void main(String[] args){
       System.out.println(isPangram("the quick brown fox jumps over the lazy dog"));//true
       System.out.println(isPangram("the quick brown fox jumped over the lazy dog"));//false, no s
       System.out.println(isPangram("ABCDEFGHIJKLMNOPQRSTUVWXYZ"));//true
       System.out.println(isPangram("ABCDEFGHIJKLMNOPQSTUVWXYZ"));//false, no r
       System.out.println(isPangram("ABCDEFGHIJKL.NOPQRSTUVWXYZ"));//false, no m
       System.out.println(isPangram("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ"));//true
       System.out.println(isPangram(""));//false
   }

}</lang> Output:

true
false
true
false
false
true
false

JavaScript

Translation of: Java

<lang javascript>function is_pangram(str) {

   var s = str.toLowerCase();
   // sorted by frequency ascending (http://en.wikipedia.org/wiki/Letter_frequency)
   var letters = "zqxjkvbpygfwmucldrhsnioate";
   for (var i = 0; i < 26; i++)
       if (s.indexOf(letters.charAt(i)) == -1)
           return false;
   return true;

}

print(is_pangram("is this a pangram")); // false print(is_pangram("The quick brown fox jumps over the lazy dog")); // true</lang>

K

<lang k>lcase  : _ci 97+!26 ucase  : _ci 65+!26 tolower : {@[x;p;:;lcase@n@p:&26>n:ucase?/:x]} panagram: {&/lcase _lin tolower x}</lang>

Example: <lang k> panagram "The quick brown fox jumps over the lazy dog" 1

 panagram "Panagram test"

0</lang>

Liberty BASIC

<lang lb>'Returns 0 if the string is NOT a pangram or >0 if it IS a pangram string$ = "The quick brown fox jumps over the lazy dog."

Print isPangram(string$)

Function isPangram(string$)

   string$ = Lower$(string$)
   For i = Asc("a") To Asc("z")
       isPangram = Instr(string$, chr$(i))
       If isPangram = 0 Then Exit Function
   Next i

End Function</lang>

Logo

<lang logo>to remove.all :s :set

 if empty? :s [output :set]
 if word? :s [output remove.all butfirst :s remove first :s :set]
 output remove.all butfirst :s remove.all first :s :set

end to pangram? :s

 output empty? remove.all :s "abcdefghijklmnopqrstuvwxyz

end

show pangram? [The five boxing wizards jump quickly.]  ; true</lang>

Lua

<lang lua>require"lpeg" S, C = lpeg.S, lpeg.C function ispangram(s)

 return #(C(S(s)^0):match"abcdefghijklmnopqrstuvwxyz") == 26

end

print(ispangram"waltz, bad nymph, for quick jigs vex") print(ispangram"bobby") print(ispangram"long sentence")</lang>

MATLAB

<lang MATLAB>function trueFalse = isPangram(string)

   %This works by histogramming the ascii character codes for lower case
   %letters contained in the string (which is first converted to all
   %lower case letters). Then it finds the index of the first letter that
   %is not contained in the string (this is faster than using the find
   %without the second parameter). If the find returns an empty array then
   %the original string is a pangram, if not then it isn't.

   trueFalse = isempty(find( histc(lower(string),(97:122))==0,1 ));

end</lang>

Sample Output: <lang MATLAB>isPangram('The quick brown fox jumps over the lazy dog.')

ans =

    1</lang>

Objeck

Translation of: Java

<lang objeck> bundle Default {

 class Pangram {
   function : native : IsPangram(test : String) ~ Bool {
     for(a := 'A'; a <= 'Z'; a += 1;) {
       if(test->Find(a) < 0 & test->Find(a->ToLower()) < 0) {
         return false;
       };
     };

     return true;
   }

   function : Main(args : String[]) ~ Nil {
     IsPangram("the quick brown fox jumps over the lazy dog")->PrintLine(); # true
     IsPangram("the quick brown fox jumped over the lazy dog")->PrintLine(); # false, no s
     IsPangram("ABCDEFGHIJKLMNOPQRSTUVWXYZ")->PrintLine(); # true
     IsPangram("ABCDEFGHIJKLMNOPQSTUVWXYZ")->PrintLine(); # false, no r
     IsPangram("ABCDEFGHIJKL.NOPQRSTUVWXYZ")->PrintLine(); # false, no m
     IsPangram("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ")->PrintLine(); # true
     IsPangram("")->PrintLine(); # false
   }
 }

} </lang>

NetRexx

<lang NetRexx>/* NetRexx */ options replace format comments java crossref savelog symbols nobinary

A2Z = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

pangrams = create_samples

loop p_ = 1 to pangrams[0]

 pangram = pangrams[p_]
 q_ = A2Z.verify(pangram.upper)
 say pangram.left(64)'\-'
 if q_ == 0 then -
   say ' [OK, a pangram]'
 else -
   say ' [Not a pangram.  Missing:' A2Z.substr(q_, 1)']'
 end p_

method create_samples public static returns Rexx

 pangrams = 

 x_ = 0
 x_ = x_ + 1; pangrams[0] = x_; pangrams[x_] = 'The quick brown fox jumps over a lazy dog.'    -- best/shortest pangram
 x_ = x_ + 1; pangrams[0] = x_; pangrams[x_] = 'The quick brown fox jumps over the lazy dog.'  -- not as short but at least it's still a pangram
 x_ = x_ + 1; pangrams[0] = x_; pangrams[x_] = 'The quick brown fox jumped over the lazy dog.' -- common misquote; not a pangram
 x_ = x_ + 1; pangrams[0] = x_; pangrams[x_] = 'The quick onyx goblin jumps over the lazy dwarf.'
 x_ = x_ + 1; pangrams[0] = x_; pangrams[x_] = 'Bored? Craving a pub quiz fix? Why, just come to the Royal Oak!' -- (Used to advertise a pub quiz in Bowness-on-Windermere)

 return pangrams

</lang> Output:

The quick brown fox jumps over a lazy dog.                       [OK, a pangram]
The quick brown fox jumps over the lazy dog.                     [OK, a pangram]
The quick brown fox jumped over the lazy dog.                    [Not a pangram.  Missing: S]
The quick onyx goblin jumps over the lazy dwarf.                 [OK, a pangram]
Bored? Craving a pub quiz fix? Why, just come to the Royal Oak!  [OK, a pangram]

OCaml

<lang ocaml>let pangram str =

 let ar = Array.make 26 false in
 String.iter (function
 | 'a'..'z' as c -> ar.(Char.code c - Char.code 'a') <- true
 | _ -> ()
 ) (String.lowercase str);
 Array.fold_left ( && ) true ar</lang>

<lang ocaml>let check str =

 Printf.printf " %b -- %s\n" (pangram str) str

let () =

 check "this is a sentence";
 check "The quick brown fox jumps over the lazy dog.";

</lang>

outputs:

false -- this is a sentence
true -- The quick brown fox jumps over the lazy dog.

Oz

<lang oz>declare

 fun {IsPangram Xs}
    {List.sub
     {List.number &a &z 1}
     {Sort {Map Xs Char.toLower} Value.'<'}}
 end

in

 {Show {IsPangram "The quick brown fox jumps over the lazy dog."}}</lang>

PARI/GP

<lang parigp>pangram(s)={

 s=vecsort(Vec(s),,8);
 for(i=97,122,
   if(!setsearch(s,Strchr(i)) && !setsearch(s,Strchr(i-32)),
     return(0)
   )
 );
 1

};

pangram("The quick brown fox jumps over the lazy dog.") pangram("The quick brown fox jumps over the lazy doe.")</lang>

Perl

<lang perl>use List::MoreUtils 'all';

sub pangram {all {$_[0] =~ /$_/i} 'a' .. 'z';}

print "Yes.\n" if pangram 'Cozy lummox gives smart squid who asks for job pen.';</lang>

Perl 6

<lang perl>sub pangram($s) {

 Set.new("a".."z").subsetorequal($s.comb);

}</lang>

PicoLisp

<lang PicoLisp>(de isPangram (Str)

  (not
     (diff
        '`(chop "abcdefghijklmnopqrstuvwxyz")
        (chop (lowc Str)) ) ) )</lang>

PHP

Translation of: D

<lang php>function isPangram($text) {

   foreach (str_split($text) as $c) {
       if ($c >= 'a' && $c <= 'z')
           $bitset |= (1 << (ord($c) - ord('a')));
       else if ($c >= 'A' && $c <= 'Z')
           $bitset |= (1 << (ord($c) - ord('A')));
   }
   return $bitset == 67108863;

}

$test = array(

   "the quick brown fox jumps over the lazy dog",
   "the quick brown fox jumped over the lazy dog",
   "ABCDEFGHIJKLMNOPQSTUVWXYZ",
   "ABCDEFGHIJKL.NOPQRSTUVWXYZ",
   "ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ"

);

foreach ($test as $str)

   echo "$str : ", isPangram($str) ? 'T' : 'F', '
';</lang>

the quick brown fox jumps over the lazy dog : T
the quick brown fox jumped over the lazy dog : F
ABCDEFGHIJKLMNOPQSTUVWXYZ : F
ABCDEFGHIJKL.NOPQRSTUVWXYZ : F
ABC.D.E.FGHI*J/KL-M+NO*PQ R STUVWXYZ : T

PL/I

<lang PL/I> test_pangram: procedure options (main);

is_pangram: procedure() returns (bit(1) aligned);

  declare text character (200) varying;
  declare c character (1);

  get edit (text) (L);
  put skip list (text);

  text = lowercase(text);

  do c = 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k',
         'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u',
         'v', 'w', 'x', 'y', 'z';
     if index(text, c) = 0 then return ('0'b);
  end;
  return ('1'b);

end is_pangram;

  put skip list ('Please type a sentence');

  if is_pangram() then
     put skip list ('The sentence is a pangram.');
  else
     put skip list ('The sentence is not a pangram.');

end test_pangram; </lang>

Output:

<lang> Please type a sentence

the quick brown fox jumps over the lazy dog The sentence is a pangram. </lang>

Prolog

Works with SWI-Prolog

<lang Prolog>pangram(L) :- numlist(0'a, 0'z, Alphabet), forall(member(C, Alphabet), member(C, L)).

pangram_example :- L1 = "the quick brown fox jumps over the lazy dog", ( pangram(L1) -> R1= ok; R1 = ko), format('~s --> ~w ~n', [L1,R1]),

L2 = "the quick brown fox jumped over the lazy dog", ( pangram(L2) -> R2 = ok; R2 = ko), format('~s --> ~w ~n', [L2, R2]). </lang> output <lang>?- pangram_example. the quick brown fox jumps over the lazy dog --> ok the quick brown fox jumped over the lazy dog --> ko true.</lang>

PureBasic

<lang PureBasic>Procedure IsPangram_fast(String$)

 String$ = LCase(string$)
 char_a=Asc("a")
 ; sets bits in a variable if a letter is found, reads string only once
 For a = 1 To Len(string$)
   char$ = Mid(String$, a, 1)
   pos   = Asc(char$) - char_a
   check.l |  1 << pos
 Next
 If check & $3FFFFFF = $3FFFFFF
   ProcedureReturn 1
 EndIf
 ProcedureReturn 0

EndProcedure

Procedure IsPangram_simple(String$)

 String$ = LCase(string$)
 found   = 1
 For a = Asc("a") To Asc("z")
 ; searches for every letter in whole string
   If FindString(String$, Chr(a), 0) = 0
     found = 0
   EndIf
 Next
 ProcedureReturn found

EndProcedure

Debug IsPangram_fast("The quick brown fox jumps over lazy dogs.") Debug IsPangram_simple("The quick brown fox jumps over lazy dogs.") Debug IsPangram_fast("No pangram") Debug IsPangram_simple("No pangram")</lang>

Python

Using set arithmetic: <lang python>import string, sys if sys.version_info[0] < 3:

   input = raw_input

def ispangram(sentence, alphabet=string.ascii_lowercase):

   alphaset = set(alphabet)
   return not alphaset - set(sentence.lower())

print ( ispangram(input('Sentence: ')) )</lang>

Sample output:

Sentence: The quick brown fox jumps over the lazy dog
True

R

Using the built-in R vector "letters": <lang R>checkPangram <- function(sentence){

 my.letters <- tolower(unlist(strsplit(sentence, "")))
 is.pangram <- all(letters %in% my.letters)
 
 if (is.pangram){
   cat("\"", sentence, "\" is a pangram! \n", sep="")
 } else {
   cat("\"", sentence, "\" is not a pangram! \n", sep="")
 }

}

</lang>

Sample output: <lang>s1 <- "The quick brown fox jumps over the lazy dog" s2 <- "The quick brown fox jumps over the sluggish dog" checkPangram(s1) "The quick brown fox jumps over the lazy dog" is a pangram! checkPangram(s2) "The quick brown fox jumps over the sluggish dog" is not a pangram! </lang>

Retro

<lang Retro>: isPangram? ( $-f )

 ^strings'toLower
 heap [ 27 allot ] preserve
 [ @ 'a - dup 0 25 within [ [ 'a + ] [ here + ] bi ! ] &drop if ]
 ^types'STRING each@ here "abcdefghijklmnopqrstuvwxyz" compare ;</lang>

REXX

<lang REXX> /*REXX program to check if a string (sentence) is a pangram. */

abc='ABCDEFGHIJKLMNOPQRSTUVWXYZ'

 do forever                 /*keep promting until a null or blank. */
 say
 say '----- Please enter a pangramic sentence:'
 say
 pull y
 if y= then leave
 ?=verify(abc,y)            /*see if all letters are present.      */

 if ?==0 then say 'Sentence is a pangram.'
         else say "Sentence isn't a pangram, missing:" substr(abc,?,1)
 say
 end

say '----- PANGRAM program ended. -----' </lang> Output:

----- Please enter a pangramic sentence:

The quick brown fox jumped over the lazy dog.
Sentence isn't a pangram, missing: S


----- Please enter a pangramic sentence:

The quick brown fox JUMPS over the lazy dog!!!
Sentence is a pangram.


----- Please enter a pangramic sentence:


----- PANGRAM program ended. -----

Ruby

<lang ruby>def pangram?(sentence)

 unused_letters = ('a'..'z').to_a - sentence.downcase.chars.to_a
 unused_letters.empty?

end

p pangram?('this is a sentence') # ==> false p pangram?('The quick brown fox jumps over the lazy dog.') # ==> true</lang>

Scala

<lang scala>def is_pangram(sentence: String) = sentence.toLowerCase.filter(c => c >= 'a' && c <= 'z').toSet.size == 26 </lang>

<lang scala> scala> is_pangram("This is a sentence") res0: Boolean = false

scala> is_pangram("The quick brown fox jumps over the lazy dog") res1: Boolean = true </lang>

Smalltalk

<lang smalltalk>!String methodsFor: 'testing'! isPangram ^((self collect: [:c | c asUppercase]) select: [:c | c >= $A and: [c <= $Z]]) asSet size = 26 </lang>

<lang smalltalk> 'The quick brown fox jumps over the lazy dog.' isPangram </lang>

SNOBOL4

Works with: Macro Spitbol

Works with: Snobol4+

Works with: CSnobol

<lang SNOBOL4> define('pangram(str)alfa,c') :(pangram_end) pangram str = replace(str,&ucase,&lcase)

       alfa = &lcase

pgr_1 alfa len(1) . c = :f(return)

       str c :s(pgr_1)f(freturn)

pangram_end

       define('panchk(str)tf') :(panchk_end)

panchk output = str

       tf = 'False'; tf = pangram(str) 'True'
       output = 'Pangram: ' tf :(return)

panchk_end

• # Test and display

       panchk("The quick brown fox jumped over the lazy dogs.")
       panchk("My girl wove six dozen plaid jackets before she quit.")
       panchk("This 41-character string: it's a pangram!")

end</lang>

Output:

The quick brown fox jumped over the lazy dogs.
Pangram: True
My girl wove six dozen plaid jackets before she quit.
Pangram: True
This 41-character string: it's a pangram!
Pangram: False

Tcl

<lang tcl>proc pangram? {sentence} {

   set letters [regexp -all -inline {[a-z]} [string tolower $sentence]]
   expr {
       [llength [lsort -unique $letters]] == 26
   }

}

puts [pangram? "This is a sentence"]; # ==> false puts [pangram? "The quick brown fox jumps over the lazy dog."]; # ==> true</lang>

TUSCRIPT

<lang tuscript> $$ MODE TUSCRIPT SET alfabet="a'b'c'd'e'f'g'h'i'j'k'l'm'n'o'p'q'r's't'u'v'w'x'y'z" SET sentences = * DATA The quick brown fox jumps over the lazy dog DATA the quick brown fox falls over the lazy dog LOOP s=sentences SET chars=STRINGS (s,":</:") SEt chars=ALPHA_SORT (chars) SET chars=REDUCE (chars) IF (chars==alfabet) THEN PRINT "pangram: ",s ELSE PRINT "no pangram: ",s ENDIF ENDLOOP </lang> Output:

pangram: The quick brown fox jumps over the lazy dog
no pangram: the quick brown fox falls over the lazy dog

Ursala

<lang Ursala>

1. import std

is_pangram = ^jZ^(!@l,*+ @rlp -:~&) ~=`A-~ letters </lang> example usage: <lang Ursala>

1. cast %bL

test =

is_pangram* <

  'The quick brown fox jumps over the lazy dog',
  'this is not a pangram'>

</lang> output:

<true,false>

VBScript

Implementation

<lang vb>function pangram( s ) dim i dim sKey dim sChar dim nOffset sKey = "abcdefghijklmnopqrstuvwxyz" for i = 1 to len( s ) sChar = lcase(mid(s,i,1)) if sChar <> " " then if instr(sKey, sChar) then nOffset = asc( sChar ) - asc("a") + 1 if nOffset > 1 then sKey = left(sKey, nOffset - 1) & " " & mid( sKey, nOffset + 1) else sKey = " " & mid( sKey, nOffset + 1) end if end if end if next pangram = ( ltrim(sKey) = vbnullstring ) end function

function eef( bCond, exp1, exp2 ) if bCond then eef = exp1 else eef = exp2 end if end function</lang>

Invocation

<lang vb>wscript.echo eef(pangram("a quick brown fox jumps over the lazy dog"), "is a pangram", "is not a pangram") wscript.echo eef(pangram(""), "is a pangram", "is not a pangram")"</lang>