Numbers in base 10 that are palindromic in bases 2, 4, and 16: Difference between revisions
Numbers in base 10 that are palindromic in bases 2, 4, and 16 (view source)
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END;
# print the numbers in decimal that are palendromic in bases 2, 4 and 16 #
# as noted by the REXX sample, even numbers ( other than 0 ) aren't #
# applicable as even numbers end in 0 in base 2 so can't be palendromic #
print( ( " 0" ) ); # clearly, 0 is palendromic in all bases #
FOR n BY 2 TO max number DO
IF PALINDROMIC ( n DIGITS 16 ) THEN
IF PALINDROMIC ( n DIGITS 4 ) THEN
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FI
OD
END
</syntaxhighlight>
{{out}}
<pre>
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
</pre>
=={{header|ALGOL W}}==
<syntaxhighlight lang="algolw">
begin % find numbers palendromic in bases 2, 4, and 16 %
% returns true if n is palendromic in the specified base, false otherwide %
logical procedure palendromic( integer value n, base ) ;
begin
integer array digit( 1 :: 32 );
integer dPos, v, lPos, rPos;
logical isPalendromic;
dPos := 0;
v := n;
while v > 0 do begin
dPos := dPos + 1;
digit( dPos ) := v rem base;
v := v div base
end while_v_gt_0 ;
isPalendromic := true;
lPos := 1;
rPos := dPos;
while rPos > lPos and isPalendromic do begin
isPalendromic := digit( lPos ) = digit( rPos );
lPos := lPos + 1;
rPos := rPos - 1
end while_rPos_gt_lPos_and_isPalendromic ;
isPalendromic
end palendromic ;
% as noted by the REXX sample, all even numbers end in 0 in base 2 %
% so 0 is the only possible even number, note 0 is palendromic in all bases %
write( " 0" );
for n := 1 step 2 until 24999 do begin
if palendromic( n, 16 ) then begin
if palendromic( n, 4 ) then begin
if palendromic( n, 2 ) then begin
writeon( i_w := 1, s_w := 0, " ", n )
end if_palendromic__n_2
end if_palendromic__n_4
end if_palendromic__n_16
end for_n
end.
</syntaxhighlight>
{{out}}
<pre>
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{{out}}
<pre>0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845</pre>
=={{header|Arturo}}==
Line 386 ⟶ 437:
<pre>0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365
3855 4095 4097 4369 12291 13107 20485 21845</pre>
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls}}
<syntaxhighlight lang="Delphi">
function GetRadixString(L: Integer; Radix: Byte): string;
{Converts integer a string of any radix}
const RadixChars: array[0..35] Of char =
('0', '1', '2', '3', '4', '5', '6', '7',
'8', '9', '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');
var I: integer;
var S: string;
var Sign: string[1];
begin
Result:='';
If (L < 0) then
begin
Sign:='-';
L:=Abs(L);
end
else Sign:='';
S:='';
repeat
begin
I:=L mod Radix;
S:=RadixChars[I] + S;
L:=L div Radix;
end
until L = 0;
Result:=Sign + S;
end;
function IsPalindrome(N, Base: integer): boolean;
{Test if number is the same forward or backward}
{For a specific Radix}
var S1,S2: string;
begin
S1:=GetRadixString(N,Base);
S2:=ReverseString(S1);
Result:=S1=S2;
end;
function IsPalindrome2416(N: integer): boolean;
{Is N palindromic for bases 2, 4 and 16}
begin
Result:=IsPalindrome(N,2) and
IsPalindrome(N,4) and
IsPalindrome(N,16);
end;
procedure ShowPalindrome2416(Memo: TMemo);
{Show all numbers Palindromic for bases 2, 4 and 16}
var S: string;
var I,Cnt: integer;
begin
S:='';
Cnt:=0;
for I:=0 to 25000-1 do
if IsPalindrome2416(I) then
begin
Inc(Cnt);
S:=S+Format('%8D',[I]);
If (Cnt mod 5)=0 then S:=S+#$0D#$0A;
end;
Memo.Lines.Add('Count='+IntToStr(Cnt));
Memo.Lines.Add(S);
end;
</syntaxhighlight>
{{out}}
<pre>
Count=23
0 1 3 5 15
17 51 85 255 257
273 771 819 1285 1365
3855 4095 4097 4369 12291
13107 20485 21845
</pre>
=={{header|Euler}}==
'''begin'''
'''new''' palendromic; '''new''' n; '''label''' forN;
palendromic
<- ` '''formal''' n; '''formal''' base;
'''begin'''
'''new''' v; '''new''' lPos; '''new''' rPos; '''new''' isPalendromic;
'''new''' digit;
'''label''' vGT0; '''label''' rGTl;
digit <- '''list''' 64;
rPos <- 0;
v <- n;
vGT0: '''if''' v > 0 '''then''' '''begin'''
rPos <- rPos + 1;
digit[ rPos ] <- v '''mod''' base;
v <- v % base;
'''goto''' vGT0
'''end''' '''else''' 0;
isPalendromic <- '''true''';
lPos <- 1;
rGTl: '''if''' rPos > lPos '''and''' isPalendromic '''then''' '''begin'''
isPalendromic <- digit[ lPos ] = digit[ rPos ];
lPos <- lPos + 1;
rPos <- rPos - 1;
'''goto''' rGTl
'''end''' '''else''' 0;
isPalendromic
'''end'''
'
;
'''out''' 0;
n <- -1;
forN: '''if''' [ n <- n + 2 ] < 25000 '''then''' '''begin'''
'''if''' '''not''' palendromic( n, 16 ) '''then''' 0
'''else''' '''if''' '''not''' palendromic( n, 4 ) '''then''' 0
'''else''' '''if''' palendromic( n, 2 ) '''then''' '''out''' n
'''else''' 0
;
'''goto''' forN
'''end''' '''else''' 0
'''end''' $
{{out}}
<pre>
NUMBER 0
NUMBER 1
NUMBER 3
NUMBER 5
NUMBER 15
NUMBER 17
NUMBER 51
NUMBER 85
NUMBER 255
NUMBER 257
NUMBER 273
NUMBER 771
NUMBER 819
NUMBER 1285
NUMBER 1365
NUMBER 3855
NUMBER 4095
NUMBER 4097
NUMBER 4369
NUMBER 12291
NUMBER 13107
NUMBER 20485
NUMBER 21845
</pre>
=={{header|F_Sharp|F#}}==
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'''Also works with gojq and fq, the Go implementations'''
'''With minor tweaks, also works with
This entry, which uses a stream-oriented approach to illustrate an
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13107 20485 21845
</pre>
=={{header|Julia}}==
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1 3 5 15 17 51 85 255 257 273 771
819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
</pre>
=={{header|Lua}}==
<syntaxhighlight lang="lua">
do -- find numbers palendromic in bases 2, 4, and 16
local function palendromic( n, base )
local digits, v = "", n
while v > 0 do
local dPos = ( v % base ) + 1
digits = digits..string.sub( "0123456789abcdef", dPos, dPos )
v = math.floor( v / base )
end
return digits == string.reverse( digits )
end
-- as noted by the REXX sample, all even numbers end in 0 in base 2
-- so 0 is the only possible even number, note 0 is palendromic in all bases
io.write( " 0" )
for n = 1, 24999, 2 do
if palendromic( n, 16 ) then
if palendromic( n, 4 ) then
if palendromic( n, 2 ) then
io.write( " ", n )
end
end
end
end
end
</syntaxhighlight>
{{out}}
<pre>
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
</pre>
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Found 22 numbers
done...
</pre>
=={{header|RPL}}==
{{works with|HP|48}}
====Brute force====
« ""
OVER SIZE 1 '''FOR''' j
OVER j DUP SUB +
'''NEXT''' SWAP DROP
» '<span style="color:blue">REVSTR</span>' STO
« → base
« ""
'''WHILE''' OVER '''REPEAT'''
SWAP base MOD LASTARG / IP
"0123456789ABCDEF" ROT 1 + DUP SUB ROT +
'''END''' SWAP DROP
» » '<span style="color:blue">D→B</span>' STO
« '''CASE'''
HEX DUP R→B →STR 3 OVER SIZE SUB DUP <span style="color:blue">REVSTR</span> ≠ '''THEN''' DROP 0 '''END'''
DUP 4 <span style="color:blue">D→B</span> DUP <span style="color:blue">REVSTR</span> ≠ '''THEN''' DROP 0 '''END'''
BIN DUP R→B →STR 3 OVER SIZE SUB DUP <span style="color:blue">REVSTR</span> ==
'''END'''
» '<span style="color:blue">PAL2416</span>' STO
« { 0 }
1 25000 '''FOR''' n
'''IF''' n <span style="color:blue">PAL2416</span> '''THEN''' n + '''END'''
2 '''STEP'''
» '<span style="color:blue">TASK</span>' STO
Runs in 42 minutes on a HP-48SX.
====Much faster approach====
The task generates palindromes in base 16, which must then be verified as palindromes in the other two bases.
« BIN 1 SF
R→B →STR 3 OVER SIZE 1 - SUB
0 1 '''FOR''' b
'''IF''' DUP SIZE b 1 + MOD '''THEN''' "0" SWAP + '''END'''
""
OVER SIZE b - 1 '''FOR''' j
OVER j DUP b + SUB +
-1 b - '''STEP'''
'''IF''' OVER ≠ '''THEN''' 1 CF 1 'b' STO '''END'''
'''NEXT''' DROP
1 FS?
» '<span style="color:blue">PAL24?</span>' STO
« HEX R→B →STR → h
« "#"
h SIZE 1 - 3 '''FOR''' j
h j DUP SUB +
-1 '''STEP'''
"h" + STR→ B→R
» » ‘<span style="color:blue">REVHEX</span>’ STO
« { }
0 15 '''FOR''' b
'''IF''' b <span style="color:blue">PAL24?</span> '''THEN''' b + '''END NEXT'''
1 2 '''FOR''' x
-1 15 '''FOR''' m
16 x 1 - ^ 16 x ^ 1 - '''FOR''' b
b
'''IF''' m 0 ≥ '''THEN''' 16 * m + '''END'''
16 x ^ *
b <span style="color:blue">REVHEX</span> +
'''IF''' DUP <span style="color:blue">PAL24?</span> '''THEN''' +
'''ELSE IF''' 25000 ≥ '''THEN''' KILL '''END''' <span style="color:grey">@ not idiomatic but useful to exit 3 nested loops</span>
'''END'''
'''NEXT NEXT NEXT'''
SORT
» '<span style="color:blue">TASK</span>' STO
<span style="color:blue">TASK</span> SORT
Runs in 2 minutes 16 on a HP-48SX: 18 times faster than brute force!
{{out}}
<pre>
1: { 0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845 }
</pre>
=={{header|Ruby}}==
<syntaxhighlight lang="ruby">res = (0..25000).select do |n|
[2, 4, 16].all? do |base|
b = n.to_s(base)
b == b.reverse
end
end
puts res.join(" ")</syntaxhighlight>
{{out}}
<pre>
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
</pre>
Line 891 ⟶ 1,223:
=={{header|Wren}}==
{{libheader|Wren-fmt}}
<syntaxhighlight lang="wren">import "./fmt" for Conv, Fmt
System.print("Numbers under 25,000 in base 10 which are palindromic in bases 2, 4 and 16:")
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}
}
System.print("\nFound %(numbers.count) such numbers.")</syntaxhighlight>
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