⚠️ Warning: This is a draft ⚠️
This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.
{{task|Text processing}} [[Category:Recursion]] [[Category:String manipulation]] [[Category:Classic CS problems and programs]] [[Category:Palindromes]]
A [[wp:Palindrome|palindrome]] is a phrase which reads the same backward and forward.
{{task heading}}
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.
{{task heading|Hints}}
- It might be useful for this task to know how to [[Reversing a string|reverse a string]].
- This task's entries might also form the subjects of the task [[Test a function]].
{{task heading|Related tasks}}
{{Related tasks/Word plays}}
360 Assembly
* 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
{{out}}
INGIRUMIMUSNOCTEETCONSUMIMURIGNI
IT IS A TRUE PALINDROME
ACL2
(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)))))
ActionScript
The following function handles non-ASCII characters properly, since charAt() returns a single Unicode character.
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;
}
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;
Ada 2012 version:
function Palindrome (Text : String) return Boolean is
(for all i in Text'Range => Text(i)= Text(Text'Last-i+Text'First));
ALGOL 68
{{trans|C}}
{{works with|ALGOL 68|Standard - no extensions to language used}} {{works with|ALGOL 68G|Any - tested with release mk15-0.8b.fc9.i386}} {{works with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386 - except for the '''FORMAT''' and ''printf'' in test}}
# 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)))
)
{{out}}
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:
{⍵≡⌽⍵} 'abc'
0
{⍵≡⌽⍵} '⍋racecar⍋'
1
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):
(⊢≡⌽) 'abc'
0
(⊢≡⌽) 'nun'
1
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:
inexact←{Aa←(⎕A,⎕a) ⋄ (⊢≡⌽)(⎕a,⎕a)[Aa⍳⍵/⍨⍵∊Aa]}
inexact 'abc,-cbA2z'
0
inexact 'abc,-cbA2'
1
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)
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
{{Out}}
true
Applesoft BASIC
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
AutoHotkey
Reversing the string:
IsPalindrome(Str){
Loop, Parse, Str
ReversedStr := A_LoopField . ReversedStr
return, (ReversedStr == Str)?"Exact":(RegExReplace(ReversedStr,"\W")=RegExReplace(Str,"\W"))?"Inexact":"False"
}
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
{{out}}
"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.
--[[User:BugFix|BugFix]] ([[User talk:BugFix|talk]]) 14:26, 13 November 2013 (UTC)
AWK
'''Non-recursive'''
See [[Reversing a string]].
function is_palindro(s)
{
if ( s == reverse(s) ) return 1
return 0
}
'''Recursive'''
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))
}
'''Testing'''
BEGIN {
pal = "ingirumimusnocteetconsumimurigni"
print is_palindro(pal)
print is_palindro_r(pal)
}
BaCon
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$, "[[:punct:]]|[[:blank:]]", TRUE)) THEN
PRINT "This is an inexact palindrome!"
ELSE
PRINT "Not a palindrome."
ENDIF
{{out}}
Enter your line... In girum imus nocte, et consumimur igni
This is an inexact palindrome!
Enter your line... Madam, I'm Adam.
This is an inexact palindrome!
Enter your line... radar
This is an exact palindrome!
Enter your line... Something else
Not a palindrome.
BASIC
{{works with|QBasic}}
' OPTION _EXPLICIT ' For QB64. In VB-DOS remove the underscore.
DIM txt$
' 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
{{out}} 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.
==={{header|IS-BASIC}}===
=
## 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.
```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"
=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.
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
=
BBC BASIC
=
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$(C% AND &5F)
ENDIF
NEXT
= B$
{{out}}
"A man, a plan, a canal: Panama!" is a palindrome
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
Batch File
@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
Or, recursive (and without setlocal enabledelayedexpansion) (compatible with ReactOS cmd.exe)
@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
Befunge
{{works with|CCBI|2.1}}
The following code reads a line from stdin and prints "True" if it is a palindrome, or False" otherwise.
v_$0:8p>:#v_:18p08g1-08p >:08g`!v
~->p5p ^ 0v1p80-1g80vj!-g5g80g5_0'ev
:a^80+1:g8<>8g1+:18pv>0"eslaF">:#,_@
[[relet]]-2010------>003-x -^"Tru"<
{{works with|Befunge|93}}
To check a string, replace "dennis sinned" with your own string.
Note that this has some limits.:
- There '''must''' be a quotation mark immediately after the string, and then nothing but spaces for the rest of that line. ** The '''v''' at the end of that same line '''must''' remain immediately above the '''2'''. (''Very'' important.) The closing quotation mark can be against the '''v''', but can't replace it.
- The potential palindrome can be no longer than 76 characters (which beats the previous version's 11), and ''everything'' (spaces, punctuation, capitalization, etc.) is considered part of the palindrome. (Best to just use lower case letters and ''nothing else''.)
"emordnilap a toN",,,,,,,,,,,,,,,,@,,,,,,,,,,,,,,,"Is a palindrome" <
2^ < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < <
4 ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v
8 ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v # ^_v
*^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v ^_v
+ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""
>"dennis sinned" v
" 2
"""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""" 0
> ^- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 9
_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ p
v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^ # v_^
v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^ v_^
^< < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < <
>09g8p09g1+09pv
|: < <
^<
Bracmat
( ( palindrome
= a
. @(!arg:(%?a&utf$!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
)
)
&
);
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
zz{ri}f[^^<-==
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).
#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;
}
More idiomatic version:
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;
}
'''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.
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);
}
'''Testing'''
#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;
}
C++
The C solutions also work in C++, but C++ allows a simpler one:
#include <string>
#include <algorithm>
bool is_palindrome(std::string const& s)
{
return std::equal(s.begin(), s.end(), s.rbegin());
}
Or, checking half is sufficient (on odd-length strings, this will ignore the middle element):
#include <string>
#include <algorithm>
bool is_palindrome(std::string const& s)
{
return std::equal(s.begin(), s.begin()+s.length()/2, s.rbegin());
}
C#
'''Non-recursive'''
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"));
}
}
'''Using LINQ operators'''
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"));
}
}
'''No string reversal'''
Reversing a string is very slow. A much faster way is to simply compare characters.
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());
}
}
Clojure
(defn palindrome? [s]
(= s (clojure.string/reverse s)))
'''lower-level, but somewhat faster'''
(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)))))
'''Test'''
user=> (palindrome? "amanaplanacanalpanama")
true
user=> (palindrome? "Test 1, 2, 3")
false
COBOL
{{works with|GnuCOBOL}}
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.
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.'
]
{{out}} 'In girum imus nocte et consumimur igni' : true 'A man, a plan, a canal: Panama!' : true 'There is no spoon.' : false
Common Lisp
(defun palindrome-p (s)
(string= s (reverse s)))
Alternate solution
I use [https://franz.com/downloads/clp/survey Allegro CL 10.1]
;; 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)
Output:
radar: palindrome
books: not palindrome
Component Pascal
BlackBox Component Builder
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.
Execute: ^Q BbtPalindrome.Do
{{out}}
'salalas' is palindrome?:> $TRUE
'madamimadam' is palindrome?:> $TRUE
'abcbda' is palindrome?:> $FALSE
Crystal
Declarative
def palindrome(s)
s == s.reverse
end
Imperative
def palindrome_imperative(s) : Bool
mid = s.size / 2
last = s.size - 1
(0...s.size).each do |i|
if s[i] != s[last - i]
return false
end
end
true
end
Performance comparison
require "benchmark"
Benchmark.ips do |x|
x.report("declarative") { palindrome("hannah") }
x.report("imperative") { palindrome_imperative("hannah")}
end
declarative 21.96M ( 45.54ns) (± 6.90%) 32 B/op 1.49× slower
imperative 32.8M ( 30.49ns) (± 3.29%) 0 B/op fastest
Delphi
uses
SysUtils, StrUtils;
function IsPalindrome(const aSrcString: string): Boolean;
begin
Result := SameText(aSrcString, ReverseString(aSrcString));
end;
D
===High-level 32-bit Unicode Version===
import std.traits, std.algorithm;
bool isPalindrome1(C)(in C[] s) pure /*nothrow*/
if (isSomeChar!C) {
auto s2 = s.dup;
s2.reverse(); // works on Unicode too, not nothrow.
return s == s2;
}
void main() {
alias pali = isPalindrome1;
assert(pali(""));
assert(pali("z"));
assert(pali("aha"));
assert(pali("sees"));
assert(!pali("oofoe"));
assert(pali("deified"));
assert(!pali("Deified"));
assert(pali("amanaplanacanalpanama"));
assert(pali("ingirumimusnocteetconsumimurigni"));
assert(pali("salà las"));
}
===Mid-level 32-bit Unicode Version===
import std.traits;
bool isPalindrome2(C)(in C[] s) pure if (isSomeChar!C) {
dchar[] dstr;
foreach (dchar c; s) // not nothrow
dstr ~= c;
for (int i; i < dstr.length / 2; i++)
if (dstr[i] != dstr[$ - 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"));
}
===Low-level 32-bit Unicode Version===
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"));
}
===Low-level ASCII Version===
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"));
}
Dart
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;
}
=={{header|Déjà Vu}}==
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"
{{out}}
true
false
Dyalect
func isPalindrom(str) {
str == str.reverse()
}
print(isPalindrom("ingirumimusnocteetconsumimurigni"))
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.
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
}
EchoLisp
;; 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)))))
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
Ela
open list string
isPalindrome xs = xs == reverse xs
isPalindrome <| toList "ingirumimusnocteetconsumimurigni"
Function reverse is taken from list module and is defined as:
reverse = foldl (flip (::)) (nil xs)
foldl f z (x::xs) = foldl f (f z x) xs
foldl _ z [] = z
Elixir
defmodule PalindromeDetection do
def is_palindrome(str), do: str == String.reverse(str)
end
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
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")
]
Link to live demo: http://dc25.github.io/palindromeDetectionElm/
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)] ).
{{out}}
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
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
=={{header|F Sharp|F#}}==
let isPalindrome (s: string) =
let arr = s.ToCharArray()
arr = Array.rev arr
Examples:
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
Factor
USING: kernel sequences ;
: palindrome? ( str -- ? ) dup reverse = ;
Falcon
'''VBA/Python programmer's approach not sure if it's the most falconic way'''
/* 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)
{{out}}
true
[Finished in 1.7s]
'''more falconic'''
/* 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"
{{out}}
Is a palindrome
[Finished in 1.5s]
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(" ","")))
}
}
FBSL
#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
{{out}}
1
1
0
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 ;
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.'''
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 ;
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
{{works with|Fortran|90 and later}}
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
'''Non-recursive'''
! 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
'''Recursive'''
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
## FreeBASIC
```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
{{out}}
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.
isPalindrome[x] := x == reverse[x]
Test in Frink with upper-plane Unicode:
isPalindrome["x\u{1f638}x"]
true
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;
GML
//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);
Palindrome detection using a [http://rosettacode.org/wiki/Loop/Downward_For#GML Downward For-Loop]
//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;
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
}
This version works with Unicode,
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
}
Or using more slicing,
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
}
Groovy
Trivial
Solution:
def isPalindrome = { String s ->
s == s?.reverse()
}
Test program:
println isPalindrome("")
println isPalindrome("a")
println isPalindrome("abcdefgfedcba")
println isPalindrome("abcdeffedcba")
println isPalindrome("abcedfgfedcb")
{{out}}
true
true
true
true
false
This solution assumes nulls are palindromes.
===Non-recursive===
Solution:
def isPalindrome = { String s ->
def n = s.size()
n < 2 || s[0..<n/2] == s[-1..(-n/2)]
}
Test program and output are the same. This solution does not handle nulls.
Recursive
Solution follows the [[#C|C palindrome_r]] recursive solution:
def isPalindrome
isPalindrome = { String s ->
def n = s.size()
n < 2 || (s[0] == s[n-1] && isPalindrome(s[1..<(n-1)]))
}
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.
is_palindrome x = x == reverse x
Or, applicative and point-free, with some pre-processing of data (shedding white space and upper case):
import Data.Char (toLower)
isPalindrome :: Eq a => [a] -> Bool
isPalindrome = (==) <*> reverse
-- Alternatively, comparing just the first half with the reversed latter half
isPal :: Eq a => [a] -> Bool
isPal s =
let (q, r) = quotRem (length s) 2
in take q s == reverse (drop (q + r) s)
sample :: String
sample = "In girum imus nocte et consumimur igni"
prepared :: String -> String
prepared = filter (' ' /=) . fmap toLower
main :: IO ()
main = print $ [isPalindrome, isPal] <*> [prepared sample]
{{Out}}
[True,True]
'''Recursive'''
See the C palindrome_r code for an explanation of the concept used in this solution.
is_palindrome_r x | length x <= 1 = True
| head x == last x = is_palindrome_r . tail. init $ x
| otherwise = False
HicEst
{{incorrect|HicEst|The stripping of spaces and case conversion should be outside the palindrome detection.}}
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
=={{header|Icon}} and {{header|Unicon}}==
procedure main(arglist)
every writes(s := !arglist) do write( if palindrome(s) then " is " else " is not", " a palindrome.")
end
The following simple procedure uses the built-in reverse. Reverse creates a transient string which will get garbage collected.
procedure palindrome(s) #: return s if s is a palindrome
return s == reverse(s)
end
{{libheader|Icon Programming Library}}
Note: The IPL procedure [http://www.cs.arizona.edu/icon/library/src/procs/strings.icn 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.
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
Ioke
Text isPalindrome? = method(self chars == self chars reverse)
J
'''Non-recursive'''
Reverse and match method
isPalin0=: -: |.
Example usage
isPalin0 'ABBA'
1
isPalin0 -.&' ' tolower 'In girum imus nocte et consumimur igni'
1
'''Recursive'''
Tacit and explicit verbs:
isPalin1=: 0:`($:@(}.@}:))@.({.={:)`1:@.(1>:#)
isPalin2=: monad define
if. 1>:#y do. 1 return. end.
if. ({.={:)y do. isPalin2 }.}:y else. 0 end.
)
Note that while these recursive verbs are bulkier and more complicated, they are also several thousand times more inefficient than isPalin0.
foo=: foo,|.foo=:2000$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
Java
'''Non-Recursive'''
public static boolean pali(String testMe){
StringBuilder sb = new StringBuilder(testMe);
return testMe.equals(sb.reverse().toString());
}
'''Recursive (this version does not work correctly with upper-plane Unicode)'''
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));
}
'''Recursive using indexes (this version does not work correctly with upper-plane Unicode)'''
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);
}
'''Regular Expression''' ([http://stackoverflow.com/questions/3664881/how-does-this-java-regex-detect-palindromes source])
public static boolean pali(String testMe){
return testMe.matches("|(?:(.)(?<=(?=^.*?(\\1\\2?)$).*))+(?<=(?=^\\2$).*)");
}
JavaScript
function isPalindrome(str) {
return str === str.split("").reverse().join("");
}
console.log(isPalindrome("ingirumimusnocteetconsumimurigni"));
ES6 implementation
str === str.split("").reverse().join("");
Or, ignoring spaces and variations in case:
(() => {
// 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();
})();
{{Out}}
true
jq
def palindrome: explode as $in | ($in|reverse) == $in;
'''Example''': "salàlas" | palindrome {{Out}} true
Jsish
/* Palindrome detection, in Jsish */
function isPalindrome(str:string, exact:boolean=true) {
if (!exact) {
str = str.toLowerCase();
str = str.replace(/[ \t,;:!?.]/g, '');
}
return str === str.match(/./g).reverse().join('');
}
;isPalindrome('BUB');
;isPalindrome('CUB');
;isPalindrome('Bub');
;isPalindrome('Bub', false);
;isPalindrome('In girum imus nocte et consumimur igni', false);
;isPalindrome('A man, a plan, a canal; Panama!', false);
;isPalindrome('Never odd or even', false);
/*
=!EXPECTSTART!=
isPalindrome('BUB') ==> true
isPalindrome('CUB') ==> false
isPalindrome('Bub') ==> false
isPalindrome('Bub', false) ==> true
isPalindrome('In girum imus nocte et consumimur igni', false) ==> true
isPalindrome('A man, a plan, a canal; Panama!', false) ==> true
isPalindrome('Never odd or even', false) ==> true
=!EXPECTEND!=
*/
Most of that code is for testing, using echo mode lines (semicolon in column 1)
{{out}}
prompt$ 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
palindrome(s) = s == reverse(s)
Non-Recursive
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
Recursive
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
k
is_palindrome:{x~|x}
Kotlin
// 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 ${if (isInexactPalindrome(candidate)) "an" else "not an"} inexact palindrome")
}
}
{{out}}
'rotor' is an exact palindrome
'rosetta' is not an exact palindrome
'step on no pets' is an exact palindrome
'été' is an exact palindrome
'In girum imus nocte et consumimur igni' is an inexact palindrome
'Rise to vote, sir' is an inexact palindrome
'A man, a plan, a canal - Panama!' is an inexact palindrome
'Ce repère, Perec' is not an inexact palindrome
LabVIEW
{{VI solution|LabVIEW_Palindrome_detection.png}}
Langur
val .ispal = f len(.s) > 0 and .s == s2s .s, len(.s)..1
val .tests = h{
"": false,
"z": true,
"aha": true,
"αηα": true,
"αννα": true,
"αννασ": false,
"sees": true,
"seas": false,
"deified": true,
"solo": false,
"solos": true,
"amanaplanacanalpanama": true,
"a man a plan a canal panama": false, # true if we remove spaces
"ingirumimusnocteetconsumimurigni": true,
}
for .word in sort(keys .tests) {
val .foundpal = .ispal(.word)
writeln .word, ": ", .foundpal, if(.foundpal == .tests[.word]: ""; " (FAILED TEST)")
}
{{out}}
: false
a man a plan a canal panama: false
aha: true
amanaplanacanalpanama: true
deified: true
ingirumimusnocteetconsumimurigni: true
seas: false
sees: true
solo: false
solos: true
z: true
αηα: true
αννα: true
αννασ: false
Lasso
define ispalindrome(text::string) => {
local(_text = string(#text)) // need to make copy to get rid of reference issues
#_text -> replace(regexp(`(?:$|\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!')
{{out}}
true
false
true
Liberty BASIC
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
{{out}}
0
0
1
LiveCode
This implementation defaults to exact match, but has an optional parameter to do inexact.
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
Logo
to palindrome? :w
output equal? :w reverse :w
end
Lua
function ispalindrome(s) return s == string.reverse(s) end
M4
'''Non-recursive'''
This uses the invert from [[Reversing a string]].
define(`palindrorev',`ifelse(`$1',invert(`$1'),`yes',`no')')dnl
palindrorev(`ingirumimusnocteetconsumimurigni')
palindrorev(`this is not palindrome')
'''Recursive'''
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(`$1'))',`no')')dnl
palindro(`ingirumimusnocteetconsumimurigni')
palindro(`this is not palindrome')
Maple
This uses functions from Maple's built-in StringTools package.
with(StringTools):
IsPalindrome("ingirumimusnocteetconsumimurigni");
IsPalindrome("In girum imus nocte et consumimur igni");
IsPalindrome(LowerCase(DeleteSpace("In girum imus nocte et consumimur igni")));
{{out}}
true
false
true
Mathematica
Custom functions:
'''Non-recursive'''
PalindromeQ[i_String] := StringReverse[i] == i
Test numbers:
PalindromeQ[i_Integer] := Reverse[IntegerDigits[i]] == IntegerDigits[i];
{{out|Examples}}
PalindromeQ["TNT"]
PalindromeRecQ["TNT"]
PalindromeQ["test"]
PalindromeRecQ["test"]
PalindromeQ["deified"]
PalindromeRecQ["deified"]
PalindromeQ["salàlas"]
PalindromeRecQ["salàlas"]
PalindromeQ["ingirumimusnocteetconsumimurigni"]
PalindromeRecQ["ingirumimusnocteetconsumimurigni"]
Note that the code block doesn't correctly show the à in salàlas. {{out}}
True
True
False
False
True
True
True
True
True
True
MATLAB
function trueFalse = isPalindrome(string)
trueFalse = all(string == fliplr(string)); %See if flipping the string produces the original string
if not(trueFalse) %If not a palindrome
string = lower(string); %Lower case everything
trueFalse = all(string == fliplr(string)); %Test again
end
if not(trueFalse) %If still not a palindrome
string(isspace(string)) = []; %Strip all space characters out
trueFalse = all(string == fliplr(string)); %Test one last time
end
end
{{out|Sample Usage}}
isPalindrome('In girum imus nocte et consumimur igni')
ans =
1
Maxima
palindromep(s) := block([t], t: sremove(" ", sdowncase(s)), sequal(t, sreverse(t)))$
palindromep("Sator arepo tenet opera rotas"); /* true */
MAXScript
'''Non-recursive'''
fn isPalindrome s =
(
local reversed = ""
for i in s.count to 1 by -1 do reversed += s[i]
return reversed == s
)
'''Recursive'''
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))
)
)
'''Testing'''
local p = "ingirumimusnocteetconsumimurigni"
format ("'%' is a palindrome? %\n") p (isPalindrome p)
format ("'%' is a palindrome? %\n") p (isPalindrome_r p)
min
{{works with|min|0.19.3}}
(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
{{out}}
false
true
false
true
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
{{out}}
Madam, I'm Adam is a palindrome
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
ML
=
mLite
=
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");
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
=
fun palindrome s =
let val cs = explode s in
cs = rev cs
end
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)
=={{header|Modula-2}}==
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.
=={{header|Modula-3}}==
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.
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"));
}
}
And a function to remove spaces and punctuation and convert to lowercase
Clean( text : string ) : string
{
def sepchars = Explode(",.;:-?!()' ");
Concat( "", Split(text, sepchars)).ToLower()
}
NetRexx
{{Trans|REXX}}
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 */
NewLISP
Works likewise for strings and for lists
(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))))
Nim
proc reverse(s: string): string =
result = newString(s.len)
for i,c in s:
result[s.high - i] = c
proc isPalindrome(s: string): bool =
s == reverse(s)
echo isPalindrome("FoobooF")
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;
}
}
}
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)
and here a function to remove the white spaces in the string:
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
and to make the test case insensitive, just use the function String.lowercase_ascii.
Oforth
## Octave
'''Recursive'''
```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
'''Non-recursive'''
function v = palindro(s)
v = all( (s == s(length(s):-1:1)) == 1);
endfunction
'''Testing'''
palindro_r("ingirumimusnocteetconsumimurigni")
palindro("satorarepotenetoperarotas")
Ol
; 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
Oz
fun {IsPalindrome S}
{Reverse S} == S
end
PARI/GP
ispal(s)={
s=Vec(s);
for(i=1,#v\2,
if(v[i]!=v[#v-i+1],return(0))
);
1
};
A version for numbers: {{works with|PARI/GP|2.6.0 and above}}
ispal(s)={
my(d=digits(n));
for(i=1,#d\2,
if(d[i]!=d[n+1=i],return(0))
);
1
};
Pascal
{{works with|Free 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;
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.
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 [[{{PAGENAME}}#C|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.
# 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
}
This example shows how to use the functions:
# 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 },
});
{{out}} 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.
Perl 6
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", $_ }
{{out}}
True A man, a plan, a canal: Panama.
False My dog has fleas
True Madam, I'm Adam.
False 1 on 1
True In girum imus nocte et consumimur igni
Phix
function is_palindrome(sequence s)
return s==reverse(s)
end function
?is_palindrome("rotator") -- prints 1
?is_palindrome("tractor") -- prints 0
constant punctuation = " `~!@#$%^&*()-=_+[]{}\\|;:',.<>/?",
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
<?php
function is_palindrome($string) {
return $string == strrev($string);
}
?>
Regular expression-based solution ([http://www.polygenelubricants.com/2010/09/matching-palindromes-in-pcre-regex.html source])
<?php
function is_palindrome($string) {
return preg_match('/^(?:(.)(?=.*(\1(?(2)\2|))$))*.?\2?$/', $string);
}
?>
PicoLisp
(de palindrome? (S)
(= (setq S (chop S)) (reverse S)) )
{{out}}
: (palindrome? "ingirumimusnocteetconsumimurigni")
-> T
Pike
int main(){
if(pal("rotator")){
write("palindrome!\n");
}
if(!pal("asdf")){
write("asdf isn't a palindrome.\n");
}
}
int pal(string input){
if( reverse(input) == input ){
return 1;
} else {
return 0;
}
}
PL/I
To satisfy the revised specification (which contradicts the preceding explanation) the following trivially solves the problem in PL/I:
is_palindrome = (text = reverse(text));
The following solution strips spaces:
text = remove_blanks(text); text = lowercase(text); return (text = reverse(text));
remove_blanks: procedure (text); declare text character (*) varying; declare (i, j) fixed binary (31); j = 0; do i = 1 to length(text); if substr(text, i, 1) = ' ' then do; j = j + 1; substr(text, j, 1) = substr(text, i, 1); end; end; return (substr(text, 1, j)); end remove_blanks; end is_palindrome;
## Potion
```Potion
# The readable recursive version
palindrome_i = (s, b, e):
if (e <= b): true.
elsif (s ord(b) != s ord(e)): false.
else: palindrome_i(s, b+1, e-1).
.
palindrome = (s):
palindrome_i(s, 0, s length - 1).
palindrome(argv(1))
PowerBASIC
The output is identical to the [[#BASIC|QBasic]] version, above.
FUNCTION isPalindrome (what AS STRING) AS LONG
DIM whatcopy AS STRING, chk AS STRING, tmp AS STRING * 1, L0 AS LONG
FOR L0 = 1 TO LEN(what)
tmp = UCASE$(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
PowerShell
An exact version based on reversing the string:
Function Test-Palindrome( [String] $Text ){
$CharArray = $Text.ToCharArray()
[Array]::Reverse($CharArray)
$Text -eq [string]::join('', $CharArray)
}
===PowerShell (Regex Version)=== This version is much faster because it does not manipulate arrays. [This is not clear; the above version was slowed down by using -join instead of [string]::join, and -like instead of -eq. After changing those it is similar, if not faster, than this version].
function Test-Palindrome
{
<#
.SYNOPSIS
Tests if a string is a palindrome.
.DESCRIPTION
Tests if a string is a true palindrome or, optionally, an inexact palindrome.
.EXAMPLE
Test-Palindrome -Text "racecar"
.EXAMPLE
Test-Palindrome -Text '"Deliver desserts," demanded Nemesis, "emended, named, stressed, reviled."' -Inexact
#>
[CmdletBinding()]
[OutputType([bool])]
Param
(
# The string to test for palindrominity.
[Parameter(Mandatory=$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]","")
}
$Text -match "^(?'char'[a-z])+[a-z]?(?:\k'char'(?'-char'))+(?(char)(?!))$"
}
Test-Palindrome -Text 'radar'
{{Out}}
True
Test-Palindrome -Text "In girum imus nocte et consumimur igni."
{{Out}}
False
Test-Palindrome -Text "In girum imus nocte et consumimur igni." -Inexact
{{Out}}
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.
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
{{Out}}
PS C:\> 'ánu-ná', 'nowt' | Test-Palindrome
Text IsPalindrome
---- ------------
ánuná True
now False
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;
}
}
{{out}}
"true" or "false" depending
Prolog
'''Non-recursive'''
From [http://www2.dcs.hull.ac.uk/NEAT/dnd/AI/prolog/tutorial2.html this tutorial].
palindrome(Word) :- name(Word,List), reverse(List,List).
'''Recursive'''
{{works with|SWI 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.
Changing '''string''' into '''atom''' makes the program run also on GNU Prolog. I.e.
{{works with|GNU 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.
PureBasic
{{works with|PureBasic|4.41}}
Procedure IsPalindrome(StringToTest.s)
If StringToTest=ReverseString(StringToTest)
ProcedureReturn 1
Else
ProcedureReturn 0
EndIf
EndProcedure
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])
def is_palindrome(s):
return s == s[::-1]
'''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.
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
'''Recursive'''
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])
Python has short-circuit evaluation of Boolean operations so a shorter and still easy to understand recursive function is
def is_palindrome_r2(s):
return not s or s[0] == s[-1] and is_palindrome_r2(s[1:-1])
'''Testing'''
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)
''' Palindrome Using Regular Expressions Python 2.7 '''
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
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.
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)
}
}
}
'''Iterative'''
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
}
'''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.
revstring <- function(stringtorev) {
return(
paste(
strsplit(stringtorev,"")[[1]][nchar(stringtorev):1]
,collapse="")
)
}
palindroc <- function(p) {return(revstring(p)==p)}
{{out}}
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
(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
>
Rascal
The most simple solution:
import String;
public bool palindrome(str text) = toLowerCase(text) == reverse(text);
A solution that handles sentences with spaces and capitals:
import String;
public bool palindrome(str text){
text = replaceAll(toLowerCase(text), " ", "");
return text == reverse(text);
}
Example:
palindrome("In girum imus nocte et consumimur igni")
bool: true
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...?
{{out}}
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
:palindrome? (s-f) dup s:hash [ s:reverse s:hash ] dip eq? ;
'ingirumimusnocteetconsumimurigni palindrome? n:put
REXX
version 1
/*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)))
'''output'''
string = In girum imus nocte et consumimur igni
The string is an inexact palindrome.
Short version
{{works with|ARexx}} {{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.
/*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)
{{out}}
ABBA is an exact palindrome.
In girum imus nocte et consumimur igni is an inexact palindrome.
djdjdj is not palindromic.
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
Ruby
'''Non-recursive'''
def palindrome?(s)
s == s.reverse
end
'''Recursive'''
def r_palindrome?(s)
if s.length <= 1
true
elsif s[0] != s[-1]
false
else
r_palindrome?(s[1..-2])
end
end
'''Testing''' Note that the recursive method is ''much'' slower -- using the 2151 character palindrome by Dan Hoey [http://www2.vo.lu/homepages/phahn/anagrams/panama.htm here], we have:
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
{{out}}
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
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"
{{out}}
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
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");
}
{{out}}
'': 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.
extern crate unicode_segmentation;
use unicode_segmentation::UnicodeSegmentation;
fn is_palindrome(string: &str) -> bool {
string.graphemes(true).eq(string.graphemes(true).rev())
}
SAS
Description
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.
Code
%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;
Example macro call and output
%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
Scala
{{libheader|Scala}} === Non-recursive, robustified===
def isPalindrome(s: String): Boolean = (s.size >= 2) && s == s.reverse
Bonus: Detect and account for odd space and punctuation
def isPalindromeSentence(s: String): Boolean =
(s.size >= 2) && {
val p = s.replaceAll("[^\\p{L}]", "").toLowerCase
p == p.reverse
}
Recursive
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)
}
'''Testing'''
// 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.")
Scheme
'''Non-recursive'''
(define (palindrome? s)
(let ((chars (string->list s)))
(equal? chars (reverse chars))))
'''Recursive'''
(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
>
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;
For palindromes where spaces shuld be ignore use:
palindrome(replace("in girum imus nocte et consumimur igni", " ", ""))
SequenceL
'''Using the Reverse Library Function'''
;
isPalindrome(string(1)) := equalList(string, reverse(string));
'''Version Using an Indexed Function'''
isPalindrome(string(1)) :=
let
compares[i] := string[i] = string[size(string) - (i - 1)] foreach i within 1 ... (size(string) / 2);
in
all(compares);
Sidef
'''Built-in'''
say "noon".is_palindrome; # true
'''Non-recursive'''
func palindrome(s) {
s == s.reverse
}
'''Recursive'''
func palindrome(s) {
if (s.len <= 1) {
true
}
elsif (s.first != s.last) {
false
}
else {
__FUNC__(s.ft(1, -2))
}
}
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.
{{out}}
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'''
s@(String traits) isPalindrome
[
(s lexicographicallyCompare: s reversed) isZero
].
'''Recursive''' Defined on Sequence since we are not using String-specific methods:
s@(Sequence traits) isPalindrome
[
s isEmpty
ifTrue: [True]
ifFalse: [(s first = s last) /\ [(s sliceFrom: 1 to: s indexLast - 1) isPalindrome]]
].
'''Testing'''
define: #p -> 'ingirumimusnocteetconsumimurigni'.
inform: 'sequence ' ; p ; ' is ' ; (p isPalindrome ifTrue: [''] ifFalse: ['not ']) ; 'a palindrome.'.
Smalltalk
{{works with|Squeak}}
isPalindrome := [:aString |
str := (aString select: [:chr| chr isAlphaNumeric]) collect: [:chr | chr asLowercase].
str = str reversed.
].
{{works with|GNU 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 ]
]
]
].
'''Testing'''
('hello' palindro) printNl.
('hello' palindroR) printNl.
('ingirumimusnocteetconsumimurigni' palindro) printNl.
('ingirumimusnocteetconsumimurigni' palindroR) printNl.
{{works with|VisualWorks Pharo Squeak}}
isPalindrome
^self reverse = self
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
{{out}}
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
SET @txt = REPLACE('In girum imus nocte et consumimur igni', ' ', '');
SELECT REVERSE(@txt) = @txt;
Swift
{{works with|Swift|1.2}}
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
}
{{works with|Swift|2.0}}
func isPal(str: String) -> Bool {
let c = str.characters
return lazy(c).reverse()
.startsWith(c[c.startIndex...advance(c.startIndex, c.count / 2)])
}
Tcl
'''Non-recursive'''
package require Tcl 8.5
proc palindrome {s} {
return [expr {$s eq [string reverse $s]}]
}
'''Recursive'''
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 $s 1 end-1]]
}
}
'''Testing'''
set p ingirumimusnocteetconsumimurigni
puts "'$p' is palindrome? [palindrome $p]"
puts "'$p' is palindrome? [palindrome_r $p]"
TUSCRIPT
$$ MODE TUSCRIPT
pal ="ingirumimusnocteetconsumimurigni"
pal_r=TURN(pal)
SELECT pal
CASE $pal_r
PRINT "true"
DEFAULT
PRINT/ERROR "untrue"
ENDSELECT
{{out}}
true
TypeScript
const detectNonLetterRegexp=/[^A-ZÀ-ÞЀ-Я]/g;
function stripDiacritics(phrase:string){
return phrase.normalize('NFD').replace(/[\u0300-\u036f]/g, "")
}
function isPalindrome(phrase:string){
const TheLetters = stripDiacritics(phrase.toLocaleUpperCase().replace(detectNonLetterRegexp, ''));
const middlePosition = TheLetters.length/2;
const leftHalf = TheLetters.substr(0, middlePosition);
const rightReverseHalf = TheLetters.substr(-middlePosition).split('').reverse().join('');
return leftHalf == rightReverseHalf;
}
console.log(isPalindrome('Sueño que esto no es un palíndromo'))
console.log(isPalindrome('Dábale arroz a la zorra el abad!'))
console.log(isPalindrome('Я иду с мечем судия'))
UNIX Shell
if [[ "${text}" == "$(rev <<< "${text}")" ]]; then
echo "Palindrome"
else
echo "Not a palindrome"
fi
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).
#import std
palindrome = ~&cixE\letters+ * -:~& ~=`A-~rlp letters
This test programs applies the function to each member of a list of three strings, of which only the first two are palindromes.
#cast %bL
examples = palindrome* <'abccba','foo ba rra bo of','notone'>
{{out}}
<true,true,false>
Vala
Checks if a word is a palindrome ignoring the case and spaces.
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;
}
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.
Public Function isPalindrome(aString as string) as Boolean
dim tempstring as string
tempstring = Lcase(Replace(aString, " ", ""))
isPalindrome = (tempstring = Reverse(tempstring))
End Function
{{out|Example}}
print isPalindrome("In girum imus nocte et consumimur igni")
True
VBScript
=Implementation=
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
=Invocation=
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")
{{out}}
0
-1
0
-1
Vedit macro language
This routine checks if current line is a palindrome:
:PALINDROME:
EOL #2 = Cur_Col-2
BOL
for (#1 = 0; #1 <= #2/2; #1++) {
if (CC(#1) != CC(#2-#1)) { Return(0) }
}
Return(1)
Testing:
Call("PALINDROME")
if (Return_Value) {
Statline_Message("Yes")
} else {
Statline_Message("No")
}
Return
Visual Basic .NET
{{trans|VBA}}
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
{{out}}
True
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"
]]
}
Returns:
[true false true]
X86 Assembly
; 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
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);
];
]
{{out}}
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).
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));
}
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
{{out}}
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