Determine whether strings are anagrams

Question

Challenge

Given two strings, work out if they both have exactly the same characters in them.

Example

Input

word, wrdo

This returns true because they are the same but just scrambled.

Input

word, wwro

This returns false.

Input

boat, toba

This returns true

Rules

Here are the rules!

• Assume input will be at least 1 char long, and no longer than 8 chars.
• No special characters, only a–z
• All inputs can be assumed to be lowercase

Test Cases

boat, boat = true
toab, boat = true
oabt, toab = true
a, aa = false
zzz, zzzzzzzz = false
zyyyzzzz, yyzzzzzy = true
sleepy, pyels = false
p,p = true

• Tom Gullen: 9 answers in 13 views... wow!
• user54200: Title request: Cod Elf, Go!
• Geobits: "Falcon Rage, go mad!"
• Esolanging Fruit: My name suggestion: "are they anagrams" → "manage the arrays"
• Deadcode: Suggested test case: aaab, bbba = false

Answer 1

Python, 32 bytes

f=lambda a,b,S=sorted:S(a)==S(b)

• gnibbler: @Debanjan, This is just the same as def f(a,b):return sorted(a)==sorted(b) The trade off is that you get to replace def+return with lambda in exchange for not using any statements
• gnibbler: @Debanjan, I think it only saves you one character. I have used a variation here, but it works out the same length as yours because I swap a newline for a comma
• Quixotic: I have one question,this approach does require calling of input and print and other stripping functions right?
• gnibbler: @Tomas, it's not rule abuse. Some questions are deliberately openended like this one appears to be. Compare with a well specified question like this. If you don't like these answers complain to the question asker
• Oliver Ni: You don't need to include the f=.

Answer 2

Golfscript, 3 chars?

$$=

usage:

'boat'$'baot'$=
1
'toab'$'boat'$=
1
'oabt'$'toab'$=
1
'a'$'aa'$=
0
'zzz'$'zzzzzzzz'$=
0
'zyyyzzzz'$'yyzzzzzy'$=
1
'sleepy'$'pyels'$=
0
'p'$'p'$=
1

• gnibbler: This is an interesting interpretation of how to supply the input :)
• st0le: Explanation please :(
• Peter Taylor: @st0le, seriously? I don't know golfscript, but it's obviously $ (sort), $ (sort), = (compare)
• Thomas Eding: Isn't this cheating a little? I mean, it isn't variable input. It has to be hard-coded. In any case, I would add 4 to the character count for the quote (') characters.
• Doorknob: This isn't valid by our current rules. You could, however, change it to @JanDvorak's 4-byte function, which would accept input via a valid input format.

Answer 3

J, 8

-:&(/:~)

Literaly, match (-:) on (&) sort up (/:~)

Sample use:

 'boat' -:&(/:~) 'boat'
1
 'toab' -:&(/:~) 'boat'
1
 'oabt' -:&(/:~) 'toab'
1
 'a' -:&(/:~) 'aa'
0
 'zzz' -:&(/:~) 'zzzzzzzz'
0
 'zyyyzzzz' -:&(/:~) 'yyzzzzzy'
1
 'sleepy' -:&(/:~) 'pyels'
0
 'p' -:&(/:~) 'p'
1

Where do the 64-bit integers come into play?

• user475: Is it not possible to write functions/subroutines in J?
• J B: @Tim Nordenfur: they're called "verbs", and take either one argument on their right as in v arg (monads) or two on both sides as in arg1 v arg2 (dyads). The one I submitted is obviously a dyad. I didn't bother to name it, since it wasn't requested and is shorter this way. Should you really want to give it a name, you'd do that like this: is_anagram_of =: -:&(/:~) and then use as 'a' is_anagram_of 'aa'.
• user475: It felt a bit cheap to substitute the arguments into the code, but I see now that it's essentially a dyad. Never mind.
• st0le: J always looks like the remains of an emoticon factory explosion.

Answer 4

Javascript, (削除) 192 (削除ここまで) (削除) 157 (削除ここまで) (削除) 152 (削除ここまで) (削除) 147 (削除ここまで) 125 bytes

Ok some of these languages are a lot more flexibile than I thought! Anyway this is the longer way I guess, but a different technique at least.

Compressed

Thanks to Peter and David for squeezing more chars out!

for(a=[j=p=2];j<123;)a[j]?p%a[++j]<1&&p++&&(j=0):(a[j]=p,j=0);function b(c,i){return c[i=i||0]?a[c.charCodeAt(i)]*b(c,++i):1}

Then do:

alert(b("hello")==b("elloh"));

Expanded Code

The compressed has had lots of changes, but this is the basic theory:

// Define dictionary of primes
a = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101];
// Returns the unique ID of the word (order irrelevant)
function b(c) {
 r = 1;
 for (i = 0; i < c.length; i++)
 r *= a[c[i].charCodeAt(0) - 97];
 return r
}
alert(b("hello") == b("hlleo"));

• user475: Great idea using primes.
• Peter Taylor: You can shave a couple of characters off the dictionary initialisation using the sieve. a=[2];for(p=3,j=0;j<26;)if(a[j]){if(p%a[j++]==0){p++;j=0}}else{a[j]=p;j=0}
• Yahel: I feel like I read about this trick on some blog talking about interview questions...
• Peter Taylor: @Tom, depends on how well optimised the sorting routines are, given that you've limited inputs to 8 characters :P
• David Murdoch: 125 characters. Recursion and ternaries FTW: for(a=[j=p=2];j<123;)a[j]?p%a[++j]<1&&p++&&(j=0):(a[j]=p,j=0);function b(c,i){return c[i=i||0]?a[c.charCodeAt(i)]*b(c,++i):1}

Answer 5

Golfscript, 8 bytes

This defines a function called A

{$\$=}:A

Test cases

;
'boat' 'boat' A
'toab' 'boat' A
'oabt' 'toab' A
'a' 'aa' A
'zzz' 'zzzzzzzz' A
'zyyyzzzz' 'yyzzzzzy' A
'sleepy' 'pyels' A
'p' 'p' A

Answer 6

Haskell, 31 bytes

function - 31

import List
f=(.sort).(==).sort

program - (削除) 81 (削除ここまで) (削除) 58 (削除ここまで) 55

import List
g=sort`fmap`getLine
main=(`fmap`g).(==)=<<g

Usage:

$ runghc anagram.hs
boat
boat
True
$ runghc anagram.hs
toab
boat
True
$ runghc anagram.hs
a
aa
False

Kudos to lambdabot and its pointfree refactoring.

• J B: Can Haskell code that only does what's wanted under runghc, but not when compiled, still be called "program"?
• Joey Adams: @J B: Can Perl code that only does what's wanted under perl still be called a "program"? :-)
• Callum Rogers: J B: Todays functional languages skew the meaning of a program making it a higher order abstraction. Rather than just a list of instruction to be executed, as haskell program can just be seen as a collection of functions, even if they are not called.
• J B: @Callum Rogers: my point is: his code behaves differently whether run under runghc or compiled, in a problem-sensitive area. The "function" is fine. The "program" does not solve the problem under anything else than runghc, and runghc is not the only legitimate way to run Haskell programs. In that context, that makes the snippet a "runghc script", not a "Haskell program". --not that I'd consider the issue important, as I said, the function is fine anyway and it's shorter.
• Rotsor: x#y=sort x==sort y is 1 character shorter

Answer 7

C#, 129 chars

namespace System.Linq{class X{static void Main(string[]a){Console.Write(a[0].OrderBy(_=>_).SequenceEqual(a[1].OrderBy(_=>_)));}}}

Readable:

namespace System.Linq
{
 class X
 {
 static void Main(string[] a)
 {
 Console.Write(a[0].OrderBy(_ => _)
 .SequenceEqual(a[1].OrderBy(_ => _)));
 }
 }
}

• Stackstuck: I think you could golf a few bytes with using System.Linq; instead of namespacing it?

Answer 8

Ruby, 34 bytes

Using the IO scheme of Peter Taylors Perl solution:

p gets.chars.sort==gets.chars.sort

• Tomas: Throws an error: -e:1:in '<main>': undefined method 'chars' for nil:NilClass (NoMethodError)

Answer 9

C program, 118

t[52],i;main(c){for(;i<52;)(c=getchar())<11?i+=26:t[i+c-97]++;
for(i=27;--i&&t[i-1]==t[i+25];);puts(i?"false":"true");}

• Mateen Ulhaq: Ever considered applying for IOCCC?
• R. Martinho Fernandes: @muntoo: have you seen anything in the IOCCC? This is way too readable for that.
• Mateen Ulhaq: @Martinho Yes, the IOCCC source codes are so beautiful. Symphonies. But he should at least try composing a small piece. :)
• Joey Adams: @muntoo: I didn't even know they were still active.
• ugoren: Just saw this one... Very nice. But can be shorter: t[256],i;main(c){for(;c+3;)(i=getchar())>10?t[i]+=c:(c-=2);for(i=257;--i&&!t[i-1];);puts(i?"false":"true");} - that's 108 characters. Very importantly, your c initialization trick is still employed.

Answer 10

JavaScript

Based on @zzzzBov's solution.

Comparison, 65 chars (40 without function)

function f(a,b){return a.split('').sort()==b.split('').sort()+""}

Comparator, 43 chars

function f(a){return a.split('').sort()+""}

• Casey Chu: Clever using the +"" to coerce to string.

Answer 11

Perl, 58 bytes

(complete program, unlike the other Perl answer which is only a function)

($c,$d)=map{[sort split//]}<>;print"@$c"eq"@$d"?true:false

49 as a function

sub f{($c,$d)=map{[sort split//]}<>;"@$c"eq"@$d"}

• Joel Berger: of course you can save 4 chars in the program by removing the " around true and false, since without using strict/warnings a bareword is a string.
• Timwi: Thanks!
• Adam Katz: I prefer this as ($c,$d)=map{[sort split//]}@ARGV;exit("@$c"ne"@$d") (51 chars) so it can take command line arguments and use the command line exit codes. It'd be 48 chars retaining <> with a multi-line input.

Answer 12

R, 54 bytes

function(x,y,`~`=sort,`+`=utf8ToInt)identical(~+x,~+y)

Try it online!

• Sumner18: I'm highly intrigued by your use of utf8ToInt, not only in this answer, but in many others that I have seen.
• J.Doe: Have you seen tips for golfing in R? utf8ToInt and its reverse tend to make for shorter string-splitting than the conventional functions.

Answer 13

Clojure - 23 chars

As an anonymous function:

#(apply = (map sort %))

Test case example:

(#(apply = (map sort %)) ["boat" "boat"])
=> true

• Chiron: Cool, I like it.
• coredump: Nice answer. I particularly like the test strings you chose ;-)

Answer 14

C++ (104 non-ws chars)

Based on counting sort. Note: Assumes strings of the same length, which seems to be implied (though not stated) by the question.

int u[123], i;
int a(char **c) {
 for(; c[0][i]; ) {
 u[c[0][i]]++;
 u[c[1][i++]]--;
 }
 i=123;
 while(i && !u[--i]);
 return !i;
}

• Joey Adams: In C, if you declare a variable in the global scope, it is initialized to zero. I guess this is also true for C++.
• Joey Adams: Local variables, on the other hand, are not initialized to zero automatically.
• Matthew Read: OK, I removed my caveat since I found ways to do without it.
• Dave Gamble: Bzzzt. You pass the test cases, but "helle" and "hollo" are apparently the same. Easy fix: change one of the ++ to a --. Then just if (u[i++]) return 0
• st0le: I haven't tested this, but the last three lines can be written as i=123;while(i&&u[--i]);return!i;

Answer 15

APL, 2 chars

≡⍦

This is the Multiset Match function from Nars2000, one of the leading-edge APL implementations. When applied to strings, it computes exactly the function required:

 'elvis' ≡⍦ 'lives'
1
 'alec guinness' ≡⍦ 'genuine class'
1

• Maltysen: Just curious, how many bytes is this? 4? 6?
• Tobia: It depends on the encoding. 6 bytes in UTF-8, 4 bytes in UCS-2, 2 bytes if any of the legacy single-byte APL charsets have the ⍦ symbol, which I doubt.

Answer 16

05AB1E, (削除) 6 (削除ここまで) 4 bytes

{I{Q

Try it online!

This took a while because of input difficulties. Golfed down due to pop.

Explanation:

{I{Q Original code
{ Takes first input e.g. word and sorts -> 'dorw'
 I Takes second input e.g. 'wrdo'
 { Sorts second input -> 'dorw'
 Q Compare if sorted 1 = sorted 2, then print result. 'dorw' = 'dorw', so prints 1.

• Kevin Cruijssen: I know this answer is 5+ years old, but by taking the input as pair of strings, it can be 3 bytes: €{Ë

Answer 17

Perl, 62 bytes

This function takes the strings as arguments and returns true or false.

sub f{my@a;for$.(1,-1){$a[ord]+=$.for split//,pop}!grep{$_}@a}

Stores the ASCII values in an array and checks if it evens out. Increments for the first word and decrements for the second word.

Answer 18

Knight, 112 bytes

;=dF;C=mB;=a 123W<96=a-a 1Id=d!E+"a"aE++"=a"a" 0";=w"+1a";C=lB;=bP:Wb;E++"=a"Ab+wAb=bGb 1Lb;=w"+~1a";Cl;=dT;CmOd

Try it online!

Knight doesn't have arrays. It does have strings, but it only allows a subset of valid ASCII character codes. So I emulated an array by constructing strings that looks like "=a{charcode} 0" and evaling it.

Knight also has blocks. A block works like a niladic function, which can change its behavior depending on global variables. This code uses two blocks: m loops over all 26 variables a97 - a122 initializing or testing something, and l handles the part of taking input and iterating over its chars to count occurrences.

; = d F d = False
; C = m B define and run block m:
 ; = a 123 a = 123
 : W < 96 = a - a 1 while 96 < (a = a - 1):
 I d if d:
 = d ! E + "a" a d = ! eval("a" + a)
 E + + "=a" a " 0" else: eval("=a" + a + " 0") (aNN = 0)
 i.e. d is false on start, so initialize all of a97..a122 to 0
; = w "+1a" w = "+1a" (which is necessary to run l)
; C = l B define and run block l:
 ; = b P b = one line of input
 : W b while b is not empty:
 ; E + eval(concatenation of
 + "=a" A b "=aNN"
 + w A b w+"NN") (aNN = 1 + aNN)
 (w = "+~1a" changes this to aNN = -1 + aNN)
 : = b G b 1 L b b = b[1:length(b)]
 i.e. count occurrences of chars in the first line
; = w "+~1a" w = "+~1a"
; C l run block l i.e. count negative occurrences in the second line
; = d T d = True
; C m run block m
 for each of a122..a97, check if aNN is zero
 if true, keep checking; otherwise do useless thing
: O d print d

Answer 19

PHP (command line, 87 characters)

function d($s){
 return array_count_values(str_split($s));
}
echo d($argv[1]) == d($argv[2]);

Answer 20

Javascript

A (very) slightly shorter version of @zzzzBov's solution, that uses .join() instead of String boxing:

function a(b,c){return b.split('').sort().join()==c.split('').sort().join()}
alert(a('abc','bca')); //true

Similarly:

function a(b){return b.split('').sort().join()}
alert(a('abc')==a('bca')); //true

• TheDoctor: this is answer ID 1337. congratz

Answer 21

Clojure REPL 41 chars

(= (sort (read-line)) (sort (read-line)))

• dmckee --- ex-moderator kitten: Welcome to the Stack Exchange network. Formatting help here.

Answer 22

Java

(everyone's favorite language apparently!)

173 chars:

import java.util.*;class g{public static void main(String[]p){String[]a=p[0].split(""),b=p[1].split("");Arrays.sort(a);Arrays.sort(b);System.out.print(Arrays.equals(a,b));}}

(Doesn't print newline char to save 2 chars from println)

Compile and run:

javac g.java
java -cp . g abcdef fedcba
true

Love to see a shorter one...

• Kevin Cruijssen: I know it's been more than 6.5 years (lol), but you can golf it by 10 bytes by adding java.util.Arrays x=null; and use x. instead of Arrays.: class g{public static void main(String[]p){java.util.Arrays x=null;String[]a=p[0].split(""),b=p[1].split("");x.sort(a);x.sort(b);System.out.print(x.equals(a,b));}} (163 bytes) And by converting it to Java 8, class g{public static void main could be interface g{static void main as well, but I guess Java 8 wasn't around yet in 2011, so keeping class is also fine. ;p

Answer 23

Scala in REPL (32)

readLine.sorted==readLine.sorted

Scala function (43)

def f(a:String,b:String)=a.sorted==b.sorted

Scala program (61)

object A extends App{println(args(0).sorted==args(1).sorted)}

These leverage a neat feature of Scala whereby a String can also be treated as a sequence of characters (Seq), with all the operations on Seq being available.

Answer 24

sed, 45 chars

It's even possible in my favourite - sed! Just one regular expression to solve the anagram! Just keep removing the corresponding letters:

:
s/(.)(.*,.*)1円/2円/
t
/\w/{i\false
d}
i\true

(to be invoked with -nE)

Perl, 48

1while s/(.)(.*,.*)1円/2円/;$_=/\w/?"false":"true"

To be invoked with -p.

Perl function, 39

sub f{,ドルwhile s/(.)(.*,.*)1円/2円/;!/\w/}

• Kjetil S: One less: sub f{1while s/(.)(.*,.*)1円/2円/;!/\w/}

Answer 25

Python 3, (削除) 107 (削除ここまで) (削除) 97 (削除ここまで) (削除) 76 (削除ここまで) 64

s=sorted;a,b=input().split(', ')
print(str(s(a)==s(b)).lower())

Obviously this can be shortened if we don't take the OP's wording literally and lowercase "true" and "false"...

• Alex Van Liew: You can shave off a few characters if you append ;s=sorted to the first line and then replace the two instances of sorted with s in the second line. Should save... 3 characters?
• Wooble: Indeed. Python 3 also saves a bit of space, and is probably reasonable to use now, 5 years after this answer was posted. Also, the .strip() was redundant, given the specified inputs.
• Alex Van Liew: Yeah, sorry. I didn't notice how old this question was when I commented, only that it was on the frontpage. ^^

Answer 26

JavaScript, 41

Comparison function (41):

a=b=>''+[...b].sort()
b=(c,d)=>a(c)==a(d)

alert(b('abc', 'cba')) // true

Comparator function (21):

a=b=>''+[...b].sort()

alert(a('abc') == a('bca')); //true

Comparator function (48):

function a(b){return String(b.split('').sort())}

alert(a('abc')==a('bca')); //true

Comparison function (78):

function a(b,c){return String(b.split('').sort())==String(c.split('').sort())}

alert(a('abc','bca')); //true

Assumes String has split and Array has sort.

• Shieru Asakoto: 38 bytes: c=>d=>(a=b=>''+[...b].sort())(c)==a(d)

Answer 27

Python, 32 bytes

p=sorted
f=lambda a,b:p(a)==p(b)

• Tomas: Does nothing in python. Are you sure it is a complete program that takes the input and produces the output as requested?
• TuxCrafting: @Tomas It's a function

Answer 28

Bash, 88 characters

diff <(grep -o .<<<1ドル|sort) <(grep -o .<<<2ドル|sort)>/dev/null && echo true || echo false

Answer 29

Regex (PCRE), 65 bytes

^((?=(.))(?!(.*?2円(?=.*
(4円?+.*?2円)))*+.*2円)(?=.*
(5円?+.)).)*
5円$

Try it online! - test cases
Try it on regex101 - try it on your own

Takes the input strings delimited by newline.

This asserts that both:

• For each character in the first string, the second string contains that character at least as many times as it occurs in the first string
• The second string is equal in length to the first

^ # Assert that we're starting at the beginning of the first word.
(
 (?=(.)) # Match and capture a character in 2,円 without consuming it.
 # Assert that the second has at least as many occurrences of 2円 as there
 # are subsequent occurrences of it in the first word (including the one we
 # just captured above).
 (?! # Negative lookahead - assert the following can't match, and also
 # conveniently erase whatever capture groups it creates when finished.
 (
 .*?2円 # Match up to the next occurrence of 2円 in the first word
 # (which will be the first one when this loop starts)
 (?=
 .*¶ # Skip to the second word.
 ( # 4円 = the following:
 4円?+ # previous value of 4,円 if any
 .*?2円 # Match up to the first subsequent occurrence of
 # 2円 in the second word (which will be the first
 # occurrence when this loop starts)
 )
 )
 )*+ # Loop the above possessively, locking in each iteration
 # atomically as it matches.
 .*2円 # Assert 2円 occurs again in the first word.
 )
 # Append one more character from the second word onto 5,円 building up 5円
 # to have the same number of characters of the second word as characters
 # we have finished processing in the first word.
 (?=
 .*¶ # Skip to the second word.
 ( # 5円 = the following:
 5円?+ # Previous value of 5,円 if any
 . # One additional character
 )
 )
 . # Skip over the 2円 just processed in the first word.
)* # Iterate the above as many times as possible.
¶ # Assert we've reached a newline, meaning we've processed the
 # entire first word, and advance to the second word.
5円$ # Assert that the second word is of equal length to the first.

It only takes +1 byte (66 bytes) to add support for zero-length words:

^((?=(.))(?!(.*?2円(?=.*
(4円?+.*?2円)))*+.*2円)(?=.*
(5円?+.)).)*
5円?$

Try it online! - test cases
Try it on regex101 - try it on your own

Regex (.NET), (削除) 70 (削除ここまで) (削除) 66 (削除ここまで) 62 bytes

^(((.))(?=(?>(?<2>2円)|.)*
((?<-2>2円)|.)*$(?(2)^)))*
(?<-3>.)*$

Try it online! - test cases
Try it on regex101 - try it on your own

This asserts that both:

• For each character in the first string, the second string contains that character at least as many times as it occurs in the first string
• The second string is not longer than the first

Although it is not required, this works even with zero-length words (for free; there is no golf sacrifice in doing so).

^ # Assert that we're starting at the beginning of the first word.
(
 ((.)) # Consume a character and capture it in both 2円 and 3円.
 
 # Assert that the second word has at least as many occurrences of 2円 as
 # there are subsequent occurrences of it in the first word (including the
 # occurrence we just consumed). This only needs to be asserted for the first
 # occurrence of each character in the first word (because on each subsequent
 # occurrence, the count of subsequent occurrences will be one fewer than
 # before) but for golf reasons, it's harmless to do it for every occurrence.
 
 (?= # Atomic lookahead
 # Step 1: Push every subsequent occurrence of 2円 onto the 2円 stack.
 (?> # Atomic group - prevent backtracking from changing what's
 # done inside it once it's been done.
 # For each subsequent character in the first word:
 (?<2>2円) # If it matches 2,円 push another copy of it onto the
 # 2円 stack, and consume the character.
 | # or
 . # Consume the character.
 )* # Iterate the above as many times as possible.
 ¶ # Consume the newline to begin reading the second word,
 # preventing the number of loop iterations in the above
 # loop from being backtracked.
 
 # Step 2: Assert that the second word has at least as many occurrences
 # of 2円 as were counted above.
 ( # For each character in the second word:
 (?<-2>2円) # If it matches 2,円 pop it off the 2円 stack, and consume
 # the character.
 | # or
 . # Consume the character.
 )*$ # Iterate the above until reaching end of string.
 (?(2)^) # Assert that the 2円 stack is empty.
 )
)* # Iterate the above as many times as possible.
¶ # Assert we've reached a newline; advance to the second word.
(?<-3>.)*$ # Assert that this word is not longer than the first word, by
 # popping every capture from the 3円 stack and matching one
 # character of the second word (it doesn't have to match 3円)
 # for each.

Regex (Perl _{^{/ PCRE}}), 71 bytes

^((?=(.))(?4)(?=.*
(3円?+.)).)*
3円$((?!(.*?2円(?=.*
(6円?+.*?2円)))*+.*2円))

Try it online! - Perl
Try it online! - PCRE2

Perl doesn't erase capture groups upon exiting from a negative lookahead. For compatibility with it, this version uses a subroutine call (?4) instead of a negative lookahead to erase the necessary capture group. The subroutine is then executed again, harmlessly, after the regex finishes and reaches the end of the string (for golf reasons, it is not hidden in a (?(DEFINE)...) block or similar construct).

Regex (_{^{PCRE / .NET}}), 74 bytes

^((?=(.))(?!(?>(.*?2円(?=.*
((?>4円?).*?2円)))*).*2円)(?=.*
((?>5円?).)).)*
5円$

Try it online! - PCRE2
Try it online! - .NET

This is the PCRE version ported to be compatible with .NET, replacing possessive quantifiers with atomic groups.

Regex (_{^{PCRE /}} Ruby), 87 bytes

^((?=(.))(?!(.*?2円(?=.*
(5円?+.*?2円))(?=.*
(4円)))*+.*2円)(?=.*
(7円?+.))(?=.*
(6円)).)*
6円?$

Try it online! - PCRE
Try it online! - Ruby

This is a port of the 65 byte PCRE version, using back-and-forth copying of 4円↔5円 and 6円↔7円 to avoid use of nested backreferences, which aren't supported by Ruby.

For a reason I have yet to determine, this doesn't work on Python^regex , even its latest version.

Regex (_{^{Perl / PCRE / Ruby / .NET}}), 147 bytes

^((?=(.))(?!(?>(.*?2円(?=((?!.*2円)|.))(?=.*
((?!.*$4円)(?>6円?).*?2円|(?=.*$4円)(?=(?>6円?).*?2円)))(?=.*
(5円)))*).*2円)(?=.*
((?>8円?).))(?=.*
(7円)).)*
7円$

Try it online! - Perl
Try it online! - PCRE2
Try it online! - Ruby
Try it online! - .NET

This is a rather interesting challenge for writing a regex polyglot.

Supporting both Perl and Ruby/.NET is rather difficult. The problem is finding a way to erase the capture group that is created inside a negative lookahead in the PCRE version. Erasing capture groups in situations like this can be a challenge, and to support this combination of regex engines, we can't use negative lookahead (Perl doesn't erase it), a subroutine call (because Ruby's syntax is incompatible with Perl, and .NET doesn't have them), a branch reset group (neither Ruby nor .NET support them), or capture group aliasing or explicit erasure (only .NET has those).

So we must signal to the place where the capture is normally made, to erase it when it needs to be erased. The method I settled upon here is rather convoluted, but it works. (It would be shorter with a lookaround conditional, but we can't use one, as Ruby doesn't have them, and .NET's syntax for them is different.) This post is already pretty long, so here's a link to the commented version.

Answer 30

VBScript

172 characters

wscript.echo "boat, boat = " & s("boat, boat")
wscript.echo "toab, boat = " & s("toab, boat")
wscript.echo "oabt, toab = " & s("oabt, toab")
wscript.echo "a, aa = " & s("a, aa")
wscript.echo "zzz, zzzzzzzz = " & s("zzz, zzzzzzzz")
wscript.echo "zyyyzzzz, yyzzzzzy = " & s("zyyyzzzz, yyzzzzzy")
wscript.echo "sleepy, pyels = " & s("sleepy, pyels")
wscript.echo "p,p = " & s("p,p")
function s(a):b=split(replace(a," ",""),","):c=0:for x=1 to len(b(0)):if instr(b(1),mid(b(0),x,1)) then c=c+1 
next:if len(b(1))-c=0 then s=true else s=false
end function

I was kinda suprised I could get it under 200.

• user775: 131 characters if I pass the values seperatly. function s(a,b):c=0:for x=1 to len(a):if instr(b,mid(a,x,1)) then c=c+1 next:if len(b)-c=0 then s=true else s=false end function