Determine if Strings are equal [closed]

Question

Your task is simple. Determine if one string equals the other (not address, the value) without the use of equality operators (such as ==, ===, or .equal()) or inequality (!=, !==) anything similar for other languages. This means anywhere! You may not use these operators anywhere in the code. You may however, you use toggles such as !exp as you're not directly comparing the exp != with something else.

In addition, you may not use any functions such as strcmp, strcasecmp, etc.

As for comparison operators (>=, <=, >, <), they are also disallowed. I realize that some answers include this, but I'd really like to see more answers that don't borderline the equality operator.

An example using PHP is shown:

<?php

$a = 'string';
$b = 'string';

$tmp = array_unique(array($a, $b));

return -count($tmp) + 2;

Simply return true or false (or something that evaluates in the language to true or false like 0 or 1) to indicate if the strings match. The strings should be hardcoded seen in the above example. The strings should not be counted in the golf, so if you declare the variable before hand, don't count the declaration.

Answer 1

C - 86 83

main(int i,char**v){return*v[1]&!(*v[1]++^*v[2]++)?main(3,v):!(*--v[1]^*--v[2]);}

Obvioulsy not the shortest, but this doesn't work with string variables and instead takes the strings as input from the console. Also, I sort of like the recursive main, even if it's obviously not the shortest version. But certainly the least advisable.

Answer 2

C++: 104 105 characters

#include<string>
#include<set>
template<typename T>
int f(T a,T b){std::set<T>S={a};return S.count(b);}

EDIT: Saved a character by making it a template function.

Function itself is 51 characters. Other C++ answers already posted are shorter, but I wanted to try one that wasn't also pure C. In fact, this requires C++11. For earlier versions of C++ you can replace std::set<T>S={a}; with std::set<T>S(&a,&a+1); at the expense of an extra 5 characters.

Answer 3

Excel VBA

Function stringmatch()
a = ""
b = ""
For i = 1 To Len(a)
If Asc(Right(Left(a, i), 1)) - Asc(Right(Left(b, i), 1)) Or Len(a) - Len(b) Then
debug.print "string unnmatch"
Exit Function
End If
Next
debug.print  "string match"
End Function

Answer 4

Groovy 347

def compare(str1,str2){
    def str1val = 0
    def str2val = 0

    MessageDigest md5 = MessageDigest.getInstance("MD5")
    md5.digest(str1.getBytes("UTF-8")).each{ str1val += it*2 }
    md5.digest(str2.getBytes("UTF-8")).each{ str2val += it*2 }

    if(!(str1val-str2val)) {
        return true    
    } else {
        return false
    }
}

Answer 5

Python, 7 chars.

b in[a]

Where b and a are strings.

Answer 6

Perl - 10

/^\Q$w\E$/

The input is in $_ and $w. The output is stored in $&.

Answer 7

Java (235 characters for golfed version at bottom).

Strings are not hardcoded, but this is a trivial change.

   public boolean areEqual(String a, String b) {
    int[] q = new int[1];
    int i = 0;
    try {
        while (true) {
            noOp(q[(int) a.charAt(i) - (int) b.charAt(i)]);
            i++;
        }
    } catch (Error e) {
        try {
            a.charAt(i);
        } catch (Error e2) {
            try {
                b.charAt(i) {catch(Error e3){
                    return true;
                }
                }
            }


            return false;
        }
        return true;
    }
}

public void noOp(int i) {
}

Traverses both strings, using the difference in chars to select an array index (which is 0-based). No in-loop comparisons or Math.max used. Using Error as a throwable instead.

Golfed for use under certain circumstances (instantiated class, used from same package):

boolean a(String a,String b){boolean t=true,f=!t;int[]q={1};int i=0;try{while(t){q[(int)a.charAt(i)-(int)b.charAt(i)];i++;}}catch(Error e){try{a.charAt(i);}catch(Error z){try{b.charAt(i){catch(Error g){return t;}}}return f;}return t;}}

Answer 8

APL (Dyalog Extended), 4 bytes^SBCS

Anonymous tacit infix function. Requires ⎕IO←0

~⍳⍥⍮

Try it online!

⍥⍮ put each argument into a list

⍳ index of right list's elements in left list (0 if match, 1 if not)

~ logical NOT

Answer 9

PowerShell, 30 bytes

param($a,$b)$a-in$b-and$b-in$a

Try it online!

Takes two string inputs as command line parameters, -a and -b.

PowerShell, 20 16 bytes

-4 bytes thanks to @mazzy!

!($args|gu).Rank

Try it online!

Alternate answer, takes two strings as commandline arguments.

Answer 10

Python, 79

a=list(a)
b=list(b)
for i in range(len(a)):
 if ord(a[i])-ord(b[i])<0:
  print 'n'

Answer 11

Python - 52 chars without strings (76 with strings)

The in operator might be skirting the edges, but the rules disallow strict equality comparison, not similarity comparison.

a="string"
b="string"
x=a in b and not(len(a)-len(b)>0 or len(a)-len(b)<0)

Answer 12

Perl, 24 bytes

$r=!(-1+keys{$a,1,$b,1})

The two strings are given in $a and $b. The result is stored in $r.

The trick is using a hash array. If the strings are equal, then only one hash entry is created.

---

A standalone variant that gets the string from STDIN and returns 0, if the strings match and 1 otherwise:

$h{<>}=$h{<>}=1;exit -1+keys%h

Answer 13

Ruby, 21

Online version

Strings have to be stored in a and b:

!!(a+b=~/^(#{a})\1$/)

Checks if the concatenation of a and b contains a twice.

Answer 14

D 30

string a="somestring";
string b="somestring";

return !reduce!"a|b"(a[]^b[]);

Answer 15

Python 15 - 2( if you hard code the string in the solution) = 13

>>> s = 'string'
>>> t = 'string'
>>> 1 is len({s,t})
True
>>> s = 'string1'
>>> t = 'string2'
>>> 1 is len({s,t})
False

Python 17

>>> s = 'string'
>>> t = 'string'
>>> s in t and t in s
True
>>> s = 'string1'
>>> t = 'string2'
>>> s in t and t in s
False

Answer 16

Perl, 25 bytes

$r="$a$b"=~/^(\Q$a\E)\1$/

The input is expected in variables $a and $b. The result is stored in $r. It uses a regular expression to check, whether the concatenation "$a$b" is a duplicate of $a.

Answer 17

Python (7)

Assuming strings are stored in variables a and b:

a<=b<=a

Answer 18

Emacs Lisp

(split-string-and-unquote a b)

returns nil when a == b, something else when a != b

(print (if (split-string-and-unquote a b) "not equal..." "equal!"))

If I'm doing string work in elisp I use the popular package s.el in which it could be written a little shorter and without a negated output

(not (s-split a b t))

Answer 19

Java - 193

new int[]{0,1,0}[new BigInteger(a.getBytes()).subtract(new BigInteger(b.getBytes())).signum()+1]

No comparison operators at all, not even a < or >.

Answer 20

C - 45 characters

r;main(){ // 9 chars here
    char *a = "str2";
    char *b = "str1";
while(r+=*a^*b,*a++**b++);return!r;} // 36 more here

Answer 21

C++ 30

#include<string>
std::string a = "test", b = a;
int main(){return!(a<b||a>b);}

---

I'm not counting the first two lines because:

The strings should not be counted in the golf.

And here are some tests.

Answer 22

Delphi (412 83 75)

Can probably be done shorter but was fun to do, going to see if i can get some off.
412 -> 83 Cant believe I missed this... Checks if s1 is in s2 and if s2 is in s1. Saved 329 chars Cheers
83 -> 75 Thanks to Manatwork. Changed check if s1 is in s2 and other way around. Saved 8 characters

const s1='string12';s2='string12';begin Write(Pos(s1,s2)*Pos(s2,s1)>0);end.

Ungolfed

const
  s1='string12';
  s2='string22';
begin
  Write(Pos(s1,s2)*Pos(s2,s1)>0);
end.

Answer 23

Dart - 15

Strings are stored in a en b.

[a].contains(b)

[a] creates a List with a.

Answer 24

Python 2 - 50 54 (including strings)

Thanks to Ace for way around the "==" issue

a='string'
b='string'
not(len(''.join(set(a+b)))-len(a))

Output

True

Non-equal strings

a='string'
b='String'
not(len(''.join(set(a+b)))-len(a))

Output

False

Answer 25

C++ - 66 57 53

int c(char*a,char*b){return *a-*b?0:*a?c(++a,++b):1;}

with the ungolfed version:

int mystrcmp( char* a, char* b )
{
  // If the characters don't match, we're done
  if( *a - *b )
  {
      return 0;
  }
  
  // If they do match, and both were nulls, we are done 
  if( !*a )
  {
      return 1;
  }
  
  return mystrcmp( ++a, ++b );
}

Driver program in original post below.

---

Old version: The golfed version:

int c(char*a, char*b){while(*a|*b)if(*a++-*b++)return 0;return 1;}

Here is a full driver program with comments:

#include <iostream>

using std::cout;
using std::endl;

int mystrcmp( char* a, char* b ) {
   // The comparison will continue as long as both *a and *b are not '\0'.
   // If one of them goes '\0' but the other does not, the strings are not 
   // of equal length and at that point we will return 0.
   while( *a | *b ) {
     // Compare the current character and return 0 if they are not equal.
     // *a++ will return the value of *a and then increment the pointer -
     // see http://stackoverflow.com/questions/9255147/does-p-increment-after-dereferencing
     if(*a++ - *b++) {
       return 0;
     }
   }

   // If we got here, both strings are equal up to and including the last '\0'.
   return 1;
}

int main()
{   
   // Prints 1
   cout << mystrcmp("Hello.", "Hello.") << endl;

   // Prints 1       
   cout << mystrcmp("Hello, world!", "Hello, world!") << endl;
   
   // Prints 0
   cout << mystrcmp("Hello", "Hello.") << endl;

   // Prints 0       
   cout << mystrcmp("Hello.", "Zello.") << endl;
   
   // Prints 0
   cout << mystrcmp("Hello", "Hallo") << endl;
   
   // Prints 0
   cout << "Hello World" << endl; 
   
   return 0;
}

Answer 26

C#

I know there's already a C# answer, I just wanted to show other options. :)

boolean:

• 51

  !Convert.ToBoolean(new[]{a,b}.Distinct().Count()-1)
  

1 or 0:

• 32

  -new[]{a,b}.Distinct().Count()+2
  

Answer 27

Q - [3 bytes]

a:"string"
b:"string"

a~b

I think this is a valid answer as ~ isn't an equals operator. From a Q tutorial:

~ asks whether two entities have the same contents.

http://kx.com/q/d/primer.htm

Answer 28

C#

void Main()
{
    Console.WriteLine(StringEqual("pie","pie")); // true
}

bool Equals(int a,int b)
{
    var aBytes = BitConverter.GetBytes(a);
    var bBytes = BitConverter.GetBytes(b);

    return aBytes.SequenceEqual(bBytes);
}

bool StringEqual(string a,string b)
{
    int i = 0;
    return a.All((c) => Equals((int)c,(int)b[i++]));
}

Answer 29

Javascrpt: 38 chars (without a-b declarations) || 31 without alert()

a="string";
b="string";

o={};o[a]=0;alert(o.hasOwnProperty(b))

sets an object key named the value of a -- checks for a key named the value of b in that object.

without alert:

o={};o[a]=0;o.hasOwnProperty(b)

Answer 30

J - 6 char

A J expression which tests for string equality without testing for string equality? Here's one:

e./a;b        NB. is a boxed in b boxed?

Not good enough? Want some more options? Go ahead, have some more, J has plenty.

<:#~.a;b      NB. (the number of the unique items among the two) - 1
1-{:~:a;b     NB. 1 - (whether the second item is unique, given the first)
2-#]/.~a;b    NB. 2 - (the number of piles the strings will sort themselves into)
+/(/:*\:)a;b  NB. is sorting them forwards the same as sorting them backwards?
A.|./:\:~a;b  NB. will sorting them forwards after sorting them backwards do anything?

The common subexpression here, a;b, is a linked to b. It puts a and b into boxes before appending them into a list. This is J's closest approximation to non-rectangular arrays: if you just tried to put the two strings in a 2-item list, the shorter would be padded with spaces on the right, which could result in you comparing 'a' and 'a ' as equal.