Manage Trash So

Question

It's become somewhat of a tradition in PPCG that some users temporarily change their names by an anagram (a new name formed by reordering the letters of the old).

Sometimes it gets difficult to find out who is who. I could use a program or function to tell if two phrases are anagrams of each other.

The challenge

The program or function should take two strings and produce a truthy result if they are anagrams of each other, and falsy otherwise.

Rules

• Input will only contain letters (ASCII 65 to 90 and 97 to 122), digits (ASCII 48 to 57) or space (ASCII 32).
• The anagram relation is independendent of case. So "Arm" and "RAM" are anagrams.
• Spaces don't count either. So "keyboard" and "Barked Yo" are anagrams
• All builtins allowed
• Input format is flexible (two strings, an array of two strings, a string containing both phrases with a suitable separator ...)

Code golf. Fewest bytes wins.

Test cases

Truthy:

Lynn, Nyl N
Digital Trauma, Tau Digital Arm
Sp3000, P S 3000
Manage Trash So, Those anagrams

Falsy

Calvins Hobbies, Helka Homba
Android, rains odd
In between days, bayed entwine
Code golf, cod elf got

Answer 1

05AB1E, 9 8 bytes

Code:

lvyð-{}Q

Explanation:

l         # Lowercase the strings
 vy   }   # Map over the list, for each...
   ð-     #   remove spaces
     {    #   and sort
       Q  # Check equality

Try it online!

Answer 2

Retina, 25

i+`(\w)(.*,.*)\1
$2
^\W*$

Try it Online! Additionally, you can run a modified multi-line version.

Delete letters from before the comma along with their matches after the comma. If we have no letters left then it was an anagram.

Answer 3

Pyth, 11 10 bytes

Thanks to @FryAmTheEggman for teaching me the power of ;!

qFmSr-d;0Q

Try it here!

Takes a list of two strings as input.

Explanation

qFmSr-d;0Q    # Q = input

  m      Q    # map Q with d as lambda variable
     -d;      # filter spaces out of the string
    r   0     # convert to lowercase
   S          # sort all characters in string
qF            # Unfold resulting list and check for equality

Answer 4

Python 2, 63 61 bytes

lambda*l:len({`sorted(s.lower())`[2::5].strip()for s in l})<2

An anonymous function that, in fact, takes n arguments and determines if all n of them are mutual palindromes! f("Lynn", "Nyl N") returns True.

This set comprehension trick is by xnor. It saved two bytes, but the old approach looked very neat:

exec"a=`sorted(input().lower())`[2::5].strip();a"*2;print a==aa

Answer 5

Jelly, 12 bytes

ḟ€⁶O&95Ṣ€QLḂ

Try it online!

How it works

ḟ€⁶O&95Ṣ€QLḂ  Main link. Input: A (list of strings)

  ⁶           Yield ' '.
ḟ€            Filter it from each string.
   O          Apply ordinal to all characters.
    &95       Take bitwise AND with 95 to make the ordinals case-insensitive.
       Ṣ€     Sort each list of ordinals.
         Q    Deduplicate the list.
          L   Get the length.
           Ḃ  Compute the length's parity (1 -> 1, 2 -> 0).

Alternate version, non-competing (9 bytes)

Jelly's uppercase atom had a bug, and Jelly still had no built-in to test lists for equality...

ḟ⁶ŒuṢµ€⁼/

Try it online!

How it works

ḟ⁶ŒuṢµ€⁼/     Main link. Input: A (list of strings)

     µ€       Map the chain to the left over A.
 ⁶            Yield ' '.
ḟ             Filter it from the string.
  Œu          Cast to uppercase.
    Ṣ         Sort.
       ⁼/     Reduce by equality.

Answer 6

MATL, 11 bytes

2:"jkXvS]X=

EDIT (May 20, 2016) The code in the link uses Xz instead of Xv, owing to recent changes in the language.

Try it online!

2:"     ]       % do this twice
   j            % read input line as a string
    k           % convert to lowercase
     Xv         % remove spaces
       S        % sort
         X=     % are they equal?

Answer 7

Javascript, 69 61 60 59 bytes

1 byte off thanks @ӍѲꝆΛҐӍΛПҒЦꝆ. 1 byte off with currying (pointed out by @apsillers)

n=>m=>(G=s=>[]+s.toLowerCase().split(/ */).sort())(n)==G(m)

f=n=>m=>
    (G=s=>[]+s.toLowerCase()
        .split(/ */)
        .sort()
    )(n)==G(m)

F=(n,m)=>document.body.innerHTML+=`<pre>f('${n}')('${m}') -> ${f(n)(m)}</pre>`

F('Luis Mendo','Don Muesli')
F('Calvins Hobbies','Helka Homba')
F('Android','rains odd')
F('In between days','bayed entwine')
F('Code golf','cod elf got')
F('Lynn','Nyl N')
F('Digital Trauma','Tau Digital Arm')
F('Sp3000','P S 3000')
F('Manage Trash So','Those anagrams')

Answer 8

CJam, 11 12 14 bytes

3 2 bytes removed thanks to @FryAmTheEggman

{lelS-$}2*=

Try it online!

{      }2*       e# do this twice
 l               e# read line as a string
  el             e# make lowercase
    S-           e# remove spaces from string
      $          e# sort
          =      e# compare strings

Answer 9

Seriously, 11 9 bytes

2`,ùSô`n=

Try It Online!

Everyone seems to be using the same algorithm. Here it is yet again.

2`    `n          Do it twice
  ,               Read a string
   ù              Make it lowercase
    S             Sort
     ô            Strip spaces.
        =         Check equality.

Edit: realized sorting does work correctly on strings, and sorts spaces to the front so strip() will work.

Answer 10

C, 165 bytes

#define d(x) int x(char*a,char*b){
d(q)return*a&224-*b&224;}
#define n(x) for(qsort(x,strlen(x),1,(__compar_fn_t)q);*x<33;x++);
d(s)n(a)n(b)return strcasecmp(a,b);}

Readable and in working context,

#include <stdlib.h>
#include <string.h>
#include <stdio.h>

// start of comparison
int q(char *a, char *b){
     return ((*a)&0xdf)-((*b)&0xdf); // case-insensitive
}
int s(char *a, char *b){
    for(qsort(a,strlen(a),1,(__compar_fn_t)q); *a<33; a++) /**/;
    for(qsort(b,strlen(b),1,(__compar_fn_t)q); *b<33; b++) /**/;
    return strcasecmp(a,b);
}
// end of comparison

int main(int i, char **v){
    printf("'%s' '%s'", v[1], v[2]);
    printf("=> %d\n", s(v[1], v[2])); // 0 if equalish
    return 0;
}

Answer 11

Baloch Gyr, 9 bytes

{ṇ₁cḷ}ᵐpᵈ

Try it online!

Truthy/falsy output is achieved through predicate success/failure, this being Brachylog.

Previously saved a byte using cṇ₁cḷḍ instead of {ṇ₁cḷ}ᵐ under the assumption that the two input strings would be the same length minus whitespace, but I realized that it would succeed where it should fail on Ah Hass, haha.

{    }ᵐ      For both strings in the input,
 ṇ₁          split on spaces,
   c         concatenate,
    ḷ        and lowercase.
       pᵈ    The strings are permutations of each other.

Answer 12

zsh, 85 bytes

[ $(for x in $@;{tr -d \ <<<$x|tr A-Z a-z|fold -1|sort|paste -sd x}|uniq|wc -l) = 1 ]

Input as command line arguments, output as return code.

The for syntax makes this Bash-incompatible.

[               # test...
$(for x in $@;  # map over arguments
{tr -d \ <<<$x  # remove spaces
|tr A-Z a-z     # lowercase
|fold -1        # put each character on its own line
|sort           # sort lines
|paste -sd x    # remove all newlines except last
}|uniq          # take only unique lines
|wc -l          # how many lines remain?
) = 1 ]         # if only 1 line left, it must have been an anagram

Answer 13

Japt, 12 bytes

N®v ¬n ¬xÃä¥

Test it online!

How it works

        // Implicit: N = array of input strings
N®    Ã // Take N, and map each item Z to:
v ¬n    //  Take Z.toLowerCase(), split into chars, and sort.
¬x      //  Join and trim off whitespace.
ä¥      // Reduce each pair of items (that's exactly one pair) X and Y to X == Y.

Answer 14

GNU Sed, 33

Score includes +2 for -rn options to sed.

This is almost a direct port of @FryAmTheEggman's Retina answer:

:
s/(\w)(.*,.*)\1/\2/i
t
/\w/Q1

Ideone.

Answer 15

Perl, 34 33 + 1 = 34 bytes

s/(.)(.*,.*)\1/$2/i?redo:say!/\w/

Requires the -n flag and the free -M5.010|-E:

$ perl -M5.010 -ne's/(.)(.*,.*)\1/$2/i?redo:say!/\w/' <<< 'hello, lloeh'
1

How it works:

                                   # '-n' make a implicit while loop around the code
 s/(.)(.*,.*)\1/$2/i               # Remove a letter that occurs on both sides of the comma.
                    ?
                     redo:         # Redo is a glorified goto statement that goes to the top of the while loop
                          say!/\w/ # Check to see if any letter is left

Thanks to msh210 for suggesting using ternary operators to save one byte

Answer 16

Japt, 10 bytes

v á øVrS v

Try it

Answer 17

PHP, 109 94 bytes

function f($x){return str_split((trim($x));}function g($x,$y){return array_diff(f($x),f($y));}

Blech, the two function/returns are killing me here.

Returns the difference between two string inputs as an array of characters. PHP considers [] falsy, satisfying the return requirements.

Answer 18

Bash + GNU utilities, 51

f()(fold -1<<<${@^^}|sort)
f $1|diff -qBw - <(f $2)

• Define a function f() which:
  • ${@^^} converts all parameters to upper case
  • fold -1 splits chars - one per line
  • sorts lines
• call diff with -q to suppress full diff output and -Bw to ignore whitespace changes

Answer 19

Pyke (commit 30, noncompetitive), 9 bytes

Fl1dk:S)q

Explanation:

F      )  -  for _ in eval_or_not(input())
 l1       -     ^.lower()
   dk:    -    ^.replace(" ", "")
      S   -   sorted(^)
        q - ^==^

Answer 20

Mathematica, 77 76 bytes

StringMatchQ[##,IgnoreCase->1>0]&@@(""<>Sort[Characters@#/." "->""]&/@{##})&

The first part is actually one of my answers to another question!

Answer 21

Pike, 54 112 109 109 96 bytes

#define a(x) sort((array)replace(lower_case(x)," ",""))
int s(mixed i){return a(i[0])==a(i[1]);}

mixed happens to be shorter than array(string).

s returns 1 if its arguments are anagrams.

Answer 22

Q, 25 Bytes

f:{~/{x@<x:x@&~^x:_x}'x}

NOTE.- counting include function name f: to facilitate tests (as lambda we can decrement 2 Bytes)

Readable version

match over {ascending not null lower x} each x

{.. x ..} is an anonymous function with arg x
_x        lowers string x
&~^x      where not null x (space is considered null)
x@..      selects elements of x according to indexes .. 
<x        ascending indexes of x (not values). Ex <"cab" is 1 2 0
x@<x      ascending values of x (x at ascending indexes of x)
~         match (diad function). Ex "one"~"one" is true
f'..      applies function f for each argument ..
f/..      applies function f over elements of sequence (fold)

Test

f("Lynn";"Nyl N")                       
f("Digital Trauma";"Tau Digital Arm")   
f("Sp3000";"P S 3000")                  
f("Manage Trash So";"Those anagrams")   
f("Calvins Hobbies";"Helka Homba")      
f("Android";"rains odd")                
f("In between days";"bayed entwine")    
f("Code golf";"cod elf got")    

generates (1b = true, 0b = false)

1b
1b
1b
1b
0b
0b
0b
0b

About Q

General-purpose language (APL derivative, specialized in data processing) developed by kx.com. Free full functional evaluation version for Windows/Linux/MacOS.

Answer 23

APL, 31 chars

{≡/{x[⍋x←('.'⎕R'\u0')⍵~' ']}¨⍵}

To be used so:

    {≡/{x[⍋x←('.'⎕R'\u0')⍵~' ']}¨⍵}'Sp3000' 'P S 3000' 
1

In English:

• { ... }¨⍵: for each of the two elements of the argument
• x←('.'⎕R'\u0')⍵~' ': transform to uppercase (using a regex...) the string without the spaces and assign the temporary result to x
• x[⍋x]: sort x
• ≡/: compare the two results of the sorting: if they match, return 1.

Answer 24

Java, 218 Bytes

First time I've ever written Java...

Golfed:

import java.util.Arrays;boolean M(String a,String b){char[]A=a.toUpperCase().replace(" ","").toCharArray();char[]B=b.toUpperCase().replace(" ","").toCharArray();Arrays.sort(A);Arrays.sort(B);return Arrays.equals(A,B);}

Ungolfed:

import java.util.Arrays;
public class ManageTrashSo {
    public boolean M(String a, String b) {
    char[] A = a.toUpperCase().replace(" ", "").toCharArray();
    char[] B = b.toUpperCase().replace(" ", "").toCharArray();
    Arrays.sort(A);
    Arrays.sort(B);
    return Arrays.equals(A, B);
   }
}

Testing:

    ManageTrashSo manageTrashSo = new ManageTrashSo();

    //True
    System.out.println(manageTrashSo.M("Lynn", "Nyl N"));
    System.out.println(manageTrashSo.M("Digital Trauma", "Tau Digital Arm"));

    //False
    System.out.println(manageTrashSo.M("Android", "rains odd"));
    System.out.println(manageTrashSo.M("In between days", "bayed entwine"));

Answer 25

PHP, 89 bytes

for(;$i++<2;)$r[]=count_chars(join(explode(" ",strtolower($argv[$i]))));echo$r[0]==$r[1];

Try it online!

PHP, 94 bytes

for(;$i++<2;sort($x),$r[]=trim(join($x)))$x=str_split(strtolower($argv[$i]));echo$r[0]==$r[1];

Try it online!

Answer 26

Excel VBA, 122 Bytes

Anonymous VBE immediate window Function that takes input from range [A1:B1] and outputs to the VBE immediate window

a=Replace([A1]," ",""):b=Replace([B1]," ",""):For i=1To Len(a):b=Replace(b,Mid(a,i,1),"|",,1,1):Next:?b=String(len(a),"|")

Answer 27

Ruby, 50 bytes

def f;gets.upcase.chars.sort.join.strip;end
p f==f

Writing f=->{...} and f[]==f[] is just as long. :(

Answer 28

PowerShell, 81 bytes

param([char[]]$a,[char[]]$b)-join($a-replace' '|sort)-eq-join($b-replace' '|sort)

A slight rewrite of my answer on the linked Anagram challenge.

Takes input as char-arrays, performs a -replace operation to remove spaces, sorts them (which sorts alphabetically, not by ASCII value), then -joins them back into a string. The -eq in PowerShell is by default case-insensitive, but here it must be performed on strings, as [char]'a' is not equal to [char]'A', hence the reason for -join.

Answer 29

Perl, 35 bytes

Include +1 for -p

Somewhat abusive since it depends on the program being given on the commandline.

perl -pe'<>=~s%\S%*_=s/$&//i?_:0%reg;$_=!//'

Then give the strings as 2 consecutive lines on STDIN

A very abusive solution is 30 bytes:

perl -ne'<>=~s%\w%1/!s/$&//i%reg;1/!//'

This crashes if the strings are not anagrams and therefore gives a false exit code from the point of view of the shell. It also gives garbage on STDERR for that case. If the strings are anagrams the program is silent and gives a "true" exit code

Answer 30

D, 99 107 103 102 99 131 116 117 bytes

Edit: forgot about Dre case insensitivity imports lambdas

I _don't like this language :D

import std.string,std.algorithm;(string[]a)=>sort(split(strip(a[0].toLower),""))==sort(split(strip(a[1].toLower),""))

To use it, assign it:

import std.string,std.algorithm;

void main () {
    auto x = (string[]a) => sort(split(strip(a[0].toLower),"")) == sort(split(strip(a[1].toLower),""))
}

Answer 31

C#, 378 bytes

I need a handicap!!

https://dotnetfiddle.net/FNDt0E

using System;
using System.Linq;
using System.Text;

public class Program
{

    public static void Main()
    {
        var l = "Hello World";

        var r = "Red Who Loll";

        var y = new Func<string,string>(s => new String(s.ToLower().Replace(" ","").OrderBy(v => v).ToArray()));
        var z = new Func<string,string,Func<string,string>,bool>((w,x,f) => f(w) == f(x));
        var o = z(l, r, y);


        Console.WriteLine("{0} & {1} are anagram: {2}",l, r, o);


                Console.WriteLine("C#, {0} bytes", Encoding.Unicode.GetByteCount(@"var y = new Func<string,string>(s => new String(s.ToLower().Replace("" "","""").OrderBy(v => v).ToArray()));
    var z = new Func<string,string,Func<string,string>,bool>((w,x,f) => f(w) == f(x));"));

    }

}

Answer 32

Factor, 68 bytes

[ [ 32 swap remove >lower natural-sort ] map duplicates length 1 = ]

An anyonymous function. Call it like { "array" "of" "strings" } ~quotation~ call.