Programming Praxis

8 Responses to “Anagrams, Again”

1. Paul said

   October 18, 2019 at 10:04 AM

   In Python. These functions work for ascii and for unicode.

   from collections import Counter
    
   def is_anagram(a, b):
       return sorted(a) == sorted(b)
    
   def is_anagram(a, b):
       return Counter(a) == Counter(b)
   
2. Zack said

   October 18, 2019 at 12:10 PM

   Here is my take on this, using Julia 1.1.1, without any dependencies: https://pastebin.com/9LgUHJr8

   Have a nice weekend!
3. Daniel said

   October 18, 2019 at 7:51 PM

   Here’s a solution in C.

   #include <stdbool.h>
   #include <stdio.h>
   #include <stdlib.h>
   
   bool check_anagrams(char* s1, char* s2) {
     int diff[256] = {0};
     for (char* s = s1; *s != '\0'; ++s) {
       ++(diff[(unsigned char)(*s)]);
     }
     for (char* s = s2; *s != '\0'; ++s) {
       --(diff[(unsigned char)(*s)]);
     }
     for (int i = 0; i < 256; ++i) {
       if (diff[i]) return false;
     }
     return true;
   }
   
   int main(int argc, char* argv[]) {
     if (argc != 3) return EXIT_FAILURE;
     bool anagram = check_anagrams(argv[1], argv[2]);
     printf("%d\n", anagram);
     return EXIT_SUCCESS;
   }
   

   Example:

   $ ./a.out programming praxis
   0
   
   $ ./a.out praxis xsairp
   1
   
4. Daniel said

   October 19, 2019 at 2:05 AM

   Here’s a sorting-based solution in C.

   #include <stdbool.h>
   #include <stdio.h>
   #include <stdlib.h>
   #include <string.h>
   
   int compare(const void* a, const void* b) {
     char a_ = *((char*)a);
     char b_ = *((char*)b);
     return (a_ > b_) - (a_ < b_);
   }
   
   // destructive
   bool check_anagrams(char* s1, char* s2) {
     qsort(s1, strlen(s1), 1, compare);
     qsort(s2, strlen(s2), 1, compare);
     return strcmp(s1, s2) == 0;
   }
   
   int main(int argc, char* argv[]) {
     if (argc != 3) return EXIT_FAILURE;
     bool anagram = check_anagrams(argv[1], argv[2]);
     printf("%d\n", anagram);
     return EXIT_SUCCESS;
   }
   

   Example:

   $ ./a.out programming praxis
   0
    
   $ ./a.out praxis xsairp
   1
   
5. matthew said

   October 19, 2019 at 8:45 AM

   @Paul: you need to be a bit careful with this approach with Unicode. Here’s an attempt to deal with combining characters, equivalences & casing:

   import unicodedata
   
   def signature(s):
       # Apply compatibility decomposition
       s = unicodedata.normalize('NFKD',s)
       # Include: Letter, Number, Symbol.
       # Exclude: Mark, Punctuation, Separator, Other
       s = [c.lower() for c in s if unicodedata.category(c)[0] in "LNS"]
       s = "".join(sorted(s))
       return(s)
   
   print(sorted("Ångström") == sorted("Ångström"))  # => False
   print(signature("Ångström") == signature("Ångström")) # => True
   print(signature("Ångström")) # => agmnorst
   print(signature("Ἀχιλλεύς")) # => αειλλςυχ
   print(signature("H̸̡̪̯ͨ͊̽̅̾̎Ȩ̬̩̾͛ͪ̈́̀́͘ ̶̧̨̱̹̭̯ͧ̾ͬC̷̙̲̝͖ͭ̏ͥͮ͟Oͮ͏̮̪̝͍M̲̖͊̒ͪͩͬ̚̚͜Ȇ̴̟̟͙̞ͩ͌͝S̨̥̫͎̭ͯ̿̔̀ͅ")) # => ceehmos
   
6. matthew said

   October 19, 2019 at 8:56 AM

   I’m a bit disappointed that Chrome on my Android phone can’t cope with the combining characters in that last post at all, FF on Linux seems to manage though.
7. matthew said

   October 19, 2019 at 9:02 AM

   Just trying something: “Ångström”, “Ἀχιλλεύς”, “H̸̡̪̯ͨ͊̽̅̾̎Ȩ̬̩̾͛ͪ̈́̀́͘ ̶̧̨̱̹̭̯ͧ̾ͬC̷̙̲̝͖ͭ̏ͥͮ͟Oͮ͏̮̪̝͍M̲̖͊̒ͪͩͬ̚̚͜Ȇ̴̟̟͙̞ͩ͌͝S̨̥̫͎̭ͯ̿̔̀ͅ”
8. matthew said

   October 19, 2019 at 9:04 AM

   OK, Chrome on Android just seems to have problems when displaying with its code font.