Problem of the Day
A new programming or logic puzzle every Mon-Fri

Comments:

• Anonymous - 10 years, 4 months ago

  ifyouarereadingthisitisbecauseyousolvedtheproblemofthedaywhichwasencodingusingarot20cipherbesuretotellallyourfriendsandpostinthecommentshowyoucameupwiththesolution

  Hmm...i got rot 10 not 20 :)

  http://rumkin.com/tools/cipher/caesar.php

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• James - 10 years, 4 months ago

  Based on the fact that almost all the characters in the encoded string are lowercase letters, I amused the encoding was some sort of 1 to 1 mapping for each letter of the alphabet. The simplest 1 to 1 mapping is a rotation cipher, so I attempted that first. Assuming the plaintext was written in English, I printed the frequency of each letter in the encoded string, and mapped the most frequent letter ('o') to the most frequent letter in English 'e' (which is rot16). Since I can read the output, I assume I've located the correct answer.

  #include <stdio.h>
  #include <string.h>
  
  char rotate_n( char c, int n )
  {
     while( n < 0 ) n += 26;
     while( n >= 26 ) n -= 26;
  
     if ( c >= 'A' && c <= 'Z' )
     {
        c = ( ( ( c - 'A' ) + n ) % 26 ) + 'A';
     }
     else if ( c >= 'a' && c <= 'z' )
     {
        c = ( ( ( c - 'a' ) + n ) % 26 ) + 'a';
     }
     return c;
  }
  
  void rotate_string_n(char * str, int n)
  {
     size_t const length = strlen( str );
     int i;
  
     for ( i = 0; i < length; i++ )
     {
        str[i] = rotate_n( str[i], n );
     }
  }
  
  void frequency( char const * str )
  {
     size_t const length = strlen( str );
     unsigned char freq['z'-'A'+1] = { 0 };
     int i;
  
     for ( i = 0; i < length; i++ )
     {
        if ( str[i] >= 'A' && str[i] <= 'z' )
        {
           freq[str[i]-'A']++;
        }
     }
  
     for ( i = 32; i <= 'z'-'A'; i++ )
     {
        printf( "%c : %d\n", i+'A', freq[i] );
     }
  
     printf( "\n" );
  }
  
  int main( void )
  {
     char encoded[] = "spiyekboboknsxqdrscsdsclomkecoiyecyvfondrozbyl"
                      "vowypdronkigrsmrgkcoxmynsxqecsxqkbyd20mszroblo"
                      "cebodydovvkvviyebpbsoxnckxnzycdsxdromywwoxdcry"
                      "giyemkwoezgsdrdrocyvedsyx";
  
  // frequency( encoded );
     rotate_string_n( encoded, -10 ); // == rot16.
  // frequency( encoded );
  
     printf( "%s\n", encoded );
     return 0;
  }
  

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