/* 

  Rosalind Counting Point mutations in Picat.

  https://rosalind.info/problems/hamm/
  """
  Problem
  [Figure 2. The Hamming distance between these two strings is 7. Mismatched symbols are colored red.]
  
  Given two strings s and t of equal length, the Hamming distance between s and t, denoted dH(s,t), 
  is the number of corresponding symbols that differ in s and t. See Figure 2.

  Given: Two DNA strings s and t of equal length (not exceeding 1 kbp).

  Return: The Hamming distance dH(s,t).

  Sample Dataset
    GAGCCTACTAACGGGAT
    CATCGTAATGACGGCCT
  Sample Output
    7
  """

  This model was created by Hakan Kjellerstrand, hakank@gmail.com
  See also my Picat page: http://www.hakank.org/picat/

*/

import util.

main => go.

go ?=>
  DNA = "GAGCCTACTAACGGGAT\nCATCGTAATGACGGCCT",
  [D1,D2] = split(strip(DNA),"\n"),
  println(ds=[D1,D2]),
  println(hamm1=hamm1(D1,D2)),

  println(hamm2=hamm2(D1,D2)),
  println(hamm3=hamm3(D1,D2)),

  nl.
go => true.


%
% The Challenge
% 
go2 ?=>
  DNA = "CGTGATTCAGCCTGCTTGAGCAAGTCTCCGCATTACGGATACTTCCAAGCCTGCAGCAGAAGCGCAACATGTTTGGTCACTTCGAATTTGTAAATAAAATTAACATTTGCCGCACTTAAGTACCTATGCATGCCCTGAAGGGCAAATCACGAAGGGAATCCGGACTCAACCTCCACAATGGGTGGTCTAACCCGCTTCCGGCGCTTGCTACAGCCTACATGAACTTTTAGGTCACTTTGGAAGCTCGGACTATGAACCTCGCGCAAATATACTAAAAGGCGCACCCCCCGCTTGTGCGTGTCGCCGAGACCTTACCGCGAGCTTGGTGTGCATCTTACTCGTAAAGAGTCACGTGAGGGCGACTCAGCTTGAGGTACAGGAGTCCACGTTATATCCAGATTTAGTTCTAAACGTTCCCCCGATTTTTGCGGCCTCCTCGCTTCAGAGTTGTCCACATGGGCCAGGTCGTCTGTCCCGACTTATGAGTGTACTTGCACCGAAGCCCTAAGGCCGCCACGAAAAACGGCGTTTGCCGATGATTGCCCGCACCAGTCCGCAGCCTCCTACTGGCTTAGATTGTGAATACGACCTGCATTTCTATCTTAAGGTACAGAATGGTTCCGGTACCCTATGATTTATTTTTTTGATGCCCGGCCACGAGACACAAGATCCGGGTACTCGCACAGCTTTTCTCCCTATCTAAGTCCCATAGTCAACGCTTAGCCTGTTTATTTCCCGGACCACTCAAGAAGCCTTCTGTATCACGTTGAGCTGCCCGCGTAGGGGCTCTAATGGCGGCGGCCTTAGGTCGACTTTCTATTGATAAGTCGATCGACATACGTCAGCTGCCTTCGGGTCACCATTTAGCCTTATGCCCCTGCACAGTGTCAGAATGATACGGGCGATATGGGTCGAACGCGTCGCTTCTTGCGCCAGGTTGTCCAAGCTTCTGGCTTTTCCCACTAT\nCGTGATACAGCGTTGACGCAATACCCTCCGCCGATTAGAAATTCCCGCGCAAGAAGCCAAAACGGTCTCACTAGAGACGTTTGGCCTTTTGACCCAAGCTGGCTATTGGACGTTCTGAGATCCCTTCGCATCGCACGATGCGCTTGTTATTCATTACGGGCCCACTATACGCCCACAAAGCTTGGCCCAAGCCACCGTCTGTAAGTGACATAGCGCATAACAACGTTAACGAAAATCTGAAATTTCCGCCTGCAGTTCTCTCACATTTATGCCACTATGGTCTCCCCCCGCTAGTCCGTGTCTGTGGGACTATACAGAAAGTTGGGTATAAAGCCTAAGGTTCAAGGTTCACTCGAGGGCAACTAACGTAATAGTCCGAACCTCTACTTTCTACCCAGATTTCGTTTTAGACAGGCGACCCAGCTTGACCGTCGCTACTTCTGCGAACTGAGCACATTGTATACCGCGCTTCAACTTCCCGCCGTGGCTACCTGCTGCGGGGGCCTGGCGACCACTCTACGGTCAGAGCTCGACGGTACATCGCCTCTAATGTCCAACGAAAACTACAGACTGCTGCTTTGAAAACGCGCCCTTTTTGGATCAGATGGCACATAAGGGTGCTGGTAACATATAATTTAAAGCTCTCATCCTCGTCAACCGGTGTCAACGTCCATGGGCGGTCACTTCGTCGCACAGGGTGTAGGAACCCTACTCAGCTGGTAGTCGGATCATTAACAGCACACCGGGAGAAGAATTCTCTGTCTCGCCAAACTTTTGGGGGCACGACCGTAGTGGCGACGGGCTTAGGCAAAATGTCCCTAGACCCCGCCATCGTGACTCTACAGTTGCTTTGTTGGCAAAATATAGTCATCTGCCTTAATGCTCAGACTGAATGCGACGAATGATTCCGGGCAACCGCGCTGTTTCGTTGTTTGCCGCTGTCAACTTACTATCGATTACGATCGT",
  [D1,D2] = split(strip(DNA),"\n"),
  % println(ds=[D1,D2]),
  println(hamm1=hamm1(D1,D2)),
  println(hamm2=hamm2(D1,D2)),
  println(hamm3=hamm3(D1,D2)),
  println(hamm4=hamm4(D1,D2)),  

  nl.
go2 => true.

%
% We assume that the two strings have the same length...
%
hamm1(D1,D2) = Hamm =>
  Hamm1 = 0,
  foreach(I in 1..D1.len)
    if D1[I] != D2[I] then
      Hamm1 := Hamm1 +1
    end
  end,
  Hamm = Hamm1.


hamm2(D1,D2) = Hamm =>
  hamm2(D1,D2,0,Hamm).
hamm2([],[],Hamm,Hamm).
hamm2([C1|D1],[C2|D2],Hamm0,Hamm) :-
  (C1 != C2 ->
     Hamm1 = Hamm0 + 1
  ;
     Hamm1 = Hamm0
  ),
  hamm2(D1,D2,Hamm1,Hamm).


hamm3(D1,D2) = [1 : {C1,C2} in zip(D1,D2), C1 != C2].len.

%
% Alternative approach using list + len.
%
hamm4(D1,D2) = Hamm.len =>
  hamm4(D1,D2,[],Hamm).
hamm4([],[],Hamm,Hamm).
hamm4([C|D1],[C|D2],Hamm0,Hamm) :-
  hamm4(D1,D2,Hamm0,Hamm).
hamm4([C1|D1],[C2|D2],Hamm0,[h|Hamm]) :-
  C1 != C2, 
  hamm4(D1,D2,Hamm0,Hamm).