subseq_count.m

《模式分析的核方法》一书中的源代码

function [result, K] = subseq_count(s,t,p)
%SUBSEQ_COUNT
%         Finds the non-contiguous subsequence match count between strings s and t
%         by implementing a dynamic matrix, where the length of the subsequence is p.
%
%         The following algorithm is used:
%         K[0](x,y) = 1 for all x,y
%         K[p](s,'empty string') = 0 for all p > 0, all s
%         K[p](sa,t) = K[p](s,t) + [Summation from 1 to i]K[p-1](s,t(1:i-1) [t(i) == a]
%
%         Simply prompting the function will return the value K(s,t), however
%         using the function as [result,K] = K(s,t) will also return the matrix K[p].
%
%         Example: subseq_count('abc','abbbbc', 3) returns a value of 4.
%            (Note that subseq_count('abc','abbbbc')=subseq_count('abbbbc','abc') since K(s,t,p) = K(t,s,p) ).
%         Example: subseq_count('a','a', 1) returns a value of 1.
%         Example: subseq_count('a','a', 0) returns a value of 1 (for the empty string).
%         Example: subseq_count('a','b', 1) returns a value of 0.
%         Example: subseq_count('a','b', 0) returns a value of 1.
%         Example: subseq_count('ab','ab', 2) returns a value of 1.
%         
%
%
%USAGE:   scalar = subseq_count('string1','string2', p);    (where p is the length of the subsequence)
%
%         [scalar, matrix] = subseq_count('string1,'string2', p);
%
%

%For more information, visit http://www.kernel-methods.net/
%
%Written and tested in Matlab 6.0, Release 12.
%Copyright 2003, Manju M. Pai 4/2003
[e-mail]
%------------------------------------------------------------------------------------------

%Obtain lengths of strings
n = length(s);
m = length(t);

%Allow one extra row & column for empty string
%Initially set every matrix index to -1 to show value has not yet been found
ans = repmat(-1, [n+1, m+1, p]);

%If p is equal to zero, just return 1 for the empty string and quit program
if p == 0
    result = 1;
    K = repmat(1, [n+1, m+1]);
    return;
end;

%Fill in the rest of the matrix using the function subseq_count(s,t)
for h=1:p
    for i=1:n+1
        for j=1:m+1      
            ans(i,j,h) = subseq_count_kernel(s(1:i-1), t(1:j-1), ans, h);
        end;
    end;
end;

result = ans(n+1,m+1,p);
K = ans(:,:,p);


%------------------------------------------------------------------------------------------

function ans = subseq_count_kernel(sa, t, K, p)
%This function is called by subseq_count(s,t,p).
%Type 'help subseq_count' for a description of the program.
%

%------------------------------------------------------------------------------------------

%Obtain lengths of both strings
n = length(sa);
m = length(t);


%Start algorithm:

 % 1) Base case:
    % K[p](sa, 'empty string') == K[p]('empty string', t) == zero
    if (m == 0) | (n == 0)
      ans = 0;
      return
    end;

  
  % 2) Split main algorithm into two parts:
    % a) K[p](s,t)
    
      %truncate last character of string
      s = sa(1:n-1);
      
      if( K( length(s)+1, length(t) + 1, p) == -1 )
        % Value has not yet been calculated
        ans = subseq_count_kernel(s,t,K,p);
      else
        % Value has already been calculated
        ans = K( length(s)+1, length(t) + 1, p);
      end;

    % b) Summation of K[p-1](s,t(1:i-1)), where t(i) == a

      %this is the letter (a) that was truncated off the string
      letter = sa(n);

      %We need this 'for' loop to go through all strings of t that end in a.
      %(If you're wondering why there are alot of '+ 1' and '- 1' in the algorithm,
      % its because the length of the rows and the columns are 1 bigger than the length
      % of the strings because of the addition of the empty string to the matrix.)
      pos_array = find(t == letter);  %array which consists of all indices of t where t(i) == a
      for index = 1:length(pos_array)
        i = pos_array(index);
        if (p-1) == 0               %This is a base case where 1 should always be returned if p = 0;
          ans = ans + 1;
        elseif( K( length(s) + 1 , length(t(1:i-1)) + 1, p-1) == -1)
          % Value has not yet been calculated
          ans = ans + subseq_count_kernel( s , t(1:i-1), K, p-1 );
        else
          % Value has already been calculated.
          ans = ans + K( length(s) + 1, length(t(1:i-1)) + 1, p-1 );
        end;
      end;   

      return
% End of algorithm