eng.kulanov.org.ua

I will publish some Matlab features. I would also very thankful for your comments and recommendations.

First of all I’d like to introduce draft for Markov Chain Builder from sequence of values:

For example we have the following sequence of values (number of jobs per seconds):

{2,4,5,6,7,3,3,3,2,3,4,5,3}

With this scripts you can build transition matrix:

Version 1:

X=unidrnd(4,1,100); % Let’s define the matrix with states
%X=[0 0 1 1 0 2 1 0 1 2 0 1 2 0 1 2 0 1 0 1 5 10]; %or just define sequence manually

X=X+1;% increment the state because if we have zero state this will make our algorithm wrong
states=unique(X);                       % states 0 – 1, 1 – 2, 2 – 3
Num_of_states=length(states);           %Number of states
F=zeros(Num_of_states,Num_of_states);   %Frequency matrix
P=zeros(Num_of_states,Num_of_states);   %Prob matrix
row=1;column=1;

for i=1:length(X)-1
pos_curr=find(states==X(i));      %use absolute values for states as to algorithm
pos_nxt=find(states==X(i+1));
F(pos_curr,pos_nxt)=F(pos_curr,pos_nxt)+1;
end;
row_sum=sum(F,2);                       %sum of row in order P<=1
for i=1:length(F)
for j=1:length(F)
if (row_sum(i)==0) continue;    % we need this check if sum of rows will be zero so no transition occur and “div by zero”
end;
P(i,j)=F(i,j)/row_sum(i);
end;
end;
%clear i;clear j; clear row_sum; clear pos_curr; clear pos_next;

Version 2: More elegant and was found at (http://www.eng-tips.com/viewthread.cfm?qid=236532&page=3)

BrCo:
This exact question came up on CSSM last year.  There are many ways to solve these sorts of problems.  My favourite is to use the sparse() function:

% Some data with repeated sequences
x=[1 6 1 6 4 4 4 3 1 2 2 3 4 5 4 5 2 6 2 6 2 6];
sparse(x(1:end-1),x(2:end),1)
ans =
(3,1)        1
(6,1)        1
(1,2)        1
(2,2)        1
(5,2)        1
(6,2)        2
(2,3)        1
(4,3)        1
(3,4)        1
(4,4)        2
(5,4)        1
(6,4)        1
(4,5)        2
(1,6)        2
(2,6)        3
Or if you want it in array form, where A(i,j) holds the number of changes from i to j:
A=full(sparse(x(1:end-1),x(2:end),1))
A =
0     1     0     0     0     2
0     1     1     0     0     3
1     0     0     1     0     0
0     0     1     2     2     0
0     1     0     1     0     0
1     2     0     1     0     0