Now, why on earth you are referring to the AI or finite automata to solve this problem is beyond me. It's a graph problem, all you want is a path from one end to the other, and if there are multiple paths, you want to find a minimal one using some measurement.

Abigail

There is also no mention of wether or not the 1's need to be adjacent..

So you seem able to do this...

1 1 0 0 0 0 0
0 0 0 0 0 1 1
0 1 1 0 0 0 0
0 0 0 0 0 1 0
0 0 1 0 1 0 0

Where the shortest path would seem to be 16263.

Update: nosbod clarified that the shortest path in the above example would rather be 26365 as you want to jump the least number of columns and not the LEFTMOST route...which would suggest the you could start anywhere on the top row..?

that would be a valid route, yes

There is no particular starting and ending point so i guess a genetic algorithm or finite-state automata would be worth looking into

The path that I gave as an example (1,3,1,1,1,1,3,4,3,1,1,1,3,1,1) is a route that uses the first 1 of each row. It is not the best route but is just an example to show a route through the matrix.

Apologies for the messy code. It is heavily commented which makes it look worse. I think this just about works but is not particularly nice. It basically loops recursively starting at the top left and working down the left hand side of the matrix. So, just using the first 4 rows as an example it would loop through the columns in this way:

1,1,1,1
1,1,1,2
1,1,1,3
1,1,1,4
......
1,1,1,21
1,1,2,1
1,1,2,2
1,1,2,3
......
1,1,3,1
1,1,3,2
.......
1,2,1,1
1,2,1,2
......
.......
21,21,21,21

It only stores the route if at every position within the observed route there are 1's

loop(0);


# the next section takes each route and counts the number of unique sn
+p's for each route

# for the num of possible routes through the matrix
for $i (0 ... $#finalroute)
{
    $number_of_unique_columns=0;
    %count=();

    
# for each step of the particular route we are looking at
    for (@{$finalroute[$i]}) 
    {
    $count{$_}++;        
    }  
# this loop counts the number of keys , therefore gives the number of 
+columns used in this route thru the matrix                      
    for (keys %count)         
    {
    $number_of_unique_columns++;                        
    }
    $route[$i]=$number_of_unique_columns;
}

#  $column_limit is user input set at the beginning and is the cut-off
+ for the num of columns used                                        
for $i ( 1 ... $column_limit )
{

#step the route array which has the number of columns used #for each r
+oute. If it equals the $i value (number of #columns) we
#are currently looking at then print out the route thru the #matrix th
+at is associated with it i.e @{$finalroute[$_]}
    for (0 ... $#route)           
    {                    
    print "@{$finalroute[$_]}\n$route[$_]\n" if  $route[$_]==$i;
    }
}






sub loop{


    my ($row,@array)=@_  ;
    
    
    for (1 ... $#number_of_columns)
    {
    push @array,$_ if $matrix[$row][$_] ==1;
    
    my $length=@array;
    my $lengthofmatrix=@matrix;

    if ($row==$lengthofmatrix)
       {
           $finalroute[$block][$y]=[@array] if                        
+$length==$lengthofmatrix;
           $y++ if $length==$lengthofmatrix;      
    }
    else
    {
        loop($row+1,$block,@array);
    }
    
    pop @array if $matrix[$row][$_]==1;
    
    }

        
}
[download]

whoops, the sub loop should be

sub loop{


    my ($row,@array)=@_  ;
    
    
    for (1 ... $#number_of_columns)
    {
    push @array,$_ if $matrix[$row][$_] ==1;
    
    my $length=@array;
    my $lengthofmatrix=@matrix;

    if ($row==$lengthofmatrix)
       {
           $finalroute[$y]=[@array] if     $length==$lengthofmatrix;
           $y++ if $length==$lengthofmatrix;      
    }
    else
    {
        loop($row+1,@array);
    }
    
    pop @array if $matrix[$row][$_]==1;
    
    }

        
}
[download]