well let's approuch one case where your array is finishing with 1 but no more than "a" times in a row, because the 1 array is dependent from array of 2 and array 2 is alwais finishing with 2 it will look like this 12.... that is y[n] = x[n - 1], than 112... that  is y[n] = x[n - 1] + x[n - 2], than 111...112.... which is y[n] = x[n] + x[n - 1] + x[n - 2] + ... + x[n - a] and is the same in the second case

I understand, thanks




Edited by author 27.09.2017 22:28

Oh, got it !
Thanks a lot :)

No information about Graphs, you mean that whether it's directed or undirected?

What? Why? You can't be serious.

I'll try to scribble the concept in pascal...

type TBoard = array[0..3, 0..3] of shortint; //the board
     TMod = array[0..2, 0..2] of shortint;   //the modifier
var md: TMod;
...
procedure ModifyBoard(x, y: shortint; var board: TBoard);
//x, y are coordinates of our move's center
//board links to the board array we're modifying
var i, j: shortint;
begin
    for i:=-1 to +1 do
        for j:=-1 to +1 do
            if (x + i >= 0) and (x + i <= 3) and
               (y + j >= 0) and (y + j <= 3) then begin //out of bounds check
                if md[i + 1, j + 1] > 0 then board[x + i, y + j]:=1 - board[x + i, y + j]; //flipping if spot is marked for that
            end;
end;

I guess not more effecient than your bit operations, but definitely easier to read. You can, like, just convert 2-byte state into a full board and convert it back with separate functions if needed.

7 2
1
1
0
0
1
1
1

В отсутствии пространственного воображения.

Edited by author 14.07.2017 12:44

Yeah you are right. It is not an algorithm. It is just an idea but i think this idea  should accept!

In topological sorting,  you are doing depth first search and memorize vertices when you leave them.
When all the vertices are considered,  you have memorized vertices  in reversed topological sorted order.

Edited by author 14.07.2017 12:45

I mean, each of these strings appear exactly once. One of possible situations is what you described.