Crackerbarrel Peg Game Help!

Hey guys, first time posting here so I hope I'm doing it right.

For my assignment, I have to create a MATLAB program in which I can get 1000 solutions from a user selected hole on this triangle solitaire peg board:

A PEG board contains 15 holes, shaped in form of a triangle. Initially 14 of the holes are filled with pegs, while there is one user-prompted empty hole somewhere on the board. During the game it is possible to take a peg, jump over a peg into an empty hole. The peg that was jumped over is then removed from the board. It is important that the jumping peg jumps 2 fields, the destination is an empty hole, and there is a peg in between. The goal of the peg solitaire game is to jump pegs in such a way that only one peg remains on the board eventually.

I need helping looping through if statements to check if a peg is in a random starting hole position, then looping through the pre-stored available moves for each hole, and checking if the conditions allow a move to be made. Then, I check if a move was made and then check if the required number of solutions has been found.

Here is my code so far:

%move {1-15} are row vectors of all the possible moves for each respectable hole
move(1,:) = [2 3 6 10 0 0 0 0]; %hole 1
move(2,:) = [3 4 7 11 0 0 0 0]; %hole 2
move(3,:) = [7 10 4 5 2 1 8 12]; %hole 3
move(4,:) = [3 2 8 11 0 0 0 0]; %hole 4
move(5,:) = [4 3 9 12 0 0 0 0]; %hole 5
move(6,:) = [7 8 10 13 0 0 0 0]; %hole 6
move(7,:) = [8 9 11 14 0 0 0 0]; %hole 7
move(8,:) = [7 6 11 13 0 0 0 0]; %hole 8
move(9,:) = [8 7 12 14 0 0 0 0]; %hole 9
move(10,:) = [6 1 7 3 11 12 13 15]; %hole 10
move(11,:) = [7 2 8 4 0 0 0 0]; %hole 11
move(12,:) = [8 3 9 5 11 10 14 15]; %hole 12
move(13,:) = [11 8 10 6 0 0 0 0]; %hole 13
move(14,:) = [12 9 11 7 0 0 0 0]; %hole 14
move(15,:) = [13 10 14 12 0 0 0 0]; %hole 15
X=input('Which peg hole do you want removed?');
sol = 0;
while sol < 1000;
    p_board = ones(15,1);
    p_board(X) = 0;
    i = 1;
    while i < 100;
        A = ceil(15*rand());%starts from random hole 1-15
        while i == 1;
            if A == X;
                continue
            end
        end
        sum = 0;%number of potential moves
        for j = 1:4;
            if move(A,2*j) > 0;
                sum = sum+1;
            end
            nnn = 0;%number of possible moves
            for k = 1:sum;
                if p_board(A)==1 && p_board(move(A,2*k-1))==1 && p_board(move(A,2*k))==0
                    nnn = nnn+1;

[Geoff Hayes]

Vinay - I'm not to clear on your move vector. In the first row, you have

move(1,:) = [2 3 6 10]; %hole 1

What exactly does this mean? Looking at the attached image, if the user moves the peg from position 1 then it can only move to 4 (jumping 2) or move to 6 (jumping 3). How can a peg move from position 1 to 10 in one jump? Same is true for the remaining pegs too. Most should have only two possible moves as they can only jump two different neighbours (4 and 6 may be the only exceptions with three possible jumps).

[Vinay]

Im very sorry. I posted the wrong picture up. Here is the correct board setup: http://i.imgur.com/QzG9G2v.gif

[Geoff Hayes]

Vinay - you realize that the code you have posted above doesn't run, right? Even if you ignore everything after the while loop (which is incomplete) the assignment of moves to your move array fails because most of the rows have four elements while other rows have eight elements. In a matrix, the number of columns in each row has to be identical. You may want to consider using a cell array which will allow each element to be a variable size. And instead of using a single row vector, you can have a matrix with each row being two columns long with the first column being the adjacent peg, and the subsequent being the destination. This will get you started:

moves = {};
moves{1} = [2 3; 6 10];

You can do something similar for the remaining 14 pegs.

[Vinay]

Yea, I haven't completely finished the rest of my code yet because I am stuck on these move arrays for all the holes. Would doing something like this also work?

move(1,:) = [2 3; 6 10; 0 0; 0 0];

That way, it is the same size as peg 3, for example, which would be

move(3,:) = [7 10; 4 5; 2 1; 8 12];

This would allow me to call certain indices within that matrix.

[Geoff Hayes]

No Vinay, that won't work because you are assigning a 4x2 matrix to one row only of matrix move. If you were to try the above, you should get the Subscripted assignment dimension mismatch error.

[Vinay]

What I ended up doing was setting it has a row vector, but adding 0's to make all the row vectors 8 elements. In order to check how many potential moves I have, I nested two for loops together to check if the number in the landing hole is greater than 1.

%move {1-15} are row vectors of all the possible moves for each respectable hole
move(1,:) = [2 3 6 10 0 0 0 0]; %hole 1
move(2,:) = [3 4 7 11 0 0 0 0]; %hole 2
move(3,:) = [7 10 4 5 2 1 8 12]; %hole 3
move(4,:) = [3 2 8 11 0 0 0 0]; %hole 4
move(5,:) = [4 3 9 12 0 0 0 0]; %hole 5
move(6,:) = [7 8 10 13 0 0 0 0]; %hole 6
move(7,:) = [8 9 11 14 0 0 0 0]; %hole 7
move(8,:) = [7 6 11 13 0 0 0 0]; %hole 8
move(9,:) = [8 7 12 14 0 0 0 0]; %hole 9
move(10,:) = [6 1 7 3 11 12 13 15]; %hole 10
move(11,:) = [7 2 8 4 0 0 0 0]; %hole 11
move(12,:) = [8 3 9 5 11 10 14 15]; %hole 12
move(13,:) = [11 8 10 6 0 0 0 0]; %hole 13
move(14,:) = [12 9 11 7 0 0 0 0]; %hole 14
move(15,:) = [13 10 14 12 0 0 0 0]; %hole 15
X=input('Which peg hole do you want removed?');
sol = 0;
while sol < 1000;
    p_board = ones(15,1);
    p_board(X) = 0;
    i = 1;
    while i < 100;
        A = ceil(15*rand());%starts from random hole 1-15
        while i == 1;
            if A == X;
                continue
            end
        end
        sum = 0;%number of potential moves
        for j = 1:4;
            if move(A,2*j) > 0;
                sum = sum+1;
            end
            nnn = 0;%number of possible moves
            for k = 1:sum;
                if p_board(A)==1 && p_board(move(A,2*k-1))==1 && p_board(move(A,2*k))==0
                    nnn = nnn+1;

I'm not sure how to now check if a move CAN be made, as the hole a peg wants to land into has to be equal to 0 on p_board

[Geoff Hayes]

You have your p_board of all ones with the exception of the X (could make this lower case and rename to something that informs the reader what the variable is for) that was removed (note that you should have some sort of check on X in case the user enters something that is not numeric or is not an integer within the interval 1:15).

Of the 14 remaining holes that have a peg in them, you have to choose one peg such that it can jump another and land in that zero hole. Your move array has that information but is not that easy to get to. You may want to consider reconstructing this array so that it is a nx3 matrix where the first column is the starting peg hole, the second column is the peg hole that is being jumped, and the third column is the destination peg hole:

pegHoleMoves = [
     1  2  3;
     1  6 10;
     2  3  4;
     2  7 11;
     % etc.
              ];

That way, given an initial empty peg hole of X, you could find all the possible jumps from a starting peg to land in that empty peg hole:

idcs = find(pegHoleMoves(:,3)==X);

idcs is a list of row indices into pegHoleMoves that allow for a jump to end up in peg hole X. If the list has one index, then there is only one possible move from pegHoleMoves(idcs,1). If the list has more than one element, they you can randomly pick a starting peg from idcs. Once you have that starting peg it gets a little easier. You make the jump, and clear out the jumped peg hole from your p_board and you could do the following: suppose that you jumped peg hole 5. It can never be jumped again, so remove it from your list of possible jumps:

pegHoleMoves(pegHoleMoves(:,2)==5) = [];

In the above, the second column represents the middle peg holes of the jump. Since peg hole 5 has been cleared, then we want to find all such moves where 5 is in the middle and remove them from the list (and you can do this by setting the row to []). Try it out and verify that it does what it should. You should also do this for the initial peg (X) that has been removed, since that peg hole can never be jumped either. This way, your list of possible moves decreases over time...

[Vinay]

Thanks for the clarity and guidance. I tried:

idcs = find(pegHoleMoves(:,3)==X);

And it is displaying values that don't correspond with those in the array, as it is spitting out numbers like 21, which are greater than 15.

[Geoff Hayes]

The 21 is expected, as it must be that the 21rst row of pegHoleMoves as the value X in its third column. find returns the indices of the rows where the third column has a value that matches X. For example if:

A = [1;1;2;2;3;3;4;4;3;3;2;2];
find(A==1)

ans = 
   [1 2]

since 1 is at A(1) and A(2). Type help find for details.

[Vinay]

Everything checks out ah-okay! Thank you so much for your help and guidance!

[Geoff Hayes]

Anytime. Glad to have been able to help!