Does anyone know how I can compare the elements in an array with the adjacent elements?
For example, if I have an array:
0 0 0 1 1 1 1 0
0 1 1 1 1 1 1 0
0 1 0 1 1 1 1 0
0 1 1 1 1 1 0 0
0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
Is there a way to cycle through each element and perform a logical test of whether the elements around it are equal to 1?
Oops, it looks like someone is doing a homework assignment. Game of life maybe?
There are many ways to do such a test. But learn to do it in a vectorized form. This involves understanding how matlab does indexing, and how the elements of a 2-d array are stored in memory. That will take some time to explain in detail, more than I want to do at this exact moment. I would definitely recommend you learn it though.
Until then, I'll just suggest that if you really are doing the game of life, then the best trick is to use conv2. Thus,
A =[0 0 0 1 1 1 1 0
0 1 1 1 1 1 1 0
0 1 0 1 1 1 1 0
0 1 1 1 1 1 0 0
0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1];
B = conv2(A,[1 1 1;1 0 1;1 1 1],'same')
B =
1 2 4 4 5 5 3 2
2 2 5 6 8 8 5 3
3 4 8 7 8 7 4 2
2 2 4 5 7 7 6 3
3 5 6 7 7 7 6 3
1 2 2 3 4 5 5 3
Loren has recently posted about this very issue: http://blogs.mathworks.com/loren/2010/01/19/mathematical-recreations-tweetable-game-of-life/ - lots of interesting things can be learned by studying the code in that post and its comments
Related
I have the following dataset about the choices of different car brands and their attributes. I would like to create a matrix based on each attribute of the cars.
RespNum Task Concept Make Exterior.Design Interior.design
1 100086500 1 1 3 2 3
2 100086500 1 2 1 3 2
3 100086500 1 3 4 1 1
4 100086500 1 4 0 0 0
5 100086500 2 1 1 3 2
6 100086500 2 2 5 1 3
Driving.performance Driving.attributes Comfort Practibility Safety
1 1 1 1 3 3
2 3 3 3 2 1
3 2 2 2 1 2
4 0 0 0 0 0
5 3 2 1 1 3
6 1 3 3 3 2
Quality Equipment Sustainability Economy Price Response
1 2 1 1 3 1 0
2 1 3 3 1 3 0
3 3 2 2 2 2 1
4 0 0 0 0 0 0
5 3 2 1 1 4 0
6 1 3 3 3 8 0
I am using the function:
Make = attribcoding(6,4,'Other')
The first input (6) is the number of levels, the second (4) is the column position in the dataset, and the last ('Other') is the name of the outside option. However, I get the following error message:
Error in dimnames(x) <- dn :
length of 'dimnames' [2] not equal to array extent
I have the following matrix:
1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2
0 0 1 2 0 0 1 2 0 0 1 2 0 0 1 2 0 0 1 2 0 0 1 2
0 0 0 0 1 1 1 1 2 2 2 2 0 0 0 0 1 1 1 1 2 2 2 2
I'd like to randomly permute the columns, with the constraint that every four numbers in the second row should contain some form of
0 0 1 2
e.g. Columns 1:4, 5:8, 9:12, 13:16, 17:20, 21:24 in the example below each contain the numbers 0 0 1 2.
0 1 0 2 2 0 1 0 0 0 2 1 1 2 0 0 2 0 1 0 1 0 0 2
Every column in the permuted matrix should have a corresponding one in the first matrix. In other words, nothing should be altered within a column.
I can't seem to think of an intuitive solution to this - Is there another way of coming up with some form of the initial matrix that both satisfies the constraint and retains the integrity of the columns? Each column represents conditions in an experiment, which is why I'd like them to be balanced.
You can compute the permutations directly in the following manner: First, permute all columns with 0 in the second row among themselves, then all 1s among themselves, and finally all 2s among themselves. This ensures that, for example, any two 0 columns are equally likely to be the first two columns in the resulting permutation of A.
The second step is to permute all columns in blocks of 4: permute columns 1-4 randomly, permute columns 5-8 randomly, etc. Once you do this, you have a matrix that maintains the (0 0 1 2) pattern for every block of 4 columns, but each set of (0 0 1 2) is equally likely to be in any given block of 4, and the (0 0 1 2) are equally likely to be in any order.
A = [1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2
0 0 1 2 0 0 1 2 0 0 1 2 0 0 1 2 0 0 1 2 0 0 1 2
0 0 0 0 1 1 1 1 2 2 2 2 0 0 0 0 1 1 1 1 2 2 2 2];
%% Find the indices of the zeros and generate a random permutation with that size
zeroes = find(A(2,:)==0);
perm0 = zeroes(randperm(length(zeroes)));
%% Find the indices of the ones and generate a random permutation with that size
wons = find(A(2,:) == 1);
perm1 = wons(randperm(length(wons)));
%% NOTE: the spelling of `zeroes` and `wons` is to prevent overwriting
%% the MATLAB builtin functions `zeros` and `ones`
%% Find the indices of the twos and generate a random permutation with that size
twos = find(A(2,:) == 2);
perm2 = twos(randperm(length(twos)));
%% permute the zeros among themselves, the ones among themselves and the twos among themselves
A(:,zeroes) = A(:,perm0);
A(:,wons) = A(:,perm1);
A(:,twos) = A(:,perm2);
%% finally, permute each block of 4 columns, so that the (0 0 1 2) pattern is preserved, but each column still has an
%% equi-probable chance of being in any position
for i = 1:size(A,2)/4
perm = randperm(4) + 4*i-4;
A(:, 4*i-3:4*i) = A(:,perm);
end
Example result:
A =
Columns 1 through 15
1 1 2 2 2 2 1 1 2 2 1 2 2 1 2
0 0 2 1 0 2 0 1 0 2 1 0 1 2 0
0 1 2 2 2 0 1 1 1 1 2 0 0 2 0
Columns 16 through 24
2 1 1 1 1 1 2 2 1
0 2 0 0 1 0 0 1 2
1 1 2 2 0 0 2 1 0
I was able to generate 100000 constrained permutations of A in about 9.32 seconds running MATLAB 2016a, to give you an idea of how long this code takes. There are certainly ways to optimize the permutation selection so you don't have to make quite so many random draws, but I always prefer the simple, straightforward approach until it proves insufficient.
You could use a rejection method: keep trying random permutations, chosen equiprobably, until one satisfies the requirement. This guarantees that all valid permutations have the same probability of being picked.
A = [ 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2
0 0 1 2 0 0 1 2 0 0 1 2 0 0 1 2 0 0 1 2 0 0 1 2
0 0 0 0 1 1 1 1 2 2 2 2 0 0 0 0 1 1 1 1 2 2 2 2 ]; % data matrix
required = [0 0 1 2]; % restriction
row = 2; % row to which the resitriction applies
sorted_req = sort(required(:)); % sort required values
done = false; % initiallize
while ~done
result = A(:, randperm(size(A,2))); % random permutation of columns of A
test = sort(reshape(result(row,:), numel(required), []), 1); % reshape row
% into blocks, each block in a column; and sort each block
done = all(all(bsxfun(#eq, test, sorted_req))); % test if valid
end
Here's an example result:
result =
2 1 1 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 1 1 2 1 2
2 0 0 1 2 1 0 0 0 1 0 2 2 0 1 0 1 2 0 0 2 0 1 0
2 1 2 2 1 2 2 0 1 1 1 2 1 1 0 0 0 0 0 0 0 2 1 2
I have 3 2d Arrays(matrix) with 0 and 1-
For each array, I will rotate 4 times clock-wise , 4 times anti clock-wise and flip the array and repeat the above and for each iteration I will repeat the steps for other array and so on to combine the array to build a symmetry or kind of Rubik's cube but with 5 elements each side. It means if I like to add 2 arrays , it means 1 of Array 1 must be fit with 0 of Array 2.
Following kind of structure-
Following is my 3 arrays
0 0 1 0 1
1 1 1 1 1
0 1 1 1 0
1 1 1 1 1
0 1 0 1 1
-------------
0 1 0 1 0
0 1 1 1 0
1 1 1 1 1
0 1 1 1 0
0 0 1 0 0
-------------
1 0 1 0 0
1 1 1 1 1
0 1 1 1 0
1 1 1 1 1
0 1 0 1 0
-------------
This problem is evolved from the problem I asked How to solve 5 * 5 Cube in efficient easy way.
Consider my rotate methods are as follows -
rotateLeft()
rotateRight()
flipSide()
for (firstArray){
element = single.rotateLeft();
for(secondArray){
element2 = single.rotateLeft();
if(element.combine(element2){
for(thirdArray){
}
}
}
}
Currently I have fixed 3 arrays , but how exactly and efficiently I must solve this problem.
I am working on a minigame called 'Pogo Painter', and I need some mathematical solutions. Below is an image (made with Paint) to illustrate a bit what it's all about.
Four players, each of different color, must claim squares to gain points. The minigame will be similar to this: http://www.youtube.com/watch?v=rKCQfAlaRrc, but slightly different. The players will be allowed to run around the playground and claim any of the squares, and points are gathered when a pattern is closed. For example, claiming blue square on A3 will create a closed blue pattern.
What kind of variables should I declare and how do I check if the pattern is closed?
Please answer if you have a solution :)
Here’s another (Discrete Optimization) way to model your problem.
Notation
View your grid as a ‘graph’ with n^2 nodes, and edges of length 1 (Edges connect two neighboring nodes.) Let the nodes be numbered 1:n^2. (For ease of notation, you can use a double array (x,y) to denote each node if you prefer.)
Decision Variables
There are k colors, one for each player (1 through 4). 0 is an unclaimed cell (white)
X_ik = 1 if player k has claimed node i. 0 otherwise.
To start out
X_i0 = 1 for all nodes i.
All nodes start out as white (0).
Neighboring sets: Two nodes i and j are ‘neighbors’ if they are adjacent to each other. (Any given node i can have at most 4 neighbors: Up down right and left.)
Edge variables:
We can now define a new set of edge variables Y_ijk that connect two adjacent nodes (i and j) with a common color k.
Y_ijk = 1 if neighboring nodes i and j are both of color k. 0 Otherwise.
(That is, X_ik = X_jk) for non-zero k.
We now have an undirected graph. Checking for ‘closed patterns’ is the same as detecting cycles.
Detecting Cycles:
A simple DFS search will do, since we have undirected cycles. Start with each colored node i, and check for cycles. If a path leads you back to a visited node, cycles exist. You can award points accordingly.
Finally, one suggestion as you design the game. You can reward points according to the “longest cycle” you detect. The shortest cycle gets 4 points, one point for each edge (or one point for each node in the cycle) whichever works best for you.
1 1
1 1 scores 4 points
1 1 1
1 1 1 scores 6 points
1 1 1
1 1 1
1 1 scores 8 points
Hope that helps.
Okay,
This is plenty of text, but it's simple.
An N-by-N square will satisfy as the game-board.
Each time a player claims a square,
If the square is not attached to any square of that player, then you must give that square a unique ID.
If the square is attached,
Count how many neighbours of each ID it has.
( See the demos I put below, to see what this means)
For each group
patterns_count += group_size - 1
If the number of groups is more than 1
Change the ID of that group as well as every other square connected to it so they all share the same ID
You must remember which IDs belong to which players.
This is what you have in your example
1 1 1 0 0 0 0 2 2
1 0 0 0 1 3 3 0 0
1 1 0 0 3 3 0 0 0
0 1 0 0 4 5 0 0 0
0 0 0 6 4 0 0 0 0
7 7 0 0 0 0 8 8 8
0 7 7 0 9 8 8 0 8
A A 7 0 9 8 0 0 8
A 0 7 0 0 0 8 8 8
And this is what it would turn out like after blue grabs A-3
1 1 1 0 0 0 0 2 2
1 0 0 0 1 3 3 0 0
1 1 0 0 3 3 0 0 0
0 1 0 0 4 5 0 0 0
0 0 0 6 4 0 0 0 0
7 7 0 0 0 0 8 8 8
0 7 7 0 9 8 8 0 8
A A 7 0 9 8 0 0 8
A 0 7 0 0 8 8 8 8
More examples of the algorithm in use
1 1 1 0
1 0 1 0
1 1 0
0 0 0 0
2 neighbours. 2x'1'
1x closed pattern.
1 1 1 0
1 0 1 0
1 1 1 0
0 0 0 0
--
1 1 1 0 0
1 0 1 0 0
1 1 0 0
1 0 1 0 0
1 1 1 0 0
3 neighbours: 3x'1'
2x closed patterns
1 1 1 0 0
1 0 1 0 0
1 1 1 0 0
1 0 1 0 0
1 1 1 0 0
--
1 1 1 0 0
1 0 1 0 0
1 1 2 2
0 0 2 0 2
0 0 2 2 2
4 neighbours: 2x'1', 2x'2'
2 Closed patterns
1 1 1 0 0
1 0 1 0 0
1 1 1 1 1
0 0 1 0 1
0 0 1 1 1
But I also consider these a closed pattern. You haven't given any description as to what should be considered one and what shouldn't be.
1 1 0
1 1 0
0 0 0
1 1 1
1 1 1
0 0 0
1 1 1
1 1 1
1 1
I'm trying to build game like http://games.yahoo.com/game/bricks-breaking in actionscript 3 (flash builder).
I am able to create an array of bricks (that are visible on game start), but I have no idea how to find a group of bricks in array.
Lets say we have array like so:
1 2 2 1 3 3 1 1 1 1 1 1 1
1 2 1 1 1 3 1 1 1 1 1 1 1
1 2 1 1 1 3 1 1 1 1 1 1 3
1 1 2 1 1 3 3 3 1 1 1 1 3
1 1 1 2 1 3 1 3 3 1 1 1 3
1 1 1 3 3 3 1 3 3 1 1 1 3
1 1 1 1 1 1 1 3 3 1 1 1 1
When the user clicks any brick colored red (in array lets say it is 3) the array after removing all 3 will look like that:
1 2 2 0 0 0 0 0 0 1 1 1 1
1 2 1 1 0 0 1 0 0 1 1 1 1
1 2 1 1 1 0 1 0 0 1 1 1 3
1 1 2 1 1 0 1 0 1 1 1 1 3
1 1 1 1 1 0 1 1 1 1 1 1 3
1 1 1 2 1 0 1 1 1 1 1 1 3
1 1 1 1 1 1 1 1 1 1 1 1 1
Basicly I want to remove all the items that are in group and are the same color.
Any suggestions how to do that?
Is there any kind of algorythm that I should use?
Thanks for advice
A simple way to remove elements is to use a recursive function. It's not the only way (or even a good one) but it should be enough for this kind of game. Basically something like this:
function breakBricks(x:int, y:int, color:int):void {
if(bricks[y][x] != color) return;
bricks[y][x] = 0;
breakBricks(x + 1, y, color);
breakBricks(x, y + 1, color);
breakBricks(x - 1, y, color);
breakBricks(x, y - 1, color);
}
Begin with the position that the user clicked and the colour of that position. If the colour matches it will set that entry to 0, if not it leaves the element alone. It recursively does this to all neighbouring elements. What is missing in this code are boundary checks which you need to add.
In the next step you could iterate over each of the arrays columns from bottom to top, keep reference of the position of the first 0 element you find and move any non-emtpy values you find after that to the lowest empty row position.