Find all and modify triplicates in an array

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Morgan Clendennin 2018-8-24

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https://ww2.mathworks.cn/matlabcentral/answers/416093-find-all-and-modify-triplicates-in-an-array

评论： Morgan Clendennin 2018-9-11

I have a 10x10 array of ones and zeros. I need to come up with a script that will search the array for any cases where a row, column, or diagonal in the array contains a triplicate (three of a kind) of either ones or zeros and replace the third number with the opposite value. For example, if row 1 is 0011100101 then because the 3rd, 4th, and 5th columns are all equal to one and as I stated previously, there can be no three of a kinds in the array, column 5 in the first row becomes 0. This then makes column 7 need to become 1 because we end up with another triplicate, and then we're done with that row. A similar process occurs with the search in the columns and diagonals. Is my code as simple as a nested for loop that says "if your current row and column equals the next column over and the next column over then the third column gets changed" and similar for the checking the columns and diagonals? Can't seem to figure this one out.

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Morgan Clendennin 2018-8-27

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Any time a triplet appears, one of the values needs to change. In the first case where the triplets are purely horizontal, the array would go from 11100 to 11001. In the second case, I guess I didn't see the problem when two opposite triplets intersect at a row and column intersection point. In this case, we could change the column to 110 but we would have to change one of the other places in the horizontal triplet (leading to either 010 or 100) as changing it to 001 would change back the vertical triplet. In keeping with changing the third (last) value of the triplet, I believe in the second case it should become

      1
      1
  1 0 0

Hopefully this makes sense.

Stephen23 2018-8-27

编辑：Stephen23 2018-8-28

@Morgan Clendennin: have you confirmed if these rules lead to a stable solution? It seems quite possible that you could end up with a kind of Game of Life situation.

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Answer 1

Malte Herrmann 2018-8-24

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https://ww2.mathworks.cn/matlabcentral/answers/416093-find-all-and-modify-triplicates-in-an-array#answer_333977

编辑：Malte Herrmann 2018-8-24

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Here you go!

function [myArray, NChanges] = ChangeTripletsInArray(myArray)
  NChanges = 0;
  % Iterate over rows
  for idxr = 1:size(myArray, 1)
      % Iterate over columns
      for idxc = 1:size(myArray, 2)-2
          threevals = myArray(idxr, idxc:idxc+2);         % Take 3 values from array
          [threevals_new, flag, NChanges] = CheckTrips(threevals, NChanges);  
          if flag == 1 % Then changes have to be included in myArray
              myArray(idxr, idxc:idxc+2) = threevals_new;
          end % End of if flag == 1
      end % End of for idxc
    end % End of for idxr
    fprintf('%1.0f changes had to be made to the array.')
    %%%Subfunction CheckTrips
    function [self, flag, n] = CheckTrips(self, n)
        if mod(sum(self), 3) == 0 % e.g. sum([1,1,1])=3 mod(3,3)=0
                                  %      sum([0,0,0])=0 mod(0,3)=0
            self(end) = abs(self(end)-1); % e.g. abs(1-1) = 0
                                          %      abs(0-1) = 1
            flag = 1;
            n = n + 1;
        else
            flag = 0;
        end
    end % End of function CheckTrips
end % End of function ChangeTripletsInArray

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Morgan Clendennin 2018-8-27

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Hi Malte, I understand and I will keep working on it myself as well. Hopefully you will notice my comment above but just in case I can repeat it here. for cases like 11100 where the triplet is purely in one direction, we can easily have the code say to change the third element of the first triplet to a zero, creating a second triplet and then change the third element of the second triplet to one. This would change 11100 to 11001. In the case where we have a triplet in the row such as this:

      1
      1
  0 0 1

Changing the third element in the column creates a triplet in the intersecting third row. Now there are 3 ways this could be accomplished. The first, changing the third element in the triplet to a 1, would re-violate the column triplet check and throw us into a never ending loop. The other two acceptable solutions are changing either the first or second element in the triple to a one, creating 100 or 010. In keeping with the pattern of change the third element, it makes sense for this condition to change the first element creating 110 on the column and 100 on the interesecting row.

Morgan Clendennin 2018-8-28

Just me again, I realized kinda by accident that if you take the output of your ChangeTripletsInArray function and transpose it and then put it back in to the function, that takes care of the columns as if they were rows. This is again testing this all with a 10x10 ones matrix as my test case. This now creates the same problem that my code has where if we change the diagonal triplets by the third value, we would create new triplets again on the rows/columns. Going to test it if we change the first member of the diagonal triplets. I'm getting closer!!

Morgan Clendennin 2018-9-11

Hi Malte, have you been able to look more into this?

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Answer 2

Stephen23 2018-8-27

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This is a simple example of how you could approach this problem. I have not dealt with the edge cases: this is a simple proof-of-concept demonstration which you can tailor as required. Note that your rules do not seem to guarantee any fixed solution in a finite number of iterations.

I = logical(randi(0:1,10));
imh = image(I,'CDataMapping','scaled');
while true
    idx = I(1:8,:)&I(2:9,:)&I(3:10,:); % rows
    I(idx) = ~I(idx);
    set(imh,'CData',I)
    pause(0.5)
    idx = I(:,1:8)&I(:,2:9)&I(:,3:10); % columns
    I(idx) = ~I(idx);
    set(imh,'CData',I)
    pause(0.5)
    idx = I(1:8,1:8)&I(2:9,2:9)&I(3:10,3:10); % diag
    I(idx) = ~I(idx);
    set(imh,'CData',I)
    pause(0.5)
    idx = I(1:8,3:10)&I(2:9,2:9)&I(3:10,1:8); % antidiag
    I(idx) = ~I(idx);
    set(imh,'CData',I)
    pause(0.5)
end

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Stephen23 2018-8-28

编辑：Stephen23 2018-8-28

" Your provided I array has a few different solutions that it shows but it infinitely loops as a correct solution where the three conditions is never met."

So far your rules do not guarantee that there is any solution. While there might be alternative ways to approach this (e.g. .using a some kind of global optimization routine to find a minimum energy state, which may or my not have triples), so far there is nothing in your rules that prevents conflict situations oscillating indefinitely, if you use a simple iterative method.

Morgan Clendennin 2018-8-28

How would you do it, if not an iterative method? Also, what if we were to change the location of the replacement in the triplet to be randomized? i.e. 111 could be replaced with either 110, 101, or 011.

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Find all and modify triplicates in an array

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Find all and modify triplicates in an array

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