I want Genetrate all possible breakage states. That is 1 become 0 but 1st,last row and 1st,last column are not going to change i.e they are allawaies 1.How can I do this? 1.

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KOUSHIK
KOUSHIK 2023 年 5 月 2 日
コメント済み: KOUSHIK 2023 年 6 月 1 日
My code is
n = 10; % Set value of n
State_Matrix = zeros(n);
s = transpose(1:n);
for i = 1:n
for j = 1:n
d = s(j) - s(i);
if d == (n/2) || d == -(n/2)
State_Matrix(i,j) = 1;
elseif d == 1 || d == -1
State_Matrix(i,j) = 2;
elseif i == n/2 && j == n/2-1 % add 010 state
State_Matrix(i,j) = 1;
else
State_Matrix(i,j) = -1; % set all other elements to -1
end
if (i == n/2 && j == n/2+1) || (i == n/2+1 && j == n/2)
State_Matrix(i,j) = -2; % set element to -2
end
end
end
% Generate all breakage states
MS = State_Matrix;
for i = 2:(n/2-1)
MS_i = Bond_Break(n,i,State_Matrix);
MS = [MS, MS_i];
end
% Store each matrix in a cell array with a meaningful name
matrices = {};
for i = 1:length(MS)/n
start_index = (i-1)*n+1;
end_index = i*n;
S_i = MS(:, start_index:end_index);
matrix_name = ['S', num2str(i)];
eval([matrix_name, ' = S_i;']);
matrices{i} = S_i;
end
% Display the number of matrices and each matrix
disp(['Number of matrices: ', num2str(length(matrices))]);
for i = 1:length(matrices)
matrix_name = ['S', num2str(i)];
disp([matrix_name, ':']);
disp(matrices{i});
end
%% Column No finding that contain zero
% column_No=[];
% for i=1:length(matrices)
% S_i=matrices{i};
% [row,column_i]=find(S_i==0);
% column_No=[column_No;column_i];
% end
%% Function to generate all breakage states for a given x value
function [S] =Break(n, x, State_Matrix)
S = [];
while x < n/2
State_Matrix(x, x+n/2) = 0;
State_Matrix(x+n/2, x) = 0;
S = [S, State_Matrix];
x = x + 1;
end
end
Now this give me 7 matrices but
-1 2 -1 -1 -1 1 -1 -1 -1 -1
2 -1 2 -1 -1 -1 0 -1 -1 -1
-1 2 -1 2 -1 -1 -1 1 -1 -1
-1 -1 2 -1 2 -1 -1 -1 0 -1
-1 -1 -1 2 -1 -2 -1 -1 -1 1
1 -1 -1 -1 -2 -1 2 -1 -1 -1
-1 1 -1 -1 -1 2 -1 2 -1 -1
-1 -1 1 -1 -1 -1 2 -1 2 -1
-1 -1 -1 0 -1 -1 -1 2 -1 2
-1 -1 -1 -1 1 -1 -1 -1 2 -1
This matrix is missing how can I do this from here
  2 件のコメント
Dyuman Joshi
Dyuman Joshi 2023 年 5 月 2 日
What should be the final output for a given n?
KOUSHIK
KOUSHIK 2023 年 5 月 2 日
If yS1:
-1 2 -1 -1 -1 1 -1 -1 -1 -1
2 -1 2 -1 -1 -1 1 -1 -1 -1
-1 2 -1 2 -1 -1 -1 1 -1 -1
-1 -1 2 -1 2 -1 -1 -1 1 -1
-1 -1 -1 2 -1 -2 -1 -1 -1 1
1 -1 -1 -1 -2 -1 2 -1 -1 -1
-1 1 -1 -1 -1 2 -1 2 -1 -1
-1 -1 1 -1 -1 -1 2 -1 2 -1
-1 -1 -1 1 -1 -1 -1 2 -1 2
-1 -1 -1 -1 1 -1 -1 -1 2 -1
S2:
-1 2 -1 -1 -1 1 -1 -1 -1 -1
2 -1 2 -1 -1 -1 0 -1 -1 -1
-1 2 -1 2 -1 -1 -1 1 -1 -1
-1 -1 2 -1 2 -1 -1 -1 1 -1
-1 -1 -1 2 -1 -2 -1 -1 -1 1
1 -1 -1 -1 -2 -1 2 -1 -1 -1
-1 0 -1 -1 -1 2 -1 2 -1 -1
-1 -1 1 -1 -1 -1 2 -1 2 -1
-1 -1 -1 1 -1 -1 -1 2 -1 2
-1 -1 -1 -1 1 -1 -1 -1 2 -1
S3:
-1 2 -1 -1 -1 1 -1 -1 -1 -1
2 -1 2 -1 -1 -1 0 -1 -1 -1
-1 2 -1 2 -1 -1 -1 0 -1 -1
-1 -1 2 -1 2 -1 -1 -1 1 -1
-1 -1 -1 2 -1 -2 -1 -1 -1 1
1 -1 -1 -1 -2 -1 2 -1 -1 -1
-1 0 -1 -1 -1 2 -1 2 -1 -1
-1 -1 0 -1 -1 -1 2 -1 2 -1
-1 -1 -1 1 -1 -1 -1 2 -1 2
-1 -1 -1 -1 1 -1 -1 -1 2 -1
S4:
-1 2 -1 -1 -1 1 -1 -1 -1 -1
2 -1 2 -1 -1 -1 0 -1 -1 -1
-1 2 -1 2 -1 -1 -1 0 -1 -1
-1 -1 2 -1 2 -1 -1 -1 0 -1
-1 -1 -1 2 -1 -2 -1 -1 -1 1
1 -1 -1 -1 -2 -1 2 -1 -1 -1
-1 0 -1 -1 -1 2 -1 2 -1 -1
-1 -1 0 -1 -1 -1 2 -1 2 -1
-1 -1 -1 0 -1 -1 -1 2 -1 2
-1 -1 -1 -1 1 -1 -1 -1 2 -1
S5:
-1 2 -1 -1 -1 1 -1 -1 -1 -1
2 -1 2 -1 -1 -1 1 -1 -1 -1
-1 2 -1 2 -1 -1 -1 0 -1 -1
-1 -1 2 -1 2 -1 -1 -1 1 -1
-1 -1 -1 2 -1 -2 -1 -1 -1 1
1 -1 -1 -1 -2 -1 2 -1 -1 -1
-1 1 -1 -1 -1 2 -1 2 -1 -1
-1 -1 0 -1 -1 -1 2 -1 2 -1
-1 -1 -1 1 -1 -1 -1 2 -1 2
-1 -1 -1 -1 1 -1 -1 -1 2 -1
S6:
-1 2 -1 -1 -1 1 -1 -1 -1 -1
2 -1 2 -1 -1 -1 1 -1 -1 -1
-1 2 -1 2 -1 -1 -1 0 -1 -1
-1 -1 2 -1 2 -1 -1 -1 0 -1
-1 -1 -1 2 -1 -2 -1 -1 -1 1
1 -1 -1 -1 -2 -1 2 -1 -1 -1
-1 1 -1 -1 -1 2 -1 2 -1 -1
-1 -1 0 -1 -1 -1 2 -1 2 -1
-1 -1 -1 0 -1 -1 -1 2 -1 2
-1 -1 -1 -1 1 -1 -1 -1 2 -1
S7:
-1 2 -1 -1 -1 1 -1 -1 -1 -1
2 -1 2 -1 -1 -1 1 -1 -1 -1
-1 2 -1 2 -1 -1 -1 1 -1 -1
-1 -1 2 -1 2 -1 -1 -1 0 -1
-1 -1 -1 2 -1 -2 -1 -1 -1 1
1 -1 -1 -1 -2 -1 2 -1 -1 -1
-1 1 -1 -1 -1 2 -1 2 -1 -1
-1 -1 1 -1 -1 -1 2 -1 2 -1
-1 -1 -1 0 -1 -1 -1 2 -1 2
-1 -1 -1 -1 1 -1 -1 -1 2 -1
ou found the output of the above code it give me 7 matrices
and I need
S8
-1 2 -1 -1 -1 1 -1 -1 -1 -1
2 -1 2 -1 -1 -1 0 -1 -1 -1
-1 2 -1 2 -1 -1 -1 1 -1 -1
-1 -1 2 -1 2 -1 -1 -1 0 -1
-1 -1 -1 2 -1 -2 -1 -1 -1 1
1 -1 -1 -1 -2 -1 2 -1 -1 -1
-1 0 -1 -1 -1 2 -1 2 -1 -1
-1 -1 1 -1 -1 -1 2 -1 2 -1
-1 -1 -1 0 -1 -1 -1 2 -1 2
-1 -1 -1 -1 1 -1 -1 -1 2 -1

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採用された回答

Saffan
Saffan 2023 年 6 月 1 日
Ran in:
Hi Koushik,
The current implementation of Bond_Break function does not incorporate all possible combinations, and therefore, it may not generate all potential breakage states. In the Bond_Break function, first get the list of all the points and then get all possible combinations of these points.
You can modify your code and Bond_Break function in the following way:
n = 10; % Set value of n
State_Matrix = zeros(n);
s = transpose(1:n);
for i = 1:n
for j = 1:n
d = s(j) - s(i);
if d == (n/2) || d == -(n/2)
State_Matrix(i,j) = 1;
elseif d == 1 || d == -1
State_Matrix(i,j) = 2;
elseif i == n/2 && j == n/2-1 % add 010 state
State_Matrix(i,j) = 1;
else
State_Matrix(i,j) = -1; % set all other elements to -1
end
if (i == n/2 && j == n/2+1) || (i == n/2+1 && j == n/2)
State_Matrix(i,j) = -2; % set element to -2
end
end
end
% Generate all breakage states
MS = State_Matrix;
MS_temp = Bond_Break(n,2,State_Matrix);
MS = [MS, MS_temp];
% Store each matrix in a cell array with a meaningful name
matrices = {};
for i = 1:length(MS)/n
start_index = (i-1)*n+1;
end_index = i*n;
S_i = MS(:, start_index:end_index);
matrix_name = ['S', num2str(i)];
eval([matrix_name, ' = S_i;']);
matrices{i} = S_i;
end
% Display the number of matrices and each matrix
disp(['Number of matrices: ', num2str(length(matrices))]);
Number of matrices: 8
for i = 1:length(matrices)
matrix_name = ['S', num2str(i)];
disp([matrix_name, ':']);
disp(matrices{i});
end
S1:
-1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 -2 -1 -1 -1 1 1 -1 -1 -1 -2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1
S2:
-1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 -2 -1 -1 -1 1 1 -1 -1 -1 -2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1
S3:
-1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 -2 -1 -1 -1 1 1 -1 -1 -1 -2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1
S4:
-1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 -2 -1 -1 -1 1 1 -1 -1 -1 -2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1
S5:
-1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 -2 -1 -1 -1 1 1 -1 -1 -1 -2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1
S6:
-1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 -2 -1 -1 -1 1 1 -1 -1 -1 -2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1
S7:
-1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 -2 -1 -1 -1 1 1 -1 -1 -1 -2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1
S8:
-1 2 -1 -1 -1 1 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 2 -1 -1 -1 0 -1 -1 -1 -1 2 -1 -2 -1 -1 -1 1 1 -1 -1 -1 -2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 0 -1 -1 -1 2 -1 2 -1 -1 -1 -1 1 -1 -1 -1 2 -1
%% Column No finding that contain zero
% column_No=[];
% for i=1:length(matrices)
% S_i=matrices{i};
% [row,column_i]=find(S_i==0);
% column_No=[column_No;column_i];
% end
%% Function to generate all breakage states for a given x value
function [S] = Bond_Break(n, x, State_Matrix)
S = [];
Points = {};
% Storing all points where breakage will occur
while x < n/2
p = [x,x+n/2];
Points{end+1} = p;
x = x + 1;
end
% Generating all possible combinations of these points:
combos = cell(length(Points), 1);
for ii = 1:length(Points)
combos{ii} = {[Points{ii}]};
end
for i = 2:length(Points)
C = combnk(1:length(Points), i);
for ii = 1:size(C, 1)
combo = [];
for jj = 1:i
combo{end+1} = Points{C(ii,jj)};
end
combos{end+1} = {combo};
end
end
%Adding matrices to S:
for i=1:length(combos)
SM = State_Matrix;
if(i<=length(Points))
p1=combos{i}{1}(1);
p2=combos{i}{1}(2);
SM(p1,p2)=0;
SM(p2,p1)=0;
else
for j=1:length(combos{i}{1})
p1=combos{i}{1}{j}(1);
p2=combos{i}{1}{j}(2);
SM(p1,p2)=0;
SM(p2,p1)=0;
end
end
S = [S,SM];
end
end

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