#MATLAB, (削除) 146 (削除ここまで)(削除) 143 (削除ここまで) 138#

(Also works on Octave online, but you need to sign in to save the function in a file).

function o=c(n,L);o=zeros(n);o(:,1)=1;for i=2:n;for j=2:i;a=o(i-1,j-1);b=o(i-1,j);c=a|b;d=a&b;c(d)=L(d);L=circshift(L,-d);o(i,j)=c;end;end

The function takes an input n and L, and returns an array o which contains the output.

For the input values, n is a scalar, and L is a column vector, which can be specified in the format [;;;]. Not quite what you show, but you say it is flexible within reason and this seems so.

The output is formatted as an n x n array containing 0's and 1's.

And an explanation:

function o=c(n,L)
%Create the initial array - an n x n square with the first column made of 1's
o=zeros(n);o(:,1)=1;
%For each row (starting with the second, as the first is done already)
for i=2:n;
 %For each column in that row, again starting with the second as the first is done
 for j=2:i;
 %Extract the current and previous elements in the row above
 a=o(i-1,j-1); %(previous)
 b=o(i-1,j); %(current)
 %Assume that min()==0, so set c to max();
 c=a|b;
 %Now check if min()==1
 d=a&b;
 %If so, set c to L(1)
 c(d)=L(d);
 %Rotate L around only if min()==1
 L=circshift(L,-d);
 %And store c back to the output matrix
 o(i,j)=c;
 end;
end

Update: I have managed to optimise away the if-else statement to save a few bytes. The input format has once again changed back to column vector.

MATLAB, (削除) 146 (削除ここまで)(削除) 143 (削除ここまで) 138

(Also works on Octave online, but you need to sign in to save the function in a file).

function o=c(n,L);o=zeros(n);o(:,1)=1;for i=2:n;for j=2:i;a=o(i-1,j-1);b=o(i-1,j);c=a|b;d=a&b;c(d)=L(d);L=circshift(L,-d);o(i,j)=c;end;end

The function takes an input n and L, and returns an array o which contains the output.

For the input values, n is a scalar, and L is a column vector, which can be specified in the format [;;;]. Not quite what you show, but you say it is flexible within reason and this seems so.

The output is formatted as an n x n array containing 0's and 1's.

And an explanation:

function o=c(n,L)
%Create the initial array - an n x n square with the first column made of 1's
o=zeros(n);o(:,1)=1;
%For each row (starting with the second, as the first is done already)
for i=2:n;
 %For each column in that row, again starting with the second as the first is done
 for j=2:i;
 %Extract the current and previous elements in the row above
 a=o(i-1,j-1); %(previous)
 b=o(i-1,j); %(current)
 %Assume that min()==0, so set c to max();
 c=a|b;
 %Now check if min()==1
 d=a&b;
 %If so, set c to L(1)
 c(d)=L(d);
 %Rotate L around only if min()==1
 L=circshift(L,-d);
 %And store c back to the output matrix
 o(i,j)=c;
 end;
end

Update: I have managed to optimise away the if-else statement to save a few bytes. The input format has once again changed back to column vector.

#MATLAB, (削除) 146 (削除ここまで) 143#

function o=c(n,L);o=zeros(n);o(:,1)=1;for i=2:n for j=2:i a=o(i-1,j-1);b=o(i-1,j);if(a&b) c=L(1);L=L([2:end 1]);else c=a|b;end;o(i,j)=c;end;end

For the input values, n is a scalar, and L is a row vector (which can be specified in the exact format you show [,,,]).

function o=c(n,L)
%Create the initial array - an n x n square with the first column made of 1's
o=zeros(n);o(:,1)=1;
%For each row (starting with the second, as the first is done already)
for i=2:n
 %For each column in that row, again starting with the second as the first is done
 for j=2:i
 %Extract the current and previous elements in the row above
 a=o(i-1,j-1); %(previous)
 b=o(i-1,j); %(current)
 if(a&b) c=L(1); %If they are both 1 (min>0), select the next random seed
 L=L([2:end 1]); %Rotate the random seed so we get the next element in it
 else
 c=a|b; %Otherwise they are different, so set to the max value
 end
 o(i,j)=c; %Store in the current column of the current row
 end
end

#MATLAB, (削除) 146 (削除ここまで) (削除) 143 (削除ここまで) 138#

function o=c(n,L);o=zeros(n);o(:,1)=1;for i=2:n;for j=2:i;a=o(i-1,j-1);b=o(i-1,j);c=a|b;d=a&b;c(d)=L(d);L=circshift(L,-d);o(i,j)=c;end;end

For the input values, n is a scalar, and L is a column vector, which can be specified in the format [;;;]. Not quite what you show, but you say it is flexible within reason and this seems so.

function o=c(n,L)
%Create the initial array - an n x n square with the first column made of 1's
o=zeros(n);o(:,1)=1;
%For each row (starting with the second, as the first is done already)
for i=2:n;
 %For each column in that row, again starting with the second as the first is done
 for j=2:i;
 %Extract the current and previous elements in the row above
 a=o(i-1,j-1); %(previous)
 b=o(i-1,j); %(current)
 %Assume that min()==0, so set c to max(); c=a|b;
 %Now check if min()==1
 d=a&b;
 %If so, set c to L(1)
 c(d)=L(d);
 %Rotate L around only if min()==1
 L=circshift(L,-d);
 %And store c back to the output matrix
 o(i,j)=c;
 end;
end

Update: I have managed to optimise away the if-else statement to save a few bytes. The input format has once again changed back to column vector.

#MATLAB, 146#

(Also works on Octave online, but you need to sign in to save the function in a file).

function o=c(n,L);o=zeros(n);o(:,1)=1;for i=2:n for j=2:i a=o(i-1,j-1);b=o(i-1,j);if(a&b) c=L(1);L=circshift(L,-1);else c=a|b;end;o(i,j)=c;end;end

The function takes an input n and L, and returns an array o which contains the output.

For the input values, n is a scalar, and L is a column vector. As you said the input format isn't strict (within reason), when calling the function L should be written for example as: [0;0;1] to ensure the correct orientation of the vector.

The output is formatted as an n x n array containing 0's and 1's.

And an explanation:

function o=c(n,L)
 %Create the initial array - an n x n square with the first column made of 1's
 o=zeros(n);o(:,1)=1;
 %For each row (starting with the second, as the first is done already)
 for i=2:n
 %For each column in that row, again starting with the second as the first is done
 for j=2:i
 %Extract the current and previous elements in the row above
 a=o(i-1,j-1); %(previous)
 b=o(i-1,j); %(current)
 if(a&b)
 c=L(1); %If they are both 1 (min>0), select the next random seed
 L=circshift(L,-1); %Rotate the random seed so we get the next element in it
 else
 c=a|b; %Otherwise they are different, so set to the max value
 end
 o(i,j)=c; %Store in the current column of the current row
 end
 end

#MATLAB, (削除) 146 (削除ここまで) 143#

(Also works on Octave online, but you need to sign in to save the function in a file).

function o=c(n,L);o=zeros(n);o(:,1)=1;for i=2:n for j=2:i a=o(i-1,j-1);b=o(i-1,j);if(a&b) c=L(1);L=L([2:end 1]);else c=a|b;end;o(i,j)=c;end;end

The function takes an input n and L, and returns an array o which contains the output.

For the input values, n is a scalar, and L is a row vector (which can be specified in the exact format you show [,,,]).

The output is formatted as an n x n array containing 0's and 1's.

And an explanation:

function o=c(n,L)
 %Create the initial array - an n x n square with the first column made of 1's
 o=zeros(n);o(:,1)=1;
 %For each row (starting with the second, as the first is done already)
 for i=2:n
 %For each column in that row, again starting with the second as the first is done
 for j=2:i
 %Extract the current and previous elements in the row above
 a=o(i-1,j-1); %(previous)
 b=o(i-1,j); %(current)
 if(a&b)
 c=L(1); %If they are both 1 (min>0), select the next random seed
 L=L([2:end 1]); %Rotate the random seed so we get the next element in it
 else
 c=a|b; %Otherwise they are different, so set to the max value
 end
 o(i,j)=c; %Store in the current column of the current row
 end
 end