Mixing
Constructing a recursion that forms a chain until no new element can be added
I have a matrix by 9x2 where the first column implies the second column, that is
A=[7 10; 1 7; 3 1; 6 9; 10 7; 2 8; 10 8; 10 4; 8 6];
I will form an implication chain using these data. Couple examples would be 3 --> 1 --> 7 --> 10 --> 8 or 3 --> 1 --> 7 --> 10 --> 4 or 2 --> 8 --> 6 --> 9
I tried to use a cell for every rule but couldn't form the while loop. The cell dimension changes with every new element but for example, I couldn't move from 3 --> 1 --> 7 --> 10 to two separate chains of 3 --> 1 --> 7 --> 10 --> 8 and 3 --> 1 --> 7 --> 10 --> 4. How can I do it?
matlab while-loop chain
asked Nov 19 '18 at 18:52
A DoeA Doe
868
add a comment |
I have a matrix by 9x2 where the first column implies the second column, that is
A=[7 10; 1 7; 3 1; 6 9; 10 7; 2 8; 10 8; 10 4; 8 6];
I will form an implication chain using these data. Couple examples would be 3 --> 1 --> 7 --> 10 --> 8 or 3 --> 1 --> 7 --> 10 --> 4 or 2 --> 8 --> 6 --> 9
I tried to use a cell for every rule but couldn't form the while loop. The cell dimension changes with every new element but for example, I couldn't move from 3 --> 1 --> 7 --> 10 to two separate chains of 3 --> 1 --> 7 --> 10 --> 8 and 3 --> 1 --> 7 --> 10 --> 4. How can I do it?
matlab while-loop chain
asked Nov 19 '18 at 18:52
A DoeA Doe
868
add a comment |
I have a matrix by 9x2 where the first column implies the second column, that is
A=[7 10; 1 7; 3 1; 6 9; 10 7; 2 8; 10 8; 10 4; 8 6];
I will form an implication chain using these data. Couple examples would be 3 --> 1 --> 7 --> 10 --> 8 or 3 --> 1 --> 7 --> 10 --> 4 or 2 --> 8 --> 6 --> 9
I tried to use a cell for every rule but couldn't form the while loop. The cell dimension changes with every new element but for example, I couldn't move from 3 --> 1 --> 7 --> 10 to two separate chains of 3 --> 1 --> 7 --> 10 --> 8 and 3 --> 1 --> 7 --> 10 --> 4. How can I do it?
matlab while-loop chain
asked Nov 19 '18 at 18:52
A DoeA Doe
868
I have a matrix by 9x2 where the first column implies the second column, that is
A=[7 10; 1 7; 3 1; 6 9; 10 7; 2 8; 10 8; 10 4; 8 6];
I will form an implication chain using these data. Couple examples would be 3 --> 1 --> 7 --> 10 --> 8 or 3 --> 1 --> 7 --> 10 --> 4 or 2 --> 8 --> 6 --> 9
I tried to use a cell for every rule but couldn't form the while loop. The cell dimension changes with every new element but for example, I couldn't move from 3 --> 1 --> 7 --> 10 to two separate chains of 3 --> 1 --> 7 --> 10 --> 8 and 3 --> 1 --> 7 --> 10 --> 4. How can I do it?
matlab while-loop chain
matlab while-loop chain
asked Nov 19 '18 at 18:52
A DoeA Doe
868
asked Nov 19 '18 at 18:52
A DoeA Doe
868
asked Nov 19 '18 at 18:52
A DoeA Doe
868
asked Nov 19 '18 at 18:52
A DoeA Doe
868
asked Nov 19 '18 at 18:52
A DoeA Doe
868
868
add a comment |
add a comment |
1 Answer
1
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oldest
votes
One option is to convert this to a directed graph. You can identify the source and sink nodes by those that only occur in the first or second column of A. You can then loop over all possible paths from each source to each sink node using shortest path.
A=[7 10; 1 7; 3 1; 6 9; 10 7; 2 8; 10 8; 10 4; 8 6];
G = digraph(A(:,1), A(:,2)); % Create graph
source = setdiff(A(:,1), A(:,2)); % Identify source and sink nodes
sink = setdiff(A(:,2), A(:,1));
for i = 1:length(source)
for j = 1:length(sink)
disp(shortestpath(G, source(i), sink(j)));
end
end
output:
2 8 6 9
3 1 7 10 4
3 1 7 10 8 6 9
answered Nov 19 '18 at 19:14
MattMatt
1,114112
That would work great! I had forgotten to do it this way. Thank you very much!
– A Doe
Nov 19 '18 at 19:17
Unfortunately, my data is larger than this matrix. For some of the analyses, it stretches up to more than 50x2. So, the shortest path turned out to be the the direct path between two nodes. I'll try the longest path, though.
– A Doe
Nov 19 '18 at 19:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
One option is to convert this to a directed graph. You can identify the source and sink nodes by those that only occur in the first or second column of A. You can then loop over all possible paths from each source to each sink node using shortest path.
A=[7 10; 1 7; 3 1; 6 9; 10 7; 2 8; 10 8; 10 4; 8 6];
G = digraph(A(:,1), A(:,2)); % Create graph
source = setdiff(A(:,1), A(:,2)); % Identify source and sink nodes
sink = setdiff(A(:,2), A(:,1));
for i = 1:length(source)
for j = 1:length(sink)
disp(shortestpath(G, source(i), sink(j)));
end
end
output:
2 8 6 9
3 1 7 10 4
3 1 7 10 8 6 9
answered Nov 19 '18 at 19:14
MattMatt
1,114112
That would work great! I had forgotten to do it this way. Thank you very much!
– A Doe
Nov 19 '18 at 19:17
Unfortunately, my data is larger than this matrix. For some of the analyses, it stretches up to more than 50x2. So, the shortest path turned out to be the the direct path between two nodes. I'll try the longest path, though.
– A Doe
Nov 19 '18 at 19:32
add a comment |
One option is to convert this to a directed graph. You can identify the source and sink nodes by those that only occur in the first or second column of A. You can then loop over all possible paths from each source to each sink node using shortest path.
A=[7 10; 1 7; 3 1; 6 9; 10 7; 2 8; 10 8; 10 4; 8 6];
G = digraph(A(:,1), A(:,2)); % Create graph
source = setdiff(A(:,1), A(:,2)); % Identify source and sink nodes
sink = setdiff(A(:,2), A(:,1));
for i = 1:length(source)
for j = 1:length(sink)
disp(shortestpath(G, source(i), sink(j)));
end
end
output:
2 8 6 9
3 1 7 10 4
3 1 7 10 8 6 9
answered Nov 19 '18 at 19:14
MattMatt
1,114112
That would work great! I had forgotten to do it this way. Thank you very much!
– A Doe
Nov 19 '18 at 19:17
Unfortunately, my data is larger than this matrix. For some of the analyses, it stretches up to more than 50x2. So, the shortest path turned out to be the the direct path between two nodes. I'll try the longest path, though.
– A Doe
Nov 19 '18 at 19:32
add a comment |
One option is to convert this to a directed graph. You can identify the source and sink nodes by those that only occur in the first or second column of A. You can then loop over all possible paths from each source to each sink node using shortest path.
A=[7 10; 1 7; 3 1; 6 9; 10 7; 2 8; 10 8; 10 4; 8 6];
G = digraph(A(:,1), A(:,2)); % Create graph
source = setdiff(A(:,1), A(:,2)); % Identify source and sink nodes
sink = setdiff(A(:,2), A(:,1));
for i = 1:length(source)
for j = 1:length(sink)
disp(shortestpath(G, source(i), sink(j)));
end
end
output:
2 8 6 9
3 1 7 10 4
3 1 7 10 8 6 9
answered Nov 19 '18 at 19:14
MattMatt
1,114112
One option is to convert this to a directed graph. You can identify the source and sink nodes by those that only occur in the first or second column of A. You can then loop over all possible paths from each source to each sink node using shortest path.
A=[7 10; 1 7; 3 1; 6 9; 10 7; 2 8; 10 8; 10 4; 8 6];
G = digraph(A(:,1), A(:,2)); % Create graph
source = setdiff(A(:,1), A(:,2)); % Identify source and sink nodes
sink = setdiff(A(:,2), A(:,1));
for i = 1:length(source)
for j = 1:length(sink)
disp(shortestpath(G, source(i), sink(j)));
end
end
output:
2 8 6 9
3 1 7 10 4
3 1 7 10 8 6 9
answered Nov 19 '18 at 19:14
MattMatt
1,114112
answered Nov 19 '18 at 19:14
MattMatt
1,114112
answered Nov 19 '18 at 19:14
MattMatt
1,114112
answered Nov 19 '18 at 19:14
MattMatt
1,114112
1,114112
That would work great! I had forgotten to do it this way. Thank you very much!
– A Doe
Nov 19 '18 at 19:17
Unfortunately, my data is larger than this matrix. For some of the analyses, it stretches up to more than 50x2. So, the shortest path turned out to be the the direct path between two nodes. I'll try the longest path, though.
– A Doe
Nov 19 '18 at 19:32
add a comment |
That would work great! I had forgotten to do it this way. Thank you very much!
– A Doe
Nov 19 '18 at 19:17
Unfortunately, my data is larger than this matrix. For some of the analyses, it stretches up to more than 50x2. So, the shortest path turned out to be the the direct path between two nodes. I'll try the longest path, though.
– A Doe
Nov 19 '18 at 19:32
That would work great! I had forgotten to do it this way. Thank you very much!
– A Doe
Nov 19 '18 at 19:17
That would work great! I had forgotten to do it this way. Thank you very much!
– A Doe
Nov 19 '18 at 19:17
Unfortunately, my data is larger than this matrix. For some of the analyses, it stretches up to more than 50x2. So, the shortest path turned out to be the the direct path between two nodes. I'll try the longest path, though.
– A Doe
Nov 19 '18 at 19:32
Unfortunately, my data is larger than this matrix. For some of the analyses, it stretches up to more than 50x2. So, the shortest path turned out to be the the direct path between two nodes. I'll try the longest path, though.
– A Doe
Nov 19 '18 at 19:32
add a comment |
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