How to write cyclic permutation as an array in matlab?

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lilly lord
lilly lord 2021-7-26
评论： lilly lord 2021-7-27
I have disjoint permutation cycles such as (1 4 6 2 5 7 8 3)(9 10) means
1 comes under 4th position, 4 will be at 6th position ,6 will be at 2nd position and so on. 3 will be at first position .(9,10) means that 9 will ne at 10 position and 10 will be placed at 9th position.(Look at the second row of the output given below)
S=[1:10; 3 6 8 1 2 4 5 7 10 9];
out put is
1 2 3 4 5 6 7 8 9 10
3 6 8 1 2 4 5 7 10 9
It is easy when we have small permutation, but for large set like
perm=[22, 35, 17, 9, 11, 23, 56, 43, 25, 19, 26, 15, 28, 30, 6, 1, 33, 24, 12, 14, 47, 18, 20, 31, 41, 39, 32, 27, 34, 13, 21, 4, 2, 16, 8, 49, 7, 38, 40, 29, 10, 3, 5, 36, 42, 37, 44, 46, 48, 50];
required=[1:50;6 4 10 21 3 30 45 16 17 29 9 24 34 12 26 2 35 47 25 18 13 50 11 33 43 19 32 15 40 28 20 39 1 27 22 5 42 7 41 38 31 36 23 37 49 44 14 46 8 48 ];
output is
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
6 4 10 21 3 30 45 16 17 29 9 24 34 12 26 2 35 47 25 18 13 50 11 33 43 19 32 15 40 28 20 39 1 27 22 5 42 7 41 38 31 36 23 37 49 44 14 46 8 48
Is there a way in matlab that we can fill up the second row for large pemutation it is difficult to write manually.
Thanks in advance

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采纳的回答

Jan
Jan 2021-7-26
编辑：Jan 2021-7-26
Ran in:
P = [1 4 6 2 5 7 8 3];
Q = zeros(size(P));
Q(circshift(P, -1)) = P
Q = 1×8
3 6 8 1 2 4 5 7
P = [22, 35, 17, 9, 11, 23, 56, 43, 25, 19, 26, 15, 28, 30, 6, 1, ...
... % ^^ Greater than length(P) ?!?
33, 24, 12, 14, 47, 18, 20, 31, 41, 39, 32, 27, 34, 13, 21, 4, ...
2, 16, 8, 49, 7, 38, 40, 29, 10, 3, 5, 36, 42, 37, 44, 46, 48, 50];
wanted = [6 4 10 21 3 30 45 16 17 29 9 24 34 12 26 2 35 47 25 18 13 50, ...
11 33 43 19 32 15 40 28 20 39 1 27 22 5 42 7 41 38 31 36 23 37 49, ...
44 14 46 8 48];
Q = zeros(size(P));
Q(circshift(P, -1)) = P
Q = 1×56
6 4 10 21 3 30 49 16 17 29 9 24 34 12 26 2 35 47 25 18 13 50 11 33 43 19 32 15 40 28
isequal(Q, wanted)
ans = logical
0
There is a typo in your data at P==56. The 45 is missing, but it must be at another location.
This is working to get your output: wanted
P = [22, 35, 17, 9, 11, 23, 43, 25, 19, 26, 15, 28, 30, 6, 1, 33, ...
24, 12, 14, 47, 18, 20, 31, 41, 39, 32, 27, 34, 13, 21, 4, 2, ...
16, 8, 49, 45, 7, 38, 40, 29, 10, 3, 5, 36, 42, 37, 44, 46, 48, 50];
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Jan
Jan 2021-7-27
编辑：Jan 2021-7-27
Ran in:
I do not know, how the input is stored for "(1 4 6 2 5 7 8 3)(9 10)". Therefore I cannot write some code to convert this unkown input to a specific output.
Bold guessing is no fair strategy in programming. But let me try it:
PList = {[1, 4, 6, 2, 5, 7, 8, 3], [9, 10]};
maxIndex = max(cellfun(@max, PList, 'UniformOutput', true));
Q = zeros(1, maxIndex);
for iP = 1:numel(PList)
P = PList{iP};
Q(circshift(P, -1)) = P;
end
Q
Q = 1×10
3 6 8 1 2 4 5 7 10 9
So simple apply the code I've suggested in a loop.
It took some time to cope the the error for P==56 in your second example. Did I oversee a point or has this been y typo?
lilly lord
lilly lord 2021-7-27
p=56 is written by mistake. Your code works excellent . Thanks a lot

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