How to write cyclic permutation as an array in matlab?

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

Answer 1

[Jan]

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];

[lilly lord]

P = [1 4 6 2 5 7 8 3];

Q = zeros(size(P));Q(circshift(P, -1)) = P. This solution works well when we have one single permutation (Thank you) but in case of disjoint cycles like I have mentioned in the question

(1 4 6 2 5 7 8 3)(9 10)

9 is under 10th position and 10 is at 9th . what should we do in this case.

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

[Jan]

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]

p=56 is written by mistake. Your code works excellent . Thanks a lot