How to write cyclic permutation as an array in matlab?

Question

lilly lord 2021-7-26

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/886454-how-to-write-cyclic-permutation-as-an-array-in-matlab

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

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Jan 2021-7-26

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/886454-how-to-write-cyclic-permutation-as-an-array-in-matlab#answer_754314

编辑：Jan 2021-7-26

在 MATLAB Online 中打开

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];

3 个评论
显示 1更早的评论隐藏 1更早的评论

Jan 2021-7-27

编辑：Jan 2021-7-27

在 MATLAB Online 中打开

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