How to calculate permutation powers

3 次查看（过去 30 天）
lilly lord
lilly lord 2021-7-31
评论： lilly lord 2021-8-2
I have to find higher powers of a permutation. For example if M is a given permutation , then to calculate M^11
M= [ 1 2 3 4 5 6 7 8 9 10; 3 6 8 1 2 4 5 7 10 9]
M^11 the answer should be [1:10;7 1 5 8 4 3 62 10 9];
The code below works well for small permutation but for large permutation and higher power gives error. If any one can help . Thanks in advance
inMat: The input permutation matrix
% n: The number of permutations
% M=[1 2 3 4 5 6; 3 4 1 5 6 2];
Temp=inMat;
Temp1=inMat;
rc=size(Temp);
outMat(1,:)=Temp(1,:);
for j=1:n
for i=1:rc(2)
r=Temp(:,i);
outMat(2,i)=Temp1(2,r(2));
end
Temp1=outMat;
end
  2 个评论
dpb
dpb 2021-8-1
For example if M is a given permutation , then to calculate M^11
M= [ 1 2 3 4 5 6 7 8 9 10; 3 6 8 1 2 4 5 7 10 9]
M^11 the answer should be [1:10;7 1 5 8 4 3 62 10 9];
I've no klew how you come up with the above as the answer???
Nor what power has to do with it given the sample code that is supposed to work ("for small permutation")...
Can't figure out what it's supposed to do as not everything is defined in the code snippet.
Image Analyst
Image Analyst 2021-8-1
I agree with dpb (above). No idea how you came up with that answer, or what a "power" of a permutation is. All I know is that M is a matrix, not a permutation.
% M is a matrix, not a permutation of anything that we know of so far.
M = [1 2 3 4 5 6 7 8 9 10; 3 6 8 1 2 4 5 7 10 9]
% With the above M, the bottom row is a permutation of the top row.
% So actually each row is a permutation of the other row.
% Make a new matrix that is a permutation of M:
randomIndexes = randperm(numel(M))
M2 = reshape(M(randomIndexes), size(M))
You can see in the last line I made a permutation of your matrix M using randperm(). But I still have no idea what you mean by the "power" of it. Do you just want to raise it to the 11th power, like
M2 = M2 .^ 11;
????

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Bruno Luong
Bruno Luong 2021-8-1
Ran in:
M= [ 1 2 3 4 5 6 7 8 9 10; 3 6 8 1 2 4 5 7 10 9]
M = 2×10
1 2 3 4 5 6 7 8 9 10 3 6 8 1 2 4 5 7 10 9
S=sparse(M(2,:),M(1,:),1);
[p11,~]=find(S^11);
M11= [M(1,:);p11']
M11 = 2×10
1 2 3 4 5 6 7 8 9 10 7 1 5 8 4 3 6 2 10 9
  3 个评论
显示 1更早的评论
lilly lord
lilly lord 2021-8-2
Thank u very much . Can u plz tell me how to take powers of disjoint cycles using the code u have shared.
lilly lord
lilly lord 2021-8-2
Your answer solved my problem . Thank u . for powers of disjoint cycles i will ask as another question

请先登录，再进行评论。

更多回答（1 个）

David Goodmanson
David Goodmanson 2021-8-2
编辑：David Goodmanson 2021-8-2
Hi lilly,
Since the 1:n in the first row is just along for the ride, you don't have to carry it along in the calculation; you can just append it at the end. For the 11th power you have to use n=10 multplciations so it might be less confusing to redifine nnew = n+1 and use nnew-1 multiplications. Anyway, the inner do loop is equivalent to the following iterative mapping process:
n = 10;
inMat = [1:10;
3 6 8 1 2 4 5 7 10 9]
Temp=inMat;
Temp1=inMat;
rc=size(Temp);
outMat(1,:)=Temp(1,:);
for j=1:n
for i=1:rc(2)
r=Temp(:,i)
outMat(2,i)=Temp1(2,r(2));
end
Temp1=outMat;
end
outMat
% alterative
p0 = inMat(2,:);
p = p0;
for j = 1:n
p = p0(p);
end
outMat2 = [1:10; p]

类别

在 Help Center 和 File Exchange 中查找有关 MATLAB 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by