Shuffle vector with constraints

2 次查看（过去 30 天）
Pieter
Pieter 2014-1-14
回答： Pieter 2014-1-14
Hi,
I have an array of 50 1's, 50 2's, etc up to 50 8's. I would like to shuffle them with the constraint that two consecutive numbers cannot be the same. I tried several options but most are very cpu intensive or did not yield a good result. Are there computation efficient solutions to this problem?
  1 个评论
Azzi Abdelmalek
Azzi Abdelmalek 2014-1-14
This is not enough as information. Because you can set them for example [1 2 3 1 2 3 1 2 3 . . .]

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

W. Owen Brimijoin
W. Owen Brimijoin 2014-1-14
A brute force method works fairly well in this scenario - 1) start with a shuffled version of the sequence, 2) find repeats, 3) swap any repeats to different locations... goto 2 (rinse and repeat). This sort of method would likely be terrible if you had fewer characters in your sequence, but for nine unique elements, it seems to halt very quickly.
Making your sequence:
seq = repmat([1:8],50,1); %make sequence
seq = [seq(:);repmat(9,30,1)]; %make sequence
Doing the reshuffling:
seq = seq(randperm(numel(seq))); %initial shuffle
old_idx= unique(find(diff(seq)==0));%find repeats
while ~isempty(old_idx), %continue until no repeats
new_idx = unique(setdiff([1:length(seq)],old_idx)); %find new spots
new_idx = new_idx((randi(length(new_idx),length(old_idx),1)))';
seq([new_idx;old_idx],:) = seq([old_idx;new_idx],:); %swap
old_idx= unique(find(diff(seq)==0));%find repeats
end

更多回答（3 个）

Mischa Kim
Mischa Kim 2014-1-14
编辑：Mischa Kim 2014-1-14
Since I do not know what techniques you have tried, here's (another) one? Start with sorting the 1's, 2's, etc. in matrix form, i.e.,
1 1 1 1...
2 2 2 2...
3 3 3 3...
and perform permutations on the individual columns (e.g using randperm). Based on the entry of the last row in the first column randomly pick a column vector with a different first row entry and make it the new second column. Using this process work your way through all the columns (you might have to do some more permutations towards the final columns). Once done, transpose and concenate all of the column vectors.
Pieter
Pieter 2014-1-14
Hi Mischa,
Thanks, this is a very good suggestion for one of my other experiments. This however has two issues I think:
1) It is not a constraint that every subvector of 8 numbers is a permutation. The only constraint is that no two consecutive numbers are the same.
2) This only works if all numbers occur equally. I forgot to mention that we have in fact 9 numbers. 1 to 8 occur 50 times, 9 occurs only 30 times.
  1 个评论
Mischa Kim
Mischa Kim 2014-1-14
编辑：Mischa Kim 2014-1-14
Hello Pieter,
  1. Understood. I used the permutation for shuffling with a certain level of randomness (while satisfying the constraint).
  2. You can still use the approach outlined above by randomly tossing in the 30 9's into the vector at the end.

请先登录，再进行评论。

Pieter
Pieter 2014-1-14
Hi Mischa,
This looks fine. I'll try in tomorrow. Thx

类别

在 Help Center 和 File Exchange 中查找有关 Mathematics 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