Shuffle vector elements such that two similar elements coming together at most twice - matlab

Question

For the sake of an example:
I have a vector named vec containing ten 1s and ten 2s. I am trying to randomly arrange it but with one condition that two same values must not come together more than twice.
What I have done till now is generating random indexes of vec using the randperm function and shuffling vec accordingly. This is what I have:
vec = [1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2];
atmost = 2;
indexes = randperm(length(vec));
vec = vec(indexes);
>> vec =
2 1 1 1 2 1 2 1 2 1 1 1 2 2 1 2 1 2 2 2
My code randomly arranges elements of vec but does not fulfill the condition of two similar values coming at most two times. How can I do this? Any ideas?

Answer 1

You can first determine the lengths of the runs of one value, and then fit the other values around them.
In the explanation I'll use values of a and b in the vector, so as not to confuse the values of the elements of the vector (1 and 2) and the lengths of the runs for each element (1 and 2).
The end goal is to use repelem to construct the shuffled vector. repelem takes a vector of the elements to repeat, and a vector of how many times to repeat each element. For example, if we have:
v = [b a b a b a b a b a b a b a b a b]
n = [1 1 1 2 1 1 2 1 2 1 1 1 1 2 1 1 0]
repelem will return:
shuffled_vec = [b a b a a b a b b a b b a b a b a a b a]
As a first step, I'll generate random values for the counts corresponding to the a values. In this example, that would be:
a_grouping = [1 2 1 1 1 1 2 1]
First, randomly select the number of 2's in the grouping vector. There can be at most n/2 of them. Then add 1's to make up the desired total.
num_total = 10; % number of a's (and b's)
% There can be 0..num_total/2 groups of two a's in the string.
two_count = randi(num_total/2 + 1) - 1;
% The remainder of the groups of a's have length one.
one_count = num_total - (2*two_count);
% Generate random permutation of two_count 2's and one_count 1's
a_grouping = [repmat(2, 1, two_count), repmat(1, 1, one_count)];
This will give us something like:
a_grouping = [2 2 1 1 1 1 1 1]
Now shuffle:
a_grouping = a_grouping(randperm(numel(a_grouping)));
With the result:
a_grouping = [1 2 1 1 1 1 2 1]
Now we need to figure out where the b values go. There must be at least one b between each run of a values (and at most two), and there may be 0, 1 or 2 b values at the beginning and end of the string. So we need to generate counts for each of the x and y values below:
all_grouping = [y 1 x 2 x 1 x 1 x 1 x 1 x 2 x 1 y]
The x values must be at least 1, so we'll assign them first. Since the y values can be either 0, 1 or 2, we'll leave them set to 0.
% Between each grouping of a's, there must be at least one b.
% There can be 0, 1, or 2 b's before and after the a's,
% so make room for them as well.
b_grouping = zeros(1, numel(a_grouping) - 1 + 2);
b_grouping(2:end-1) = 1; % insert one b between each a group
For each of the the remaining counts we need to assign, just select a random slot. If it's not filled yet (i.e. if it's < 2), increment the count, otherwise find a different slot.
% Assign location of remaining 2's
for s = numel(a_grouping):num_total
unassigned = true;
while unassigned
% generate random indices until we find one that's open
idx = randi(numel(b_grouping));
if b_grouping(idx) < 2
b_grouping(idx) = b_grouping(idx) + 1;
unassigned = false;
end
end
end
Now we've got separate counts for the a's and b's:
a_grouping = [1 2 1 1 1 1 2 1]
b_grouping = [1 1 1 2 2 1 1 1 0]
We'll build the value vector (v from the start of the example) and interleave the groupings (the n vector).
% Interleave the a and b values
group_values = zeros(1, numel(a_grouping) + numel(b_grouping));
group_values(1:2:end) = 2;
group_values(2:2:end) = 1;
% Interleave the corresponding groupings
all_grouping = zeros(size(group_values));
all_grouping(2:2:end) = a_grouping;
all_grouping(1:2:end) = b_grouping;
Finally, repelem puts everything together:
shuffled_vec = repelem(group_values, all_grouping)
The final result is:
shuffled_vec =
1 2 2 1 1 2 1 1 2 2 1 1 2 2 1 1 2 2 1 2
Full code:
num_total = 10; % number of a's (and b's)
% There can be 0..num_total/2 groups of two a's in the string.
two_count = randi(num_total/2 + 1) - 1;
% The remainder of the groups of a's have length one.
one_count = num_total - (2*two_count);
% Generate random permutation of two_count 2's and one_count 1's
a_grouping = [repmat(2, 1, two_count), repmat(1, 1, one_count)];
a_grouping = a_grouping(randperm(numel(a_grouping)));
% disp(a_grouping)
% Between each grouping of a's, there must be at least one b.
% There can be 0, 1, or 2 b's before and after the a's,
% so make room for them as well.
b_grouping = zeros(1, numel(a_grouping) - 1 + 2);
b_grouping(2:end-1) = 1; % insert one b between each a group
% Assign location of remaining 2's
for s = numel(a_grouping):num_total
unassigned = true;
while unassigned
% generate random indices until we find one that's open
idx = randi(numel(b_grouping));
if b_grouping(idx) < 2
b_grouping(idx) = b_grouping(idx) + 1;
unassigned = false;
end
end
end
% Interleave the a and b values
group_values = zeros(1, numel(a_grouping) + numel(b_grouping));
group_values(1:2:end) = 2;
group_values(2:2:end) = 1;
% Interleave the corresponding groupings
all_grouping = zeros(size(group_values));
all_grouping(2:2:end) = a_grouping;
all_grouping(1:2:end) = b_grouping;
shuffled_vec = repelem(group_values, all_grouping)

Answer 2

This should generate fairly random non-consecutive vectors. Whether it covers all possibilities uniformly, I'm not sure.
out=[];
for i=1:10
if randi(2)==1
out=[out,1,2];
else
out=[out,2,1];
end
end
disp(out)
Example results
1,2,1,2,1,2,1,2,2,1,2,1,1,2,2,1,2,1,1,2,
2,1,2,1,1,2,2,1,2,1,1,2,2,1,2,1,1,2,1,2,
1,2,2,1,2,1,1,2,1,2,2,1,2,1,2,1,2,1,2,1,
2,1,2,1,2,1,1,2,1,2,2,1,1,2,2,1,2,1,1,2,
2,1,1,2,1,2,1,2,2,1,2,1,1,2,1,2,1,2,1,2,
2,1,1,2,2,1,2,1,1,2,1,2,2,1,1,2,1,2,2,1,
1,2,2,1,1,2,1,2,2,1,2,1,2,1,1,2,2,1,1,2,
2,1,1,2,2,1,2,1,1,2,2,1,1,2,1,2,1,2,1,2,
1,2,2,1,1,2,1,2,2,1,2,1,1,2,1,2,1,2,1,2,
1,2,1,2,2,1,2,1,2,1,2,1,1,2,1,2,2,1,2,1,