How to generate a vector of numbers picked from two groups with the condition that there cannot be more than 2 consecutive numbers from the same group?

Dear people,

I created a vector with 26 ones, 26 twos, 26 threes and so on until the number 12. Then I shuffle these numbers in another vector so now the vector is like this a = [1 5 2 10 9 3 5...]. From these 12 different numbers some of them (let's say 1 2 3 and 4) come from the same group (group A) and the rest come from group B.

The problem is that I need them to not be more than 2 of the same group in a row (for example if I have [1 3 2..] is wrong, but if I have [1 3 6...] it's okay).

I created a loop in which I check if there are more than 2 numbers of the same group in a row, and if it is the case I reshuffle the main vector. But Matlab never stops. Do you know another possibility to achieve what I need?

Thank you so much in advance

Answer 1

[Jeff Miller]

If I understand what you are asking, this should be close to what you want:

vecLength = 20;  % how many numbers you want to generate
group1Numbers = 1:4;  % specify which numbers belong to each group
group2Numbers = 5:12;

% first make a vector with group numbers 1 or 2, with a max of 2 in a row
% from the same group
groupVec = zeros(vecLength,1);
groupVec(1:2) = randi([1, 2],2,1);
for ipos=3:vecLength
    if groupVec(ipos-2) == groupVec(ipos-1)
        groupVec(ipos) = 3 - groupVec(ipos-1);  % switch group 1 to 2 or vice versa
    else
        groupVec(ipos) = randi([1, 2],1,1);
    end
end
groupVec' % this is the sequence of group numbers

% now pick numbers randomly from the pre-selected groups:
finalVec = zeros(vecLength,1);
for ipos=1:vecLength
    switch groupVec(ipos)
        case 1
            randpos = randi([1 numel(group1Numbers)],1,1);
            finalVec(ipos) = group1Numbers(randpos);
        case 2
            randpos = randi([1 numel(group2Numbers)],1,1);
            finalVec(ipos) = group2Numbers(randpos);
    end
end

finalVec'  % this is the final sequence of numbers that I think you wanted

[Trisia Cinca Tomás]

Thanks a lot Jeff for your answer.

What I meant was a little bit different. Instead of make an example with what I want to do, I'll try to explain exactly what I need.

I need to generate exactly 416 numbers. From these 416, 26 I need them to be ones, 52 I need them to be twos, and so on like in the following code:

Block2_condition = [1 2 3 4 5 6 7 8 9 10 11 12];
Block2_Number_trials_event =  [26 52 26 26 52 26 26 52 26 26 52 26];

So what I did was to create 26 ones, 52 twos...and so on like in the following code:

Block2_event1 = ones(1, Block2_Number_trials_event(1)).*Block2_condition(1);
Block2_event2 = ones(1, Block2_Number_trials_event(2)).*Block2_condition(2);
Block2_event3 = ones(1, Block2_Number_trials_event(3)).*Block2_condition(3);
Block2_event4 = ones(1, Block2_Number_trials_event(4)).*Block2_condition(4);
Block2_event5 = ones(1, Block2_Number_trials_event(5)).*Block2_condition(5);
Block2_event6 = ones(1, Block2_Number_trials_event(6)).*Block2_condition(6);
Block2_event7 = ones(1, Block2_Number_trials_event(7)).*Block2_condition(7);
Block2_event8 = ones(1, Block2_Number_trials_event(8)).*Block2_condition(8);
Block2_event9 = ones(1, Block2_Number_trials_event(9)).*Block2_condition(9);
Block2_event10 = ones(1, Block2_Number_trials_event(10)).*Block2_condition(10);
Block2_event11 = ones(1, Block2_Number_trials_event(11)).*Block2_condition(11);
Block2_event12 = ones(1, Block2_Number_trials_event(12)).*Block2_condition(12);

I need the 4 first and 4 last numbers to force them to be 3, 6, 9 & 12 so I erased 2 from each Block2_event group in the following manner:

Block2_event3(1:2) = []; % We remove 2 events of condition 3 from the vector
Block2_event6(1:2) = []; % We remove 2 events of condition 6 from the vector
Block2_event9(1:2) = []; % We remove 2 events of condition 9 from the vector
Block2_event12(1:2) = []; % We remove 2 events of condition 12 from the vector

Then I created a vector with all event I want and add the 4 first and 4 last events I'm forcing them to be 3, 6, 9 & 12 in a random position within these 4 first and last positions

% Combining all vectors in one sequence:
Block2_event = [Block2_event1, Block2_event2, Block2_event3, Block2_event4, Block2_event5, Block2_event6 Block2_event7, Block2_event8, Block2_event9, Block2_event10, Block2_event11, Block2_event12];

% Adding the 4 first and last trials previously generated:
Block2_event_random = zeros(1, length(Block2_event)+8);
Block2_event_random(1:4) = Shuffle([3 6 9 12]); % We add conditions 3, 6, 9 and 12 randomly at the beginning
Block2_event_random(end-3:end) = Shuffle([3 6 9 12]); % We add conditions 3, 6, 9 and 12 randomly at the end

Next, I shuffle the numbers from 5:end-4

% Randomizing all remaining events:
Block2_event_random(5:end-4) = Shuffle(Block2_event);

As we need to have at most two of each "group" of numbers followed in a row, I made the following:

% Groups of numbers
group1 = [1,4,7,10,3,6,9,12];
group2 = [2,5,8,11];

% Check if they follow the rule
for itwo = 3:length(Block2_event_random)-4
    if ~isempty(intersect(Block2_event_random(itwo),group1)) && ~isempty(intersect(Block2_event_random(itwo+1),group1))
        if ~isempty(intersect(Block2_event_random(itwo+2),group1))
            btwo(itwo-2) = ~isempty(intersect(Block2_event_random(itwo+2),group1));
        end
    elseif ~isempty(intersect(Block2_event_random(itwo),group2)) && ~isempty(intersect(Block2_event_random(itwo+1),group2))
        if ~isempty(intersect(Block2_event_random(itwo+2),group2))
            btwo(itwo-2) = ~isempty(intersect(Block2_event_random(itwo+2),group2));
        end
    else
        btwo(itwo-2) = false;
    end
end
% As long as the rule is not true, re-shuffle the vector
while ~isempty(find(btwo))
    Block2_event_random(5:end-4) = Shuffle(Block2_event)
    for itwo = 3:length(Block2_event_random)-4
        if ~isempty(intersect(Block2_event_random(itwo),group1)) && ~isempty(intersect(Block2_event_random(itwo+1),group1))
            if ~isempty(intersect(Block2_event_random(itwo+2),group1))
                btwo(itwo-2) = ~isempty(intersect(Block2_event_random(itwo+2),group1));
            end
        elseif ~isempty(intersect(Block2_event_random(itwo),group2)) && ~isempty(intersect(Block2_event_random(itwo+1),group2))
            if ~isempty(intersect(Block2_event_random(itwo+2),group2))
                btwo(itwo-2) = ~isempty(intersect(Block2_event_random(itwo+2),group2));
            end
        else
            btwo(itwo-2) = false;
        end
    end
end

What happened here is that Matlab seems to never reach the final. I thought maybe it was because there is no possible combination to follow the rule with this combination of numbers (being 26 ones, 52 twos and so on). However I thought that maybe I was doing something wrong.

Hence I decided to, instead of using "while ~isempty(find(btwo))", change it by "length(find(btwo))> 105" and try to reach a number of "no followed rules" that allows matlab to find a combination. So by letting Matlab create a combination with at most 105 "no followed rules" it reaches a combination. In reality I would need none "no followed rules" to exist, however it seems to be impossible.

But, I thought that maybe I'm still wrong and I missed something.

It is because of that I thought of doing something similar to what you explained to me, however in my case the problem is that I don't know how to be sure that I'm creating a vector with 26 ones, 52 twos...etc. if I do it your way.

Do you know a way to do it?

Thanks so much in advance

[Jeff Miller]

That's a much tougher problem, which I do not know how to solve.

I am not surprised that MATLAB never finds a solution, since there are so many more random sequences that are not solutions than there are sequences that are solutions.

I would try to generate an acceptable sequence of 1's and 2's first. Once you had that sequence, you could replace each 1 or 2 with a number from its corresponding group. But getting a 1's & 2's sequence satisfying your constraints does not look easy.

Sorry I cannot be more help.

[Trisia Cinca Tomás]

That's what I thought.

Thank you so much Jeff for your help.