How do I create a set of 20 random binary numbers (only 0 and 1) but such that 0 or 1 does not repeat 4 times in a row?

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moizz
moizz 2012-11-7
How do I create a set of 20 random binary numbers (only 0 and 1) but such that 0 or 1 does not repeat 4 times in a row?
for example the set should not have ( 0 , 0, 1, 0, 1,1,1,1) because I dont want 4 of the same number in a row since 1 repeats 4 times

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回答（6 个）

Walter Roberson
Walter Roberson 2012-11-7
Create a set of random binary numbers. Test whether it has the required properties. If it does, accept it; otherwise loop back and generate another set.
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Walter Roberson
Walter Roberson 2012-11-7
It doesn't have to scale well; if larger matrices had been indicated then I would have given different code. The problem with Order Analysis is that it is for the asymptoptic case and ignores that at low orders there might be more efficient code.

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Jonathan Epperl
Jonathan Epperl 2012-11-7
Create them step-by-step, and if you have 3 0 or 1 in a row, force a 1 or 0 for the next. What I mean:
rbits = false(1,20);
rbits(1:3) = rand(3,1)>.5; % The first 3 can be whatever
for k=4:20
rbits(k) = ~all(rbits(k-3:k-1)) & ( all(~rbits(k-3:k-1)) | rand()>.5);
end
rbits
I am fairly certain that is equivalent to just filling the vector randomly and then changing the forth 1 to a zero, the 4th zero to a 1:
rbits = rand(1,20)>.5
rbits_old = ~rbits
while any(rbits~=rbits_old)
rbits_old = rbits;
RB = [rbits' circshift(rbits',1 ) circshift(rbits',2) circshift(rbits',3)]';
rbits(all(RB,1)) = 0
rbits(all(~RB,1)) = 1
end
I am not 100% certain about the second syntax and whether it could get you into an infinite loop, but if your number of random bits was much more than 20, a syntax along the lines of the second suggestion should be faster.
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moizz
moizz 2012-11-7
Thank You very much, also how would I be able to add to this script such that there are equal amount of 1 and 0, So for example there will always be ten 1 and ten 0.
Jonathan Epperl
Jonathan Epperl 2012-11-7
Jeez, that is way easier, why didn't you say so from the start ;-)
% 10 ones, 10 zeros
p = [true(1,10) false(1,10)];
% permute them randomly
rbits = p(randperm(20));
RB = [rbits' circshift(rbits',1 ) circshift(rbits',2) circshift(rbits',3)]';
% Now check whether your requirements are satisfied
while any(all(RB)|all(~RB))
% If they are, you won't get in here, if they ain't, then try another
% permutation
rbits = p(randperm(20));
RB = [rbits' circshift(rbits',1 ) circshift(rbits',2), circshift(rbits',3)]';
end
rbits

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Azzi Abdelmalek
Azzi Abdelmalek 2012-11-7
x=zeros(1,20)
x(1:10)=1
y=x(randperm(20))
c1=3:4:20
y(c1+1)=not(y(c1))
Sean de Wolski
Sean de Wolski 2012-11-7
编辑：Sean de Wolski 2012-11-7
How about:
x = [0 0 0 0 0 1 0 1 0 0 1 1 1 1 0 1 1 1 1 0]; %to test it
x = strrep(x,[0 0 0 0],[0 0 1 0]);
x = strrep(x,[1 1 1 1],[1 1 0 1]);
In general, x would just be:
x = rand(sz)>0.5;
Andrei Bobrov
Andrei Bobrov 2012-11-8
编辑：Andrei Bobrov 2012-11-8
a = randi([0 1],1,20);
p = conv(a,[1 1 1 1],'same');
a(p == 4) = 0;
a(p == 0) = 1;
add
a = ones(1,20);
a(randi(4,1,5) + 4*(0:4)) = 0;
a(conv(a,[1 1 1 1],'same') == 4) = 0;
or
ii = randi(3,1,20);
ii = ii(1:find(cumsum(ii) >= 20,1,'first'));
out1 = cell(size(ii));
out1(1:2:end) = arrayfun(@(x)ones(1,x),ii(1:2:end),'un',0);
out1(2:2:end) = arrayfun(@(x)zeros(1,x),ii(2:2:end),'un',0);
out = cell2mat(out1);
out = out(1:20);
Matt Fig
Matt Fig 2012-11-8
编辑：Matt Fig 2012-11-8
Some methods proposed here don't seem to generate 'random' vectors. Here is an approach that is perhaps 'more random' than some, yet faster than Walter's brute force (especially for larger vectors). As all of the functions called are built-ins, it works pretty fast....
X = rand(1,10000)>.5;
S = {strfind(X,[1 1 1 1]),strfind(X,[0 0 0 0])};
while ~all(cellfun('isempty',S))
for ii = 0:1
for jj = 1:length(S{ii+1})
X(S{ii+1}(jj) + randperm(4,1)-1) = ii;
end
end
S = {strfind(X,[1 1 1 1]),strfind(X,[0 0 0 0])};
end

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