By : user2950511
Date : November 17 2020, 11:55 AM

With these it helps The following solution finds all the places in B where zeroes can go (T0), and where ones can go (T1). Then it loops through all those permutations, producing each mapping. code :
A = [0 0 0 1 1 0 1 0];
B = [1 1 0 0 0 0 0 1];
l = length(A);
T0 = perms(find(B==0)); % all targets for zeroes
T1 = perms(find(B==1)); % all targets for ones
s0 = find(A==0); % source of zeroes
s1 = find(A==1); % source of ones
nT0 = size(T0,1);
nT1 = size(T1,1);
q = zeros(1,l);
for i = 1:nT0
t0 = T0(i,:); % current target for zeroes
for j = 1:nT1
t1 = T1(j,:); % current target for ones
q(s0) = t0;
q(s1) = t1
% q has a mapping
end
end
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Find all possible injective mappings of 2 vectors
By : NotSoGreat
Date : March 29 2020, 07:55 AM
should help you out I have 2 vectors: , You basically need to compute all permutations of the range vector B: code :
perms(B).'
aux = perms(B);
mappings = unique(aux(:,1:length(A)),'rows').';
>> reshape([repmat(A.',1,size(mappings,2)); mappings],length(A),[])
ans =
1 5 1 5 1 4 1 4 1 3 1 3
2 4 2 3 2 5 2 3 2 4 2 5

find the vector with the largest size among group of vectors c++
By : stendu
Date : March 29 2020, 07:55 AM
will be helpful for those in need The comparison functor object passed to std::max_element should return true if the first operand is less than the second one. Your comparison has this the wrong way around. You need code :
bool Longest(const vector<int> &A, const vector<int> &B)
{
return A.size() < B.size();
}

Is LSH about transforming vectors to binary vectors for hamming distance?
By : Ajay Acharya
Date : March 29 2020, 07:55 AM

Efficient solution to find the number of unique binary vectors
By : Johnny Richie Kasali
Date : March 29 2020, 07:55 AM
I wish this helpful for you I'd use the inclusionexclusion principle. You want to know the cardinality of the union of the set. For your example you have: code :
N(X11X  1XX1  11XX) = N(X11X) + N(1XX1) + N(11XX) 
N(X11X && 1XX1)  N(X11X && 11XX)  N(1XX1 && 11XX) +
N(X11X && 1XX1 && 11XX)
N(X11X && 1XX1) = 1 * 1 * 1 * 1 = 1.

What is the fastest way to find orphans between two large (size ~900K ) Vectors of Strings in Java?
By : Bruno Accorsi
Date : March 29 2020, 07:55 AM
Hope that helps I'm currently working on a Java program that is required to handle large amounts of data. I have two Vectors... , Use a HashSet for the lookups. Explanation:



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