Referring to the source code of scipy.misc.comb, the update routine of the result is: val = 1 for j in xrange(min(k, N-k)): val = (val*(N-j))//(j+1) return val whereas the update routine you suggested is: ntok = 1 ktok = 1 for t in xrange(1, min(k, n – k) + 1): ntok *= n ktok *= … Read more
(Because it was unclear in the original version of the question whether you meant to remove a subsequence or an unordered list, I’ve updated my answer to address both cases.) 1. Removing a sub-sequence in order We get a sequence of values, e.g. [1,5,1,3,4,2,3,2,1,3,1,2], and a sub-sequence of values to be removed from the first … Read more
What you’re looking for is called a derangement (another lovely Latinate word to know, like exsanguination and defenestration). The fraction of all permutations which are derangements approaches 1/e = approx 36.8% — so if you are generating random permutations, just keep generating them, and there’s a very high probability that you’ll find one within 5 … Read more
Here’s an ancient algorithm which is exact and doesn’t overflow unless the result is to big for a long long unsigned long long choose(unsigned long long n, unsigned long long k) { if (k > n) { return 0; } unsigned long long r = 1; for (unsigned long long d = 1; d <= … Read more
itertools.product is made for this: >>> import itertools >>> [“”.join(seq) for seq in itertools.product(“01”, repeat=2)] [’00’, ’01’, ’10’, ’11’] >>> [“”.join(seq) for seq in itertools.product(“01”, repeat=3)] [‘000’, ‘001’, ‘010’, ‘011’, ‘100’, ‘101’, ‘110’, ‘111’]
This is a well-studied problem of generating all k-subsets, or k-combinations, which can be easily done without recursion. The idea is to have array of size k keeping sequence of indices of elements from the input array (which are numbers from 0 to n – 1) in increasing order. (Subset then can be created by … Read more
Consider this solution using the NDGRID function: sets = {[1 2], [1 2], [4 5]}; [x y z] = ndgrid(sets{:}); cartProd = [x(:) y(:) z(:)]; cartProd = 1 1 4 2 1 4 1 2 4 2 2 4 1 1 5 2 1 5 1 2 5 2 2 5 Or if you want … Read more
K (Wikipedia), 15 characters: p:{x{+’:x,0}\1} Example output: p 10 (1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 … Read more
Your overall approach is sound. I’m pretty sure the problem lies with your make_canonical function. You can try printing out the hands with num_cards set to 3 or 4 and look for equivalencies that you’ve missed. I found one, but there may be more: # The inputs are equivalent and should return the same value … Read more