The solution I propose to solve this challenge is based on repeating the algorithm several times while recordig valid solutions. As solution is not unique, I expect the algorithm to throw more than 1 solution. Each of them will have a score based on neighbours affinity. I’ll call an ‘attempt‘ to a complete run trying … Read more
To generate one permutation use random.shuffle and store a copy of the result. Repeat this operation in a loop and each time check for duplicates (there probably won’t be any though). Once you have 5000 items in your result set, stop. To address the point in the comment, Python’s random module is based on the … Read more
This web page looks promising. def next_permutation(seq, pred=cmp): “””Like C++ std::next_permutation() but implemented as generator. Yields copies of seq.””” def reverse(seq, start, end): # seq = seq[:start] + reversed(seq[start:end]) + \ # seq[end:] end -= 1 if end <= start: return while True: seq[start], seq[end] = seq[end], seq[start] if start == end or start+1 == … 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
itertools.permutations(my_list, 3)
Here is a simple solution to compute the nth permutation of a string: function string_nth_permutation(str, n) { var len = str.length, i, f, res; for (f = i = 1; i <= len; i++) f *= i; if (n >= 0 && n < f) { for (res = “”; len > 0; len–) { … Read more
In C++ you can use std::next_permutation to go through permutations one by one. You need to sort the characters alphabetically before calling std::next_permutation for the first time: cin>>anagrama; int len = strlen(anagrama); sort(anagrama, anagrama+len); do { cout << anagrama << endl; } while (next_permutation(anagrama, anagrama+len)); Here is a demo on ideone. If you must implement … Read more
You need to understand the mathematical theory of permutation cycles, also known as “orbits” (it’s important to know both “terms of art” since the mathematical subject, the heart of combinatorics, is quite advanced, and you may need to look up research papers which could use either or both terms). For a simpler introduction to the … Read more
Python Number of characters: 548 482 425 421 416 413 408 from operator import * n=len def C(N,T): R=range(1<<n(N));M=[{}for i in R];p=1 for i in range(n(N)):M[1<<i][1.*N[i]]=”%d”%N[i] while p: p=0 for i in R: for j in R: m=M[i|j];l=n(m) if not i&j:m.update((f(x,y),”(“+s+o+t+”)”)for(y,t)in M[j].items()if y for(x,s)in M[i].items() for(o,f)in zip(‘+-*/’,(add,sub,mul,div))) p|=l<n(m) return min((abs(x-T),e)for t in M for(x,e)in t.items())[1] … Read more