The number of permutations of ten women is Of course the same technique works with ten ladies, except that we need ten slots and ten steps instead of three. By the multiplication principle, the number of ways of completing the entire procedure by doing all three steps is Finally, after filling the first two slots, there is only one lady left, so there is just one way to fill the last slot. After this first step is complete, there are then two ways of filling the second slot, as there are only two ladies left. We have three ways to place a lady in the first slot. In arranging the ladies there are three corresponding steps we must fill the first slot, then the second slot, and then the third. Going back to the case of three ladies, we imagine that there are three slots where the ladies may stand: The way to do this is to divide the procedure of lining up the women into steps, and then use the multiplication principle. Thus it appears that we must devise a method of counting the permutations without actually writing them all down. Possibilities referring to the ladies by the letters A, B, C, we discover theĮach one of the arrangements of the women is called a permutation altogether there are six possible permutations.īut what if there were ten ladies? Now the problem becomes more serious, as it would take forever to write down all the possibilities, and even if you tried you would never know whether you left any out. Three women, it is not difficult to ponder a moment and write down all the In how many different ways can you line up the women? As there are only Photographer you can arrange the women from left to right in the order you Set j = last-2 and find first j such that a =.Of your three friends, Alice, Betty, and Cindy, lined up in a row. So 6 is next larger and 2345(least using numbers other than 6) Find lexicogrpahically least way to extend the new aĬonsider example array state of for sorted Īfter 56432(treat as number) ->nothing larger than 6432(using 6,4,3,2) beginning with 5.Increase a by smallest feasible amount.Find largest j such that a can be increased.There are n! permutations at most and hasNextPermutation(.) runs in O(n) time complexity This is the asymptotically optimal way O(n*n!) of generating permutations after initial sorting. It was enough when I needed it, but it's no itertools.permutations by a long shot. This method is non-recursive, but it is slightly slower on my computer and xrange raises an error when n! is too large to be converted to a C long integer (n=13 for me). NewPermutation.append(available.pop(index)) This way the numbers 0 through n!-1 correspond to all possible permutations in lexicographic order. You have n choices for the first item, n-1 for the second, and only one for the last, so you can use the digits of a number in the factorial number system as the indices.
I used an algorithm based on the factorial number system- For a list of length n, you can assemble each permutation item by item, selecting from the items left at each stage. Indices, indices = indices, indicesĪnd another, based on itertools.product: def permutations(iterable, r=None):įor indices in product(range(n), repeat=r): If len(elements) AB AC AD BA BC BD CA CB CD DA DB DC Use itertools.permutations from the standard library: import itertoolsĪdapted from here is a demonstration of how itertools.permutations might be implemented: def permutations(elements):
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