![]() ![]() Therefore we start with all digits in ascending order and permute until all they reverse direction. Hence we only change the position of a “digit” when everything to the right is in descending order. And when are there no more permutations of the last 3 digits? When the last 3 digits are in descending order. So when do we finally “use” the 1? When there are only no more permutations of the last 3 digits. In the example above we see that 1 stays as the first number for a long time as there are many reorderings of the last 3 digits which increase the permutation by a smaller amount. What can we do when we are given other elements? Well we can simply transform them into numbers as “indexes” into the elements given (like array indices). This is easy to do with our examples so far – we’ve used digits and so we can think of each permutation as numbers with all the ordering goodness that comes from that. Once we do that we can just start with the “ smallest” permutation and increment it minimally till we reach the “ largest” permutation. Public void permutation1(int a, int n) !īoth problems can be solved by defining an ordering of permutations. ![]() This is pretty much a direct definition of n!=n × (n-1)! and is very simple to implement: Remove each element from the n elements one at a time, then append it to the (n-1)! remaining permutations.Given we know there are n! permutations of elements we are lead directly to a basic backtracking algorithm for permutations – This is, of course, the definition of n!. For example these are all the permutations of three elements:īasically we pick the first element from the n items, the second from the remaining (n-1) items, the third from the remaining (n-2) items and so on. Brief RecapĪ “permutation”, as we may remember from high school, is an re-ordering of elements. The last algorithm I will describe will be my favorite – simple, elegant, and efficient to use. Right now we will take a look at a few key algorithms and, by the end, we will have a good grasp of what’s out there and how to think about permutation algorithms. Also there are dozens of algorithms which are both fascinating and fun! Practically speaking, we encounter permutations less often so why should we spend time on it? Well because it is a fundamental problem in computing, it provides a basis for backtracking algorithms, and we can use it for computing exact answers to some problems. Continuing on last week’s theme, this week I’d like to share my favorite methods for generating permutations of elements! ![]()
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