So the total steps in this invocation equal (IV) => The sequence is unchanged. The recursive calls "generate(n - 1, A)" result in a right-rotation. The swaps that get executed in this invocation result in a right-rotation, see (III). The recursive calls "generate(n - 1, A)" leave the ordering unchanged (because it invokes generate with first argument being 1). Trivially, the ordering is unchanged after invoking the function. Now we can see why (I) and (II) are true: Taking the element at 0 out before the rotation, then after the rotation swapping it with the new element at 0 does not change the outcome (if rotating n times). For example, 1,1,2 have the following unique. If you have a sequence of length 5, then rotate it 5 times, it ends up unchanged. Given a collection of numbers that might contain duplicates, return all possible unique permutations. (IV) This series of steps leaves the sequence in the exact same ordering as before: Repeat n times: (III) This series of swaps result in a rotation to the right by one position: A Aįor example try it with sequence ABCD: A A: DBCA Just swap the elements at 0 and (n-1) in each iteration to produce a unique set of elements.įinally, let's see why the initial statements are true: Rotate-right So the element at index 0 will always be a different element automatically. The recursive call "generate(n - 1, A)" has rotated the elements right. So the for-loop can iteratively swap the element at i=0.(n-1) with the element at (n - 1) and will have called "generate(n - 1, A)" each time with another element missing. The recursive call "generate(n - 1, A)" does not change the order. (II) where n is even, rotates the elements to the right, for example ABCD becomes DABC. (I) where n is odd, leaves the elements in the exact same ordering when it is finished. Please just assume that these statements are true for a moment (i'll show that later): Each invocation of the "generate" function However, I think it is hard to understand it, so came up with an explanation that is hopefully easier to understand: I found an article that tries to explain it here: Why does Heap's algorithm work? Looking at the diagram on the Wikipedia page might help. You can prove that both of these facts are true using induction, although that doesn't provide any intuition as to why it's true. Whereas if the vector is of odd length, it will be simply swap the first and last elements: → Given a vector of even length, a full iteration of Heap's algorithm will rearrange the elements according to the rule → To see how Heap's algorithm works, you need to look at what a full iteration of the loop does to the vector, in both even and odd cases. See for the correct algorithm (at least, it's correct today) or see the discussion at Heap's algorithm permutation generator In either case, it does unnecessary work. Reverse the part of the permutation to the right of where that element was.īoth steps (1) and (3) are worst-case O(n), but it is easy to prove that the average time for those steps is O(1).Īn indication of how tricky Heap's algorithm is (in the details) is that your expression of it is slightly wrong because it does one extra swap the extra swap is a no-op if n is even, but significantly changes the order of permutations generated when n is odd. Swap that element with the smallest element to its right which is larger than it. Given some permutation, find the next one by:įinding the rightmost element which is smaller than the element to its right. The lexicographic order algorithm is extremely simple to describe. There is a much more intuitive algorithm which will produce permutations in lexicographical order although it is amortized O(1) (per permutation) instead of O(1), it is not noticeably slower in practice, and it is much easier to derive on the fly. The equation is nPr = (n!)/(n-k)!, where n is the number of participants, and k is the number of people that receive medals.įor more information regarding your Texas Instruments TI-Nspire graphing calculator, please refer to the guidebooks section.Heap's algorithm is probably not the answer to any reasonable interview question. This example would be calculated using permutations, since the order of the results matter. How many possible permutations of first, second, and third place be awarded in a race between 10 people? The equation is nCr = (n!)/((n-k)!k!), where n is the number of participants, and k is the number of people we are choosing. This example would be calculated using combinations, since the order of the people being chosen doesn't matter. How many possible combinations of picking a team of 3 people out of a group of 10 people? The examples below will demonstrate how to calculate combinations and permutations using the TI-Nspire family handhelds and computer software. How do I calculate combinations and permutations on the TI-Nspire family products? Solution 29710: Calculating Combinations and Permutations on the TI-Nspire™ Family Products.
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