The incumbent was found after a bit less than four minutes (which will be relevant as I explore other models, in future posts). Permutations II - Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order. For example, the three elements 1,2,3 can be arranged into 3 6 unique permutations: 1,2,3, 1,3,2, 2,1,3. A permutation is an arrangement of elements. Update: A longer run, using MIP emphasis 3 (which focuses on improving the lower bound), still had a gap of 33% after four hours. LeetCode 47: Permutations II (Medium) Problem Statement: Given a list of integers that may contain duplicates, return all possible unique permutations of those integers, in any order. Given an array nums of distinct integers, return all the possible permutations. So we're contending with a somewhat weak bound. At each level, the branching factor is reduced by 1 so it will go like N, N-1, N-2.1 starting from root. In this case, each path will go upto N depth in solution tree. Time - O(n (P(N, k))) - In worst case, all numbers in the input array will be unique. (The last two minutes of that five minute run only closed the gap from about 57.5% to 56.5%.) I'm pretty sure the actual optimal value will be a lot closer to 5.5 million that to the last lower bound in the five minute run (2,479,745). Space - O(N) - Each call stack depth would be N where N is the length of input list. ![]() Unfortunately, after five minutes the gap was still 56.55%, and closing very slowly. It may be that Nate's frequency data, which I'm using, differs slightly from the frequency data Hardmath123 used.) I get a value of 5,510,008 for that solution. (Speaking of which, Hardmath123 quoted an objective value of 5,499,341 and posted a layout. That's a bit worse than the solution Hardmath123 got in the original post. After five minutes, the incumbent solution had objective value 5,706,873. At any rate, I did the five minute run with MIP emphasis 2, which emphasizes proving optimality. Someday maybe I'll figure out why it's ignoring those carefully defined SOS1 weights. It did that even if skipped the branching priorities, which irks me a bit. I included both the branching priorities and the SOS1 constraints, but the CPLEX presolver eliminated all the SOS1 constraints as "redundant". How does this model do? I ran it for five minutes on a decent desktop PC (using four threads). The starting point for all the math programming formulations is a matrix of binary variables $x_.$$ So I'll start by discussing a couple of mixed integer linear program formulations. I'll in fact try out a quadratic model subsequently, but my inclination is always to try to linearize anything that can't outrun or outfight me. As I noted there, Nate Brixius correctly characterized the problem as a quadratic assignment problem (QAP). Self.This continues my previous post about the problem of optimally laying out a one-dimensional typewriter keyboard, where "optimally" is taken to mean minimizing the expected amount of lateral movement to type a few selected books. A midjourney generated image of a fruit salad bird with a 3:2 aspect ratio. We already know that 3 out of 16 gave us 3,360 permutations. You can use Permutation Prompts to create combinations and permutations. Going back to our pool ball example, let's say we just want to know which 3 pool balls are chosen, not the order. Note: The reason that we have only 2 choices instead of 3, is that there is a duplicate in the given input. ![]() Suppose that we pick the number 1, now the remaining numbers would become 1, 2. # Recursively search for more permutations assume that the order does matter (ie permutations), then alter it so the order does not matter. Given the input of 1, 1, 2, at the first stage, we have 2 choices to pick a number as the first number in the final permutation, i.e. # Skip this number if it is already used or if it is a duplicate of the previous number and the previous number is not used continue # Mark the number as used and add it to the current permutation ![]() If used or (i > 0 and nums = nums and not used): Return # Otherwise, try adding each unused number to the current permutation and recursively search for more permutations for i in range(len(nums)): Self.backtrack(permutations,, nums, * len(nums))ĭef backtrack( self, permutations: List], currentPermutation: List, nums: List, used: List): # If the current permutation is the same length as the original array, we have found a permutation if len(currentPermutation) = len(nums): # Use a backtracking search to generate the permutations # Sort the array so that we can skip duplicates when generating permutations Output: ,]ġ List]: # Initialize an empty list to store the permutations Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order*.*
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