Python uses the Backtracking Method subset tree template to solve the Sorting Problem example, python backtracking

Python uses the Backtracking Method subset tree template to solve the Sorting Problem example, python backtracking

This example describes how to use a subset tree template based on the Backtracking Method to Solve the Sorting Problem in Python. We will share this with you for your reference. The details are as follows:

Problem

Select m elements from n elements for arrangement. Each element can be repeated for up to r times. Here, m is [2, n], r is [1, m].

For example, three elements can be selected from four elements for arrangement. Each element can be repeated for r times at most.

Analysis

The length of the solution x is fixed, m.

For the solution x, the first row is the element x [0] With 0th positions, and the second row is the element x [1] with 1st positions. We regard the latter as a state of the former, that is, x [1] is a state of x [0 !!

Generally, we think of x [k] as a state in state space a of x [k-1], and all we have to do is traverse all States of a [k-1.

Apply the subset tree template.

Code

''' Selection problem: Select m elements from n elements for sorting. Each element can be repeated up to r times. Here, m is [2, n], r is [1, m]. Author: hhh5460 time: June 2, 2017 statement: This algorithm copyright belongs to hhh5460 all '''n' = 4a = ['A', 'B', 'C', 'C ', 'D'] m = 3 # pick 3 r = 2 from 4 # Each element can be repeated 2x = [0] * m # A solution (m yuan 0-1 array) X = [] # A group of solutions # conflict Detection def conflict (k): global n, r, x, X, a # The element x [k] in the decomposition cannot exceed r if x [: k + 1]. count (x [k])> r: return True return False # No conflict # use a subset tree template to implement the selection problem def perm (k): # Reach the k element global n, m, a, x, X if k = m: # print (x) # X. append (x [:]) # Save (a solution) else: for I in a: # traverse X [k-1] State Space a, other things are handed over to the pruning function! X [k] = I if not conflict (k): # pruning perm (k + 1) # testing perm (0) # sorting from x [0]