#### Category : backtracking

i am new to programing an trying to learn it. As I am always looking for new intressting python tutorials i have come accross an tutorial for a sudoku solver using backtracking technique. https://www.techwithtim.net/tutorials/python-programming/sudoku-solver-backtracking/ Since this sudoku solver only shows one single solution, I was wondering if there was a way to set a count ..

I am working on N Queens Problem, I encountered this strange phenomenon that my algorithm works fine when the input matrix is [ [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0] ] but fails to output when my matrix is [ [0]*4 ]*4 Code: https://github.com/Surya-kiran-paluru/N-Queens-Problem/blob/main/test.py "My code ..

The problem description is as follows: The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other. Given an integer n, return all distinct solutions to the n-queens puzzle. You may return the answer in any order. So no two queens can ..

I have a piece of Python code that implements a recursive backtracking algorithm for solving the famous N-Queens problem in chess. def Backtrack(board, col): if col >= N: return True for i in range(N): if (ld[i – col + N – 1] != 1 and rd[i + col] != 1) and cl[i] != 1: board[i][col] ..

I was trying to build a solution to the problem of generating all possible combination of k of the length n ex: k = 4, n = 2 output : [1, 1] [1, 2] [1, 3] [1, 4] [2, 1] [2, 2] [2, 3] [2, 4] [3, 1] [3, 2] [3, 3] [3, 4] [4, ..

I am writing a Python program that will generate a random table, remove numbers so that it only has one solution and then solving the sudoku program using the backtracking algorithm. However, it is not removing numbers correctly so that it only has one solution. I think that the problem is somewhere in the solve_function ..

I am finding it hard to solve this problem that returns an optimized list of coins >>> backtracking_exchange(56, [20, 10, 5, 1]) [2, 1, 1, 1] This is my code so far: def backtracking_exchange(amount, denominations): res = [] def soln(amount,denominations,c): used_coins = [] if amount == 0: return c for coin in denominations: if amount ..