How to resolve the algorithm Knight's tour step by step in the Sidef programming language

Published on 12 May 2024 09:40 PM
#Sidef

How to resolve the algorithm Knight's tour step by step in the Sidef programming language

Table of Contents

Problem Statement

Problem: you have a standard 8x8 chessboard, empty but for a single knight on some square. Your task is to emit a series of legal knight moves that result in the knight visiting every square on the chessboard exactly once. Note that it is not a requirement that the tour be "closed"; that is, the knight need not end within a single move of its start position. Input and output may be textual or graphical, according to the conventions of the programming environment. If textual, squares should be indicated in algebraic notation. The output should indicate the order in which the knight visits the squares, starting with the initial position. The form of the output may be a diagram of the board with the squares numbered according to visitation sequence, or a textual list of algebraic coordinates in order, or even an actual animation of the knight moving around the chessboard. Input: starting square Output: move sequence

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Knight's tour step by step in the Sidef programming language

Source code in the sidef programming language

var board = []
var I = 8
var J = 8
var F = (I*J > 99 ? '%3d' : '%2d')

var (i, j) = (I.irand, J.irand)

func from_algebraic(square) {
     if (var match = square.match(/^([a-z])([0-9])\z/)) {
         return(I - Num(match[1]), match[0].ord - 'a'.ord)
     }
     die "Invalid block square: #{square}"
}

func possible_moves(i,j) {
    gather {
        for ni,nj in [
            [i-2,j-1], [i-2,j+1], [i-1,j-2], [i-1,j+2],
            [i+1,j-2], [i+1,j+2], [i+2,j-1], [i+2,j+1],
        ] {
            if ((ni ~~ ^I) && (nj ~~ ^J) && !board[ni][nj]) {
                take([ni, nj])
            }
        }
    }
}

func to_algebraic(i,j) {
    ('a'.ord + j).chr + Str(I - i)
}

if (ARGV[0]) {
    (i, j) = from_algebraic(ARGV[0])
}

var moves = []
for move in (1 .. I*J) {
    moves << to_algebraic(i, j)
    board[i][j] = move
    var min = [9]
    for target in possible_moves(i, j) {
        var (ni, nj) = target...
        var nxt = possible_moves(ni, nj).len
        if (nxt < min[0]) {
            min = [nxt, ni, nj]
        }
    }

    (i, j) = min[1,2]
}

say (moves/4 -> map { .join(', ') }.join("\n") + "\n")

for i in ^I {
    for j in ^J {
        (i%2 == j%2) && print "\e[7m"
        F.printf(board[i][j])
        print "\e[0m"
    }
    print "\n"
}

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