Solve a Hidato puzzle: Difference between revisions
Rename Perl 6 -> Raku, alphabetize, minor clean-up
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17 7 6 3
5 4</pre>
=={{header|C sharp}}==
The same solver can solve Hidato, Holy Knight's Tour, Hopido and Numbrix puzzles.<br/>
The input can be an array of strings if each cell is one character. The length of the first row must be the number of columns in the puzzle.<br/>
Any non-numeric value indicates a no-go.<br/>
If there are cells that require more characters, then a 2-dimensional array of ints must be used. Any number < 0 indicates a no-go.<br/>
The puzzle can be made circular (the end cell must connect to the start cell). In that case, no start cell needs to be given.
<lang csharp>using System.Collections;
using System.Collections.Generic;
using static System.Console;
using static System.Math;
using static System.Linq.Enumerable;
public class Solver
{
private static readonly (int dx, int dy)[]
//other puzzle types elided
hidatoMoves = {(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)};
private (int dx, int dy)[] moves;
public static void Main()
{
Print(new Solver(hidatoMoves).Solve(false, new [,] {
{ 0, 33, 35, 0, 0, -1, -1, -1 },
{ 0, 0, 24, 22, 0, -1, -1, -1 },
{ 0, 0, 0, 21, 0, 0, -1, -1 },
{ 0, 26, 0, 13, 40, 11, -1, -1 },
{ 27, 0, 0, 0, 9, 0, 1, -1 },
{ -1, -1, 0, 0, 18, 0, 0, -1 },
{ -1, -1, -1, -1, 0, 7, 0, 0 },
{ -1, -1, -1, -1, -1, -1, 5, 0 }
}));
}
public Solver(params (int dx, int dy)[] moves) => this.moves = moves;
public int[,] Solve(bool circular, params string[] puzzle)
{
var (board, given, count) = Parse(puzzle);
return Solve(board, given, count, circular);
}
public int[,] Solve(bool circular, int[,] puzzle)
{
var (board, given, count) = Parse(puzzle);
return Solve(board, given, count, circular);
}
private int[,] Solve(int[,] board, BitArray given, int count, bool circular)
{
var (height, width) = (board.GetLength(0), board.GetLength(1));
bool solved = false;
for (int x = 0; x < height && !solved; x++) {
solved = Range(0, width).Any(y => Solve(board, given, circular, (height, width), (x, y), count, (x, y), 1));
if (solved) return board;
}
return null;
}
private bool Solve(int[,] board, BitArray given, bool circular,
(int h, int w) size, (int x, int y) start, int last, (int x, int y) current, int n)
{
var (x, y) = current;
if (x < 0 || x >= size.h || y < 0 || y >= size.w) return false;
if (board[x, y] < 0) return false;
if (given[n - 1]) {
if (board[x, y] != n) return false;
} else if (board[x, y] > 0) return false;
board[x, y] = n;
if (n == last) {
if (!circular || AreNeighbors(start, current)) return true;
}
for (int i = 0; i < moves.Length; i++) {
var move = moves[i];
if (Solve(board, given, circular, size, start, last, (x + move.dx, y + move.dy), n + 1)) return true;
}
if (!given[n - 1]) board[x, y] = 0;
return false;
bool AreNeighbors((int x, int y) p1, (int x, int y) p2) => moves.Any(m => (p2.x + m.dx, p2.y + m.dy).Equals(p1));
}
private static (int[,] board, BitArray given, int count) Parse(string[] input)
{
(int height, int width) = (input.Length, input[0].Length);
int[,] board = new int[height, width];
int count = 0;
for (int x = 0; x < height; x++) {
string line = input[x];
for (int y = 0; y < width; y++) {
board[x, y] = y < line.Length && char.IsDigit(line[y]) ? line[y] - '0' : -1;
if (board[x, y] >= 0) count++;
}
}
BitArray given = Scan(board, count, height, width);
return (board, given, count);
}
private static (int[,] board, BitArray given, int count) Parse(int[,] input)
{
(int height, int width) = (input.GetLength(0), input.GetLength(1));
int[,] board = new int[height, width];
int count = 0;
for (int x = 0; x < height; x++)
for (int y = 0; y < width; y++)
if ((board[x, y] = input[x, y]) >= 0) count++;
BitArray given = Scan(board, count, height, width);
return (board, given, count);
}
private static BitArray Scan(int[,] board, int count, int height, int width)
{
var given = new BitArray(count + 1);
for (int x = 0; x < height; x++)
for (int y = 0; y < width; y++)
if (board[x, y] > 0) given[board[x, y] - 1] = true;
return given;
}
private static void Print(int[,] board)
{
if (board == null) {
WriteLine("No solution");
} else {
int w = board.Cast<int>().Where(i => i > 0).Max(i => (int?)Ceiling(Log10(i+1))) ?? 1;
string e = new string('-', w);
foreach (int x in Range(0, board.GetLength(0)))
WriteLine(string.Join(" ", Range(0, board.GetLength(1))
.Select(y => board[x, y] < 0 ? e : board[x, y].ToString().PadLeft(w, ' '))));
}
WriteLine();
}
}</lang>
{{out}}
<pre>
32 33 35 36 37 -- -- --
31 34 24 22 38 -- -- --
30 25 23 21 12 39 -- --
29 26 20 13 40 11 -- --
27 28 14 19 9 10 1 --
-- -- 15 16 18 8 2 --
-- -- -- -- 17 7 6 3
-- -- -- -- -- -- 5 4
</pre>
=={{header|C++}}==
Line 634 ⟶ 780:
02 16 19 22 23 27 37 36 32
01 20 21 24 25 26 35 34 33
</pre>
Line 2,757:
17 7 6 3
5 4
=={{header|Phix}}==
Line 3,495 ⟶ 3,387:
_ _ _ _ 17 7 6 3
_ _ _ _ _ _ 5 4</pre>
=={{header|Raku}}==
(formerly Perl 6)
This uses a Warnsdorff solver, which cuts down the number of tries by more than a factor of six over the brute force approach. This same solver is used in:
* [[Solve a Hidato puzzle#Perl_6|Solve a Hidato puzzle]]
* [[Solve a Hopido puzzle#Perl_6|Solve a Hopido puzzle]]
* [[Solve a Holy Knight's tour#Perl_6|Solve a Holy Knight's tour]]
* [[Solve a Numbrix puzzle#Perl_6|Solve a Numbrix puzzle]]
* [[Solve the no connection puzzle#Perl_6|Solve the no connection puzzle]]
<lang perl6>my @adjacent = [-1, -1], [-1, 0], [-1, 1],
[ 0, -1], [ 0, 1],
[ 1, -1], [ 1, 0], [ 1, 1];
solveboard q:to/END/;
__ 33 35 __ __ .. .. ..
__ __ 24 22 __ .. .. ..
__ __ __ 21 __ __ .. ..
__ 26 __ 13 40 11 .. ..
27 __ __ __ 9 __ 1 ..
.. .. __ __ 18 __ __ ..
.. .. .. .. __ 7 __ __
.. .. .. .. .. .. 5 __
END
sub solveboard($board) {
my $max = +$board.comb(/\w+/);
my $width = $max.chars;
my @grid;
my @known;
my @neigh;
my @degree;
@grid = $board.lines.map: -> $line {
[ $line.words.map: { /^_/ ?? 0 !! /^\./ ?? Rat !! $_ } ]
}
sub neighbors($y,$x --> List) {
eager gather for @adjacent {
my $y1 = $y + .[0];
my $x1 = $x + .[1];
take [$y1,$x1] if defined @grid[$y1][$x1];
}
}
for ^@grid -> $y {
for ^@grid[$y] -> $x {
if @grid[$y][$x] -> $v {
@known[$v] = [$y,$x];
}
if @grid[$y][$x].defined {
@neigh[$y][$x] = neighbors($y,$x);
@degree[$y][$x] = +@neigh[$y][$x];
}
}
}
print "\e[0H\e[0J";
my $tries = 0;
try_fill 1, @known[1];
sub try_fill($v, $coord [$y,$x] --> Bool) {
return True if $v > $max;
$tries++;
my $old = @grid[$y][$x];
return False if +$old and $old != $v;
return False if @known[$v] and @known[$v] !eqv $coord;
@grid[$y][$x] = $v; # conjecture grid value
print "\e[0H"; # show conjectured board
for @grid -> $r {
say do for @$r {
when Rat { ' ' x $width }
when 0 { '_' x $width }
default { .fmt("%{$width}d") }
}
}
my @neighbors = @neigh[$y][$x][];
my @degrees;
for @neighbors -> \n [$yy,$xx] {
my $d = --@degree[$yy][$xx]; # conjecture new degrees
push @degrees[$d], n; # and categorize by degree
}
for @degrees.grep(*.defined) -> @ties {
for @ties.reverse { # reverse works better for this hidato anyway
return True if try_fill $v + 1, $_;
}
}
for @neighbors -> [$yy,$xx] {
++@degree[$yy][$xx]; # undo degree conjectures
}
@grid[$y][$x] = $old; # undo grid value conjecture
return False;
}
say "$tries tries";
}</lang>
=={{header|REXX}}==
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