Solve a Hopido puzzle: Difference between revisions
| Line 1,472: | Line 1,472: | ||
if M[R,C] = 0 then print(" ") else printf("%2d ", M[R,C]) end, |
if M[R,C] = 0 then print(" ") else printf("%2d ", M[R,C]) end, |
||
if C = NC then nl end |
if C = NC then nl end |
||
end. |
end. |
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| ⚫ | |||
| ⚫ | |||
Output: |
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| ⚫ | |||
6 9 12 5 8 11 4 |
6 9 12 5 8 11 4 |
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14 17 20 25 16 19 22 |
14 17 20 25 16 19 22 |
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13 18 21 |
13 18 21 |
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1 |
1 |
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CPU time 0.019 seconds |
CPU time 0.019 seconds</pre> |
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| ⚫ | |||
=={{header|Prolog}}== |
=={{header|Prolog}}== |
||
Revision as of 13:22, 30 May 2022
You are encouraged to solve this task according to the task description, using any language you may know.
Hopido puzzles are similar to Hidato. The most important difference is that the only moves allowed are: hop over one tile diagonally; and over two tiles horizontally and vertically. It should be possible to start anywhere in the path, the end point isn't indicated and there are no intermediate clues. Hopido Design Post Mortem contains the following:
"Big puzzles represented another problem. Up until quite late in the project our puzzle solver was painfully slow with most puzzles above 7×7 tiles. Testing the solution from each starting point could take hours. If the tile layout was changed even a little, the whole puzzle had to be tested again. We were just about to give up the biggest puzzles entirely when our programmer suddenly came up with a magical algorithm that cut the testing process down to only minutes. Hooray!"
Knowing the kindness in the heart of every contributor to Rosetta Code, I know that we shall feel that as an act of humanity we must solve these puzzles for them in let's say milliseconds.
Example:
. 0 0 . 0 0 . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . 0 0 0 0 0 . . . 0 0 0 . . . . . 0 . . .
Extra credits are available for other interesting designs.
- Related tasks
- A* search algorithm
- Solve a Holy Knight's tour
- Knight's tour
- N-queens problem
- Solve a Hidato puzzle
- Solve a Holy Knight's tour
- Solve a Numbrix puzzle
- Solve the no connection puzzle
11l
<lang 11l>V neighbours = [[2, 2], [-2, 2], [2, -2], [-2, -2], [3, 0], [0, 3], [-3, 0], [0, -3]] V cnt = 0 V pWid = 0 V pHei = 0
F is_valid(a, b)
R -1 < a & a < :pWid & -1 < b & b < :pHei
F iterate(&pa, x, y, v)
I v > :cnt
R 1
L(i) 0 .< :neighbours.len
V a = x + :neighbours[i][0]
V b = y + :neighbours[i][1]
I is_valid(a, b) & pa[a][b] == 0
pa[a][b] = v
V r = iterate(&pa, a, b, v + 1)
I r == 1
R r
pa[a][b] = 0
R 0
F solve(pz, w, h)
V pa = [[-1] * h] * w
V f = 0
:pWid = w
:pHei = h
L(j) 0 .< h
L(i) 0 .< w
I pz[f] == ‘1’
pa[i][j] = 0
:cnt++
f++
L(y) 0 .< h
L(x) 0 .< w
I pa[x][y] == 0
pa[x][y] = 1
I 1 == iterate(&pa, x, y, 2)
R (1, pa)
pa[x][y] = 0
R (0, pa)
V r = solve(‘011011011111111111111011111000111000001000’, 7, 6) I r[0] == 1
L(j) 6
L(i) 7
I r[1][i][j] == -1
print(‘ ’, end' ‘’)
E
print(‘ #02’.format(r[1][i][j]), end' ‘’)
print()
E
print(‘No solution!’, end' ‘’)</lang>
- Output:
01 25 17 03
27 13 10 07 14 11 08
24 21 18 02 22 19 16
06 26 12 09 04
23 20 15
05
AutoHotkey
<lang AutoHotkey>SolveHopido(Grid, Locked, Max, row, col, num:=1, R:="", C:=""){ if (R&&C) ; if neighbors (not first iteration) { Grid[R, C] := ">" num ; place num in current neighbor and mark it visited ">" row:=R, col:=C ; move to current neighbor }
num++ ; increment num if (num=max) ; if reached end return map(Grid) ; return solution
if locked[num] ; if current num is a locked value { row := StrSplit((StrSplit(locked[num], ",").1) , ":").1 ; find row of num col := StrSplit((StrSplit(locked[num], ",").1) , ":").2 ; find col of num if SolveHopido(Grid, Locked, Max, row, col, num) ; solve for current location and value return map(Grid) ; if solved, return solution } else { for each, value in StrSplit(Neighbor(row,col), ",") { R := StrSplit(value, ":").1 C := StrSplit(value, ":").2
if (Grid[R,C] = "") ; a hole or out of bounds || InStr(Grid[R, C], ">") ; visited || Locked[num+1] && !(Locked[num+1]~= "\b" R ":" C "\b") ; not neighbor of locked[num+1] || Locked[num-1] && !(Locked[num-1]~= "\b" R ":" C "\b") ; not neighbor of locked[num-1] || Locked[num] ; locked value || Locked[Grid[R, C]] ; locked cell continue
if SolveHopido(Grid, Locked, Max, row, col, num, R, C) ; solve for current location, neighbor and value return map(Grid) ; if solved, return solution } } num-- ; step back for i, line in Grid for j, element in line if InStr(element, ">") && (StrReplace(element, ">") >= num) Grid[i, j] := 0 }
- --------------------------------
- --------------------------------
- --------------------------------
Neighbor(row,col){ return Trim( "" . "," row ":" col-3 . "," row ":" col+3 . "," row-3 ":" col . "," row+3 ":" col
. "," row+2 ":" col+2 . "," row+2 ":" col-2 . "," row-2 ":" col+2 . "," row-2 ":" col-2 , ",") }
- --------------------------------
map(Grid){ for i, row in Grid { for j, element in row line .= (A_Index > 1 ? "`t" : "") element map .= (map<>""?"`n":"") line line := "" } return StrReplace(map, ">") }</lang> Examples:<lang AutoHotkey>;-------------------------------- Grid := [["",0 ,0 ,"",0 ,0 ,""] ,[0 ,0 ,0 ,0 ,0 ,0 ,0] ,[0 ,0 ,0 ,0 ,0 ,0 ,0] ,["",0 ,0 ,0 ,0 ,0 ,""] ,["","",0 ,0 ,0 ,"",""] ,["","","",0 ,"","",""]]
- --------------------------------
- find locked cells, find max value
Locked := [] max := 1 for i, line in Grid for j, element in line if (element >= 0) max++ , list .= i ":" j "`n"
random, rnd, 1, %max% loop, parse, list, `n, `r if (A_Index = rnd) { row := StrSplit(A_LoopField, ":").1 col := StrSplit(A_LoopField, ":").2 Grid[row,col] := 1 Locked[1] := row ":" col "," Neighbor(row, col) break }
- --------------------------------
MsgBox, 262144, ,% SolveHopido(Grid, Locked, Max, row, col) return</lang>
Outputs:
17 24 16 25 22 8 11 21 7 10 20 13 2 5 14 1 4 15 18 23 9 19 26 12 3 6 27
C#
The same solver can solve Hidato, Holy Knight's Tour, Hopido and Numbrix puzzles.
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.
Any non-numeric value indicates a no-go.
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.
<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
hopidoMoves = {(-3,0),(0,-3),(0,3),(3,0),(-2,-2),(-2,2),(2,-2),(2,2)},
private (int dx, int dy)[] moves;
public static void Main()
{
Print(new Solver(hopidoMoves).Solve(false,
".00.00.",
"0000000",
"0000000",
".00000.",
"..000..",
"...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>
- Output:
-- 1 8 -- 2 7 -- 25 22 19 26 23 20 27 9 16 13 10 17 14 11 -- 4 24 21 3 6 -- -- -- 18 15 12 -- -- -- -- -- 5 -- -- --
C++
<lang cpp>
- include <vector>
- include <sstream>
- include <iostream>
- include <iterator>
- include <stdlib.h>
- include <string.h>
using namespace std;
struct node {
int val; unsigned char neighbors;
};
class nSolver { public:
nSolver()
{
dx[0] = -2; dy[0] = -2; dx[1] = -2; dy[1] = 2; dx[2] = 2; dy[2] = -2; dx[3] = 2; dy[3] = 2; dx[4] = -3; dy[4] = 0; dx[5] = 3; dy[5] = 0; dx[6] = 0; dy[6] = -3; dx[7] = 0; dy[7] = 3;
}
void solve( vector<string>& puzz, int max_wid )
{
if( puzz.size() < 1 ) return; wid = max_wid; hei = static_cast<int>( puzz.size() ) / wid; int len = wid * hei, c = 0; max = len; arr = new node[len]; memset( arr, 0, len * sizeof( node ) );
for( vector<string>::iterator i = puzz.begin(); i != puzz.end(); i++ ) { if( ( *i ) == "*" ) { max--; arr[c++].val = -1; continue; } arr[c].val = atoi( ( *i ).c_str() ); c++; }
solveIt(); c = 0; for( vector<string>::iterator i = puzz.begin(); i != puzz.end(); i++ ) { if( ( *i ) == "." ) { ostringstream o; o << arr[c].val; ( *i ) = o.str(); } c++; } delete [] arr;
}
private:
bool search( int x, int y, int w )
{
if( w > max ) return true;
node* n = &arr[x + y * wid]; n->neighbors = getNeighbors( x, y );
for( int d = 0; d < 8; d++ ) { if( n->neighbors & ( 1 << d ) ) { int a = x + dx[d], b = y + dy[d]; if( arr[a + b * wid].val == 0 ) { arr[a + b * wid].val = w; if( search( a, b, w + 1 ) ) return true; arr[a + b * wid].val = 0; } } } return false;
}
unsigned char getNeighbors( int x, int y )
{
unsigned char c = 0; int a, b; for( int xx = 0; xx < 8; xx++ ) { a = x + dx[xx], b = y + dy[xx]; if( a < 0 || b < 0 || a >= wid || b >= hei ) continue; if( arr[a + b * wid].val > -1 ) c |= ( 1 << xx ); } return c;
}
void solveIt()
{
int x, y, z; findStart( x, y, z ); if( z == 99999 ) { cout << "\nCan't find start point!\n"; return; } search( x, y, z + 1 );
}
void findStart( int& x, int& y, int& z )
{
for( int b = 0; b < hei; b++ ) for( int a = 0; a < wid; a++ ) if( arr[a + wid * b].val == 0 ) { x = a; y = b; z = 1; arr[a + wid * b].val = z; return; }
}
int wid, hei, max, dx[8], dy[8]; node* arr;
};
int main( int argc, char* argv[] ) {
int wid; string p;
p = "* . . * . . * . . . . . . . . . . . . . . * . . . . . * * * . . . * * * * * . * * *"; wid = 7;
istringstream iss( p ); vector<string> puzz;
copy( istream_iterator<string>( iss ), istream_iterator<string>(), back_inserter<vector<string> >( puzz ) );
nSolver s; s.solve( puzz, wid );
int c = 0;
for( vector<string>::iterator i = puzz.begin(); i != puzz.end(); i++ )
{
if( ( *i ) != "*" && ( *i ) != "." ) { if( atoi( ( *i ).c_str() ) < 10 ) cout << "0"; cout << ( *i ) << " "; } else cout << " "; if( ++c >= wid ) { cout << endl; c = 0; }
} cout << endl << endl; return system( "pause" );
} </lang>
- Output:
01 04 12 03
27 16 19 22 15 18 21
05 08 11 02 07 10 13
23 26 17 20 25
06 09 14
24
D
From the refactored C++ version with more precise typing. This tries all possible start positions. The HopidoPuzzle struct is created at compile-time, so its pre-conditions can catch most malformed puzzles at compile-time. <lang d>import std.stdio, std.conv, std.string, std.range, std.algorithm, std.typecons;
struct HopidoPuzzle {
private alias InputCellBaseType = char;
private enum InputCell : InputCellBaseType { available = '#', unavailable = '.' }
private alias Cell = uint;
private enum : Cell { unknownCell = 0, unavailableCell = Cell.max } // Special Cell values.
// Neighbors, [shift row, shift column].
private static immutable int[2][8] shifts = [[-2, -2], [2, -2], [-2, 2], [2, 2],
[ 0, -3], [0, 3], [-3, 0], [3, 0]];
private immutable size_t gridWidth, gridHeight; private immutable Cell nAvailableCells; private /*immutable*/ const InputCell[] flatPuzzle; private Cell[] grid; // Flattened mutable game grid.
@disable this();
this(in string[] rawPuzzle) pure @safe
in {
assert(!rawPuzzle.empty);
assert(!rawPuzzle[0].empty);
assert(rawPuzzle.all!(row => row.length == rawPuzzle[0].length)); // Is rectangular.
// Has at least one start point.
assert(rawPuzzle.join.representation.canFind(InputCell.available));
} body {
//immutable puzzle = rawPuzzle.to!(InputCell[][]);
immutable puzzle = rawPuzzle.map!representation.array.to!(InputCell[][]);
gridWidth = puzzle[0].length;
gridHeight = puzzle.length;
flatPuzzle = puzzle.join;
nAvailableCells = flatPuzzle.representation.count!(ic => ic == InputCell.available);
grid = flatPuzzle
.representation
.map!(ic => ic == InputCell.available ? unknownCell : unavailableCell)
.array;
}
Nullable!(string[][]) solve() pure /*nothrow*/ @safe
out(result) {
if (!result.isNull)
assert(!grid.canFind(unknownCell));
} body {
// Try all possible start positions.
foreach (immutable r; 0 .. gridHeight) {
foreach (immutable c; 0 .. gridWidth) {
immutable pos = r * gridWidth + c;
if (grid[pos] == unknownCell) {
immutable Cell startCell = 1; // To lay the first cell value.
grid[pos] = startCell; // Try.
if (search(r, c, startCell + 1)) {
auto result = zip(flatPuzzle, grid)
//.map!({p, c} => ...
.map!(pc => (pc[0] == InputCell.available) ?
pc[1].text :
InputCellBaseType(pc[0]).text)
.array
.chunks(gridWidth)
.array;
return typeof(return)(result);
}
grid[pos] = unknownCell; // Restore.
}
}
}
return typeof(return)(); }
private bool search(in size_t r, in size_t c, in Cell cell) pure nothrow @safe @nogc {
if (cell > nAvailableCells)
return true; // One solution found.
foreach (immutable sh; shifts) {
immutable r2 = r + sh[0],
c2 = c + sh[1],
pos = r2 * gridWidth + c2;
// No need to test for >= 0 because uint wraps around.
if (c2 < gridWidth && r2 < gridHeight && grid[pos] == unknownCell) {
grid[pos] = cell; // Try.
if (search(r2, c2, cell + 1))
return true;
grid[pos] = unknownCell; // Restore.
}
}
return false; }
}
void main() @safe {
// enum HopidoPuzzle to catch malformed puzzles at compile-time.
enum puzzle = ".##.##.
#######
#######
.#####.
..###..
...#...".split.HopidoPuzzle;
immutable solution = puzzle.solve; // Solved at run-time.
if (solution.isNull)
writeln("No solution found.");
else
writefln("One solution:\n%(%-(%2s %)\n%)", solution);
}</lang>
- Output:
One solution: . 1 4 . 12 3 . 27 16 19 22 15 18 21 5 8 11 2 7 10 13 . 23 26 17 20 25 . . . 6 9 14 . . . . . 24 . . .
Elixir
This solution uses HLPsolver from here <lang elixir># require HLPsolver
adjacent = [{-3, 0}, {0, -3}, {0, 3}, {3, 0}, {-2, -2}, {-2, 2}, {2, -2}, {2, 2}]
board = """ . 0 0 . 0 0 . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . 0 0 0 0 0 . . . 0 0 0 . . . . . 1 . . . """ HLPsolver.solve(board, adjacent)</lang>
- Output:
Problem:
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0
0 0 0
1
Solution:
5 25 17 3
27 13 10 7 14 11 8
24 21 18 4 22 19 16
6 26 12 9 2
23 20 15
1
Go
<lang go>package main
import (
"fmt" "sort"
)
var board = []string{
".00.00.", "0000000", "0000000", ".00000.", "..000..", "...0...",
}
var moves = [][2]int{
{-3, 0}, {0, 3}, {3, 0}, {0, -3},
{2, 2}, {2, -2}, {-2, 2}, {-2, -2},
}
var grid [][]int
var totalToFill = 0
func solve(r, c, count int) bool {
if count > totalToFill {
return true
}
nbrs := neighbors(r, c)
if len(nbrs) == 0 && count != totalToFill {
return false
}
sort.Slice(nbrs, func(i, j int) bool {
return nbrs[i][2] < nbrs[j][2]
})
for _, nb := range nbrs {
r = nb[0]
c = nb[1]
grid[r][c] = count
if solve(r, c, count+1) {
return true
}
grid[r][c] = 0
}
return false
}
func neighbors(r, c int) (nbrs [][3]int) {
for _, m := range moves {
x := m[0]
y := m[1]
if grid[r+y][c+x] == 0 {
num := countNeighbors(r+y, c+x) - 1
nbrs = append(nbrs, [3]int{r + y, c + x, num})
}
}
return
}
func countNeighbors(r, c int) int {
num := 0
for _, m := range moves {
if grid[r+m[1]][c+m[0]] == 0 {
num++
}
}
return num
}
func printResult() {
for _, row := range grid {
for _, i := range row {
if i == -1 {
fmt.Print(" ")
} else {
fmt.Printf("%2d ", i)
}
}
fmt.Println()
}
}
func main() {
nRows := len(board) + 6
nCols := len(board[0]) + 6
grid = make([][]int, nRows)
for r := 0; r < nRows; r++ {
grid[r] = make([]int, nCols)
for c := 0; c < nCols; c++ {
grid[r][c] = -1
}
for c := 3; c < nCols-3; c++ {
if r >= 3 && r < nRows-3 {
if board[r-3][c-3] == '0' {
grid[r][c] = 0
totalToFill++
}
}
}
}
pos, r, c := -1, 0, 0
for {
for {
pos++
r = pos / nCols
c = pos % nCols
if grid[r][c] != -1 {
break
}
}
grid[r][c] = 1
if solve(r, c, 2) {
break
}
grid[r][c] = 0
if pos >= nRows*nCols {
break
}
}
printResult()
}</lang>
- Output:
1 22 14 21
18 10 7 17 11 8 16
5 24 27 4 23 26 13
2 19 9 15 20
6 25 12
3
Icon and Unicon
Minor variant of Solve_a_Holy_Knight's_tour. Works in Unicon only.
<lang unicon>global nCells, cMap, best record Pos(r,c)
procedure main(A)
puzzle := showPuzzle("Input",readPuzzle())
QMouse(puzzle,findStart(puzzle),&null,0)
showPuzzle("Output", solvePuzzle(puzzle)) | write("No solution!")
end
procedure readPuzzle()
# Start with a reduced puzzle space
p := [[-1],[-1]]
nCells := maxCols := 0
every line := !&input do {
put(p,[: -1 | -1 | gencells(line) | -1 | -1 :])
maxCols <:= *p[-1]
}
every put(p, [-1]|[-1])
# Now normalize all rows to the same length
every i := 1 to *p do p[i] := [: !p[i] | (|-1\(maxCols - *p[i])) :]
return p
end
procedure gencells(s)
static WS, NWS
initial {
NWS := ~(WS := " \t")
cMap := table() # Map to/from internal model
cMap["#"] := -1; cMap["_"] := 0
cMap[-1] := " "; cMap[0] := "_"
}
s ? while not pos(0) do {
w := (tab(many(WS))|"", tab(many(NWS))) | break
w := numeric(\cMap[w]|w)
if -1 ~= w then nCells +:= 1
suspend w
}
end
procedure showPuzzle(label, p)
write(label," with ",nCells," cells:")
every r := !p do {
every c := !r do writes(right((\cMap[c]|c),*nCells+1))
write()
}
return p
end
procedure findStart(p)
if \p[r := !*p][c := !*p[r]] = 1 then return Pos(r,c)
end
procedure solvePuzzle(puzzle)
if path := \best then {
repeat {
loc := path.getLoc()
puzzle[loc.r][loc.c] := path.getVal()
path := \path.getParent() | break
}
return puzzle
}
end
class QMouse(puzzle, loc, parent, val)
method getVal(); return val; end method getLoc(); return loc; end method getParent(); return parent; end method atEnd(); return nCells = val; end
method visit(r,c)
if /best & validPos(r,c) then return Pos(r,c)
end
method validPos(r,c)
v := val+1
xv := (0 <= puzzle[r][c]) | fail
if xv = (v|0) then { # make sure this path hasn't already gone there
ancestor := self
while xl := (ancestor := \ancestor.getParent()).getLoc() do
if (xl.r = r) & (xl.c = c) then fail
return
}
end
initially
val := val+1 if atEnd() then return best := self QMouse(puzzle, visit(loc.r-3,loc.c), self, val) QMouse(puzzle, visit(loc.r-2,loc.c-2), self, val) QMouse(puzzle, visit(loc.r, loc.c-3), self, val) QMouse(puzzle, visit(loc.r+2,loc.c-2), self, val) QMouse(puzzle, visit(loc.r+3,loc.c), self, val) QMouse(puzzle, visit(loc.r+2,loc.c+2), self, val) QMouse(puzzle, visit(loc.r, loc.c+3), self, val) QMouse(puzzle, visit(loc.r-2,loc.c+2), self, val)
end</lang>
Sample run:
->hopido <hopido1.in
Input with 27 cells:
_ _ _ _
_ _ _ _ _ _ _
_ _ _ _ _ _ _
_ _ _ _ _
_ _ _
1
Output with 27 cells:
3 21 13 22
25 9 6 26 10 7 27
20 17 14 2 18 15 12
4 24 8 5 23
19 16 11
1
->
Java
<lang java>import java.util.*;
public class Hopido {
final static String[] board = {
".00.00.",
"0000000",
"0000000",
".00000.",
"..000..",
"...0..."};
final static int[][] moves = {{-3, 0}, {0, 3}, {3, 0}, {0, -3},
{2, 2}, {2, -2}, {-2, 2}, {-2, -2}};
static int[][] grid;
static int totalToFill;
public static void main(String[] args) {
int nRows = board.length + 6;
int nCols = board[0].length() + 6;
grid = new int[nRows][nCols];
for (int r = 0; r < nRows; r++) {
Arrays.fill(grid[r], -1);
for (int c = 3; c < nCols - 3; c++)
if (r >= 3 && r < nRows - 3) {
if (board[r - 3].charAt(c - 3) == '0') {
grid[r][c] = 0;
totalToFill++;
}
}
}
int pos = -1, r, c;
do {
do {
pos++;
r = pos / nCols;
c = pos % nCols;
} while (grid[r][c] == -1);
grid[r][c] = 1;
if (solve(r, c, 2))
break;
grid[r][c] = 0;
} while (pos < nRows * nCols);
printResult(); }
static boolean solve(int r, int c, int count) {
if (count > totalToFill)
return true;
List<int[]> nbrs = neighbors(r, c);
if (nbrs.isEmpty() && count != totalToFill)
return false;
Collections.sort(nbrs, (a, b) -> a[2] - b[2]);
for (int[] nb : nbrs) {
r = nb[0];
c = nb[1];
grid[r][c] = count;
if (solve(r, c, count + 1))
return true;
grid[r][c] = 0;
}
return false; }
static List<int[]> neighbors(int r, int c) {
List<int[]> nbrs = new ArrayList<>();
for (int[] m : moves) {
int x = m[0];
int y = m[1];
if (grid[r + y][c + x] == 0) {
int num = countNeighbors(r + y, c + x) - 1;
nbrs.add(new int[]{r + y, c + x, num});
}
}
return nbrs;
}
static int countNeighbors(int r, int c) {
int num = 0;
for (int[] m : moves)
if (grid[r + m[1]][c + m[0]] == 0)
num++;
return num;
}
static void printResult() {
for (int[] row : grid) {
for (int i : row) {
if (i == -1)
System.out.printf("%2s ", ' ');
else
System.out.printf("%2d ", i);
}
System.out.println();
}
}
}</lang>
1 22 14 21
18 10 7 17 11 8 16
5 24 27 4 23 26 13
2 19 9 15 20
6 25 12
3
Julia
Uses the Hidato puzzle solver module, which has its source code listed here in the Hadato task. <lang julia>using .Hidato # Note that the . here means to look locally for the module rather than in the libraries
const hopid = """
. 0 0 . 0 0 . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . 0 0 0 0 0 . . . 0 0 0 . . . . . 0 . . . """
const hopidomoves = [[-3, 0], [0, -3], [-2, -2], [-2, 2], [2, -2], [0, 3], [3, 0], [2, 2]]
board, maxmoves, fixed, starts = hidatoconfigure(hopid) printboard(board, " 0", " ") hidatosolve(board, maxmoves, hopidomoves, fixed, starts[1][1], starts[1][2], 1) printboard(board)
</lang>
- Output:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 04 15 5 16 1 22 25 2 21 24 27 14 11 8 17 12 9 6 3 20 23 26 19 13 10 7 18
Kotlin
<lang scala>// version 1.2.0
val board = listOf(
".00.00.", "0000000", "0000000", ".00000.", "..000..", "...0..."
)
val moves = listOf(
-3 to 0, 0 to 3, 3 to 0, 0 to -3,
2 to 2, 2 to -2, -2 to 2, -2 to -2
)
lateinit var grid: List<IntArray> var totalToFill = 0
fun solve(r: Int, c: Int, count: Int): Boolean {
if (count > totalToFill) return true
val nbrs = neighbors(r, c)
if (nbrs.isEmpty() && count != totalToFill) return false
nbrs.sortBy { it[2] }
for (nb in nbrs) {
val rr = nb[0]
val cc = nb[1]
grid[rr][cc] = count
if (solve(rr, cc, count + 1)) return true
grid[rr][cc] = 0
}
return false
}
fun neighbors(r: Int, c: Int): MutableList<IntArray> {
val nbrs = mutableListOf<IntArray>()
for (m in moves) {
val x = m.first
val y = m.second
if (grid[r + y][c + x] == 0) {
val num = countNeighbors(r + y, c + x) - 1
nbrs.add(intArrayOf(r + y, c + x, num))
}
}
return nbrs
}
fun countNeighbors(r: Int, c: Int): Int {
var num = 0
for (m in moves)
if (grid[r + m.second][c + m.first] == 0) num++
return num
}
fun printResult() {
for (row in grid) {
for (i in row) {
print(if (i == -1) " " else "%2d ".format(i))
}
println()
}
}
fun main(args: Array<String>) {
val nRows = board.size + 6
val nCols = board[0].length + 6
grid = List(nRows) { IntArray(nCols) { -1} }
for (r in 0 until nRows) {
for (c in 3 until nCols - 3) {
if (r in 3 until nRows - 3) {
if (board[r - 3][c - 3] == '0') {
grid[r][c] = 0
totalToFill++
}
}
}
}
var pos = -1
var rr: Int
var cc: Int
do {
do {
pos++
rr = pos / nCols
cc = pos % nCols
}
while (grid[rr][cc] == -1)
grid[rr][cc] = 1
if (solve(rr, cc, 2)) break
grid[rr][cc] = 0
}
while (pos < nRows * nCols)
printResult()
}</lang>
- Output:
1 22 14 21
18 10 7 17 11 8 16
5 24 27 4 23 26 13
2 19 9 15 20
6 25 12
3
Mathematica/Wolfram Language
Uses shortest tours on graphs to solve it: <lang Mathematica>puzz = ".00.00.\n0000000\n0000000\n.00000.\n..000..\n...0..."; puzz //= StringSplit[#, "\n"] & /* Map[Characters]; puzz //= Transpose /* Map[Reverse]; pos = Position[puzz, "0", {2}]; moves = Select[Select[Tuples[pos, 2], MatchQ[EuclideanDistance @@ #, 2 Sqrt[2] | 3] &], OrderedQ]; g = Graph[UndirectedEdge @@@ moves]; ord = Most[FindShortestTour[g]2]; Graphics[MapThread[Text, {Range[Length[ord]], VertexList[g]ord}]]</lang>
- Output:
Shows a graphical solution.
Nim
<lang Nim>import algorithm, sequtils, strformat
const Moves = [(-3, 0), (0, 3), ( 3, 0), ( 0, -3),
( 2, 2), (2, -2), (-2, 2), (-2, -2)]
type
Hopido = object grid: seq[seq[int]] nRows, nCols: int totalToFill : Natural
Neighbor = (int, int, int)
proc initHopido(board: openArray[string]): Hopido =
result.nRows = board.len + 6 result.nCols = board[0].len + 6 result.grid = newSeqWith(result.nRows, repeat(-1, result.nCols))
for row in 0..board.high:
for col in 0..board[0].high:
if board[row][col] == '0':
result.grid[row + 3][col + 3] = 0
inc result.totalToFill
proc countNeighbors(hopido: Hopido; row, col: Natural): int =
for (x, y) in Moves:
if hopido.grid[row + y][col + x] == 0:
inc result
proc neighbors(hopido: Hopido; row, col: Natural): seq[Neighbor] =
for (x, y) in Moves:
if hopido.grid[row + y][col + x] == 0:
let num = hopido.countNeighbors(row + y, col + x) - 1
result.add (row + y, col + x, num)
proc solve(hopido: var Hopido; row, col, count: Natural): bool =
if count > hopido.totalTofill: return true
var nbrs = hopido.neighbors(row, col) if nbrs.len == 0 and count != hopido.totalToFill: return false
nbrs.sort(proc(a, b: Neighbor): int = cmp(a[2], b[2]))
for (row, col, _) in nbrs:
hopido.grid[row][col] = count
if hopido.solve(row, col, count + 1):
return true
hopido.grid[row][col] = 0
proc findSolution(hopido: var Hopido) =
var pos = -1 var row, col: Natural
while true:
while true:
inc pos
row = pos div hopido.nCols
col = pos mod hopido.nCols
if hopido.grid[row][col] != -1:
break
hopido.grid[row][col] = 1
if hopido.solve(row, col, 2):
break
hopido.grid[row][col] = 0
if pos >= hopido.nRows * hopido.nCols:
break
proc print(hopido: Hopido) =
for row in hopido.grid:
for val in row:
stdout.write if val == -1: " " else: &"{val:2} "
echo()
when isMainModule:
const Board = [".00.00.",
"0000000",
"0000000",
".00000.",
"..000..",
"...0..."]
var hopido = initHopido(Board) hopido.findSolution() hopido.print()</lang>
- Output:
1 22 14 21
18 10 7 17 11 8 16
5 24 27 4 23 26 13
2 19 9 15 20
6 25 12
3
Perl
<lang perl>#!/usr/bin/perl
use strict; # http://www.rosettacode.org/wiki/Solve_a_Hopido_puzzle use warnings;
$_ = do { local $/; }; s/./$&$&/g; # double chars my $w = /\n/ && $-[0]; my $wd = 3 * $w + 1; # vertical gap my $wr = 2 * $w + 8; # down right gap my $wl = 2 * $w - 8; # down left gap place($_, '00'); die "No solution\n";
sub place
{
(local $_, my $last) = @_;
(my $new = $last)++;
/$last.{10}(?=00)/g and place( s/\G00/$new/r, $new ); # right
/(?=00.{10}$last)/g and place( s/\G00/$new/r, $new ); # left
/$last.{$wd}(?=00)/gs and place( s/\G00/$new/r, $new ); # down
/(?=00.{$wd}$last)/gs and place( s/\G00/$new/r, $new ); # up
/$last.{$wr}(?=00)/gs and place( s/\G00/$new/r, $new ); # down right
/(?=00.{$wr}$last)/gs and place( s/\G00/$new/r, $new ); # up left
/$last.{$wl}(?=00)/gs and place( s/\G00/$new/r, $new ); # down left
/(?=00.{$wl}$last)/gs and place( s/\G00/$new/r, $new ); # up right
/00/ and return;
print "Solution\n\n", s/ / /gr =~ s/0\B/ /gr;
exit;
}
__DATA__ . 0 0 . 0 0 . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . 0 0 0 0 0 . . . 0 0 0 . . . . . 0 . . .</lang>
- Output:
Solution .. 2 24 .. 1 25 .. 7 10 13 6 9 12 5 15 22 19 16 23 20 17 .. 3 8 11 4 26 .. .. .. 14 21 18 .. .. .. .. .. 27 .. .. ..
Phix
Simple brute force approach.
with javascript_semantics sequence board integer limit, tries constant ROW = 1, COL = 2 constant moves = {{-2,-2},{-2,2},{2,-2},{2,2},{-3,0},{3,0},{0,-3},{0,3}} function solve(integer row, integer col, integer n) integer nrow, ncol tries+= 1 if n>limit then return 1 end if for move=1 to length(moves) do nrow = row+moves[move][ROW] ncol = col+moves[move][COL]*3 if nrow>=1 and nrow<=length(board) and ncol>=1 and ncol<=length(board[row]) and board[nrow][ncol]=' ' then board[nrow][ncol-1..ncol] = sprintf("%2d",n) if solve(nrow,ncol,n+1) then return 1 end if board[nrow][ncol-1..ncol] = " " end if end for return 0 end function procedure Hopido(sequence s, integer w, integer h) integer rx, ry atom t0 = time() board = split(s,'\n') limit = 0 for x=1 to h do for y=3 to w*3 by 3 do if board[x][y]='0' then board[x][y] = ' ' limit += 1 end if end for end for while 1 do rx = rand(h) ry = rand(w)*3 if board[rx][ry]=' ' then exit end if end while board[rx][ry] = '1' tries = 0 if solve(rx,ry,2) then puts(1,join(board,"\n")) printf(1,"\nsolution found in %d tries (%3.2fs)\n",{tries,time()-t0}) else puts(1,"no solutions found\n") end if end procedure constant board1 = """ . 0 0 . 0 0 . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . 0 0 0 0 0 . . . 0 0 0 . . . . . 0 . . .""" Hopido(board1,7,6)
The best and worse cases observed were:
. 13 22 . 14 11 . 6 3 25 7 4 1 27 23 20 17 12 21 18 15 . 8 5 2 26 10 . . . 24 19 16 . . . . . 9 . . . solution found in 46 tries (0.00s) . 20 11 . 19 22 . 2 5 8 1 4 7 27 10 13 16 21 12 15 18 . 25 3 6 26 23 . . . 9 14 17 . . . . . 24 . . . solution found in 67702 tries (0.09s)
Picat
<lang Picat> import sat. main =>
Grid = {{0,1,1,0,1,1,0},
{1,1,1,1,1,1,1}, {1,1,1,1,1,1,1}, {0,1,1,1,1,1,0}, {0,0,1,1,1,0,0}, {0,0,0,1,0,0,0}},
NR = len(Grid), NC = len(Grid[1]),
Es = [{(R,C), (R1,C1), _} : R in 1..NR, C in 1..NC, R1 in 1..NR, C1 in 1..NC, % Edges
((R1 = R, abs(C1 - C) = 3); (C1 = C, abs(R1 - R) = 3); % horizontal hop
(abs(R1 - R) = 2, abs(C1 - C) = 2)), % diagonal hop
Grid[R,C] = 1, Grid[R1,C1] = 1],
hcp_grid(Grid, Es), % find a Hamiltion cylce on the vertices in Grid through the edges in Es
solve(vars(Es)),
% Write the solution as a matrix, starting at the base tile 6|4:
M = {{0: _ in 1..NC} : _ in 1..NR},
R1 = 6, C1 = 4, I = 0,
do
I := I + 1, M[R1,C1] := I, Stop = 0,
foreach({(R1,C1), (R2,C2), 1} in Es, break(Stop == 1))
R1 := R2, C1 := C2, Stop := 1
end
while ((R1,C1) != (6,4)),
foreach(R in 1..NR, C in 1..NC)
if M[R,C] = 0 then print(" ") else printf("%2d ", M[R,C]) end,
if C = NC then nl end
end.
</lang> Output:
24 15 23 26
6 9 12 5 8 11 4
14 17 20 25 16 19 22
2 7 10 3 27
13 18 21
1
CPU time 0.019 seconds
Prolog
This is a pure prolog implementation (no cuts,etc..), the only libary predicate used is select/3 witch is pure.
<lang prolog>hopido(Grid,[C|Solved],Xs,Ys) :- select(C,Grid,RGrid), solve(RGrid,C,Solved,Xs,Ys).
solve([],_,[],_,_). solve(Grid,p(X,Y),[p(X1,Y1)|R],Xs,Ys) :- valid_move(X,Y,Xs,Ys,X1,Y1), select(p(X1,Y1),Grid,NextGrid), solve(NextGrid,p(X1,Y1),R,Xs,Ys).
valid_move(X,Y,Xs,_,X1,Y) :- j3(X,X1,Xs). % right (3,0) valid_move(X,Y,Xs,_,X1,Y) :- j3(X1,X,Xs). % left (-3,0) valid_move(X,Y,_,Ys,X,Y1) :- j3(Y,Y1,Ys). % up (0,3). valid_move(X,Y,_,Ys,X,Y1) :- j3(Y1,Y,Ys). % down (0,-3). valid_move(X,Y,Xs,Ys,X1,Y1) :- j2(X,X1,Xs), j2(Y,Y1,Ys). % up-right (2,2). valid_move(X,Y,Xs,Ys,X1,Y1) :- j2(X1,X,Xs), j2(Y,Y1,Ys). % up-left (-2,2). valid_move(X,Y,Xs,Ys,X1,Y1) :- j2(X1,X,Xs), j2(Y1,Y,Ys). % down-left (-2,-2). valid_move(X,Y,Xs,Ys,X1,Y1) :- j2(X,X1,Xs), j2(Y1,Y,Ys). % down-right (2,-2).
j2(O,N,[O,_,N|_]). j2(O,N,[_|T]) :- j2(O,N,T).
j3(O,N,[O,_,_,N|_]). j3(O,N,[_|T]) :- j3(O,N,T).</lang> To test send in a list of p/2 terms that represent points that can be hopped to (order is not important).
The grid coords can be anything so need to specify the valid coordinates as a list (in this case numbers between 0 and 6). <lang prolog>puzzle([
p(1,0),p(2,0) ,p(4,0),p(5,0),
p(0,1),p(1,1),p(2,1),p(3,1),p(4,1),p(5,1),p(6,1),
p(0,2),p(1,2),p(2,2),p(3,2),p(4,2),p(5,2),p(6,2),
p(1,3),p(2,3),p(3,3),p(4,3),p(5,3),
p(2,4),p(3,4),p(4,4),
p(3,5)
]).
test :- puzzle(P), XYs = [0,1,2,3,4,5,6], hopido(P,S,XYs,XYs), maplist(writeln,S).</lang>
- Output:
?- time(test). p(1,0) p(4,0) p(4,3) p(1,3) p(3,5) p(5,3) p(5,0) p(2,0) p(4,2) p(1,2) p(3,4) p(5,2) p(2,2) p(4,4) p(6,2) p(3,2) p(0,2) p(2,4) p(2,1) p(5,1) p(3,3) p(1,1) p(4,1) p(2,3) p(0,1) p(3,1) p(6,1) % 356,635 inferences, 0.062 CPU in 0.081 seconds (78% CPU, 5715268 Lips) true
Python
<lang python> from sys import stdout
neighbours = [[2, 2], [-2, 2], [2, -2], [-2, -2], [3, 0], [0, 3], [-3, 0], [0, -3]] cnt = 0 pWid = 0 pHei = 0
def is_valid(a, b):
return -1 < a < pWid and -1 < b < pHei
def iterate(pa, x, y, v):
if v > cnt:
return 1
for i in range(len(neighbours)):
a = x + neighbours[i][0]
b = y + neighbours[i][1]
if is_valid(a, b) and pa[a][b] == 0:
pa[a][b] = v
r = iterate(pa, a, b, v + 1)
if r == 1:
return r
pa[a][b] = 0
return 0
def solve(pz, w, h):
global cnt, pWid, pHei
pa = [[-1 for j in range(h)] for i in range(w)]
f = 0
pWid = w
pHei = h
for j in range(h):
for i in range(w):
if pz[f] == "1":
pa[i][j] = 0
cnt += 1
f += 1
for y in range(h):
for x in range(w):
if pa[x][y] == 0:
pa[x][y] = 1
if 1 == iterate(pa, x, y, 2):
return 1, pa
pa[x][y] = 0
return 0, pa
r = solve("011011011111111111111011111000111000001000", 7, 6) if r[0] == 1:
for j in range(6):
for i in range(7):
if r[1][i][j] == -1:
stdout.write(" ")
else:
stdout.write(" {:0{}d}".format(r[1][i][j], 2))
print()
else:
stdout.write("No solution!")
</lang>
- Output:
01 25 17 03
27 13 10 07 14 11 08
24 21 18 02 22 19 16
06 26 12 09 04
23 20 15
05
Racket
This solution uses the module "hidato-family-solver.rkt" from Solve a Numbrix puzzle#Racket. The difference between the two is essentially the neighbourhood function.
<lang racket>#lang racket (require "hidato-family-solver.rkt")
(define hoppy-moore-neighbour-offsets
'((+3 0) (-3 0) (0 +3) (0 -3) (+2 +2) (-2 -2) (-2 +2) (+2 -2)))
(define solve-hopido (solve-hidato-family hoppy-moore-neighbour-offsets))
(displayln
(puzzle->string
(solve-hopido
#(#(_ 0 0 _ 0 0 _)
#(0 0 0 0 0 0 0)
#(0 0 0 0 0 0 0)
#(_ 0 0 0 0 0 _)
#(_ _ 0 0 0 _ _)
#(_ _ _ 0 _ _ _)))))
</lang>
- Output:
_ 2 20 _ 3 19 _ 7 10 13 6 9 12 5 15 22 25 16 21 24 27 _ 1 8 11 4 18 _ _ _ 14 23 26 _ _ _ _ _ 17 _ _ _
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
- Solve a Hopido puzzle
- Solve a Holy Knight's tour
- Solve a Numbrix puzzle
- Solve the no connection puzzle
<lang perl6>my @adjacent = [3, 0],
[2, -2], [2, 2],
[0, -3], [0, 3],
[-2, -2], [-2, 2],
[-3, 0];
solveboard q:to/END/;
. _ _ . _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ . . . _ _ _ . . . . . 1 . . . 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>
- Output:
21 4 20 3
26 12 15 25 11 14 24
17 6 9 18 5 8 19
22 27 13 23 2
16 7 10
1
59 tries
REXX
This REXX program is a slightly modified version of the REXX Hidato program.
No particular effort was made to reduce the elapsed time in solving the puzzle. <lang rexx>/*REXX program solves a Hopido puzzle, it also displays the puzzle and the solution. */ call time 'Reset' /*reset the REXX elapsed timer to zero.*/ maxR=0; maxC=0; maxX=0; minR=9e9; minC=9e9; minX=9e9; cells=0; @.= parse arg xxx /*get the cell definitions from the CL.*/ xxx=translate(xxx, , "/\;:_", ',') /*also allow other characters as comma.*/
do while xxx\=; parse var xxx r c marks ',' xxx
do while marks\=; _=@.r.c
parse var marks x marks
if datatype(x,'N') then x=x/1 /*normalize X. */
minR=min(minR,r); maxR=max(maxR,r); minC=min(minC,c); maxC=max(maxC,c)
if x==1 then do; !r=r; !c=c; end /*the START cell. */
if _\== then call err "cell at" r c 'is already occupied with:' _
@.r.c=x; c=c+1; cells=cells+1 /*assign a mark. */
if x==. then iterate /*is a hole? Skip*/
if \datatype(x,'W') then call err 'illegal marker specified:' x
minX=min(minX,x); maxX=max(maxX,x) /*min and max X. */
end /*while marks¬= */
end /*while xxx ¬= */
call show /* [↓] is used for making fast moves. */ Nr = '0 3 0 -3 -2 2 2 -2' /*possible row for the next move. */ Nc = '3 0 -3 0 2 -2 2 -2' /* " column " " " " */ pMoves=words(Nr) /*the number of possible moves. */
do i=1 for pMoves; Nr.i=word(Nr, i); Nc.i=word(Nc,i); end /*i*/
if \next(2,!r,!c) then call err 'No solution possible for this Hopido puzzle.' say 'A solution for the Hopido exists.'; say; call show etime= format(time('Elapsed'), , 2) /*obtain the elapsed time (in seconds).*/ if etime<.1 then say 'and took less than 1/10 of a second.'
else say 'and took' etime "seconds."
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ err: say; say '***error*** (from Hopido): ' arg(1); say; exit 13 /*──────────────────────────────────────────────────────────────────────────────────────*/ next: procedure expose @. Nr. Nc. cells pMoves; parse arg #,r,c; ##=#+1
do t=1 for pMoves /* [↓] try some moves. */
parse value r+Nr.t c+Nc.t with nr nc /*next move coördinates*/
if @.nr.nc==. then do; @.nr.nc=# /*let's try this move. */
if #==cells then leave /*is this the last move?*/
if next(##,nr,nc) then return 1
@.nr.nc=. /*undo the above move. */
iterate /*go & try another move.*/
end
if @.nr.nc==# then do /*this a fill-in move ? */
if #==cells then return 1 /*this is the last move.*/
if next(##,nr,nc) then return 1 /*a fill-in move. */
end
end /*t*/
return 0 /*This ain't working. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ show: if maxR<1 | maxC<1 then call err 'no legal cell was specified.'
if minX<1 then call err 'no 1 was specified for the puzzle start'
w=max(2,length(cells)); do r=maxR to minR by -1; _=
do c=minC to maxC; _=_ right(@.r.c,w); end /*c*/
say _
end /*r*/
say; return</lang>
output when the input is:
1 4 1 \2 3 . . . \3 2 . . . . . \4 1 . . . . . . . \5 1 . . . . . . . \6 2 . . \6 5 . .
. . . .
. . . . . . .
. . . . . . .
. . . . .
. . .
1
A solution for the Hopido exists.
5 12 4 11
8 22 25 7 21 24 27
13 16 19 2 15 18 3
6 9 23 26 10
14 17 20
1
and took less than 1/10 of a second.
Ruby
This solution uses HLPsolver from here <lang ruby>require 'HLPsolver'
ADJACENT = [[-3, 0], [0, -3], [0, 3], [3, 0], [-2, -2], [-2, 2], [2, -2], [2, 2]]
board1 = <<EOS . 0 0 . 0 0 . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . 0 0 0 0 0 . . . 0 0 0 . . . . . 1 . . . EOS t0 = Time.now HLPsolver.new(board1).solve puts " #{Time.now - t0} sec"</lang> Which produces:
Problem:
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0
0 0 0
1
Solution:
3 12 4 11
8 18 21 7 17 20 6
13 24 27 14 23 26 15
2 9 19 5 10
22 25 16
1
0.001 sec
Tcl
<lang tcl>package require Tcl 8.6
oo::class create HopidoSolver {
variable grid start limit
constructor {puzzle} {
set grid $puzzle for {set y 0} {$y < [llength $grid]} {incr y} { for {set x 0} {$x < [llength [lindex $grid $y]]} {incr x} { if {[set cell [lindex $grid $y $x]] == 1} { set start [list $y $x] } incr limit [expr {$cell>=0}] } } if {![info exist start]} { return -code error "no starting position found" }
}
method moves {} {
return { 0 -3 -2 -2 -2 2 -3 0 3 0
-2 2 2 2
0 3 }
}
method Moves {g r c} {
set valid {} foreach {dr dc} [my moves] { set R [expr {$r + $dr}] set C [expr {$c + $dc}] if {[lindex $g $R $C] == 0} { lappend valid $R $C } } return $valid
}
method Solve {g r c v} {
lset g $r $c [incr v] if {$v >= $limit} {return $g} foreach {r c} [my Moves $g $r $c] { return [my Solve $g $r $c $v] } return -code continue
}
method solve {} {
while {[incr i]==1} { set grid [my Solve $grid {*}$start 0] return } return -code error "solution not possible"
}
method solution {} {return $grid}
}
proc parsePuzzle {str} {
foreach line [split $str "\n"] {
if {[string trim $line] eq ""} continue lappend rows [lmap {- c} [regexp -all -inline {(.)\s?} $line] { string map {" " -1 "." -1} $c }]
}
set len [tcl::mathfunc::max {*}[lmap r $rows {llength $r}]]
for {set i 0} {$i < [llength $rows]} {incr i} {
while {[llength [lindex $rows $i]] < $len} { lset rows $i end+1 -1 }
} return $rows
} proc showPuzzle {grid name} {
foreach row $grid {foreach cell $row {incr c [expr {$cell>=0}]}}
set len [string length $c]
set u [string repeat "_" $len]
puts "$name with $c cells"
foreach row $grid {
puts [format " %s" [join [lmap c $row { format "%*s" $len [if {$c==-1} list elseif {$c==0} {set u} {set c}] }]]]
}
} set puzzle [parsePuzzle { . 0 0 . 0 0 . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . 0 0 0 0 0 . . . 0 0 0 . . . . . 1 . . . }] showPuzzle $puzzle "Input" HopidoSolver create hop $puzzle hop solve showPuzzle [hop solution] "Output"</lang>
- Output:
Input with 27 cells
__ __ __ __
__ __ __ __ __ __ __
__ __ __ __ __ __ __
__ __ __ __ __
__ __ __
1
Output with 27 cells
3 6 23 7
27 11 14 26 10 13 25
5 17 20 4 16 19 22
2 9 12 24 8
15 18 21
1
Wren
<lang ecmascript>import "/sort" for Sort import "/fmt" for Fmt
var board = [
".00.00.", "0000000", "0000000", ".00000.", "..000..", "...0..."
]
var moves = [
[-3, 0], [0, 3], [ 3, 0], [ 0, -3], [ 2, 2], [2, -2], [-2, 2], [-2, -2]
]
var grid = [] var totalToFill = 0
var countNeighbors = Fn.new { |r, c|
var num = 0 for (m in moves) if (grid[r + m[1]][c + m[0]] == 0) num = num + 1 return num
}
var neighbors = Fn.new { |r, c|
var nbrs = []
for (m in moves) {
var x = m[0]
var y = m[1]
if (grid[r + y][c + x] == 0) {
var num = countNeighbors.call(r + y, c + x) - 1
nbrs.add([r + y, c + x, num])
}
}
return nbrs
}
var solve // recursive solve = Fn.new { |r, c, count|
if (count > totalToFill) return true
var nbrs = neighbors.call(r, c)
if (nbrs.isEmpty && count != totalToFill) return false
var cmp = Fn.new { |n1, n2| (n1[2] - n2[2]).sign }
Sort.insertion(nbrs, cmp) // stable sort
for (nb in nbrs) {
var rr = nb[0]
var cc = nb[1]
grid[rr][cc] = count
if (solve.call(rr, cc, count + 1)) return true
grid[rr][cc] = 0
}
return false
}
var printResult = Fn.new {
for (row in grid) {
for (i in row) {
if (i == -1) {
System.write(" ")
} else {
Fmt.write("$2d ", i)
}
}
System.print()
}
}
var nRows = board.count + 6 var nCols = board[0].count + 6 grid = List.filled(nRows, null) for (r in 0...nRows) {
grid[r] = List.filled(nCols, -1)
for (c in 3...nCols - 3) {
if (r >= 3 && r < nRows - 3) {
if (board[r - 3][c - 3] == "0") {
grid[r][c] = 0
totalToFill = totalToFill + 1
}
}
}
} var pos = -1 var r var c while (true) {
while (true) {
pos = pos + 1
r = (pos / nCols).truncate
c = pos % nCols
if (grid[r][c] != -1) break
}
grid[r][c] = 1
if (solve.call(r, c, 2)) break
grid[r][c] = 0
if (pos >= nRows * nCols) break
} printResult.call()</lang>
- Output:
1 22 14 21
18 10 7 17 11 8 16
5 24 27 4 23 26 13
2 19 9 15 20
6 25 12
3
zkl
This solution uses the code from Solve_a_Numbrix_puzzle#zkl <lang zkl>hi:= // 0==empty cell, X==not a cell
- <<<
" X 0 0 X 0 0 X
0 0 0 0 0 0 0 0 0 0 0 0 0 0 X 0 0 0 0 0 X X X 0 0 0 X X X X X 0 X X X";
- <<<
adjacent:=T( T(-3,0),
T(-2,-2), T(-2,2),
T(0,-3), T(0,3),
T(2,-2), T(2,2),
T(3,0) );
puzzle:=Puzzle(hi,adjacent); puzzle.print_board(); puzzle.solve(); println(); puzzle.print_board(); println();</lang>
- Output:
Number of cells = 27
__ __ __ __
__ __ __ __ __ __ __
__ __ __ __ __ __ __
__ __ __ __ __
__ __ __
__
1 8 2 9
12 24 21 13 25 22 14
19 6 3 18 7 4 27
16 11 23 15 10
20 5 26
17