Solve a Hopido puzzle

From Rosetta Code
Task
Solve a Hopido puzzle
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.

Realated Tasks:


AutoHotkey[edit]

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, ">")
}
Examples:
;--------------------------------
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
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++[edit]

 
#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" );
}
 
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[edit]

Translation of: C++

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.

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);
}
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  .  .  .

Icon and Unicon[edit]

Minor variant of Solve_a_Holy_Knight's_tour. Works in Unicon only.

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

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[edit]

Works with: Java version 8
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();
}
}
}
             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 6[edit]

Using the solver from Solve_a_Hidato_puzzle.

my @adjacent = [3, 0],
[2, -2], [2, 2],
[0, -3], [0, 3],
[-2, -2], [-2, 2],
[-3, 0];
 
solveboard q:to/END/;
. _ _ . _ _ .
_ _ _ _ _ _ _
_ _ _ _ _ _ _
. _ _ _ _ _ .
. . _ _ _ . .
. . . 1 . . .
END
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

Phix[edit]

Simple brute force approach.

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 x, y
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
x = rand(h)
y = rand(w)*3
if board[x][y]=' ' then exit end if
end while
board[x][y] = '1'
tries = 0
if solve(x,y,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)

Racket[edit]

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
(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 _ _ _)))))
 
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  _  _  _

REXX[edit]

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.

/*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

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[edit]

This solution uses HLPsolver from here

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"

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[edit]

Works with: Tcl version 8.6
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"
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         

zkl[edit]

This solution uses the code from Solve_a_Numbrix_puzzle#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();
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