Solve a Hidato puzzle

From Rosetta Code
Jump to: navigation, search
Task
Solve a Hidato puzzle
You are encouraged to solve this task according to the task description, using any language you may know.

The task is to write a program which solves Hidato (aka Hidoku) puzzles.

The rules are:

  • You are given a grid with some numbers placed in it. The other squares in the grid will be blank.
    • The grid is not necessarily rectangular.
    • The grid may have holes in it.
    • The grid is always connected.
    • The number “1” is always present, as is another number that is equal to the number of squares in the grid. Other numbers are present so as to force the solution to be unique.
    • It may be assumed that the difference between numbers present on the grid is not greater than lucky 13.
  • The aim is to place a natural number in each blank square so that in the sequence of numbered squares from “1” upwards, each square is in the wp:Moore neighborhood of the squares immediately before and after it in the sequence (except for the first and last squares, of course, which only have one-sided constraints).
    • Thus, if the grid was overlaid on a chessboard, a king would be able to make legal moves along the path from first to last square in numerical order.
    • A square may only contain one number.
  • In a proper Hidato puzzle, the solution is unique.

For example the following problem

Sample Hidato problem, from Wikipedia

has the following solution, with path marked on it:

Solution to sample Hidato problem

Realated Tasks:


Contents

[edit] Bracmat

(
( hidato
= Line solve lowest Ncells row column rpad
, Board colWidth maxDigits start curCol curRow
, range head line cellN solution output tail
. out$!arg
& @(!arg:? ((%@:>" ") ?:?arg))
& 0:?row:?column
& :?Board
& ( Line
= token
. whl
' ( @(!arg:?token [3 ?arg)
& ( ( @(!token:? "_" ?)
& :?token
| @(!token:? #?token (|" " ?))
)
& (!token.!row.!column) !Board:?Board
|
)
& 1+!column:?column
)
)
& whl
' ( @(!arg:?line \n ?arg)
& Line$!line
& 1+!row:?row
& 0:?column
)
& Line$!arg
& ( range
= hi lo
. (!arg+1:?hi)+-2:?lo
& '($lo|$arg|$hi)
)
& ( solve
= ToDo cellN row column head tail remainder
, candCell Solved rowCand colCand pattern recurse
.  !arg:(?ToDo.?cellN.?row.?column)
& range$!row:(=?row)
& range$!column:(=?column)
&
' (  ?head ($cellN.?rowCand.?colCand) ?tail
& (!rowCand.!colCand):($row.$column)
& !recurse
|  ?head
(.($row.$column):(?rowCand.?colCand))
(?tail&!recurse)
. ((!rowCand.!colCand).$cellN)
 : ?candCell
& (  !head !tail:
& out$found!
& !candCell
| solve
$ ( !head !tail
. $cellN+1
. !rowCand
. !colCand
)
 : ?remainder
& !candCell+!remainder
)
 : ?Solved
)
 : (=?pattern.?recurse)
& !ToDo:!pattern
& !Solved
)
& infinity:?lowest
& (  !Board
 : ? (<!lowest:#%?lowest.?start) (?&~)
| solve$(!Board.!lowest.!start):?solution
)
& :?output
& 0:?curCol
& !solution:((?curRow.?).?)+?+[?Ncells
& @(!Ncells:? [?maxDigits)
& 1+!maxDigits:?colWidth
& ( rpad
= len
.  !arg:(?arg.?len)
& @(str$(!arg " "):?arg [!len ?)
& !arg
)
& whl
' ( !solution:((?row.?column).?cellN)+?solution
& (  !row:>!curRow:?curRow
& !output \n:?output
& 0:?curCol
|
)
& whl
' ( !curCol+1:~>!column:?curCol
& !output rpad$(.!colWidth):?output
)
&  !output rev$(rpad$(rev$(str$(!cellN " ")).!colWidth))
 : ?output
& !curCol+1:?curCol
)
& str$!output
)
& "
__ 33 35 __ __
__ __ 24 22 __
__ __ __ 21 __ __
__ 26 __ 13 40 11
27 __ __ __ 9 __ 1
__ __ 18 __ __
__ 7 __ __
5 __"
 : ?board
& out$(hidato$!board)
);

Output:


 __ 33 35 __ __
 __ __ 24 22 __
 __ __ __ 21 __ __
 __ 26 __ 13 40 11
 27 __ __ __  9 __  1
       __ __ 18 __ __
             __  7 __ __
                    5 __
found!
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

[edit] C

Depth-first graph, with simple connectivity check to reject some impossible situations early. The checks slow down simpler puzzles significantly, but can make some deep recursions backtrack much earilier.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
 
int *board, *flood, *known, top = 0, w, h;
 
static inline int idx(int y, int x) { return y * w + x; }
 
int neighbors(int c, int *p)
/*
@c cell
@p list of neighbours
@return amount of neighbours
*/

{
int i, j, n = 0;
int y = c / w, x = c % w;
 
for (i = y - 1; i <= y + 1; i++) {
if (i < 0 || i >= h) continue;
for (j = x - 1; j <= x + 1; j++)
if (!(j < 0 || j >= w
|| (j == x && i == y)
|| board[ p[n] = idx(i,j) ] == -1))
n++;
}
 
return n;
}
 
void flood_fill(int c)
/*
fill all free cells around @c with “1” and write output to variable “flood”
@c cell
*/

{
int i, n[8], nei;
 
nei = neighbors(c, n);
for (i = 0; i < nei; i++) { // for all neighbours
if (board[n[i]] || flood[n[i]]) continue; // if cell is not free, choose another neighbour
 
flood[n[i]] = 1;
flood_fill(n[i]);
}
}
 
/* Check all empty cells are reachable from higher known cells.
Should really do more checks to make sure cell_x and cell_x+1
share enough reachable empty cells; I'm lazy. Will implement
if a good counter example is presented. */

int check_connectity(int lowerbound)
{
int c;
memset(flood, 0, sizeof(flood[0]) * w * h);
for (c = lowerbound + 1; c <= top; c++)
if (known[c]) flood_fill(known[c]); // mark all free cells around known cells
 
for (c = 0; c < w * h; c++)
if (!board[c] && !flood[c]) // if there are free cells which could not be reached from flood_fill
return 0;
 
return 1;
}
 
void make_board(int x, int y, const char *s)
{
int i;
 
w = x, h = y;
top = 0;
x = w * h;
 
known = calloc(x + 1, sizeof(int));
board = calloc(x, sizeof(int));
flood = calloc(x, sizeof(int));
 
while (x--) board[x] = -1;
 
for (y = 0; y < h; y++)
for (x = 0; x < w; x++) {
i = idx(y, x);
 
while (isspace(*s)) s++;
 
switch (*s) {
case '_': board[i] = 0;
case '.': break;
default:
known[ board[i] = strtol(s, 0, 10) ] = i;
if (board[i] > top) top = board[i];
}
 
while (*s && !isspace(*s)) s++;
}
}
 
void show_board(const char *s)
{
int i, j, c;
 
printf("\n%s:\n", s);
 
for (i = 0; i < h; i++, putchar('\n'))
for (j = 0; j < w; j++) {
c = board[ idx(i, j) ];
printf(!c ? " __" : c == -1 ? " " : " %2d", c);
}
}
 
int fill(int c, int n)
{
int i, nei, p[8], ko, bo;
 
if ((board[c] && board[c] != n) || (known[n] && known[n] != c))
return 0;
 
if (n == top) return 1;
 
ko = known[n];
bo = board[c];
board[c] = n;
 
if (check_connectity(n)) {
nei = neighbors(c, p);
for (i = 0; i < nei; i++)
if (fill(p[i], n + 1))
return 1;
}
 
board[c] = bo;
known[n] = ko;
return 0;
}
 
int main()
{
make_board(
#define USE_E 0
#if (USE_E == 0)
8,8, " __ 33 35 __ __ .. .. .."
" __ __ 24 22 __ .. .. .."
" __ __ __ 21 __ __ .. .."
" __ 26 __ 13 40 11 .. .."
" 27 __ __ __ 9 __ 1 .."
" . . __ __ 18 __ __ .."
" . .. . . __ 7 __ __"
" . .. .. .. . . 5 __"
#elif (USE_E == 1)
3, 3, " . 4 ."
" _ 7 _"
" 1 _ _"
#else
50, 3,
" 1 _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . 74"
" . . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ ."
" . . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ ."
#endif
);
 
show_board("Before");
fill(known[1], 1);
show_board("After"); /* "40 lbs in two weeks!" */
 
return 0;
}
Output:
 Before:
 __ 33 35 __ __         
 __ __ 24 22 __         
 __ __ __ 21 __ __      
 __ 26 __ 13 40 11      
 27 __ __ __  9 __  1   
       __ __ 18 __ __   
             __  7 __ __
                    5 __
 
 After:
 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

[edit] C++

 
#include <iostream>
#include <sstream>
#include <iterator>
#include <vector>
 
//------------------------------------------------------------------------------
using namespace std;
 
//------------------------------------------------------------------------------
struct node
{
int val;
unsigned char neighbors;
};
//------------------------------------------------------------------------------
class hSolver
{
public:
hSolver()
{
dx[0] = -1; dx[1] = 0; dx[2] = 1; dx[3] = -1; dx[4] = 1; dx[5] = -1; dx[6] = 0; dx[7] = 1;
dy[0] = -1; dy[1] = -1; dy[2] = -1; dy[3] = 0; dy[4] = 0; dy[5] = 1; dy[6] = 1; dy[7] = 1;
}
 
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 = 0;
arr = new node[len]; memset( arr, 0, len * sizeof( node ) );
weHave = new bool[len + 1]; memset( weHave, 0, len + 1 );
 
for( vector<string>::iterator i = puzz.begin(); i != puzz.end(); i++ )
{
if( ( *i ) == "*" ) { arr[c++].val = -1; continue; }
arr[c].val = atoi( ( *i ).c_str() );
if( arr[c].val > 0 ) weHave[arr[c].val] = true;
if( max < arr[c].val ) max = arr[c].val;
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;
delete [] weHave;
}
 
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 );
if( weHave[w] )
{
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 == w )
if( search( a, b, w + 1 ) ) return true;
}
}
return false;
}
 
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 m = -1, a, b;
for( int yy = -1; yy < 2; yy++ )
for( int xx = -1; xx < 2; xx++ )
{
if( !yy && !xx ) continue;
m++; a = x + xx, b = y + yy;
if( a < 0 || b < 0 || a >= wid || b >= hei ) continue;
if( arr[a + b * wid].val > -1 ) c |= ( 1 << m );
}
return c;
}
 
void solveIt()
{
int x, y; findStart( x, y );
if( x < 0 ) { cout << "\nCan't find start point!\n"; return; }
search( x, y, 2 );
}
 
void findStart( int& x, int& y )
{
for( int b = 0; b < hei; b++ )
for( int a = 0; a < wid; a++ )
if( arr[a + wid * b].val == 1 ) { x = a; y = b; return; }
x = y = -1;
}
 
int wid, hei, max, dx[8], dy[8];
node* arr;
bool* weHave;
};
//------------------------------------------------------------------------------
int main( int argc, char* argv[] )
{
int wid;
string p = ". 33 35 . . * * * . . 24 22 . * * * . . . 21 . . * * . 26 . 13 40 11 * * 27 . . . 9 . 1 * * * . . 18 . . * * * * * . 7 . . * * * * * * 5 ."; wid = 8;
//string p = "54 . 60 59 . 67 . 69 . . 55 . . 63 65 . 72 71 51 50 56 62 . * * * * . . . 14 * * 17 . * 48 10 11 * 15 . 18 . 22 . 46 . * 3 . 19 23 . . 44 . 5 . 1 33 32 . . 43 7 . 36 . 27 . 31 42 . . 38 . 35 28 . 30"; wid = 9;
//string p = ". 58 . 60 . . 63 66 . 57 55 59 53 49 . 65 . 68 . 8 . . 50 . 46 45 . 10 6 . * * * . 43 70 . 11 12 * * * 72 71 . . 14 . * * * 30 39 . 15 3 17 . 28 29 . . 40 . . 19 22 . . 37 36 . 1 20 . 24 . 26 . 34 33"; wid = 9;
 
istringstream iss( p ); vector<string> puzz;
copy( istream_iterator<string>( iss ), istream_iterator<string>(), back_inserter<vector<string> >( puzz ) );
hSolver 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:

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 09 10 01
      15 16 18 08 02
            17 07 06 03
                  05 04

56 58 54 60 61 62 63 66 67
57 55 59 53 49 47 65 64 68
09 08 52 51 50 48 46 45 69
10 06 07          44 43 70
05 11 12          72 71 42
04 14 13          30 39 41
15 03 17 18 28 29 38 31 40
02 16 19 22 23 27 37 36 32
01 20 21 24 25 26 35 34 33

[edit] Curry

Works with: PAKCS

Probably not efficient.

import CLPFD
import Constraint (andC, anyC)
import Findall (unpack)
import Integer (abs)
 
 
hidato :: [[Int]] -> Success
hidato path =
test path inner
& domain inner 1 40
& allDifferent inner
& andFD [x `near` y | x <- cells, y <- cells]
& labeling [] (concat path)
where
andFD = solve . foldr1 (#/\#)
cells = enumerate path
inner free
 
near :: (Int,Int,Int) -> (Int,Int,Int) -> Constraint
(x,rx,cx) `near` (y,ry,cy) = x #<=# y #/\# dist (y -# x)
#\/# x #># y #/\# dist (x -# y)
#\/# x #=# 0
#\/# y #=# 0
where
dist d = abs (rx - ry) #<=# d
#/\# abs (cx - cy) #<=# d
 
enumerate :: [[Int]] -> [(Int,Int,Int)]
enumerate xss = [(x,row,col) | (xs,row) <- xss `zip` [1..]
, (x ,col) <- xs `zip` [1..]
]
 
test [[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
,[ 0, A, 33, 35, B, C, 0, 0, 0, 0]
,[ 0, D, E, 24, 22, F, 0, 0, 0, 0]
,[ 0, G, H, I, 21, J, K, 0, 0, 0]
,[ 0, L, 26, M, 13, 40, 11, 0, 0, 0]
,[ 0, 27, N, O, P, 9, Q, 1, 0, 0]
,[ 0, 0, 0, R, S, 18, T, U, 0, 0]
,[ 0, 0, 0, 0, 0, V, 7, W, X, 0]
,[ 0, 0, 0, 0, 0, 0, 0, 5, Y, 0]
,[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]
[ A, 33, 35, B, C
, D, E, 24, 22, F
, G, H, I, 21, J, K
, L, 26, M, 13, 40, 11
, 27, N, O, P, 9, Q, 1
, R, S, 18, T, U
, V, 7, W, X
, 5, Y
] = success
 
main = unpack hidato
Output:
Execution time: 1440 msec. / elapsed: 2270 msec.
[[0,0,0,0,0,0,0,0,0,0],[0,32,33,35,36,37,0,0,0,0],[0,31,34,24,22,38,0,0,0,0],[0,30,25,23,21,12,39,0,0,0],[0,29,26,20,13,40,11,0,0,0],[0,27,28,14,19,9,10,1,0,0],[0,0,0,15,16,18,8,2,0,0],[0,0,0,0,0,17,7,6,3,0],[0,0,0,0,0,0,0,5,4,0],[0,0,0,0,0,0,0,0,0,0]]
More values? [y(es)/N(o)/a(ll)]

[edit] D

[edit] More C-Style Version

This version retains some of the characteristics of the original C version. It uses global variables, it doesn't enforce immutability and purity. This style is faster to write for prototypes, short programs or less important code, but in larger programs you usually want more strictness to avoid some bugs and increase long-term maintainability.

Translation of: C
import std.stdio, std.array, std.conv, std.algorithm, std.string;
 
int[][] board;
int[] given, start;
 
void setup(string s) {
auto lines = s.splitLines;
auto cols = lines[0].split.length;
auto rows = lines.length;
given.length = 0;
 
board = new int[][](rows + 2, cols + 2);
foreach (row; board)
row[] = -1;
 
foreach (r, row; lines) {
foreach (c, cell; row.split) {
switch (cell) {
case "__":
board[r + 1][c + 1] = 0;
break;
case ".":
break;
default:
int val = cell.to!int;
board[r + 1][c + 1] = val;
given ~= val;
if (val == 1)
start = [r + 1, c + 1];
}
}
}
given.sort();
}
 
bool solve(int r, int c, int n, int next = 0) {
if (n > given.back)
return true;
 
if (board[r][c] && board[r][c] != n)
return false;
 
if (board[r][c] == 0 && given[next] == n)
return false;
 
int back = board[r][c];
 
board[r][c] = n;
foreach (i; -1 .. 2)
foreach (j; -1 .. 2)
if (solve(r + i, c + j, n + 1, next + (back == n)))
return true;
 
board[r][c] = back;
return false;
}
 
void printBoard() {
foreach (row; board) {
foreach (c; row)
writef(c == -1 ? " . " : c ? "%2d " : "__ ", c);
writeln;
}
}
 
void main() {
auto hi = "__ 33 35 __ __ . . .
__ __ 24 22 __ . . .
__ __ __ 21 __ __ . .
__ 26 __ 13 40 11 . .
27 __ __ __ 9 __ 1 .
. . __ __ 18 __ __ .
. . . . __ 7 __ __
. . . . . . 5 __"
;
 
hi.setup;
printBoard;
"\nFound:".writeln;
solve(start[0], start[1], 1);
printBoard;
}
Output:
 .  .  .  .  .  .  .  .  .  . 
 . __ 33 35 __ __  .  .  .  . 
 . __ __ 24 22 __  .  .  .  . 
 . __ __ __ 21 __ __  .  .  . 
 . __ 26 __ 13 40 11  .  .  . 
 . 27 __ __ __  9 __  1  .  . 
 .  .  . __ __ 18 __ __  .  . 
 .  .  .  .  . __  7 __ __  . 
 .  .  .  .  .  .  .  5 __  . 
 .  .  .  .  .  .  .  .  .  . 

Found:
 .  .  .  .  .  .  .  .  .  . 
 . 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  . 
 .  .  .  .  .  .  .  .  .  . 

[edit] Stronger Version

Translation of: C

This version uses a little stronger typing, performs tests a run-time with contracts, it doesn't use global variables, it enforces immutability and purity where possible, and produces a correct text output for both larger ad small boards. This style is more fit for larger programs, or when you want the code to be less bug-prone or a little more efficient.

With this coding style the changes in the code become less bug-prone, but also more laborious. This version is also faster, its total runtime is about 0.02 seconds or less.

import std.stdio, std.conv, std.ascii, std.array, std.string,
std.algorithm, std.exception, std.range, std.typetuple;
 
struct Hidato {
// alias Cell = RangedValue!(int, -1, int.max);
alias Cell = int;
alias Pos = size_t;
enum : Cell { emptyCell = -1, unknownCell = 0 }
 
immutable Cell boardMax;
immutable size_t nCols, nRows;
Cell[] board;
Pos[] known;
bool[] flood;
 
this(in string input) pure @safe
in {
assert(!input.strip.empty);
} out {
assert(nCols > 0 && nRows > 0);
immutable size = nCols * nRows;
assert(board.length == size);
assert(known.length == size + 1);
assert(flood.length == size);
assert(boardMax > 0 && boardMax <= size);
assert(board.reduce!max == boardMax);
assert(board.canFind(1) && board.canFind(boardMax));
assert(flood.all!(f => f == 0));
assert(known.all!(rc => rc >= 0 && rc < size));
 
foreach (immutable i, immutable cell; board) {
assert(cell == Hidato.emptyCell ||
cell == Hidato.unknownCell ||
(cell >= 1 && cell <= size));
if (cell > 0)
assert(i == known[size_t(cell)]);
}
} body {
bool[Cell] pathSeen; // A set.
immutable lines = input.splitLines;
this.nRows = lines.length;
this.nCols = lines[0].split.length;
 
immutable size = nCols * nRows;
this.board.length = size;
this.board[] = emptyCell;
this.known.length = size + 1;
this.flood.length = size;
 
auto boardMaxMutable = Cell.min;
Pos i = 0;
 
foreach (immutable row; lines) {
assert(row.split.length == nCols,
text("Wrong cols n.: ", row.split.length));
 
foreach (immutable cell; row.split) {
switch (cell) {
case "_":
this.board[i] = Hidato.unknownCell;
break;
case ".":
this.board[i] = Hidato.emptyCell;
break;
default: // Known.
immutable val = cell.to!Cell;
enforce(val > 0, "Path numbers must be > 0.");
enforce(val !in pathSeen,
text("Duplicated path number: ", val));
pathSeen[val] = true;
this.board[i] = val;
this.known[val] = i;
boardMaxMutable = max(boardMaxMutable, val);
}
i++;
}
}
 
this.boardMax = boardMaxMutable;
}
 
 
private Pos idx(in size_t r, in size_t c) const pure nothrow @safe @nogc {
return r * nCols + c;
}
 
private uint nNeighbors(in Pos pos, ref Pos[8] neighbours)
const pure nothrow @safe @nogc {
immutable r = pos / nCols;
immutable c = pos % nCols;
typeof(return) n = 0;
 
foreach (immutable sr; TypeTuple!(-1, 0, 1)) {
immutable size_t i = r + sr; // Can wrap-around.
if (i >= nRows)
continue;
foreach (immutable sc; TypeTuple!(-1, 0, 1)) {
immutable size_t j = c + sc; // Can wrap-around.
if ((sc != 0 || sr != 0) && j < nCols) {
immutable pos2 = idx(i, j);
neighbours[n] = pos2;
if (board[pos2] != Hidato.emptyCell)
n++;
}
}
}
 
return n;
}
 
/// Fill all free cells around 'cell' with true and write
/// output to variable "flood".
private void floodFill(in Pos pos) pure nothrow @safe @nogc {
Pos[8] n = void;
 
// For all neighbours.
foreach (immutable i; 0 .. nNeighbors(pos, n)) {
// If pos is not free, choose another neighbour.
if (board[n[i]] || flood[n[i]])
continue;
flood[n[i]] = true;
floodFill(n[i]);
}
}
 
/// Check all empty cells are reachable from higher known cells.
private bool checkConnectity(in uint lowerBound) pure nothrow @safe @nogc {
flood[] = false;
 
foreach (immutable i; lowerBound + 1 .. boardMax + 1)
if (known[i])
floodFill(known[i]);
 
foreach (immutable i; 0 .. nCols * nRows)
// If there are free cells which could not be
// reached from floodFill.
if (!board[i] && !flood[i])
return false;
return true;
}
 
private bool fill(in Pos pos, in uint n) pure nothrow @safe @nogc {
if ((board[pos] && board[pos] != n) ||
(known[n] && known[n] != pos))
return false;
 
if (n == boardMax)
return true;
 
immutable ko = known[n];
immutable bo = board[pos];
board[pos] = n;
 
Pos[8] p = void;
if (checkConnectity(n))
foreach (immutable i; 0 .. nNeighbors(pos, p))
if (fill(p[i], n + 1))
return true;
 
board[pos] = bo;
known[n] = ko;
return false;
}
 
void solve() pure nothrow @safe @nogc
in {
assert(!known.empty);
} body {
fill(known[1], 1);
}
 
string toString() const pure {
immutable d = [Hidato.emptyCell: ".",
Hidato.unknownCell: "_"];
immutable form = "%" ~ text(boardMax.text.length + 1) ~ "s";
 
string result;
foreach (immutable r; 0 .. nRows) {
foreach (immutable c; 0 .. nCols) {
immutable cell = board[idx(r, c)];
result ~= format(form, d.get(cell, cell.text));
}
result ~= "\n";
}
return result;
}
}
 
void solveHidato(in string problem) {
auto game = problem.Hidato;
writeln("Problem:\n", game);
game.solve;
writeln("Solution:\n", game);
}
 
void main() {
solveHidato(" _ 33 35 _ _ . . .
_ _ 24 22 _ . . .
_ _ _ 21 _ _ . .
_ 26 _ 13 40 11 . .
27 _ _ _ 9 _ 1 .
. . _ _ 18 _ _ .
. . . . _ 7 _ _
. . . . . . 5 _"
);
 
solveHidato(". 4 .
_ 7 _
1 _ _"
);
 
solveHidato(
"1 _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . 74
. . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ .
. . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ ."

);
}
Output:
Problem:
  _ 33 35  _  _  .  .  .
  _  _ 24 22  _  .  .  .
  _  _  _ 21  _  _  .  .
  _ 26  _ 13 40 11  .  .
 27  _  _  _  9  _  1  .
  .  .  _  _ 18  _  _  .
  .  .  .  .  _  7  _  _
  .  .  .  .  .  .  5  _

Solution:
 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

Problem:
 . 4 .
 _ 7 _
 1 _ _

Solution:
 . 4 .
 3 7 5
 1 2 6

Problem:
  1  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  . 74
  .  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .  _  .
  .  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .  .  _  _  .

Solution:
  1  2  3  .  .  8  9  .  . 14 15  .  . 20 21  .  . 26 27  .  . 32 33  .  . 38 39  .  . 44 45  .  . 50 51  .  . 56 57  .  . 62 63  .  . 68 69  .  . 74
  .  .  4  .  7  . 10  . 13  . 16  . 19  . 22  . 25  . 28  . 31  . 34  . 37  . 40  . 43  . 46  . 49  . 52  . 55  . 58  . 61  . 64  . 67  . 70  . 73  .
  .  .  .  5  6  .  . 11 12  .  . 17 18  .  . 23 24  .  . 29 30  .  . 35 36  .  . 41 42  .  . 47 48  .  . 53 54  .  . 59 60  .  . 65 66  .  . 71 72  .

[edit] Erlang

To simplify the code I start a new process for searching each potential path through the grid. This means that the default maximum number of processes had to be raised ("erl +P 50000" works for me). The task takes about 1-2 seconds on a low level Mac mini. If faster times are needed, or even less performing hardware is used, some optimisation should be done.

 
-module( solve_hidato_puzzle ).
 
-export( [create/2, solve/1, task/0] ).
 
-compile({no_auto_import,[max/2]}).
 
create( Grid_list, Number_list ) ->
Squares = lists:flatten( [create_column(X, Y) || {X, Y} <- Grid_list] ),
lists:foldl( fun store/2, dict:from_list(Squares), Number_list ).
 
print( Grid_list ) when is_list(Grid_list) -> print( create(Grid_list, []) );
print( Grid_dict ) ->
Max_x = max_x( Grid_dict ),
Max_y = max_y( Grid_dict ),
Print_row = fun (Y) -> [print(X, Y, Grid_dict) || X <- lists:seq(1, Max_x)], io:nl() end,
[Print_row(Y) || Y <- lists:seq(1, Max_y)].
 
solve( Dict ) ->
{find_start, [Start]} = {find_start, dict:fold( fun start/3, [], Dict )},
Max = dict:size( Dict ),
{stop_ok, {Max, Max, [Stop]}} = {stop_ok, dict:fold( fun stop/3, {Max, 0, []}, Dict )},
My_pid = erlang:self(),
erlang:spawn( fun() -> path(Start, Stop, Dict, My_pid, []) end ),
receive
{grid, Grid, path, Path} -> {Grid, Path}
end.
 
task() ->
%% Square is {X, Y}, N}. N = 0 for empty square. These are created if not present.
%% Leftmost column is X=1. Top row is Y=1.
%% Optimised for the example, grid is a list of {X, {Y_min, Y_max}}.
%% When there are holes, X is repeated as many times as needed with two new Y values each time.
Start = {{7,5}, 1},
Stop = {{5,4}, 40},
Grid_list = [{1, {1,5}}, {2, {1,5}}, {3, {1,6}}, {4, {1,6}}, {5, {1,7}}, {6, {3,7}}, {7, {5,8}}, {8, {7,8}}],
Number_list = [Start, Stop, {{1,5}, 27}, {{2,1}, 33}, {{2,4}, 26}, {{3,1}, 35}, {{3,2}, 24},
{{4,2}, 22}, {{4,3}, 21}, {{4,4}, 13}, {{5,5}, 9}, {{5,6}, 18}, {{6,4}, 11}, {{6,7}, 7}, {{7,8}, 5}],
Grid = create( Grid_list, Number_list ),
io:fwrite( "Start grid~n" ),
print( Grid ),
{New_grid, Path} = solve( create(Grid_list, Number_list) ),
io:fwrite( "Start square ~p, Stop square ~p.~nPath ~p~n", [Start, Stop, Path] ),
print( New_grid ).
 
 
create_column( X, {Y_min, Y_max} ) -> [{{X, Y}, 0} || Y <- lists:seq(Y_min, Y_max)].
 
is_filled( Dict ) -> [] =:= dict:fold( fun keep_0_square/3, [], Dict ).
 
keep_0_square( Key, 0, Acc ) -> [Key | Acc];
keep_0_square( _Key, _Value, Acc ) -> Acc.
 
max( Position, Keys ) ->
[Square | _T] = lists:reverse( lists:keysort(Position, Keys) ),
Square.
 
max_x( Dict ) ->
{X, _Y} = max( 1, dict:fetch_keys(Dict) ),
X.
 
max_y( Dict ) ->
{_X, Y} = max( 2, dict:fetch_keys(Dict) ),
Y.
 
 
neighbourhood( Square, Dict ) ->
Potentials = neighbourhood_potential_squares( Square ),
neighbourhood_squares( dict:find(Square, Dict), Potentials, Dict ).
 
neighbourhood_potential_squares( {X, Y} ) -> [{Sx, Sy} || Sx <- [X-1, X, X+1], Sy <- [Y-1, Y, Y+1], {X, Y} =/= {Sx, Sy}].
 
neighbourhood_squares( {ok, Value}, Potentials, Dict ) ->
Square_values = lists:flatten( [neighbourhood_square_value(X, dict:find(X, Dict)) || X <- Potentials] ),
Next_value = Value + 1,
neighbourhood_squares_next_value( lists:keyfind(Next_value, 2, Square_values), Square_values, Next_value ).
 
neighbourhood_squares_next_value( {Square, Value}, _Square_values, Value ) -> [{Square, Value}];
neighbourhood_squares_next_value( false, Square_values, Value ) -> [{Square, Value} || {Square, Y} <- Square_values, Y =:= 0].
 
neighbourhood_square_value( Square, {ok, Value} ) -> [{Square, Value}];
neighbourhood_square_value( _Square, error ) -> [].
 
path( Square, Square, Dict, Pid, Path ) -> path_correct( is_filled(Dict), Pid, [Square | Path], Dict );
path( Square, Stop, Dict, Pid, Path ) ->
Reversed_path = [Square | Path],
Neighbours = neighbourhood( Square, Dict ),
[erlang:spawn( fun() -> path(Next_square, Stop, dict:store(Next_square, Value, Dict), Pid, Reversed_path) end ) || {Next_square, Value} <- Neighbours].
 
path_correct( true, Pid, Path, Dict ) -> Pid ! {grid, Dict, path, lists:reverse( Path )};
path_correct( false, _Pid, _Path, _Dict ) -> dead_end.
 
print( X, Y, Dict ) -> print_number( dict:find({X, Y}, Dict) ).
 
print_number( {ok, 0} ) -> io:fwrite( "~3s", ["."] ); % . is less distracting than 0
print_number( {ok, Value} ) -> io:fwrite( "~3b", [Value] );
print_number( error ) -> io:fwrite( "~3s", [" "] ).
 
start( Key, 1, Acc ) -> [Key | Acc]; % Allow check that we only have one key with value 1.
start( _Key, _Value, Acc ) -> Acc.
 
stop( Key, Max, {Max, Max_found, Stops} ) -> {Max, erlang:max(Max, Max_found), [Key | Stops]}; % Allow check that we only have one key with value Max.
stop( _Key, Value, {Max, Max_found, Stops} ) -> {Max, erlang:max(Value, Max_found), Stops}. % Allow check that Max is Max.
 
store( {Key, Value}, Dict ) -> dict:store( Key, Value, Dict ).
 
Output:
2> solve_hidato_puzzle:task().
Start grid
  . 33 35  .  .         
  .  . 24 22  .         
  .  .  . 21  .  .      
  . 26  . 13 40 11      
 27  .  .  .  9  .  1   
        .  . 18  .  .   
              .  7  .  .
                    5  .
Start square {{7,5},1}, Stop square {{5,4},40}.
Path [{7,5}, {7,6}, {8,7}, {8,8}, {7,8}, {7,7}, {6,7}, {6,6}, {5,5}, {6,5}, {6,4}, {5,3}, {4,4}, {3,5}, {3,6}, {4,6}, {5,7}, {5,6}, {4,5}, {3,4},
      {4,3}, {4,2}, {3,3}, {3,2}, {2,3}, {2,4}, {1,5},{2,5}, {1,4}, {1,3}, {1,2}, {1,1}, {2,1}, {2,2}, {3,1}, {4,1}, {5,1}, {5,2}, {6,3}, {5,4}]
 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

[edit] Haskell

{-# LANGUAGE TupleSections #-}
{-# LANGUAGE Rank2Types #-}
import qualified Data.IntMap as I
import Data.IntMap (IntMap)
import Data.List
import Data.Maybe
import Data.Time.Clock
 
data BoardProblem = Board { cells :: IntMap (IntMap Int)
, endVal :: Int
, onePos :: (Int,Int)
, givens :: [Int] } deriving (Show,Eq)
 
tupIns x y v m = I.insert x (I.insert y v (I.findWithDefault I.empty x m)) m
tupLookup x y m = I.lookup x m >>= I.lookup y
 
makeBoard = (\x -> x{givens = dropWhile (<=1) $ sort $ givens x})
. foldl' f (Board I.empty 0 (0,0) []) . concatMap (zip [0..])
. zipWith (\y w -> map (y,) $ words w) [0..]
where f bd (x,(y,v)) = if v=="." then bd else
Board (tupIns x y (read v) (cells bd))
(if read v > endVal bd then read v else endVal bd)
(if v=="1" then (x,y) else onePos bd)
(read v:givens bd)
 
hidato brd = listToMaybe $ h 2 (cells brd) (onePos brd) (givens brd) where
h nval pmap (x,y) gs | nval == endVal brd = [pmap]
| nval == head gs = if null nvalAdj then []
else h (nval+1) pmap (fst $ head nvalAdj) (tail gs)
| not $ null nvalAdj = h (nval+1) pmap
(fst $ head nvalAdj) gs
| otherwise = hEmptyAdj
where around = [(x-1,y-1),(x,y-1),(x+1,y-1), (x-1,y),(x+1,y)
,(x-1,y+1),(x,y+1),(x+1,y+1)]
lkdUp = map (\(x,y) -> ((x,y),tupLookup x y pmap)) around
nvalAdj = filter ((==Just nval) . snd) lkdUp
hEmptyAdj = concatMap (\((nx,ny),_) -> h (nval+1)
(tupIns nx ny nval pmap) (nx,ny) gs)
$ filter ((==Just 0) . snd) lkdUp
 
printCellMap cellmap = putStrLn $ concat strings
where maxPos = xyBy I.findMax maximum
minPos = xyBy I.findMin minimum
xyBy :: (forall a. IntMap a -> (Int,a)) -> ([Int] -> Int) -> (Int, Int)
xyBy a b = (fst (a cellmap)
, b $ map (fst . a . snd) $ I.toList cellmap)
strings = map f [(x,y) | y<-[snd minPos..snd maxPos]
, x<-[fst minPos..fst maxPos]]
f (x,y) = let z = if x == fst maxPos then "\n" else " " in
case tupLookup x y cellmap of
Nothing -> " " ++ z
Just n -> (if n<10 then '
':show n else show n) ++ z
 
main = do
let sampleBoard = makeBoard sample
printCellMap $ cells sampleBoard
printCellMap $ fromJust $ hidato sampleBoard
 
sample = [" 0 33 35 0 0"
," 0 0 24 22 0"
," 0 0 0 21 0 0"
," 0 26 0 13 40 11"
,"27 0 0 0 9 0 1"
,". . 0 0 18 0 0"
,". . . . 0 7 0 0"
,". . . . . . 5 0"]

Output:

 0 33 35  0  0         
 0  0 24 22  0         
 0  0  0 21  0  0      
 0 26  0 13 40 11      
27  0  0  0  9  0  1   
       0  0 18  0  0   
             0  7  0  0
                   5  0

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

[edit] Icon and Unicon

This is an Unicon-specific solution but could easily be adjusted to work in Icon.

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]]
nCells := maxCols := 0
every line := !&input do {
put(p,[: -1 | gencells(line) | -1 :])
maxCols <:= *p[-1]
}
put(p, [-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) = puzzle[loc.r][loc.c]; end
method goNorth(); return visit(loc.r-1,loc.c); end
method goNE(); return visit(loc.r-1,loc.c+1); end
method goEast(); return visit(loc.r, loc.c+1); end
method goSE(); return visit(loc.r+1,loc.c+1); end
method goSouth(); return visit(loc.r+1,loc.c); end
method goSW(); return visit(loc.r+1,loc.c-1); end
method goWest(); return visit(loc.r, loc.c-1); end
method goNW(); return visit(loc.r-1,loc.c-1); end
 
method visit(r,c)
if /best & validPos(r,c) then return Pos(r,c)
end
 
method validPos(r,c)
xv := puzzle[r][c]
if xv = (val+1) then return
if xv = 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 +:= 1
if atEnd() then return best := self
QMouse(puzzle, goNorth(), self, val)
QMouse(puzzle, goNE(), self, val)
QMouse(puzzle, goEast(), self, val)
QMouse(puzzle, goSE(), self, val)
QMouse(puzzle, goSouth(), self, val)
QMouse(puzzle, goSW(), self, val)
QMouse(puzzle, goWest(), self, val)
QMouse(puzzle, goNW(), self, val)
end

Sample run:

->hd <hd.in 
Input with 40 cells:
                              
     _ 33 35  _  _            
     _  _ 24 22  _            
     _  _  _ 21  _  _         
     _ 26  _ 13 40 11         
    27  _  _  _  9  _  1      
           _  _ 18  _  _      
                 _  7  _  _   
                       5  _   
                              
Output with 40 cells:
                              
    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   
                              
->

[edit] Java

Translation of: D
Works with: Java version 7
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
 
public class Hidato {
 
private static int[][] board;
private static int[] given, start;
 
public static void main(String[] args) {
String[] input = {"_ 33 35 _ _ . . .",
"_ _ 24 22 _ . . .",
"_ _ _ 21 _ _ . .",
"_ 26 _ 13 40 11 . .",
"27 _ _ _ 9 _ 1 .",
". . _ _ 18 _ _ .",
". . . . _ 7 _ _",
". . . . . . 5 _"};
 
setup(input);
printBoard();
System.out.println("\nFound:");
solve(start[0], start[1], 1, 0);
printBoard();
}
 
private static void setup(String[] input) {
/* This task is not about input validation, so
we're going to trust the input to be valid */

 
String[][] puzzle = new String[input.length][];
for (int i = 0; i < input.length; i++)
puzzle[i] = input[i].split(" ");
 
int nCols = puzzle[0].length;
int nRows = puzzle.length;
 
List<Integer> list = new ArrayList<>(nRows * nCols);
 
board = new int[nRows + 2][nCols + 2];
for (int[] row : board)
for (int c = 0; c < nCols + 2; c++)
row[c] = -1;
 
for (int r = 0; r < nRows; r++) {
String[] row = puzzle[r];
for (int c = 0; c < nCols; c++) {
String cell = row[c];
switch (cell) {
case "_":
board[r + 1][c + 1] = 0;
break;
case ".":
break;
default:
int val = Integer.parseInt(cell);
board[r + 1][c + 1] = val;
list.add(val);
if (val == 1)
start = new int[]{r + 1, c + 1};
}
}
}
Collections.sort(list);
given = new int[list.size()];
for (int i = 0; i < given.length; i++)
given[i] = list.get(i);
}
 
private static boolean solve(int r, int c, int n, int next) {
if (n > given[given.length - 1])
return true;
 
if (board[r][c] != 0 && board[r][c] != n)
return false;
 
if (board[r][c] == 0 && given[next] == n)
return false;
 
int back = board[r][c];
if (back == n)
next++;
 
board[r][c] = n;
for (int i = -1; i < 2; i++)
for (int j = -1; j < 2; j++)
if (solve(r + i, c + j, n + 1, next))
return true;
 
board[r][c] = back;
return false;
}
 
private static void printBoard() {
for (int[] row : board) {
for (int c : row) {
if (c == -1)
System.out.print(" . ");
else
System.out.printf(c > 0 ? "%2d " : "__ ", c);
}
System.out.println();
}
}
}

Output:

 .  .  .  .  .  .  .  .  .  . 
 . __ 33 35 __ __  .  .  .  . 
 . __ __ 24 22 __  .  .  .  . 
 . __ __ __ 21 __ __  .  .  . 
 . __ 26 __ 13 40 11  .  .  . 
 . 27 __ __ __  9 __  1  .  . 
 .  .  . __ __ 18 __ __  .  . 
 .  .  .  .  . __  7 __ __  . 
 .  .  .  .  .  .  .  5 __  . 
 .  .  .  .  .  .  .  .  .  . 

Found:
 .  .  .  .  .  .  .  .  .  . 
 . 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  . 
 .  .  .  .  .  .  .  .  .  .

[edit] Mathprog

/*Hidato.mathprog, part of KuKu by Nigel Galloway
 
Find a solution to a Hidato problem
 
[email protected]
April 1st., 2011
*/
 
param ZBLS;
param ROWS;
param COLS;
param D := 1;
set ROWSR := 1..ROWS;
set COLSR := 1..COLS;
set ROWSV := (1-D)..(ROWS+D);
set COLSV := (1-D)..(COLS+D);
param Iz{ROWSR,COLSR}, integer, default 0;
set ZBLSV := 1..(ZBLS+1);
set ZBLSR := 1..ZBLS;
 
var BR{ROWSV,COLSV,ZBLSV}, binary;
 
void0{r in ROWSV, z in ZBLSR,c in (1-D)..0}: BR[r,c,z] = 0;
void1{r in ROWSV, z in ZBLSR,c in (COLS+1)..(COLS+D)}: BR[r,c,z] = 0;
void2{c in COLSV, z in ZBLSR,r in (1-D)..0}: BR[r,c,z] = 0;
void3{c in COLSV, z in ZBLSR,r in (ROWS+1)..(ROWS+D)}: BR[r,c,z] = 0;
void4{r in ROWSV,c in (1-D)..0}: BR[r,c,ZBLS+1] = 1;
void5{r in ROWSV,c in (COLS+1)..(COLS+D)}: BR[r,c,ZBLS+1] = 1;
void6{c in COLSV,r in (1-D)..0}: BR[r,c,ZBLS+1] = 1;
void7{c in COLSV,r in (ROWS+1)..(ROWS+D)}: BR[r,c,ZBLS+1] = 1;
 
Izfree{r in ROWSR, c in COLSR, z in ZBLSR : Iz[r,c] = -1}: BR[r,c,z] = 0;
Iz1{Izr in ROWSR, Izc in COLSR, r in ROWSR, c in COLSR, z in ZBLSR : Izr=r and Izc=c and Iz[Izr,Izc]=z}: BR[r,c,z] = 1;
 
rule1{z in ZBLSR}: sum{r in ROWSR, c in COLSR} BR[r,c,z] = 1;
rule2{r in ROWSR, c in COLSR}: sum{z in ZBLSV} BR[r,c,z] = 1;
rule3{r in ROWSR, c in COLSR, z in ZBLSR}: BR[0,0,z+1] + BR[r-1,c-1,z+1] + BR[r-1,c,z+1] + BR[r-1,c+1,z+1] + BR[r,c-1,z+1] + BR[r,c+1,z+1] + BR[r+1,c-1,z+1] + BR[r+1,c,z+1] + BR[r+1,c+1,z+1] - BR[r,c,z] >= 0;
 
solve;
 
for {r in ROWSR} {
for {c in COLSR} {
printf " %2d", sum{z in ZBLSR} BR[r,c,z]*z;
}
printf "\n";
}
data;
 
param ROWS := 8;
param COLS := 8;
param ZBLS := 40;
param
Iz: 1 2 3 4 5 6 7 8 :=
1 . 33 35 . . -1 -1 -1
2 . . 24 22 . -1 -1 -1
3 . . . 21 . . -1 -1
4 . 26 . 13 40 11 -1 -1
5 27 . . . 9 . 1 -1
6 -1 -1 . . 18 . . -1
7 -1 -1 -1 -1 . 7 . .
8 -1 -1 -1 -1 -1 -1 5 .
 ;
 
end;

Using the data in the model produces the following:

Output:
>glpsol --minisat --math Hidato.mathprog
GLPSOL: GLPK LP/MIP Solver, v4.47
Parameter(s) specified in the command line:
 --minisat --math Hidato.mathprog
Reading model section from Hidato.mathprog...
Reading data section from Hidato.mathprog...
64 lines were read
Generating void0...
Generating void1...
Generating void2...
Generating void3...
Generating void4...
Generating void5...
Generating void6...
Generating void7...
Generating Izfree...
Generating Iz1...
Generating rule1...
Generating rule2...
Generating rule3...
Model has been successfully generated
Will search for ANY feasible solution
Translating to CNF-SAT...
Original problem has 5279 rows, 4100 columns, and 33359 non-zeros
2520 covering inequalities
2719 partitioning equalities
Solving CNF-SAT problem...
Instance has 7076 variables, 24047 clauses, and 77735 literals
==================================[MINISAT]===================================
| Conflicts |     ORIGINAL     |              LEARNT              | Progress |
|           | Clauses Literals |   Limit Clauses Literals  Lit/Cl |          |
==============================================================================
|         0 |   21432    75120 |    7144       0        0     0.0 |  0.000 % |
==============================================================================
SATISFIABLE
Objective value =  0.000000000e+000
Time used:   0.0 secs
Memory used: 14.5 Mb (15192264 bytes)
 32 33 35 36 37  0  0  0
 31 34 24 22 38  0  0  0
 30 25 23 21 12 39  0  0
 29 26 20 13 40 11  0  0
 27 28 14 19  9 10  1  0
  0  0 15 16 18  8  2  0
  0  0  0  0 17  7  6  3
  0  0  0  0  0  0  5  4
Model has been successfully processed

Modelling Evil Case 1:

data;
param ROWS := 3;
param COLS := 3;
param ZBLS := 7;
param
Iz: 1   2   3 :=
 1 -1   4  -1
 2  .   7   .
 3  1   .   .
 ;
end;

Produces:

>glpsol --minisat --math Hidato.mathprog --data Evil1.data
GLPSOL: GLPK LP/MIP Solver, v4.47
Parameter(s) specified in the command line:
 --minisat --math Hidato.mathprog --data Evil1.data
Reading model section from Hidato.mathprog...
Hidato.mathprog:47: warning: data section ignored
47 lines were read
Reading data section from Evil1.data...
11 lines were read
Generating void0...
Generating void1...
Generating void2...
Generating void3...
Generating void4...
Generating void5...
Generating void6...
Generating void7...
Generating Izfree...
Generating Iz1...
Generating rule1...
Generating rule2...
Generating rule3...
Model has been successfully generated
Will search for ANY feasible solution
Translating to CNF-SAT...
Original problem has 256 rows, 200 columns, and 935 non-zeros
56 covering inequalities
193 partitioning equalities
Solving CNF-SAT problem...
Instance has 337 variables, 1237 clauses, and 4094 literals
==================================[MINISAT]===================================
| Conflicts |     ORIGINAL     |              LEARNT              | Progress |
|           | Clauses Literals |   Limit Clauses Literals  Lit/Cl |          |
==============================================================================
|         0 |    1060     3917 |     353       0        0     0.0 |  0.000 % |
==============================================================================
SATISFIABLE
Objective value =  0.000000000e+000
Time used:   0.0 secs
Memory used: 0.8 Mb (861188 bytes)
  0  4  0
  3  7  5
  1  2  6
Model has been successfully processed

Modelling Evil Case 2 - The Snake in the Grass:

data;
param ROWS := 3;
param COLS := 50;
param ZBLS := 74;
param
Iz:  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 :=
 1   1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1 74
 2  -1 -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1  . -1
 3  -1 -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1 -1  .  . -1
;
end;

Produces:

G:\IAJAAR4.47>glpsol --minisat --math Hidato.mathprog --data Evil2.data
GLPSOL: GLPK LP/MIP Solver, v4.47
Parameter(s) specified in the command line:
 --minisat --math Hidato.mathprog --data Evil2.data
Reading model section from Hidato.mathprog...
Hidato.mathprog:47: warning: data section ignored
47 lines were read
Reading data section from Evil2.data...
Evil2.data:11: warning: final NL missing before end of file
11 lines were read
Generating void0...
Generating void1...
Generating void2...
Generating void3...
Generating void4...
Generating void5...
Generating void6...
Generating void7...
Generating Izfree...
Generating Iz1...
Generating rule1...
Generating rule2...
Generating rule3...
Model has been successfully generated
Will search for ANY feasible solution
Translating to CNF-SAT...
Original problem has 25500 rows, 19500 columns, and 147452 non-zeros
11026 covering inequalities
14400 partitioning equalities
Solving CNF-SAT problem...
Instance has 31338 variables, 98310 clauses, and 305726 literals
==================================[MINISAT]===================================
| Conflicts |     ORIGINAL     |              LEARNT              | Progress |
|           | Clauses Literals |   Limit Clauses Literals  Lit/Cl |          |
==============================================================================
|         0 |   84134   291550 |   28044       0        0     0.0 |  0.000 % |
|       101 |   31135   126809 |   30848      98     5496    56.1 | 65.521 % |
|       251 |   31135   126809 |   33933     244    12470    51.1 | 66.552 % |
|       476 |   27353   115512 |   37327     446    23819    53.4 | 68.160 % |
|       814 |   26574   113330 |   41059     770    42161    54.8 | 69.586 % |
|      1321 |   25432   110534 |   45165    1262    83658    66.3 | 70.056 % |
==============================================================================
SATISFIABLE
Objective value =  0.000000000e+000
Time used:   1.0 secs
Memory used: 60.9 Mb (63862624 bytes)
  1  2  3  0  0  8  9  0  0 14 15  0  0 20 21  0  0 26 27  0  0 32 33  0  0 38 39  0  0 44 45  0  0 50 51  0  0 56 57  0  0 62 63  0  0 68 69  0  0 74
  0  0  4  0  7  0 10  0 13  0 16  0 19  0 22  0 25  0 28  0 31  0 34  0 37  0 40  0 43  0 46  0 49  0 52  0 55  0 58  0 61  0 64  0 67  0 70  0 73  0
  0  0  0  5  6  0  0 11 12  0  0 17 18  0  0 23 24  0  0 29 30  0  0 35 36  0  0 41 42  0  0 47 48  0  0 53 54  0  0 59 60  0  0 65 66  0  0 71 72  0
Model has been successfully processed

Modelling Evil Case 3 - A fatter snake in the Grass:

data;
param ROWS := 4;
param COLS := 46;
param ZBLS := 82;
param
Iz:  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 :=
 1   1  0 -1 -1 -1  0  0 -1 -1 -1  0  0 -1 -1 -1  0  0 -1 -1 -1  0  0 -1 -1 -1  0  0 -1 -1 -1  0  0 -1 -1 -1  0  0 -1 -1 -1  0  0 -1 -1 -1 82
 2  -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 
 3  -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1  0 -1  0 -1 -1 
 4   0  0  0 -1 -1  0  0  0 -1 -1  0  0  0 -1 -1  0  0  0 -1 -1  0  0  0 -1 -1  0  0  0 -1 -1  0  0  0 -1 -1  0  0  0 -1 -1  0  0  0 -1 -1 -1
 ;
end;

Produces:

>glpsol --minisat --math Hidato.mathprog --data Evil3.data
GLPSOL: GLPK LP/MIP Solver, v4.47
Parameter(s) specified in the command line:
 --minisat --math Hidato.mathprog --data Evil3.data
Reading model section from Hidato.mathprog...
Hidato.mathprog:47: warning: data section ignored
47 lines were read
Reading data section from Evil3.data...
12 lines were read
Generating void0...
Generating void1...
Generating void2...
Generating void3...
Generating void4...
Generating void5...
Generating void6...
Generating void7...
Generating Izfree...
Generating Iz1...
Generating rule1...
Generating rule2...
Generating rule3...
Model has been successfully generated
Will search for ANY feasible solution
Translating to CNF-SAT...
Original problem has 32684 rows, 23904 columns, and 198488 non-zeros
15006 covering inequalities
17596 partitioning equalities
Solving CNF-SAT problem...
Instance has 39792 variables, 130040 clauses, and 407222 literals
==================================[MINISAT]===================================
| Conflicts |     ORIGINAL     |              LEARNT              | Progress |
|           | Clauses Literals |   Limit Clauses Literals  Lit/Cl |          |
==============================================================================
|         0 |  112710   389892 |   37570       0        0     0.0 |  0.000 % |
==============================================================================
SATISFIABLE
Objective value =  0.000000000e+000
Time used:   0.0 secs
Memory used: 80.2 Mb (84067912 bytes)
  1  2  0  0  0 10 11  0  0  0 19 20  0  0  0 28 29  0  0  0 37 38  0  0  0 46 47  0  0  0 55 56  0  0  0 64 65  0  0  0 73 74  0  0  0 82
  0  0  3  0  9  0  0 12  0 18  0  0 21  0 27  0  0 30  0 36  0  0 39  0 45  0  0 48  0 54  0  0 57  0 63  0  0 66  0 72  0  0 75  0 81  0
  0  4  0  8  0  0 13  0 17  0  0 22  0 26  0  0 31  0 35  0  0 40  0 44  0  0 49  0 53  0  0 58  0 62  0  0 67  0 71  0  0 76  0 80  0  0
  5  6  7  0  0 14 15 16  0  0 23 24 25  0  0 32 33 34  0  0 41 42 43  0  0 50 51 52  0  0 59 60 61  0  0 68 69 70  0  0 77 78 79  0  0  0
Model has been successfully processed

[edit] Nimrod

import strutils, algorithm
 
var board: array[0..19, array[0..19, int]]
var given, start: seq[int] = @[]
var rows, cols: int = 0
 
proc setup(s: string) =
var lines = s.splitLines()
cols = lines[0].split().len()
rows = lines.len()
 
for i in 0 .. rows + 1:
for j in 0 .. cols + 1:
board[i][j] = -1
 
for r, row in pairs(lines):
for c, cell in pairs(row.split()):
case cell
of "__" :
board[r + 1][c + 1] = 0
continue
of "." : continue
else :
var val = parseInt(cell)
board[r + 1][c + 1] = val
given.add(val)
if (val == 1):
start.add(r + 1)
start.add(c + 1)
given.sort(cmp[int], Ascending)
 
proc solve(r, c, n: int, next: int = 0): Bool =
if n > given[high(given)]:
return True
if board[r][c] < 0:
return False
if (board[r][c] > 0 and board[r][c] != n):
return False
if (board[r][c] == 0 and given[next] == n):
return False
 
var back = board[r][c]
board[r][c] = n
for i in -1 .. 1:
for j in -1 .. 1:
if back == n:
if (solve(r + i, c + j, n + 1, next + 1)): return True
else:
if (solve(r + i, c + j, n + 1, next)): return True
board[r][c] = back
result = False
 
 
proc printBoard() =
for r in 0 .. rows + 1:
for cellid,c in pairs(board[r]):
if cellid > cols + 1: break
if c == -1:
write(stdout, " . ")
elif c == 0:
write(stdout, "__ ")
else:
write(stdout, "$# " % align($c,2))
writeln(stdout, "")
 
var hi: string = """__ 33 35 __ __ . . .
__ __ 24 22 __ . . .
__ __ __ 21 __ __ . .
__ 26 __ 13 40 11 . .
27 __ __ __ 9 __ 1 .
. . __ __ 18 __ __ .
. . . . __ 7 __ __
. . . . . . 5 __"""
 
setup(hi)
printBoard()
echo("")
echo("Found:")
discard solve(start[0], start[1], 1)
printBoard()
Output:
 .  .  .  .  .  .  .  .  .  . 
 . __ 33 35 __ __  .  .  .  . 
 . __ __ 24 22 __  .  .  .  . 
 . __ __ __ 21 __ __  .  .  . 
 . __ 26 __ 13 40 11  .  .  . 
 . 27 __ __ __  9 __  1  .  . 
 .  .  . __ __ 18 __ __  .  . 
 .  .  .  .  . __  7 __ __  . 
 .  .  .  .  .  .  .  5 __  . 
 .  .  .  .  .  .  .  .  .  . 

Found:
 .  .  .  .  .  .  .  .  .  . 
 . 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  . 
 .  .  .  .  .  .  .  .  .  . 

[edit] Perl

use strict;
use List::Util 'max';
 
our (@grid, @known, $n);
 
sub show_board {
for my $r (@grid) {
print map(!defined($_) ? ' ' : $_
? sprintf("%3d", $_)
: ' __'
, @$r), "\n"
}
}
 
sub parse_board {
@grid = map{[map(/^_/ ? 0 : /^\./ ? undef: $_, split ' ')]}
split "\n", shift();
for my $y (0 .. $#grid) {
for my $x (0 .. $#{$grid[$y]}) {
$grid[$y][$x] > 0
and $known[$grid[$y][$x]] = "$y,$x";
}
}
$n = max(map { max @$_ } @grid);
}
 
sub neighbors {
my ($y, $x) = @_;
my @out;
for ( [-1, -1], [-1, 0], [-1, 1],
[ 0, -1], [ 0, 1],
[ 1, -1], [ 1, 0], [ 1, 1])
{
my $y1 = $y + $_->[0];
my $x1 = $x + $_->[1];
next if $x1 < 0 || $y1 < 0;
next unless defined $grid[$y1][$x1];
push @out, "$y1,$x1";
}
@out
}
 
sub try_fill {
my ($v, $coord) = @_;
return 1 if $v > $n;
 
my ($y, $x) = split ',', $coord;
my $old = $grid[$y][$x];
 
return if $old && $old != $v;
return if exists $known[$v] and $known[$v] ne $coord;
 
$grid[$y][$x] = $v;
print "\033[0H";
show_board();
 
try_fill($v + 1, $_) && return 1
for neighbors($y, $x);
 
$grid[$y][$x] = $old;
return
}
 
parse_board
# ". 4 .
# _ 7 _
# 1 _ _";
 
# " 1 _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . 74
# . . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _
# . . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _
# ";
 
"__ 33 35 __ __ .. .. .. .
__ __ 24 22 __ .. .. .. .
__ __ __ 21 __ __ .. .. .
__ 26 __ 13 40 11 .. .. .
27 __ __ __ 9 __ 1 .. .
. . __ __ 18 __ __ .. .
. .. . . __ 7 __ __ .
. .. .. .. . . 5 __ ."
;
 
print "\033[2J";
try_fill(1, $known[1]);
Output:
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

[edit] 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 appraoch. Rather than recalculating degree over and over, we maintain an array of known degrees for each node.

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";
}

[edit] PicoLisp

(load "@lib/simul.l")
 
(de hidato (Lst)
(let Grid (grid (length (maxi length Lst)) (length Lst))
(mapc
'((G L)
(mapc
'((This Val)
(nond
(Val
(with (: 0 1 1) (con (: 0 1))) # Cut off west
(with (: 0 1 -1) (set (: 0 1))) # east
(with (: 0 -1 1) (con (: 0 -1))) # south
(with (: 0 -1 -1) (set (: 0 -1))) # north
(set This) )
((=T Val) (=: val Val)) ) )
G L ) )
Grid
(apply mapcar (reverse Lst) list) )
(let Todo
(by '((This) (: val)) sort
(mapcan '((Col) (filter '((This) (: val)) Col))
Grid ) )
(let N 1
(with (pop 'Todo)
(recur (N Todo)
(unless (> (inc 'N) (; Todo 1 val))
(find
'((Dir)
(with (Dir This)
(cond
((= N (: val))
(if (cdr Todo) (recurse N @) T) )
((not (: val))
(=: val N)
(or (recurse N Todo) (=: val NIL)) ) ) ) )
(quote
west east south north
((X) (or (south (west X)) (west (south X))))
((X) (or (north (west X)) (west (north X))))
((X) (or (south (east X)) (east (south X))))
((X) (or (north (east X)) (east (north X)))) ) ) ) ) ) ) )
(disp Grid 0
'((This)
(if (: val) (align 3 @) " ") ) ) ) )

Test:

(hidato
(quote
(T 33 35 T T)
(T T 24 22 T)
(T T T 21 T T)
(T 26 T 13 40 11)
(27 T T T 9 T 1)
(NIL NIL T T 18 T T)
(NIL NIL NIL NIL T 7 T T)
(NIL NIL NIL NIL NIL NIL 5 T) ) )

Output:

   +---+---+---+---+---+---+---+---+
 8 | 32  33  35  36  37|   |   |   |
   +   +   +   +   +   +---+---+---+
 7 | 31  34  24  22  38|   |   |   |
   +   +   +   +   +   +---+---+---+
 6 | 30  25  23  21  12  39|   |   |
   +   +   +   +   +   +   +---+---+
 5 | 29  26  20  13  40  11|   |   |
   +   +   +   +   +   +   +---+---+
 4 | 27  28  14  19   9  10   1|   |
   +---+---+   +   +   +   +   +---+
 3 |   |   | 15  16  18   8   2|   |
   +---+---+---+---+   +   +   +---+
 2 |   |   |   |   | 17   7   6   3|
   +---+---+---+---+---+---+   +   +
 1 |   |   |   |   |   |   |  5   4|
   +---+---+---+---+---+---+---+---+
     a   b   c   d   e   f   g   h

[edit] Prolog

Works with SWI-Prolog and library(clpfd) written by Markus Triska.
Puzzle solved is from the Wilkipedia page : http://en.wikipedia.org/wiki/Hidato

:- use_module(library(clpfd)).
 
hidato :-
init1(Li),
% skip first blank line
init2(1, 1, 10, Li),
my_write(Li).
 
 
init1(Li) :-
Li = [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, A, 33, 35, B, C, 0, 0, 0, 0,
0, D, E, 24, 22, F, 0, 0, 0, 0,
0, G, H, I, 21, J, K, 0, 0, 0,
0, L, 26, M, 13, 40, 11, 0, 0, 0,
0, 27, N, O, P, 9, Q, 1, 0, 0,
0, 0, 0, R, S, 18, T, U, 0, 0,
0, 0, 0, 0, 0, V, 7, W, X, 0,
0, 0, 0, 0, 0, 0, 0, 5, Y, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 
LV = [ A, 33, 35, B, C,
D, E, 24, 22, F,
G, H, I, 21, J, K,
L, 26, M, 13, 40, 11,
27, N, O, P, 9, Q, 1,
R, S, 18, T, U,
V, 7, W, X,
5, Y],
 
 
LV ins 1..40,
all_distinct(LV).
 
% give the constraints
% Stop before the last line
init2(_N, Col, Max_Col, _L) :-
Col is Max_Col - 1.
 
% skip zeros
init2(N, Lig, Col, L) :-
I is N + Lig * Col,
element(I, L, 0),
!,
V is N+1,
( V > Col -> N1 = 1, Lig1 is Lig + 1; N1 = V, Lig1 = Lig),
init2(N1, Lig1, Col, L).
 
 
% skip first column
init2(1, Lig, Col, L) :-
!,
init2(2, Lig, Col, L) .
 
% skip last column
init2(Col, Lig, Col, L) :-
!,
Lig1 is Lig+1,
init2(1, Lig1, Col, L).
 
% V5 V3 V6
% V1 V V2
% V7 V4 V8
% general case
init2(N, Lig, Col, L) :-
I is N + Lig * Col,
element(I, L, V),
 
I1 is I - 1, I2 is I + 1, I3 is I - Col, I4 is I + Col,
I5 is I3 - 1, I6 is I3 + 1, I7 is I4 - 1, I8 is I4 + 1,
 
maplist(compute_BI(L, V), [I1,I2,I3,I4,I5,I6,I7,I8], VI, BI),
 
sum(BI, #=, SBI),
 
( ((V #= 1 #\/ V #= 40) #/\ SBI #= 1) #\/
(V #\= 1 #/\ V #\= 40 #/\ SBI #= 2)) #<==> 1,
 
labeling([ffc, enum], [V | VI]),
 
N1 is N+1,
init2(N1, Lig, Col, L).
 
compute_BI(L, V, I, VI, BI) :-
element(I, L, VI),
VI #= 0 #==> BI #= 0,
( VI #\= 0 #/\ (V - VI #= 1 #\/ VI - V #= 1)) #<==> BI.
 
% display the result
my_write([0, A, B, C, D, E, F, G, H, 0 | T]) :-
maplist(my_write_1, [A, B, C, D, E, F, G, H]), nl,
my_write(T).
 
my_write([]).
 
my_write_1(0) :-
write(' ').
 
my_write_1(X) :-
writef('%3r', [X]).
Output:
?- hidato.
 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
true 

[edit] Python

board = []
given = []
start = None
 
def setup(s):
global board, given, start
lines = s.splitlines()
ncols = len(lines[0].split())
nrows = len(lines)
board = [[-1] * (ncols + 2) for _ in xrange(nrows + 2)]
 
for r, row in enumerate(lines):
for c, cell in enumerate(row.split()):
if cell == "__" :
board[r + 1][c + 1] = 0
continue
elif cell == ".":
continue # -1
else:
val = int(cell)
board[r + 1][c + 1] = val
given.append(val)
if val == 1:
start = (r + 1, c + 1)
given.sort()
 
def solve(r, c, n, next=0):
if n > given[-1]:
return True
if board[r][c] and board[r][c] != n:
return False
if board[r][c] == 0 and given[next] == n:
return False
 
back = 0
if board[r][c] == n:
next += 1
back = n
 
board[r][c] = n
for i in xrange(-1, 2):
for j in xrange(-1, 2):
if solve(r + i, c + j, n + 1, next):
return True
board[r][c] = back
return False
 
def print_board():
d = {-1: " ", 0: "__"}
bmax = max(max(r) for r in board)
form = "%" + str(len(str(bmax)) + 1) + "s"
for r in board[1:-1]:
print "".join(form % d.get(c, str(c)) for c in r[1:-1])
 
hi = """\
__ 33 35 __ __ . . .
__ __ 24 22 __ . . .
__ __ __ 21 __ __ . .
__ 26 __ 13 40 11 . .
27 __ __ __ 9 __ 1 .
. . __ __ 18 __ __ .
. . . . __ 7 __ __
. . . . . . 5 __"""

 
setup(hi)
print_board()
solve(start[0], start[1], 1)
print
print_board()
Output:
 __ 33 35 __ __         
 __ __ 24 22 __         
 __ __ __ 21 __ __      
 __ 26 __ 13 40 11      
 27 __ __ __  9 __  1   
       __ __ 18 __ __   
             __  7 __ __
                    5 __

 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

[edit] Racket

[edit] Standalone

Algorithm is depth first search for each number, repeating for all numbers in ascending order. It currently runs slowish due to temporary shortcomings in untyped Racket's array indexing, but finished immediately when tested with custom 2d vector library.

 
#lang racket
(require math/array)
 
;#f = not a legal position, #t = blank position
(define board
(array
#[#[#t 33 35 #t #t #f #f #f]
#[#t #t 24 22 #t #f #f #f]
#[#t #t #t 21 #t #t #f #f]
#[#t 26 #t 13 40 11 #f #f]
#[27 #t #t #t 9 #t 1 #f]
#[#f #f #t #t 18 #t #t #f]
#[#f #f #f #f #t 7 #t #t]
#[#f #f #f #f #f #f 5 #t]]))
 
;filters elements with the predicate, returning the element and its indices
(define (array-indices-of a f)
(for*/list ([i (range 0 (vector-ref (array-shape a) 0))]
[j (range 0 (vector-ref (array-shape a) 1))]
#:when (f (array-ref a (vector i j))))
(list (array-ref a (vector i j)) i j)))
 
;returns a list, each element is a list of the number followed by i and j indices
;sorted ascending by number
(define (goal-list v) (sort (array-indices-of v number?) (λ (a b) (< (car a) (car b)))))
 
;every direction + start position that's on the board
(define (legal-moves a i0 j0)
(for*/list ([i (range (sub1 i0) (+ i0 2))]
[j (range (sub1 j0) (+ j0 2))]
 ;cartesian product -1..1 and -1..1, except 0 0
#:when (and (not (and (= i i0) (= j j0)))
 ;make sure it's on the board
(<= 0 i (sub1 (vector-ref (array-shape a) 0)))
(<= 0 j (sub1 (vector-ref (array-shape a) 1)))
 ;make sure it's an actual position too (the real board isn't square)
(array-ref a (vector i j))))
(cons i j)))
 
;find path through array, returning list of coords from start to finish
(define (hidato-path a)
 ;get starting position as first goal
(match-let ([(cons (list n i j) goals) (goal-list a)])
(let hidato ([goals goals] [n n] [i i] [j j] [path '()])
(match goals
 ;no more goals, return path
['() (reverse (cons (cons i j) path))]
 ;get next goal
[(cons (list n-goal i-goal j-goal) _)
(let ([move (cons i j)])
 ;already visiting a spot or taking too many moves to reach the next goal is no good
(cond [(or (member move path) (> n n-goal)) #f]
 ;taking the right number of moves to be at the goal square is good
 ;so go to the next goal
[(and (= n n-goal) (= i i-goal) (= j j-goal))
(hidato (cdr goals) n i j path)]
 ;depth first search using every legal move to find next goal
[else (ormap (λ (m) (hidato goals (add1 n) (car m) (cdr m) (cons move path)))
(legal-moves a i j))]))]))))
 
;take a path and insert it into the array
(define (put-path a path)
(let ([a (array->mutable-array a)])
(for ([n (range 1 (add1 (length path)))] [move path])
(array-set! a (vector (car move) (cdr move)) n))
a))
 
;main function
(define (hidato board) (put-path board (hidato-path board)))
 
Output:
> (hidato board)
(mutable-array
 #[#[32 33 35 36 37 #f #f #f]
   #[31 34 24 22 38 #f #f #f]
   #[30 25 23 21 12 39 #f #f]
   #[29 26 20 13 40 11 #f #f]
   #[27 28 14 19 9 10 1 #f]
   #[#f #f 15 16 18 8 2 #f]
   #[#f #f #f #f 17 7 6 3]
   #[#f #f #f #f #f #f 5 4]])

[edit] Using Hidato Family Solver from Numbrix

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 moore-neighbour-offsets
'((+1 0) (-1 0) (0 +1) (0 -1) (+1 +1) (-1 -1) (-1 +1) (+1 -1)))
 
(define solve-hidato (solve-hidato-family moore-neighbour-offsets))
 
(displayln
(puzzle->string
(solve-hidato
#(#( 0 33 35 0 0)
#( 0 0 24 22 0)
#( 0 0 0 21 0 0)
#( 0 26 0 13 40 11)
#(27 0 0 0 9 0 1)
#( _ _ 0 0 18 0 0)
#( _ _ _ _ 0 7 0 0)
#( _ _ _ _ _ _ 5 0)))))
 
Output:
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

[edit] REXX

Programming note:   the coördinates for the cells used are the same as an   X Y   grid, that is,
the bottom left-most cell is   1 1   and the tenth cell on row 2 is   2 10
If any marker is negative, then it's assumed to be a Numbrix puzzle (and the absolute value is used).
Over half of the REXX program deals with validating the input and displaying the puzzle.

Hidato and Numbrix are registered trademarks.

/*REXX pgm solves a Hidato or Numbrix puzzle, displays puzzle & solution*/
maxr=0; maxc=0; maxx=0; minr=9e9; minc=9e9; minx=9e9; cells=0; @.=
parse arg xxx; PZ='Hidato puzzle' /*get cell definitions from C.L. */
xxx=translate(xxx, , "/\;:_", ',') /*also allow other chars 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 do; x=x/1 /*normalize X*/
if x<0 then PZ='Numbrix puzzle'
x=abs(x) /*use │x│ */
end
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 /*start cell.*/
if _\=='' then call err "cell at" r c 'is already occupied with:' _
@.r.c=x; c=c+1; cells=cells+1 /*assign mark*/
if x==. then iterate /*hole? Skip.*/
if \datatype(x,'W') then call err 'illegal marker specified:' x
minx=min(minx,x); maxx=max(maxx,x) /*min & max X*/
end /*while marks¬='' */
end /*while xxx ¬='' */
call showGrid /* [↓] used for making fast moves*/
Nr = '0 1 0 -1 -1 1 1 -1' /*possible row for the next move.*/
Nc = '1 0 -1 0 1 -1 1 -1' /* " col " " " " */
pMoves=words(Nr) -4*(left(PZ,1)=='N') /*is this to be a Numbrix puzzle?*/
do i=1 for pMoves; Nr.i=word(Nr,i); Nc.i=word(Nc,i); end /*fast moves*/
if \next(2,!r,!c) then say 'No solution possible for this' PZ "puzzle."
else say 'A solution for the' PZ "exists."
say; call showGrid
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────ERR subroutine──────────────────────*/
err: say; say '***error!*** (from' PZ"): " arg(1); say; exit 13
/*──────────────────────────────────NEXT subroutine─────────────────────*/
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=# /*a move.*/
if #==cells then leave /*last 1?*/
if next(##,nr,nc) then return 1
@.nr.nc=. /*undo the above move. */
iterate /*go & try another move*/
end
if @.nr.nc==# then do /*is this a fill-in ? */
if #==cells then return 1 /*last 1.*/
if next(##,nr,nc) then return 1 /*fill-in*/
end
end /*t*/
return 0 /*This ain't working. */
/*──────────────────────────────────SHOWGRID subroutine─────────────────*/
showGrid: 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'
if maxx\==cells then call err 'no' cells "was specified for the puzzle end"
w=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 using the following as input:
1 7 5 .\2 5 . 7 . .\3 3 . . 18 . .\4 1 27 . . . 9 . 1\5 1 . 26 . 13 40 11\6 1 . . . 21 . .\7 1 . . 24 22 .\8 1 . 33 35 . .

  . 33 35  .  .
  .  . 24 22  .
  .  .  . 21  .  .
  . 26  . 13 40 11
 27  .  .  .  9  .  1
        .  . 18  .  .
              .  7  .  .
                    5  .


A solution for the Hidato puzzle exists.

 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

[edit] Ruby

[edit] Without Warnsdorff

The following class provides functionality for solving a hidato problem:

# Solve a Hidato Puzzle
#
class Hidato
Cell = Struct.new(:value, :used, :adj)
ADJUST = [[-1, -1], [-1, 0], [-1, 1], [0, -1], [0, 1], [1, -1], [1, 0], [1, 1]]
 
def initialize(board, pout=true)
@board = []
board.each_line do |line|
@board << line.split.map{|n| Cell[Integer(n), false] rescue nil} + [nil]
end
@board << [] # frame (Sentinel value : nil)
@board.each_with_index do |row, x|
row.each_with_index do |cell, y|
if cell
@sx, @sy = x, y if cell.value==1 # start position
cell.adj = ADJUST.map{|dx,dy| [x+dx,y+dy]}.select{|xx,yy| @board[xx][yy]}
end
end
end
@xmax = @board.size - 1
@ymax = @board.map(&:size).max - 1
@end = @board.flatten.compact.size
puts to_s('Problem:') if pout
end
 
def solve
@zbl = Array.new(@end+1, false)
@board.flatten.compact.each{|cell| @zbl[cell.value] = true}
puts (try(@board[@sx][@sy], 1) ? to_s('Solution:') : "No solution")
end
 
def try(cell, seq_num)
return true if seq_num > @end
return false if cell.used
value = cell.value
return false if value > 0 and value != seq_num
return false if value == 0 and @zbl[seq_num]
cell.used = true
cell.adj.each do |x, y|
if try(@board[x][y], seq_num+1)
cell.value = seq_num
return true
end
end
cell.used = false
end
 
def to_s(msg=nil)
str = (0...@xmax).map do |x|
(0...@ymax).map{|y| "%3s" % ((c=@board[x][y]) ? c.value : c)}.join
end
(msg ? [msg] : []) + str + [""]
end
end

Test:

# Which may be used as follows to solve Evil Case 1:
board1 = <<EOS
. 4
0 7 0
1 0 0
EOS

Hidato.new(board1).solve
 
# Which may be used as follows to solve this tasks example:
board2 = <<EOS
0 33 35 0 0
0 0 24 22 0
0 0 0 21 0 0
0 26 0 13 40 11
27 0 0 0 9 0 1
. . 0 0 18 0 0
. . . . 0 7 0 0
. . . . . . 5 0
EOS

Hidato.new(board2).solve
 
# Which may be used as follows to solve The Snake in the Grass:
board3 = <<EOS
1 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 74
. . 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 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 .
EOS

t0 = Time.now
Hidato.new(board3).solve
puts " #{Time.now - t0} sec"
Output:
Problem:
     4   
  0  7  0
  1  0  0

Solution:
     4   
  3  7  5
  1  2  6

Problem:
  0 33 35  0  0         
  0  0 24 22  0         
  0  0  0 21  0  0      
  0 26  0 13 40 11      
 27  0  0  0  9  0  1   
        0  0 18  0  0   
              0  7  0  0
                    5  0

Solution:
 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

Problem:
  1  0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0       74
        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  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0   

Solution:
  1  2  3        8  9       14 15       20 21       26 27       32 33       38 39       44 45       50 51       56 57       62 63       68 69       74
        4     7    10    13    16    19    22    25    28    31    34    37    40    43    46    49    52    55    58    61    64    67    70    73   
           5  6       11 12       17 18       23 24       29 30       35 36       41 42       47 48       53 54       59 60       65 66       71 72   

 40.198299 sec

[edit] With Warnsdorff

I modify method as follows to implement Warnsdorff like

# Solve a Hidato Like Puzzle with Warnsdorff like logic applied
#
class HLPsolver
Cell = Struct.new(:value, :used, :adj)
 
def initialize(board, pout=true)
@board = []
frame = ADJACENT.flatten.map(&:abs).max
board.each_line do |line|
@board << line.split.map{|n| Cell[Integer(n), false] rescue nil} + [nil]*frame
end
frame.times {@board << []} # frame (Sentinel value : nil)
@board.each_with_index do |row, x|
row.each_with_index do |cell, y|
if cell
@sx, @sy = x, y if cell.value==1 # start position
cell.adj = ADJACENT.map{|dx,dy| [x+dx,y+dy]}.select{|xx,yy| @board[xx][yy]}
end
end
end
@xmax = @board.size - frame
@ymax = @board.map(&:size).max - frame
@end = @board.flatten.compact.size
@format = " %#{@end.to_s.size}s"
puts to_s('Problem:') if pout
end
 
def solve
@zbl = Array.new(@end+1, false)
@board.flatten.compact.each{|cell| @zbl[cell.value] = true}
puts (try(@board[@sx][@sy], 1) ? to_s('Solution:') : "No solution")
end
 
def try(cell, seq_num)
value = cell.value
return false if value > 0 and value != seq_num
return false if value == 0 and @zbl[seq_num]
cell.used = true
if seq_num == @end
cell.value = seq_num
return true
end
a = []
cell.adj.each_with_index do |(x, y), n|
cl = @board[x][y]
a << [wdof(cl.adj)*10+n, x, y] unless cl.used
end
a.sort.each do |key, x, y|
if try(@board[x][y], seq_num+1)
cell.value = seq_num
return true
end
end
cell.used = false
end
 
def wdof(adj)
adj.count {|x,y| not @board[x][y].used}
end
 
def to_s(msg=nil)
str = (0...@xmax).map do |x|
(0...@ymax).map{|y| @format % ((c=@board[x][y]) ? c.value : c)}.join
end
(msg ? [msg] : []) + str + [""]
end
end

Which may be used as follows to solve Hidato Puzzles:

require 'HLPsolver'
 
ADJACENT = [[-1, -1], [-1, 0], [-1, 1], [0, -1], [0, 1], [1, -1], [1, 0], [1, 1]]
 
# solve Evil Case 1:
board1 = <<EOS
. 4
0 7 0
1 0 0
EOS

HLPsolver.new(board1).solve
 
boardx = <<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 1 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 0 0 0
0 0 0 0 0 0 0 0
EOS

HLPsolver.new(boardx).solve
 
# solve this tasks example:
board2 = <<EOS
0 33 35 0 0
0 0 24 22 0
0 0 0 21 0 0
0 26 0 13 40 11
27 0 0 0 9 0 1
. . 0 0 18 0 0
. . . . 0 7 0 0
. . . . . . 5 0
EOS

HLPsolver.new(board2).solve
 
#solve The Snake in the Grass:
board3 = <<EOS
1 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 74
. . 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 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 . . 0 0 .
EOS

t0 = Time.now
HLPsolver.new(board3).solve
puts " #{Time.now - t0} sec"

Which produces:

Problem:
   4  
 0 7 0
 1 0 0

Solution:
   4  
 3 7 5
 1 2 6

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  1  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  0  0  0
  0  0  0  0  0  0  0  0

Solution:
 33 34 36 37 41 42 43 44
 32 35 38 40 56 55 46 45
  2 31 39 57 59 60 54 47
  3  1 30 58 61 62 53 48
  4  6 18 29 63 64 52 49
  5  7 17 19 28 51 50 25
  8 11 13 16 20 27 26 24
  9 10 12 14 15 21 22 23

Problem:
  0 33 35  0  0         
  0  0 24 22  0         
  0  0  0 21  0  0      
  0 26  0 13 40 11      
 27  0  0  0  9  0  1   
        0  0 18  0  0   
              0  7  0  0
                    5  0

Solution:
 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

Problem:
  1  0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0       74
        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  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0        0  0   

Solution:
  1  2  3        8  9       14 15       20 21       26 27       32 33       38 39       44 45       50 51       56 57       62 63       68 69       74
        4     7    10    13    16    19    22    25    28    31    34    37    40    43    46    49    52    55    58    61    64    67    70    73   
           5  6       11 12       17 18       23 24       29 30       35 36       41 42       47 48       53 54       59 60       65 66       71 72   

 0.003001 sec

HLPsolver may be used to solve Knight's tour:

[edit] Seed7

$ include "seed7_05.s7i";
 
var set of integer: given is {};
var array array integer: board is 0 times 0 times 0;
var integer: startRow is 0;
var integer: startColumn is 0;
 
const proc: setup (in array string: input) is func
local
var integer: r is 0;
var integer: c is 0;
var array string: row is 0 times "";
var string: cell is "";
var integer: value is 0;
begin
board := (length(input) + 2) times 0 times 0;
for key r range input do
row := split(input[r], " ");
board[r + 1] := (length(row) + 2) times - 1;
for key c range row do
cell := row[c];
if cell = "_" then
board[r + 1][c + 1] := 0;
elsif cell[1] in {'0' .. '9'} then
value := integer parse cell;
board[r + 1][c + 1] := value;
incl(given, value);
if value = 1 then
startRow := r + 1;
startColumn := c + 1;
end if;
end if;
end for;
end for;
board[1] := (length(row) + 2) times - 1;
board[length(input) + 2] := (length(row) + 2) times - 1;
end func;
 
const func boolean: solve (in integer: r, in integer: c, in integer: n) is func
result
var boolean: solved is FALSE;
local
var integer: back is 0;
var integer: i is 0;
var integer: j is 0;
begin
if n > max(given) then
solved := TRUE;
elsif board[r][c] = 0 and n not in given or board[r][c] = n then
back := board[r][c];
board[r][c] := n;
for i range -1 to 1 until solved do
for j range -1 to 1 until solved do
solved := solve(r + i, c + j, n + 1);
end for;
end for;
if not solved then
board[r][c] := back;
end if;
end if;
end func;
 
const proc: printBoard is func
local
var integer: r is 0;
var integer: c is 0;
begin
for key r range board do
for c range board[r] do
if c = -1 then
write(" . ");
elsif c > 0 then
write(c lpad 2 <& " ");
else
write("__ ");
end if;
end for;
writeln;
end for;
end func;
 
const proc: main is func
local
const array string: input is [] ("_ 33 35 _ _ . . .",
"_ _ 24 22 _ . . .",
"_ _ _ 21 _ _ . .",
"_ 26 _ 13 40 11 . .",
"27 _ _ _ 9 _ 1 .",
". . _ _ 18 _ _ .",
". . . . _ 7 _ _",
". . . . . . 5 _");
begin
setup(input);
printBoard;
writeln;
if solve(startRow, startColumn, 1) then
writeln("Found:");
printBoard;
end if;
end func;
Output:
 .  .  .  .  .  .  .  .  .  . 
 . __ 33 35 __ __  .  .  .  . 
 . __ __ 24 22 __  .  .  .  . 
 . __ __ __ 21 __ __  .  .  . 
 . __ 26 __ 13 40 11  .  .  . 
 . 27 __ __ __  9 __  1  .  . 
 .  .  . __ __ 18 __ __  .  . 
 .  .  .  .  . __  7 __ __  . 
 .  .  .  .  .  .  .  5 __  . 
 .  .  .  .  .  .  .  .  .  . 

Found:
 .  .  .  .  .  .  .  .  .  . 
 . 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  . 
 .  .  .  .  .  .  .  .  .  . 

[edit] Tcl

proc init {initialConfiguration} {
global grid max filled
set max 1
set y 0
foreach row [split [string trim $initialConfiguration "\n"] "\n"] {
set x 0
set rowcontents {}
foreach cell $row {
if {![string is integer -strict $cell]} {set cell -1}
lappend rowcontents $cell
set max [expr {max($max, $cell)}]
if {$cell > 0} {
dict set filled $cell [list $y $x]
}
incr x
}
lappend grid $rowcontents
incr y
}
}
 
proc findseps {} {
global max filled
set result {}
for {set i 1} {$i < $max-1} {incr i} {
if {[dict exists $filled $i]} {
for {set j [expr {$i+1}]} {$j <= $max} {incr j} {
if {[dict exists $filled $j]} {
if {$j-$i > 1} {
lappend result [list $i $j [expr {$j-$i}]]
}
break
}
}
}
}
return [lsort -integer -index 2 $result]
}
 
proc makepaths {sep} {
global grid filled
lassign $sep from to len
lassign [dict get $filled $from] y x
set result {}
foreach {dx dy} {-1 -1 -1 0 -1 1 0 -1 0 1 1 -1 1 0 1 1} {
discover [expr {$x+$dx}] [expr {$y+$dy}] [expr {$from+1}] $to \
[list [list $from $x $y]] $grid
}
return $result
}
proc discover {x y n limit path model} {
global filled
# Check for illegal
if {[lindex $model $y $x] != 0} return
upvar 1 result result
lassign [dict get $filled $limit] ly lx
# Special case
if {$n == $limit-1} {
if {abs($x-$lx)<=1 && abs($y-$ly)<=1 && !($lx==$x && $ly==$y)} {
lappend result [lappend path [list $n $x $y] [list $limit $lx $ly]]
}
return
}
# Check for impossible
if {abs($x-$lx) > $limit-$n || abs($y-$ly) > $limit-$n} return
# Recursive search
lappend path [list $n $x $y]
lset model $y $x $n
incr n
foreach {dx dy} {-1 -1 -1 0 -1 1 0 -1 0 1 1 -1 1 0 1 1} {
discover [expr {$x+$dx}] [expr {$y+$dy}] $n $limit $path $model
}
}
 
proc applypath {path} {
global grid filled
puts "Found unique path for [lindex $path 0 0] -> [lindex $path end 0]"
foreach cell [lrange $path 1 end-1] {
lassign $cell n x y
lset grid $y $x $n
dict set filled $n [list $y $x]
}
}
 
proc printgrid {} {
global grid max
foreach row $grid {
foreach cell $row {
puts -nonewline [format " %*s" [string length $max] [expr {
$cell==-1 ? "." : $cell
}]]
}
puts ""
}
}
 
proc solveHidato {initialConfiguration} {
init $initialConfiguration
set limit [llength [findseps]]
while {[llength [set seps [findseps]]] && [incr limit -1]>=0} {
foreach sep $seps {
if {[llength [set paths [makepaths $sep]]] == 1} {
applypath [lindex $paths 0]
break
}
}
}
puts ""
printgrid
}

Demonstrating (dots are “outside” the grid, and zeroes are the cells to be filled in):

solveHidato "
0 33 35 0 0 . . .
0 0 24 22 0 . . .
0 0 0 21 0 0 . .
0 26 0 13 40 11 . .
27 0 0 0 9 0 1 .
. . 0 0 18 0 0 .
. . . . 0 7 0 0
. . . . . . 5 0
"
Output:
Found unique path for 5 -> 7
Found unique path for 7 -> 9
Found unique path for 9 -> 11
Found unique path for 11 -> 13
Found unique path for 33 -> 35
Found unique path for 18 -> 21
Found unique path for 1 -> 5
Found unique path for 35 -> 40
Found unique path for 22 -> 24
Found unique path for 24 -> 26
Found unique path for 27 -> 33
Found unique path for 13 -> 18

 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

More complex cases are solvable with an extended version of this code, though that has more onerous version requirements.

Personal tools
Namespaces

Variants
Actions
Community
Explore
Misc
Toolbox