Solve a Numbrix puzzle

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Task
Solve a Numbrix puzzle
You are encouraged to solve this task according to the task description, using any language you may know.

Numbrix puzzles are similar to Hidato. The most important difference is that it is only possible to move 1 node left, right, up, or down (sometimes referred to as the Von Neumann neighborhood). Published puzzles also tend not to have holes in the grid and may not always indicate the end node. Two examples follow:

Example 1

Problem.

 0  0  0  0  0  0  0  0  0
 0  0 46 45  0 55 74  0  0
 0 38  0  0 43  0  0 78  0
 0 35  0  0  0  0  0 71  0
 0  0 33  0  0  0 59  0  0
 0 17  0  0  0  0  0 67  0
 0 18  0  0 11  0  0 64  0
 0  0 24 21  0  1  2  0  0
 0  0  0  0  0  0  0  0  0

Solution.

 49 50 51 52 53 54 75 76 81
 48 47 46 45 44 55 74 77 80
 37 38 39 40 43 56 73 78 79
 36 35 34 41 42 57 72 71 70
 31 32 33 14 13 58 59 68 69
 30 17 16 15 12 61 60 67 66
 29 18 19 20 11 62 63 64 65
 28 25 24 21 10  1  2  3  4
 27 26 23 22  9  8  7  6  5
Example 2

Problem.

 0  0  0  0  0  0  0  0  0
 0 11 12 15 18 21 62 61  0
 0  6  0  0  0  0  0 60  0
 0 33  0  0  0  0  0 57  0
 0 32  0  0  0  0  0 56  0
 0 37  0  1  0  0  0 73  0
 0 38  0  0  0  0  0 72  0
 0 43 44 47 48 51 76 77  0
 0  0  0  0  0  0  0  0  0

Solution.

  9 10 13 14 19 20 63 64 65
  8 11 12 15 18 21 62 61 66
  7  6  5 16 17 22 59 60 67
 34 33  4  3 24 23 58 57 68
 35 32 31  2 25 54 55 56 69
 36 37 30  1 26 53 74 73 70
 39 38 29 28 27 52 75 72 71
 40 43 44 47 48 51 76 77 78
 41 42 45 46 49 50 81 80 79
Task

Write a program to solve puzzles of this ilk, demonstrating your program by solving the above examples. Extra credit for other interesting examples.

Related Tasks:

Contents

[edit] C++

 
#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] = -1; dy[0] = 0; dx[1] = 1; dy[1] = 0;
dx[2] = 0; dy[2] = -1; dx[3] = 0; dy[3] = 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 = len;
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 ) == "*" ) { max--; arr[c++].val = -1; continue; }
arr[c].val = atoi( ( *i ).c_str() );
if( arr[c].val > 0 ) weHave[arr[c].val] = true;
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, int dr )
{
if( ( w > max && dr > 0 ) || ( w < 1 && dr < 0 ) || ( w == max && weHave[w] ) ) return true;
 
node* n = &arr[x + y * wid];
n->neighbors = getNeighbors( x, y );
if( weHave[w] )
{
for( int d = 0; d < 4; 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 + dr, dr ) ) return true;
}
}
return false;
}
 
for( int d = 0; d < 4; 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 + dr, dr ) ) 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 < 4; 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, 1 );
if( z > 1 ) search( x, y, z - 1, -1 );
}
 
void findStart( int& x, int& y, int& z )
{
z = 99999;
for( int b = 0; b < hei; b++ )
for( int a = 0; a < wid; a++ )
if( arr[a + wid * b].val > 0 && arr[a + wid * b].val < z )
{
x = a; y = b;
z = arr[a + wid * b].val;
}
 
}
 
int wid, hei, max, dx[4], dy[4];
node* arr;
bool* weHave;
};
//------------------------------------------------------------------------------
int main( int argc, char* argv[] )
{
int wid; string p;
//p = ". . . . . . . . . . . 46 45 . 55 74 . . . 38 . . 43 . . 78 . . 35 . . . . . 71 . . . 33 . . . 59 . . . 17 . . . . . 67 . . 18 . . 11 . . 64 . . . 24 21 . 1 2 . . . . . . . . . . ."; wid = 9;
//p = ". . . . . . . . . . 11 12 15 18 21 62 61 . . 6 . . . . . 60 . . 33 . . . . . 57 . . 32 . . . . . 56 . . 37 . 1 . . . 73 . . 38 . . . . . 72 . . 43 44 47 48 51 76 77 . . . . . . . . . ."; wid = 9;
p = "17 . . . 11 . . . 59 . 15 . . 6 . . 61 . . . 3 . . . 63 . . . . . . 66 . . . . 23 24 . 68 67 78 . 54 55 . . . . 72 . . . . . . 35 . . . 49 . . . 29 . . 40 . . 47 . 31 . . . 39 . . . 45"; wid = 9;
 
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:
49 50 51 52 53 54 75 76 81
48 47 46 45 44 55 74 77 80
37 38 39 40 43 56 73 78 79
36 35 34 41 42 57 72 71 70
31 32 33 14 13 58 59 68 69
30 17 16 15 12 61 60 67 66
29 18 19 20 11 62 63 64 65
28 25 24 21 10 01 02 03 04
27 26 23 22 09 08 07 06 05

09 10 13 14 19 20 63 64 65
08 11 12 15 18 21 62 61 66
07 06 05 16 17 22 59 60 67
34 33 04 03 24 23 58 57 68
35 32 31 02 25 54 55 56 69
36 37 30 01 26 53 74 73 70
39 38 29 28 27 52 75 72 71
40 43 44 47 48 51 76 77 78
41 42 45 46 49 50 81 80 79

17 16 13 12 11 10 09 60 59
18 15 14 05 06 07 08 61 58
19 20 03 04 65 64 63 62 57
22 21 02 01 66 79 80 81 56
23 24 69 68 67 78 77 54 55
26 25 70 71 72 75 76 53 52
27 28 35 36 73 74 49 50 51
30 29 34 37 40 41 48 47 46
31 32 33 38 39 42 43 44 45

[edit] D

From the refactored C++ version with more precise typing. The NumbrixPuzzle struct is created at compile-time, so its asserts and exceptions can catch most malformed puzzles at compile-time.

Translation of: C++
import std.stdio, std.conv, std.string, std.range, std.array, std.typecons, std.algorithm;
 
struct {
alias BitSet8 = ubyte; // A set of 8 bits.
alias Cell = uint;
enum : string { unavailableInCell = "#", availableInCell = "." }
enum : Cell { unavailableCell = Cell.max, availableCell = 0 }
 
this(in string inPuzzle) pure @safe {
const rawPuzzle = inPuzzle.splitLines.map!(row => row.split).array;
assert(!rawPuzzle.empty);
assert(!rawPuzzle[0].empty);
assert(rawPuzzle.all!(row => row.length == rawPuzzle[0].length)); // Is rectangular.
 
gridWidth = rawPuzzle[0].length;
gridHeight = rawPuzzle.length;
immutable nMaxCells = gridWidth * gridHeight;
grid = new Cell[nMaxCells];
auto knownMutable = new bool[nMaxCells + 1];
uint nAvailableMutable = nMaxCells;
bool[Cell] seenCells; // To avoid duplicate input numbers.
 
uint i = 0;
foreach (const piece; rawPuzzle.join) {
if (piece == unavailableInCell) {
nAvailableMutable--;
grid[i++] = unavailableCell;
continue;
} else if (piece == availableInCell) {
grid[i] = availableCell;
} else {
immutable cell = piece.to!Cell;
assert(cell > 0 && cell <= nMaxCells);
assert(cell !in seenCells);
seenCells[cell] = true;
knownMutable[cell] = true;
grid[i] = cell;
}
 
i++;
}
 
known = knownMutable.idup;
nAvailable = nAvailableMutable;
}
 
@disable this();
 
 
auto solve() pure nothrow @safe @nogc
out(result) {
if (!result.isNull) {
// Can't verify 'result' here because it's const.
// assert(!result.get.join.canFind(availableCell.text));
 
assert(!grid.canFind(availableCell));
auto values = grid.filter!(c => c != unavailableCell);
auto interval = iota(reduce!min(values.front, values.dropOne),
reduce!max(values.front, values.dropOne) + 1);
assert(values.walkLength == interval.length);
assert(interval.all!(c => values.count(c) == 1)); // Quadratic.
}
} body {
auto result = grid
.map!(c => (c == unavailableCell) ? unavailableInCell : c.text)
.chunks(gridWidth);
alias OutRange = Nullable!(typeof(result));
 
const start = findStart;
if (start.isNull)
return OutRange();
 
search(start.r, start.c, start.cell + 1, 1);
if (start.cell > 1) {
immutable direction = -1;
search(start.r, start.c, start.cell + direction, direction);
}
 
if (grid.any!(c => c == availableCell))
return OutRange();
else
return OutRange(result);
}
 
private:
 
 
bool search(in uint r, in uint c, in Cell cell, in int direction)
pure nothrow @safe @nogc {
if ((cell > nAvailable && direction > 0) || (cell == 0 && direction < 0) ||
(cell == nAvailable && known[cell]))
return true; // One solution found.
 
immutable neighbors = getNeighbors(r, c);
 
if (known[cell]) {
foreach (immutable i, immutable rc; shifts) {
if (neighbors & (1u << i)) {
immutable c2 = c + rc[0],
r2 = r + rc[1];
if (grid[r2 * gridWidth + c2] == cell)
if (search(r2, c2, cell + direction, direction))
return true;
}
}
return false;
}
 
foreach (immutable i, immutable rc; shifts) {
if (neighbors & (1u << i)) {
immutable c2 = c + rc[0],
r2 = r + rc[1],
pos = r2 * gridWidth + c2;
if (grid[pos] == availableCell) {
grid[pos] = cell; // Try.
if (search(r2, c2, cell + direction, direction))
return true;
grid[pos] = availableCell; // Restore.
}
}
}
return false;
}
 
 
BitSet8 getNeighbors(in uint r, in uint c) const pure nothrow @safe @nogc {
typeof(return) usable = 0;
 
foreach (immutable i, immutable rc; shifts) {
immutable c2 = c + rc[0],
r2 = r + rc[1];
if (c2 >= gridWidth || r2 >= gridHeight)
continue;
if (grid[r2 * gridWidth + c2] != unavailableCell)
usable |= (1u << i);
}
 
return usable;
}
 
 
auto findStart() const pure nothrow @safe @nogc {
alias Triple = Tuple!(uint,"r", uint,"c", Cell,"cell");
Nullable!Triple result;
 
auto cell = Cell.max;
foreach (immutable r; 0 .. gridHeight) {
foreach (immutable c; 0 .. gridWidth) {
immutable pos = gridWidth * r + c;
if (grid[pos] != availableCell &&
grid[pos] != unavailableCell && grid[pos] < cell) {
cell = grid[pos];
result = Triple(r, c, cell);
}
}
}
 
return result;
}
 
static immutable int[2][4] shifts = [[0, -1], [0, 1], [-1, 0], [1, 0]];
immutable uint gridWidth, gridHeight;
immutable int nAvailable;
immutable bool[] known; // Given known cells of the puzzle.
Cell[] grid; // Flattened mutable game grid.
}
 
 
void main() {
// enum NumbrixPuzzle to catch malformed puzzles at compile-time.
enum puzzle1 = ". . . . . . . . .
. . 46 45 . 55 74 . .
. 38 . . 43 . . 78 .
. 35 . . . . . 71 .
. . 33 . . . 59 . .
. 17 . . . . . 67 .
. 18 . . 11 . . 64 .
. . 24 21 . 1 2 . .
. . . . . . . . ."
.NumbrixPuzzle;
 
enum puzzle2 = ". . . . . . . . .
. 11 12 15 18 21 62 61 .
. 6 . . . . . 60 .
. 33 . . . . . 57 .
. 32 . . . . . 56 .
. 37 . 1 . . . 73 .
. 38 . . . . . 72 .
. 43 44 47 48 51 76 77 .
. . . . . . . . ."
.NumbrixPuzzle;
 
enum puzzle3 = "17 . . . 11 . . . 59
. 15 . . 6 . . 61 .
. . 3 . . . 63 . .
. . . . 66 . . . .
23 24 . 68 67 78 . 54 55
. . . . 72 . . . .
. . 35 . . . 49 . .
. 29 . . 40 . . 47 .
31 . . . 39 . . . 45"
.NumbrixPuzzle;
 
 
foreach (puzzle; [puzzle1, puzzle2, puzzle3]) {
auto solution = puzzle.solve; // Solved at run-time.
if (solution.isNull)
writeln("No solution found for puzzle.\n");
else
writefln("One solution:\n%(%-(%2s %)\n%)\n", solution);
}
}
Output:
One solution:
49 50 51 52 53 54 75 76 81
48 47 46 45 44 55 74 77 80
37 38 39 40 43 56 73 78 79
36 35 34 41 42 57 72 71 70
31 32 33 14 13 58 59 68 69
30 17 16 15 12 61 60 67 66
29 18 19 20 11 62 63 64 65
28 25 24 21 10  1  2  3  4
27 26 23 22  9  8  7  6  5

One solution:
 9 10 13 14 19 20 63 64 65
 8 11 12 15 18 21 62 61 66
 7  6  5 16 17 22 59 60 67
34 33  4  3 24 23 58 57 68
35 32 31  2 25 54 55 56 69
36 37 30  1 26 53 74 73 70
39 38 29 28 27 52 75 72 71
40 43 44 47 48 51 76 77 78
41 42 45 46 49 50 81 80 79

One solution:
17 16 13 12 11 10  9 60 59
18 15 14  5  6  7  8 61 58
19 20  3  4 65 64 63 62 57
22 21  2  1 66 79 80 81 56
23 24 69 68 67 78 77 54 55
26 25 70 71 72 75 76 53 52
27 28 35 36 73 74 49 50 51
30 29 34 37 40 41 48 47 46
31 32 33 38 39 42 43 44 45

[edit] Icon and Unicon

This is a Unicon-specific solution, based on the Unicon Hidato problem solver:

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 := []
nCells := maxCols := 0
every line := !&input do {
put(p,[: gencells(line) :])
maxCols <:= *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["_"] := 0; 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] = (val|0)); end
method visit(r,c); return (/best, validPos(r,c), Pos(r,c)); end
 
method validPos(r,c)
v := val+1 # number we're looking for
xv := puzzle[r,c] | fail
if (xv ~= 0) & (xv != v) then fail
if xv = (0|v) then {
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-1,loc.c) , self, val) # North
QMouse(puzzle, visit(loc.r, loc.c+1), self, val) # East
QMouse(puzzle, visit(loc.r+1,loc.c), self, val) # South
QMouse(puzzle, visit(loc.r, loc.c-1), self, val) # West
end
Output:
Sample runs:
->numbrix <numbrix1.in
Input with 81 cells:
                                 
     _  _  _  _  _  _  _  _  _   
     _  _ 46 45  _ 55 74  _  _   
     _ 38  _  _ 43  _  _ 78  _   
     _ 35  _  _  _  _  _ 71  _   
     _  _ 33  _  _  _ 59  _  _   
     _ 17  _  _  _  _  _ 67  _   
     _ 18  _  _ 11  _  _ 64  _   
     _  _ 24 21  _  1  2  _  _   
     _  _  _  _  _  _  _  _  _   
                                 
Output with 81 cells:
                                 
    49 50 51 52 53 54 75 76 81   
    48 47 46 45 44 55 74 77 80   
    37 38 39 40 43 56 73 78 79   
    36 35 34 41 42 57 72 71 70   
    31 32 33 14 13 58 59 68 69   
    30 17 16 15 12 61 60 67 66   
    29 18 19 20 11 62 63 64 65   
    28 25 24 21 10  1  2  3  4   
    27 26 23 22  9  8  7  6  5   
                                 
->numbrix <numbrix2.in
Input with 81 cells:
                                 
     _  _  _  _  _  _  _  _  _   
     _ 11 12 15 18 21 62 61  _   
     _  6  _  _  _  _  _ 60  _   
     _ 33  _  _  _  _  _ 57  _   
     _ 32  _  _  _  _  _ 56  _   
     _ 37  _  1  _  _  _ 73  _   
     _ 38  _  _  _  _  _ 72  _   
     _ 43 44 47 48 51 76 77  _   
     _  _  _  _  _  _  _  _  _   
                                 
Output with 81 cells:
                                 
     9 10 13 14 19 20 63 64 65   
     8 11 12 15 18 21 62 61 66   
     7  6  5 16 17 22 59 60 67   
    34 33  4  3 24 23 58 57 68   
    35 32 31  2 25 54 55 56 69   
    36 37 30  1 26 53 74 73 70   
    39 38 29 28 27 52 75 72 71   
    40 43 44 47 48 51 76 77 78   
    41 42 45 46 49 50 81 80 79   
                                 
->

[edit] Perl 6

Using the Warnsdorff solver from Solve_a_Hidato_puzzle:

my @adjacent =           [-1, 0],
[ 0, -1], [ 0, 1],
[ 1, 0];
 
solveboard q:to/END/;
__ __ __ __ __ __ __ __ __
__ __ 46 45 __ 55 74 __ __
__ 38 __ __ 43 __ __ 78 __
__ 35 __ __ __ __ __ 71 __
__ __ 33 __ __ __ 59 __ __
__ 17 __ __ __ __ __ 67 __
__ 18 __ __ 11 __ __ 64 __
__ __ 24 21 __ 1 2 __ __
__ __ __ __ __ __ __ __ __
END
Output:
49 50 51 52 53 54 75 76 81
48 47 46 45 44 55 74 77 80
37 38 39 40 43 56 73 78 79
36 35 34 41 42 57 72 71 70
31 32 33 14 13 58 59 68 69
30 17 16 15 12 61 60 67 66
29 18 19 20 11 62 63 64 65
28 25 24 21 10  1  2  3  4
27 26 23 22  9  8  7  6  5
1275 tries

And

solveboard q:to/END/;
0 0 0 0 0 0 0 0 0
0 11 12 15 18 21 62 61 0
0 6 0 0 0 0 0 60 0
0 33 0 0 0 0 0 57 0
0 32 0 0 0 0 0 56 0
0 37 0 1 0 0 0 73 0
0 38 0 0 0 0 0 72 0
0 43 44 47 48 51 76 77 0
0 0 0 0 0 0 0 0 0
END
Output:
 9 10 13 14 19 20 63 64 65
 8 11 12 15 18 21 62 61 66
 7  6  5 16 17 22 59 60 67
34 33  4  3 24 23 58 57 68
35 32 31  2 25 54 55 56 69
36 37 30  1 26 53 74 73 70
39 38 29 28 27 52 75 72 71
40 43 44 47 48 51 76 77 78
41 42 45 46 49 50 81 80 79
4631 tries

Oddly, reversing the tiebreaker rule that makes hidato run twice as fast causes this last example to run four times slower. Go figure...

[edit] Racket

This is a general "Hidato" style solver (which is why there is a search for a 0 start point (which supports Hopido). There is already a Racket implementation of Hidato, so to allow a variety of approaches to be demonstrated, the main library for this set of problems is here.

hidato-family-solver.rkt

#lang racket
;;; Used in my solutions of:
;;; "Solve a Hidato Puzzle"
;;; "Solve a Holy Knights Tour"
;;; "Solve a Numbrix Puzzle"
;;; "Solve a Hopido Puzzle"
 
;;; As well as the solver being common, the solution renderer and input formats are common
(provide
 ;; Input: list of neighbour offsets
 ;; Output: a solver function:
 ;; Input: a puzzle
 ;; Output: either the solved puzzle or #f if impossible
solve-hidato-family
 ;; Input: puzzle
 ;; optional minimum cell width
 ;; Output: a pretty string that can be printed
puzzle->string)
 
;; Cell values are:
;; zero? - unvisited
;; positive? - nth visitied
;; else - unvisitable. In the puzzle layout, it's a _. In the hash it's a -1, so we can care less
;; about number type checking.
;; A puzzle is a sequence of sequences of cell values
;; We work with a puzzle as a hash keyed on (cons row-num col-num)
 
;; Take a puzzle and get a working hash of it
(define (puzzle->hash p)
(for*/hash
(((r row-num) (in-parallel p (in-naturals)))
((v col-num) (in-parallel r (in-naturals)))
#:when (integer? v))
(values (cons row-num col-num) v)))
 
;; Takes a hash and recreates a vector of vectors puzzle
(define (hash->puzzle h# (blank '_))
(define keys (hash-keys h#))
(define n-rows (add1 (car (argmax car keys))))
(define n-cols (add1 (cdr (argmax cdr keys))))
(for/vector #:length n-rows ((r n-rows))
(for/vector #:length n-cols ((c n-cols))
(hash-ref h# (cons r c) blank))))
 
;; See "provide" section for description
(define (puzzle->string p (w #f))
(match p
[#f "unsolved"]
[(? sequence? s)
(define (max-n-digits p)
(and p (add1 (order-of-magnitude (* (vector-length p) (vector-length (vector-ref p 0)))))))
(define min-width (or w (max-n-digits p)))
(string-join
(for/list ((r s))
(string-join
(for/list ((c r)) (~a c #:align 'right #:min-width min-width))
" "))
"\n")]))
 
(define ((solve-hidato-family neighbour-offsets) board)
(define board# (puzzle->hash board))
 ;; reverse mapping, will only take note of positive values
(define targets# (for/hash ([(k v) (in-hash board#)] #:when (positive? v)) (values v k)))
 
(define (neighbours r.c)
(for/list ((r+.c+ neighbour-offsets))
(match-define (list r+ c+) r+.c+)
(match-define (cons r c ) r.c)
(cons (+ r r+) (+ c c+))))
 
 ;; Count the moves, rather than check for "no more zeros" in puzzle
(define last-move (length (filter number? (hash-values board#))))
 
 ;; Depth first solution of the puzzle (we have to go deep, it's where the solutions are!
(define (inr-solve-pzl b# move r.c)
(cond
[(= move last-move) b#] ; no moves needed, so solved
[else
(define m++ (add1 move))
(for*/or ; check each neighbour as an option
((r.c+ (in-list (neighbours r.c)))
#:when (equal? (hash-ref targets# move r.c) r.c) ; we're where we should be!
#:when (match (hash-ref b# r.c+ -1) (0 #t) ((== m++) #t) (_ #f)))
(inr-solve-pzl (hash-set b# r.c+ m++) m++ r.c+))]))
 
(define (solution-starting-at n)
(define start-r.c (for/first (((k v) (in-hash board#)) #:when (= n v)) k))
(and start-r.c (inr-solve-pzl board# n start-r.c)))
 
(define sltn
(cond [(solution-starting-at 1) => values]
 ;; next clause starts from 0 for hopido
[(solution-starting-at 0) => values]))
 
(and sltn (hash->puzzle sltn)))
#lang racket
(require "hidato-family-solver.rkt")
 
(define von-neumann-neighbour-offsets
'((+1 0) (-1 0) (0 +1) (0 -1)))
 
(define solve-numbrix (solve-hidato-family von-neumann-neighbour-offsets))
 
(displayln
(puzzle->string
(solve-numbrix
#(#(0 0 0 0 0 0 0 0 0)
#(0 0 46 45 0 55 74 0 0)
#(0 38 0 0 43 0 0 78 0)
#(0 35 0 0 0 0 0 71 0)
#(0 0 33 0 0 0 59 0 0)
#(0 17 0 0 0 0 0 67 0)
#(0 18 0 0 11 0 0 64 0)
#(0 0 24 21 0 1 2 0 0)
#(0 0 0 0 0 0 0 0 0)))))
 
(newline)
 
(displayln
(puzzle->string
(solve-numbrix
#(#(0 0 0 0 0 0 0 0 0)
#(0 11 12 15 18 21 62 61 0)
#(0 6 0 0 0 0 0 60 0)
#(0 33 0 0 0 0 0 57 0)
#(0 32 0 0 0 0 0 56 0)
#(0 37 0 1 0 0 0 73 0)
#(0 38 0 0 0 0 0 72 0)
#(0 43 44 47 48 51 76 77 0)
#(0 0 0 0 0 0 0 0 0)))))
Output:
49 50 51 52 53 54 75 76 81
48 47 46 45 44 55 74 77 80
37 38 39 40 43 56 73 78 79
36 35 34 41 42 57 72 71 70
31 32 33 14 13 58 59 68 69
30 17 16 15 12 61 60 67 66
29 18 19 20 11 62 63 64 65
28 25 24 21 10  1  2  3  4
27 26 23 22  9  8  7  6  5

 9 10 13 14 19 20 63 64 65
 8 11 12 15 18 21 62 61 66
 7  6  5 16 17 22 59 60 67
34 33  4  3 24 23 58 57 68
35 32 31  2 25 54 55 56 69
36 37 30  1 26 53 74 73 70
39 38 29 28 27 52 75 72 71
40 43 44 47 48 51 76 77 78
41 42 45 46 49 50 81 80 79

[edit] REXX

This solution is essentially same as the (REXX) Hidato puzzle solver.

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).

/*REXX program solves a  Numbrix (R) puzzle, displays puzzle & solution.*/
maxr=0; maxc=0; maxx=0; minr=9e9; minc=9e9; minx=9e9; cells=0; @.=
parse arg xxx; PZ='Numbrix 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 x=abs(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 /*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 call err 'No solution possible for this' PZ"."
say; 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 return 1 /*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'
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:
when using the input of:

1 1 . . . . . . . . ./2 1 . . 24 21 . 1 2 . ./3 1 . 18 . . 11 . . 64 ./4 1 . 17 . . . . . 67 ./5 1 . . 33 . . . 59 . ./6 1 . 35 . . . . . 71 ./7 1 . 38 . . 43 . . 78 ./8 1 . . 46 45 . 55 74 . ./9 1 . . . . . . . . .

  .  .  .  .  .  .  .  .  .
  .  . 46 45  . 55 74  .  .
  . 38  .  . 43  .  . 78  .
  . 35  .  .  .  .  . 71  .
  .  . 33  .  .  . 59  .  .
  . 17  .  .  .  .  . 67  .
  . 18  .  . 11  .  . 64  .
  .  . 24 21  .  1  2  .  .
  .  .  .  .  .  .  .  .  .


A solution for the Numbrix puzzle exists.

 49 50 51 52 53 54 75 76 81
 48 47 46 45 44 55 74 77 80
 37 38 39 40 43 56 73 78 79
 36 35 34 41 42 57 72 71 70
 31 32 33 14 13 58 59 68 69
 30 17 16 15 12 61 60 67 66
 29 18 19 20 11 62 63 64 65
 28 25 24 21 10  1  2  3  4
 27 26 23 22  9  8  7  6  5
Output:
when using the input of:

1 1 . . . . . . . . .\2 1 . 43 44 47 48 51 76 77 .\3 1 . 38 . . . . . 72 .\4 1 . 37 . 1 . . . 73 .\5 1 . 32 . . . . . 56 .\6 1 . 33 . . . . . 57 .\7 1 . 6 . . . . . 60 .\8 1 . 11 12 15 18 21 62 61 .\9 1 . . . . . . . . .

  .  .  .  .  .  .  .  .  .
  . 11 12 15 18 21 62 61  .
  .  6  .  .  .  .  . 60  .
  . 33  .  .  .  .  . 57  .
  . 32  .  .  .  .  . 56  .
  . 37  .  1  .  .  . 73  .
  . 38  .  .  .  .  . 72  .
  . 43 44 47 48 51 76 77  .
  .  .  .  .  .  .  .  .  .


A solution for the Numbrix puzzle exists.

  9 10 13 14 19 20 63 64 65
  8 11 12 15 18 21 62 61 66
  7  6  5 16 17 22 59 60 67
 34 33  4  3 24 23 58 57 68
 35 32 31  2 25 54 55 56 69
 36 37 30  1 26 53 74 73 70
 39 38 29 28 27 52 75 72 71
 40 43 44 47 48 51 76 77 78

[edit] Ruby

This solution uses HLPsolver from here

require 'HLPsolver'
 
ADJACENT = [[-1, 0], [0, -1], [0, 1], [1, 0]]
 
board1 = <<EOS
0 0 0 0 0 0 0 0 0
0 0 46 45 0 55 74 0 0
0 38 0 0 43 0 0 78 0
0 35 0 0 0 0 0 71 0
0 0 33 0 0 0 59 0 0
0 17 0 0 0 0 0 67 0
0 18 0 0 11 0 0 64 0
0 0 24 21 0 1 2 0 0
0 0 0 0 0 0 0 0 0
EOS

HLPsolver.new(board1).solve
 
board2 = <<EOS
0 0 0 0 0 0 0 0 0
0 11 12 15 18 21 62 61 0
0 6 0 0 0 0 0 60 0
0 33 0 0 0 0 0 57 0
0 32 0 0 0 0 0 56 0
0 37 0 1 0 0 0 73 0
0 38 0 0 0 0 0 72 0
0 43 44 47 48 51 76 77 0
0 0 0 0 0 0 0 0 0
EOS

HLPsolver.new(board2).solve

Which produces:

Problem:
  0  0  0  0  0  0  0  0  0
  0  0 46 45  0 55 74  0  0
  0 38  0  0 43  0  0 78  0
  0 35  0  0  0  0  0 71  0
  0  0 33  0  0  0 59  0  0
  0 17  0  0  0  0  0 67  0
  0 18  0  0 11  0  0 64  0
  0  0 24 21  0  1  2  0  0
  0  0  0  0  0  0  0  0  0

Solution:
 49 50 51 52 53 54 75 76 81
 48 47 46 45 44 55 74 77 80
 37 38 39 40 43 56 73 78 79
 36 35 34 41 42 57 72 71 70
 31 32 33 14 13 58 59 68 69
 30 17 16 15 12 61 60 67 66
 29 18 19 20 11 62 63 64 65
 28 25 24 21 10  1  2  3  4
 27 26 23 22  9  8  7  6  5

Problem:
  0  0  0  0  0  0  0  0  0
  0 11 12 15 18 21 62 61  0
  0  6  0  0  0  0  0 60  0
  0 33  0  0  0  0  0 57  0
  0 32  0  0  0  0  0 56  0
  0 37  0  1  0  0  0 73  0
  0 38  0  0  0  0  0 72  0
  0 43 44 47 48 51 76 77  0
  0  0  0  0  0  0  0  0  0

Solution:
  9 10 13 14 19 20 63 64 65
  8 11 12 15 18 21 62 61 66
  7  6  5 16 17 22 59 60 67
 34 33  4  3 24 23 58 57 68
 35 32 31  2 25 54 55 56 69
 36 37 30  1 26 53 74 73 70
 39 38 29 28 27 52 75 72 71
 40 43 44 47 48 51 76 77 78
 41 42 45 46 49 50 81 80 79
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