Solve a Hidato puzzle

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 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.
• 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

Mathprog

<lang mathprog> /*Hidato.mathprog, part of KuKu by Nigel Galloway

 Find a solution to a Hidato problem

 Nigel_Galloway@operamail.com
 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;

</lang>

Produces:

GLPSOL: GLPK LP/MIP Solver, v4.47
Parameter(s) specified in the command line:
 --math Hidato.mathprog

...

INTEGER OPTIMAL SOLUTION FOUND
Time used:   0.0 secs
Memory used: 6.4 Mb (6712828 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

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 <lang Prolog>:- 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]).

</lang> 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 

Tcl

<lang 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

}</lang> Demonstrating (dots are “outside” the grid, and zeroes are the cells to be filled in): <lang tcl>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 

"</lang> 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