Solve a Hidato puzzle: Difference between revisions

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printgrid
printgrid
}</lang>
}</lang>
Demonstrating (dots are “outside” the grid, and zeroes are the cells to be filled in):
Demonstrating:
<lang tcl>solveHidato "
<lang tcl>solveHidato "
0 33 35 0 0 . . .
0 33 35 0 0 . . .

Revision as of 11:48, 16 January 2012

Solve a Hidato puzzle is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

The task is to write a program which solves Hidato puzzles.

Extra credit: show that the code can be reused to solve the Knight's Tour.

Sample Hidato problem, from Wikipedia
Sample Hidato problem, from Wikipedia

The above problem has the following solution, with path marked on it:

Solution to sample Hidato problem
Solution to sample Hidato problem

Mathprog

For the Knight's Tour see: http://rosettacode.org/wiki/Knight%27s_tour#Mathprog <lang mathprog>/*Hidato.mathprog, part of KuKu by Nigel Galloway 

 Find a solution to a Hidato problem

 Nigel_Galloway
 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 := 7; param COLS := 7; param ZBLS := 49; param Iz: 1 2 3 4 5 6 7 := 

1  .   .   6   .  23   .   . 
2  .  40   .   .   9   .   . 
3  .  39   .   .   .   .  21 
4  1  38   .   .  12   .  19 
5 36   .  30   .   .  18   . 
6  .  32   .   .  14   .  16 
7  .  33   .   .   .  48  49 
;

end;</lang>

Produces: 

GLPSOL: GLPK LP/MIP Solver, v4.47
Parameter(s) specified in the command line:
 --math H20110503.mprog
Reading model section from H20110503.mathprog...
Reading data section from H20110503.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
GLPK Integer Optimizer, v4.47
4318 rows, 4050 columns, 30631 non-zeros
4050 integer variables, all of which are binary
Preprocessing...
38 hidden packing inequaliti(es) were detected
220 rows, 223 columns, 1099 non-zeros
223 integer variables, all of which are binary
Scaling...
 A: min|aij| = 1.000e+000  max|aij| = 1.000e+000  ratio = 1.000e+000
Problem data seem to be well scaled
Constructing initial basis...
Size of triangular part = 220
Solving LP relaxation...
GLPK Simplex Optimizer, v4.47
220 rows, 223 columns, 1099 non-zeros
      0: obj =  0.000000000e+000  infeas = 3.100e+001 (0)
*   167: obj =  0.000000000e+000  infeas = 9.430e-015 (0)
OPTIMAL SOLUTION FOUND
Integer optimization begins...
+   167: mip =     not found yet >=              -inf        (1; 0)
+   181: >>>>>  0.000000000e+000 >=  0.000000000e+000   0.0% (1; 0)
+   181: mip =  0.000000000e+000 >=     tree is empty   0.0% (0; 1)
INTEGER OPTIMAL SOLUTION FOUND
Time used:   0.0 secs
Memory used: 5.9 Mb (6168823 bytes)
  4  5  6  8 23 24 25
  3 40  7 10  9 22 26
  2 39 41 11 28 27 21
  1 38 42 29 12 20 19
 36 37 30 43 13 18 17
 35 32 31 44 14 15 16
 34 33 45 46 47 48 49
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
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