Solve a Hidato puzzle: Difference between revisions
Thundergnat (talk | contribs) (→{{header|Perl 6}}: Syntax fix, "0" is true now. (didn't actually affect this particular problem)) |
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</pre> |
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=={{header|Go}}== |
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{{trans|Java}} |
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<lang go>package main |
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import ( |
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"fmt" |
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"sort" |
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"strconv" |
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"strings" |
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) |
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var board [][]int |
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var start, given []int |
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func setup(input []string) { |
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/* This task is not about input validation, so |
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we're going to trust the input to be valid */ |
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puzzle := make([][]string, len(input)) |
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for i := 0; i < len(input); i++ { |
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puzzle[i] = strings.Fields(input[i]) |
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} |
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nCols := len(puzzle[0]) |
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nRows := len(puzzle) |
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list := make([]int, nRows*nCols) |
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board = make([][]int, nRows+2) |
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for i := 0; i < nRows+2; i++ { |
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board[i] = make([]int, nCols+2) |
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for j := 0; j < nCols+2; j++ { |
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board[i][j] = -1 |
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} |
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} |
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for r := 0; r < nRows; r++ { |
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row := puzzle[r] |
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for c := 0; c < nCols; c++ { |
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switch cell := row[c]; cell { |
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case "_": |
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board[r+1][c+1] = 0 |
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case ".": |
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break |
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default: |
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val, _ := strconv.Atoi(cell) |
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board[r+1][c+1] = val |
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list = append(list, val) |
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if val == 1 { |
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start = append(start, r+1, c+1) |
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} |
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} |
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} |
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} |
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sort.Ints(list) |
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given = make([]int, len(list)) |
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for i := 0; i < len(given); i++ { |
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given[i] = list[i] |
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} |
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} |
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func solve(r, c, n, next int) bool { |
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if n > given[len(given)-1] { |
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return true |
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} |
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back := board[r][c] |
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if back != 0 && back != n { |
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return false |
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} |
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if back == 0 && given[next] == n { |
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return false |
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} |
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if back == n { |
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next++ |
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} |
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board[r][c] = n |
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for i := -1; i < 2; i++ { |
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for j := -1; j < 2; j++ { |
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if solve(r+i, c+j, n+1, next) { |
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return true |
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} |
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} |
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} |
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board[r][c] = back |
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return false |
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} |
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func printBoard() { |
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for _, row := range board { |
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for _, c := range row { |
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switch { |
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case c == -1: |
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fmt.Print(" . ") |
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case c > 0: |
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fmt.Printf("%2d ", c) |
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default: |
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fmt.Print("__ ") |
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} |
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} |
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fmt.Println() |
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} |
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} |
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func main() { |
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input := []string{ |
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"_ 33 35 _ _ . . .", |
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"_ _ 24 22 _ . . .", |
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"_ _ _ 21 _ _ . .", |
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"_ 26 _ 13 40 11 . .", |
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"27 _ _ _ 9 _ 1 .", |
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". . _ _ 18 _ _ .", |
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". . . . _ 7 _ _", |
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". . . . . . 5 _", |
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} |
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setup(input) |
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printBoard() |
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fmt.Println("\nFound:") |
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solve(start[0], start[1], 1, 0) |
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printBoard() |
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}</lang> |
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{{out}} |
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<pre> |
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. . . . . . . . . . |
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. __ 33 35 __ __ . . . . |
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. __ __ 24 22 __ . . . . |
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. __ __ __ 21 __ __ . . . |
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. __ 26 __ 13 40 11 . . . |
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. 27 __ __ __ 9 __ 1 . . |
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. . . __ __ 18 __ __ . . |
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. . . . . __ 7 __ __ . |
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. . . . . . . 5 __ . |
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. . . . . . . . . . |
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Found: |
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. . . . . . . . . . |
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. 32 33 35 36 37 . . . . |
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. 31 34 24 22 38 . . . . |
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. 30 25 23 21 12 39 . . . |
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. 29 26 20 13 40 11 . . . |
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. 27 28 14 19 9 10 1 . . |
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. . . 15 16 18 8 2 . . |
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. . . . . 17 7 6 3 . |
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. . . . . . . 5 4 . |
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. . . . . . . . . . |
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</pre> |
</pre> |
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Revision as of 14:33, 27 June 2018
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
has the following solution, with path marked on it:
- Related tasks
- A* search algorithm
- N-queens problem
- Solve a Holy Knight's tour
- Solve a Knight's tour
- Solve a Hopido puzzle
- Solve a Numbrix puzzle
- Solve the no connection puzzle;
AutoHotkey
<lang AutoHotkey>SolveHidato(Grid, Locked, Max, row, col, num:=1, R:="", C:=""){ if (R&&C) ; if neighbors (not first iteration) { Grid[R, C] := ">" num ; place num in current neighbor and mark it visited ">" row:=R, col:=C ; move to current neighbor }
num++ ; increment num if (num=max) ; if reached end return map(Grid) ; return solution
if locked[num] ; if current num is a locked value { row := StrSplit((StrSplit(locked[num], ",").1) , ":").1 ; find row of num col := StrSplit((StrSplit(locked[num], ",").1) , ":").2 ; find col of num if SolveHidato(Grid, Locked, Max, row, col, num) ; solve for current location and value return map(Grid) ; if solved, return solution } else { for each, value in StrSplit(Neighbor(row,col), ",") { R := StrSplit(value, ":").1 C := StrSplit(value, ":").2
if (Grid[R,C] = "") ; a hole or out of bounds || InStr(Grid[R, C], ">") ; visited || Locked[num+1] && !(Locked[num+1]~= "\b" R ":" C "\b") ; not neighbor of locked[num+1] || Locked[num-1] && !(Locked[num-1]~= "\b" R ":" C "\b") ; not neighbor of locked[num-1] || Locked[num] ; locked value || Locked[Grid[R, C]] ; locked cell continue
if SolveHidato(Grid, Locked, Max, row, col, num, R, C) ; solve for current location, neighbor and value return map(Grid) ; if solved, return solution } } num-- ; step back for i, line in Grid for j, element in line if InStr(element, ">") && (StrReplace(element, ">") >= num) Grid[i, j] := "Y" }
- --------------------------------
- --------------------------------
- --------------------------------
Neighbor(row,col){ R := row-1 loop, 9 { DeltaC := Mod(A_Index, 3) ? Mod(A_Index, 3)-2 : 1 res .= (R=row && !DeltaC) ? "" : R ":" col+DeltaC "," R := Mod(A_Index, 3) ? R : R+1 } return Trim(res, ",") }
- --------------------------------
map(Grid){ for i, row in Grid { for j, element in row line .= (A_Index > 1 ? "`t" : "") . element map .= (map<>""?"`n":"") line line := "" } return StrReplace(map, ">") }</lang> Examples:<lang AutoHotkey>;-------------------------------- Grid := [[ "Y" , 33 , 35 , "Y" , "Y"] ,[ "Y" , "Y" , 24 , 22 , "Y"] ,[ "Y" , "Y" , "Y" , 21 , "Y" , "Y"] ,[ "Y" , 26 , "Y" , 13 , 40 , 11 ] ,[ 27 , "Y" , "Y" , "Y" , 9 , "Y" , 1 ] ,[ "" , "" , "Y" , "Y" , 18 , "Y" , "Y"] ,[ "" , "" , "" , "" , "Y" , 7 , "Y" , "Y"] ,[ "" , "" , "" , "" , "" , "" , 5 , "Y"]]
- --------------------------------
- find locked cells, find row and col of first value "1" and max value
Locked := [] for i, line in Grid for j, element in line { if element = 1 row :=i , col := j if element is integer Locked[element] := i ":" j "," Neighbor(i, j) ; save locked elements position and neighbors , max := element > max ? element : max ; find max value }
- --------------------------------
MsgBox, 262144, ,% SolveHidato(Grid, Locked, Max, row, col) return</lang>
Outputs:
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
Bracmat
<lang 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) );</lang> 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
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. <lang c>#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; }</lang>
- 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
C++
<lang cpp>
- 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" );
} //-------------------------------------------------------------------------------------------------- </lang> 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
Curry
Probably not efficient. <lang curry>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</lang>
- 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)]
D
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.
<lang d>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;
}</lang>
- 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 . . . . . . . . . . .
Stronger Version
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. <lang d>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
. . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . . . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ ." );
}</lang>
- 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 .
Elixir
<lang elixir># Solve a Hidato Like Puzzle with Warnsdorff like logic applied
defmodule HLPsolver do
defmodule Cell do
defstruct value: -1, used: false, adj: []
end
def solve(str, adjacent, print_out\\true) do
board = setup(str)
if print_out, do: print(board, "Problem:")
{start, _} = Enum.find(board, fn {_,cell} -> cell.value==1 end)
board = set_adj(board, adjacent)
zbl = for %Cell{value: n} <- Map.values(board), into: %{}, do: {n, true}
try do
solve(board, start, 1, zbl, map_size(board))
IO.puts "No solution"
catch
{:ok, result} -> if print_out, do: print(result, "Solution:"),
else: result
end
end
defp solve(board, position, seq_num, zbl, goal) do
value = board[position].value
cond do
value > 0 and value != seq_num -> nil
value == 0 and zbl[seq_num] -> nil
true ->
cell = %Cell{board[position] | value: seq_num, used: true}
board = %{board | position => cell}
if seq_num == goal, do: throw({:ok, board})
Enum.each(wdof(board, cell.adj), fn pos ->
solve(board, pos, seq_num+1, zbl, goal)
end)
end
end
defp setup(str) do
lines = String.strip(str) |> String.split(~r/(\n|\r\n|\r)/) |> Enum.with_index
for {line,i} <- lines, {char,j} <- Enum.with_index(String.split(line)),
:error != Integer.parse(char), into: %{} do
{n,_} = Integer.parse(char)
{{i,j}, %Cell{value: n}}
end
end
defp set_adj(board, adjacent) do
Enum.reduce(Map.keys(board), board, fn {x,y},map ->
adj = Enum.map(adjacent, fn {i,j} -> {x+i, y+j} end)
|> Enum.reduce([], fn pos,acc -> if board[pos], do: [pos | acc], else: acc end)
Map.update!(map, {x,y}, fn cell -> %Cell{cell | adj: adj} end)
end)
end
defp wdof(board, adj) do # Warnsdorf's rule
Enum.reject(adj, fn pos -> board[pos].used end)
|> Enum.sort_by(fn pos ->
Enum.count(board[pos].adj, fn p -> not board[p].used end)
end)
end
def print(board, title) do
IO.puts "\n#{title}"
{xmin, xmax} = Map.keys(board) |> Enum.map(fn {x,_} -> x end) |> Enum.min_max
{ymin, ymax} = Map.keys(board) |> Enum.map(fn {_,y} -> y end) |> Enum.min_max
len = map_size(board) |> to_char_list |> length
space = String.duplicate(" ", len)
Enum.each(xmin..xmax, fn x ->
Enum.map_join(ymin..ymax, " ", fn y ->
case Map.get(board, {x,y}) do
nil -> space
cell -> to_string(cell.value) |> String.rjust(len)
end
end)
|> IO.puts
end)
end
end</lang>
Test: <lang elixir>adjacent = [{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}]
"""
. 4 0 7 0 1 0 0
""" |> HLPsolver.solve(adjacent)
"""
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
""" |> HLPsolver.solve(adjacent)
"""
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 0 . 0 0 0 . 0 0 0 . 0 0 0 . 0 0 0 . 0
""" |> HLPsolver.solve(adjacent)</lang>
- 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 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 9 10 11 17 18 19 25 26 27 33 34 35 41 42 43 49 50 51
4 8 12 16 20 24 28 32 36 40 44 48 52
5 6 7 13 14 15 21 22 23 29 30 31 37 38 39 45 46 47 53
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. <lang Erlang> -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 ). </lang>
- 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
Go
<lang go>package main
import (
"fmt" "sort" "strconv" "strings"
)
var board [][]int var start, given []int
func setup(input []string) {
/* This task is not about input validation, so
we're going to trust the input to be valid */
puzzle := make([][]string, len(input))
for i := 0; i < len(input); i++ {
puzzle[i] = strings.Fields(input[i])
}
nCols := len(puzzle[0])
nRows := len(puzzle)
list := make([]int, nRows*nCols)
board = make([][]int, nRows+2)
for i := 0; i < nRows+2; i++ {
board[i] = make([]int, nCols+2)
for j := 0; j < nCols+2; j++ {
board[i][j] = -1
}
}
for r := 0; r < nRows; r++ {
row := puzzle[r]
for c := 0; c < nCols; c++ {
switch cell := row[c]; cell {
case "_":
board[r+1][c+1] = 0
case ".":
break
default:
val, _ := strconv.Atoi(cell)
board[r+1][c+1] = val
list = append(list, val)
if val == 1 {
start = append(start, r+1, c+1)
}
}
}
}
sort.Ints(list)
given = make([]int, len(list))
for i := 0; i < len(given); i++ {
given[i] = list[i]
}
}
func solve(r, c, n, next int) bool {
if n > given[len(given)-1] {
return true
}
back := board[r][c]
if back != 0 && back != n {
return false
}
if back == 0 && given[next] == n {
return false
}
if back == n {
next++
}
board[r][c] = n
for i := -1; i < 2; i++ {
for j := -1; j < 2; j++ {
if solve(r+i, c+j, n+1, next) {
return true
}
}
}
board[r][c] = back return false
}
func printBoard() {
for _, row := range board {
for _, c := range row {
switch {
case c == -1:
fmt.Print(" . ")
case c > 0:
fmt.Printf("%2d ", c)
default:
fmt.Print("__ ")
}
}
fmt.Println()
}
}
func main() {
input := []string{
"_ 33 35 _ _ . . .",
"_ _ 24 22 _ . . .",
"_ _ _ 21 _ _ . .",
"_ 26 _ 13 40 11 . .",
"27 _ _ _ 9 _ 1 .",
". . _ _ 18 _ _ .",
". . . . _ 7 _ _",
". . . . . . 5 _",
}
setup(input)
printBoard()
fmt.Println("\nFound:")
solve(start[0], start[1], 1, 0)
printBoard()
}</lang>
- 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 . . . . . . . . . . .
Haskell
<lang 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" ]</lang>
- 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
Icon and Unicon
This is an Unicon-specific solution but could easily be adjusted to work in Icon. <lang unicon>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</lang>
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
->
Java
<lang java>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();
}
}
}</lang>
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 . . . . . . . . . . .
Kotlin
<lang scala>// version 1.2.0
lateinit var board: List<IntArray> lateinit var given: IntArray lateinit var start: IntArray
fun setUp(input: List<String>) {
val nRows = input.size
val puzzle = List(nRows) { input[it].split(" ") }
val nCols = puzzle[0].size
val list = mutableListOf<Int>()
board = List(nRows + 2) { IntArray(nCols + 2) { -1 } }
for (r in 0 until nRows) {
val row = puzzle[r]
for (c in 0 until nCols) {
val cell = row[c]
if (cell == "_") {
board[r + 1][c + 1] = 0
}
else if (cell != ".") {
val value = cell.toInt()
board[r + 1][c + 1] = value
list.add(value)
if (value == 1) start = intArrayOf(r + 1, c + 1)
}
}
}
list.sort()
given = list.toIntArray()
}
fun solve(r: Int, c: Int, n: Int, next: Int): Boolean {
if (n > given[given.lastIndex]) return true
val back = board[r][c]
if (back != 0 && back != n) return false
if (back == 0 && given[next] == n) return false
var next2 = next
if (back == n) next2++
board[r][c] = n
for (i in -1..1)
for (j in -1..1)
if (solve(r + i, c + j, n + 1, next2)) return true
board[r][c] = back
return false
}
fun printBoard() {
for (row in board) {
for (c in row) {
if (c == -1)
print(" . ")
else
print(if (c > 0) "%2d ".format(c) else "__ ")
}
println()
}
}
fun main(args: Array<String>) {
var input = listOf(
"_ 33 35 _ _ . . .",
"_ _ 24 22 _ . . .",
"_ _ _ 21 _ _ . .",
"_ 26 _ 13 40 11 . .",
"27 _ _ _ 9 _ 1 .",
". . _ _ 18 _ _ .",
". . . . _ 7 _ _",
". . . . . . 5 _"
)
setUp(input)
printBoard()
println("\nFound:")
solve(start[0], start[1], 1, 0)
printBoard()
}</lang>
- 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 . . . . . . . . . . .
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>
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
Nim
<lang nim>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))
writeLine(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()</lang>
- 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 . . . . . . . . . . .
Perl
<lang 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, $known[1]);</lang>
- Output:
32 33 35 36 37
31 34 24 22 38
30 25 23 21 12 39
29 26 20 13 40 11
27 28 14 19 9 10 1
15 16 18 8 2
17 7 6 3
5 4
Perl 6
This uses a Warnsdorff solver, which cuts down the number of tries by more than a factor of six over the brute force approach. This same solver is used in:
- Solve a Hidato puzzle
- Solve a Hopido puzzle
- Solve a Holy Knight's tour
- Solve a Numbrix puzzle
- Solve the no connection puzzle
<lang perl6>my @adjacent = [-1, -1], [-1, 0], [-1, 1],
[ 0, -1], [ 0, 1],
[ 1, -1], [ 1, 0], [ 1, 1];
solveboard q:to/END/;
__ 33 35 __ __ .. .. ..
__ __ 24 22 __ .. .. ..
__ __ __ 21 __ __ .. ..
__ 26 __ 13 40 11 .. ..
27 __ __ __ 9 __ 1 ..
.. .. __ __ 18 __ __ ..
.. .. .. .. __ 7 __ __
.. .. .. .. .. .. 5 __
END
sub solveboard($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";
}</lang>
Phix
<lang Phix>sequence board, warnsdorffs, knownx, knowny
integer width, height, limit, nchars, tries string fmt, blank
constant ROW = 1, COL = 2 constant moves = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}}
function onboard(integer row, integer col)
return row>=1 and row<=height and col>=nchars and col<=nchars*width
end function
procedure init_warnsdorffs() integer nrow,ncol
for row=1 to height do
for col=nchars to nchars*width by nchars do
for move=1 to length(moves) do
nrow = row+moves[move][ROW]
ncol = col+moves[move][COL]*nchars
if onboard(nrow,ncol)
and board[nrow][ncol]='_' then
warnsdorffs[nrow][ncol] += 1
end if
end for
end for
end for
end procedure
function solve(integer row, integer col, integer n) integer nrow, ncol
tries+= 1
if n>limit then return 1 end if
if knownx[n] then
for move=1 to length(moves) do
nrow = row+moves[move][ROW]
ncol = col+moves[move][COL]*nchars
if nrow = knownx[n]
and ncol = knowny[n] then
if solve(nrow,ncol,n+1) then return 1 end if
exit
end if
end for
return 0
end if
sequence wmoves = {}
for move=1 to length(moves) do
nrow = row+moves[move][ROW]
ncol = col+moves[move][COL]*nchars
if onboard(nrow,ncol)
and board[nrow][ncol]='_' then
wmoves = append(wmoves,{warnsdorffs[nrow][ncol],nrow,ncol})
end if
end for
wmoves = sort(wmoves)
-- avoid creating orphans
if length(wmoves)<2 or wmoves[2][1]>1 then
for m=1 to length(wmoves) do
{?,nrow,ncol} = wmoves[m]
warnsdorffs[nrow][ncol] -= 1
end for
for m=1 to length(wmoves) do
{?,nrow,ncol} = wmoves[m]
board[nrow][ncol-nchars+1..ncol] = sprintf(fmt,n)
if solve(nrow,ncol,n+1) then return 1 end if
board[nrow][ncol-nchars+1..ncol] = blank
end for
for m=1 to length(wmoves) do
{?,nrow,ncol} = wmoves[m]
warnsdorffs[nrow][ncol] += 1
end for
end if
return 0
end function
procedure Hidato(sequence s, integer w, integer h, integer lim) integer y, ch, ch2, k atom t0 = time()
s = split(s,'\n')
width = w
height = h
nchars = length(sprintf(" %d",lim))
fmt = sprintf(" %%%dd",nchars-1)
blank = repeat('_',nchars)
board = repeat(repeat(' ',width*nchars),height)
knownx = repeat(0,lim)
knowny = repeat(0,lim)
limit = 0
for x=1 to height do
for y=nchars to width*nchars by nchars do
if y>length(s[x]) then
ch = '.'
else
ch = s[x][y]
end if
if ch='_' then
limit += 1
elsif ch!='.' then
k = ch-'0'
ch2 = s[x][y-1]
if ch2!=' ' then
k += (ch2-'0')*10
board[x][y-1] = ch2
end if
knownx[k] = x
knowny[k] = y
limit += 1
end if
board[x][y] = ch
end for
end for
warnsdorffs = repeat(repeat(0,width*nchars),height)
init_warnsdorffs()
tries = 0
if solve(knownx[1],knowny[1],2) then
puts(1,join(board,"\n"))
printf(1,"\nsolution found in %d tries (%3.2fs)\n",{tries,time()-t0})
else
puts(1,"no solutions found\n")
end if
end procedure
constant board1 = """
__ 33 35 __ __ .. .. .. __ __ 24 22 __ .. .. .. __ __ __ 21 __ __ .. .. __ 26 __ 13 40 11 .. .. 27 __ __ __ 9 __ 1 .. .. .. __ __ 18 __ __ .. .. .. .. .. __ 7 __ __ .. .. .. .. .. .. 5 __"""
Hidato(board1,8,8,40)
constant board2 = """
. 4 . _ 7 _ 1 _ _"""
Hidato(board2,3,3,7)
constant board3 = """
1 _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . 74 . . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . . . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ ."""
Hidato(board3,50,3,74)
constant board4 = """
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"""
Hidato(board4,9,9,72)
constant board5 = """
__ 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"""
Hidato(board5,9,9,72)
constant board6 = """
1 __ .. .. .. __ __ .. .. .. __ __ .. .. .. __ __ .. .. .. __ __ .. .. .. __ __ .. .. .. __ __ .. .. .. __ __ .. .. .. __ __ .. .. .. 82 .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ .. __ .. .. __ __ __ .. .. __ __ __ .. .. __ __ __ .. .. __ __ __ .. .. __ __ __ .. .. __ __ __ .. .. __ __ __ .. .. __ __ __ .. .. __ __ __ .. .. .."""
Hidato(board6,46,4,82)</lang>
- 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 solution found in 760 tries (0.00s) . 4 . 3 7 5 1 2 6 solution found in 10 tries (0.00s) 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 . solution found in 74 tries (0.00s) 54 53 60 59 58 67 66 69 70 52 55 61 57 63 65 68 72 71 51 50 56 62 64 . . . . 49 12 13 14 . . 17 21 . 48 10 11 . 15 16 18 20 22 47 46 9 . 3 2 19 23 24 45 44 8 5 4 1 33 32 25 41 43 7 6 36 34 27 26 31 42 40 39 38 37 35 28 29 30 solution found in 106 tries (0.00s) 56 58 54 60 61 62 63 66 67 57 55 59 53 49 47 65 64 68 9 8 52 51 50 48 46 45 69 10 6 7 . . . 44 43 70 5 11 12 . . . 72 71 42 4 14 13 . . . 30 39 41 15 3 17 18 28 29 38 31 40 2 16 19 22 25 27 37 36 32 1 20 21 24 23 26 35 34 33 solution found in 495 tries (0.00s) 1 2 . . . 10 11 . . . 19 20 . . . 28 29 . . . 37 38 . . . 46 47 . . . 55 56 . . . 64 65 . . . 73 74 . . . 82 . . 3 . 9 . . 12 . 18 . . 21 . 27 . . 30 . 36 . . 39 . 45 . . 48 . 54 . . 57 . 63 . . 66 . 72 . . 75 . 81 . . 4 . 8 . . 13 . 17 . . 22 . 26 . . 31 . 35 . . 40 . 44 . . 49 . 53 . . 58 . 62 . . 67 . 71 . . 76 . 80 . . 5 6 7 . . 14 15 16 . . 23 24 25 . . 32 33 34 . . 41 42 43 . . 50 51 52 . . 59 60 61 . . 68 69 70 . . 77 78 79 . . . solution found in 82 tries (0.02s)
PicoLisp
<lang 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 @) " ") ) ) ) )</lang>
Test: <lang PicoLisp>(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) ) )</lang>
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
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
Python
<lang 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()</lang>
- 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
Racket
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>
- 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))) </lang>
- 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]])
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>#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)))))
</lang>
- 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
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. <lang rexx>/*REXX program solves a Numbrix (R) puzzle, it also displays the puzzle and solution. */ maxR=0; maxC=0; maxX=0; minR=9e9; minC=9e9; minX=9e9; cells=0; @.= parse arg xxx; PZ='Hidato puzzle' /*get the cell definitions from the CL.*/ xxx=translate(xxx, , "/\;:_", ',') /*also allow other characters as comma.*/
do while xxx\=; parse var xxx r c marks ',' xxx
do while marks\=; _=@.r.c
parse var marks x marks
if datatype(x,'N') then 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 /*the START cell. */
if _\== then call err "cell at" r c 'is already occupied with:' _
@.r.c=x; c=c+1; cells=cells+1 /*assign a mark. */
if x==. then iterate /*is a hole? Skip*/
if \datatype(x,'W') then call err 'illegal marker specified:' x
minX=min(minX,x); maxX=max(maxX,x) /*min and max X. */
end /*while marks¬= */
end /*while xxx ¬= */
call show /* [↓] is used for making fast moves. */ Nr = '0 1 0 -1 -1 1 1 -1' /*possible row for the next move. */ Nc = '1 0 -1 0 1 -1 1 -1' /* " column " " " " */ 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 /*for fast moves. */
if \next(2,!r,!c) then call err 'No solution possible for this' PZ "puzzle." say 'A solution for the' PZ "exists."; say; call show exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ err: say; say '***error*** (from' PZ"): " arg(1); say; exit 13 /*──────────────────────────────────────────────────────────────────────────────────────*/ next: procedure expose @. Nr. Nc. cells pMoves; parse arg #,r,c; ##=#+1
do t=1 for pMoves /* [↓] try some moves. */
parse value r+Nr.t c+Nc.t with nr nc /*next move coördinates.*/
if @.nr.nc==. then do; @.nr.nc=# /*let's try this move. */
if #==cells then leave /*is this the last move?*/
if next(##,nr,nc) then return 1
@.nr.nc=. /*undo the above move. */
iterate /*go & try another move.*/
end
if @.nr.nc==# then do /*this a fill-in move ? */
if #==cells then return 1 /*this is the last move.*/
if next(##,nr,nc) then return 1 /*a fill-in move. */
end
end /*t*/
return 0 /*this ain't working. */
/*──────────────────────────────────────────────────────────────────────────────────────*/ show: if maxR<1 | maxC<1 then call err 'no legal cell was specified.'
if minX<1 then call err 'no 1 was specified for the puzzle start'
w=max(2,length(cells)); do r=maxR to minR by -1; _=
do c=minC to maxC; _=_ right(@.r.c,w); end /*c*/
say _
end /*r*/
say; return</lang>
output when 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
Ruby
Without Warnsdorff
The following class provides functionality for solving a hidato problem: <lang ruby># 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</lang>
Test: <lang ruby># 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"</lang>
- 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
With Warnsdorff
I modify method as follows to implement Warnsdorff like <lang ruby># Solve a Hidato Like Puzzle with Warnsdorff like logic applied
class HLPsolver
attr_reader :board
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</lang> Which may be used as follows to solve Hidato Puzzles: <lang ruby>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"</lang>
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:
Seed7
<lang 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;</lang>
- 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 . . . . . . . . . . .
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
More complex cases are solvable with an extended version of this code, though that has more onerous version requirements.
zkl
<lang zkl>hi:= // 0==empty cell, X==not a cell
- <<<
"0 33 35 0 0 X X X 0 0 24 22 0 X X X 0 0 0 21 0 0 X X 0 26 0 13 40 11 X X 27 0 0 0 9 0 1 X X X 0 0 18 0 0 X X X X X 0 7 0 0 X X X X X X 5 0";
- <<<
board,given,start:=setup(hi); print_board(board); solve(board,given, start.xplode(), 1); println(); print_board(board);</lang> <lang zkl>fcn print_board(board){
d:=D(-1," ", 0,"__");
foreach r in (board[1,-1]){
r[1,-1].pump(String,'wrap(c){ "%2s ".fmt(d.find(c,c)) }).println();
}
} fcn setup(s){
lines:=s.split("\n");
ncols,nrows:=lines[0].split().len(),lines.len();
board:=(nrows+2).pump(List(), (ncols+2).pump(List(),-1).copy);
given,start:=List(),Void;
foreach r,row in (lines.enumerate()){
foreach c,cell in (row.split().enumerate()){
if(cell=="X") continue; // X == not in play, leave at -1
val:=cell.toInt(); board[r+1][c+1]=val; given.append(val); if(val==1) start=T(r+1,c+1);
} } return(board,given.filter().sort(),start);
} fcn solve(board,given, 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;
foreach i,j in ([-1..1],[-1..1]){
if(solve(board,given, r+i,c+j,n+1, next)) return(True);
}
board[r][c]=back;
False
}</lang>
- 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