How to resolve the algorithm Solve a Numbrix puzzle step by step in the Icon and Unicon programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Solve a Numbrix puzzle step by step in the Icon and Unicon programming language
Table of Contents
Problem Statement
Numbrix puzzles are similar to Hidato. The most important difference is that it is only possible to move 1 node left, right, up, or down (sometimes referred to as the Von Neumann neighborhood). Published puzzles also tend not to have holes in the grid and may not always indicate the end node. Two examples follow: Problem. Solution. Problem. Solution. Write a program to solve puzzles of this ilk, demonstrating your program by solving the above examples. Extra credit for other interesting examples.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Solve a Numbrix puzzle step by step in the Icon and Unicon programming language
Source code in the icon programming language
global nCells, cMap, best
record Pos(r,c)
procedure main(A)
puzzle := showPuzzle("Input",readPuzzle())
QMouse(puzzle,findStart(puzzle),&null,0)
showPuzzle("Output", solvePuzzle(puzzle)) | write("No solution!")
end
procedure readPuzzle()
# Start with a reduced puzzle space
p := []
nCells := maxCols := 0
every line := !&input do {
put(p,[: gencells(line) :])
maxCols <:= *p[-1]
}
# Now normalize all rows to the same length
every i := 1 to *p do p[i] := [: !p[i] | (|-1\(maxCols - *p[i])) :]
return p
end
procedure gencells(s)
static WS, NWS
initial {
NWS := ~(WS := " \t")
cMap := table() # Map to/from internal model
cMap["_"] := 0; cMap[0] := "_"
}
s ? while not pos(0) do {
w := (tab(many(WS))|"", tab(many(NWS))) | break
w := numeric(\cMap[w]|w)
if -1 ~= w then nCells +:= 1
suspend w
}
end
procedure showPuzzle(label, p)
write(label," with ",nCells," cells:")
every r := !p do {
every c := !r do writes(right((\cMap[c]|c),*nCells+1))
write()
}
return p
end
procedure findStart(p)
if \p[r := !*p][c := !*p[r]] = 1 then return Pos(r,c)
end
procedure solvePuzzle(puzzle)
if path := \best then {
repeat {
loc := path.getLoc()
puzzle[loc.r][loc.c] := path.getVal()
path := \path.getParent() | break
}
return puzzle
}
end
class QMouse(puzzle, loc, parent, val)
method getVal(); return val; end
method getLoc(); return loc; end
method getParent(); return parent; end
method atEnd(); return (nCells = val, puzzle[loc.r,loc.c] = (val|0)); end
method visit(r,c); return (/best, validPos(r,c), Pos(r,c)); end
method validPos(r,c)
v := val+1 # number we're looking for
xv := puzzle[r,c] | fail
if (xv ~= 0) & (xv != v) then fail
if xv = (0|v) then {
ancestor := self
while xl := (ancestor := \ancestor.getParent()).getLoc() do
if (xl.r = r) & (xl.c = c) then fail
return
}
end
initially
val := val+1
if atEnd() then return best := self
QMouse(puzzle, visit(loc.r-1,loc.c) , self, val) # North
QMouse(puzzle, visit(loc.r, loc.c+1), self, val) # East
QMouse(puzzle, visit(loc.r+1,loc.c), self, val) # South
QMouse(puzzle, visit(loc.r, loc.c-1), self, val) # West
end
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