Maze generation: Difference between revisions

Added solution for Action!
m (Updated description and link for Fōrmulæ solution)
(Added solution for Action!)
Line 71:
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
</pre>
 
=={{header|Action!}}==
Action! language does not support recursion. Therefore an iterative approach with a stack has been proposed.
<lang Action!>DEFINE TOP="0"
DEFINE RIGHT="1"
DEFINE BOTTOM="2"
DEFINE LEFT="3"
DEFINE WIDTH="160"
DEFINE HEIGHT="96"
 
DEFINE STACK_SIZE="5000"
BYTE ARRAY stack(STACK_SIZE)
INT stackSize
 
PROC InitStack()
stackSize=0
RETURN
 
BYTE FUNC IsEmpty()
IF stackSize=0 THEN
RETURN (1)
FI
RETURN (0)
 
BYTE FUNC IsFull()
IF stackSize>=STACK_SIZE THEN
RETURN (1)
FI
RETURN (0)
 
PROC Push(BYTE x,y)
IF IsFull() THEN Break() RETURN FI
stack(stackSize)=x stackSize==+1
stack(stackSize)=y stackSize==+1
RETURN
 
PROC Pop(BYTE POINTER x,y)
IF IsEmpty() THEN Break() RETURN FI
stackSize==-1 y^=stack(stackSize)
stackSize==-1 x^=stack(stackSize)
RETURN
 
PROC FillScreen()
BYTE POINTER ptr ;pointer to the screen memory
INT screenSize=[3840]
 
ptr=PeekC(88)
SetBlock(ptr,screenSize,$55)
 
Color=0
Plot(0,HEIGHT-1) DrawTo(WIDTH-1,HEIGHT-1) DrawTo(WIDTH-1,0)
RETURN
 
PROC GetNeighbors(BYTE x,y BYTE ARRAY n BYTE POINTER count)
DEFINE WALL="1"
 
count^=0
IF y>2 AND Locate(x,y-2)=WALL THEN
n(count^)=TOP count^==+1
FI
IF x<WIDTH-3 AND Locate(x+2,y)=WALL THEN
n(count^)=RIGHT count^==+1
FI
IF y<HEIGHT-3 AND Locate(x,y+2)=WALL THEN
n(count^)=BOTTOM count^==+1
FI
IF x>2 AND Locate(x-2,y)=WALL THEN
n(count^)=LEFT count^==+1
FI
RETURN
 
PROC Maze(BYTE x,y)
BYTE ARRAY stack,neighbors
BYTE dir,nCount
 
FillScreen()
 
Color=2
InitStack()
Push(x,y)
WHILE IsEmpty()=0
DO
Pop(@x,@y)
GetNeighbors(x,y,neighbors,@nCount)
IF nCount>0 THEN
Push(x,y)
Plot(x,y)
dir=neighbors(Rand(nCount))
IF dir=TOP THEN
y==-2
ELSEIF dir=RIGHT THEN
x==+2
ELSEIF dir=BOTTOM THEN
y==+2
ELSE
x==-2
FI
DrawTo(x,y)
Push(x,y)
FI
OD
RETURN
 
PROC Main()
BYTE CH=$02FC,COLOR0=$02C4,COLOR1=$02C5
BYTE x,y
 
Graphics(7+16)
COLOR0=$0A
COLOR1=$04
 
x=Rand((WIDTH RSH 1)-1) LSH 1+1
y=Rand((HEIGHT RSH 1)-1) LSH 1+1
Maze(x,y)
 
DO UNTIL CH#$FF OD
CH=$FF
RETURN</lang>
{{out}}
[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Maze_generation.png Screenshot from Atari 8-bit computer]
 
=={{header|Ada}}==
Anonymous user