; begin Linear Pachinko
;
;!
Function a($)
dim $c, $d, $h, $l, $x, $y

; Get the length for division.
$l = $^0

; Add a hole to the left and right.
$  = '.' + $ + '.'

; Put all relevant occurences in an array.
$x = _, '.', './', '\.', '.|.', '.|', '|.'

; Assign points for occurences in array $x (floor = 0, mountain side or 
; wall with two holes = 10, wall with single hole = 5). These will be 
;multiplied by 10 later to get 100% or 50%.
$y = 0, 10, 10, 10, 10, 5, 5

; Loop through array $x.
For $c = 0 to 6

	; Split on occurence. The ubound of the array is the number of occurences.
	$  = Split($,$x[$c])
	
	; Add the points to the total (occurences X value X 10).
	$h = $h + UBound($)*$y[$c]*10

	; Join $ again. On the first occurence (floor) it is replaced 
	; with nothing ($d = empty), because we don't need them. 
	$  = Join($,$d)

	; The rest of the occurences are replaced with a single hole.
	$d = '.'
	
	; Divide the total points (minus 200 because of the added left and right
	; holes) by the length of the original $.
	$a = ($h-200) / $l
	
EndFunction
;!
;!
; end Linear Pachinko

; begin Linear Pachinko
;
;!
; I'm setting up an array of all possible occurences. I've
; put the length of the original string in an extra field,
; because that saves me a variable. The use of the other
; fields will be explained later.

; I'm looping from 3 to 9 because of the linear result in
; the calculation ($c/4). this will return the following
; range: 0, 1, 1, 1, 1, 2, 2. 

; In the first run I'm splitting on floor parts. they don't
; have any value (the calculation 100/($c/4) will result in
; 0 (see range above) and I don't want them back in the 
; string, so they will be replaced with an empty var ($h).
; When I've removed the floor parts, the only remaining
; possibilities are the ones that are in the array $o.

; The second run I'm splitting on holes. They always have a
; 100% value, so the calculation will be 100/1 and this is
; added to the total $s. Because the variable $h is now
; filled with an x, all the holes will be replaced with an x.

; the x-es in the split line represent the left and Right
; exits so they will be included in the next evaluations.

; The third to fifth fields in $o are 100% values so they
; are also added to the total. The sixth and seventh fields
; are 50% values, so the calculation results in 2 which gives
; 100/2 = 50.

; After each calculation of $s I generate a $a, but this is 
; just done because of shortness of code. Only after the
; seventh run is $s complete and will the calculation 
; $s / $o[7] ($o[7] being the original length of the string)
; result in the correct return value.
; This explains why I can do without the floor parts. After
; all, if it's not a 100% or a 50% hit, it must be a 0% hit
; (ie a floor part).

Function a($)
dim $c, $h, $s, $o

$o = _, '.', 'x/', '\x', 'x|x', 'x|', '|x', $^0

For $c = 3 to 9
	$  = Split(x+$+x, $o[$c-3])
	$s = $s + UBound($) * 100/($c/4)
	$a = $s / $o[7]
	$  = Join($, $h)
	$h = x
EndFunction
;!
;!
; end Linear Pachinko