CODEWITHSUNDEEP
recursionassemblyx86flickerflood-fill
Sep Roland
Sep Roland

Reputation: 39506

Avoiding flicker while using a recursive flood fill algorithm

The wikipedia article about flood fill describes the following recursive algorithm:

Flood-fill (node):

  1. If node is not Inside return.
  2. Set the node
  3. Perform Flood-fill one step to the south of node.
  4. Perform Flood-fill one step to the north of node
  5. Perform Flood-fill one step to the west of node
  6. Perform Flood-fill one step to the east of node
  7. Return.

Because I couldn't find too much examples of flood-filling using the assembly language, I implemented this myself (using FASM) and made - I think - a rather decent version. There's much worse you can find here and here!

For those that are inquisitive, performance-wise the order that the neighboring pixels are visited does not change much. So it does not need to be South->North->West->East. Any of the 24 permutations will do, but better not pick an ordering where you don't immediately follow a direction by its opposite direction, because those will double the stack usage! The good-ones are SNWE, NSWE, SNEW, NSEW, WESN, EWSN, WENS, and EWNS.

The program works correctly but also produces flicker on screen. That's probably due to using a recursive solution for flood-filling... But is it?

; ------------------------------
; BL is OldColor
; BH is NewColor
; CX is X
; DX is Y
; IN (bx,cx,dx)
FloodWhile:
    pusha
    mov   al, bl                ; NewColor
    mov   ah, 0Ch
    mov   si, ax                ; CONST
    shr   bx, 8                 ; -> BL is OldColor, BH is DisplayPage

    cmp   al, bl                ; NewColor must be different from OldColor
    je    .z
    mov   ah, 0Dh               ; BIOS.ReadPixel
    int   10h                   ; -> AL
    cmp   al, bl                ; First pixel must have OldColor
    jne   .z
    call  .Flood                ; -> (AX)
.z: popa
    ret
; - - - - - - - - - - - - - - -
; IN (bx,cx,dx,si) OUT () MOD (ax)
.Flood:
    push  di
    mov   di, 12
    mov   ax, si                ; -> AL is NewColor, AH is FunctionID
    int   10h                   ; BIOS.WritePixel

.a: add   cx, [.z+di]           ; Try a neighboring pixel (SNWE)
    cmp   cx, 320
    jnb   .b
    add   dx, [.z+di+2]
    cmp   dx, 200
    jnb   .b
    mov   ah, 0Dh               ; BIOS.ReadPixel
    int   10h                   ; -> AL
    cmp   al, bl
    jne   .b
    call  .Flood
.b: sub   di, 4
    jns   .a
    dec   cx                    ; Return from East

    pop   di
    ret
;         East  West   North South  <-- SNWE
.z: dw    +2,0, -1,+1, 0,-2, 0,+1
; ------------------------------

Upvotes: 0

Views: 30

Answers (1)

Sep Roland
Sep Roland

Reputation: 39506

That's probably due to using a recursive solution for flood-filling... But is it?

It is true that recursion is infamous for its potential for overflowing the stack, but recursion need not be slow per se. The first code snippet that I'll show below will use a re-worked recursive algorithm and it will run about 62 times faster than yours. Its iterative friend comes in at just 10% faster.

The program works correctly but also produces flicker on screen.

Flicker can go away by relying on very fast graphics routines, or by applying graphics routines of any speed to an offscreen memory buffer followed by updating the real screen in one very fast copy operation. The three code snippets that I show below will be using a hybrid approach: simultaneously writing to an offscreen buffer and to the real screen.

The major changes include:

  • Not using BIOS for pixel input/output for a 5x speed increase. BIOS is a generalist and so there's simply too much overhead involved in addressing the screen. If we address the screen ourselves, we can handpick the best way to do it and cut corners however we like.
  • Not reading from the video memory for a 12x speed increase. Accessing video memory and especially reading from it, is a very costly operation. I maintain a so-called double buffer which mirrors what is on the real screen, and whenever I need to read a pixel I read from that buffer. Conversely, whenever I need to write a pixel I write it to both that buffer and the real screen.
  • Not exclusively dealing with single pixels for a 2x speed increase. The optimal solution gathers as many as possible complying pixels from the same scanline, and then uses an efficient line-drawing code speeding up the output.

The tests ran on the legacy 320x200 256-color screen.

Examples of flood-filled shapes

Flood-filling a circle (R=75) producing a plain disc:

asm stack heap reads writes pixels/second
Question 90 35300 0 69925 17481 86865
Good 103 17650 0 69925 17481 5323081 61x
Better 150 0 1192 69925 17481 5943896 68x
Best 206 0 12 35557 17481 9578630 110x

Flood-filling a labyrinth of concentric circles:

asm stack heap reads writes pixels/second
Question 90 38156 0 58273 14568 86791
Good 103 19078 0 58273 14568 5524459 63x
Better 150 0 156 58273 14568 5924359 68x
Best 206 0 168 38092 14568 9410852 108x

62x faster than the original

This is the recursive solution but with optimizations applied, eg. the code does but a single address calculation based on X and Y, and then applies suitable increments to the scanline address and to the offset within the scanline.

; ------------------------------
; AL is OldColor
; AH is NewColor
; BX is X
; CX is Y
; GS is DoubleBuffer (64KB)
; IN (ax,bx,cx,GS)
GoodFloodWhile:
    push  es
    pusha
    mov   di, 0A000h
    mov   es, di
    mov   di, bx                ; X
    imul  bp, cx, 320           ; Y * 320

    cmp   al, ah                ; NewColor must be different from OldColor
    je    .z
    cmp   [GS:bp+di], ah        ; First pixel must have OldColor
    jne   .z
    call  .Flood
.z: popa
    pop   es
    ret
; - - - - - - - - - - - - - - -
    ALIGN 16
; IN (ax,di,bp,es,GS) OUT ()
.Flood:
    mov   [es:bp+di], al        ; Center pixel on Screen
    mov   [GS:bp+di], al        ; Center pixel in Double Buffer

    inc   di                    ; Try pixel to the right
    cmp   di, 320
    jnb   .NotR
    cmp   [GS:bp+di], ah        ; OldColor
    jne   .NotR
    call  .Flood
.NotR:
    dec   di

    dec   di                    ; Try pixel to the left
    js    .NotL
    cmp   [GS:bp+di], ah        ; OldColor
    jne   .NotL
    call  .Flood
.NotL:
    inc   di

    add   bp, 320               ; Try pixel downward
    cmp   bp, 200*320
    jnb   .NotD
    cmp   [GS:bp+di], ah        ; OldColor
    jne   .NotD
    call  .Flood
.NotD:

    sub   bp, 640               ; Try pixel upward
    jb    .NotU
    cmp   [GS:bp+di], ah        ; OldColor
    jne   .NotU
    call  .Flood
.NotU:
    add   bp, 320

    ret
; ------------------------------

68x faster than the original

This iterative solution uses a small (2KB) ringbuffer where it doesn't just record X and Y coordinates, but rather stores scanline addresses and offsets within the scanline. The motivation here was to avoid the many address calculations that require multiplication with the bytes-per-scanline value.
Always choose a power-of-two for the size of the ringbuffer. That way, the wraparound for the read- and write pointers can be just one simple and operation. The ringbuffer works FirstInFirstOut (FIFO) as opposed to the stack that works LastInFirstOut (LIFO). This is an important fact because if you were to use a LIFO buffer for your iterative solution that buffer would also consume a lot of bytes (just like it was with the recursion's stack).
Tip: provided that you insert a suitable delay, this can be a good code to animate a liquid running through pipes...

; ------------------------------
; AL is OldColor
; AH is NewColor
; BX is X
; CX is Y
; GS is DoubleBuffer (64KB)
; IN (ax,bx,cx,GS)
BetterFloodWhile:
    push  es
    pusha
    mov   di, 0A000h
    mov   es, di
    mov   di, bx                ; X
    imul  bp, cx, 320           ; Y * 320

    cmp   al, ah                ; NewColor must be different from OldColor
    je    .z
    cmp   [GS:bp+di], ah        ; First pixel must have OldColor
    jne   .z
    xor   si, si                ; SI is OffsetReadPointer in FIFO buffer
    xor   bx, bx                ; BX is OffsetWritePointer in FIFO buffer
    call  .Store                ; -> BX
    call  .Flood                ; -> (BX SI DI BP)
.z: popa
    pop   es
    ret
; - - - - - - - - - - - - - - -
    ALIGN 16
; IN (ax,bx,si,di,bp,es,GS) OUT () MOD (bx,si,di,bp)
.Flood:
    mov   di, [Buffer+si]       ; Offset in scanline
    mov   bp, [Buffer+si+2]     ; Address of scanline
    add   si, 4
    and   si, 2047              ; 2KB Ringbuffer (FIFO)

    inc   di                    ; Try pixel to the right
    cmp   di, 320
    jnb   .NotR
    cmp   [GS:bp+di], ah        ; OldColor
    jne   .NotR
    call  .Store                ; -> BX
.NotR:
    dec   di

    dec   di                    ; Try pixel to the left
    js    .NotL
    cmp   [GS:bp+di], ah        ; OldColor
    jne   .NotL
    call  .Store                ; -> BX
.NotL:
    inc   di

    add   bp, 320               ; Try pixel downward
    cmp   bp, 200*320
    jnb   .NotD
    cmp   [GS:bp+di], ah        ; OldColor
    jne   .NotD
    call  .Store                ; -> BX
.NotD:

    sub   bp, 640               ; Try pixel upward
    jb    .NotU
    cmp   [GS:bp+di], ah        ; OldColor
    jne   .NotU
    call  .Store                ; -> BX
.NotU:

    cmp   si, bx
    jne   .Flood
    ret    
; - - - - - - - - - - - - - - -
    ALIGN 16
; IN (al,bx,di,bp,es,GS) OUT (bx)
.Store:
    mov   [es:bp+di], al        ; Center pixel on Screen
    mov   [GS:bp+di], al        ; Center pixel in Double Buffer
    mov   [Buffer+bx], di
    mov   [Buffer+bx+2], bp
    add   bx, 4
    and   bx, 2047              ; 2KB Ringbuffer (FIFO)
    ret
; ------------------------------

109x faster than the original

The focus here is to find long runs of pixels (aka spans) that can be drawn with an efficient line-drawing method.
Code alignment proved to be important in this code.
It was the excellent work of Lode Vandevenne that inspired me to write this optimal solution.

; ------------------------------
; AL is OldColor
; AH is NewColor
; BX is X
; CX is Y
; GS is DoubleBuffer (64KB)
    ALIGN 16
    DB 4 dup 0
; IN (ax,bx,cx,GS)
BestFloodWhile:
    push  es
    pusha
    mov   dx, cx                ; Y
    cmp   al, ah                ; NewColor must be different from OldColor
    je    .Done
    xor   si, si
    xor   cx, cx
; - - - - - - - - - - - - - - -
.Again:
    imul  di, dx, 320
    cmp   [GS:di+bx+0], ah      ; OldColor
    jne   .z                    ; Got colored in the mean time
    lea   bp, [word bx+0]
    add   bx, cx                ; Avoids retesting CX pixels
    
.a: add   bx, 1                 ; X++ Extend to the right
    cmp   bx, 320
    jnb   .b
    cmp   [GS:di+bx], ah        ; OldColor
    je    .a
.b: xchg  bp, bx

.c: dec   bx                    ; X-- Extend to the left
    js    .d
    cmp   [GS:di+bx], ah        ; OldColor
    je    .c
.d: inc   bx                    ; X++

    sub   bp, bx                ; -> BP is length of this line
    mov   cx, 0A000h            ; NewColor line on screen
    call  .Line                 ; -> CX=0 (ES)
    mov   cx, GS                ; NewColor line in Double Buffer
    call  .Line                 ; -> CX=0 (ES)

    sub   di, 320
    dec   dx                    ; Y-- Check above
    js    .e
    push  bp                    ; (1)
    call  .Scan                 ; -> BX SI (BP=0)
    pop   bp                    ; (1)
    sub   bx, bp
.e: inc   dx                    ; Y++

    add   di, 640
    inc   dx                    ; Y++ Check below
    cmp   dx, 200
    jnb   .f
    call  .Scan                 ; -> BX SI (BP=0)
.f: dec   dx                    ; Y--
; - - - - - - - - - - - - - - -
.z: sub   si, 6
    jb    .Done
    mov   bx, [Buffer+si]       ; X
    mov   dx, [Buffer+si+2]     ; Y
    mov   cx, [Buffer+si+4]     ; L-1
    jmp   .Again
; - - - - - - - - - - - - - - -
.Done:
    popa
    pop   es
    ret
; - - - - - - - - - - - - - - -
    ALIGN 16
; IN (al,bx,cx,di,bp) OUT (cx=0) MOD (es)
.Line:                          ; Draw BP long line at [cx:di+bx]
    mov   es, cx
    add   di, bx
    mov   cx, bp                ; BP is 1+
    test  di, 1                 ; Use aligned words
    jz    .y
    stosb
    dec   cx                    ; -> CX is 0+
.y: shr   cx, 1                 ; -> CF
    push  ax
    mov   ah, al
    rep stosw
    pop   ax
    jnc   .x
    stosb
.x: sub   di, bp
    sub   di, bx
    ret
; - - - - - - - - - - - - - - -
    ALIGN 16
; IN (ah,bx,cx=0,dx,si,di,bp,GS) OUT (bx,si) MOD (bp=0)
.Scan:                          ; Scan BP long line at [GS:di+bx]
    cmp   [GS:di+bx], ah        ; OldColor
    je    .v                    ; Part of a span
.w: add   bx, 1                 ; X++
    dec   bp
    jnz   .Scan
    ret
.v: add   si, word 6            ; Begin a new span
    mov   [Buffer+si-6], bx     ; X
    mov   [Buffer+si-4], dx     ; Y
    mov   [Buffer+si-2], cx     ; L-1 Counts the additional pixels on the right
    inc   bx                    ; X++
    dec   bp
    jz    .t
.u: cmp   [GS:di+bx], ah        ; OldColor
    jne   .w                    ; End current span
    inc   word [Buffer+si-2]    ; Cont current span, add to 'additional pixels'
    inc   bx                    ; X++
    dec   bp
    jnz   .u
.t: ret
; ------------------------------

Upvotes: 1

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