| | Oswald
Registered: Apr 2002
Posts: 5086 |
Sorting
are sorters really so slow in games? :) I have made my unrolled version for my theoretical game (;), and it takes 132 rlines to sort 32 numbers o_O worst case is ~200 lines. tho when its the case of 4 numbers have to be swapped only it does the job in ~10 lines. wastes a lot of memory but I like it :) |
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| | ChristopherJam
Registered: Aug 2004
Posts: 1408 |
Color Bar, that is indeed a fast way to get a sorted list of Y values, but what is needed is a sorted list of the indices of the sprites that have those Y positions.
(eg if sprites 1, 2, 3 and 4 are at positions 10, 46, 73 and 32, the required output is 1,4,2,3, not 10,32,46,73 ) |
| | Oswald
Registered: Apr 2002
Posts: 5086 |
35-40 lines for 32 sprites is crazy good guys 1.x line per sprite. |
| | Rastah Bar
Account closed
Registered: Oct 2012
Posts: 336 |
Quote: Color Bar, that is indeed a fast way to get a sorted list of Y values, but what is needed is a sorted list of the indices of the sprites that have those Y positions.
(eg if sprites 1, 2, 3 and 4 are at positions 10, 46, 73 and 32, the required output is 1,4,2,3, not 10,32,46,73 )
Oh, I see, thanks.
The fact that multiple sprites can have the same Y values makes the problem considerably harder, it seems. |
| | Perplex
Registered: Feb 2009
Posts: 255 |
Here is some old sort code I have used, I think it's originally from Armalyte, or at least based on that code (by Gary Liddon and Dan Phillips.)
It uses the stack for buckets, so lower your stack pointer to somewhere below 50 (or whatever is your lowest Y-value for sorting), then initialize the stack area above that with $ff values.
NOBJECTS = 24 ; Number of objects to be sorted
OBJ_YPOS = $80 ; Y-values of all objects
OBJ_SORT = $A0 ; Indices of all objects sorted by Y-value
TEMP_SP = $FF ; Temporary stack pointer
StackSort:
; Save stack pointer
tsx
stx TEMP_SP
; Phase 1 - Store Y-values in buckets
.repeat NOBJECTS, i
lda OBJ_YPOS + i
;lsr
tax
txs
pla
bpl *-1 ; Already value here, try next
lda #i
pha
.endrep
; Phase 2 - Gather (indices of) sorted values
;ldx #(50/2)-1
ldx #50-1
txs
.repeat NOBJECTS, i
pla
bmi *-1
sta OBJ_SORT + i
lda #$ff ; Clean up after ourselves
pha
pla
.endrep
; Restore stack pointer
ldx TEMP_SP
txs
rts
The two commented out lines are for speeding up the gathering phase by halving each Y-value. Depending on how your multiplexer works, you may not need exact order for very close objects if you handle those at the same time.
In circumstances where limiting the amount of available stack is not viable, it can easily be modified to use some other memory area at the cost of a few cycles per object. |
| | The Human Code Machine
Registered: Sep 2005
Posts: 112 |
While working on Vincenct 2 in the nineties I had the same problem finding a fast solution for sorting sprites and for most purposes this sorter here: http://codebase64.org/doku.php?id=base:flexible_32_sprite_multi..
was the best I came up with. Inspired by Rune Gram-Madsen's Bubble a la Rune sorting routine, published at the Asm-One manual, I wrote a fast implementation for the c64. It's a modified bubble sort working bidirectional and for presorted arrays it's very fast. |
| | Rastah Bar
Account closed
Registered: Oct 2012
Posts: 336 |
Perplex, that is interesting, thanks for sharing the code. Do you know of an efficient modification that ensures exact ordering? |
| | Perplex
Registered: Feb 2009
Posts: 255 |
If you need exact ordering even with many values clustered close together, then simply moving on to the next unoccupied bucket won't really work, so you'd need a quite different approach, I think. |
| | Hein
Registered: Apr 2004
Posts: 942 |
Using bucket sort ($10 size) and insertion sort to sort the indices of a list of y-values. In this case it takes around 40 rasterlines for 32 sprites when placed on zp. It assumes there's no need for more than 8 sprites per bucket, but 15 are possible.
The multiplexer checks per bucket based on the counter at the end of each bucket.
*=$10
clear_bucket_counter
lda #0
sta buckets+15
sta buckets+15+16*1
sta buckets+15+16*2
sta buckets+15+16*3
sta buckets+15+16*4
sta buckets+15+16*5
sta buckets+15+16*6
sta buckets+15+16*7
sta buckets+15+16*8
sta buckets+15+16*9
sta buckets+15+16*10
sta buckets+15+16*11
sta buckets+15+16*12
sta buckets+15+16*13
sta buckets+15+16*14
sta buckets+15+16*15
ldx #32
set_element_index
stx element_index
lda sprite_y,x
ldx #$f0
sbx #$00 ;select right bucket
ldy buckets+15,x ;get bucket counter (# elements in bucket)
inc buckets+15,x ;increase bucket counter
stx set_2+1 ;set high nybble bucket index
dey
bmi place_current_element ;element count=0 skip sort
stx get_1+1
stx get_2+1
inx
stx set_1+1
sty slot_hi+1 ;highest index
compare_element_y
get_1 ldx buckets,y
cmp sprite_y,x
dey
bcc insert_lower_element ;woohoo, found a spot
bpl compare_element_y
insert_lower_element
sty slot_lo+1
slot_hi ldy #0
move_higher_elements
get_2 lda buckets,y ;move higher elements up
set_1 sta buckets,y
dey
slot_lo cpy #$ff
bcc move_higher_elements
place_current_element ;finally, insert the lower element index in the right place
iny
lax element_index
set_2 sta buckets,y
dex
bne set_element_index
rts
element_index
.byte 0
*=$1000
buckets
.fill 256,0
|
| | Fresh
Registered: Jan 2005
Posts: 101 |
I use a slightly modified version of "Ocean" sorter, with a (somewhat) optimized backward sorting routine.
Its worst case is significantly better than plain Ocean.
Code in kickass syntax.
.const NSPRITES = 32
.const indexes = $60
.const ypos = $80
// SORT FORWARD
// ************
// Check if values are sorted, if necessary call sortbw routine to
// put an unsorted value in the correct position
sortfw:
.for(var i=0;i<NSPRITES;i++) {
over:
.if (i!=NSPRITES-1) {
ldy indexes+1+i
ldx indexes+i
lda ypos,y
cmp ypos,x
bcs sortfw[i+1].over
jsr sortbw[NSPRITES-2-i].back
}
else rts
}
// SORT BACKWARD
// *************
// Move backward a value in .A which is less than the one in current position.
// All values at the left of the current position are ordered so, once we found
// the correct spot for .A, we don't need to check forward, we can get back
// to where the function was called
sortbw:
.for(var i=0;i<NSPRITES-1;i++) {
back:
stx indexes+NSPRITES-1-i
.if (i!=NSPRITES-2) {
ldx indexes+NSPRITES-3-i
cmp ypos,x
bcc sortbw[i+1].back
}
sty indexes+NSPRITES-2-i
rts
}
|
| | Rastah Bar
Account closed
Registered: Oct 2012
Posts: 336 |
Quote: Color Bar, that is indeed a fast way to get a sorted list of Y values, but what is needed is a sorted list of the indices of the sprites that have those Y positions.
(eg if sprites 1, 2, 3 and 4 are at positions 10, 46, 73 and 32, the required output is 1,4,2,3, not 10,32,46,73 )
I think that can be done in less than 37 rasterlines.
Consider the following unrolled code for outputting sprite indices, sorted according to Y-value (phase 2)
.bucketK
bcc .nextBucket ;Branches always: decrease this value by 3 with selfmodifying code in phase 1 for every new sprite in the bucket.
lda #sprite_index ;index of last sprite
pha
...
lda #sprite_index ;index of first sprite.
pha
sty .bucketK+1 ;Restore branch (to skip all sprites for empty buckets). Call with y=$1b for max 8 sprites per bucket.
.nextBucket ;unrolled
-------------
3 cycles for an empty bucket, 12 voor buckets with 1 sprite => 3*(220-32)+12*32 = 948 cycli
Phase1: The following unrolled code jumps to the code that modifies the above phase 2 code for the right bucket:
dex ;counts sprite index. Initialized with $20.
bmi .phase2
lda Yvalues,x
asl
bcs *+8 ;There are more than 128 buckets, so I use 2 jump tables
sta *+4
jmp ($xxxx) ;jump table for buckets <128 ;jumps to code that fills out phase2 Bucket code (see below)
sta *+4
jmp ($xxxx) ;jump table for buckets >=128 ;21/22 cycles sofar
.phase2
clc ;to branch always (either to next bucket or to fetch the sprite indices in the current bucket).
ldy #27 ;to restore the branch value of the bucket
jmp .Bucket0
;code to fill out bucketK
ldy .bucketK+1 ;The trick is that I use this as an index for the number of sprites already in the bucket
lda newBranchValue,y ;fetch value to branch to the code for one more sprite
sta .bucketK+1 ;since this is unrolled code, the address of .bucketK is known
tay
txa
sta .bucketK+3,y ;Store new sprite index. This works since the branch value decreases by 3 for each new sprite, but the address where to store the sprite index as well!
... ;Unrolled => repeat all the above code starting with dex.
--------------
41/42 cycles ->41.5*32 = 1328
In total 1328+948 =2276 -> 36.1 lines.
But there will be at least 25 additional cycles from page boundary crossings. Slightly less than 37 lines I think if I did not overlook something.
Table with branch values (for max 8 sprites per bucket) looks like (x=don't care)
newBranchValue
BYTE 0 x x 0 x x 3 x x 6 x x 9 x x 12 x x 15 x x 18 x x x x x 21 x
The first zero makes that a full bucket replaces the last sprite with the new one. |
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