Compact /10 and %10 operations?

[pubby]

I just use 100 byte lookup tables.

[Bregalad]

You'll only get automatic wrap around detection in one direction anyway (either when adding or subtracting).

You won't - automatic overflow detection would work only when using 0-9 or when using 246-255; the second case being less intuitive but not more optimal than the first.

[Jarhmander]

I see optimality in having numbers in tiles 246-255, at least for things like score points: you rarely substract points, if ever, in a game. From the look of it, you genuinely optimized your tile for scorekeeping. I find it actually clever.

[qalle]

I just use 100 byte lookup tables.

This division routine with remainder is quite fast and needs 100 bytes' worth of tables. (I didn't test it extensively.)

Code: Select all

; Start with 0-99 in A.
; Get the quotient and the remainder of A divided by 10.
; Needs 9 bytes and 14 cycles per operation and 100 bytes for the tables.

lsr  ; keep LSB in carry
tax
lda unsigned_divide_by_5,x
; now A = the quotient
lda unsigned_modulo_5,x
rol  ; get LSB from carry
; now A = the remainder

unsigned_divide_by_5:
    .byte 0, 0, 0, 0, 0, 1, 1, 1, 1, 1
    .byte 2, 2, 2, 2, 2, 3, 3, 3, 3, 3
    .byte 4, 4, 4, 4, 4, 5, 5, 5, 5, 5
    .byte 6, 6, 6, 6, 6, 7, 7, 7, 7, 7
    .byte 8, 8, 8, 8, 8, 9, 9, 9, 9, 9

unsigned_modulo_5:
    .byte 0, 1, 2, 3, 4, 0, 1, 2, 3, 4
    .byte 0, 1, 2, 3, 4, 0, 1, 2, 3, 4
    .byte 0, 1, 2, 3, 4, 0, 1, 2, 3, 4
    .byte 0, 1, 2, 3, 4, 0, 1, 2, 3, 4
    .byte 0, 1, 2, 3, 4, 0, 1, 2, 3, 4

Edit: optimized the modulo routine.
Edit: combined the routines.

[calima]

Your division table is 50 bytes long. What happens when x >= 50?

[tokumaru]

Your division table is 50 bytes long. What happens when x >= 50?

It won't ever be >= 50, because of the LSR. If the largest number is 99, X will be at most 49. By getting rid of the least significant bit (and putting it back later) the tables can be made half the size.

[Zutano]

I just use 100 byte lookup tables.

This division routine with remainder is quite fast and needs 100 bytes' worth of tables. (I didn't test it extensively.)

Code: Select all

; Start with 0-99 in A.
; Get the quotient and the remainder of A divided by 10.
; Needs 9 bytes and 14 cycles per operation and 100 bytes for the tables.

lsr  ; keep LSB in carry
tax
lda unsigned_divide_by_5,x
; now A = the quotient
lda unsigned_modulo_5,x
rol  ; get LSB from carry
; now A = the remainder

unsigned_divide_by_5:
    .byte 0, 0, 0, 0, 0, 1, 1, 1, 1, 1
    .byte 2, 2, 2, 2, 2, 3, 3, 3, 3, 3
    .byte 4, 4, 4, 4, 4, 5, 5, 5, 5, 5
    .byte 6, 6, 6, 6, 6, 7, 7, 7, 7, 7
    .byte 8, 8, 8, 8, 8, 9, 9, 9, 9, 9

unsigned_modulo_5:
    .byte 0, 1, 2, 3, 4, 0, 1, 2, 3, 4
    .byte 0, 1, 2, 3, 4, 0, 1, 2, 3, 4
    .byte 0, 1, 2, 3, 4, 0, 1, 2, 3, 4
    .byte 0, 1, 2, 3, 4, 0, 1, 2, 3, 4
    .byte 0, 1, 2, 3, 4, 0, 1, 2, 3, 4

Edit: optimized the modulo routine.
Edit: combined the routines.

Wow, that's a great solution. The same concept should apply when dividing by any number for which a power-of-two is a factor.
Like, dividing by 12 for example:

Code: Select all

; Value from 0-143 in A.

lsr
rol temp
lsr
tax
lda unsigned_divide_by_3, x
; now A = the quotient
lda unsigned_modulo_3, x
rol
lsr temp
rol
; now A = the remainder

unsigned_divide_by_3:
    .byte 0, 0, 0, 1, 1, 1, 2, 2, 2
    .byte 3, 3, 3, 4, 4, 4, 5, 5, 5
    .byte 6, 6, 6, 7, 7, 7, 8, 8, 8
    .byte 9, 9, 9, 10, 10, 10, 11, 11, 11

unsigned_modulo_3:
    .byte 0, 1, 2, 0, 1, 2, 0, 1, 2
    .byte 0, 1, 2, 0, 1, 2, 0, 1, 2
    .byte 0, 1, 2, 0, 1, 2, 0, 1, 2
    .byte 0, 1, 2, 0, 1, 2, 0, 1, 2

It requires an extra temporary variable (since we initially divide by 4), but it's still a neat concept.

EDIT: Also this website has a ton of math implementations for 6502.

[calima]

Ah indeed, thanks for pointing that out.