Scott Dattalo wrote:
...
>> ab-cde+fgh
>> a-2*bcd+efgh
>> -abcd+efgh

> Actually, the last one should be
>  -2*abcd + efgh

Yes, thanks for noticing.

> This is the one I was thinking that would be most optimal. That's why I
> said hint 16 = 18 - 2.

> Another way to look at this is to write the number explicitly:

>   abcdefgh = (abcd)*16^1 + (efgh)*16^0

> If you want to know if this is divisible by 9, then:

>  abcdefgh % 9 = (abcd)*16 % 9  + efgh % 9
>               = (abcd)*(-2) % 9 + efgh % 9

> because, 16 is congruent to -2 mod 9. It's also congruent to 7, so we
> could write:

>  abcdefgh % 9 = abcd*7 % 9 + efgh % 9

> if you know 'a' is 0, then you can determine if the number is divisible in
> 11 instructions using this trick:

> movwf temp
> swapf temp,w
> subwf temp,f
> skpdc
> addlw 7
> subwf temp,w
> skpdc
> addlw 9
> andlw 0x0f
> skpz
> addlw -9

This calculates -2*abcd + efgh, doesn't it? I can't see a
multiplication by 7. But this is a neat idea! Using the DC flag for
keeping the sum in the lower 4 bits. Extending it further:

> movwf temp
> swapf temp,w
> subwf temp,f
> skpdc
;> addlw 7 - comment this out
   addlw -2        ;equivalent to adding 7, but gives us a
                   ;convenient DC flag polarity, as after subtraction
  skpdc            ;if overflowed again, subtract 16 % 9 = 7 or
   addlw 2         ;add 2, which is the same in that case.

> subwf temp,w
> skpdc
> addlw 9
> andlw 0x0f
> skpz
> addlw -9

That's 1 instruction too much though.

 Nikolai

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