> >> As we only need 6 bits of every byte, we can overlap the cards in
> >> memory, every four cards in three bytes:
> >>
> >> AAAAAABB BBBBCCCC CCDDDDDD EEEEEEFF FFFFGGGG GGHHHHHH
>
> >Don't forget that someone's also gonna have to *draw* a screen with all
> >these cards. I doubt you'd have time to extract card values out of that
> >packing during a screen kernel, and you can forget about using Andrew
> >Davie's packing method for that :) I'd use one byte per card to make that
> >a good deal simpler.
>
> I take it doing by nibbles isn't that bad though?
That's what I'd guess. All you have to do to get a nibble value out of a
byte is AND it with $0F or $F0. To get values out of your packing scheme,
you need to read from two memory locations and do some tricky shifting and
stuff to combine the fragments together.
> >Actually, you need but _one_ bit, not two, for each card after the first.
> >After the first card, you know what color each card below will be, so you
> >only need one bit to distinguish suit. E.g., if the top card is the queen
> >of spades, you only need one bit to tell whether the next one is the jack
> >of hearts or the jack of diamonds.
>
> Ah! Did I miss that, or are you refining?
I was refining.. didn't notice that till now.
> >Of course, you need some counter to keep track of how many cards in the
> >stack (your terminator won't work anymore.) Still, though, that's 6 bits
> >for the top card plus 4 for the counter plus 12 for the other cards;
> >smaller than your 6 bits for the top card plus 2 each for 12 other cards.
>
> True. Of course, to avoid packing & unpacking time, we'll just use one
> byte for the top card, and exactly two bytes for the up stack. I assume
> we'd start those two bytes with the counter itself?
You could probably do that - 12 bits for the stack and 4 for the counter
should compress neatly into two bytes.
> Well, actually, I had the (uncompressed) ace piles at 4 bytes, but if
> we use one byte to store the order of suits (you'd have to juggle a
> terminator state, starting with/alternating 00000000 and 11111111 states
> as you added), then you'd only need two bytes (4 nibbles) for the top
> card values, saving one more byte.
What I'd do is just hardcode the ace piles into a fixed order.. that
solves everything there.
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