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Forums > C64 Coding > Fast way to rotate a char?

2017-01-04 08:32

Rudi
Account closed

Registered: May 2010
Posts: 125

Fast way to rotate a char?

Im not talking about rol or ror, but swap bits so that they are rotated 90 degrees:

Example:

a char (and the bits can be random):

10110010 byte 1..
11010110 byte 2.. etc..
00111001
01010110
11011010
10110101
00110011
10110100

after "rotation" (rows and columns are swapped):

is it possible to use lookup tables for this or would that lookup table be too big?
or other lookuptable for getting and setting bits?

-Rudi

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2017-01-07 15:51

Rudi
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Registered: May 2010
Posts: 125

Krill: No. I ONLY want to swap x and y 90 degrees. No other degrees. Im working on Axis's method with masks and so on.. (bcoz it seems to be the one technique that use fewest cycles).

2017-01-07 19:58

Skate

Registered: Jul 2003
Posts: 494

I think Rudi wants to rotate dynamically created chars, not a static charset. Rudi, did i get you right?

2017-01-07 21:06

Frantic

Registered: Mar 2003
Posts: 1647

I think so, yes, as he wrote that "the bits can be random". (But that doesn't exclude the possibility of playing with other types of representations as such.)

2017-01-07 21:10

Rudi
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Registered: May 2010
Posts: 125

Quote: I think Rudi wants to rotate dynamically created chars, not a static charset. Rudi, did i get you right?

Yes Skate! That is right.

2017-01-08 10:53

Rastah Bar
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Registered: Oct 2012
Posts: 336

I may have found a method that takes 432 cycles. An 8x8 rotation can be performed by moving around 4x4 blocks, then 2x2 subblocks and then the bits in each subblock. Now you can identify for each resulting destination bit pair (that is, bist 7+6, or 5+4, ..., or 1+0) the nybbles of the source bytes it came from. So an approach could be to take two source bytes, merge the desired nybbles into one byte, and then use look up tables to transform it to the desired bit pairs (at the right location, i.e., a bit pair 7+6 from the merged source nybbles could be transformed to a destination bit pair 3+2, for example).

ldx first_source_byte
lda second_source_byte
and #mask   ;mask=f0 if you need the high nybble and 0f for the low nybble
ora byte2DesiredNybble,x  ;look up table to merge the right nybble in the right way
tax
lda first_destination
ora bits76toRightPosition,x  ; Table filters out the desired bits and puts them in the right order at the right position.
sta first_destination
lda second_destination
ora bits54toRightPosition,x
sta second_destination
...
lda fourth_destination
ora bits10toRightPosition,x
sta fourth_destination
-----------------------+
54 cycles (source and destination bytes in ZP).

This fills in 4 times 2 bits, so you have to do that 8 times to fill the entire destination matrix. So that is 432 cycles in total. But you need a bunch of look-up tables, so it is not very memory efficient.

I hope I did not overlook something ...

2017-01-08 11:38

Krill

Registered: Apr 2002
Posts: 2969

While 8x8 tile transpose is an entertaining academic exercise, i'm dubious of its practical applications for C-64 purposes.

In fact, i'd be hard-pressed to find an effect that is rendered faster with generating "horizontal" 8x8 tiles first and then rotating them, as opposed to rendering "vertical" 8x8 tiles right away.

Also, an actual chunky to char approach seems more feasible. Render to linear chunky bitmap (1 byte per pixel), then collect the bits and render final 8x8 tiles.

2017-01-08 12:30

Rudi
Account closed

Registered: May 2010
Posts: 125

Krill: Bear in mind im not too an experienced coder on C64 that I see solutions right away. I figured out this might be an nice intellectual exercise for myself, hopefully for others too(?) but wether or not it has any practical usage, that I have no idea..

Btw, I have a working 336 cycles. It consists of twelve "copies" of this code (pseudocode):

lda $<zp1>
tay
and #<mask1>
ldx $<zp2>
eor $<tab1>, x
sta $<zp3>
txa
and #<mask2>
eor $<tab2>, y
sta $<zp4>

2017-01-08 13:22

Axis/Oxyron
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Registered: Apr 2007
Posts: 91

Rudi, except of the fact that you used EOR instead of ORA and didnt use LAX for the first read on zp (would be possible if you swap x and y registers), this looks identical to the code I posted pretty early in this thread. Is there a special reason to use EOR?

2017-01-08 13:37

Rudi
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Registered: May 2010
Posts: 125

Quote: Rudi, except of the fact that you used EOR instead of ORA and didnt use LAX for the first read on zp (would be possible if you swap x and y registers), this looks identical to the code I posted pretty early in this thread. Is there a special reason to use EOR?

The eor was a consequence of the formulas i used for masking and swapping (I derived this from Kalms tutor):

Example for the 4x4 swapping:

tmp0 = byte0 & 0xf0; //xxxx----
tmp1 = byte1 & 0xf0; //xxxx----
tmp2 = byte2 & 0xf0; //xxxx----
tmp3 = byte3 & 0xf0; //xxxx----
tmp4 = byte4 & 0x0f; //----xxxx
tmp5 = byte5 & 0x0f; //----xxxx
tmp6 = byte6 & 0x0f; //----xxxx
tmp7 = byte7 & 0x0f; //----xxxx
data0 = byte0 << 4;
data1 = byte1 << 4;
data2 = byte2 << 4;
data3 = byte3 << 4;
data4 = byte4 >> 4;
data5 = byte5 >> 4;
data6 = byte6 >> 4;
data7 = byte7 >> 4;
data0 ^= tmp4;
data1 ^= tmp5;
data2 ^= tmp6;
data3 ^= tmp7;
data4 ^= tmp0;
data5 ^= tmp1;
data6 ^= tmp2;
data7 ^= tmp3;

Sorry for the long code..

EOR is used for the last xor-swapping. Since I cannot use lookup-table for two different values (fex. data0 and tmp4). The last eight operations in the above are done with the EOR-instruction.

Some of the lookup-tables i derived are doing EOR, AND and SHIFTS at the same time:

shl2_eor_cc[i] = (i ^ (i & 0xcc)) << 2;
shr2_eor_33[i] = (i ^ (i & 0x33)) >> 2;
shl1_eor_aa[i] = (i ^ (i & 0xaa)) << 1;
shr1_eor_55[i] = (i ^ (i & 0x55)) >> 1;

I scratched my head around how your ORA worked. And since I didnt understand that Dreamass-macrocode I wrote mine from scratch. But maybe I should look at your LAX-method next.

Edit: Now I see that it doesnt really matter if one use ora or eor for this technique.

2017-01-08 13:46

Axis/Oxyron
Account closed

Registered: Apr 2007
Posts: 91

But you know that:
(i ^ (i & 0xcc))

is the same as:
i & 0x33

;o)