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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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2017-01-04 20:35

Oswald

Registered: Apr 2002
Posts: 5017

axis' one is close to max possible I guess, go for it if understandable version is not enough :)

2017-01-04 20:42

Skate

Registered: Jul 2003
Posts: 490

Table lookup doesn't necessarily be faster than bit shifting. What i was saying is even using large tables to get rid of bit shifting doesn't save cycles unless you start thinking out of the box.

2017-01-04 20:56

Skate

Registered: Jul 2003
Posts: 490

here is an algo which looks similar to c2p method without 8 bit restrictions.

void transpose8(unsigned char A[8], int m, int n, 
                unsigned char B[8]) {
   unsigned x, y, t; 

   // Load the array and pack it into x and y. 

   x = (A[0]<<24)   | (A[m]<<16)   | (A[2*m]<<8) | A[3*m]; 
   y = (A[4*m]<<24) | (A[5*m]<<16) | (A[6*m]<<8) | A[7*m]; 

   t = (x ^ (x >> 7)) & 0x00AA00AA;  x = x ^ t ^ (t << 7); 
   t = (y ^ (y >> 7)) & 0x00AA00AA;  y = y ^ t ^ (t << 7); 

   t = (x ^ (x >>14)) & 0x0000CCCC;  x = x ^ t ^ (t <<14); 
   t = (y ^ (y >>14)) & 0x0000CCCC;  y = y ^ t ^ (t <<14); 

   t = (x & 0xF0F0F0F0) | ((y >> 4) & 0x0F0F0F0F); 
   y = ((x << 4) & 0xF0F0F0F0) | (y & 0x0F0F0F0F); 
   x = t; 

   B[0]=x>>24;    B[n]=x>>16;    B[2*n]=x>>8;  B[3*n]=x; 
   B[4*n]=y>>24;  B[5*n]=y>>16;  B[6*n]=y>>8;  B[7*n]=y; 
}

2017-01-05 07:37

HCL

Registered: Feb 2003
Posts: 716

Ah, that's just beautiful Axis!!

..as usual, i'm the last one to read the thread.. but if anyone still wonders what that code does, it is mirroring the 8x8-matrix by starting with 2x2-bit-blocks, then 4x4-blocks, then finally the whole 8x8-block. Mirroring and rotation is of course just a matter of configuring the macros :).

..and of course Graham did this 15+ years ago, hope that makes you all feel as retard as i do ;).

2017-01-05 08:50

Pex Mahoney Tufvesson

Registered: Sep 2003
Posts: 50

I used Rudi's 472-cycle approach "This takes ~472 cycles, and is basically similar to Oswalds approach" in Storebror
There was no need to make it any faster, since it was a one-time calculation just before the android starts rotating. My routine included not only rotation, but mirroring across horizontal, vertical and diagonals as well.

And HCL... I'm even later than you are. So don't worry, you're ok! ;)
---
Have a noise night!
http://mahoney.c64.org

2017-01-05 10:19

CreaMD

Registered: Dec 2001
Posts: 3034

I wonder if this would significantly speed up my proportional routine and help me "posthumously" win the internal compo I had with Wotnau ;-)

So which one is the fastest? I need to rotate full line of bitmap 320 bytes char by char, +90 degrees each

2017-01-05 11:01

Dano

Registered: Jul 2004
Posts: 226

Haha, did my a proportial Routine for NewsPress back then the same way as CreaMD did. Switched to a standard one some time later as this was even faster. Guess i did the rotation the Oswald-way too back then.

2017-01-05 11:39

Oswald

Registered: Apr 2002
Posts: 5017

don think you can beat a bitshifter with this c2p, couple hundred cycles even with axis method, while bitshifter worst case is about 170 (lsr a ror zp * 8 lines * 3 times)

2017-01-05 12:52

Rudi
Account closed

Registered: May 2010
Posts: 125

Im trying to figure my head around this:
Since the rotation is just a bunch of bitswaps. For example swapping bit 0 and 7, bit 1 and 15, 2 and 23 etc.

00 01 02 03 04 05 06 07
08 09 10 11 12 13 14 15
etc..
56 57 58 59 60 61 62 63.

How would a swapbit-table look for this? The number of swaps are only 32. only 32 lda's and sta's? lets say ~8 cycles in total per swap: 8*32 = 256 cycles for all the swaps. but still I dont know if its possible to have a table for this..
Edit: hm, woops. i think perhaps 32 is wrong, since both locations need to be written to, so maybe two sta's instead of just one.

2017-01-05 13:11

Cruzer

Registered: Dec 2001
Posts: 1048

Oswald: Why 3 times and not 8?