Re: SEAForth questions again

I think I can answer one of your questions.  The comments around the
definition of f*f say that it is for multiplying fixed point fractions,
not integers.  These fractions are represented, according to the
comment, as follows:

ss ifff fff ffff ffff

where ss are sign bits (not sure why there are two at this point), i is
an integer, either 0 or 1, and the f's are the bits that represent a
fraction.  For example, the number $8000 would represent 1.0.  $4000
would represent 0.5.  $2000 would be 0.25.  When you multiplied 2 by 5
you were really multiplying a very small fraction by another very small
fraction, and the answer was so small that it was truncated to zero.

I added this:

: asFraction ( num den - fraction) $8000 rot rot */ ;

to help me enter fractions and watch the results of f*f.  This
definition needs to come outside the 'machine' '[' pair, it is not
machineforth.  It is used as follows:

machine
: test-f*f
   [ 1 2 asFraction ]# [ 1 2 asFraction ]# . .
   f*f test-f*f -;

You should see $4000 (0.5) pushed onto the data stack twice, and the
result of f*f is $2000, meaning 0.25.

I assume f*f is only defined in the ROM for the cores with A/D and D/A
is because it will be useful for scaling A/D inputs.

As to how you would multiply the integers in your matrix example, I
don't have an easy answer to that yet, but I'll work on.

Charley.

On Fri, 2006-11-03 at 13:31 -0800, John Drake wrote:
> I've been trying to write my first SEAForth demo.
> To try and exploit the parrallelism I settled on
> matrix multiplication as a test program.  
> I'm trying to multiply a 2 row matrix by a two
> column matrix.  These are the steps I'm attempting.
> 
> 1) Data initially is in node 12
> 2) Row1 is passed to node 18
> 3) Row2 is passed to node 13
> 4) Column1 is passed to nodes 18 and 13 
>    simultaneously
> 5) Multiplications R1*C1 and R2*C1 done 
>    simultaneously
> 6) Column2 is passed to nodes 18 and 13 
>    simultaneously
> 7) Multiplications R1*C2 and R2*C2 done 
>    simultaneously
> 8) Results R1*C1, R1*C2, R2*C1, R2*C2 are 
>    returned to node 12
> 
> I looked at the code runvmram.f as an
> example.  It uses these steps.
> 
> 1) Data initially in node 12
> 2) Data passed through node 13 to node 14
> 3) Node 14 does a difference between the
>    previous and current data value and passes
>    result to node 15
> 4) Node 15 does a "running average" and passes
>    result to node 16
> 5) Node 16 "catches" the data into a buffer
> 
> Here's my first question.  The following code
> is from node 14.
> 
> \ ******* node 14 ********************
> \ differentiator
> \ pass "difference" between current and 
> \ previous to the next node
> 
> decimal
> 14 node !
> 0 org  
> machine
> 
> '--l-  a! . .   \ point reg a to left node
> 'r---  b! . .   \ point reg b to right node
> dup xor         \ init t to zero
> begin  
>   @a over over .  ( curr prev curr )
>   + not ( c c-p )  !b .  ( c ) \ will become p
>   4  drop . .
> again  
> 
> The "+ not" would seem to me to return the
> value -(c+p) rather than (c-p).  Am I missing
> something here?  (This isn't essential to my
> problem, but I'm just trying to understand the
> code.)
> 
> Anyway, here's my own code.  (Note: for the
> sake of brevity I'm only including code for
> nodes 12 and 13 as node 18 is a clone of
> node 13's code.)
> 
> \ ******* node 12 *************************
> \ Pass row1 to node 13, row2 to node 18
> \ and columns 1 and 2 to nodes 13 and 18 
> \ simultaneously
> decimal
> 
> 12 node!  
> $0 org  
> machine
> 
> : send-data
>    begin
>       'r--- b! . .   \ point reg b to right node
>       7 for
>         @p+ !b . unext
>         [  1 ,  2 , 3 , 4 , 5 ,  6 ,  7 ,  8 , ]
>       '-d-- b! . .   \ point reg b to down node
>       7 for
> 	@p+ !b . unext
>         [  9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , ]
>       'rd-- b! 	     \ point reg b to r and d nodes
>       0 !b . . \ dummy write for synch purposes
>       15 for
> 	@p+ !b . unext
>         [ 17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 
>           25 , 26 , 27 , 28 , 29 , 30 , 31 , 32 , ]
>       $20 a! 'r--- b! 
>       @b !a+ @b !a+
>       '-d-- b! . .
>       @b !a+ @b !a+
>    again
> [
> 
> \ ******* node 13 *****************************
> \ Receive row1 then multiply it by columns 1 and 2 
> \ and return results.
> 
> decimal
> 13 node !  
> 0 org  
> machine
> 
> 'r---  b! . .  \ point b to right node
> $20 dup a! push  \ point a to buffer - push adr
> 
> 7 for 
>   @b !a+ . unext
> 
> @b drop . . 	\ dummy read for synchronization purposes
> 
> pop dup a! push	\ reset a to start of buffer 
> dup xor		\ initilize TOS to 0
> 7 for
>   @b @a+ f*f .
>   + next . .
> 
> dup dup xor .	\ preserve result: 0 TOS
> 
> pop a! . .      \ reset a to start of buffer
> 7 for
>   @b @a+ f*f .
>   + next . .
> 
> !b !b . .	\ Send results back to r node
> 
> Now for my next question.  After running this I
> didn't get correct results.  I'm not sure I'm 
> using f*f correctly.  I put the following test
> code in.
> 
> \ ******* node 23 **********************
> \ Test multiplication
> 
> decimal
> 23 node !
> 0 org  
> machine
> 
> 2 5 f*f .
> 
> But the end result was:
> 
> 23
> 0 .
> a=00000
> b=155
> p=005
> r=15555
> t=00000
> s=15555
> 
> That can't be right.  So I'm wondering what I'm
> doing wrong?  Here is a link to my entire sourcefile
> so someone else can look at this.  
> 
> http://www.quartus.net/twiki/pub/Main/VentureForth/matrix.f
> 
> Note that in order to run this you have to change 
> the rombios.f file.  By default only nodes 18 and 
> 23 have the f*f word compiled.  I copied and 
> pasted that to node 13.
> 
> Other than problems with f*f my code seems to
> work correctly.  It's interesting to watch the
> processors activate-deactivate and to see stuff
> running in parrallel.  While this is a trivial
> example, I did have to deal with solving
> deadlock and race conditions.  I think the 
> VentureForth environment is very helpful for
> immediately spotting such things.
> 
> Regards,
> 
> John M. Drake
> 
> 
> 
>  
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