Array splitted in 4 and used in 4 cogs.

darkxceed · 2008-09-02 17:15

Hello,

I have made a program that calculates alot.
I only used one cog for it but I want it to be faster, so why not use more cogs.

The calculations must me splitted by 4 and send to 4 different cogs.

The problem now is that I want the array that is defined in the main program to be used in the for cogs.
ie.
VAR
long·· B[noparse][[/noparse]512]······· 'this array has to be used by 4 cogs

The above declaration will be used by the 4 cogs so cog#1 has to know [noparse][[/noparse]1-128] to do that part of the calculation
Cog#2 has to get [noparse][[/noparse]129-256] to do it's calculation etc...

Only thing I found how to do it is with @B to get the start of B adres but don't know how to get further on.

And how do I pol to know that all of the cogs are finished with the calculations

Bart

·

Paul Baker · 2008-09-02 17:19

Are you doing this in spin or assembly?

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Paul Baker
Propeller Applications Engineer

Parallax, Inc.

Ken Peterson · 2008-09-02 17:49

Darkxceed: Pass your routine the address of the first element and the length of segment to be calculated. Like the following:


ptr1 := @B
ptr2 := @B + 128 * 4 'long
ptr3 := @B + 256 * 4
ptr4 := @B + 384 * 4

cognew(calculate(ptr1, 128 * 4), @stack1)
cognew(calculate(ptr2, 128 * 4), @stack2)
cognew(calculate(ptr3, 128 * 4), @stack3)
cognew(calculate(ptr4, 128 * 4), #stack4)

pub calculate(ptr, count) | value
repeat count
  value := LONG[noparse][[/noparse]ptr]
  <calculate here>
  ptr += 4

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·"I have always wished that my computer would be as easy to use as my telephone.· My wish has come true.· I no longer know how to use my telephone."

- Bjarne Stroustrup

Post Edited (Ken Peterson) : 9/2/2008 5:57:01 PM GMT

darkxceed · 2008-09-02 17:50

I'm doing this in spin.

Ken Peterson · 2008-09-02 17:53

You can speed up your calculation drastically if you do it in PASM. Granted, the coding is not as easy. Is the calculation you are doing complicated?

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·"I have always wished that my computer would be as easy to use as my telephone.· My wish has come true.· I no longer know how to use my telephone."

- Bjarne Stroustrup

Phil Pilgrim (PhiPi) · 2008-09-02 17:53

Ken,

It's an array of longs, but addresses are in bytes, so you want:

ptr1 := @B
ptr2 := @B + 512
ptr3 := @B + 1024
ptr4 := @B + 1536

-Phil

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'Still some PropSTICK Kit bare PCBs left!

Ken Peterson · 2008-09-02 17:56

Phil,

I probably posted before I corrected it for longs. Should be correct now.

My example shows hard-coded numbers, but this is of course not the best way to design a routine.· The different indexes should be calculated based on the size of the array and the number of cogs doing the calculation.

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·"I have always wished that my computer would be as easy to use as my telephone.· My wish has come true.· I no longer know how to use my telephone."

- Bjarne Stroustrup

darkxceed · 2008-09-02 17:57

Ken Peterson said...
Darkxceed: Pass your routine the address of the first element and the length of segment to be calculated. Like the following:


ptr1 := @B
ptr2 := @B + 128 * 4 'long
ptr3 := @B + 256 * 4
ptr4 := @B + 384 * 4

cognew(calculate(ptr1, 128 * 4), @stack1)
cognew(calculate(ptr2, 128 * 4), @stack2)
cognew(calculate(ptr3, 128 * 4), @stack3)
cognew(calculate(ptr4, 128 * 4), #stack4)

pub calculate(ptr, count) | value
repeat count
  value := LONG[noparse][[/noparse]ptr]
  <calculate here>
  ptr += 4

Hi Ken,

Thanks for the reply, I will try it now, but do you know how big the stack array has to? 9 long's or?
Can not find it in the manuel how big have to define the stack for a cog.

Ken Peterson · 2008-09-02 18:01

darkxceed: somebody posted some code a while back on the forum that would measure stack use. Can't recall who posted it. Another approach is to start with a large stack (32 or 64) and then keep paring it down until the program no longer works, then add back some padding for good measure. How much you need will probably depend on what calculations you are doing.· If your calculation is recursive then you might need a lot.

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·"I have always wished that my computer would be as easy to use as my telephone.· My wish has come true.· I no longer know how to use my telephone."

- Bjarne Stroustrup

darkxceed · 2008-09-02 18:04

Ken Peterson said...
You can speed up your calculation drastically if you do it in PASM. Granted, the coding is not as easy. Is the calculation you are doing complicated?

My calculations are not that complicated but the prop·has to do it 512 times.

And after it another 512 times with another equation.

In the future I will try PASM, but the prop with it·8 cogs is fast enough in spin.
·

darkxceed · 2008-09-02 18:49

·Hi Ken,

The code below is what I·tried.

But doesn't work some how.

I used 3 extra cogs plus the one I of the main loop.

This way I know that after the 3 cogs are called to calculate, the "main" cog wil perform the last calculation and knows for sure that the other cogs are finished.

Maybe I have to make 3 methodes BufCalc1, BufCalc2 and BufCalc3 and Wavecalc1 ..2 ..3?

Pub Start

repeat

· cognew(BufCalc(ptr1, 512), @Stack1)
· cognew(BufCalc(ptr2, 512), @Stack2)
· cognew(BufCalc(ptr3, 512), @Stack3)

· repeat T from 384 to 512
··· Buffer[noparse]/noparse]T]:=B[noparse][[/noparse]Xp[noparse][[/noparse]T-B[noparse]/noparse]T]+B[noparse][[/noparse]Xn[noparse][[/noparse]T··· 'after this repeat loop is finished, other 3 cogs are finished for sure

· cognew(WaveCalc(ptr1, 512), @Stack1)
· cognew(WaveCalc(ptr2, 512), @Stack2)
· cognew(WaveCalc(ptr3, 512), @Stack3)
·
· repeat T from 384 to 512················· 'again after this is finished the previous cogs are also finished

···· B[noparse][[/noparse]T] := B[noparse][[/noparse]T]+A[noparse][[/noparse]T]/C

pub BufCalc(ptr ,count ) | value
repeat count
· value := LONG[noparse][[/noparse]ptr]
· Buffer[noparse]/noparse]T]:=B[noparse][[/noparse]Xp[noparse][[/noparse]T-B[noparse]/noparse]T]+B[noparse][[/noparse]Xn[noparse][[/noparse]T
· ptr += 4

pub WaveCalc(ptr ,count) | value
repeat count
· value := LONG[noparse][[/noparse]ptr]
···· B[noparse][[/noparse]T] := B[noparse][[/noparse]T]+A[noparse][[/noparse]T]/C
· ptr += 4

Paul Baker · 2008-09-02 19:54

At least one reason you code doesn't work is that hub memory pointers are aligned to BYTE, what this means is that to access sucessive longs you increment the pointer by 4. And since each cog is processing every 4th element, each pointer should be incremented by 16 instead of 4 if I understand what you are trying to do.

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Paul Baker
Propeller Applications Engineer

Parallax, Inc.

Ken Peterson · 2008-09-02 20:56

In each loop I had the line "value := LONG[noparse][[/noparse]ptr]". This is how you dereference the pointer to obtain the long value at that location. It doesn't appear that you're using that value in your calculation but instead trying to index it with T. Also, you are using more than one array. That would require more pointers.

For example, your WaveCalc calculation might be

pub WaveCalc(b_ptr, a_ptr ,count)
repeat count
  LONG[noparse][[/noparse]b_ptr] := LONG[noparse][[/noparse]b_ptr]+LONG[noparse][[/noparse]a_ptr]/C
  a_ptr += 4
  b_ptr += 4

Paul: I imagined that the array is divided into quarters, rather than each cog doing every forth value. Ptr1, ptr2, ptr3 and ptr4 each point to 1/4 of the array.

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·"I have always wished that my computer would be as easy to use as my telephone.· My wish has come true.· I no longer know how to use my telephone."

- Bjarne Stroustrup

Ken Peterson · 2008-09-02 21:02

You know...I keep thinking PASM when it comes to launching cogs. You only really need pointers for PASM.· Spin always works on global VAR space so you can use the same indexed arrays in all cogs.

Pub Start

repeat

  cognew(BufCalc(0, 128), @Stack1) 
  cognew(BufCalc(128, 128), @Stack2)
  cognew(BufCalc(256, 128), @Stack3)
  BufCalc(384, 128)

  cognew(WaveCalc(0, 128), @Stack1)  
  cognew(WaveCalc(128, 128), @Stack2)
  cognew(WaveCalc(256, 128), @Stack3)
  WaveCalc(384, 128)

pub BufCalc(idx ,count )
repeat count
  Buffer[noparse][[/noparse]idx]:=B[noparse][[/noparse]Xp[noparse][[/noparse]idx]]-B[noparse][[/noparse]idx]+B[noparse][[/noparse]Xn[noparse][[/noparse]idx]]
  idx++
 

pub WaveCalc(idx ,count)
repeat count
  B[noparse][[/noparse]idx] := B[noparse][[/noparse]idx]+A[noparse][[/noparse]idx]/C
  idx++

How's this?

Don't forget, there's latency when starting a cog.· That third cog might not be going by the time the local function call starts so you can't necessarily assume it's done when the local call is done.

Another idea that might gain a smidg of extra speed is to just have each cog repeat internally on its own quarter of the array instead of re-launching cogs every time through your main loop.

I noticed another thing:· Your BufCalc function may or may not access elements of B that are outside of it's 1/4 array domain.· I see you are using·other values·(Xp & Xn) to index B as well, and these may be outside of the index domain for that cog.· This could complicate things if you are changing more than one value in B at a time.

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·"I have always wished that my computer would be as easy to use as my telephone.· My wish has come true.· I no longer know how to use my telephone."

- Bjarne Stroustrup

Post Edited (Ken Peterson) : 9/2/2008 9:21:17 PM GMT

darkxceed · 2008-09-03 17:17

Ken,

I tried it but no succes.

Can you tell me how big the stack has to be?

And where does the size depends on.

Timmoore · 2008-09-03 17:29

To start with just increase the stack for each cog to ~200 longs, worry about actual needed size once its working. There is an object in object exchange to help find the stack size needed.
One problem I see is the 3 cogs are running in parallel the main cog does not know when they have finished, your assumption is that if you also run bufCal on the main cg, the others should be finished before the main cog but dont forget starting a cog takes some time so you dont know when they are finished.
Try adding a waitcnt(clkfreq+cnt) between each cognew, that should make them all run one after the other. If that works remove each waitcnt, one at a time from the first, I would expect you need a delay after the main call to BufCal and WaveCall, though you shold be able to reduce it from 1 sec.
The other thing is from the code its not obvious what you think is wrong, are you using Buffer and its not got the right answer?

darkxceed · 2008-09-03 17:36

Hi Ken,

It works now I made a fault.

As I see in the example the array's are easy share by the cogs.
I thought that it had to be done with pointers or something like that.
But thats more assembly which I don't understand... yet.

My question still stands for the stack size.

·

darkxceed · 2008-09-03 17:49

Timmoore said...
To start with just increase the stack for each cog to ~200 longs, worry about actual needed size once its working. There is an object in object exchange to help find the stack size needed.
One problem I see is the 3 cogs are running in parallel the main cog does not know when they have finished, your assumption is that if you also run bufCal on the main cg, the others should be finished before the main cog but dont forget starting a cog takes some time so you dont know when they are finished.
Try adding a waitcnt(clkfreq+cnt) between each cognew, that should make them all run one after the other. If that works remove each waitcnt, one at a time from the first, I would expect you need a delay after the main call to BufCal and WaveCall, though you shold be able to reduce it from 1 sec.
The other thing is from the code its not obvious what you think is wrong, are you using Buffer and its not got the right answer?

Hi Tim,

It doesn't matter when one of the BufCalc is finished.
This also counts for the WaveCalc.

I will add an byte int, "sync".

After each bufCalc is finished it will add 1.
If the "sync"·equals 3 then the WaveCalc's procedure's may run they also have.

My concerns are maybe the·array's, they will be accessed by 4 cogs at the same time.
As Tim mentioned I believe.

·

Array splitted in 4 and used in 4 cogs.

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