Oops, for size of y.

But the calibration argument still stands.
The same expression needs to get the same
CPB, but clock rate does not reflect that.
Instead it relects CPU architectures, not J expressions.

I hope this is right

   (#y) %~ 1.45e9 * 6!:2 '+./y'
0.00210641


--- Roger Hui <[EMAIL PROTECTED]> wrote:

> Huh?
> 
> a. In this sequence of msgs y has been a Boolean
> vector, in which case you need to divide by
> (#y) rather than (4*#y).  If you are using a
> different y please tell me what it is.
> 
> b. I don't understand how your results show that
> calibration is required.  If anything it shows
> that calibration gives the wrong answers.
> 
> 
> 
> ----- Original Message -----
> From: Oleg Kobchenko <[EMAIL PROTECTED]>
> Date: Friday, December 22, 2006 8:54 am
> Subject: Re: [Jprogramming] Cycles per Byte
> 
> > OK. Next is N]\y,   N<0
> > 
> >   (4*#y) %~ 1.45e9 * 6!:2 '_2]\y'   NB. 2.8 GHz calibrated
> > 0.737346
> > 
> > I specifically reproduced the results for
> > $ and + to show that the clock rate does not
> > reflect time without calibration.
> > 
> > 
> > --- Roger Hui <[EMAIL PROTECTED]> wrote:
> > 
> > > "The first thing we do, let's kill all the lawyers."
> > > -- W. Shakespeare, Henry VI, Part II, Act 4, 
> > >    Scene 2, 71-78.
> > > 
> > > f depends on the value of every byte of its
> > > arguments.  Moreover, functions that merely
> > > increments or decrements reference counts are
> > > excluded, in particular the monads < > ] [ and
> > > the dyads ] [ ; are excluded.
> > > 
> > > You should not have to calibrate the rate.
> > > A higher GHz rating g should have resulted in
> > > a smaller time result from 6!:2 and the product  
> > > g * 6!:2 'blah'   should work out to be the same.
> > > Numbers for   n %~ g * 6!:2 'blah'  on different
> > > machines permit comparisons on how efficient
> > > the machines are.
> > > 
> > > 
> > > ----- Original Message -----
> > > From: Oleg Kobchenko <[EMAIL PROTECTED]>
> > > Date: Thursday, December 21, 2006 9:12 pm
> > > Subject: Re: [Jprogramming] Cycles per Byte
> > > 
> > > > My champion so far is ]y
> > > > 
> > > >   (1*m) %~ 1.485e9 * 6!:2 'm$0'   [ m=: 1e7
> > > > 2.3605
> > > >   (4*#x) %~ 1.45e9 * 6!:2 'x+y'
> > > > 2.68411
> > > >   (4*#x) %~ 1.45e9 * 6!:2 ']y'
> > > > 0.000273429
> > > > 
> > > > Note to callibrate the rate, I have to use
> > > > g=1.485e9 or so on P4 2.8GHz.
> > > > 
> > > > 
> > > > --- Roger Hui <[EMAIL PROTECTED]> wrote:
> > > > 
> > > > > Let f be a function that depends on every byte of
> > > > > its argument(s).  >:y and x+y are examples of
> > > > > such a function, and {.y is not.  The number of 
> > > > > cycles per byte (CPB) for f is 
> > > > >    n %~ g * 6!:2 '  f y'
> > > > >    n %~ g * 6!:2 'x f y'
> > > > > where n is the number of bytes in dense x, y, 
> > > > > or the result x f y, whichever is the greatest;
> > > > > and g is the clock rate of the CPU.
> > > > > 
> > > > > For example, on an ordinary (not overclocked) 
> > > > > 2.2 Ghz AMD 3200+ machine,
> > > > > 
> > > > >    (1*m) %~ 2.2e9 * 6!:2 'm$0'   [ m=: 1e7
> > > > > 2.34726
> > > > >    (4*m) %~ 2.2e9 * 6!:2 'm$2'   [ m=: 1e7
> > > > > 2.3531
> > > > >    (8*m) %~ 2.2e9 * 6!:2 'm$0.2' [ m=: 1e7
> > > > > 2.35242
> > > > > 
> > > > > The CPB for $ is approximately 2.35.
> > > > > 
> > > > > Some others:
> > > > > 
> > > > >    x=: 1e7 [EMAIL PROTECTED] 2
> > > > >    y=: 1e7 [EMAIL PROTECTED] 2
> > > > >    (#x)   %~ 2.2e9 * 6!:2 'x+.y' 
> > > > > 4.85849
> > > > >    (4*#x) %~ 2.2e9 * 6!:2 'x+y'
> > > > > 2.69899
> > > > >    (4*#x) %~ 2.2e9 * 6!:2 '/:x'
> > > > > 6.04693
> > > > >    (#x)   %~ 2.2e9 * 6!:2 '/:~x'
> > > > > 3.09869



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