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Subject: Re: SSDF Deep Fritz - Junior 6: 0,5 - 2,5 Now: 1 - 6 !!

Author: Christophe Theron

Date: 05:43:49 01/02/01

Go up one level in this thread


On January 02, 2001 at 07:53:49, CLiebert wrote:

>On January 02, 2001 at 06:59:48, Djordje Vidanovic wrote:
>
>>On January 02, 2001 at 03:38:36, Jouni Uski wrote:
>>
>>>On January 02, 2001 at 03:32:29, Helmut Conrady wrote:
>>>
>>>>On January 02, 2001 at 03:22:46, Jouni Uski wrote:
>>>>
>>>>>From CSS magazine 6/00: Deep Fritz makes no sense to use with slow ~400 Mhz PC
>>>>>and single prosessor!
>>>>>
>>>>>Jouni
>>>>
>>>>On which page have you read this?
>>>>
>>>>Helmut
>>>
>>>Sorry don't remember, but something like that was written in Deep Fritz article.
>>>Actually I was surprised, when article was so objective!
>>>
>>>Jouni
>>
>>
>>The closest reference in CSS 6/00 was on page 21 where Mathias Feist says the
>>following: "...Aus Performance-Gruenden ist mindestens ein Pentium II/Celeron
>>anzuraten".  (At least PII or Celeron recommended)
>>
>>***  Djordje
>
>
>
>1:6 is still possible, I have seen a lot of such rows from nearly
>equal programs. But Fritz/K6 against Junior/K6 on a 450mhz pc is indeed a little
>bit unlucky combination: Fritz6 dislike the Amd and is much faster on a P3. It
>it likes fast pc´s with bigger internal cache.
>With Junior it is just the opposite, it likes the k6 (despite Amir never used
>it). If you replay this match on intel-pcs you will get most problably very
>different results. I am shure that this result doesn´t reflect the real relation
>between these engines...
>
>But nevertheless, thats life.
>
>Christian



What is the speed difference in nps between DF on K6-2 and DF on PIII (using the
same clock frequency)? From this we could compute the estimated elo difference
and see if it is really significant.

For example, a speed difference of 10% accounts for a 10 elo points difference.
20% would give 18 elo points difference.

The general formula, which has been verified so far in all the past SSDF lists,
is:

  elo_diff = 70 * log(speed_ratio) / log(2)

For 10%, speed_ratio=1.1. For 20%, speed_ratio=1.2 and so on...



    Christophe



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