Author: K. Burcham
Date: 07:51:57 11/10/01
as you can see in this position, with 1500 mhz using shredder5, we can get
values for kns, and total nodes for a posted time.
so these values for 58 minutes are: 307 kns, and 1,087,429,783 nodes.
what if we used a multiplier of 13.5. 13.5 x 1500 mhz = 20000 mhz.
using this value of 13.5, we can divide 58 minutes by 13.5 = 4.29 minutes.
lets say that Qe5 was a winning move.
so from this we could say that with shredder5 on 20 gig, could find Qe5
in 4.29 minutes. usually a long standard time control for 4 minutes.
what about blitz. 20 x 1500 mhz = 30000 mhz. 58 divided 20 = 2.9 minutes.
what about bullet. 40 x 1500 mhz = 60000 mhz. 58 divided 40 = 1.45 minutes.
way too much time for blitz and bullet.
i know there are many variables here. and the future will get crazy with
the machines that will play chess. i read the other day that we might have
24 volts running to our fish bowl. that the water molecules with the current
passing through will actually do the switching. the chiphead that gave this
theory, said with this molecule switching, everything as we know it, as an
enduser would be instant.
anyway i am changing my way of thinking about these mhz increases, and how our
depth changes with each. it seems that it will take an enourmous amount of
todays mhz to get shredder5 to play Qe5 in a 3/0 or a 3/3 game.
i conclude that with todays cpu, as we know them, even with smaller dies, we
cannot achieve the necessary mhz for shredder5 to find Qe5.
[D] r2r2k1/1R2qp2/p5pp/2P5/b1PN1b2/P7/1Q3PPP/1B1R2K1 b - -
15.01 15:31 +3.21 1...Rab8 2.Rxe7 Rxb2 3.Rf1 Rxd4 4.Re4 Rxe4 5.Bxe4 Kg7
6.g3 Be5 7.Kg2 f5 8.Bd5 Kf6 (296.898.463) 318.8
15.03 58:55 +3.22++ 1...Qe5 2.Bc2 Bxc2 3.Qxc2 Rxd4 4.g3 Rad8 5.Rxd4 Rxd4
6.Kg2 Rd2 7.gxf4 Qe1 8.Rb8+ Kh7 9.Qxd2 Qxd2 10.Kg3 g5 11.c6 (1.087.429.783)
307.5
15.03 77:16 +3.57++ 1...Qe5 2.Bc2 Bxc2 3.Qxc2 Rxd4 4.g3 Rad8 5.Rxd4 Rxd4
6.Kg2 Rd2 7.gxf4 Qe1 8.Rb8+ Kh7 9.Qxd2 Qxd2 10.Kg3 Qc3+ (1.420.005.345) 306.2
15.03 104:53 +3.80 1...Qe5 2.Bc2 Bxc2 3.Qxc2 Rxd4 4.g3 Rad8 5.Rbb1 Rd2 6.Qb3
Qxc5 7.Qf3 Bc7 8.Rxd2 Rxd2 9.Qe3 Qxe3 (1.920.599.184) 305.1
please dont post where your tunnel ram and two fours running Klotz finds Qe5 in
.0000001 second. that is not what this post is about. it is about a given
position, with a certain program finding a better score, with a better move,
in a shorter time.
kburcham
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