30.Kg1 Bh2+ 31.Kh1 Bg3+ and mate or
30.Ke2 Bxg4+ 31.Kd2 Bf4+

Right, but it takes some calculation to see that Black doesn't get perpetual check:

29. Kf1 Qh3+ 30.Ke2 Bxg4+ 31. Kd2 Bf4+ 32.Re3 Bxe3+ 33.fxe3

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33...Qg2+ 34.Kc1 Qf1+ 35.Kb2 Qb5+ 36.Bb3 Qe2+ 37.Qc2 and White is winning.

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I can identify with all those who thought this was tough for a Monday, maybe CG is making up for last week. Looking at ir later I tried to find an improvement for W after 25 ... Qf4. I don't think there is one.

<28. Bf5 Nxf5 29. Qg7+ Kxg7 30. Nh5+>.

Ok, good start, but what if <29...Nxg7>?

Then how about <28. Re5 Bxe5 29. Qxg7+ Kxg7 30. Nh5+>?

Ok, but that loses a rook.

Ah, it must be:

1. (4.98): 28.Re3 Qxh2+ 29.Kf1 Re7 30.Qh7[] Qxh7 31.Nxh7+ Ke8 32.Rae1[] Bd6 33.Nf6+ Kd8 34.Nh5[] Rf7 35.Rxe6 Nxe6 36.Rxe6 Kd7 37.Re2 Be7 38.Bf5+ Kd8 39.Re6

2. (4.98): 28.Kf1 Qxh2 29.Re3[] Re7 30.Qh7[] Qxh7 31.Nxh7+ Ke8 32.Rae1[] Bd6 33.Nf6+ Kd8 34.Nh5[] Rf7 35.Rxe6 Nxe6 36.Rxe6 Kd7 37.Re2 Be7 38.Bf5+ Kd8 39.Re6

LOL.

I looked at both 28 Re3 and 28 Kf1, but actually got further with 28 Nh7+ Kg8 29 Re5 and ensuing swap-off.

A somewhat longer Rybka run (I finally gave in and got it) yields this :

1) 28.Nh7+ Kg8 29.Kf1 Qxh2 30.Nf6+ Kf8 31.Re3 Re7 32.Qh7 5.04/20

2) 28.Re3 Qxh2+ 29.Kf1 Re7 30.Qh7 5.04/20

3) 28.Kf1 Qxh2 29.Re3 Re7 30.Qh7 5.04/20

4) 28.Qxg7+ Kxg7 29.Nh5+ Kf7 30.Nxf4 Bxf4 31.Bf5 Bxf5 32.Rxe8 Kxe8 33.gxf5 Kf7 34.Kf1 Bxh2 35.Rb1 b6 36.Re1 Kf6 37.Re6+ Kxf5 38.Rxc6 Bb8 39.Ke2 Bf4 40.Kd3 Bb8 41.c4 dxc4+ 42.Kxc4 Bf4 43.d5 4.53/20

The first three are clearly all transpositions of the same line: essentially, let black take h2, force a queen exchange on the h file then double up rooks on the open e file winning an exchange on e6.

Variation 4 is Smyslov's line played out with further equal exchanges and slightly behind on Rybka score.

A third contender emerges if we rerun Rybka after 28. Rh7+ Kg8

B1) 29.Nf6+ Kf8 30.Re3 Qxh2+ 31.Kf1 Re7 32.Rae1 Bd6 33.Qh7 Qxh7 34.Nxh7+ Ke8 35.Nf6+ Kd8 36.Nh5 Rf7 37.Rxe6 Nxe6 38.Rxe6 Be7 39.Bf5 Rf8 40.Ke2 Ba3 41.f4 Rh8 42.Bc2 Kd7 43.f5 Be7 5.01/20

B2) 29.Kf1 Qxh2 30.Nf6+ Kf8 31.Re3 Re7 32.Rae1 Bd6 33.Qh7 Qxh7 34.Nxh7+ Ke8 35.Nf6+ Kd8 36.Nh5 Rf7 37.Rxe6 Nxe6 38.Rxe6 Be7 39.Bf5 Rf8 40.Ke2 Ba3 41.f4 Rh8 42.Bc2 Kd7 43.f5 Be7 5.01/20

B3) 29.Re5 Bxe5 30.dxe5 Qxe5 31.Nf6+ Kf8 32.Rd1 a5 33.Rd3 Bf7 34.Nd7+ Kg8 35.Nxe5 Bxg6 36.Nxg6 Re2 37.Bb3 Re4 38.Bd1 Ne8 39.h4 Kg7 40.h5 Re1+ 41.Kg2 Nd6 4.85/20

Variations B1) and B2) are transpositions of the main line.

Variation B3) sacrifices a rook and pawn on e5 for a bishop, but preserves h2 and sets up a winning exchange via a royal fork from d7 (I thought this worked but needed silicon help here to work it through).

Anyway, time to get to the point.

There are at least three lines vying to be solutions to 'White's best at move 28'.

The positions following move 38 are in order of Rybka score:

Variation B1/B2

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Variation B3

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and Smyslov's line

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All three are clearly winning positions for white.

Against opposition who is determined to play on, all require a similar amount of work to force a result.

Therefore, all (and probably a few others) are more or less equal aren't they?

I've noticed that computers can give a 0.00 evaluation to a move which repeats the position even if one side or another is clearly winning. I first spotted this when I ran Fritzie in the background during a live game. When a player deliberately repeats moves to gain time on the clock, Fritzie will occasionally throw out a 0.00 evaluation for these repetitions, regardless of the rest of the board.

For a bit of fun, I made up a position to demonstrate. It's white to move:

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Of course, 1. Rc8# is a no brainer, but let's see what happens if we deliberately repeat moves...

1. Bf3+ Kb8 2. Bd1

And now Fritzie insists that the eval is 2...Ka8 0.00, even though Rc8# is still an option and we haven't had a threefold repetition.

You get a similar effect stepping through today's game with Fritz on infinite analysis. After 27...Kf8, Fritzie evaluates 28. Nh7+ as 0.00 because it repeats a position that we have seen before.

Let's play the apparently drawing 28. Nh7+ to see what happens. Now Fritz says that 28...Ke7 is a whopping -7.65, but 28...Kg8 is the same comforting 0.00.

Carrying on, we play 28...Kg8. Now Fritz offers three moves which give white a hefty advantage (Kf1, Re5 and Re3), but the repetition 29. Nf6+ is again exactly equal with 0.00.

Again, we play into this "drawn" line with 29. Nf6+ and again, Fritz says that 29...Kg8 is 0.00.

Now we repeat with 30. Qxg7+ and the eval jumps to +3.2.

I think the moral is this: when a computer says that a position is 0.00, it could mean that the position is dead drawn (ie K v K). It could mean that one side can force a draw by perpetual. Or, surprisingly, it could mean that the position has been repeated just once, but is still winnnable.