The British itself!

[Simon Brown]

Matthew, can you show us the idea? I though Rc5 first but he has played f6+

[Matthew Turner]

Infiltrate with the Rook, attack things on the seventh and get yourself a passed pawn into the bargain. Nothing too complicated, but Ward has maybe found a tricky defence.

[John Moore]

Andrew Lewis has just allowed, I think, 34 .. Qh3ch

[Simon Brown]

Thanks. I liked the idea of keeping the rook passive on b6 but couldn't see how to make progress. This position presumably tailor-made for Keith though.

[Simon Brown]

Andrew Lewis has just allowed, I think, 34 .. Qh3ch

Except that 34 f6 is check unfortunately....

[Christopher Kreuzer]

Nick Pert now has a passed pawn against Howell. If Nick Pert and Arkell win, would Arkell or Richard Pert play Hawkins?

[Matthew Turner]

Simon,
I wondered about that too with a4 at some point to try and trap the Rook, Keith would be happy to try that so he must be very confident with this more forcing line.

[Christopher Kreuzer]

Andrew Lewis has just allowed, I think, 34 .. Qh3ch

Except that 34 f6 is check unfortunately....

Yes, but go back a few moves. Could Black have played 32...Kh8?

[Simon Brown]

Certainly better than 32...Rg7 but just a draw I think after Qh6

[Christopher Kreuzer]

Looking at the colours, I think Arkell (currently 5 Whites and 4 Blacks) will play Hawkins as Black. This keeps Hawkins on the right colour (White). Richard Pert has had two Blacks in row, so giving him a third Black wouldn't really work, neither would breaking Hawkins's colour run. Having said that, Arkell still needs to win against Ward... and I've not taken into account the other pairings, which will probably mess things up.

[IM Jack Rudd]

Pert R v Hawkins, Howell v Hebden looks pretty much forced whatever the result of Arkell-Ward or Pert N-Howell.

[Christopher Kreuzer]

Why not Hawkins (8)-Arkell (7) on one and Howell (7.5)-Hebden (7.5) on two? And R. Pert (7) vs someone else on three and N. Pert (7) vs someone else as well? Is it because it is difficult to find people to pair with the Pert brothers? (I've presumed Arkell wins and Howell draws). It does seem a bit perverse that the leader gets given Black to finish with, breaking the colour sequence, but in the last round of an event with an odd number of rounds, that maybe matters less.

[IM Jack Rudd]

Because you don't break score-groups unless absolutely necessary; that takes priority over giving people the correct colour.

[Christopher Kreuzer]

Ah yes, you would pair Hawkins and R. Pert; and then Howell and Hebden, and then Arkell and N. Pert. No-one else on 7 points. And avoiding three Blacks in a row for R. Pert takes precedence over keeping the leader on the correct colour. Pity though. Now I'm going to have to ask when the British ever lacked a GM on top board in the last round!

[Simon Brown]

OK, going home now and out of contact for about 90 minutes. I'll predict a draw on 2 and a win for Keith (and I won't speculate about tomorrow's draw.....)