[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 595
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both have the same amount of material-
Black has a rook (5 pawns)+3 pawns
white has a bishop (3 pawns)+5 pawns...
positional difference only :-) I'll try to gobble up material according to botvinnik :-)
Alsou... what if not a pawn to move - but a rook to attack the pawns?
well, eat the queens - you draw further... with a black pawn forward :-)
After black's move with any pawn white goes b6, threatening to put the queen. Play with yourself, you'll see that Black has no options here.
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F1-F5
Listen... how about one more time... :-) I'm having second thoughts... - your king's path to black pawns is blocked by a rook on line 5 - pawns in a row and will bump into an elephant
d row blocked by a black pawn....
you could also make it a draw... or even crush your king :-)
The South has ordered a mizzer. The East's move. How will the draw end with all opponents playing optimally?
The draw goes clockwise, i.e. after East goes South.
"Optimal play" means that South tends to end up taking as few bribes as possible (i.e. it can sometimes override the initiative and take them on purpose so as not to get even more of them). In turn, East and West cooperate openly to get the South to take as much as possible. In short, the game.
West and East see each other's cards, but not South's (he doesn't show what he's taken down). But this is usually easy to calculate. South sees East's cards and West's cards.
P.S. Yeah, I see, no one has ever painted the bullet here. All right, I'll decide. In principle, I see two bribes South. Possible and three, but that is if the South will play suboptimally.
A+B=...
Is that too weak?
Is that not enough?