[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 :-)

 
Aleksander:

Alsou... what if not a pawn to move - but a rook to attack the pawns?

Aleksander:
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


We go to b6 in exactly the same way, and on the next rook move we put the pawn on b7. Since b7 is protected by the white bishop, the black rook cannot beat the pawn, and will be forever bound to control b8 to prevent the pawn from turning into a queen. But then the white king has complete freedom of action and can easily get rid of the remaining black pawns, while also retaining the ability to lead forward pawns on lines a and d
 

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.

 
Anything else you'd like me to throw in?
 
3 bribes if not too much brandy drunk.
 
Better a puzzling one.
 
One question remains.
A+B=...
Is that too weak?
 
There's a very interesting problem to solve. I'll remember the condition in a moment.
 
MikeM:
Is that not enough?
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