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Crazy situation, have anyone can solve this?

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cocacolapepsi
Remellion wrote:

I'll leave (e) up for grabs as that seems very combinatoric based, and I hate combinatorics. Although at a glance it's obvious that if the answer is that there are 10 ways, there must be only 5 other black pieces that can be removed in conjunction with [bQb6 or bRg4].

Can you check Uncia_Uncia's solution?

bravewolf9x
Remellion wrote:

BigDoggProblem got (a) and (b) right. (Of course he did.) Solution is unique too.

(c): Focus on the impossible double check first. To resolve that, there are only 3 possibilities: remove -wNe4, -bQb6 or -bRg4. In each case, white must still have promoted one pawn to a rook (the wRa1 died in the cage still). Black in the diagram position is missing 2 bishops (useless; died on c8 and f8), 2 pawns (both promoted to light bishops) and nothing else. So white only captured the bishops on c8 and f8. Even removing bQb6 or bRg4 only accounts for white's doubled h-pawns - white's f-pawn doesn't have anything to capture and promote to a rook. Since in all 3 of these cases the position remains impossible, it is impossible to remove a single unit to yield a legal position.

 

(d) Again, to deal with the double check, one of the removed pieces must be the wNe4. We're not allowed to remove black pieces, so we know for a fact both black light bishops are promoted from the c7 and h7 pawns, and that white did not capture any black pieces but the c8- and f8-bishops (since there are the other black units on the board). So we must deal with the doubled h-pawns; remove one of the white h-pawns (say -wPh2). But we still have the fact that one of the white rooks is promoted (since wRa1 was captured in its cage) and white again cannot have captured anything with the f- or g-pawns to promote them. Therefore a contradiction, and removing any 2 white units will not result in a legal position.

 

I'll leave (e) up for grabs as that seems very combinatoric based, and I hate combinatorics. Although at a glance it's obvious that if the answer is that there are 10 ways, there must be only 5 other black pieces that can be removed in conjunction with [bQb6 or bRg4].

In part d) What will happen if I remove wR-G5 and wN-E4

bravewolf9x
BigDoggProblem wrote:
bravewolf9x wrote:
BigDoggProblem wrote:

I'll leave c) - e) to other solvers. :)

Do you agree with Remellion 

Looks like solid reasoning to me.

I have the same Question for you, bro. In part d) What will happen if I remove wR-G5 and wN-E4

BigDoggProblem
bravewolf9x wrote:
BigDoggProblem wrote:
bravewolf9x wrote:
BigDoggProblem wrote:

I'll leave c) - e) to other solvers. :)

Do you agree with Remellion 

Looks like solid reasoning to me.

I have the same Question for you, bro. In part d) What will happen if I remove wR-G5 and wN-E4

I still have a doubled white h-pawn that had nothing available to capture to get there.

bravewolf9x

thanks for BigDoggProblemRemellionUncia_Uncia. Your answers are correct.

Please enjoy my new puzzle at http://www.chess.com/forum/view/more-puzzles/the-69-position

HumongusChungus1234

Sup I just selected a random old thread just to give random people an unnecessary notification. Also OP, have anyone can solve this is not a correct sentence. Have a good day.