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Only 2 moves to "end", very hard!

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bravewolf9x

Black go first.

- Show that there are atleast 7 different ways that Black can "finish" this game with maximum 2 moves.

- 2 ways is different if they are able to satisfy 1 of 2 conditions:

 + First moves of Black are different.

 + Same Black first moves. But with any White first move, then Black have 2 methods to deal at his 2nd move.

Example: Qe4-b1

Alec_CA

Hello

bravewolf9x
Alec_CA wrote:

Hello

hi, do you have solution for this puzzle?

vmsfinale

i couldnt understand the conditons.  that sort of grammar ...

bravewolf9x
Oran_perrett wrote:

also as a bonus does this count?

 

no :)

bravewolf9x
Oran_perrett wrote:
 

(1... gxf2+2. Kf1Qe1#)

This is wrong, 1... gxf2+2. Kxf2

bravewolf9x

you just have 6 acceptable solutions for this puzzle. Thank you :)

Alec_CA

Many ways

bravewolf9x
Oran_perrett wrote:

i still have 7:

1...Qe1#

1...Qb1# 

1...Qg2#

1...gxh2+ 2.Kf1 h1=Q# 

1...gxh2+ 2.Kf1 h1=R#

1...g2 2.f3 Qe1# 

1...Qf3 2. Kf1 Qd1#

 

And why does white winning not count?

1...gxh2+ 2.Kf1 h1=Q# 

1...gxh2+ 2.Kf1 h1=R#

those solutions are only 2 cases of 1...gxh2+ 2.Kf1. So they are the same.

My final way is:

1...Qh1 2.Kh1 g2 3.Kg1 but Black doesn't have any way to go, so no one win

 

bravewolf9x
Oran_perrett wrote:

your stalemate is nice but so is my white mating. really your puzzle is flawed

1... Qh72. d4Qh43. d5#

Because in this solution, Black still have ability to deal with White. Maybe in first move, Black make a mistake, but he can fix it absolutely.

1... Qh72. d4Qh73. d5 Qd5