White to move and win! No computer lines please!
6Q1/8/7q/2B5/4p3/2K1pp2/p6p/k7 w – – 0 1
Pervakov, 1986
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Maybe 1.Qe1 attacking Qd1+ checkmate is winning for white..
1. KC2 Qc6 2. Qc4 followed by BD4++
1. Kc2
Qb3
1) Qb3 (threat Qb2#), Qf6+
2) Bd4, Qc6+
3) Kb4#
1. Kc2
Black cannot prevent mate with the bishop or queen on the a1 – h8 diagonal.
1.Bd4 Qb6
White was threatening 2.Kc2#
(1…. Kb1 2.Qb3+ Kc1 3.Qc2#)
(1… Qc6+ 2.Kb4 Qc3+ 3.Bxc3+ Kb1 4.Qb3+ Kc1 5.Qb2+ Kd1 6.Qa1+ Kc2 7.Qxa2+ Kd1 8.kb3 & 9.Qc2#)
2.Bxb6 h1=Q
3.Bd4 Qc1+
4.Kb4+ Qb2+
5.Bxb2+ Kxb2
6.Qb3+ Kc1
(6… Ka1 7.Kc3)
7.Qxe3+ Kd1
White should be able to be able to gobble up remaining pawns.
LOL!!! I remember this beastly problem. Hard to believe it was first posted here 4 years ago.
Note to the rest- the problem is hideously difficult to prove out, and requires a deep analysis. I spent an entire afternoon working on it the last time, and still failed to fully solve it.
Yancey,
You can say that again! I have spent half a day on this, discarded 2 solutions after frightening analysis and now I am stuck on the tail end of a Q vs Ps endgame and the end seems to be near but ever elusive!
This is the first time I am trying this problem!
Harry
My gut feel was that 1. Qd5 was the right move. I analysed several variations but finally in one variation it did not win for white; even with improvements it lost for white. I struggled and finally it dawned on me to play the same theme but with a different move order. After almost 6 hrs of analysis I got it! And it fell in place! Beautifully simple! Finally!
Eureka ! Eureka! Eureka!
1.Bd4!!
>A-1….Qc6+.2.Kb4+!.Kb1.3.Qb3+.Kc1.4.Bxe3#!!
>B-1….Kb1.2.Qb3+.Ka1.(if Kc1.3.Qc2#).3.Kc2#
>C-1….Qb6!.2.Bxb6!.h1=Q.3.Bd4!
>>C1-3…..Qb1.4.Kc4+Qb2.5.Qg1#
>>C2-3…..Qc1+.4.Kb4+.Qb2+.5.Bxb2+.Kxb2.
>>>C21-6.Qb3+.Kc1.(if Ka1.7.Ka3 and 8.Qb2#).7.Qxa2.
>>>>C211-7….Kd1.8.Kb3!. and transposes into one of the following variations.
>>>>C212-7….f2.8.Qe2 ! stops the pawns from queening, confines BK to first rank and mates shortly!.Kb1.9.Kb3! and mate next move.
>>>>C213-7…e2.8.Kb3!.
>>>>>C2131-8…e1=Q/e3/f2.9.Qc2#
>>>>>C2132-8…e1=N.9.Kc3! and mates next move
>>>>>C2133-8….Kd1.9.Qa1+!.Kd2.10.Qc3+!.Kd1.11.Kb2!
>>>>>>C21331-11….e1=Q/R/B.12.Qc2#
>>>>>>C21332-11….e3/f2.12.Qc1#
>>>>>>C21333-11….e1=N!.12.Qe3!!!!
>>>>>>>C213331-12….Nd3+.13.Kc3! and 14.Qd2#
>>>>>>>C213332-12…Nc2.13.Qc1#.Ke2.14.Kxc2!! and wins
>>>>>>>>C2133321-14…f2.15.Qd1+!.Ke3.16.Qf1!! and wins easily!
>>>>>>>>C2133322-14…e3.15.Qg1!!.f2.16.Qd1#
>>>>>>>>C2133323-14…Kf2.15.Qd2+.
>>>>>>>>>C21333231-15….Kf1.16.Kd1.
>>>>>>>>>>C213332311-16….Kg1.17.Ke1.wins easily as King and Queen control the pawns.
>>>>>>>>>>C213332312-16….f2.17.Qe3!!.Kg2.18.Qxe4+.Kg1.19.Qg4+.Kh1.20.Qf3+.Kg1. 21.Ke2! and wins the last f pawn and mates shortly!
>>>>>>>>>C21333232-15….Kg1.16.Kd1.
>>>>>>>>>>C213332321-16….f2.17.Qe3!! and wins as in C213332312 above.
>>>>>>>>>>C213332322-16….e3.17.Qxe3+.f2.18.Ke2! and wins.
>>>>>>>>>>C213332323-16….Kh1.17.Ke1! and Qe3!! and wins easily as King and Queen control the pawns.
I believe that the above proves and demonstrates that White wins starting with 1.Bd4!
At last! That was the toughest! Ever!
Harry
1.Bd4 Qc6+ 2.Kb4+…(…Kb1 3.Qb3+ Kc1 4.Bxe3#) 2…Qc3+ 3.Bxc3+ Kb1 4.Qb3+ Kc1 5.Bd4 Kd2 6.Qc3+ Kd1 7.Qa1+ Kd2 8.Qxa2+ Kd3 9.Qxh2…(…Kxd4 10.Qd6#) 9…f2 10.Qf4 /+-/
Bd4! win for white.
In my earlier post, where I had given the solution variation C21 starting with 1.Bd4!!..Qb6!. 2.Bxb6!.h1=Q.3.Bd4!.Qc1+.4.Kb4+.Qb2+.5.Bxb2+.Kxb2.6.Qb3+.Kc1.(if Ka1.7.Ka3 and 8.Qb2#)., I had given the continuation with 7.Qxa2. where my only slight discomfort was the reply 7…Kd1
I analysed and found that 7.Qxe3+! is more convincing and complete without an iota of doubt and more realistically demonstrable. here goes the continuing solution…For completeness I present the solution in full so that you need not refer to my earlier comment.
1.Bd4!!
>A-1….Qc6+.2.Kb4+!.Kb1.3.Qb3+.Kc1.4.Bxe3#!!
>B-1….Kb1.2.Qb3+.Ka1.(if Kc1.3.Qc2#).3.Kc2#
>C-1….Qb6!.2.Bxb6!.h1=Q.3.Bd4!
>>C1-3…..Qb1.4.Kc4+Qb2.5.Qg1#
>>C2-3…..Qc1+.4.Kb4+.Qb2+.5.Bxb2+.Kxb2.
>C21-1.Bd4.Qb6!.2.Bxb6!.h1=Q.3.Bd4!.Qc1+.4.Kb4+.Qb2+.5.Bxb2+.Kxb2.6.Qb3+.Kc1.(if Ka1.7.Ka3 and 8.Qb2#).7.Qxe3+!
>>C211-7….Kb2.8.Qd2+.Kb1.9.Kb3!.a1=N+! (else 10.Qd1#).10.Kc3! and mate in 2 moves.Nc2.11.Qxc2+.Ka1.12.Qb7/Qa7#
>>C212-7…Kb1.8.Kb3!.a1=N+!.(else 10.Qe1#).9.Kc3!.Nc2.10.Qg8+.Ka2.11.Qc1 and 12.Qb2#
>>C213-7…Kc2.8.Qxe4+.
>>>C2131-8…Kd1.9.Qxf3+ and wins easily!
>>>C2132-8…Kd2.9.Qd5+. followed by 10.Qxa2 and white wins easily as the lone king cannot defend the f pawn.
>>>C2133-8….Kb2.9.Qe1!!.a1=Q.10.Qd2+.Kb1.11.Kb3!! and mates in 2.
>>>C2134-8….Kc1.9.Kb3!
>>>>C21341-9…f2/a1=Q/R/B.10.Qc2#
>>>>C21342-9…a1=N+.10.Kc3!.f2.(if ..Kd1.11.Qxf3+ wins easily).11.Qh1+.f1=Q.12.Qxf1#
>>C214-7…Kd1.8.Qd4+!
>>>C2141-8….Ke1/e2.9.Qxe4+.Kf2.10.Qc2+.Kg3.11.Qxa2.f2.12.Qg7+ and wins
>>>C2142-8….Kc1.9.Qa1+.Kd2.10.Qxa2+.Ke3.11.Kc3!.f2.12.Qd2+.Kf3.13.Qd1+.Kg2.14.Qe2.
>>>>C21421-14…e3.15.Kd3.Kg1.16.Qg4+.
>>>>>C214211-16…Kf1.17.Kxe3.Ke1.18.Qe2#
>>>>>C214212-16…Kh1/2.17.Ke2!. mating shortly
>>>>C21422-14…Kg1.15.Qe3!
>>>>>C214221-15….Kg2.16.Qxe4+ wins easily against lone P+K
>>>>>C214222-15…Kf1.16.Qxe4 winning easily.
>>>C2143-8…Kc2.9.Qxe4+. as in C213 above
I am satisfied after a whole days work that this is cracked!
Harry
Qc4
Harry, your solution is very good, but I did not see the winning line on move 3. ….. Qe1+
Appreciate it very much if you can post the solution.. thanks..
There is some mistake.My posting has appeared as Anonymous at 8:04:00 AM CDT
Dear Anonymous,
The
1.Bd4!!
>A-1….Qc6+.2.Kb4+!.Kb1.3.Qb3+.Kc1.4.Bxe3#!!
>B-1….Kb1.2.Qb3+.Ka1.(if Kc1.3.Qc2#).3.Kc2#
>C-1….Qb6!.2.Bxb6!.h1=Q.3.Bd4!
>>C1-3…..Qb1.4.Kc4+Qb2.5.Qg1#
>>C2-3…..Qc1+.4.Kb4+.Qb2+.5.Bxb2+.Kxb2. 6.Qb3+.Kc1. 7.Qxe3+!.etc.
The balance solution
>>C3-3…..Qe1+.4.Kc2+!.Qc3.5.Bxc3# and
>>C4-3…..Qd1.4.Qg1!!.Qxg1.(if any other move 5.Qxd1#).5.Kc2#
Hope you got the full solution now. In my long analysis C3 and C4 were left out by mistake. Thanks for pointing it out.
Harry
Dear Harry,
Hats off to you for your patience.
In the unintended role of Anonymous at 8:04:00 AM CDT July 18 I had given 1.Bd4 Qb6 2.Bxb6 h1=Q+ 3.Bd4 Qc1+ 4.Kb4 Qb2+ 5.Bxb2 Kxb2 6.Qb3+ Kc1 7.Qxe3+ and did not continue further.If7… Kd1 8.Qd4+ Kc1 I would prefer 9.Kb3 (instead of your 9.Qa1+) Black has only two ways
(A)9…a1=Q 10.Qxa1+ Kd2 11.Qd4+
(B)9…f2 10.Qxf2 a1=Q 11.Qc2#
1.Bd4 Qb6 2.Bxb6 h1=Q 3.Bd4 Qd1 4.Qg1 e2 5.Bh8 (why h8 only? evident from next move.) f2/e3 6.Qg7 e1=Q+ 7.Kc4+ Kb1 8.Qb2#
Last night I was too sleepy to give (full?Never. Only more) explanation.1.Bd4 Qb6
2.Bxb6 h1=Q
3.Bd4 Qd1
4.Qg1 e2
Trying to buy some time and maintain status co.
5.Bh8 f2/e3
This has more to do with blocking the g1-a7 diagonal rather than attacking Q.
(5… Kb1 6.Qb6+ Kc1 7.Qb2#)
(If 5…e1=Q+ 6.Qxe1 Qxe1+ 7.Kc2#)
6.Qg7 e1=Q+
(6….Kb1 7.Qb7+)
7.Kc4+ Kb1
8.Qb2#
Dear Prof. S.G. Bhat,
I have a more consistent solution for the 4….e2 line. Please below.
1.Bd4. Qb6.2.Bxb6.h1=Q.3..Bd4.Qd1.4.Qg1.e2. 5.Qe1!!.f2./e3..6.Kc4+.Kb1.7.Qb4+.Qb3+.8..Qxb3+.Kc1. 9.Be3#!
Of corse 5…Qxe1+ allows 6.Kc2#!
Harry
Dear Prof Bhat,
Your solution of 5.Bh8, 6.Qg7 works but it takes 10 moves to mate in total for black can deviate from your given moves by inter posing Queen twice!
7….Qd4+.8.Qxd4+.Qc3+.9.Qxc3+.Kb1.10.Qb2#.
Whereas my solution with 5.Qe1! Forces mate in 9 moves and as stated is more consistent with the queen pinned to the first rank after 4.Qg1!
Harry
dear Harry,
Again hats off to you for your analysis stretching my scheme by 2 more moves.In my notes i drew attention to the idea black had,to block white Q’s access to b6 square by 5…f2 order to play 6…Kb1.This concern has not been addressed in 5.Qe1??.Now black does play 5… Kb1 and white is lost for ever.As a matter of fact 5.Qh1? would have served the same purpose as 5.Qe1?
Dear Prof. Bhat,
I checked out and found you were right!
In the variation
1.Bd4. Qb6.2.Bxb6.h1=Q.3..Bd4.Qd1.4.Qg1.e2. 5.Qe1.loses to 5….Kb1!
The alternative 5. Qe3! Wins in all variations except this line 5….Qd3+!!.6.Kb4!.Kb1!!
Hence colorectal is 5.Bh8 etc as you pointed out.
Cheers!
Harry
What a beautiful study and What a drama in the last variation!
4.Qg1 pins the black Q.White Q does not need support but black Q needs it just to stay there.Hence 4… e2.
White Q however can not stay on first rank without access to a square on b file.Black voluntarily cleared G1-A7 diagonal by 4… e2 but white B is coming in the way.So white plays Bh8 getting access to b6 for the Q.Black tries to close the diagonal by 5….f2.White plays 6.Qg7 getting access to b7 square.Instead of5.Bh8, 5.Bg7 would be wrong because g7 is to be reserved for Q.5.Be5 or 5.Bf6 are also wrong because of inverted battery of B and Q.For propylaxis(not to allow … Kb1) access to b file is sufficient but for positive action of discovered check K has to move to b file(since c2 is not available.) and check on b file will not be available for Q and only B and Q battery will do the job.Thanks to Harry for his relentless effort without which I would have forgotten the study after first posting.
FYI
Link1
And:
Link2
A link to the problem from YACPD and the comments section of the previous posting here.
Dear Yancey.
Thanks for the 2 links.
Why is MarieLucie not commenting nowadays? Is she too busy with other things?Her comments are very illustrative.
you had tough time with 7… Kc2.
in my opinion QRP can be taken care of by K and Q together whereas it is dangerous to leave the two connected pawns with black K nearby.Perhaps you were too anxious about the QRP on 7th rank.it seems to be partly borne out of your variation 8.Qc3 kb1 when 9.Kb3 is not very comfortable because of 9… a1=N+ and c3 is not available for the K and he has to go away from the scene.You seem to have dismissed 8.Qxe4+ because of close proximity of black K to QRP.8.Qd4+ Kc1 is a simple affair in spite of going after QRP with both e and f pawns present by keeping black K away from these pawns.vide my comments on July 19 11:55 AM CDT