# No.1077,1078,1079 (KM)

 No.1077-1079  Karol Mlynka (Slovakia) Original Problems, Julia’s Fairies – 2016 (I): January – June    →Previous ; →Next ; →List 2016(I) Please send your original fairy problems to: julia@juliasfairies.com

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 No.1077 Karol Mlynka Slovakia original – 24.05.2016 Solution: (click to show/hide) white Ka3 black royal LBe8 black Pb7 #5                                                (1+2) White Super-Transmuting King Royal Bishop-Locust e8 1.Ka3-b3 ? zugzwang. but 1...b7-b5 ! 1.Ka3-b4 ? {A} threat: but 1...b7-b6 ! {a} 1.Ka3-a4 ! threat: 2.Ka4-a5 zugzwang. 2...b7-b5 3.Ka5-b6 threat: 4.Kb6-c7 threat: 5.Kc7-d8 # 2...b7-b6 + 3.Ka5-a6=P threat: 4.a6-a7 threat: 5.a7-a8=Q # 5.a7-a8=R # 1...b7-b6 {a} 2.Ka4-b4 {A} zugzwang. 2...b6-b5 3.Kb4-c5 threat: 4.Kc5-d6 threat: 5.Kd6-e7 # { (C+ by Popeye 4.75)} Tries with different refutations of b.Pb7. Key allowing check and transformation into Pawn. Vladimirov paradox and two royal mates. Rex solus in Three men problem. (Note: Cooked by normal King!) (Author) No.1078 Karol Mlynka Slovakia original – 24.05.2016 Solutions: (click to show/hide) White Pd5 Kf5 Bh1 Black Ra8 Ke8 Pa7 Pe7 Pg7 Pd6 h#3              2 solutions            (3+6) White Super-Transmuting King 1.e7-e6+? Kf5*e6=P! 1.e7-e5 d5*e6 ep. 2.0-0-0 e6-e7 3.Kc8-b8 e7*d8=Q # 1.g7-g6 + Kf5*g6=P 2.Ra8-d8 g6-g7 3.Rd8-d7 g7-g8=Q # { (C+ by Popeye 4.75)} Valladao theme with transformation of Pressburger King. (Author) No.1079 Karol Mlynka Slovakia original – 24.05.2016 Solutions: (click to show/hide) White Pb5 Ke4 Pf3 Black Pa7 Pb7 Pf7 Kb6 Pe5 Pf4 ser-h#6        2 solutions           (3+6) White Super-Transmuting King Annan Chess 1.a7-a5 2.a5-a4 3.a4-a3 4.a3-a2 5.a2-a1=R 6.Ra1-e1 + Ke4-b4=R # 1.a7-a6 2.a6*b5 3.b5-b4 4.b4-b3 5.b3-b2 6.b2-b1=Q + Ke4*b1=Q # { (C+ by Popeye 4.75)} 2-fold Excelsior of b. Pa7 with different promotions in a Kindergarten problem. (Author)

### 4 Responses to No.1077,1078,1079 (KM)

1. seetharaman says:

1078 is nice with two related solutions… but idle WB in one solution is difficult to digest.

2. seetharaman says:

In 1079, it is a fairy condition which is idle in the second solution !

3. Karol Mlynka says:

It seems to me, that both fairy conditions are necessary in each solution of 1079. No solution without Annan Chess or Pressburger king!.

• seetharaman says:

I am sorry. You are right of course. Black king has too many flight squares but for Annan condition.