November 17th

Михаил Владимирович Тронов (17-11-1892 - 07-12-1978) Russian composer (Mikhail Vladimirovich Tronov)

Mikhail Tronov
[Wikipedia]

Mikhail Tronov was a climatologist and composed direct mates in the Bohemian style.

Тронов, Михаил Владимирович
Шахматы 1926
3^rd Prize

#3

10 + 9

Show Solution

1.Qf1-d3 ! (2.Qd3-e4 [3.Bd5-b7#] 2...d6xc5 3.Qe4-e6#)

1...Rh3-h6/Rh3xh7 2.c5xd6 ~ 3.c4-c5#
1...d6xc5 2.Qd3-g6+ e7-e6 3.Qg6xe6#
1...e7-e5 2.Qd3-c2 (3.Qc2-a4#) 2...d6xc5 3.Qc2-g6#
1...e7-e6 2.Bd5-a8 (3.Qd3xd6#) 2...d6-d5 3.c4xd5#

A fine set of mates.

Enguelberto Berlingozzo (17-11-1905) Brazilian composer

Enghelberto Berlingozzo composed mostly helpmates, but also direct mates or selfmates, some of his problems in collaboration with Felix Sonnenfeld.

Berlingozzo, Enguelberto & Sonnenfeld, Felix Alexander
Memorial J. Figueiredo, 1980
1^st Prize

h#2

3.1.1...

6 + 11

Show Solution

1.Sd3-e1 Qf2-f5 2.Se1-c2 Rc4-d4#
1.Sd3-f4 Qf2-c2 2.Sf4-d5 Rc4-e4#
1.Sd3xc5 Qf2-d4 2.Sc5-e4 Rc4-c2#
In each solution the black Knight interferes with one of the black pieces Bb1, Qg6 or Rd5; one of these three pieces is interfered by the white Queen and finally the white Rook interferes with the last of these third pieces: cyclic interferences.

Николай Дмитриевич Надеждин (17-11-1938) Russian composer (Nikolay Dmitryevich Nadezhdin)

Nikolay Nadezhdin composes direct mates, mostly in two moves.

Надеждин, Николай Дмитриевич
E Heinonen-50 JT, 1974
2^nd Prize

#2

9 + 8

Show Solution

1.Sd4? (2.Be6#[B])
1...Bxd7[a] 2.c6#[D]
1...e4[c] 2.Qh5#[C]
1...exd4 2.Qe6#
but 1...c6[b]!
1.Qe3? (2.Se7#[A])
1...e4[c] 2.Qd4#
but 1...Bxd7[a]!
1.Qg4? (2.Se3#[E])
1...Bxd7[a] 2.Se7#[A]
but 1...e4[c]!
1.Kf6! (2.Qxe5#)
1...Bxd7[a] 2.Se7#[A]
1...c6[b] 2.Be6#[B]
1...e4[c] 2.Ne3#[E]
Dombrovskis.

Vladimir Janál (17-11-1964) Czech composer

Vladimir Janál
[Chess-db]

Vladimir Janál is specialized in long problems, especially series movers, where the aim may be mate, stalemate or capture. He is thus the author of a world record minimal for series-helpstalemate stipulation: sh=99 with a lone black King.
In the below problem two black units can move and the move order is logically determined:

Janál, Vladimír
Blog zlínského problemisty, 21^st Jun 2009

Ser-h=75

10 + 2

Show Solution

In order to reach the final position with bKg4, Black must eliminate wPh3, therefore also wRh1, and wSd4. This is achieved in several steps and systematic manoeuvres:
- the bB reaches g1 to interfere with wRh1
1.Ke1-d2 2.Kd2-c3 3.Kc3-b4 4.Kb4-a5 5.Ka5-a6 6.Ka6-b7 7.Kb7-c8 8.Kc8-d8 9.Kd8-e8 10.Ke8-f8 11.Kf8-g7 12.Kg7-f6 13.Kf6-g5 14.Kg5-f4 15.Bh4-g3 16.Bg3-h2 17.Bh2-g1 
- bK goes back to capture wRh1
18.Kf4-g5 19.Kg5-f6 20.Kf6-g7 21.Kg7-f8 22.Kf8-e8 23.Ke8-d8 24.Kd8-c8 25.Kc8-b7 26.Kb7-a6 27.Ka6-a5 28.Ka5-b4 29.Kb4-c3 30.Kc3-d2 31.Kd2-e1 32.Ke1-f1 33.Kf1-g2 34.Kg2xh1 
- bK goes back again to allow the switchback of bB to h4
35.Kh1-g2 36.Kg2-f1 37.Kf1-e1 38.Ke1-d2 39.Kd2-c3 40.Kc3-b4 41.Kb4-a5 42.Ka5-a6 43.Ka6-b7 44.Kb7-c8 45.Kc8-d8 46.Kd8-e8 47.Ke8-f8 48.Kf8-g7 49.Kg7-f6 50.Kf6-g5 51.Kg5-f4 52.Bg1-h2 53.Bh2-g3 54.Bg3-h4
- bK goes back again to capture wPh3
55.Kf4-g5 56.Kg5-f6 57.Kf6-g7 58.Kg7-f8 59.Kf8-e8 60.Ke8-d8 61.Kd8-c8 62.Kc8-b7 63.Kb7-a6 64.Ka6-a5 65.Ka5-b4 66.Kb4-c3 67.Kc3-d2 68.Kd2-e1 69.Ke1-f1 70.Kf1-g2 71.Kg2xh3 
- finally a few moves are needed to reach the stalemate position:
72.Kh3-g4 73.Bh4-f6 74.Bf6xd4 75.Bd4-e3 Ba7xe3 =