Original Problems (46)

Original Problems (page 46)

Original fairy problems published during 2012 will participate in the informal tourney JF-2012

The site is mostly about fairies, but h# and s# are also welcomed for publication! Please send your problems to my e-mail: julia@juliasfairies.com

Go to →List of Problems; →Page 45 ;  →Page 47

I’d like to present you another series of problems by Emmanuel Manolas – this time with Maximummer condition! I’d like to leave the commenting of these problems to the readers who knows the mystery of Maximummer and selfmates, please!

See comlicated Emmanuel‘s problems:

No.96 – s#17

No.97 – s#11

Definition:

Maximummer: Black must play the geometrically longst move or may choose from among longest moves of equal length, distances being measured from the center of each square. Diagonal and oblique distances are measured from the orthogonal coordinates by using Pythagora’s theorem (take the square root of the sum of the squares of the orthogonal distances). All other orthodox chess rules apply.

Degradation Chess (Relegation Chess): Any piece (not King) entering its own side’s pawn-rank is changed into a pawn.

You can сlick on “Solutions” to show or hide the solutions!

 No.96 Emmanuel Manolas Greece original – 14.08.2012   s#17*                                            (2+7) Maximummer     Solutions: (click to show/hide)   Set play : 1…Qh1#   Try : {1.d5? Qa1!}    Key : 1.Kh7!  1…Qh1+ 2.Kg6 Qa8 3.Kg5 Qh1 4.Kf4 Qa8 5.Ke3 Qh1 6.Kd3 Qa8 7.Kc2 Qh1 8.Kxb2 Qa8 9.Kc2 Qh1 10.Kd3 Qa8 11.Ke3 Qh1 12.Kf4 Qa8 13.Kg5 Qh1 14.Kg6 Qa8 15.d5 Qa1 16.Kh7 Qh8+ 17.Kxh8 Qh4#   Switchbacks of bQ visiting 4 corners, Round-trip of wK.  (Author) No.97 Emmanuel Manolas Greece original – 14.08.2012   s#11*                                           (2+5) Maximummer     Solutions: (click to show/hide)   Set play : 1…Rh4#   Key : 1.Kh7! 1…Rh4+ 2.Kg6 Ra4 3.bxa4 Qb1+ 4.Kg5 Qh7 5.a5 Qb1 6.a6 Qh7 7.a7 Qb1 8.a8=R+ Qb8 9.Kh6 Qe8 10.Ra6 Qa4 11.Rg6 Qh4#    Switchbacks of bQ, Pelle movement, under-promotion, black Chernous theme. (Author)
 No.98 Emmanuel Manolas Greece original – 14.08.2012   sr#25                                           (3+9) Degradation Сhess Maximummer     Solutions: (click to show/hide)   1.c4 2.c5 3.c6 4.c7 5.c8=B 6.Ba6 7.Bxe2(=wPe2) 8.exf3 9.f4 10.f5 11.f6 12.f7 13.f8=B 14.Bc5 15.Bxf2(=wPf2) 16.f4 17.f5 18.f6 19.f7 20.f8=B 21.Ba3 22.Bxb2(=wPb2) 23.b4 24.b5 25.b6 Bxb6#   Three times Excelsior, three under-promotions to Bishop, three inverse promotions. (Author)

The diagrams are made on WinChloe and its Echecs font is used for Logo design

7 Responses to Original Problems (46)

1. seetharaman says:

Very nice problems from Manolas! I like the pointed ideas in these three problems (96, 97, 98). The queen visiting four corners is rare I think, even with Maximummer. The second queen unfortunately seems necessary to restrict the white King.
The play of white in 98 is surprising, with tripled minor promotion and conversion of Bishop to pawn ! But I wonder if the reflex and maximummer condition are necessary? Can it be done as a series-helpmate without the reflex or maximummer conditions?

2. Thank you Mr. seetharaman for your comment!
The problems 96 & 97 are positionally related. In 96, bQb4 makes unique the approach of wK to b2 *and* gives mate. In 97, line piece on b4 is replaced by a bR which in 3rd move is captured.
In 98, without Maximummer conditioning, the Black can mate wK by promoting a pawn to bQ (actually there are 54081 solutions). With stipulation ss#25 there is no solution.
Correct me if I am wrong, but in a series-helpmate the bK is mated. Do you suggest a reversal of colors?

• seetharaman says:

Congrats again for the nice idea of your No.98. In this problem the reflex condition is not used in the solution …. that is, white’s underpromotions or the order of moves are not determined by the reflex condition. Therefore my suggestion was indeed to reverse the colors and set it as SH#25. The moves will be same. With Maximummer condition to avoid the cooks, I think it is sound. Alternatively, you can try to force the underpromotion for the reason that White will have to mate black if promotion is to queen.

• With position turned 180 degrees, color of pieces reversed, [1BB3R1/2PP1pP1/2P5/7p/3P3k/8/8/7K], conditions Degradation & Blank Maximummer, stipulation sh#25, the problem has similar solution, [1.f5 2.f4 3.f3 4.f2 5.f1=B 6.Bh3 7.Bxd7(=bP) 8.dxc6 9.c5 10.c4 11.c3 12.c2 13.c1=B 14.Bf4 15.Bxc7(=bP) 16.c5 17.c4 18.c3 19.c2 20.c1=B 21.Bh6 22.Bxg7(=bP) 23.g5 24.g4 25.g3 Bxg3#], as you correctly have stated. Thanks!

• correction : Reverse Promotion & White Maximummer

• seetharaman says:

Reverse promotion ?