Grasshopper Chess: Maximummers

In Maximummer problems, Black is restricted to making only his longest moves. The moves must of course be legal, so they cannot be selfchecks (in problems with Gs this device is often used to cut a maxi-move short), and if the BK is in check the move must be the longest move to stop the check. If there are several equal length moves any of them may be taken. This device is often used to generate branching lines of play (variations) in direct-play problems.

A three-step diagonal move (length √18, where 18 = 3² + 3²) is longer than a four-step lateral move (4 = √16). A four-step diagonal move (length √32, where 32 = 4² + 4²) is shorter than a six-step lateral move (6 = √36). A five-step diagonal move (length √50, where 50 = 5² + 5²) is longer than a seven-step lateral move (7 = √49).

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Maximummer Mate in 3

10 men

P. R. Nielsen
Fairy Chess Review #10707 1957
Maximummer Mate in 3

Solution: 1.Bb8
1....Qa2 2.Ba7 Qa6 3.Rb8‡
1....Qg8 2.Gh8 Qc8 3.Ra7‡
1....Q×g2 2.Gh1 Qc6 3.Gg8‡

Comments: Three variations. Three WG pins of BQ


Maximummer Mate in 4

6 men

C. M. Fox
Chess Amateur 1930
Maximummer Mate in 4, (b) Se2 to f1

Solution:
(a) 1.Ke7 Gf7 2.Sd4 Ge8 3.Kd7 Ga8 4.Sc6‡
(b) 1.Sd2 Gc3 2.Sc4 Gc5 3.Sb6 Ga7 4.Sd7‡

Comments: Changed mate. Widespread pieces.

F. Hansson
Problemist Fairy Chess Supplement #47 October 1930
Maximummer. Mate in 4

Solution:
1.Sd5 Gd4 2.Rf2
Ga1 3.Rc2 Ge1 4.Se3‡
Gg1 3.Re2 Gc1 4.Sc3‡

Comments: Echo. Reflection in d file.

12 men

P. C. Taylor
Problemist Fairy Chess Supplement #587 October 1932
Maximummer. Mate in 4

Solution:
1.Gh1
1....Qb1 2.Pe4 Qb8 3.Gcf1 Qg3† 4.Gg2‡
1....Gg2 2.Ghf1† Gg6 3.Gf4 Gag4 4.Gh6‡

Comments: Crosscheck in first line. Trapped knight hurdle h4 in second line.

15 men

A. Belleli
Greetings from Egypt #19 1939
Maximummer Mate in 4

Solution: 1.Ge8
1....R×e3 2.Gd3 R×e7 3.Gd1 Re2 4.Gf3‡
1....Rg1 2.Bb4 Rg4 3.B×d2 Ra4 4.Gf3‡
1....Pc5 2.Gd6 R×e7 3.Gb4 Ra7 4.Bg2‡

Comments: Three variations and some shorter maxi moves.


Maximummer Mate in 5

12 men

S. Segenreich
feenschach #1335 1974
Maximummer Mate in 5

Solution: 1.Pxg4 Gh1 2.Sf3
2....Gc1 3.Sd4 Ge5 4.Se2 Ge1 5.Sc3‡
2....Ge1 3.Sd4 Gc5 4.Sc2 Gc1 5.Se3‡

Comments: Two variations. Symmetry and Asymmetry


Maximummer Selfmate in 3

8 men

G.A.Ekestubbe
Eskilstuna Kuriren 1943
Maximummer Selfmate in 3

Solution: 1.Gh1 Qb6 2.Gb5 Qg1 3.Gb1 (pinning Q) Qc1‡

Comment: Queen move along pin line, stopping short before capture of hurdle.

B.Rehm
Feenschach 1957
Maximummer Selfmate in 3

Solution: 1.Qd4 (double unpin) Bh4 2.Kf4 Qh7 3.Be7† Qe4‡

Comment: Antipin (stops Q×Qe4)


Maximummer Selfmate in 4

5 men

C.M.Fox
Die Schwalbe 1934
Maximummer Selfmate in 4 (a) diagram (b) Gh1 - c7

Solution:
(a) 1.Bg2 Gf3 2.Bf1 Kb3 3.Bd3 Gc3 4.Bb1 Gb2‡
(b) 1.Bc6† Kb3 2.Bb5 Kc2 3.Bc4 Gc3 4.Ba2 Gb2‡

Comment: Exact echo. Some short maximum moves.


Maximummer Stalemates

4 men

C. M. Fox (solution shortened from 22 to 19 by D. H. Hersom)
Problemist Fairy Chess Supplement #505 August 1932
Maximummer Stalemate in 19

Solution: 1.Kd3 Ka2 2.Kd4 Ge5 3.Kd5 Gc5 4.Gd6 Ge7 5.Kd4 Gc5 6.Ke4 Ge7 7.Gf8 Ge3 8.Kd3 Gc3 9.Kd2 Ge1 10.Kc3 Gb4 11.Kc2 Ka1 12.Kc3 Gd2 13.Kd4 Gd5 14.Ke4 Gf3 15.Gf2 Gd5 16.Kd3 Gd2 17.Kc2 Gg2 18.Gh2 Gb2 19.Kc1 stalemate.

Comments: Cycle by BG.

Notes: All Black moves are forced.

C. M. Fox (solution shortened from 22 to 20 by D. H. Hersom)
Problemist Fairy Chess Supplement #506 August 1932
Maximummer Stalemate in 20

Solution: 1.Kc4 Gc5 2.Kd5 Ge5 3.Gd6 Gc7 4.Kc4 Gc3 5.Kd3 Ge3 6.Kd4 Gc5 and the rest as in the previous problem from 6.Ke4 onwards (adding 1 to the move number)

Comments: Cycle by WG.

Notes: All Black moves forced. Diagram rotated 90 degrees clockwise from position as published.


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Composer
Source Problem number and date
Stipulation

Solution: 1. ‡

Comments: