In Maximummer problems, Black is restricted to making only his longest moves. The moves must of course be legal, so they cannot be selfchecks
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.
A threestep diagonal move (length √18, where 18 = 3² + 3²) is longer than a fourstep lateral move (4 = √16).
A fourstep diagonal move (length √32, where 32 = 4² + 4²) is shorter than a sixstep lateral move (6 = √36).
A fivestep diagonal move (length √50, where 50 = 5² + 5²) is longer than a sevenstep lateral move (7 = √49).
Maximummer Selfmates are now on a separate page.
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P. R. Nielsen
Solution: 1.Bb8
Comments: Three variations. Three WG pins of BQ 

C. M. Fox
Solution:
Comments: Changed mate. Widespread pieces. 

F. Hansson
Solution:
Comments: Echo. Reflection in d file. 

P. C. Taylor
Solution:
Comments: Crosscheck in first line. Trapped knight hurdle h4 in second line. 

A. Belleli
Solution: 1.Ge8
Comments: Three variations and some shorter maxi moves. 

S. Segenreich
Solution: 1.Pxg4 Gh1 2.Sf3
Comments: Two variations. Symmetry and Asymmetry 

C. M. Fox (solution shortened from 22 to 19 by D. H. Hersom)
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)
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. 

Composer
Solution: 1. ‡ Comments: 