
E. Holladay
Solution: afile: 1Kb4 Bd5 2.Kc3Ka3 3.Kc2 Ga4 4.Kb1 Ba2 5.Ka1 Kb3‡
Comments: Ideal mates. 

T. R. Dawson
Solution: 1.Kg5 Gh5 2.Kh4 Kf2 3.Kh3 Gd1 4.Kh2 Kf1+ 5.Kh1 Gf3‡ Comments: Asymmetry 

G. P. Jelliss
Solution:
Comments: Asymmetry in (a), Gf1 can be at g1 or h1 in (b)


T. R. Dawson and C. M. Fox
Solution: (a) 1.Kd3 Gc2 2.Kc3 Kb5 3.Kb2 Ka4 4.Gb1 Ga2 5.Ka1 Kb3‡ (b) 1.Ke5 Kb5 2.Kd6 Gb1 3.Kc7 Gc8 4.Kb8 Ga6 5.Ga7 Kc6‡ (c) 1.Kd4 Kd7 2.Kc5 Gc6 3.Kb6 Ge4 4.Ka7 Kc8+ 5.Ka8 Gc6‡ (d) 1.Kf5 Kxd5 2.Kg6 Kc4 3.Kh7 Kb3 4.Kh8 Ka2 5.Gh7 Ga1‡ Comments: Two other twins in this set proved to be unsound. 

T.R.Dawson & C.M.Fox
Solution: 1.Kd5 Kf3 2.Ke6 Ke4 3.Kf6 Kd5 4.Kg7 Ke6 5.Kh7 Kf7 6.Kh8 Ng6‡ Comment: Same mate as Kemp HM4, but with WG further away. 

C.M.Fox
Solution: 1.Kb2 Ga1 2.Ga2 Ga3 3.Gd1 Gc1 4.Kb3 Ke5 5.Ga4 Kd4 6.Ka3 Kc3 7.Gb1 Ga1‡ Effect: Black Gformation rotated 90° from diagram to finale. 

Composer
Solution: 1. ‡ Comments: 