Orthodox Helpmates

G. P. Jelliss The Problemist Apr-Jun 1976

Helpmate in 2 (duplex)

My first composed helpmate in orthodox chess received a 'Commendation' in the Tourney for Helpmates in 2 (duplex) (The Problemist April-June 1976, pp.37-8) for which it was composed.

"Duplex" means that there is one solution with Black moving first and White giving mate, as is usual in helpmates, and another solution with White moving first leading to a mate by Black.

Solution:Black 1.B×g3 R×g7 2.Bh2 Bd4‡
White 1.R×g7 Bd4 2.Rg8 Rh2‡

Comments: 4th Commend. "Ultimately the masked half-pin is being exploited. Who will show a complete half-pin in a duplex on both sides?" Shlomo Seider.

G. P. Jelliss The British Chess Magazine Nov 1976 #11085 p.528
(solution Apr 1977 p.191)

Helpmate in 2½ (2 ways)

Solution:1.Re6 N3d4 2.f4 Nd6 3.Re5‡
and 1.Bd3 Ne5 2.f3 Nd4 3.Be4‡

Comments: "Marvellous find; variation of self-blocks is neat, but what really makes the problem is the differentiation of WP moves" D. A. Smedley.

Modal change (i.e. interchange of orthogonal and diagonal aspects) is one of my favourite themes.

G. P. Jelliss The Problemist Jan-Feb 1978 p.194 #H579
(solution May-Jun 1978 p.228-9)

Helpmate in 3½

Solution:1.b4 Ra3 2.Kc1 Ra5 3.Kd2 Re5 4.c3‡

Comments: Unfortunately this problem was completely anticipated by Eugene Albert, Schach 1966.
It also appears as Problem 432A in Albert's Ideal Mate Chess Problems 1966.

This anticipation, and the one in the next composition, discouraged me from composing further orthodox helpmates, because of the difficulty of knowing what has been done.

G. P. Jelliss The Problemist Jan-Feb 1978, p.194 #H580
(solution May-Jun 1978 p.229)

Helpmate in 4

Solution:1.Bf1 Ra8 2.Bb5 Rxh8 3.Ka4 Rh1 4.Rb4 Ra1‡

Comments: "Rundlauf of considerable elegance" B. P. Barnes. "Flying fortress" R. Brain. Also quoted in lecture by C. G. Rains, Sep 1979 p.359.

Unfortunately anticipated by A. Havasi, Fairy Chess Review 1944 (BKd6, Rh3, Bh8, WKc2 Qe1, Ra1) showing the task with only 6 pieces but using BQ. Also anticipated by an earlier example by T. R. Dawson Bolton Football Field 15 Jun 1912, version (BKf5, Rf1, Rg4, Ba8, Pg3, WKb3, Rh1).

G. P. Jelliss Problem Observer 1984, First Prize.

Helpmate in 3, (b) a7 to e8

Solution:(a) 1.Nb5 Ba7 2.Qb6 Re6+ 3.Kc5 Bxb6‡
(b) 1.Nf6 Re8 2.Qe7 Bc5+ 3.Ke6 Rxe7‡

Quoted in The Problemist May 1986, #20 p.175 "Recently honoured work by British composers".

Comments: Line clearance ... very elegant and well-matched" P. S.Valois.

G. P. Jelliss The Problemist Sep 1989 p.113 #H1371
(solution Jan 1990 p.174)

Helpmate in 3 (2 ways)

Solution:1.Kc5 Ke5 2.Rb4 Sd4 3.Rcc4 Sd7‡
and 1.Rd6+ Kf5 2.Kd5 Se6 3.Rcc6 Se3‡

Comments: Exact echo, showing a "transflection" (translation combined with reflection).
"Very good that all pieces move in both solutions" K. Valtonen.