Improving a fairy helpstalemate – Part 1

5 Nov. 2016 | by Peter Wong

The helpstalemate problem below by the Indian composer recently caught my eye. It employs a fairy piece in effecting three different promotions in a very light setting. The unorthodox piece is a royal knight, which moves normally but is subject to check and mate like a king. For instance, after 1.rSb8 d8(Q)+, the royal knight is checked by the queen and it must move to a6 or c6 but not d7 where it would still be in check. The helpstalemate task means that Black plays first and helps White to give stalemate in the specified two moves. The solutions are 1.rSe5 d8(Q) 2.rSg4 Qd6, 1.rSa5 d8(R) 2.rSb7 Rd5, and 1.rSe7 d8(B)+ 2.rSg8 Bg5. The three promoted pieces deliver a variety of model stalemates by trapping the knight. While this is an appealing idea, I wondered if the problem could be tweaked to include a knight promotion, thereby completing the Allumwandlung theme. Moreover, the only purpose of the h4-pawn is to prevent a dual in the queen-promotion solution, which could have also finished with 2…Qh8 if the h-file were clear. Would it be possible to dispense with this uneconomical pawn? Let’s investigate, bearing in mind that because of our clear-cut goals and the small number of units involved, we can make use of trial and error rather effectively – more so than usual in problem construction. And naturally we will be aided by the solving program Popeye, capable of instantly confirming if a proposed position is sound or not.

Ramaswamy Ganapathi

The Problemist Supplement 2016

Helpstalemate in 2, Royal knight c6, 3 solutions

A simple way to look for additional play in such a problem is to consider what would occur if White has the move. Such set play does indeed exist in the diagram and it happens to be a unique line that entails a queen promotion: 1…d8(Q) rSa7 Qe8. Now if we adopt this set play in place of the full-length queen-promotion solution, we can actually remove the king and the h4-pawn, both of which are needed only in that one solution. The result is a two-unit problem – see diagram A – that still manages to show three different promotions. (In such a fairy problem with no black king, the "illegality" of a position lacking the white king is moot.) Admittedly, the original queen-promotion solution is more interesting in that the royal knight is stalemated not on an edge square, and the play utilises the board more fully as well. Still, I personally would have preferred the ultra-economy of having just the two thematic units. In any case, let’s continue with our attempt to incorporate a knight promotion, by building on this two-unit position.

Version A

Helpstalemate in 2, Royal knight c6, 2 solutions and set play

As the knight is a relatively weak piece, it's usually not trivial to “force” a white knight promotion aimed at confining Black. A prospective scheme here is to stalemate the royal knight on a5, from which its access to b7 and c6 would be immediately covered by a knight promotion on d8. That would leave only two other flights on b3 and c4 to be guarded, most naturally by a white king on c3. Hence if the king were to be placed one step away from c3, we can envisage an additional set play, 1…Kc3 2.rSa5 d8(S). The best square for the king is c2, where it doesn’t affect the other set play and solutions, or create any cooks (diagram B). Unfortunately, this position is unsound because of the dual, 1...d8(S)+ 2.rSa5 Kc3. Clearly we have to ensure that the knight-promotion set line has only one move order. How we go about this will be examined in the next instalment.

Version B (unsound)

Helpstalemate in 2, Royal knight c6, 2 solutions and 2 set play