First prize problem by Geoff Foster and Ian Shanahan
7 May 2014 | by Peter Wong
The Aussie duo of Geoff Foster and Ian Shanahan have carried off another major First Prize, this time in the 2011 Fairies section of The Problemist. The winning problem is a series-mover showing a favourite theme of the composers: multiple pins and unpins. Earlier renditions of the idea can be found in the Walkabout columns dated 8 Nov. 2010 and 7 Aug. 2011. Here we see some extra fairy elements, like the nightriders and the absence of the white king, but they are well justified by the problem’s strategic intensity (nine unpins in ten moves) and pure construction.
Let’s consider the two types of unorthodox pieces used in this composition. A reflecting bishop, travelling on diagonal lines, is able to bounce off a board edge at 90-degrees and continue its move. Thus in the diagram, the piece on g6 has access to f7, e8 and also d7, c6 and b5 by reflecting off the top edge; in fact the reflecting bishop is pinning the nightrider on b5, without which the black king would be in check via the g6-e8-a4-c2 line. Likewise, the h5-nightrider is pinned along the g6-h5-d1-c2 line, and the h7-nightrider along g6-h7-g8-a2-b1-c2. The nightrider is a long-range piece, analogous to the rook and the bishop, that can make any number of knight-steps in a straight line as one move. For example, the h6-nightrider can move to g8, f7, d8, g4, and f2, but its access to d4 and b3 is obstructed by the f5-piece.
Geoff Foster & Ian Shanahan
The Problemist 2011, 1st Prize
Series-helpstalemate in 10, Reflecting bishop g6, 7 Nightriders
The series-helpstalemate objective indicates that Black will play ten consecutive moves to reach a position where White can deliver stalemate. What sort of stalemate position is possible with the material present? Assuming that White’s stalemating move will be a capture, we need to immobilise six black nightriders, all of which can be pinned – four by the reflecting bishop and two by the rooks. The black king has eight flight-squares, six of which can be blocked by the pinned nightriders, and if White plays Bxa3 at the end, the bishop will control the remaining flights on b2 and c1. However, this scheme has a snag, because pinning a nightrider on b1 requires the path g6-h7-g8-a2-b1 to be clear, meaning the b3-flight cannot be blocked. To get around this, we will pin a nightrider on a4 instead of b3 (via the g6-e8-a4-c2 line) – workable since b3 will be guarded directly by the reflecting bishop. (Note that pinning a nightrider on b5 instead would be ineffective for the stalemate, because the piece could move to f7 without allowing the reflecting bishop to check!)
Even after determining the final position, it’s far from easy to work out how it can be reached within ten moves. The solution shows an intricate sequence in which the nightriders unpin one another nine consecutive times. 1.Ng8 2.Nhf3 3.Nhd3 4.Nfd1 5.Nb1 6.Ngd2 7.Ngc4 8.Na4 9.N5c3 10.Nca3 Bxa3 stalemate. The judge writes, “There are no inactive units and no cookstoppers, and all nightriders move during the solution. A unique setting and strategy and my unambiguous first pick from the start.”