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 Problem of the Week


315. William Whyatt
problem 1957
1st Prize
Mate in 2

The weekly problem’s solution will appear in the following week, when a new work is quoted.
See last week’s problem with solution: No.314.
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 Walkabout
Archives: 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016
Chess and problem rambles by PW


20 Nov. 2016 – Improving a fairy helpstalemate – Part 2


In the previous column, we looked at a helpstalemate featuring a royal knight and some potential ways to enhance the problem. First, by utilising the set play inherent in the position which contains a queen promotion, we saved a couple of units while preserving the triple-promotion theme. Then we attempted to add a knight-promotion set play by reinserting the white king, but the resulting position was spoiled by a dual. So would it be possible to compel a single move order in this set play? One option is to replace the orthodox d7-pawn with a royal one, a unit that is susceptible to check like its black knight counterpart. Given that a royal pawn promotes to a royal piece, an immediate 1…d8(rS) in the set play would be illegal because d8 is attacked by the black knight. Hence the pawn promotes only on the second move, after the black piece has moved to a5.

Two issues arise from this scheme, however: (1) the white king can no longer be used because of a convention against a player having more than one royal unit, and (2) the queen-promotion set play, intended to start with 1…d8(rQ), is rendered illegal as well. On the first issue, a good replacement for the white king is a fairy piece known as the wazir; it's a close relative of the king that moves only one orthogonal step at a time. Such a wazir on c3 would (like the king) guard b3 and c4 and so help to trap the royal knight on a5 (though unlike the king, it wouldn’t control b4 or any diagonally-adjacent square). Now we can test various placements of the wazir that are one step away from c3, and see if any of them would also resolve the second issue, the lack of a queen-promotion phase. Adding the wazir on c2 (where we put the king in Version B) produces numerous cooks in which the royal knight is stalemated on a2, e.g. 1.rSb4 d8(rQ/rR/rB) 2.rSa2 rQb6/rRb8/rBe7 (these lines were ineffective with the king on c2 because then 1.rSb4 would check). If we place the wazir on d3, the resulting position (diagram C) comes very close to fulfilling our aims. The set play 1…Wc3 2.rSa5 d8(rS) and two original solutions (rook and bishop promotions) all work as intended, and the wazir’s placement has generated new play that is precise and involves a queen promotion. But two such solutions have been created – 1.rSa7 d8(rQ) 2.rSb5 rQe7 and 1.rSb4 d8(rQ) 2.rSc2 rQa5 – one too many for our purposes!

Version C
Helpstalemate in 2
Royal knight c6
Royal pawn d7, Wazir d3
4 solutions and set play

Besides deploying a royal pawn, is there another way to force the move order of the knight-promotion set play? Yes, if the white unit whose function is to control b3 and c4 happens to be attacking a5 initially, then this unit must move first before the pawn promotes, so as to allow the royal knight to access that square. A white king cannot handle this task, however, because the only square from which it could both attack a5 and move to c3 is b4, but this square is guarded by the black knight. So let's return to the idea of using a wazir, and note that it has an alternative square from which to control b3 and c4, namely b4 (where unlike the king it maintains the stalemate by not checking the royal knight on a5). Now by adding the wazir on b5 or a4, it does the job of attacking a5 while being able to move to b4. If placed on b5, the piece would cause many cooks, including even a one-move solution, 1.rSa7 Wb6/Wc5. But if the wazir starts on a4, the ensuing play seems accurate while meeting all of our requirements (see diagram below). The knight-promotion set play is dual-free: 1...Wb4 2.rSa5 d8(S). Since only a conventional pawn is needed, the queen-promotion set play becomes viable again: 1...d8(Q) 2.Sa7 Qe8. And the full-length solutions, 1.rSe7 d8(B)+ 2.rSg8 Bg5 and 1.rSb8 d8(R)+ 2.rSa6 Rc8, complete an economical Allumwandlung.

Ramaswamy Ganapathi
The Problemist Supplement 2016
Version by Peter Wong
Helpstalemate in 2
Royal knight c6, Wazir a4
2 solutions and 2 set play

There is something curious about the rook-promotion solution, though. It is different from the one in the original problem: 1.rSa5 d8(R) 2.rSb7 Rd5. The wazir has prevented this solution by attacking a5, but in a remarkable stroke of luck, by also guarding b4 the piece has enabled another precise solution that necessitates a rook promotion!


5 Nov. 2016 – Improving a fairy helpstalemate – Part 1


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.

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


2 Oct. 2016 – What’s New


Alexander Goldstein (1911-1988) was one of the greatest Australian problem composers, best known for his three-movers. Bob Meadley has compiled all of Alex’s works in an e-book, titled Alexander Goldstein: His Life, His Chess Problems. (I helped with its editing process mainly by formatting the text and designing the layout.) This wonderful collection starts with ‘An Anecdotal Introduction’ by Ian Shanahan and ‘An Interview with Sophie Goldstein’, the latter a moving account of Alex’s life as conveyed by his widow. The next chapters feature more than two hundreds of Alex’s compositions – mostly directmates but also some helpmates and selfmates. These problems are accompanied by expert comments from Geoff Foster, Andy Sag, and Arthur Willmott. Alex’s chess writings are then presented, including a piece called ‘Miraculous Escape from a Siberian Mine’ which begins memorably with the lines, “Two chess problems saved my life. This is how it happened.” The book concludes with a large section of scans, comprising more of Alex’s articles and a variety of materials about him.

Alexander Goldstein
Chwila 1931
Mate in 3

To download this free e-book, use the link above or go to the Problemists and History page of this site. Here are two splendid examples of Alex’s works from the compilation. In the first, White must be mindful of Black’s bishop or rook giving check if either is unpinned, e.g. if 1.Sc1? aiming for 2.Sd3 and 3.Sf2, then 1…Bxb6+! refutes. The key 1.Sc3! waits for Black to self-obstruct with a pawn move. 1…cxb6 allows 2.Sd1 and 3.Sf2 since 2…Bxb6+ by the freed bishop is ruled out. Likewise 1…b2 means White can unpin the black rook with 2.Se4, and 3.Sf2 is unstoppable as 2…Ra2+ is blocked. One more variation involves a nice switchback, 1…h3 2.Se2 and 3.Sg3.

Alexander Goldstein
Parallèle 50 1949
1st Prize
Mate in 3

The second problem shows even more impressive strategy. First note that if White tries to unpin the d6-knight, then Sxb7 (a pin-mate) becomes a threat, but 1.Qf6? is defeated by 1…Bxd5! The key 1.Qf5! instead threatens 2.Qc8 and 3.Qa8. Black has three defences but they err by preventing …Bxd5, after which White’s unpinning plan becomes viable. If 1…Bh3, then White plays 2.Qe6 and 3.Sxb7 – not 2.Qf6?/Qg6? Bc8! After 1…e4, White chooses 2.Qf6 and 3.Sxb7 – not 2.Qe6?/Qg6? Bg7+! And 1…Rf3 must be followed by 2.Qg6 and 3.Sxb7 – not 2.Qe6?/Qf6? Rf6+! So three white queen moves with a similar motive are subtly differentiated. The by-play makes further use of the queen: 1…Sd7 2.Qxd7 and 3.Qxb5 – another pin-mate.

The two-mover by J.T. Eaton in the previous Walkabout column is solved by 1.Rb2! (releasing both black knights). However, Geoff Foster advises that this problem is exactly anticipated by Gerardus Drese, Elk Wat Wils 1935.