Welcome to OzProblems.com, a site devoted to the chess problem art in Australia! Whether you’re a player who is new to composition chess or an experienced solver looking for challenging problems, we have something for you. Our aim is to promote the enjoyment of chess problems, which are at once interesting puzzles and the most artistic form of chess.

 Problem of the Week

200. Peter Wong
U.S. Problem Bulletin 1988

White retracts 1 move,
and then mates 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. See previous Problems of the Week without solutions: Page 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8.

Use the contact form on the About page to comment on a problem you’ve solved. You can also subscribe to OzProblems updates – get an email notification when a new problem or Walkabout column is added.

Archives: 2010 | 2011 | 2012 | 2013 | 2014
Chess and problem rambles by PW

30 Aug. 2014 – “Open problems” in proof games

The latest issue of the German problem magazine feenschach contains an article named “A compilation of some fascinating open problems in the Proof Game genre,” by the French expert Nicolas Dupont. He proposes more than fifty “open problems” or composing tasks indicative of the ultimate ideas that could be achieved in shortest proof games. These are challenges inspired by existing proof games that demonstrate some sorts of record effects, and your goal is to better them or to accomplish some related tasks. The article cites sixty-two such “extreme” proof games, and here I present two of them.

Many established themes in proof games relate to pawn promotions. A classic example is the Pronkin theme, in which a piece seemingly on its game-array square turns out to be a promoted pawn, replacing the original piece that has been captured. The highest number of such Pronkin pieces to be rendered in a proof game is four, and the best problem to have attained that figure is perhaps the one shown below. Here the four white thematic pieces are remarkably all of different types, so that the Allumwandlung theme is featured as well. The paradoxical white promotions are necessary to assist Black in positioning its pieces in time. Thus White must sacrifice two pieces to the b5- and c6-pawns quite early to allow Black to develop on the queen-side, and this necessitates two promotions to replace the captured pieces. Further, Black has promoted to two minor pieces, and each has time to make only one move after the promotion, viz. …Sg1-h3 and …Bd1-h5. To facilitate this plan, White must leave the original g1-knight and d1-queen to be captured on their home squares by the promoting pawns. Consequently, White must promote two more pawns to substitute for these pieces.

Nicolas Dupont & Gerd Wilts
Probleemblad 2009
Ded. to Andrey Frolkin
& Dimitri Pronkin

SPG in 31½

The solution is 1.e4 a6 2.Bb5 axb5 3.h4 Ra6 4.h5 Rg6 5.h6 Sf6 6.hxg7 h5 7.a4 h4 8.a5 h3 9.a6 h2 10.a7 hxg1(S) 11.Ra6 Sh3 12.Rc6 dxc6 13.e5 Kd7 14.e6+ Kd6 15.exf7 e5 16.f4 e4 17.f5 Ke5 18.g8(B) Bc5 19.f8(S) e3 20.Bc4 Be6 21.a8(R) Sd7 22.Ra1 Qa8 23.Sh7 Rd8 24.Bf1 Se8 25.f6 e2 26.f7 exd1(B) 27.f8(Q) Bh5 28.Qf3 Bb3 29.Qd1 Kf4 30.Sg5 Se5 31.Sf3 Rdd6 32.Sg1. So the a1-rook, d1-queen, f1-bishop, and g1-knight are all promoted – an incredible conception showing tremendous technical skill on the part of the composers. The open problem in this case is not to improve on the composition (that might be impossible!), but to realise other combinations of four-fold Pronkins, such as two white and two black rook promotions.

Not all of the proof games discussed in the article are “blockbusters” like this promotion extravaganza. Others display more elegant types of themes but which still involve a maximum task of some sort, as in our second selection. In proof games with two solutions, a difficult changed-play scheme consists of a player castling in one sequence of play but not in the other, with both options leading to the same diagram position. Mark Kirtley’s problem doubles this idea, impressively bringing about a reciprocal change of castling between White and Black. 1.c4 Sf6 2.Qc2 Sh5 3.Qxh7 f5 4.Qxg7 Bxg7 5.Sf3 Bxb2 6.Bxb2 Kf7 7.Bd4 Re8 8.Bxa7 b6 9.Sd4 Bb7 10.Sb3 Bxg2 11.Bxg2 Kg8 12.Bc6 Sxc6 13.0-0 Rb8 14.Kg2, and 1.c3 Sf6 2.Qc2 Sh5 3.Qxh7 f5 4.Qxg7 Bxg7 5.c4 Bxb2 6.Bxb2 0-0 7.Bd4 Re8 8.Bxa7 b6 9.Sf3 Bb7 10.Sd4 Bxg2 11.Sb3 Bxf1 12.Kxf1 Sc6 13.Kg2 Rb8 14.Rf1. The challenge inspired by this work also seems very hard to fulfil: construct a two-solution proof game in which White makes an en passant capture in one phase, and Black does so in the other.

Mark Kirtley
Die Schwalbe 2013

SPG in 13½
2 solutions
You can view the complete “open problems” article as a free PDF-file from the feenschach site.

14 Aug. 2014 – What’s New

Canberra resident Molham Hassan has contributed a profile article to the Australian Problemists section. A surgeon by profession, he got involved in chess problems in the 1970s, and has been composing regularly since. He produces mostly traditional two- and three-movers, and eight of his best works are presented here.

Nigel Nettheim has provided another instalment of problem columns from the Australasian Chess Review. Covering the year 1944, the file can be viewed or downloaded from the Problem Magazines and Columns page. Previously Nigel has also re-processed some existing materials from the Oz Archives to make the PDF-files more accessible in size, such as the ACR 1930a/b and 1931a/b, and all of the Australasian Chess Magazine. Thanks, Nigel!

And here I quote
another amusing
comic from xkcd,
titled “Explorers.”

15 Jun. 2014 – Solving a four-move directmate, and the ‘Check!’ magazine

Two problemists who are regular solvers of the Weekly Problems, Dennis Hale and Nigel Nettheim, have sent in materials to share on this site. Dennis forwarded the four-mover below, composed by a Spanish expert (whose quality works have appeared in the FIDE Albums), for me to solve. The position looks rather heavy and daunting at first sight, but it turns out to be not too hard to unravel. And the problem has a sparkling main variation that makes it very quotable. I encourage you to try tackling it before reading on.

The black king has two orthogonal flights, suggesting that White will aim for a queen or rook mate on the h-file. The white queen seems out of play, and 1.Qa1! looks promising in view of 1…Bxa1 2.Rc1 and 3.Rh1. The threat is 2.Qxb1 Bc1 3.Q/Rxc1 and 4.Q/Rh1. Black’s defence 1…Rxd3 initiates the thematic variation, 2.Rc1 Bxc1 3.Rxg5 fxg5 4.Qh8. Five pieces are impressively diverted from the long diagonal, to allow the queen to pass through and mate with the longest possible move on a chessboard. If 2…Rd1, White answers with 3.Rxd1 and 4.Rh1. Notice how White cannot shuffle the move order, e.g. 1…Rxd3 2.Rxg5? fxg5 3.Rc1 is stopped by 3…Rd1! A second full-length variation goes 1…c5/c6 2.Qa7 Sc7+ 3.Qxc7 and 4.Qh7.

Valentin Marin y Llovet
Deutsche Schachzeitung 1902

Mate in 4
The Oz Archives section of this site brings together almost all of the problem columns found in early Australian chess magazines. Gaps exist, however, and Nigel has kindly offered to provide some materials that were missing and which he has scanned from his magazine collection. The first lot of these files is now available for download. It is a complete run of the problem columns conducted by Frederick Hawes in the magazine Check!, which appeared from July 1944 to December 1945. Check it out on the Problem Magazines and Columns page.

7 May 2014 – First prize problem by Geoff Foster and Ian Shanahan

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 8/11/10 and 7/8/11. 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.

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!)

Geoff Foster &
Ian Shanahan
The Problemist 2011
1st Prize

Series-helpstalemate in 10
Reflecting bishop g6
7 Nightriders
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.”