Home
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


218. Juan Kloostra &
Denis Saunders

The Problemist Supplement 1999
Mate in 3

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.217.
See previous Problems of the Week without solutions: Page 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9.

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.


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


26 Dec. 2014 – Helpmates of the ‘distant’ future – Part 2


This month I’ll provide two early examples of ‘helpmates of the distant future’ (somewhat oxymoronically), followed by an amazing update to the previous column. The problemist Chris Feather originated the term ‘Helpmates of the Future’ in 2000 when he produced a booklet by that name, which surveys the field with more than one hundred illustrations. As a composer he not only pioneered the basic HOTF scheme (in the 1970s!), but he also created a very early rendition of the idea in its expanded form, as given below.

Chris Feather
The Problemist 1997
Special Prize
Helpmate in 2
3 solutions
(b) Swap Kc4 & Rf4

The focus of this problem is the half-battery arrangement on the fourth rank. In the three solutions of the diagram position, the d4-bishop and e4-knight are either moved or captured to activate the white rook. 1.Kxd4 Sxc3+ 2.Kc5 Sxa4, 1.Qxe4 Bxb6 2.Qd4 Rxd4 & 1.Qc6 Bxe3 2.Kxb5 Sxc3. The twin exchanges the black king and white rook so that the half-battery is pointing the other way. The resulting play shows strategic effects that precisely match those seen in the first triplet, to generate three pairs of corresponding solutions. (b) 1.Kxe4 Bxe3+ 2.Kxe5 Bf4, 1.Qxd4 Sd2 2.Qe4 Rxe4 & 1.Qe6 Bxc3 2.Kf5 Sd6.

János Csák
Orbit 2002
1st Prize
Helpmate in 2
6 solutions

The second example makes an interesting comparison with the six-phase helpmate we looked at last month. Though both works feature the same couple of direct batteries (R + P and B + P) the thematic pieces are employed quite differently, so this older problem is only a precursor and not an anticipation. 1.c1(B) Bf6 2.Bd2 cxb5 & 1.f1(S) Rc7 2.Sd2 d5. 1.Rxc6 bxa8(Q) 2.Kxc4 Qxc6 & 1.Bxe5 b8(Q) 2.Kxd4 Qxe5. 1.Sd5 Rxg6 2.Kxc4 Rc6 & 1.Se3 Bxb8 2.Kxd4 Be5. Elements of play include the mixed promotions, white switchbacks, and reciprocal captures by the black and white rook/bishop pairs.

Emil Klemanič
Ladislav Salai Jr.
Michal Dragoun
Kalyan Seetharaman &
Nikola Predrag
The Problemist 2014
Helpmate in 2
8 solutions

In a remarkable development, the ‘distant future’ helpmate cited in the previous column has been improved to show no less than four pairs of analogous solutions. Another two composers contributed to the collaborative effort that accomplishes this wonderful feat. Though the two added phases don’t have a formal connection with the rest of the problem (lacking moves that recur elsewhere with changed functions), that is of little importance. The new play involves Black capturing the e5- and c4-pawns for the purpose of self-block; this is nicely differentiated from another pair of solutions in which the captures of the same pawns are motivated by square-clearance. I will refrain from concocting a name for this super format of 4 x 2 related phases!

1.Rb5 axb5 2.a4 c5 (A) & 1.Qh7 gxh7 2.g6 e6 (C).
1.gxf6 c5+ (A) 2.Kxe5 f4 (B) & 1.axb4 e6+ (C) 2.Kxc4 b3 (D).
1.Sf7 f4 (B) 2.Sxe5+ Bxe5 & 1.Ba6 b3 (D) 2.Bxc4 Rxc4.
1.Bg3 Be7 2.Bxe5 Bc5 & 1.Sxb2 Rb3 2.Sxc4 Rd3.


29 Nov. 2014 – Helpmates of the ‘distant’ future – Part 1


The helpmate was invented in the mid-19th century, and early examples of the form usually consist of a single line of play with modest thematic content. Over the years the genre evolves and the problems become more elaborate and varied. Now the modern helpmate (at least in two- and three-movers) typically involves two solutions that are strategically rich and interrelated. The trend towards greater complexity continues with the arrival of ‘helpmates of the future’ (HOTF). This is the name given to a scheme in which a problem contains not one but two pairs of analogous solutions. Helpmates of this sort are of course quite demanding to compose, and it’s probably not a coincidence that they came to prominence around the turn of the millennium, when computer-testing of problems became commonplace.

The British Chess Problem Society held a HOTF tourney in 2001-03, and the First Prize went to the problem shown below. It features two white indirect batteries (R + B on the file and B + R on the diagonal), which are utilised or dismantled in a variety of ways. The diagram position is solved by 1.fxe5 Rc5 2.exd4 Bc4 and 1.Rxd5 Bf4 2.Rxd4 Re3. The twin replaces the d4-pawn with a black queen, which converts the solutions to 1.Ke3 Bg6 2.Qf4 Bd4 and 1.Kc4 Rb8 2.Qc5 Rd4. While the two solutions of each pair show matching play, the two pairs are themselves strategically distinct – such an element of contrast is considered desirable or even necessary in HOTF. The judge of the tourney praised this work for its creative twinning, originality and harmonious play.

Franz Pachl
HOTF Tourney 2001-03
1st Prize
Helpmate in 2
2 solutions
(b) BQd4

What could be the next step in this development of increasingly intricate and dense helpmates? One easy answer – which is anything but easy to arrange – is to have a problem incorporate three pairs of related solutions. We might call this type, ‘helpmates of the distant future,’ and a rare example recently appeared in The Problemist. The play of this superb composition revolves around two white batteries (R + P and B + P) aimed at the black king. In each of the three pairs of solutions, these batteries are exploited in analogous ways, to bring about a three-fold orthogonal-diagonal transformation.

Emil Klemanič
Ladislav Salai Jr. &
Michal Dragoun
The Problemist 2014
Helpmate in 2
6 solutions

1.Sb5 axb5 2.a4 c5 (A) & 1.Rh7 gxh7 2.g6 e6 (C).
1.gxf6 c5+ (A) 2.Kxe5 f4 (B) & 1.axb4 e6+ (C) 2.Kxc4 b3 (D).
1.Re8 f4 (B) 2.Rxe5 Bxe5 & 1.Qe2 b3 (D) 2.Qxc4+ Rxc4.

What makes this helpmate even more special is how certain moves recur in different solutions with new functions, to create another kind of pattern. In the three solutions listed on the left-hand side, the white pawn moves c5 and f4 (labelled ‘A’ and ‘B’ respectively) act variously as a first move and as a mating move, and a similar role reversal of e6 and b3 (‘C’ and ‘D’) occur in the right-hand group. That means the three solutions of each group, while showing different strategy, have a formal connection that enhances the coherence of the problem.


6 Oct. 2014 – Composing event in the Australian Junior Chess Championships


The national Junior Chess Championships, to be held in Canberra early next year, include a new problem composing event which has just been launched. This competition, an initiative of Nigel Nettheim, is conducted online and open to everyone, not only juniors. Entrants can download the paper from the AJCC site now and work on the questions at their leisure. There are four composing tasks, which I have arranged at Nigel’s request. For each task, you are given an incomplete (or incorrect) position of a mate-in-two problem (or mate-in-three in one case), and your aim is to finish its construction. Since hints are provided and you don’t start the composing process from scratch, we have called this a Guided Chess Problem Composing Competition.

To illustrate, here’s the position for one of the slightly harder tasks. It’s a three-mover that is solved by 1.Qg4!, with the threat of 2.Qe4. Black has three defences, 1…c4, 1…Kc6 and 1…Ke5, but one results in a flawed variation. 1…Ke5 produces a dual in that White can continue with 2.Sc8+, 2.Kg6, or 2.Kg7, any of which would force mate on the third move. How could you modify the position to remove the dual, so that only one of these white second moves works? The rest of the problem’s play must be preserved. Further details about this task, including the full solution of the diagrammed three-mover, are provided in the paper.

Mate in 3 (with a dual)

The closing date for entries to this event is the 7th December 2014. For more information, go to the Guided Problem Composing page of the AJCC site, where you will also find a link to a useful article by Nigel, “A Quick Introduction to Chess Problem Composition”.


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.

Mark Kirtley
Die Schwalbe 2013
SPG in 13½
2 solutions

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.

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