28 Jan. 2014 – ‘Problem Potpourri’ draws to a close, and two selections
In a sign of the times, the magazine Australasian Chess has ceased publication at the end of 2013, apparently a print media casualty of the encroaching Internet. Regrettably this also means the closure of ‘Problem Potpourri’, the excellent column run by Geoff Foster that’s been showcasing new works from both Australian and overseas composers. Since we don’t have a successor to the column for now, I propose to use this site as an outlet for original problems, particularly from Australian composers. If you’d like to submit your work for publication as a Problem of the Week, please contact me. Solvers are also encouraged to send their comments on the problems, and I will start adding such feedback to the solution pages.
Let’s look at a couple of highlights from the final year of ‘Problem Potpourri’. The most attractive directmate, in my view, is a twomover by the UK problemist, David Shire. It combines two themes: multiple mates on the same square and the pseudo le Grand, the latter being a type of reciprocal change between try and key. The thematic try 1.Rf6? threatens 2.d6, and if 1…Rxe5 then 2.Sfd6. More virtual play follows with 1…Sd4 2.Sbd6, 1…Qa2 2.Qxc2, and 1…Sf4 2.Rxf4, but 1…Qd1! refutes. The try thus leads to three different mating moves on d6. These mates are seen again in the actual play, yet they function in new ways. The flightgiving key 1.Re6! threatens 2.Sfd6, and if 1…Rf5 then 2.d6 – the moves Sfd6 and d6 thereby reverse their roles as the threat and a mating response when compared with the try play. Strikingly, the third thematic mate 2.Sbd6 also recurs but is transferred to a new defence, 1…Kf5. There’s byplay with 1…Qf1 allowing 2.Qxc2.
Fittingly a trio of International Masters conceived the helpmate triplet, which features impressive cyclic play with perfect construction. In the three parts, White’s pieces on the first rank rotate their functions by taking turns to (1) be sacrificed, (2) guard flights, and (3) deliver mate. The solutions run (a) 1.dxc1(S) Rd4+ 2.Ka3 Rb4 3.Sa2 Sc2, (b) 1.dxe1(R) Bxe3 2.Rg1 Bf4 3.Rg4 Rh1, and (c) 1.exd1(B) Sf3 2.Bb3 Sg5 3.Bg8 Bb2. Each time Black promotes to a different piece, one that matches White’s eventual mating unit. A spectacular composition and it’s a testimonial to the high standards of ‘Problem Potpourri’.
David Shire Australasian Chess 2013 

Mate in 2 
Let’s look at a couple of highlights from the final year of ‘Problem Potpourri’. The most attractive directmate, in my view, is a twomover by the UK problemist, David Shire. It combines two themes: multiple mates on the same square and the pseudo le Grand, the latter being a type of reciprocal change between try and key. The thematic try 1.Rf6? threatens 2.d6, and if 1…Rxe5 then 2.Sfd6. More virtual play follows with 1…Sd4 2.Sbd6, 1…Qa2 2.Qxc2, and 1…Sf4 2.Rxf4, but 1…Qd1! refutes. The try thus leads to three different mating moves on d6. These mates are seen again in the actual play, yet they function in new ways. The flightgiving key 1.Re6! threatens 2.Sfd6, and if 1…Rf5 then 2.d6 – the moves Sfd6 and d6 thereby reverse their roles as the threat and a mating response when compared with the try play. Strikingly, the third thematic mate 2.Sbd6 also recurs but is transferred to a new defence, 1…Kf5. There’s byplay with 1…Qf1 allowing 2.Qxc2.
Jorge Lois Jorge Kapros & Christer Jonsson Australasian Chess 2013 

Helpmate in 3 (b) Kb4 to h4 (c) Kb4 to h8 
Fittingly a trio of International Masters conceived the helpmate triplet, which features impressive cyclic play with perfect construction. In the three parts, White’s pieces on the first rank rotate their functions by taking turns to (1) be sacrificed, (2) guard flights, and (3) deliver mate. The solutions run (a) 1.dxc1(S) Rd4+ 2.Ka3 Rb4 3.Sa2 Sc2, (b) 1.dxe1(R) Bxe3 2.Rg1 Bf4 3.Rg4 Rh1, and (c) 1.exd1(B) Sf3 2.Bb3 Sg5 3.Bg8 Bb2. Each time Black promotes to a different piece, one that matches White’s eventual mating unit. A spectacular composition and it’s a testimonial to the high standards of ‘Problem Potpourri’.
19 Feb. 2014 – Australian Junior Chess ProblemSolving Championship
For the eighth consecutive year, a problemsolving competition took place as part of the national Junior Chess Championships. Held at the Knox Grammar School in Sydney this year, it attracted 85 solvers (70 boys and 15 girls), the highest number ever for the event. Nigel Nettheim was again the main organiser, and you can read his comprehensive Report on the Championship on this site. Having arranged the competition successfully for so many years, Nigel has also put together a useful guide on how to run a junior problemsolving event, and it’s included as an Appendix in the Report. For a list of this year’s winners, go to the Prize List page of the Australian Junior Chess Championships site.
The contestants had two hours to deal with fifteen problems, which were capably set by Geoff Foster and Nigel. The selectors posed mostly directmates and endgame studies, though a selfmate and a proof game were added to the mix. Incidentally, these less common problem types are explained in Nigel’s updated Quick Introduction to Chess Problems and Endgame Studies, an invaluable read for any prospective entrant.
Above are two directmates from the Championship for you to solve. Although Formanek’s twomover is placed early in the paper (No.5) – where the tasks are arranged roughly from easy to hard – it still held me up considerably. De Jong’s threemover is ordered last and indeed it’s quite challenging (only one participant managed to crack it within the time limit). With no time pressure bearing down on me, I solved it in a few minutes and had tremendous fun deciphering its many variations. To see the two solutions, along with the rest of the problems used in the event, check out the Report mentioned above.
The contestants had two hours to deal with fifteen problems, which were capably set by Geoff Foster and Nigel. The selectors posed mostly directmates and endgame studies, though a selfmate and a proof game were added to the mix. Incidentally, these less common problem types are explained in Nigel’s updated Quick Introduction to Chess Problems and Endgame Studies, an invaluable read for any prospective entrant.
Bedrich Formanek Pionýrské Noviny 1961 

Mate in 2 
Leonard de Jong Magyar Sakkvilág 1930 1st Prize 

Mate in 3 
Above are two directmates from the Championship for you to solve. Although Formanek’s twomover is placed early in the paper (No.5) – where the tasks are arranged roughly from easy to hard – it still held me up considerably. De Jong’s threemover is ordered last and indeed it’s quite challenging (only one participant managed to crack it within the time limit). With no time pressure bearing down on me, I solved it in a few minutes and had tremendous fun deciphering its many variations. To see the two solutions, along with the rest of the problems used in the event, check out the Report mentioned above.
30 Mar. 2014 – ‘The Original Christopher Reeves’
The recent January issue of The Problemist includes a supplement on Chris Reeves (19392012), one of the best British problem composers. As the title of the booklet suggests, Chris was a highly original and inventive problemist – demanding qualities especially in the wellworked genre of twomovers, his favoured field. He came to prominence in the 1960s and produced many masterful works for a decade or so, before his other, “real life,” commitments brought about a period of inactivity. He returned to the problem world in the 1990s, prolifically as a composer, editor, tourney judge, and team leader of his country in international competitions. His extended break from chess, quite unusual for a composer of his calibre, no doubt contributed to his relatively small output of about two hundred problems.
The booklet is introduced by David Shire in ‘Chris Reeves: Composer and Editor extraordinary’, which discusses Chris’s perfectionist style and the way he draws the best from his collaborators. The main section presents about one hundred of Chris’s twomovers, selected with comments by David, followed by two small chapters on his threemovers and helpmates.
Here are two sample works that are illustrative of his standard. The first twomover features the Pickaninny theme: a black pawn on its initial square has four available moves and each induces a different mate. Thus the d7pawn generates these set variations: 1…dxc6+ 2.Bxc6, 1…dxe6 2.Bc8, 1…d6 2.Sd5, and 1…d5 2.Qb4. The key 1.Qxe5! threatens 2.exd7, and because the queen has opened the dfile for the d3rook – as well as lost control of b4 – none of the set mates work anymore against the pawn defences. Instead, the actual play becomes 1…dxc6+ 2.Sxc6, 1…dxe6 2.Qxe6, 1…d6 2.Qf6, and 1…d5 2.Qc7. The problem hence achieves the remarkable task of a completely changed Pickaninny. Two other variations are 1…Rd5+/Re3 2.Sxd5 and 1…Rd6 2.Qf6.
The second selection is even more impressive, showing a cycle of white selfinterferences. First note the set play, 1…Sh5 2.e4, 1…Bxd3 2.Rxd3, and 1…Rxb4 2.Sxb4. If White moves one of the three thematic pieces on e2, f2, and h3 to e3, the black d2bishop is cut off and White threatens 2.Sf4. The piece landing on e3, however, would also interfere with the remaining two of the white trios, and thereby disrupt two of the set variations. Thus the try 1.e3? impedes the h3rook and f2bishop, but 1…Bxd3 allows the changed mate 2.Qxd3, and only 1…Rxb4! refutes (2.Sxb4+ Kc5!). The second try 1.Re3? obstructs the f2bishop and e2pawn, but 1…Rxb4 now enables 2.Bxc6, and 1…Sh5! is the only spoiler (2.e4??). The last try 1.Be3? blocks the e2pawn and h3rook, but another change takes place with 1…Sh5 2.Qf3, and Black must answer with 1…Bxd3! (2.Rxd3??). In these try phases, the cyclic play combined with changed mates runs beautifully like clockwork. The postkey phase utilises the queen in a new way, with 1.Qd1! threatening the pinmate 2.Sf4. Since no white selfinterference occurs, the set play is retained: 1…Sh5 2.e4, 1…Bxd3 2.Rxd3, and 1…Rxb4 2.Sxb4. Good byplay follows with 1…Rxd1 2.c4, 1…Se5 2.Rxe5, and 1…Qc7/Qb8 2.Se7.
The booklet is introduced by David Shire in ‘Chris Reeves: Composer and Editor extraordinary’, which discusses Chris’s perfectionist style and the way he draws the best from his collaborators. The main section presents about one hundred of Chris’s twomovers, selected with comments by David, followed by two small chapters on his threemovers and helpmates.
Christopher Reeves Die Schwalbe 1965 

Mate in 2 
Here are two sample works that are illustrative of his standard. The first twomover features the Pickaninny theme: a black pawn on its initial square has four available moves and each induces a different mate. Thus the d7pawn generates these set variations: 1…dxc6+ 2.Bxc6, 1…dxe6 2.Bc8, 1…d6 2.Sd5, and 1…d5 2.Qb4. The key 1.Qxe5! threatens 2.exd7, and because the queen has opened the dfile for the d3rook – as well as lost control of b4 – none of the set mates work anymore against the pawn defences. Instead, the actual play becomes 1…dxc6+ 2.Sxc6, 1…dxe6 2.Qxe6, 1…d6 2.Qf6, and 1…d5 2.Qc7. The problem hence achieves the remarkable task of a completely changed Pickaninny. Two other variations are 1…Rd5+/Re3 2.Sxd5 and 1…Rd6 2.Qf6.
Christopher Reeves problem 1969 1st Prize 

Mate in 2 
The second selection is even more impressive, showing a cycle of white selfinterferences. First note the set play, 1…Sh5 2.e4, 1…Bxd3 2.Rxd3, and 1…Rxb4 2.Sxb4. If White moves one of the three thematic pieces on e2, f2, and h3 to e3, the black d2bishop is cut off and White threatens 2.Sf4. The piece landing on e3, however, would also interfere with the remaining two of the white trios, and thereby disrupt two of the set variations. Thus the try 1.e3? impedes the h3rook and f2bishop, but 1…Bxd3 allows the changed mate 2.Qxd3, and only 1…Rxb4! refutes (2.Sxb4+ Kc5!). The second try 1.Re3? obstructs the f2bishop and e2pawn, but 1…Rxb4 now enables 2.Bxc6, and 1…Sh5! is the only spoiler (2.e4??). The last try 1.Be3? blocks the e2pawn and h3rook, but another change takes place with 1…Sh5 2.Qf3, and Black must answer with 1…Bxd3! (2.Rxd3??). In these try phases, the cyclic play combined with changed mates runs beautifully like clockwork. The postkey phase utilises the queen in a new way, with 1.Qd1! threatening the pinmate 2.Sf4. Since no white selfinterference occurs, the set play is retained: 1…Sh5 2.e4, 1…Bxd3 2.Rxd3, and 1…Rxb4 2.Sxb4. Good byplay follows with 1…Rxd1 2.c4, 1…Se5 2.Rxe5, and 1…Qc7/Qb8 2.Se7.
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 seriesmover showing a favourite theme of the composers: multiple pins and unpins. Earlier renditions of the idea can be found in the Walkabout columns dated 8/11/2010 and 7/8/2011. 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 90degrees 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 g6e8a4c2 line. Likewise, the h5nightrider is pinned along the g6h5d1c2 line, and the h7nightrider along g6h7g8a2b1c2. The nightrider is a longrange piece, analogous to the rook and the bishop, that can make any number of knightsteps in a straight line as one move. For example, the h6nightrider can move to g8, f7, d8, g4, and f2, but its access to d4 and b3 is obstructed by the f5piece.
The serieshelpstalemate 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 flightsquares, 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 g6h7g8a2b1 to be clear, meaning the b3flight cannot be blocked. To get around this, we will pin a nightrider on a4 instead of b3 (via the g6e8a4c2 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.”
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 90degrees 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 g6e8a4c2 line. Likewise, the h5nightrider is pinned along the g6h5d1c2 line, and the h7nightrider along g6h7g8a2b1c2. The nightrider is a longrange piece, analogous to the rook and the bishop, that can make any number of knightsteps in a straight line as one move. For example, the h6nightrider can move to g8, f7, d8, g4, and f2, but its access to d4 and b3 is obstructed by the f5piece.
Geoff Foster & Ian Shanahan The Problemist 2011 1st Prize 

Serieshelpstalemate in 10 Reflecting bishop g6 7 Nightriders 
The serieshelpstalemate 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 flightsquares, 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 g6h7g8a2b1 to be clear, meaning the b3flight cannot be blocked. To get around this, we will pin a nightrider on a4 instead of b3 (via the g6e8a4c2 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.”
15 Jun. 2014 – Solving a fourmove 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 fourmover 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 hfile. 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 fulllength variation goes 1…c5/c6 2.Qa7 Sc7+ 3.Qxc7 and 4.Qh7.
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.
Valentin Marin y Llovet Deutsche Schachzeitung 1902 

Mate in 4 
The black king has two orthogonal flights, suggesting that White will aim for a queen or rook mate on the hfile. 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 fulllength variation goes 1…c5/c6 2.Qa7 Sc7+ 3.Qxc7 and 4.Qh7.
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.
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 threemovers, 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 reprocessed some existing materials from the Oz Archives to make the PDFfiles 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.”
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 reprocessed some existing materials from the Oz Archives to make the PDFfiles 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.”
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 sixtytwo 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 gamearray 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 c6pawns quite early to allow Black to develop on the queenside, 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. …Sg1h3 and …Bd1h5. To facilitate this plan, White must leave the original g1knight and d1queen to be captured on their home squares by the promoting pawns. Consequently, White must promote two more pawns to substitute for these pieces.
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 a1rook, d1queen, f1bishop, and g1knight 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 fourfold 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 changedplay 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 13.00 Rb8 14.Kg2, and 1.c3 Sf6 2.Qc2 Sh5 3.Qxh7 f5 4.Qxg7 Bxg7 5.c4 Bxb2 6.Bxb2 00 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 twosolution 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 PDFfile from the feenschach site.
Many established themes in proof games relate to pawn promotions. A classic example is the Pronkin theme, in which a piece seemingly on its gamearray 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 c6pawns quite early to allow Black to develop on the queenside, 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. …Sg1h3 and …Bd1h5. To facilitate this plan, White must leave the original g1knight and d1queen 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 a1rook, d1queen, f1bishop, and g1knight 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 fourfold 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 changedplay 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 13.00 Rb8 14.Kg2, and 1.c3 Sf6 2.Qc2 Sh5 3.Qxh7 f5 4.Qxg7 Bxg7 5.c4 Bxb2 6.Bxb2 00 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 twosolution 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 PDFfile from the feenschach site.
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 mateintwo problem (or mateinthree 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 threemover 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 threemover, are provided in the paper.
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”.
To illustrate, here’s the position for one of the slightly harder tasks. It’s a threemover 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 threemover, 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”.
29 Nov. 2014 – Helpmates of the ‘distant’ future – Part 1
The helpmate was invented in the mid19th 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 threemovers) 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 computertesting of problems became commonplace.
The British Chess Problem Society held a HOTF tourney in 200103, 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 d4pawn 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.
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 threefold orthogonaldiagonal transformation.
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 lefthand 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 righthand group. That means the three solutions of each group, while showing different strategy, have a formal connection that enhances the coherence of the problem.
The British Chess Problem Society held a HOTF tourney in 200103, 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 d4pawn 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 200103 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 threefold orthogonaldiagonal 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.
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 seen below.
The focus of this problem is the halfbattery arrangement on the fourth rank. In the three solutions of the diagram position, the d4bishop and e4knight 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 halfbattery 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.
The second example makes an interesting comparison with the sixphase 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.
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 c4pawns for the purpose of selfblock; this is nicely differentiated from another pair of solutions in which the captures of the same pawns are motivated by squareclearance. I will refrain from concocting a name for this super format of 4x2 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.
Chris Feather The Problemist 1997 Special Prize 

Helpmate in 2 3 solutions (b) Swap Kc4 & Rf4 
The focus of this problem is the halfbattery arrangement on the fourth rank. In the three solutions of the diagram position, the d4bishop and e4knight 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 halfbattery 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 sixphase 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 c4pawns for the purpose of selfblock; this is nicely differentiated from another pair of solutions in which the captures of the same pawns are motivated by squareclearance. I will refrain from concocting a name for this super format of 4x2 related phases!
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.
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