11 Feb. 2011 – Shortest proof games with 3 solutions
In a shortest proof game problem, the aim is to reconstruct the game that leads to the diagram position (from the normal opening array), using the fewest possible moves. By convention, the solution of a SPG will consist of a precise sequence of moves, without any dual or alternative move order. Because of this requirement, it’s fairly difficult to compose a SPG problem with two precise solutions, where both arrive at the same final position. A SPG that involves three solutions, then, would be particularly challenging to set up, and the first problem to achieve this task is given below.
Its stipulation of “SPG in 6½” means that the diagram has to be reached after White’s 7th move. The theme shown is a threefold change of white “raiders”, or pieces that capture various enemy units before returning to their home squares.
This problem of mine was mentioned in Part II of Mark Kirtley’s ‘Proof Games Shorties’, a series of articles on SPGs that are at most seven moves in length. The author of the article asked if anyone knew of, or could create, another SPG with three solutions. This challenge was not met until a few years later, when in Part V of the series (which for some reason is not included on the Retrograde Analysis Corner site linked above) another problem with the same task was uncovered.
This marvellous work by Göran Wicklund has three quite varied solutions, one of which is especially surprising by diverging from the symmetrical play seen in the other two parts. I am withholding the solutions to both SPGs to encourage you to solve them!
Peter Wong Problem Observer 1995 Version 1st Prize 

SPG in 6½ 3 solutions 
Its stipulation of “SPG in 6½” means that the diagram has to be reached after White’s 7th move. The theme shown is a threefold change of white “raiders”, or pieces that capture various enemy units before returning to their home squares.
This problem of mine was mentioned in Part II of Mark Kirtley’s ‘Proof Games Shorties’, a series of articles on SPGs that are at most seven moves in length. The author of the article asked if anyone knew of, or could create, another SPG with three solutions. This challenge was not met until a few years later, when in Part V of the series (which for some reason is not included on the Retrograde Analysis Corner site linked above) another problem with the same task was uncovered.
Göran Wicklund Springaren 1998 

SPG in 7 3 solutions 
This marvellous work by Göran Wicklund has three quite varied solutions, one of which is especially surprising by diverging from the symmetrical play seen in the other two parts. I am withholding the solutions to both SPGs to encourage you to solve them!
27 Feb. 2011 – What’s New
“Frank Ravenscroft and His Letters” by Bob Meadley has been added to the Oz Archives section. This large PDF file (73MB) comprises scans of letters written by the late problemist Frank Ravenscroft to Bob Meadley and Rurik Bergmann, as well as other material about Ravenscroft.
On the Archives’ Magazines and Columns page, five previously missing (or incomplete) problem columns of the Australasian Chess Magazine are now included, filling in all the gaps.
And don’t miss the recent updates of the William Whyatt material, such as a collection of his joint works with Alex Goldstein.
And don’t miss the recent updates of the William Whyatt material, such as a collection of his joint works with Alex Goldstein.
9 Apr. 2011 – Australian Junior Chess ProblemSolving Championship
As in previous years, a problem solving competition took place as part of the 2011 Australian Junior Chess Championship, held in the city of Melbourne. The Report for the solving championship by one of its organisers, Nigel Nettheim, is now available. Seventyseven junior solvers participated in this successful event, along with five adults, and you can see a list of the prizewinners on the Australian Junior Chess Championships website [no longer available].
Shown here is one of the problems used in this solving competition. See how long it takes for you to solve this difficult twomover. The solution can be found in the Report (Problem 9 in the U18 section).
G. Baumgartner The SunHerald 1961 Commended 

Mate in 2 
Shown here is one of the problems used in this solving competition. See how long it takes for you to solve this difficult twomover. The solution can be found in the Report (Problem 9 in the U18 section).
14 May 2011 – Two problem conventions re castling and capturing en passant
Two special moves in chess, castling and the en passant capture, differ from other moves in that their legality depends on not only the current position, but the prior play as well. An issue arises in composed problems when these special moves are an option in the diagram, since we’re not given the play that lead to it. Two problem conventions deal with such situations. In the case of castling, if the king and a rook of one side are on their original squares, castling is deemed legal in subsequent play, unless it can be proved that the king or the rook must have moved previously in a hypothetical game.
The twomover above is a neat illustration of this rule. Since White is to play in the diagram, Black must have made the last move, with a piece that is still on the board. Neither of the pawns could have made this move, since they are still on their initial squares, so it must have been made by the king or the rook. Thus we’ve shown that Black has disturbed at least one of the two pieces previously, a fact that renders the castling move illegal. The key here is 1.Ra8! (threat: 2.B~) 1…Kf8 2.Be5, 1…Rg8+ 2.Bg3, and Black cannot play 1…00, which otherwise would be a refutation. In part (b) of this twin, with a black pawn added on g2, Black’s last move could have been made by this pawn. That means castling is now considered legal, and it would defeat the try 1.Ra8? The new key is 1.Be5! (2.Ra8) 1…00 2.Rg3.
The convention for en passant captures applies to problem positions where a pawn is on its fifth rank while an enemy pawn is adjacent to it on the same rank. In such cases, capturing the enemy pawn en passant is deemed illegal, unless it can be proved that the only possible last move was a doublestep by that pawn.
The second problem exemplifies this kind of proof. Black’s last move wasn’t made by the king, since if …Kg6h6 was just played that would imply the two kings were standing next to each other, while …Kg7h6 as the last move would mean that the f6pawn had just given check, but that’s impossible because the squares where that pawn could have come from (e5, f5, and g5) are all occupied. Hence Black’s last move was made by the g5pawn. This move wasn’t …g6g5, because that would mean White was in check while Black has the move – an illegal situation. And since f6 and h6 are occupied, that rules out the alternatives …fxg5 and …hxg5, leaving …g7g5 as the pawn’s only possible last move. Therefore 1.hxg6 e.p.! is legalised as the problem’s key, which forces 1…Kh5 2.Rxh7. For more examples of problems involving such retroanalysis, see Dennis Hale’s article.
Wolfgang Pauly Chess Amateur 1913 

Mate in 2 (b) Add BPg2 
The twomover above is a neat illustration of this rule. Since White is to play in the diagram, Black must have made the last move, with a piece that is still on the board. Neither of the pawns could have made this move, since they are still on their initial squares, so it must have been made by the king or the rook. Thus we’ve shown that Black has disturbed at least one of the two pieces previously, a fact that renders the castling move illegal. The key here is 1.Ra8! (threat: 2.B~) 1…Kf8 2.Be5, 1…Rg8+ 2.Bg3, and Black cannot play 1…00, which otherwise would be a refutation. In part (b) of this twin, with a black pawn added on g2, Black’s last move could have been made by this pawn. That means castling is now considered legal, and it would defeat the try 1.Ra8? The new key is 1.Be5! (2.Ra8) 1…00 2.Rg3.
The convention for en passant captures applies to problem positions where a pawn is on its fifth rank while an enemy pawn is adjacent to it on the same rank. In such cases, capturing the enemy pawn en passant is deemed illegal, unless it can be proved that the only possible last move was a doublestep by that pawn.
Friedrich Amelung Düna Zeitung 1897 

Mate in 2 
The second problem exemplifies this kind of proof. Black’s last move wasn’t made by the king, since if …Kg6h6 was just played that would imply the two kings were standing next to each other, while …Kg7h6 as the last move would mean that the f6pawn had just given check, but that’s impossible because the squares where that pawn could have come from (e5, f5, and g5) are all occupied. Hence Black’s last move was made by the g5pawn. This move wasn’t …g6g5, because that would mean White was in check while Black has the move – an illegal situation. And since f6 and h6 are occupied, that rules out the alternatives …fxg5 and …hxg5, leaving …g7g5 as the pawn’s only possible last move. Therefore 1.hxg6 e.p.! is legalised as the problem’s key, which forces 1…Kh5 2.Rxh7. For more examples of problems involving such retroanalysis, see Dennis Hale’s article.
25 Jun. 2011 – ‘FIDE Album 20012003’
The FIDE Albums are anthologies of the world's best chess compositions (or at least that’s the aim of these collections!), brought out once every three years. The latest edition has just been published, covering the period 200103. Given that ten years have passed since the original appearance of some of the selected problems, most would agree that the timeliness of the Album's publication leaves some room for improvement. But the delay is partly explained by the massive tasks assigned to the expert judges, who undertook to appraise, in some cases, more than a thousand problems. At any rate, the wait is well worth it, as the book itself is a firstclass production.
In the same polished format as the previous five volumes, this hardbound Album contains 1349 problems of all types (16% of submitted entries), with trilingual theme descriptions and detailed indexes. From the Australian perspective, only one homegrown work was selected, a fairy helpmate by yours truly, but it's a bit too unorthodox to discuss here. Instead I quote a twomover by Viktor Chepizhny that scored the maximum 12 points, and hence has nearofficial status as a masterpiece. The scoring system for inclusion in the Albums is fairly straightforward. In each of eight sections divided according to problem genre, three judges separately give a score from 0 to 4 points to each entry (halfpoints are allowed), and a sum total of 8 or more points is required for the problem to be selected. It is rare to attain the maximum 12 points, and Chepizhny's twomover is one of just two problems in that section to have achieved the feat. Yet this outstanding work is surprisingly accessible.
The problem involves two pairs of closely related phases of play, triggered by three white tries and the key. The four parts utilise the white batteries in different ways and show changed mates in response to two thematic defences, 1…Sxd3 and 1…Sxg4. The first try 1.Bf8? (threat: 2.Bc5), by guarding c5, permits 1…Sxd3 2.Qxd3 and activates the Q + S battery, 1…Sed7 2.Sc1, and 1…Ra5 2.Sdxe5; but there is no answer to 1…Sbd7! An analogous try 1.Bh6? (2.Be3) controls e3, enabling both 1…Sxg4 2.Rxg4 and the R + S battery to operate, 1…Sxc4 2.Sxh2; but 1…f1(S)! refutes. The third try 1.e7? (2.Rd6) prepares to fire the B + R battery when the black knight selfpins, 1…Sxd3 2.Rc6, and 1…Sxg4 2.Re6; now 1…Ra5! defeats the try. The key 1.Kg5! (2.Rf4) makes room for the B + R battery to fire again – against the same knight defences – but in a new direction, 1…Sxd3 2.Rf5, and 1…Sxg4 2.Rxf3; also 1…Sf7+ 2.Rxf7. A splendid rendition of the Zagoruiko theme, this twomover shows helpmatelike congruity between the various phases.
In the same polished format as the previous five volumes, this hardbound Album contains 1349 problems of all types (16% of submitted entries), with trilingual theme descriptions and detailed indexes. From the Australian perspective, only one homegrown work was selected, a fairy helpmate by yours truly, but it's a bit too unorthodox to discuss here. Instead I quote a twomover by Viktor Chepizhny that scored the maximum 12 points, and hence has nearofficial status as a masterpiece. The scoring system for inclusion in the Albums is fairly straightforward. In each of eight sections divided according to problem genre, three judges separately give a score from 0 to 4 points to each entry (halfpoints are allowed), and a sum total of 8 or more points is required for the problem to be selected. It is rare to attain the maximum 12 points, and Chepizhny's twomover is one of just two problems in that section to have achieved the feat. Yet this outstanding work is surprisingly accessible.
Viktor Chepizhny A. Dombrovskis Memorial Tourney 2002 1st Prize 

Mate in 2 
The problem involves two pairs of closely related phases of play, triggered by three white tries and the key. The four parts utilise the white batteries in different ways and show changed mates in response to two thematic defences, 1…Sxd3 and 1…Sxg4. The first try 1.Bf8? (threat: 2.Bc5), by guarding c5, permits 1…Sxd3 2.Qxd3 and activates the Q + S battery, 1…Sed7 2.Sc1, and 1…Ra5 2.Sdxe5; but there is no answer to 1…Sbd7! An analogous try 1.Bh6? (2.Be3) controls e3, enabling both 1…Sxg4 2.Rxg4 and the R + S battery to operate, 1…Sxc4 2.Sxh2; but 1…f1(S)! refutes. The third try 1.e7? (2.Rd6) prepares to fire the B + R battery when the black knight selfpins, 1…Sxd3 2.Rc6, and 1…Sxg4 2.Re6; now 1…Ra5! defeats the try. The key 1.Kg5! (2.Rf4) makes room for the B + R battery to fire again – against the same knight defences – but in a new direction, 1…Sxd3 2.Rf5, and 1…Sxg4 2.Rxf3; also 1…Sf7+ 2.Rxf7. A splendid rendition of the Zagoruiko theme, this twomover shows helpmatelike congruity between the various phases.
7 Aug. 2011 – Two firstprize problems by Geoff Foster and Ian Shanahan
Congratulations to Geoff Foster and Ian Shanahan whose works have gained First Prizes in the respective tourneys of two major problem publications, feenschach and StrateGems. Both problems are serieshelpstalemates, a type of seriesmover in which Black plays a number of consecutive moves to reach a position where White can deliver stalemate. The two compositions also share a theme: multiple selfunpins by the black pieces, impressively rendered in each case.
Let’s analyse the first, joint work. Black has six units besides the king that must be immobilised. Four of them can be pinned by the available white pieces, one can be captured by White’s stalemating move, and if that move is Sxh3, the knight will block the remaining black unit, the h4pawn. Such a knight move will also cover the f4flight. This plan requires Black to promote the bpawn and rearrange the pinned pieces – the aim is to obstruct the flights on d4 and d5, and to relocate the f5rook which is pinning the white knight. 1.b4 2.Sxb2 3.Sc4 4.b3 5.b2 6.b1(S) 7.Sc3 8.Sd5 9.Re6 10.Sc6 11.Sde7 12.Rg6 13.Rd5 14.Sd4 15.Se5 16.Sf5 17.Rg3 18.Rh3 Sxh3. The black pieces unpin one another by interposition eleven times (the underlined moves), a record number when this problem was published.
Geoff’s solo effort extends the number of unpins to fourteen, which is the current record. In this position, not counting the black king and the blocked f6pawn, Black needs to disable six units. Again, four of them may be pinned by the white linepieces, and one captured by the stalemating move. The only way to deal with the last black unit is to pin it with a promoted piece, i.e. Black will place a piece on e7 in anticipation of White’s final move, fxe8(Q). Note also that Black’s whitesquared bishop cannot be fully restrained by a diagonal pin; nor can it be pinned while blocking the king’s orthogonal flightsquares, which are black. Therefore the only role left for this bishop is to be the captive on e8. 1.Rd6 2.b5 3.b4 4.b3 5.Sb6 6.Rd5 7.b2 8.bxa1(B) 9.Be5 10.Sg4 11.Be4 12.Bd6 13.Sc4 14.Rf5 15.Sge5 16.Bc6 17.Be7 18.Rd5 19.Sd6 20.Bxe8 fxe8(Q). Another attractive sequence where the precise move order is determined by intricate pinning/unpinning effects.
Geoff Foster & Ian Shanahan feenschach 2008 1st Prize 

Serieshelpstalemate in 18 
Let’s analyse the first, joint work. Black has six units besides the king that must be immobilised. Four of them can be pinned by the available white pieces, one can be captured by White’s stalemating move, and if that move is Sxh3, the knight will block the remaining black unit, the h4pawn. Such a knight move will also cover the f4flight. This plan requires Black to promote the bpawn and rearrange the pinned pieces – the aim is to obstruct the flights on d4 and d5, and to relocate the f5rook which is pinning the white knight. 1.b4 2.Sxb2 3.Sc4 4.b3 5.b2 6.b1(S) 7.Sc3 8.Sd5 9.Re6 10.Sc6 11.Sde7 12.Rg6 13.Rd5 14.Sd4 15.Se5 16.Sf5 17.Rg3 18.Rh3 Sxh3. The black pieces unpin one another by interposition eleven times (the underlined moves), a record number when this problem was published.
Geoff Foster StrateGems 2009 1st Prize 

Serieshelpstalemate in 20 
Geoff’s solo effort extends the number of unpins to fourteen, which is the current record. In this position, not counting the black king and the blocked f6pawn, Black needs to disable six units. Again, four of them may be pinned by the white linepieces, and one captured by the stalemating move. The only way to deal with the last black unit is to pin it with a promoted piece, i.e. Black will place a piece on e7 in anticipation of White’s final move, fxe8(Q). Note also that Black’s whitesquared bishop cannot be fully restrained by a diagonal pin; nor can it be pinned while blocking the king’s orthogonal flightsquares, which are black. Therefore the only role left for this bishop is to be the captive on e8. 1.Rd6 2.b5 3.b4 4.b3 5.Sb6 6.Rd5 7.b2 8.bxa1(B) 9.Be5 10.Sg4 11.Be4 12.Bd6 13.Sc4 14.Rf5 15.Sge5 16.Bc6 17.Be7 18.Rd5 19.Sd6 20.Bxe8 fxe8(Q). Another attractive sequence where the precise move order is determined by intricate pinning/unpinning effects.
30 Sep. 2011 – What’s New
Paul Dunn, the games archivist of Ozbase (the Australian Chess Games Archive), is extracting problem tourney reports from early 20th century Australian papers, and he has kindly offered to share them with readers of this site. The first such award, the Australian Problem Tourney 190910 for originals published in various chess columns, was adjudicated by the chess problem giant, Alain C. White.
Meanwhile, here's another comic strip from the xkcd site…
Meanwhile, here's another comic strip from the xkcd site…
6 Oct. 2011 – ‘Chris J. Feather – Selected Helpmates’
The British problemist Chris Feather is one of the foremost experts on helpmates in the world. He's also my favourite problem composer, so I was delighted to hear that he has published an anthology of his works, titled Chris J. Feather – Selected Helpmates. It contains over 300 of his problems, all accompanied by his comments and that of coauthor GM Zivko Janevski. The book also includes Chris's articles on various helpmate themes and motifs, and a section on his composing influences and inspirations. As to be expected, the quality of the problems collected here is fantastic. His works are marked by great artistry combined with strategic intensity. Here’s a taste of what you will find in this excellent anthology.
The first selection involves a wealth of strategic effects in its two analogous solutions: 1.Qxc4 Qxg4+ 2.Kd5 Qxc4, and 1.Qxf3 Qxc6+ 2.Kf4 Qxf3. In both phases, the black queen unpins its counterpart and captures a knight, with the aim of clearing that square for White's mating move. The white queen captures a pinned piece, “doubling” with the initial pinner in the control of a line. After a king move (to a square that the black queen had crossed over), the white queen moves along said line and gives a pinmate, supported by the linepiece. A striking example of an orthogonaldiagonal transformation.
In helpmates with two parts, we often see a reciprocal change of functions between two pieces, e.g. the two white knights in the first problem exchange their roles of being a guard and a sacrificed piece. The second problem extends this kind of linking effect and presents a cyclic change of functions among three white pieces. The three solutions are: 1.Kg5 Rc5+ 2.Sf5 Qh4, 1.Kf5 Qxd7+ 2.Se6 Rxh5, and 1.Ke3 Re1+ 2.Se2 Rc3. Thus the queen and two rooks take turns to (1) pin the knight, (2) give mate, and (3) guard flights. Such a cyclic relationship between the three phases confers both unity and diversity to the problem. Note also that the black knight is already pinned in the diagram, and is repinned on three different squares – a pleasing idea in itself. Masterfully constructed, this helpmate has ideal white economy in requiring solely the thematic pieces of that colour besides the king.
Chris J. Feather – Selected Helpmates is available for 20 euros (+ postage) from Ralf Kraetschmer (ralf.kraetschmer{at}tonline.de), who's the bookseller of the German chess composition society.
Chris Feather Sakkélet 1991 

Helpmate in 2 2 solutions 
The first selection involves a wealth of strategic effects in its two analogous solutions: 1.Qxc4 Qxg4+ 2.Kd5 Qxc4, and 1.Qxf3 Qxc6+ 2.Kf4 Qxf3. In both phases, the black queen unpins its counterpart and captures a knight, with the aim of clearing that square for White's mating move. The white queen captures a pinned piece, “doubling” with the initial pinner in the control of a line. After a king move (to a square that the black queen had crossed over), the white queen moves along said line and gives a pinmate, supported by the linepiece. A striking example of an orthogonaldiagonal transformation.
Chris Feather Hatchings2 1995 

Helpmate in 2 3 solutions 
In helpmates with two parts, we often see a reciprocal change of functions between two pieces, e.g. the two white knights in the first problem exchange their roles of being a guard and a sacrificed piece. The second problem extends this kind of linking effect and presents a cyclic change of functions among three white pieces. The three solutions are: 1.Kg5 Rc5+ 2.Sf5 Qh4, 1.Kf5 Qxd7+ 2.Se6 Rxh5, and 1.Ke3 Re1+ 2.Se2 Rc3. Thus the queen and two rooks take turns to (1) pin the knight, (2) give mate, and (3) guard flights. Such a cyclic relationship between the three phases confers both unity and diversity to the problem. Note also that the black knight is already pinned in the diagram, and is repinned on three different squares – a pleasing idea in itself. Masterfully constructed, this helpmate has ideal white economy in requiring solely the thematic pieces of that colour besides the king.
Chris J. Feather – Selected Helpmates is available for 20 euros (+ postage) from Ralf Kraetschmer (ralf.kraetschmer{at}tonline.de), who's the bookseller of the German chess composition society.
20 Oct. 2011 – What’s New
A second award judged by Alain C. White, the Australian Problem Tourney 191011, is now available for download. Thanks to Paul Dunn for this report, which can be found in the Oz Archives section (along with future instalments).
Recently I came across an interesting online article about Brian Tomson, the late problemist from Newcastle. It's an entry from a site called IRLchess (Irish Chess History and Records), discussing Brian's chess activities both as a player and a composer. Check it out!
And here's another cartoon from Tony Lurie's Toonopia site…
Recently I came across an interesting online article about Brian Tomson, the late problemist from Newcastle. It's an entry from a site called IRLchess (Irish Chess History and Records), discussing Brian's chess activities both as a player and a composer. Check it out!
And here's another cartoon from Tony Lurie's Toonopia site…
3 Dec. 2011 – A prizewinning problem by Arthur Willmott
Congratulations to Arthur Willmott whose threemove problem has been placed equal second in the prestigious Brian Harley Award. This competition, organised by the British Chess Problem Society, is for the best threemover published in the UK by a Commonwealth composer, over a three year period (20062008). Arthur's problem also received a Second Prize in Problem Observer's informal tourney. The solution of this classy directmate can be found on Arthur's page of this site (Problem 4).
Arthur also celebrated his 90th birthday a few weeks ago! Still an active composer and solver, he certainly deserves our compliments in attaining this milestone.
Arthur Willmott Problem Observer 200708 2nd Prize 

Mate in 3 
Arthur also celebrated his 90th birthday a few weeks ago! Still an active composer and solver, he certainly deserves our compliments in attaining this milestone.
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