Dennis K. Hale

(b. 11th June, 1945)

It was Gregory Koshnitsky’s incredibly long-running column in the Sun-Herald newspaper which introduced me to the art of chess problems. Over the years he published a wonderful variety of superb direct mates in two. I recall in the mid-fifties (having been taught the moves by an Hungarian migrant and his son) initially wondering how players got themselves into such convoluted positions, but it gradually dawned on me that the positions were “made up” or, as I later learned was the correct term, “composed”. My first written attempt to “Dear Mr. Koshnitsky”, as he always was to be, referred to “horizontal rows” and “vertical rows” – it must have been a waiter – and needless to say I got it wrong. Borrowing a book from the local library soon set me right on the basic terminology of the game. Having ever been only a mere dabbler in both problems and the game (but what wonderful enjoyment they have given), my attempt to differentiate direct mates in two (always my favourite problem genre) and the game is that mates in two exhibit complexity in breadth, whereas the game (though not mine) is complexity in depth.

It was John Kellner in his Sunday Mirror chess column who first tempted me to try to compose a problem. He conducted the Frank Ravenscroft Memorial Tournament in 1968 – a composing tourney for retrograde analysis (RA) problems. This type attracted me as it seemed solving them was akin to solving a mystery with strict logic given only the position and stipulation as clues. I submitted a handful of entries then nothing further was composed until that tireless worker for chess problems, Bob Meadley, later a truly deserving recipient of the Australian Chess Federation’s Purdy Medal for chess journalism and then running the ‘Problem Corner’ column in Bernie Johnson’s Chess In Australia magazine, wrote me an encouraging letter concerning my earlier attempts to compose. This set me off again for about three to four years. Virtually all of my meagre output of problems – all RA – were composed during these two periods, 1968 and 1977 to 1980. The problems in the later period were somewhat of an improvement on my earlier efforts. The best was the dedicatory problem which Bob kindly accepted for his book on Bill Whyatt.

Writing a comprehensive introduction to the vast field of retrograde analysis problems is frankly beyond my knowledge and capability – as I said, I am a mere dabbler in this highly complex genre. I will thus content myself with a few hopefully helpful remarks. The question of whether a particular position (no matter how odd or how oddly reached) could have occurred in a legally played game of chess is central to RA. The history of the position – how it occurred – is basic. These matters can be very difficult to determine due to complicating factors such as pawn structure (is the number of captures required to achieve that structure consistent with the number of men available for capture?), the possibility of promotions, en passant captures, castling, etc. These complicating factors can interact with each other, e.g. there may be enough captures available for say White to have a promoted white-square bishop, but not enough for a promoted black-square bishop. Enough of mere generalisations – let’s look at some examples aimed at giving a feel for what RA problems are about. In a nutshell, they are about deduction. The clues are in the position, are you a good enough sleuth to observe and draw the correct conclusions?

The stipulation (what you as the solver are required to do) can be quite varied:

  • Legal Position? That is, could the problem position have occurred in a legally played game of chess?

  • Last Move? That is, what was the last move in this legally played game of chess?

  • Shortest Proof Game in ‘n’ moves. That is, how could the position have been reached in the specified number of moves from the standard initial position? Normally the sequence is required to be unique, unless there are two or three very different solutions – the composer will specify this. The originator of this website, Peter Wong, is an acknowledged expert in the proof game genre.

There are many other possible stipulations, but all involve looking back to a position’s antecedents – how could it have come about?

To give an initial idea of the type of thinking required for RA problems, consider the following little problem from Raymond Smullyan’s book on RA, The Chess Mysteries of Sherlock Holmes. In RA problems the usual convention of White’s pawns moving up the board from the bottom is observed.

Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979

White to move. What did Black just play in this legally played game?

The reader is asked to make a determined effort to solve this without looking at the solution.

The solution is as follows. The move must have been made by the BK (the only black piece on the board). The BK cannot have moved to a8 from b7 or b8 as the BK and the WK can never be on adjacent squares. Hence, it must have moved from a7 to a8, to escape the check from the WB on g1, but how could the WB have moved to g1 to give check? It couldn’t – the only possibility from h2 is ruled out because h2 is occupied (by a WP). Seemingly a dead-end has been reached, but what if Black’s last move was a capture? A WB as the captured piece is no help, nor is a WQ or WR which both involve impossible double checks without the possibility of discovery. What about a WS? Eureka – we have it! White’s last move could have been, in fact could only have been S(b6) to a8 discovering check from the WB. Interestingly, though not relevant to the problem’s solution, before S(b6) to a8, a8 may have been vacant or occupied by BQ, BB, or BS, but not a BR (why not?).

To reiterate, the solution is that Black’s last move was K(a7)xS(a8).

One excellent website on RA problems is The Retrograde Analysis Corner. Please go to that site and attempt, under “A Collection Of Instructive Problems”, the following problems:

  • No.6 (Proof Game in 4 moves.)

  • No.12 (You have to determine the last move in the game. Hint: don’t forget that there must be a possible legal move before that last move.)

  • No.15 (Hint: how can the WB have given check?)

  • No.20 (White to move – what did Black just play?)

The extent and depth of this website indicate the extent of RA problems. Another useful website on RA is that of The British Chess Problem Society – select “Retrograde Analysis”.

Now for some of my composing efforts. I like to give titles to my problems. Some were adopted by the publications in which they appeared, some were not – I will give them regardless. Starting with four simple, hopefully easy, problems (which primarily were composed with the aim of making RA more accessible to the uninitiated), I will end with four more complex ones. The solution to each problem follows its diagram immediately, except for No.7 and No.8 which have links to their solutions, provided in diagram form.

1. Dennis K. Hale
“Impasse”
Tasmanian Chess Magazine, Mar. 1979

Could either of these two positions have occurred in a legally played game of chess?
(a) The diagram position
(b) Transfer BPb5 to g5

In both positions, all of White’s missing men must have been captured by the black pawns still on the board, and all of Black’s missing men must have been captured by the white pawns still on the board.

Position (a) is possible as both the Wd-P [original white d-pawn] and the Be-P [original black e-pawn] must have promoted without capturing (there are none spare) and then as a promoted piece been captured by a pawn (given the pawn configuration, neither the Wd-P nor the Be-P, both non-capturing, could’ve been captured as a pawn by an opposing pawn, hence the necessity to promote to a piece then move to a square where it can be captured by an opposing pawn).

Position (b) is impossible. The only way for it to come about is for both the Wd-P and the Bd-P to have promoted without capturing and then be captured as a promoted piece by an opposing pawn, but this is impossible as the Wd-P and the Bd-P, both non-capturing, could not have bypassed each other.

2. Dennis K. Hale
“Irresistible Force”
Chess in Australia, Jan. 1979

Could either of these two positions have occurred in a legally played game of chess?
(a) The diagram position
(b) Transfer WPg2 to c4

In both positions:
(1) The last move could only have been White’s 0-0-0+.
(2) All of White’s missing men must have been captured by the black pawns still on the board, and all of Black’s missing men must have been captured by the white pawns.

Position (a): This position is possible. The We-P must have captured the Bf-P on f3.

Position (b): This position is impossible. The Bf-P must have been captured either before or after promoting to a piece. It couldn’t have made a capture before itself being captured (there are none spare), nor could it have been captured on its original file (the present white pawn configuration makes this impossible). Therefore it must have promoted and then been captured by one of the white pawns, but this is inconsistent with the only possible last move (White’s 0-0-0+) as the WK would have to have moved (making 0-0-0 impossible) when the Bf-P was on f2, giving check.

3. Dennis K. Hale
“Numbers Game”
Chess in Australia, May 1977

Which of these two positions could have occurred in a legally played game of chess?
(a) The diagram position
(b) Transfer BPg7 to f7

In both positions:
(1) The last move must have been White’s B(h2)-g1+ as there is no possibility of a discovered check.
(2) B(h2)-g1 could have been a capturing or non-capturing move. If:
(i) a capturing move then the BR(h1) poses no difficulty for the possibility of the position from which B(h2)-g1 was played.
(ii) a non-capturing move then Black’s move which preceded B(h2)-g1 must have been P(g2)xh1=R+ . Otherwise in the position from which B(h2)-g1 was played the WK would have been in check from the BR with no possibility that Black’s checking move was legal.
(3) Therefore it is possible for the position to have occurred in a legally played game of chess if either:
(i) White’s last move was B(h2)xg1+, or
(ii) White’s last move was B(h2)-g1+ preceded by P(g2)xh1=R+.
(4) In neither position (a) nor (b) is B(h2)xg1 a possibility for White’s last move as White’s pawn configuration requires a minimum of eight pawn captures and Black has lost eight men – there are no spare captures available. Hence the possibility of either position depends on 3(ii).

Position (a): If Black’s last move was P(g2)xh1=R+ then the minimum pawn captures required by Black to achieve both this and his pawn configuration would be seven (either his f-P or h-P would have made two captures). But White has only lost six men. Therefore, position (a) could not have occurred in a legally played game.

Position (b): With the pawn on f7, Black would only require a minimum of six pawn captures as his P(g2) which promoted to a rook could have been his g-P which itself would need only one capture. Therefore position (b) could have occurred in a legally played game.

4. Dennis K. Hale
“Monistic Mitre”
Chess in Australia, Jun. 1984

Could either of these three positions have occurred in a legally played game of chess?
(a) The diagram position
(b) Transfer WPc5 to g3
(c) In position (b), transfer BPf4 to b6

In all three positions, the two bishops (one black, one white) must be promoted pawns as neither original black-square bishop could have moved from its initial square – the pawn structure prevents this.

Position (a): This position is impossible. The Wg-P must have promoted to a bishop on b8 (b8 is the only possible square as had the Wg-P promoted on another square, the black pawn configuration would have prevented it from moving off the 8th rank), and the Bb-P must have promoted to a bishop on g1. Although there are enough potential captures to achieve this (six each being required – five for each of the Wg-P and the Bb-P and one for each of the Wh-P and the Ba-P), the position is impossible as the Wg-P and the Bb-P could not have bypassed each other in their paths along the g2 to b7 diagonal – only one of the pawns could have promoted, not both.

Position (b): This position is impossible. The Bb-P could have promoted to a bishop on g1 and the Wc-P could have promoted to a bishop on b8. The WP(g3) is either the g-P or the h-P. If the g-P then it must have been on g3 when the Bb-P promoted (couldn’t have been on g2 as the Bb-P had to occupy that square) and so the resulting promoted bishop could never have moved other than to h2 – certainly not to b4. If the WP(g3) is the h-P then it must have always been on either h2 or g3 so again the promoted BB could never have moved to b4.

Position (c): This position is possible. The Wc-P could have promoted to a bishop on b8 from a7 (with the Bb-P still on b7 and only moving to b6 after the promoted WB moves away). Similarly the Bf-P could have promoted to a bishop on g1 from h2 (with the Wg-P still on g2 and only moving to g3 after the promoted BB moves away).

5. Dennis K. Hale
“Cleric’s Cell”, Ded. to Bill Morris
Chess in Australia, May 1981

Which of these positions could have occurred in a legally played game of chess?
(a) The diagram position
(b) BPf6 to f7
(c) In position (b), replace WSf3 by WP

In all positions:

(1) The last move could only have been by the Bb-P, i.e. P(g2)-g1=S+ as there are six black pawns in the c7-h7-h2 triangle. This requires Black to have made at least eight pawn captures (the Bb-P having made five, and the Ba-P(d2) having made three). As there are only eight White men off the board, Black has no spare pawn captures. Hence the BP(g3) could not have made any captures during the game and is therefore the g-P, and so the WP(g4) must have made at least one capture.

(2) The white pawn configuration requires White to have made at least six pawn captures (considering that the WP(g4) must have made at least one). As there are seven black men off the board, White has one spare pawn capture.

(3) The possibility of each position having occurred in a legally played game depends on whether all the missing white men could have been captured by the two black pawns. The difficulty is that the Wg-P could not have been captured on its original square as then the WB from f1 would not have been available for useful capture.

(4) The Wh-P could not have been captured as a pawn by the two black pawns (as the Wh-P could not have moved to a square in either of their possible paths), nor could it have been captured by them as a promoted piece for the following reasons:
(A) Positions (a) and (b): The Wh-P would have had to use White’s spare pawn capture to promote (to pass the Bh-P) so the WP(g4) would have to be the Wf-P as the Wg-P would have required two captures to reach g4 around the non-capturing Bg-P. Hence the Wg-P could only have been captured on g2 as it could not have made a capture (White’s spare pawn capture had to be used by the Wh-P) to either:
(i) be captured on f3 by the Bb-P on its way to g2, or
(ii) promote around the non-capturing Bg-P and then be captured. So the Wh-P could not have promoted. Hence the WP(g4) must be the h-P, as the Wf-P and the Wg-P (promoted or otherwise) must have been among the eight white men required to be captured by the two black pawns.
(B) Position (c): The Wh-P could not have promoted (using White’s spare pawn capture) as the Wg-P would have required two captures to reach g4 around the non-capturing Bg-P (the WP(f3) must be the Wf-P as otherwise the Bb-P could not have bypassed the Wg-P to reach g2). Hence the WP(g4) must be the h-P.

(5) The Wh-P(g4) must have made its capture from h3(xg4) after the non-capturing Bg-P had moved to g3.

Now to consider the possibility of each of the positions:

Position (a): Yes – the Wg-P could have promoted after making only one capture, i.e. from g6xf7 with the Bg-P on g7 (later the Bg-P moved to g3 and then the Wh-P(h3)xg4). Note that the Wg-P could not have captured from g2xf3 and then itself have been captured on f3 by the Bb-P because then the Wf-P could not have been captured by either of the two black pawns either as a pawn or after promoting (both of which alternatives would have involved an extra capture).

Position (b): No – the Wg-P could neither have promoted after making only one capture (the BP(f7) precludes this) nor have been captured on f3.

Position (c): Yes – now the Wg-P could have been captured on f3 – after the Bb-P reached g2 the Wf-P would have moved to f3 and then the BK to f2.

6. Dennis K. Hale
“Endless Errant”
Chess in Australia, Jun. 1979, 1st Hon. Mention

Which of these positions could have occurred in a legally played game of chess?
(a) The diagram position
(b) Remove BPb4
(c) In position (b), transfer BQb2 to b1
(d) In position (c), replace WRc1 by WB
(e) In position (d), transfer BQb1 to b2

All positions:
(1) The P(f6) must be the Wb-P as there are six white pawns in the c2-h2-h7 triangle.
(2) The P(b3) must be the Wa-P.
(3) The P(d3) must be the We-P.
(4) The P(h6) must be the Wh-P.
(5) The last move must have been Black’s P(e2)xf1=R+.
(6) The BB(a2) must be a promoted pawn.

Position (a): Could not have occurred.

White has a last move, i.e. R-c1, but there are insufficient captures available for Black’s pawns to provide the white pawns with six captures. Black has four captures instead of the required five (which includes one by either the Ba-P or the Bc-P as the BB(a2) must have promoted on b1).

Considering Black’s pawns:
(i) The Bb-P is either on b4 or is the promoted BB(a2). Either the Ba-P (in which case the non-capturing Bc-P was captured on the c-file) or the Bc-P (in which case the non-capturing Ba-P promoted on a1) made one capture.
(ii) The non-capturing Bd-P was captured on the d-file.
(iii) The non-capturing Bf-P was captured on f6 (could not itself have promoted because of insufficient captures).
(iv) The Be-P made one capture – e2xf1=R+.
(v) White had to make six captures so either the Bg-P or Bh-P (the other is on h5) promoted on h1 (neither of these could have captured as pawns). Considering that the non-capturing Wh-P is on h6, this would require three additional captures by Black as to promote on h1 the Bg-P would require one capture (in which case the Bh-P would require two to reach h5) or the Bh-P would require two captures (in which case the Bg-P would require one to reach h5). As there are two required black pawn captures in (i) to (iv) above, a total of five black pawn captures would be required, but there are only four white men off the board.

Position (b): Could have occurred.

White has a last move, i.e. R-c1. The Bb-P promoted to a bishop on b1 before a2xb3 – the black pawns now require one less capture than in (a).

Position (c): Could not have occurred.

White does not have a last move to precede Black’s P(e2)xf1=R+:
(i) g5xh6 is eliminated as White would require eight captures (only seven available).
(ii) f5-f6 is eliminated as the P(f6) is the Wb-P.
(iii) b2-b3 is eliminated as the P(b3) is the Wa-P.
(iv) e2xd3 and Q-e1 are eliminated as Black’s last move must have been P(e2)xf1=R+.
(v) S(e3 or h2)-f1 is eliminated as the BK would have been in check with White to move.
(vi) K(h2)xS(g1) (preceded by S(h3)-g1 disc.+ which entails that either the Bg-P or Bh-P promoted to a S) is eliminated as this would involve an infinite sequence of retrograde checks by the BS(g1) between h3 and f4 or g5. [Note on (vi): It is essential that S(h3)xS(g1) is not possible as Black’s second last, otherwise the WS(g1) could have made White’s second last – the possibility is eliminated by the lack of captures available to Black.]

Position (d): Could have occurred.

White has a last move, e.g. B-c1.

Position (e): Could have occurred.

White has a last move, i.e. K(h2)xS(g1). White’s second last move must have been R(g1)-f1 (and so Black’s last P(e2)xR(f1)=R+) to avoid the infinite retrograde checks by the BS(g1). The non-capturing Bb-P promoted to a bishop on b1 then White played a2xb3 then the non-capturing Ba-P promoted (so could not be to a knight]. So for this position to be possible in a legally played game, four black pawns must have promoted – the Ba-P on a1, the Bb-P to a bishop on b1, the Be-P to a rook on f1, and either the Bg-P or Bh-P to a knight on h1. The four captures required by Black and the seven required by White are available.

Note: In (d) White’s last move could also have been K(h2)xS(g1) – the point of the transition from (d) to (e) is that this possibility could be overlooked in (d) whereas it must be seen in (e).

For the following two problems I will give the actual solutions only (without reasons). I will leave the reasoning to you. The first “Prelates’ Partie” was my best problem from the 1968 period.

7. Dennis K. Hale
“Prelates’ Partie”
Australian Chess Problem Magazine, Mar. 1997
(republished from The Sunday Mirror, 1968)

Black to move. The position preceding Black’s second last move contains three pieces which must have resulted from pawn promotions. It is possible to prove on which three squares the pieces stand. What is the position? There are two completely different solutions.

8. Dennis K. Hale
“Penultimate”
Chess in Australia, Aug. 1977, 2nd Prize

Beginning with Black’s second last move, what were the moves that led to this position from a legally played game?

I hope you have found something interesting and challenging in the above problems. If you enjoyed the more complex problems, you might like to attempt my most complex efforts:

  1. My dedicatory problem to Bill Whyatt which Bob Meadley kindly used in his book, W. A. Whyatt’s Chess Problems (1979) – it is a 63-position RA twin (budget for at least a few hours to solve it).

  2. “Lucky Last” which Ian Shanahan kindly published in his ‘Chess Billabong’ column in Australasian Chess, Sep.-Oct. 2007 (the solution to which is over 5,000 positions).

  3. “Less Majeste” in Chess in Australia, Apr. 1977, which was awarded First Prize in the second Chess In Australia Problem Tourney.

  4. My three RA problems in The Problemist (The British Chess Problem Society) will also hopefully present some challenges – they are R36, R40, and R57 in the Sep.-Oct. 1977, Jul. 1978, and May 1980 issues, respectively.

Well that’s all I can say about my limited dabbling in the vast field of Retrograde Analysis.