Proof Games and more Proof Games

This is a thread for Proof Game problems. For those who aren't familiar with them, you're given a diagram and asked to produce a legal game that results in that diagram in the specificed amount of moves. For example, "PG in 20.5 moves" means that you create a game that reaches the diagram after white's 21st move.

1st PG:

Composer: M.Caillaud

PG in 19.5 moves

Nothing yet, I'm three moves off:

SOLV'D

Theme : 2x black bishop circuit ['rundlauf']

PG #2

Composer: K.Prentos

PG in 20.0 moves

Ugh. Almost there, except the white a-pawn is driving me crazy. I can't get a rook around it.

Ok, here's some analysis in white (don't look if you want to try to solve it):

In order to capture all 8 white pieces, black needs to use cross-captures for every pawn. Black barely has enough time to put his pieces into position. The problem is, I don't see how white can give black enough tempi with cross-captures in time.

Still no answer. :(

Black's move count is precise: each pawn is one move, Ra8-a1 is one, Bc8-b7-a8 is two, Qd8-a5/b6/c7/d4/d7-a7 is two, Ke8-d7 is one, Bf8-a3 is one, Ng8-...-g7 is three, and Rh8-b8/h3-b3 (to be captured by White's pawn) is two. (Moves might be captures.)

So the pawn moves must be a7xb6 in order to let Ra8-a1, and consequently b7xa6 to put an a-pawn. Now that b-file is never free of pawns, we know the kingside rook moves Rh8-h3-b3, so there must be h7xg6 and consequently g7xh6. Also, the queen moved Qd8-c7/d7-a7 (the other routes are blocked by the a/b-pawns, and Qd8-a5, a7xb6, Qa5-a7 is impossible because then the rook cannot go out), and White does a2xb3 to free the rook's path a8-a1.

Note that a2xb3 accounts for the single White capture, so the remaining pawns c-g march forward to rank 6 to get eaten by Black pawns, except for probably one on rank 5 captured by Black's kingside knight. This amounts to at least 4*3 + 2 = 14 moves. White's a2xb3 gives one more move, for a total of 15.

Since c-g pawns don't leave their files, the missing pieces queenside rook and two bishops must get captured on a6, b6, and h6. The bishops can go Bc1-h6 and Bf1-a6 to get captured in one move each, but this forces White's rook to get captured on b6, for two moves. And since this amounts to seven Black pawn captures, we need one more to restore the one-pawn-per-file structure, so White's remaining pawn is not captured by the knight after all; this accounts for all 20 moves.

Thus, White made the following:

• c2-c4-c5-c6
• d2-d4-d5-d6
• e2-e4-e5-e6
• f2-f4-f5-f6
• g2-g4-g5-g6
• a2xb3
• Ra1-a6-b6
• Bc1-h6
• Bf1-a6

And Black made the following:

• a7xb6
• b7xa6
• c7xd6
• d7xc6
• e7xf6
• f7xe6
• g7xh6
• h7xg6
• Ra8-a1
• Bc8-b7-a8
• Qd8-c7/d7-a7
• Ke8-d7
• Bf8-a3
• Ng8-...-g7 (three moves)
• Rh8-h3-b3

And we know a couple of move orders:

1. wPg2-g4-g5-g6 (Black has two free plies)
2. bPh7xg6
3. bRh8-h3-b3 (White has two free plies)
4. wPa2xb3
5. wRa1-a6-b6 (Black has two free plies)
6. bPa7xb6
7. bRa8-a1 (White has one free ply)
8. bPe7xf6 (White has one free ply, White must have done wPf2-f6)
9. bBf8-a3 (White has one free ply)

So let's try constructing around this...

SOLV'D

For the new solvers, see that counting thing he did? Very important for solving PGs.

PG #3:

Composer: U.Heinonen

PG in 22.0 moves

White is missing one pawn, black is missing one pawn too. Black made a hxg capture - therefore the white b-pawn promoted, and may or may not be on the board now. Black's b-pawn did not swerve off the b-file, although it may have promoted on b1 and is now an impostor somewhere, the original piece captured by white's b-pawn before promoting on a8 or c8, likely the latter.

Assuming all pieces on the board are original, white made at least 15 moves with pieces on the board. Black made at least 13 moves with pieces on the board. In addition, it took white at least 5 more moves to promote that pawn, and 2 more to get it or another white piece sacrificed to black's h-pawn.

Now we assume black really did promote the b-pawn, because now we see that white's movement is extremely tight - all 22 moves are accounted for (15 + 5 + 2) - and no time is spared to capture the b-pawn. Therefore black promoted the b-pawn, and this impostor is now on the board. It is likely to be the b1-knight, both because it looks right and because it's the easiest piece for the white b-pawn to have captured.

Whither white's g4-rook originate? Ra1-b1-b4-g4 looks good, but upon closer inspection it fails. Rb1 comes after Nd2, after d3, after Bc4... which prevents Rb4-g4. So white's only other option is to scrounge up another move by castling long, and play Rd1-e1-e3-g3-g4.

That in itself poses a final problem. If white's king was on c1, black can't promote a pawn from b2 as that's check. So the natural reaction that black's knight promoted, left b1 and returned later. Thanks to the clutter around b1, the entrance square must be d2, cleared off after white played Ndf3 after Re3-g3... here we get into a realm of move order that is only clear when one attempts to construct a solution.

So without further ado, a 22.0 solution. Unto Heinonen makes tough ones; an 8-move promoted bN Rundlauf.

I find this extremely frightening:

Dr. L. Ceriani and Dr. K. Fabel
Am Rande des Schachbretts, 1947
SPG in 183.0 moves

I'd surely like to see if one can make a "logical" solution with how you get that, as the source doesn't list any, just a single game.

Well, this one is so long it has to be worked from both ends. It's probably a bit more work than I care to do, but if I were inclined to actually solve it, I'd start by releasing the position - retracting moves to get the bK out from behind those pawns. Then I'd count the number of moves needed to get from the start position to the intermediate one. Then, I'd try to cut the fat and bring the total count down to 183.

White is missing one queen, which is captured for dxc, so Black's e-pawn never leaves its file.

White made at least four pawn captures. Since Black has 11 pieces plus the captured 4, this gives 15; no more White pawn captures.

White pawns' moves, the quickest routes: a2xb3, b2xc3, c2-c4, d2-d4, e2xf3, h2xg3. Those are six moves. Also note that White's d-pawn never leaves its file.

White bishops+queen+king's moves: Bc1-a3-f8, Bf1-c4-d5, Qd1-...-c5/c6 (two moves), Ke1-d2/e2-e3. Those are eight moves.

Rc8 clearly comes from d-file, visiting d8 before going to c8. But White's d-pawn never leaves its file, so the rook must have come from the center, so we have the path Ra1/h1-a5/h5-d5-d8-c8 for four moves.

The other rook must go Ra1/h1-a6/h6-f6, for two moves. This leaves three moves for the knights, which thus must be Ng1-e2-c3-a4.

Additionally, since Nb1 never moves, we know that we need Ra1-a5/a6-d5/d6-d8-c8 (instead of Rh1-h5-...), and Rh1-h6-f6 (instead of Ra1-a6-f6).

Note that d2-d4 and e2xf3 blocks two paths of White queen from d1, so either White queen remain on the first rank for its first move or it goes along the diagonal d1-a4.

Suppose it remains on the first rank. Thus the second move must be from c1 (to c5 or c6), g1 (to c5), or h1 (to c6); no other square allows a single move. But all of them are blocked; c1-c5 is blocked by White's c-pawn remaining on its file, g1-c5 is blocked by the stationary f2, and h1-c6 is blocked by the stationary g2. Thus it moves along d1-a4, and so c2-c4 is before White queen's first move.

Now, what square along the d1-a4 diagonal allows a move to c5/c6? Only c2 (to c5 or c6) and a4 (to c6) allows it. But again, c2 is blocked because there's a pawn on c4, so White queen moves Qd1-a4-c6.

So now we have a complicated dependency system:

• wPe2xf3 before wBf1-c4-d5 before wPc2-c4 before wQd1-a4-c6 before bPd7xc6 and wPa2xb3 before wRa1-a6
• wPe2xf3 before wNg1-e2-c3-a4 before wPb2xc3 before wBc1-a3-f8 before bPc6-c5
• wRa1-a6 before wNc3-a4

Note the italicized text. It's not wRa1-a5 because wBf1-c4-d5 occupies the critical d5 square, so wRa1 has to go via a6-d6-d8-c8. This also means we need some more dependency:

• bPc6-c5 before wRa6-d6-d8-c8

But with so many dependencies like this, what's White's first move?

It cannot be any of the pawn captures for obvious reasons. It can neither be the knights because Nb1 doesn't move while Ng1 goes to e2, currently occupied. So either c2-c4 or d2-d4. But c2-c4 needs Bf1-c4-d5, so White's first move is d2-d4.

But then what is the second? White still has no pawn capture, c2-c4 is still bound, Bc1 and Qd1 don't need to go via the freed d2 square...

...except Ke1-d2, which can still reach e3 in one move.

What is the third? White still has no pawn capture (the only captures possible on White's third move are b2xBa3 and g2xBh3, neither of which is required), and the freed e1 square is useless, so Kd2-e3 is the move.

Now White runs out of moves, so he must have a pawn capture. But axb3 and bxc3 need to wait, so either we do exf3 or hxg3.

If we do hxg3, we only give White three moves: hxg3 and Rh1-h6-f6. Meanwhile, if we do exf3, we have lots of the dependencies freed up, so let's do e2xf3 as White's fourth move.

What piece can feed f3 there? The rooks are impossible. Bc8 cannot be freed before bPd7xc6, which needs wPe2xf3 to be performed first. Bf8 is of the wrong color. Ng8 has the wrong parity. Either bNb8-c6-e5-f3 or bPe7-e6 followed by bQd8-f6-f3.

Let's take a look again at the dependencies:

• wPe2xf3 before wBf1-c4-d5 before wPc2-c4 before wQd1-a4-c6 before bPd7xc6 and wPa2xb3 before wRa1-a6
• wPe2xf3 before wNg1-e2-c3-a4 before wPb2xc3 before wBc1-a3-f8 before bPc6-c5 before wRa6-d6-d8-c8
• wRa1-a6 before wNc3-a4

There is a long chain of White moves here:

1. e2xf3
2. Bf1-c4
3. Bc4-d5
4. c2-c4
5. Qd1-a4
6. Qa4-c6
7. a2xb3
8. Ra1-a6
9. Nc3-a4
10. b2xc3
11. Bc1-a3
12. Ba3-f8
13. Ra6-d6
14. Rd6-d8
15. Rd8-c8

And since e2xf3 occurs as White's 4th move, we have very little time available.

Okay it's 4am over here so I should continue this later.

chaotic_iak, Ra6-d6 can't happen after Qc6...

After Qc6, Black does d7xc6, so the queen has vanished. We only need Black to do c6-c5 before White can do Ra6-d6.

There's still a problem. Black can't sacrifice a piece on c3 without playing c5 first, which would prevent Bc1-a3-f8.

Oh wait, you have bxc3 as the third capture. So the last one is on g3.

Just to show a few problems: