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The sequel you've all (er, at least some of you have) been waiting for!

Part 1 is here if you want to check it out.

In this game, we have an entire army against a single piece yet again. This time, white has the single piece, an Angel that begins on d1.

The Angel has a queen's move, but can jump as a knight (so a possible first move is 1. Axd8).

(Note: The Angel does NOT get the knight's normal 2x1 move.)

Whenever the Angel is captured, she misses her next turn, then resurrects on d1 at the start of the following turn and gets her turn as usual. This can only happen up to six times.

Once again, the Black king acts as just another piece and may be captured.

Black's objective is deceptively simple: queen a pawn or capture the Angel for the seventh time.

White's objective is to capture all of Black's pawns.

Assuming best play, what is the outcome of this game?

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  • $\begingroup$ Could you clarify what exactly "can jump as a knight" means here? I think you're saying that the angel can make any move a queen could make on an otherwise empty board, capturing an opposing piece if it lands on one; is that right? (Or, e.g., do you intend that it can't jump over more than one piece?) $\endgroup$
    – Gareth McCaughan ♦
    Commented Oct 9, 2018 at 16:20
  • $\begingroup$ Do you know the answer to this? If not, do you have specific reason to believe it's possible to work out the answer? (Note that the corresponding question for chess remains unanswered after something like 1500 years of intense scrutiny by humans and machines.) $\endgroup$
    – Gareth McCaughan ♦
    Commented Oct 9, 2018 at 16:21
  • $\begingroup$ @GarethMcCaughan it can make any move a queen can on an open board (as you said. that's definitely a better way to put it than how I phrased it) $\endgroup$
    – Excited Raichu
    Commented Oct 9, 2018 at 16:21
  • $\begingroup$ and I definitely feel like it's possible to work out. $\endgroup$
    – Excited Raichu
    Commented Oct 9, 2018 at 16:22
  • $\begingroup$ I'm pretty sure I know which way I'd bet on this, but it seems like proving it (whether I'm right about the answer or not) will involve quite a lot of calculation. $\endgroup$
    – Gareth McCaughan ♦
    Commented Oct 9, 2018 at 16:23

1 Answer 1

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Thoughts for white:

White can open by exchanging one angel to a queen, a rook and a minor piece.

1. Axd8 Kxd8
2. (skip) b6
3. Aa1 Bb7
4. Axh8
and white gets either the kingside knight or bishop on the next move.

There are 8 pawns and white only has 6 angels to use, so white needs to find two unprotected pawns. Ideally white would want to get rid of black's pieces and attack black's pawn chain from behind.

Thoughts for black:

I think black needs to try to form a pawn chain and mostly ignore threats on pieces. If black can get a protected pawn on the 2nd rank, white can't stop it from promoting. For example, with pawns on d2 and e3:

1. Axd2 exd2 (Ad1 e2 with d1# or e1# next)
2. (skip) d1#

Black can force a protected pawn on the 4th rank.

1. Axd8 Kxd8
2. (skip) g6
3. Aa1 f5
4. Axh8 d6
5. Axg8 e5
6. Axh7 f4

If white ignores the pawn, black can get a protected pawn on the 3rd rank by using an outside pawn as a distraction.

7. Axg6 a5
8. Ag2 a4
9. Axa8 a3
10. Axa3 e4
with e3 or f3 to follow.

It might be best for white to give up an angel as soon as black gets to the 4th.

7. Axg6 a5
8. Ag4 a4
9. Axf4 exf4
10. (skip) b5
11. Axa4 bxa4

Two isolated pawns on the 4th may not be enough... Not sure.

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