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Consider this Martin Gardner/Raymond Smullyan puzzle just asked by Favst over on Chess.SE (Source: The Colossal Book of Puzzles and Games, problem A.12 "Doubled Knights"):

... Assume that a game is played with the new rule that a knight can on a single turn move twice. [I.e., both players' both knights are replaced by fairy pieces that move like knights but after their first knight-move may optionally make a second knight-move; either or both of those knight-moves may be capturing.] The first player, White, [is confident that he'll be able to win].

"Actually, I think I have a slight advantage," says Black, "so just to even the playing field, I'll play without my king pawn."

Is White still guaranteed to win?

"In fact," says Black, "I'll even play without my king and queen pawns!"

Is White still guaranteed to win?

Finally, an open problem: Is there any set of pawns/pieces that Black can give up, to escape the easy mate by White? (Define "escape" as "Black gets to make at least 4 moves.")

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full pieces

1. Nc4#

missing e7 and/or d7 pawns

1. Nc4+ Ke7 2. Nf4#

Note that positioned like that the two knights cover all but 6 squares in black's half.

enter image description here
red: covered by either knight in sinlge move
yellow: covered by left knight in double move
green: covered by right knight in double move

To avoid this being check mate black would have to

open up g8 i.e. sacrifice at least one super knight. But this makes e4 available to white:

1. Ne4+ Kf8/d8/e7 2. Nf4#

So white wins in max

2 moves

no matter what.

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"Actually, I think I have a slight advantage," says Black, "so just to even the playing field, I'll play without my king pawn."

Is White still guaranteed to win?

Yes. One might think 1. Nb1-d5 still wins easily, because whether the king remains on e8 or moves to e7, the white knight can still capture it. But Black can reply 1... Ng8xd5. However, 1. Nb1-b5 forces 1... Ke8-e7. Now White can move 2. Ng1-h4, after which Black is still in check; the b5 knight threatens d6 and e6 (in two moves), so 2... Ke7-f6 is forced. But then 3. Nh4-f4 is mate.

I'm not sure about the other claims ...

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Answer to the two non-"open problem"s:

With the BKP removed,

W opens 1. N-c3-b5+ (threatening, say, N-d6-e8). B must respond by playing Ke7. Now W plays 2. N-f3-h4+ (threatening, say, N-g6-e7). B can't capture the Nh4, and his only K moves are to e8 (Nb5-d6-e8), d6 (which is even in check normally), e6 (Nb5-d4-e6), and f6. He must choose the last of these. But now 3. Nb5-c3-e4 is mate: again B can't capture either knight, the four squares {f,g}{5,6} are all in normal check, e7 is still threatened, e6 is covered by Ne4-g5-e6, and e5 is covered by Nh4-g6-e4.

With the BQP removed as well,

on the face of it B has two new ideas: he can play Kd7 immediately in response to 1. Nb5, or he can play Ke7 as before and then try Kd7 after 2. Nh4. Let's look at the second of those first. After 3. Nb5-a3-c4 the Nc4 is uncapturable and B's only move that leaves his K out of immediate range of the WNs is Ke6. Then after 4. Nc4-e5-f3 B's only moves are Kd5 and Kf6. After 4... Kd5 we have 5. Nh4-f5-e3 forcing 5... Kc5, then 6. Ne3-d5-c3 is finally mate. After 4... Kf6 we have 5. Nh4-f5-g3 which is mate.

I should confess that

this takes five moves for W to deliver mate; I suspect, especially given the parenthesis at the end of the question, that there is a more efficient way, but a mate is a mate.

The other try

for B is, as mentioned above, 1... Kd7 instead of 1... Ke7. Then 2. Nb5-a3-c4 forces either 2... Ke7 or 2... Ke6. After either of those, 3. Ng1-h3-f4 is mate.

The third question is obviously more difficult. I don't think I would trust my own analysis. I might write a little computer program to try to solve it...

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