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1. What was the last move played?

The piece on h4 is a white black-squared bishop because all captures must have been made on white squares by black pawns (the h pawn must capture on g2 otherwise it could not be a white piece on h4 since the only other way to queen is to capture twice to the f-file, but this means white has no other pieces).

The reason it cannot be a black piece is because:

i)One pawn is needed to promote to a knight. There is only 1 promoted black pawn so the other two knights must be the knights at the start of the game.

ii)It cannot be a rook or queen otherwise white would be in check on his/her move

iii) Black's black-squared bishop could not have moved.

Alive without breath,
As cold as death;
Never thirsty, ever drinking,
All in mail clinking/.
What am I?

52 of us know only light,
36 of us know only dark,
None of us have the same voice,
But all of us have one purpose
What are we?

I do not exist, I'm just a word.
and When someone mentions me.. I break.
What am I?

For the record, I'm not the greatest riddle solver so yeh I kind of don't know how hard these are meant to be (i give up easily).

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Visible Analysis
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A1. White is missing seven things: (a) the "light-eligible" group of 5: QRNN + Bf1, (b) the "dark-only" Bc1, and (c) Pf2.

A2. By minimality of diagonals, Black's Pa6 is its b7-pawn, which made 1 capture on a light square, and Black's doubled Pc4 is its f7-pawn, which has made 3 captures on light squares (e6-d5-c4). Black has 4 captures on light squares.

A3. Given [A2], Black is missing six things: (a) the "f-victim group" of 4: QRR + Ph7, (b) Bc8, (c) Bf8.

A4. Black's Bf8 is stillborn: it never got out (due to Pe7/Pg7), so it must have been captured on f8.

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Given
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G1. White's last move must have been cxd8=R+.

G2. Exactly one of [A1abc] and [A3abc] is the mystery piece on h4. This could include any result of a promotion.

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Shallow
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S1. By [A1c], White is missing only one pawn, at f2. By [G1], White's last move was a pawn promotion from c7. Ergo, that must be White's f2-pawn, and it has captured its way to c7. Then it has just captured again on d8. So White has made 4 captures.

S2. Of Black's six missing things, two cannot be captured on d8: [A2c] its Bf8, by [A4], and [A2b] obviously its Bc8 can never stand on d8. Therefore, White's 4 captures by [S1] exactly covers Black's "f-victim" group [A2a] of 4: QRR + Ph7.

T2. Consider the fates of White's missing things. [A2] pinpoints 4/5 of [A1a], White's "white-eligible" group of 5. [G1] pinpoints [A1c], White's Pf2. 1 of [A1a], and [A1b] the Bc1, are unknown.

T3. One of [T1] and [T2] is the mystery piece. But Black's light-squared Bc8 is the wrong color to stand on dark h4. Therefore, the mystery piece on h4 is White.

T4. From [S2], Black's h-pawn (before or after promotion) is an f-victim, i.e. it must have been captured by White's f-pawn. We prove in two ways that Black's h-pawn has promoted.

T4-A. Suppose White's f-pawn captured Black's h-pawn while it was still a pawn. But this requires at least 3 additional captures total on both sides (e.g. Black captures once to g and White captures twice f->g and g->f; or Black captures three times h->f->g->e). But by [T3], there are only three total missing things of unknown fate, and one of them is still alive on h4. So there's not enough corpses to admit 3 more captures. Ergo, the assumption is false, and the h-pawn must have promoted.

T4-B. Consider: What Black piece was captured on d8? Lemma: It cannot have been a Black Qd8/Rd8. <Proof: It would be checking White's Kh8, and Black has no way to deliver that check. (a) c8/d7/e7 are blocked, and remember that c7 is blocked by White's pawn just before it captures. So Black couldn't move a Q or R to d8 with check. (b) Black might have moved the mystery piece from e8/f8/g8 to h4, with discovered check. But that's not a legal move for any piece. QED.> Therefore, White's Pc7 must have captured a Black Bd8 or Nd8. By [A4], Black's original Bf8 never got out, and Black still has two Ns in play. Hence, Black's piece on d8 is a second dark-square B, or a third N, ergo Black has promoted. By [A3], Black is missing only one pawn, its Ph7, so that's the one that promoted.

T5. How does Black's h7-pawn promote? It must have (a) captured once and promoted on g1, or (b) captured twice and promoted on f1, or (c) captured more than twice. But by [T2], White has only 2 unknown things left, and one's still alive on h4. Ergo, (b) and (c) are impossible, and Black's h-pawn captured exactly once, and promoted on g1.

T6. From the pawn structure, the only square on which Black's h-pawn could have captured to reach g1 is on g2, a light square. With [A2], this means Black made 5 total captures on light squares, which exactly covers [A1a], White's "light-eligible" group of 5. From [T2], that pinpoints all of [A1a].

T7. From [T2] and [T3], the only White unknown left is [A1b], its dark-square Bc1.

- If "wife" is X's own wife: X is a parasitic male?

Anglerfish: http://rmbr.nus.edu.sg/news/index.p...

Sea devil: http://www.austmus.gov.au/fishes/fi...

<Gilmoy> Surrogate mother is getting warm ...

To give you some background, this is an anglo-saxon riddle (ie very old english) which is around about 2,000 years old. It is most definitely not one that I made up, nor is it from the hobbit or lord of the rings.