Rebuild-it puzzles

The king can always capture the Rook on g1, so it's not mate. But black still has a move left. Which one and where is the black king?

Puzzle #18: No clue about the solution

KlekleLegacy wrote:

(Puzzle #19):

Pieces to place: .
Resulting position: White to play and mate in one for Black.

+bKb4, forcing 1. Kxg1 and allowing 1… Bxe3# (1… Ne2#?? is illegal because of pin by Rg4).

Other squares would do at least one of the following:

i) puts the bK in check (so, illegal position)

ii) blocks Rf7 from doing double check with Rg1 (allows 1. B/Rxg1 for white and no mate for black)

iii) both Nd4 and Bg5 would be pinned to the king, so after the forced 1. Kxg1 neither piece could deliver the mate.

Nice problem! Liked the way of how the pins on the black king worked once he was added.

Here is a hint (trying not to give away too much):

Both the f2 and the e3 pawn are essential to making this solution possible. They are key to the puzzle. Removing either one of them has the following consequence: the problem would have no more solution.

Another hint:

Also, after the and the are placed, White is in a position where it is his turn to play and he has four possible responses, one of which is a capture, but all four of them fail to the same following move by Black which checkmates White.

Excellent! There is even a more or less unique black last move before white's start! Puzzles 18 en 19 are probably the best in the whole set. They have more depth in the reasoning for the placements!

Btw, if you adapt the diagram slightly you can probably prove which piece black captured on his last move or you can prove that black captured nothing. Perhaps you can even prove uniqueness of white's move before black's last move! Since placement puzzles are typically considered retrograde in nature, you score pluses for everything you can add in that department!

I see how to add white pieces in a way to be able to prove (if this was what I wanted) Black captured nothing on his prior move and how a promotion into a rook with capture as the only explanation as to why the white rook is on g8 can, in the case where Black's prior move was not a capture, that White's move before Black's prior move was unique and can be proven to be unique.

There is a cost, however, in the nature of the pieces chosen to be added to the position in puzzle #18: these are bishops. It means that, if White had lost no pieces, there could very likely be (unless White promoted earlier one of two pawns from the position of the beginning of the game into a on a square of the same color as the one of the to be placed back) information about the square color of the to be placed back in the position to be rebuilt, which was not the case in the original #18 puzzle. In that case (all the on the board except the one to be placed back are on the same square color), it gives extra information about the to be placed back. I can argue doing so makes the problem easier, which can be detrimental to the puzzle. Giving too much information for the sake of making the puzzle more retrograde can theoretically, because of the nature of the bishops and of the nature of the starting position, be also detrimental to a rebuild-it puzzle where a bishop needs to be put back on the board, and make the solving of the puzzle way easier or «too easy».

Taking that into account, and taking into account that adding extra pieces in the position which then needs to be rebuild can reduce the amount of squares where the person solving the problem can put the required pieces to be placed back on the board (finding the solution of the puzzle or not), I think sometimes it is preferable to not push too hard for a puzzle to be retrograde in order to keep the challenge of the problem/puzzle.

Regardless of that, regarding your last comment, I will take the challenge of making a problem more retrograde for one of my future puzzles while still keeping the problem challenging .

Comment above updated.

Good reasoning! Indeed, you do not wish to make the solving easier by adding extra retro content. So that would be part of the challenge. It could be simply based on "insufficient material to make a(ny) capture on c8". The point I wanted to make is that - if unique retraction moves can be added at negligible cost - that would be a valuable addition to this composition type. Much more than for instance for a common endgame study or directmate problem.

Btw, everyone can see that only certain squares are candidates for a solution. No one will claim that adding a pawn on b2 makes solving easier while it is a small price for a unique retraction move. Note that your diagram already violates the efficiency principle in several ways. Like, the extra queen on a3 takes away a potential solution square for one of the bishops while it plays no role in the solution. If you wanted maximum space for the placements you would have cleaned up the original diagram!

Here is how to solve puzzle #22:

Once the is placed back on the board, Black has all eight pawns from the beginning of the game, which means all of his pieces on the board are from the beginning of the game. Therefore White can't have captured any of Black's pieces since Black also has the knight pair, the bishop pair, the rook pair and his original queen. Therefore no white pawn or piece has captured any black piece. It means every white pawn from the incomplete board pictured above comes from its original file.

White only has 7 pawns. When the 2 bishops are placed back on the incomplete board pictured above, White has 7 pawns, a rook pair, a bishop pair, knight pair, and a queen. It means that Black has captured at most one of White's pieces (if any). Since White is missing a pawn, it may mean it promoted, or else it means it has been captured earlier in the game. But could White have promoted his d?

From the fact that Black captured at most one of White's pieces (if any), it is true for all of Black's pawns. So we can deduce the a3 originated from a7, the b7 from b7, the e7 from e7, the f6 from f7, the g5 from g7, and the h5 from h7. Weither or not the d6 originated from c7 or d7, the file would still be closed for his d. It means White's d pawn has not been promoted and therefore has been captured by Black. Ergo the 2 required to be placed back are White's original white-squared and White's original black-squared .

White is said to be mated when all pieces are placed back on the incomplete board. But on said board, White seems checked twice, once by the on a7 and once by the on c1. Indeed, the Black pawn required to be placed back on the board can't get past the d file, so it can't be on f2 checking White's . But the double-check is impossible because the c2 is in the way. So the checkmate comes from a single check, but which piece delivered it?

If the check comes from the c1 , how does it happen? The c2 blocks the c file, so the c1 couldn't have come from higher ranks. What about a capture, then? xc1# is not possible: which piece would the black rook have captured? It couldn't have been a white pawn since they start on the 2nd rank. When the two are placed back, White has all his minor and major pieces from the beginning of the game... Ergo the checkmate couldn't have come from Black's c1 .

But, if it doesn't come from the c1 , something has to block a check from said rook: at least one . But where to put it? The d1, e1, and f1 squares are all available...

If the checkmate doesn't come from the c1 , it has, by process of elimination, to come from the a7 . But can't the a5 prevent the checkmate by simply capturing it? Yes, unless there is something blocking its path on a6. It can't be a , so it has to be a . Here is our first placed back piece solved: a on a6.

By placing a on a6, we placed White's light-squared bishop. We can then deduce the bishop preventing the c1 from checking the is on e1 because it has to be black-squared (property of the bishops from the position of the beginning of the game). Here, we placed our second piece. The only piece left to place is the .

Here, we have to ask ourselves: how did the bishop on a7 come to check White? It couldn't have come from the a7-g1 diagonal, and the b8 square is occupied. Ergo, we can deduce the checkmate comes from a discovered check or by en passant. None of Black's pieces on the incomplete board could have delivered a discovered checkmate. So the checkmate has to be from en passant or from a discovered check from the yet not placed . But it couldn't have come from en passant neither, since it would have required a piece able to make a discovered check before the en passant, which Black already does not have on the incomplete board.

The two only candidates for a discovered check by a move are from c5 to c4 and from d4 to d3. c4+ is not checkmate since White has the only move e3. Therefore, the needs to be placed on d3 to block the queen's path to e3: the checkmate is delivered by from d4 to d3.

#23 has multiple solutions:

I) +wRc5 (or c6), +wBa6, +wKc7 --> 1.Bb7#

II) +wBc6, +wKc7, +wRb7 --> 1.Rb8#

Diagram at time of solving:

@BigDoggProblem Here is the problem updated. Can you solve it now? (change: black knight added on e8)

Pieces to place: , , .
Resulting position: White to play and mate in one by White.

Puzzle 24

+wBe4, +wNd5, +wKa6. Seems unique as far as I can tell: the double check is key.