New answers tagged

3

There certainly is more than meets the eye here. First of all, Now that we have that established, we need to figure out Now this is interesting, because it means the pawn was somewhere else. If it were on d3, the black knight would have nowhere to have come from, So three half-moves ago, the board must necessarily have looked something like this: From ...


3

To add on to Glorfindel's answer,


2

Solution:


5

is mate because


4



7

I present to you, pure inefficiency: It can definitely be done faster, but this works.


1

In this position that I composed long ago (this position is legal), every move is completely forced and it is checkmate in !8 moves (15 ply): 1. Qxc7+ Qxc7+ 2. Qxc7+ Qxc7+ 3. Qxc7+ Qxc7+ 4. Bxc7+ Kb7+ 5. Bb8+ Qxe7+ 6. Qxe7+ Qxe7+ 7. Qxe7+ Qxe7+ 8. Rxe7#


1

EDIT: Actually, seeing the "only one line" requirement, here is a 16 that is a slight modification of a position made by Bernd Schwarzkopf and Karl Scherer (Urdruck) im Feensnach 1980 on page 13. A PDF can be found here. ! FEN: 1KN4r/r7/k7/pr2R3/5b2/8/8/8 w KQkq - 0 9 (Picture Please!) ! 1. Qxa7+ Qxa7+ 2. Rxa7+ Qxa7+ 3. Rxa7+ Qxa7+ 4. Rxa7+ Qxa7+ 5. ...


3

Wrong answer (I made a boneheaded mistake, kindly pointed out in comments by @RShields, to which I have drawn attention below.)


2

I initially thought the answer was something like Given the poster's comment confirming the first four moves: I believe the continuation is as follows, for the following reasons.


2

Considering black always play the best move: Prove: From comment add variant: From OP clarification: "I mean to say that by White to win I meant for White to reach a winning position. I also forgot to mention that the solution is 7 moves long The title is a bit of hint as to what the solution has." Possibly the solution was as Sconibulus answer suggests ...


2

This is not a (new) answer to the original question, but I don't have enough reputation to comment. I tried to address the call for generalization using a similar technique as Jaap. Below the results for the board sizes that fit in my main memory. Unfortunately, 6 x 6 does not fit. size # configs w b ========================= 3 x 2 180 12 13 ...


4

According to this answer by @bof on Mathematics Stack Exchange:


11

The answer is Reasoning: So This could only happen White's f and g pawns So would be


6

My goal is to describe my solution strategy in detail, even though my final solution isn't that different from that of @Bass. At multiple points, I will ask a question, which I encourage you to answer on your own before reading further. So, let's start. The black rook on a1 is the most interesting part of the position. The black rook on h8 is gone and it ...


34

I wrote a computer program and it showed that $18$ moves is the optimum. Here is one such solution: Oddly enough, even if you relax the condition of alternating white and black moves, it cannot be done in fewer moves. For $3\times3$ the optimal number of moves is $16$. Without the need to alternate moves the optimum is $14$ moves, for example just by ...


0

EDIT: As @greenturtle pointed out in a comment, it seems that everyone else is doing the count by ply, and not the whole moves. The question is unclear to me about this on how the count is done. So thus my count is wrong by the majority's decision. As such, just for fun, here is a symmetrical solution of 20 moves that uses the same notations as my below ...


8

Got 19 by moving around... might be possible to do better:


3

Found a 19 move solution, but no idea about optimum. Where the columns are a, b, c, d and rows are 1, 2, 3, 4, starting from bottom left.


3

Found a solution in 20, though I have no idea if it's optimal. One of my assumptions was that "Chess rules apply" meant I had to alternate black and white moves.


66

Now that we have three increasingly complex proofs (two deleted, one of them mine) that it's impossible, it's pretty clear that it must be Here's why: This is, without doubt, the most refreshing chess problem I've ever tried to solve. Thanks, OP!


30

(Edit: so this is wrong..) Brilliant puzzle! The answer is: [Spoiler alert! Scroll down at your own risk] We proceed by contradiction. Assume that indeed, White can castle. We have the following undeniable facts: Neither the White King nor the White h-Rook have moved. Since neither of them have moved, the only way the Black Rook could have gotten to ...


2

Well, this one is much easier than the last one, with more extra clues. I did all this without looking at either Daniel Duque's answer or my previous partial answer. Step-by-step deduction Now we have: Now we have: Done! Final solution


4

I am not entirely sure about what the OP means by needing knowledge from previous puzzles to understand the meaning behing the colours; nevertheless I figured a path the knight can follow though his realm: Unfortunately there is not much I can add about how I got there, and it would be very extensive to explain all 64 squares:


3

Step-by-step deductions (partial) Filled grid (partial) (A big number means that number is definitely in that square. Two or three small numbers means that square must contain one of those numbers: e.g. G1 must be either 33 or 63, although either 33 or 63 might also be elsewhere. A number with a question mark means that square is one of only a few ...


0

Getting the ball rolling with some initial thoughts... Notation: Bold - Certain Not Bold - Hints, like those in Sudokus 1. 2. 3. 4. 5. Grid after Step 1-5:


2

Sorry for reviving a 5 years old question, but I can fit: I hope to avoid downvotes by pointing out that this troll solution satisfies all the conditions of the original question.


7

I was not the first, but my answer is Edit: I later spotted two solutions:


6

the correct path is Image:


6

First to answer :) This is the answer, here you go :) Method:


1

The white king is on and the last few moves were Proof: Let's consider these possibilities in order. Therefore Now, This is no good yet, because


1

To submit something that no one has yet suggested, Rb2..Nc7# is valid but not mandatory for black because the option of Bxf7 is always in play unless black's king is placed into check. The solution hinges on F7 because outside of A1, it is the subject square of which Black has the most options. A1 tells us two things when combined with F7: You must check ...


1

I'm fairly certain Going through all possibly moves: The only piece I'm confused about the purpose of is the white pawn on h4. EDIT:


3

Since we're encouraged to think outside the box, This sort of "solution" is not without precedent. For example,


1

Alongwith what Zymurge found ,


4

The answer is: This works because


8

I must be missing something because I'm getting a lot more than 54 mates-in-two for the position below? I have 116 listed, although I'm doing this by hand so there may be some errors included. Is there some rule I haven't considered?


2



1

Assuming black pawns move upwards, I think promoting either pawn into a Lolcat should do the trick in one move.


5

My best solution for two equal-sized armies gives each army an area of exactly $\frac{7}{48} \approx 0.1458$. 3 I am not aware of any proof that this is the best solution for two armies. However, you can approximate our point-sized soldiers with regular chess queens on an NxN board, and it turns out that many provably best solutions to that discrete ...


2

One of the solutions (maybe non optimal) Update: Some explanation


0

Just for fun, castling queenside for checkmate was the original idea here. But there are heaps of duals in such an open position, so it’s not really that strong of a puzzle. I also managed to get that idea out in two less moves then stipulated. The Intended Solution: ! ! 1. Ke4 Bc4 2. bxc4 g5 3. Kd4 g4 4. Ke4 h5 5. Kd4 h4 6. Ke4 Rh5 7. Kd4 Re5 8. ...


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