110
White must
Indeed
Note that
87
Since the puzzle turned out to be way hairier than expected, my earlier answer got quite messy and hard to follow, so here's a complete rewrite. It's still very clunky, but I hope it's much clearer than the earlier one. Also, there's an executive summary at the end, if you are not interested in all the gory details.
First, we'll need to make some general ...
75
I see Evargalo has found the right answer, but the first thing that occurred to me was
75
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!
31
(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 ...
28
I claim that
The reason:
26
The word that doesn't belong is
The crossword looks like this:
How I solved it:
26
This seems to work:
And the position looks like this:
Apart from the symmetrical solution, this might very well be unique:
22
Featured in the past, compositions by musical uncle Francis, at:
Handbill:
La-Fa-Sol-Do-Re
Ti-Fa-Sol-Do-Re Ti-Ti-Do-Sol-Mi-Re Sol-Do-Do-Mi-Sol Sol-Ti-La-Fa-Mi-Re-Mi-Re
Ti-Fa-Sol-Do-Re La-Ti-Do-Fa-Mi-Re Fa-La-Ti-Mi-Sol Sol-Ti-La-Fa-Mi-Re-Mi-Re
Mi-Mi-Sol-Mi-Sol La-Ti-Do-Sol-Mi-Re Fa-Do-Ti-Mi-Sol Sol-Do-Mi-Re Do-Mi-Re
...
19
It is currently:
because
All that means it is now the
19
Oooh, that's a nice one. First of all,
Further more, it's a mating move by
I first tried (in vain) to
So it must be
It turns out that
with the following moves:
Final position:
18
HTM's answer is very nice and must be what's intended here. But I'm not quite seeing why the puzzle isn't cooked as follows:
17
Not exactly a checkmate, but there is an option with a similar effect that can be performed in zero moves:
17
Note that without their king, White’s plan would be
Thus, the aforementioned sequence of moves will
The only square White’s king can be in that satisfies the constraint is
15
In the position shown,
Now,
Before that
Black
White
So we have the following sequence of events.
Now,
We still haven't figured out
Suppose that
Now
So let's consider
We must now be quite close to having nailed down
So far, so good. But black
Aha!
Therefore white's last move was
It's possible that some of the reasoning above can be pruned ...
13
rand al'thor points out that
and since
the numbers are
but after the square marked 16, Robert's got this (image courtesy of OP)
and we can now see why Robert is distressed:
... which is a situation that always frustrates me, though I don't get quite as angry...
12
This solution is no longer valid. The original posting of the problem had a typo (and lacked the restriction of "no auctions"); with the typo, at least one auction was required to solve, and this answer provided such a solution.
For the corrected problem, @dcfyj's post provides the correct answer.
It is Player 1: Battleship's turn
Here's how the game plays ...
12
Well, for the mate in 1,
For a mate in 2,
An assured path to victory:
As to the final question:
EDITED after the sudden appearance of the corner knight:
The stuff in the spoiler tags got shuffled around too, so let's recap:
12
I suspect the broken rule is
Allowing the checkmates
A hint pointing to this answer:
12
The answer is
Reasoning:
So
This could only happen
White's f and g pawns
So
would be
11
Since the rook and the bishop are in a threatening position for both kings, they must belong to the same color. I can see but one explanation for this:
and the last step was
As to how a mate in two was possible from here, let's assume the colors were like this:
Then Megan could have been mated with:
(From the story we can assume that Megan probably didn'...
11
To add on to Glorfindel's answer,
10
6 rounds have been played and no-one has any voids before you sit down. There are therefore at most 1 round of spades and hearts (if there were 2 then 8 cards are played + 4 visible + 2 with the opponents =14), and 2 rounds of diamonds and clubs.
The only way to get to 6 is if exactly those were played, meaning that before you sat down there had been 1 ...
10
It is:
Reasoning:
10
The other guys already got the answer. (Maybe not with completely airtight arguments, but nevertheless.) The explanations would be a lot easier to follow if they had pictures, so here's one:
Legend:
Red square: after turn 1, you are in one of these.
Green square: after turn 2, you are in one of these, or in some red square (except square 2).
Red circle ...
9
As requested by @KeyboardWielder in a comment on my earlier answer
After much back and forth, and investigation of the structure of the code using the print-to-console debugging technique passed down through the ages, Ananda and the Master had come to the following conclusions:
The lines of the input file were being read in the correct order.
The input ...
9
Challenging for how few moves are involved.
9
I present to you, pure inefficiency:
It can definitely be done faster, but this works.
9
is mate because
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