-1
$\begingroup$

You go into a room, and the door closes. Some sign says:

You must find a forced checkmate on the black king in two moves. The key for the Vigenere is ONLY your moves. In other words, if you played Qc3 followed by Re1 (regardless of black's moves) it would make the key Queencthreerookeone.

The chessboard:

enter image description here

A sign on the left says:

Here is your first digit: Wwfcz evj aaylh hvmaxn eev seegp.

There are 3 signs on the right says:

T = 19

Here are the five last numbers of the code: recognize the opening and encode the first letter of each opening to the cipher above. Read from top to bottom and line by line, the 1 indicates the opening's first move.

  1. Nh3 d5, 1. d4 d5, 1. d4 e5, 1. f3 f5.

And as always, there is a codepad.

1  2  3
4  5  6
7  8  9
 ENT  0

Question: What is the code and why?

$\endgroup$
1
  • 3
    $\begingroup$ Apparently this is from a puzzle game named Untitled Door Game. See comments here. $\endgroup$
    – Jafe
    May 12 at 11:29
2
$\begingroup$

The mate-in-two is

1. Rb1+ Kxb1
2. Qb2#

So the key is

ROOKBONEQUEENBTWO, which produces... no intelligible result when used.

However, the board seems upside down -- white start on ranks 1-2 but the position makes a lot more sense if black pawns move up (although I think it's a legal position either way). So assuming the markings are wrong, we turn the board markings around. Now the moves are

1. Rg8+ Kxg8
2. Qg7#

So we get the key ROOKGEIGHTQUEENGSEVEN, which does give an intelligible result: First and third digits are equal.

Now to figure out the listed openings:

1. Nh3 d5 is the Amar opening.
1. d4 d5 is the closed game.
1. d4 e5 is the Englund gambit.
1. f3 f6 is the Duras gambit.

Given that T = 19,

This could be a simple A1Z26 conversion, except we start from zero instead of one. Converting the openings' first letters (ACED) gives either 1-3-5-4 or, if we use padding zeroes, 01-03-05-04. Given the hint from before that the first and third digits are equal, maybe we're looking for the latter combination, so 01030504?

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.