Cast Key - Hanayama Metal Puzzle

Product Information

Let’s imagine our squares to be arranged into a 2x2x2x2 tesseract — a 4 dimensional hypercube of side length 2. The rules for B columns and C and D rows are very similar. English is a little vague when it comes to logic but X unless Y could be regarded as X and not Y. So if X and Y are invited to a party but really don’t like each other we might have the rule X unless Y and Y unless X, more formally (X and not Y) and (Y and not X) which is exclusive OR. Because our Xs and Ys are disjoint, the full rule of X will give us one half of the exclusive OR and the full rule for Y will give us the other half. Learning should be fun and on Wordgames.com you can find lots of games like Free The Key that make practice a joy. Play now one our our best puzzle games! As with all vector spaces, a general vector here is a linear combination of the independent basis vectors. However the scaling factors are all either 0 or 1 which implies that each basis vector is part of the linear combination or it is not. In effect, this vector space describes the 2x2x2x2 tesseract from Approach A. Your companion re-enters the room, without having any opportunity to see or communicate with you. He observes the chessboard and the arrangement of coins and points to the square where he believes the key and freedom reside.

They have a chessboard where each square is covered by a coin — either heads or tails. Moreover it’s a special chess board with a hidden compartment in each square. A single one of these squares contains a symbolic key to the jail and freedom for you and your cell companion. You will know which square contains the key and your fellow prisoner has to guess. The prisoner warden gets to set all the coins on the board and we only get to flip one to convey our message- a tiny signal in all that noise. At first glance this seems completely impossible but let’s start by understanding what is relevant in our puzzle and what we can discard. Then we’ll see if we can start with a simpler version of the problem. Nothing sneaky is allowed on pain of immediate death i.e. This is a pure logic problem- there is no meta game. Paper and Pencil, calculator and plenty of time are available to you and your cellmate if necessary.On the chessboard in Diagram 13, we use combinations of letters to label the rows and columns. For example A by itself represents the 2nd column; B by itself represents the the third column; A and B together represent the 4th columns and no A or B indicates the first column. This means that A can indicate either the second or the forth column. The squares are also labeled with combinations of letters. By analogy with the leap year rule we have the following (involving all the squares that contain A in their label): So all we need to do is the following. Scan the board from top to bottom calculating a cumulative value which starts as the vector 0000. Each time we encounter a dark coin, we perform an exclusive OR of our square’s location label with the current cumulative value to get the new cumulative value. At the end, this cumulative value represents the location of a square. If we use the board in Diagram 9, this turns out to be 1011. However we need the board to represent the third column of the second row or 0110. To list the bits that need to be corrected, we simply exclusive OR these two vectors together to get 1101 which is the location of the coin that we need to flip. When my friend Peter came up with this approach, knowing my love of Pascal’s triangle, he was keen to point out the numbers of squares are 1 with no letters, 4 with one letter, 6 with 2 letters, 4 with 3 letters and 1 with 4 letters which is a row of the triangle: 1 4 6 4 1. Diagram 2: The chessboard from Diagram 1, now covered in counters after the secret compartment has been closed Some Properties of the Problem Let’s start by solving the two square case which, based on the hints above, is relatively simple. We need one square that does nothing when we flip its coin (aka change its counter )- let’s choose the one on the left. This means that the right square needs to encode the location of the key. Let’s choose light for left and dark for right. When the warden has finished with the board, if the right-hand coin indicates the incorrect square for the key, we flip it. However if it is already correct, we flip the left-hand coin which has no effect on our key location message.

Exploring different approaches to this seemingly impossible problem reveals some interesting connections between various branches of mathematics including logic, set theory, multidimensional vector spaces and even category theory. Outline of the Puzzle In some sense, the chessboard must encode the 4 bits of information required to specify the row and column of the key square. Given that the prison warden will have chosen a configuration either at random or maliciously, some or all the bits encoded may be wrong. So it must be possible to choose a coin to flip (represented here by the counter changing color) that will fix the wrong bits without impacting the correct ones. In the case where the chessboard already encodes 4 correct bits, there must be at least one square that has no effect on the 4 bits when its coin is flipped.

Game Controls & Info

You and your cellmate can discuss how to encode a message using the chessboard but the prison warden can hear and understand everything that you say. You observe the prison warden hiding the key in one square and then arranging the 64 coins as heads or tails however they deem fit — presumably trying to frustrate your system.