Ilmari's answer covers the logic of the answer perfectly, but in case anyone's still confused, here's a version of the diagram I find more intuitive (I made it myself while solving the puzzle). The Pink squares are rooms where the princess could be on the given day, the blue squares are where the prince knocks that day, and the black squares are rooms in ...


Maybe this diagram will help you visualize the solution: In this diagram, the vertical axis shows the room number (1–17) and the horizontal axis shows the day (1–30). The $\color{red}{\text{red}}$ dots show the rooms on whose door the prince knocks on each day, and the $\color{darkgreen}{\text{green}}$ dots and arrows show the possible ...


My most sincere apologies for this. Really.


Sure you can. There's finitely many possible mazes, so solve each one in sequence. To solve a maze, imagine you're in that maze. Figure out where you are in the maze by simulating starting on the start space and following the instructions corresponding to the sequence of steps you've taken so far. Then, make the moves that would take you from there to the ...


Give these names to all the squares: 163 4 8 725 Each number can only be accessed by way of the numbers before and after it (where 8 wraps around to 1). That means they form a loop. Since they can never pass each other up on the loop, their relative ordering cannot change. Therefore it is impossible.


In the worst-case scenario, it requires to locate the radioactive rods. Several answers already describe strategies for locating the radioactive rods. I will give another. Testing strategy: Start by testing two rods. If neither of these rods is radioactive, use the five remaining tests on five of the six remaining rods, one at a time. If two of these rods ...


I found a solution that uses 16 moves. After exhaustively checking that there is no solution in 14 moves, I conclude that 16 moves is optimal, because after any odd number of moves the number of white and black squares occupied by knights cannot be equal.


Alain Remillard has given the mathematician's answer. Here's the physicist's one: Step 1: Obviously, in such a universe, regardless of their speed, the cannonballs will travel in a straight line and hit each other in the middle. Step 2: Assume "Step 1" does not exist. Therefore


You need at least 16 Moves. Let's make the task visually more simple. The initial board is: a4 b4 c4 a3 b3 c3 a2 b2 c2 a1 b1 c1 We cut it into 12 cells and connect only those, which are separated exactly by one move of a knight. Easy to check that the result is the following: c4 - a3 - c2 - a1 | | | | b2 b1 b4 b3 ...


Perform tests of nine sheep on all but one sheep according to the illustrated patterns: The two important properties exhibited are The claim is that given a set of test results there is at most one possible group of five wolves. Suppose instead that some set of test results could have been produced by two different groups of five wolves A and B. Then both ...


Ok, I think I got something. The answer should be : The ancient civilization The operation *|* is performed : Now, the real problem. In order to obtain the result for 2 numbers, we have to And now, all the values in the question : With, of course, the final answer :


As atonement for my insolent lateral-thinking answer, I offer an optimality proof. If you keep repeating the correct code, the are six possible different orders: 1 abcdabcdabcd 2 abdcabdcabdc 3 acbdacbdacbd 4 acdbacdbacdb 5 adbcadbcadbc 6 adcbadcbadcb Each of the orders contains four possible codes. The orders are important, since after testing one code ...


This is a pretty common puzzle. Warm up Answer: Advanced Answer Explanation:


There are only 16 different possible state combinations of the four ancestor cells, and you can find them all in the image, so there is a unique answer. The rule is as follows:


The pattern is self-similar, and can be formed by repeatedly scaling and rotating copies of itself: An alternate dissection that fits in a diamond:


The knights need a maximum of: I made a quick drawing to show my strategy. The yellow squares are the rooms the knights look into that night, the black squares are rooms in which the princess logically cannot be.


The solution assumes that she cannot ever stay in the the same room. Either on even days she is in even rooms or on even days she is in odd rooms. As he is going from room 2 to room 3 he is going from an odd day to an even day. She can go from room 3 to 2 if she is in even rooms on even days. She can never pass him if she is in odd rooms on even days as he ...


Treat each soldier as a binary digit, $1$ if facing east and $0$ if facing west, and order them from west to east (i.e. left to right). We start off with a random $n$-digit binary number, and the algorithm is: The process can only end when the number is a string of $0$'s followed by a string of $1$'s, i.e. it is one less than a power of two. Its digit sum ...


A series of variants of this puzzle came up in one of the trade magazines - possibly Communications of the ACM. When two soldiers face each other, they are required to turn around. Assuming the soldiers are considered to be interchangeable, this is equivalent to the soldiers marching 'through' each other (or swapping positions, if you prefer). In this ...


Edit: Now that @GOTO 0 got it in 16, I can at least prove that his solution is optimal. Proof: My best was:


It seems like you are trying to find One such example that satisfies this is In particular, As mentioned by armb in the comments, there is a good answer here discussing the maximum orders for an $n \times n \times n$ Rubiks cube.


Here is a revised solution, for... ...which  (again) seems like the maximum to me.  has been verified by Molhan as being maximal.   Trivial steps have been condensed. These steps may be reversed, exchanging the roles of pegs 1 and 6, to complete moving the whole tower from peg 1 to peg 6. This approach was derived by ...


Any commutative hash function will do.  Using RSA makes this relatively easy, I think. So Alice and Beth both establish their secret primes, and, in a twist, keep everything secret.  $ % Make EA, EB, DA and DB look like functions; i.e., *not* italic: \def\EA{\operatorname{EA}} \def\EB{\operatorname{EB}} \def\DA{\operatorname{DA}} \def\DB{\operatorname{DB}} $...


Your link to Diophantine equations shows that you already have an idea of the answer. The only actions available are filling a jug completely from the tub, emptying a jug completely, or pouring one jug into another. If we fill empty jug $k$ from the tub, then the amount of water we have is increased by $v_k$. Similarly, if we empty jug $k$ when it is full, ...


Not exactly answering this question, but given 9 infections still on an 8x8 board, it is actually possible to delay the inevitable until 40 days. Pretty counter-intuitive huh? Locations are A1, A5, A8, B3, B7, D1, E8, G1, H8 Also, I believe that this is the maximum for an 8x8 board with any number of initial infections. As for the edges argument I made ...


I have a solution with a success rate of 93.5%, according to my simulations. The reason this solution works so well is Here's my code that I used to verify my solution:


First of all, let's see why your brute-forcing fails. (This is the puzzle part, the rest is plain old math.) No matter which you chose, the number at the bottom right would have to be both odd and even at the same time, so there's no integer solution. However, there are four equations and four unknowns, so we should have at least one solution (unless the ...


The standard solution is that all clones signal their first passage using some state of the bulbs to an elected clone, the "counter", who counts how many there are. You need to address 3 things. 1. How to transmit the information of one's passage reliably, 2. How to deal with the initial state, 3. How to elect a counter. Note that the lights don't ...


Thinking out loud, not a solution yet, but spoilery enough that I didn't want to put it in a comment: However, Still-not-an-answer UPDATE: However, I also notice that the situation is not symmetrical: we may know a way to find $k$ wolves among $n$ sheep using $t$ tests, but that won't help us at all to find $n-k$ wolves among $n$ sheep. (Under the ...


Here is a permutation that works upto n=12: 1 6 11 4 5 10 15 8 9 14 19 12 13 18 23 16 17 22 27 20 21 26 31 24 25 30 35 28 29 34 39 32 33 38 43 36 37 42 47 40 41 46 51 44 45 50 3 48 49 2 7 52 Essentially, keep the first and fourth cards with same rank in its place. And permute the second cards of all ranks cyclically, and the third cards of all ranks ...

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