isaacg
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I can make the robot take The robot must take this many steps because

Simpler explanation: First, notice that B cannot be the knight, because then for his statement to be true, A would also have to be a knight, and we know there is only one knight. Second, notice that ...

I have a solution using 448 consecutive integers. The integers are centered on a number which I will call $c$, which is: ...

Suppose, instead of maximum length or maximum length difference, you care about maximum edit distance (levenshtein). Here are the winners: 9 BETTERING DUNLAP -....--..-...-.--. 9 HICCUP HEARTENING ......

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:

I wrote a straightforward depth-first-search program, and found a sequence of length namely This took about five minutes to find. I don't see any obvious pattern to extend this indefinitely. I'll ...

Cleaned up answer: To make things concrete, let's say that the cop's strategy is represented as a function c(t), which is dependent on the thief's trajectory w(t), and vice-versa. The cop's strategy ...

Let's start with a simpler question. Suppose there were three blue-eyed people - then what information do the three people learn on the first night? Before the first night, blue-eyed people see two ...

I can prove that the answer is exactly Several people, including Jaap Scherphuis, have shown that the square can be covered with this many pentominoes, so it only remains to show that at least this ...

The optimal solution is This set of coins allows N = This set of coins is given in this paper: Some Extremal Postage Stamp Bases, by Michael F. Challis and John P. Robinson. This paper was found by ...

We're going to There's a single odd letter out in each of the clues, and for all but the hex clue, the odd letter out is based on how the word is typed on a QWERTY keyboard. OLD FLAGS: FISHY DISH: ...

As others (ex: Gareth McCaughan) have mentioned, by looking at divisibility by primes up to 13 we can narrow the list down to 19 possibilities: 11111, 11441, 11551, 11771, 11881, 33113, 33223, 33443, ...

I found/figured out the solution. Starting with the reference to this problem here: http://www.chesshistory.com/winter/winter44.html#CN_5449 A few entries down the page, a more illuminating further ...

Solution, using a very specific definition of a rule of chess I can solve this puzzle, if I use Assuming I use that rule, here is a solution:

Another option (besides "I can lie"): Say "I always lie." This statement is false, but would be true for the liar, so your friend can pick you out. In contrast, "I can lie" is true, but would be ...

I can do it in $6 + \epsilon$ square meters, where $\epsilon$ is any number greater than zero. Let the rectangle be of dimensions $w \times (6/w + 3/2)$, where $w$ is a very, very small width. Lay it ...

I have found a $6156/1505 \approx 4.09$ solution. Here is how it goes: Label the faces A, B, C, D, E, F. Roll the dice 3 times. Each permutation of a given set of three faces is equally likely, ...

I can prove that is minimal. The difficulty comes in distinguishing each of the 14 pairs of neighboring slips of paper, as well as the first and last papers. There are 15 such pairs of slips of ...

The best in my word list is 13 letters:

Current best list, as far as I know: a: agama, 3/5 b: bobby, 3/5 c: coccic, 4/6 d: dodded, 4/6 e: peewee, 4/6 f: fluff, 3/5 g: glogg, 3/5 h: hashish, 3/7 i: bikini, 3/6 j: hajji, 2/5 k: kakkak, 4/6 l: ...

We need to reason about two worlds: reality, and what the prisoners believe is happening. Based on A's statement, Therefore, From B's comment, From C's comment, Now, something important happens: ...

I wrote a program to find all subdivisions of the morse code into english words. The minimum number of words possible is 3. There are 96 subdivisions into 3 words in my dictionary. None of them look ...

I solved the puzzle, and found that the solution is not quite unique: In the bottom right corner, there are three possible arrangements of three pieces: Either J, L, O, or L, J, O or J, Z, L. These ...

If the pencils are cylinders, and do not have to have equal radii, nine mutually touching cylingers is possible. See page 15 of this paper.

I got that the woman must have at least: coins at the end of the game. Reasoning:

My intuition is that the answer is D, for the following reasons: As we go from left to right, the number of highlight squares goes ?, 3, 2, 1. So we would expect the answer to have 4 highlighted ...

A starting point for part (b): Let's consider some smaller boards. I'm going to normalize the average coin value to 0, and try to analyze arbitrary starting configurations where the coins sum to 0. ...

I tried solving this puzzle, and have come to a concise proof that no valid solution exists. I invite everyone to try to find any point where this proof is not valid, because I would greatly prefer to ...

The secret is that we are As a result, we could However, the puzzle asks us to