isaacg
  • Member for 7 years, 5 months
  • Last seen this week
Creating the hardest 6x6 maze
Accepted answer
27 votes

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

View answer
Knights , Knaves and Spies - Part 1
26 votes

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 ...

View answer
Consecutive integers around a circle
Accepted answer
24 votes

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

View answer
Open-ended challenge: Find the longest morsagram word pair
19 votes

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 ......

View answer
Blindfold Bingo
Accepted answer
14 votes

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:

View answer
A robot surviving on top of a 3x3 platform
12 votes

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 ...

View answer
Can the cop catch the thief?
11 votes

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 ...

View answer
Blue eyes riddle: what information is gleaned after the first night?
11 votes

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 ...

View answer
Two Men - The Physicist and Plumber
11 votes

View answer
Covering an 8x8 grid with X pentominoes
Accepted answer
10 votes

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 ...

View answer
Eight coins for the fair king
Accepted answer
10 votes

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 ...

View answer
Treasure hunt 'round the world! (clue 4)
Accepted answer
7 votes

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: ...

View answer
Nice no-computers way to find limerick primes?
5 votes

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, ...

View answer
Checkmate with two knights, but with a twist
Accepted answer
5 votes

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 ...

View answer
Find Those Chess Notations! #4
Accepted answer
5 votes

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:

View answer
Knight and knave impostors!
5 votes

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 ...

View answer
Optimal present wrapping with a rectangle
Accepted answer
5 votes

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 ...

View answer
Fair results with unfair die
5 votes

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, ...

View answer
A mini game with 15 pieces of paper
4 votes

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 ...

View answer
The Survivors of Dictionaria
Accepted answer
4 votes

The best in my word list is 13 letters:

View answer
Find words with the highest proportion of a single letter
3 votes

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: ...

View answer
Guess the hat colors - Riddle 3
Accepted answer
3 votes

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: ...

View answer
How to solve morse code without spaces?
3 votes

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 ...

View answer
Let's Play Tetris!
3 votes

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 ...

View answer
How do I arrange pencils so they all touch each other?
3 votes

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.

View answer
The Trickster's Game
3 votes

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

View answer
Mensa sample question - squares with shaded and unshaded parts
Accepted answer
2 votes

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 ...

View answer
Is equality possible?
2 votes

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. ...

View answer
Logic puzzle: Driving around the United States
1 votes

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 ...

View answer
Something Simple
Accepted answer
1 votes

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

View answer