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3 votes

The universal ticket

Update: Honing in the parameters allowed for a score of 153. This is much closer than I expected to get to the 165 mentioned on the website. original: I decided to go for a brute force approach and ...
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  • 31
1 vote

If the door in Monty Hall is opened by a person without knowledge of where the prize is, and also does not reveal prize, do the odds change?

The problem with this scenario is that it totally ignores the case where the uninformed person opens the door and reveals the big prize, something that will happen ⅓ of the time. Then, having ...
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1 vote
Accepted

If the door in Monty Hall is opened by a person without knowledge of where the prize is, and also does not reveal prize, do the odds change?

This is the "ignorant Monty" variant of the problem and the probability is Proof:
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  • 6,454
0 votes

Multiple Choice Puzzle

The "question" is meaningless. Consider the following Python function: ...
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  • 624
0 votes

Multiple Choice Puzzle

There is no paradox, just an illusion in the concept. That is, you are all deluded with the right answer because you are seeing the answers. If you assume that you don't know what the answer is, that ...
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0 votes

Find the last number in the pattern

A web search turns up https://dokumen.tips/documents/exemple-de-probleme-pentru-testul-de-competente-de-gandire-.html?page=1 and http://www.cs.ubbcluj.ro/wp-content/uploads/Subiect-test-admitere-...
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  • 266
13 votes

The universal ticket

Very unlikely to be optimal, but got to 120 on my first go: Approach: mess around with the problem until it becomes clear that connectivity of the squares will be the main problem. invent glue, ...
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3 votes

The Pawn March Problem

A simple fact: The reachable rank is unbounded. Proof: An asymptotic bound: The reachable rank is O(sqrt n log n) Proof (sketch): It seems there is still room for improvement here, cf. last sentence ...
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1 vote

The Pawn March Problem

Here is an upper bound. Here is how This strategy is not optimal though.
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  • 19.9k
1 vote

Taking turns adding a number 1,2,3 to a 3x3 matrix without repeating numbers in the rows or columns: does the first player always win?

Without regard for strategy, considering cases:
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  • 1,591
5 votes

Taking turns adding a number 1,2,3 to a 3x3 matrix without repeating numbers in the rows or columns: does the first player always win?

Alice wins. Strategy:
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  • 12.2k
5 votes

Cover an n times n grid with non-diagonal non-intersecting n-1 shortest paths

Terminology: There are two general "orientations" a path can have: bottom left to top right; let's call that "slashy" top left to bottom right; let's call that "backslashy&...
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2 votes

Cover an n times n grid with non-diagonal non-intersecting n-1 shortest paths

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  • 1,591
-1 votes

Cover an n times n grid with non-diagonal non-intersecting n-1 shortest paths

If we assume that the minimal length of a path is independent of the rest of the grid, then Auriboros' result is optimal. However, if the minimal length of a path is defined with regards to the other ...
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-1 votes

Cover an n times n grid with non-diagonal non-intersecting n-1 shortest paths

The minimum amount of paths you need to fill an $n*n$ grid is: Proof:
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  • 5,322
6 votes

Create all of the integers from 1 to 100 using 1, 5, 5, and 7

89:
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  • 68.9k
2 votes

Multi-colored polyominoes inside a 7x7 grid

An obvious upper bound for the maximum number of distinct shapes is $2+4+4=10$, and...
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5 votes

Polyominoes inside a 10x10 grid

I can do it with no straight polyominoes (except dominoes): An obvious upper bound for the maximum number of distinct shapes is $1+2+5+5+5=18$, and...
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4 votes
Accepted

Polyominoes inside a 10x10 grid

Unfortunately, this is also too easy. This solution is trivial:
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8 votes
Accepted

The alien quantum-tunnelling hyperking labyrinth escape

The answer: Note on the existence of the main loop:
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  • 6,977
2 votes

The alien quantum-tunnelling hyperking labyrinth escape

As a warm-up, can you find the path the hyperking can take from the upper left corner to reach another corner?
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  • 6,454
4 votes
Accepted

Shocking Singing Logic

I think the answer is Because there are two different arrangements that make all of the statements true, which are (in order of A, B, C, D, E, *)
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  • 15k
10 votes

Multi-colored polyominoes inside a 7x7 grid

Here is a solution in which the red and green do not touch.
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10 votes
Accepted

Multi-colored polyominoes inside a 7x7 grid

I think this would work as a possibility
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  • 123k
4 votes
Accepted

A Rotten Riddle

The process: The resulting text (with spaces added) is: And so the answer is:
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  • 6,454
6 votes

Flipping Platonic solids

I think I worked these out right (by computer search). Solutions given by describing moves according to absolute direction (degrees). Tetrahedron Octahedron Icosahedron Dodecahedron
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  • 1,591
6 votes
Accepted

Tetromino in a Pentomino Lair

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3 votes
Accepted

Ten tetrominoes inside an 8x8 grid

As promised, here are the three solutions that my computer found. The first was already found by franck vivien (and Bass), the second by JLee.
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5 votes

Fitting pentominoes inside a 10x10 grid

Rob Pratt beat me to it, but I'll post anyway because my computer found a couple of other solutions to the bonus question. I used my own program to solve it. I ran it overnight, and after 15 hours it ...
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6 votes
Accepted

Fitting pentominoes inside a 10x10 grid

Bonus: I used integer linear programming as follows. Introduce binary decision variable $x_p$ for each possible placement of a pentomino in the grid. Let binary decision variable $y_{ij}$ indicate ...
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3 votes
Accepted

Flipping through the faces of a cube?

This can be done in Since this puzzle The most straightforward flips are But @Penguino has found a series of flips that
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  • 6,879
0 votes

Create all of the integers from 1 to 100 using 1, 5, 5, and 7

2 partial solutions for 99:
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  • 226
5 votes

Fitting pentominoes inside a 10x10 grid

As a quick baseline solution: Bonus: (Edit) I also found a nice symmetric solution for the first question: I wonder what Hexomino's solution is...
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  • 10.6k
5 votes

Ten tetrominoes inside an 8x8 grid

Now that I realize we don't have to use only T's, here is one solution:
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  • 15k
1 vote

What is 12 in this same pattern?

Partial answer. Googling OEIS plus some of the numbers, I found which led me to: so we're looking for values A, B, C such that and apparently Running this through a spreadsheet gives the following ...
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  • 266
11 votes

Ten tetrominoes inside an 8x8 grid

what about the following arrangement of I found it manually by searching arrangement of same pieces, then had to change a bit strategy
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1 vote

Create all of the integers from 1 to 100 using 1, 5, 5, and 7

Here is my answer for $89$
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9 votes

Create all of the integers from 1 to 100 using 1, 5, 5, and 7

Here's an answer for 67: I'll edit in any others as I find them.
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3 votes
Accepted

Sum Like It Simple

Mark the numbers like this: a c b d e f g
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7 votes
Accepted

Four pipes on a 8x8 grid

Assuming you pick all 8 start/end points and then all the pipes get laid: I think I've got but not sure if I can prove it's the shortest possible route, with these starting points:
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  • 186
0 votes

Four pipes on a 8x8 grid

Do you need to select all four pairs of cells before the workers start, then they pick paths that minimize the total number of sections used across all four pipes? Or can you select one pair, they ...
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  • 266
1 vote
Accepted

Who is most likely to win all the chips?

I would love to be able to add this as a comment as this answer isn't really in the spirt of puzzles but I lack the reputation. I wrote some python code to brute force solving the problem. From this I ...
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7 votes
Accepted

Reverse the bit order

Extending the exact same method to 10 bits I get The way it works:
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  • 19.9k
6 votes

Who is most likely to win all the chips?

Some observations: Given that general strategy,
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  • 6,454
16 votes
Accepted

The Game of Golden Squares

I've achieved tiles, and can prove that this is the optimal solution. Reasoning: Golly 4.0+ pastable RLE of this solution: ...
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  • 4,454
2 votes

The Game of Golden Squares

While I cannot beat loopywalt's answer, my construction got just over halfway there. I'm posting it in case it sparks ideas for better constructions.
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1 vote

The Game of Golden Squares

I can get up to How do:
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  • 5,322
8 votes

The Game of Golden Squares

Update: Up to by introducing a small asymmetry Update ends. Original answer: I can do using this setup. This is essentially a big Over time (1275 turns) this will fill up to a First few groups ...
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0 votes

Ages of Widow and Her Children

I'm going to say the solution is This fulfills all requirements. Though,
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  • 10.9k
0 votes

Ages of Widow and Her Children

This feels like an easy solution or am I missing something? This doesn't exactly work in real life as pointed out in the comments, although there are multiple similar solutions:
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