## Search found 28 matches

- Mon Feb 23, 2015 12:03 pm
- Forum: Geometry
- Topic: Flipping objects in a bounded region
- Replies:
**1** - Views:
**9877**

### Flipping objects in a bounded region

(Sorry, this problem is kind of computational geometry, not math-oriented...) A convex polygonal region R with k vertices is given on a 2D plane, and there are n convex polygons P i , with k vertices each, which are pairwise disjoint and completely inside R. These polygons do not sum to R; in other ...

- Mon Jan 26, 2015 1:51 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 118
- Replies:
**20** - Views:
**11202**

### Re: Problem 118

What language are you using?

- Sat Jan 24, 2015 10:12 am
- Forum: Discrete Mathematics
- Topic: Partial orders and expectations
- Replies:
**3** - Views:
**7263**

### Re: Partial orders and expectations

... but I suspect E(n, k) = n!/2 k . While coding a brute-force counting algorithm to check some more values, I think I found a proof to this: The expected value E(n,k) is N = (the total number of the tuple (one of n! permutations and its k matching orderings)) divided by D = (the total number of k...

- Fri Jan 23, 2015 10:11 am
- Forum: Discrete Mathematics
- Topic: Partial orders and expectations
- Replies:
**3** - Views:
**7263**

### Re: Partial orders and expectations

Yes, the correct value of E(3,2) is 3/2, not 4/3. I edited the original post.

- Thu Jan 22, 2015 10:53 am
- Forum: Discrete Mathematics
- Topic: Partial orders and expectations
- Replies:
**3** - Views:
**7263**

### Partial orders and expectations

There are n cards numbered 1 to n, and you have a machine that will pick orderings between k pairs of cards. For example, if n = 5 and k = 3, you might see something like this: 1 -> 3 4 -> 1 5 -> 2 (left -> right: place "right" somewhere on the right side of "left") In this case,...

- Wed Oct 22, 2014 10:38 am
- Forum: Recreational
- Topic: The Tetris Problem
- Replies:
**6** - Views:
**5112**

### Re: The Tetris Problem

That's much more than I expected. So 4-width 7-line problem can be solved using only one Single, one Double and one Tetris, without any fancy rotations... It seems like the problem (both variations) will fit well as the final boss for a Tetris-like puzzle game, if I ever have a chance to make one......

- Tue Oct 21, 2014 4:20 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 047
- Replies:
**49** - Views:
**14108**

### Re: Problem 047

Yes, 3^3 can be a factor of one number in the family.

- Tue Oct 21, 2014 2:56 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 047
- Replies:
**49** - Views:
**14108**

### Re: Problem 047

You count the same prime only once. So 644 has 3 distinct prime factors (2, 7, 23).

Power of each prime factor does not matter; otherwise the problem would have no solution because one number in 4 consecutive numbers must be a multiple of 4 = 2^2.

Power of each prime factor does not matter; otherwise the problem would have no solution because one number in 4 consecutive numbers must be a multiple of 4 = 2^2.

- Sun Oct 19, 2014 5:13 pm
- Forum: Combinatorics
- Topic: Combinations under constraint
- Replies:
**0** - Views:
**8426**

### Re: Combinations under constraint

Do you mean, sum(E(k)) is divisible by max(E(k)), not max(S(n))? If so, one property of the sequence F(n) will be that dF(n) = F(n) - F(n-1) = # of subsets of S(n-1) whose sum is divisible by n, including the empty set. Each dF(n) can be calculated using DP, having array size of n(n-1)/2 = O(n^2) an...

- Sun Oct 19, 2014 2:19 am
- Forum: Recreational
- Topic: The Tetris Problem
- Replies:
**6** - Views:
**5112**

### Re: The Tetris Problem

It was really fast!

Yes, that was what I was looking for. Thanks!

Yes, that was what I was looking for. Thanks!

- Sat Oct 18, 2014 5:41 pm
- Forum: Recreational
- Topic: The Tetris Problem
- Replies:
**6** - Views:
**5112**

### The Tetris Problem

Once I was playing Tetris several months ago, and I suddenly came up with this question: Is it possible to clear 4 lines of width 7 using exactly one piece of each type of tetriminoes? Background information: The tetrominoes used in the Tetris game are called tetriminoes. (Beware one alphabet) And t...

- Mon Sep 29, 2014 10:01 am
- Forum: News, Suggestions, and FAQ
- Topic: New awards.
- Replies:
**68** - Views:
**33506**

### Re: New awards.

How about a contiguous award on top of centurion (i.e. Block of 200/300/400)? It is way harder to solve consecutive problems compared to hopping around and solving the low hanging fruits. Getting 200 consecutive would be much harder than 100 or even 150, mainly due to a few high tiers at 150-200 ra...

- Thu Sep 18, 2014 5:09 am
- Forum: Combinatorics
- Topic: Combinatorial problem
- Replies:
**1** - Views:
**7367**

### Re: Combinatorial problem

The problem is a variation of Set Cover( http://en.wikipedia.org/wiki/Set_cover_problem ) whose optimization is NP-hard: Take the set S of combinations of 5 numbers from 1-10, which will have 252 elements in total. Take a new set of subsets of S where each subset contains all reachable 5-combination...

- Sat Sep 13, 2014 1:39 pm
- Forum: News, Suggestions, and FAQ
- Topic: Errors/Warnings/Bugs
- Replies:
**591** - Views:
**197781**

### Re: Errors/Warnings/Bugs

I encountered a weird bug on the Problems page.

I was viewing Unsolved problems, sorted by Solved By in Descending order.

I clicked to see All problems, and then the page suddenly showed:

After I clicked All again, the page returned to normal.

I was viewing Unsolved problems, sorted by Solved By in Descending order.

I clicked to see All problems, and then the page suddenly showed:

After I clicked All again, the page returned to normal.

- Fri Sep 12, 2014 6:17 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 017
- Replies:
**83** - Views:
**32858**

### Re: Problem 017

No, the question says the right thing. I get the correct answer only after adding 11("one thousand") at the end, so if your solution is accepted without counting it, you are actually getting a wrong answer which is off by 11. ...since many other problems involved half-open intervals rather...

- Wed Aug 27, 2014 3:21 pm
- Forum: Geometry
- Topic: Finding offset and angle of planes
- Replies:
**1** - Views:
**10042**

### Re: Finding offset and angle of planes

Before talking about the solutions, a few assumptions are needed to have a unique solution: (1) The two triangles seen from both tools must be identical (have same side lengths). (2) The triangle must be scalene (different side lengths) and non-degenerate (three points not on a straight line). Here ...

- Sun Aug 17, 2014 5:14 am
- Forum: News, Suggestions, and FAQ
- Topic: Project Euler Returns
- Replies:
**80** - Views:
**34492**

### Re: Project Euler Returns

It might be unrelated to the renewal of the site, but an image is missing on

**Problem 377**(View Problem). The source says 'sod-13.gif'.- Wed Aug 13, 2014 9:11 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 051
- Replies:
**60** - Views:
**29604**

### Re: Problem 051

Basically I test all permutations of all the digits in no particular order. Obviously having some similar digits there is some repetition. In the end I have output all the families with the minimum number of digits required to get 8 primes. I wonder if there is a way to know when my answer is THE a...

- Thu Aug 07, 2014 10:56 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 022
- Replies:
**60** - Views:
**23611**

### Re: Problem 022

First of all, don't post code for a problem here, even if the code is incorrect. When you can't tell what went wrong, you can test your code with small inputs, for example AAA A AB AA which should evaluate to 1+4+9+12 = 26. Errors can be anywhere, from sorting algorithm to lexicographic comparison a...

- Wed Aug 06, 2014 3:31 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 051
- Replies:
**60** - Views:
**29604**

### Re: Problem 051

You are correct. The answer is not too high, so solving the problem is computationally tractable.