Project Euler Forum

I think one useful tool is simply to classify. A large chunk of problems are fundamentally about prime factorisations of integers from 1 to N: i.e. number theory. (They can be further classified: e.g. problems which rely on efficient counting of primes up to N). A smaller chunk of problems are about...

The question states The rules do not place any restriction on the number of occurrences of M, so all integers have a valid representation. but doesn't explain how to get negative integers, and I got the tick with an answer which assumes that negative integers don't have a valid representation. I thi...

There's an error in the problem statement. It currently says

the denominator is a power of 3 i.e. $b=3^k, k>0$.

It should say

the denominator is a power of 3 other than 1, i.e. $b=3^k, k>0$.

(The other possible correction, $>$ to $\ge$, doesn't match up with the test cases).

For those interested in knowing, the minimal information script has been re-implemented. I have updated the first post to reflect any changes that may have taken place. There seems to be one change, possibly more recent than that, which isn't reflected in the first post: when logged in, I'm seeing ...

On another site, I recently tackled a question asked about an "idle" computer game which boiled down to the expected time to reach the absorbing state in a Markov process. As I didn't know the standard approach (I have now learnt it, so don't feel obliged to explain it :wink:), I reasoned as follows...

Aha. Thanks.

I was expecting to see either Problem 253 or at least a countdown to it today. Have I missed an announcement? Is something up?

Another obvious extension is the regular "solids" of infinite volume. There are three, with respectively 6 triangles, 4 squares, or 3 hexagons meeting at each vertex, and tiling the surface of an infinitely large sphere.

tiny wrote:The text says: whose perimeters do not exceed 1.000.000.000

Does this mean, that for every triangle the perimeter is less than 1.000.000.000?

Almost. It means less than or equal to 1.000.000.000.

I leave it as an exercise for the reader to prove that the un ordered partition of n health between r gladiators is (n+r-1)Cn. I think there is a typo there. Yes. For n=160, r=16, I find the number of ordered partitions to be much larger : 175 C 160 =1823828930470225269045 and the number of unorder...

Huh, that looks interesting. How did you calculate that number? It's the ordered partition* of 160 health between 16 gladiators. I leave it as an exercise for the reader to prove that the unordered partition of n health between r gladiators is (n+r-1)Cn. * The partition into n distinguishing the n....

I make it a Markov model with 236236542585120 states. Fortunately the transition matrix is sparse

#!/bin/sh tr A-Z a-z </usr/share/dict/british-english | sort -u >/tmp/wordlist cat /tmp/wordlist | tr a-z phxdtl%brjzfvn%aqiyeum%csk | grep -v "%" | sort >/tmp/translated cat /tmp/wordlist /tmp/translated | sort | uniq -d Not that many. There are a handful of 5-letter ones: buffy hulls bulls huffy ...

The spoiler seems to be broken, so I've unspoilered this. Sorry. There's an easily proved lower bound of 2.52; it's achieved by the following 38 sets. I used a brute force search considering only coin values up to 100. The puzzle is so closely related to Golomb rulers that I thought offsetting one o...

hk,forget the direct picking of numbers. What's intended is that if Ben has B hp and Hur has H hp, then Ben wins with probability B/(B+H) and the only other possible event is that Hur wins.

[spoiler]59, at which point the probability is .8333...[/spoiler]

There's an efficient way to do it for sets not larger than 63 (64 if you have unsigned longs in your language). HAKMEM #169. Alternatively, see Knuth's preprint of TAoCP, Vol 4, Fascicle 1a, or check out my code in the discussion fora for problem 215*. *Or for another problem, but I've edited that o...

1 is not a prime.

Simple code, but it could be made somewhat faster. As far as I know, Math.round(Math.sqrt(s)) is never off by more than one in the range of long. If you use StrictMath.sqrt then you're guaranteed an error of no more than 1 ulp. Since the largest long is 2^63-1, the largest sqrt of a long is about 1...