Project Euler Forum

When a problem I've suggested is published, should I wait with entering the correct the answer so to give other people a chance to compete for being the first ? It sounds reasonable, and I also have the feeling that this is happening since I doubt Robert suggested all the new problems ( ), however I don't see any message (yet) saying it's polite to wait a while..

edit: Starting at post 8 you'll find some clarifications about problem 194

Also, why is problem 194 solved by 12 people the very moment it is released

edit: I guess the solved by column is incorrect

It was a symptom of the technical problems we had last week. I had to exchange all references for #194 to #193; it looks like I forgot that one. It should fix itself when the first person solves it.

I don't even understand the problem...

Yes, for problems you've proposed yourself, politeness suggests that you wait a while before you enter your answer. But once there are a few posts, it is perfectly understandable if you enter the answer to be able to view them.
Of course you have to guess when that is the case, so I'd say wait a couple of hours or so

sfabriz wrote:I don't even understand the problem...

Could you be more specific?

For some reason I expected this problem this week. Am I correct that it wasn't published last week because of the pictures of the graphs?

Btw, couldn't you have published it at some horrible time like 1am or 5am. That would've made waiting a few hours before entering the answer a lot easier

What part is unclear, fabriz?

Well, I don't understand why the compound graph represents (2,2,6) or (2,2,4)...
Why not (2,2,5) or (2,2,n) where n is up to the number of vertices? If I have 20 vertices and I use 20 colors, I don't paint neighbours with the same color for sure.
I mean, given the graph I could use more than 6 colors to paint it... or not?
Also, I guess we can glue the A and B pieces together in different ways, like ABAB, ABBA, BAAB, and so on, right?
It's really akward, but I don't understand what I have to do...

It represents a configuration of type (2,2,n) for n > 3, since it uses 4 colours. All configurations of (2,2,4) are also configurations of type (2,2,5) however (2,2,5) has even more configurations, namely those that use 5 colours.

You have to count all possible configurations of (25,75,1984). Each configuration consist of 25 times graph A and 75 times graph B. These are glued together in some way and then coloured using up to 1984 colours.

Stijn already posted the first part of my answer, so:
Yes, AABB, ABAB, BAAB all have type (2,2,x).

euler wrote:It was a symptom of the technical problems we had last week. I had to exchange all references for #194 to #193; it looks like I forgot that one. It should fix itself when the first person solves it.

Yup, it's fixed Congrats aleksey

That (2,2,4) or (2,2,6) is misleading I guess. My first thought was: why not (2,2,5) or (2,2,7), and so on...
So I was looking for a good reason not to paint the compound graph with 5, 7 or more colors.
All right, thank you, I'll try to solve it.

When I saw 12 people solved it at 9.02 with the problem coming out at 9.00 I thought it was something like:
Given 2 natural numbers a and b let's define f(a,b) = a+b. Find f(11,13) ... or something like that...

Cheers,
sfabriz

sfabriz wrote: Given 2 natural numbers a and b let's define f(a,b) = a+b. Find f(11,13) ... or something like that...

Cheers,
sfabriz

Nah, that's too easy. Should at least be f(a,b) = a*b and f(37,61)

Daniel, perhaps something like this is more clear:

I would ask for two clarifications:
"where the units are glued along the vertical edges as in the graph"
May unit B be rotated by 180 degrees ?

"so that no two adjacent vertices have the same colour"
Does it mean, that no two coloured points directly connected
by a line may have the same colour ?

Thank you very much in advance.

tom314 wrote:I would ask for two clarifications:
"where the units are glued along the vertical edges as in the graph"
May unit B be rotated by 180 degrees ?.

I guess no. Stick with the example provided by the problem.

"so that no two adjacent vertices have the same colour"
Does it mean, that no two coloured points directly connected
by a line may have the same colour ?

Yes. If they share and edge (the line you talk about) they are adjacent vertices. In that case they can't be of the same color.

Cheers

@tom314: No. Yes.

problem 195 will be accessible friday morning right?

Yep.
Just set it live, updated RSS feed and sent email notification.


Edit: email received.

When will the next problem be accessible?