Problem 373

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omg
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Problem 373

Post by omg »

373 is really hard,
I guess writing code can take less than 3 hours only if you have solved almost all previous problems,
otherwise it can easily take more than 12 hours even if you see the algorithm for the 1-minute solution.
And brute-force solution will obviously work for months...
Dear Project Euler team members, please don't publish problem 374 until problem 373 is solved by 50 Eulerians.
Thank you!
sivakd
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Re: Problem 373

Post by sivakd »

How does it matter that there should be 50 solvers before publishing the next problem? Most of these consecutive problems are not even related to each other. Also, you seem to have interesting statistics about the complexity and the problem and the time it takes to code it. Actually "seeing" the algorithm is the hardest part and usually not the coding unless there are precision and overflow issues or other corner case issues that make it hard to debug. And if you do see the algorithm, it shouldn't matter that you solved the previous problems to code the current one.
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hk
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Re: Problem 373

Post by hk »

@omg:
every now and then we publish a problem that is somewhat harder than the other ones.
If you look at the following fastest solvers lists you can see that #373 falls in that category but not exceptionally so
http://projecteuler.net/fastest=361
http://projecteuler.net/fastest=344
http://projecteuler.net/fastest=331
http://projecteuler.net/fastest=325
http://projecteuler.net/fastest=319
In the past we did not pause after such a problem.
That gives the users the time to distract from that difficult one and solve a problem they find easier to solve.
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thomas24
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Problem 373

Post by thomas24 »

_Edit_: removed possible spoilers.

Sorry.
Last edited by thomas24 on Fri Mar 02, 2012 2:38 pm, edited 1 time in total.
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hk
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Re: Problem 373

Post by hk »

@Thomas24
As you can see I merged the topic you created with an already existing one.
To keep thing manageable we urgently request people not to create duplicate topics.
Personally I'm not a great fan of the kind of questions you're asking for a recent problem.
To be honest:
your question is on the verge of violating forum rules.
I'm under the impression you're quite new at Project Euler, but you know: this problem is less than a week old.
Please come back in a week or five or six when everybody has had full opportunity to solve the problem entirely on its own without additional clues.
In the meantime I would think it a good idea if your question remained unanswered, neither positive nor negative.
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War ruins the life and health of untold numbers of innocent children.
thomas24
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Re: Problem 373

Post by thomas24 »

@hk
thanks for pointing that out, no problem. I don't mind if you remove the post, either, or could do that myself if you tell me how to do that.

Regards,

Thomas
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hk
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Re: Problem 373

Post by hk »

At the upper part of your post there are a few buttons at the right.
The buttons [edit] and [quote] are clear by themselves.
For the others:
If you hover over them with your mouse you can see what function they have.
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War ruins the life and health of untold numbers of innocent children.
dnosrc
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Re: Problem 373

Post by dnosrc »

hk wrote:At the upper part of your post there are a few buttons at the right.
The buttons [edit] and
are clear by themselves.
For the others:
If you hover over them with your mouse you can see what function they have.
like this?

edit: doesnt work properly if you wrote
in a post :P
fistach
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Re: Problem 373

Post by fistach »

Hi,

can you prove that S(100) = 4950 by listing these triangles?
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hk
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Re: Problem 373

Post by hk »

fistach wrote:Hi,

can you prove that S(100) = 4950 by listing these triangles?
You don't need to list them to prove it.
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sbbone
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Re: Problem 373

Post by sbbone »

If you have an integer sided triangle with sides (a,b,c) and integer radius r circumscribed circle, should you
also count e.g. the mirrored (or is "congruent" the mathematically correct word?) version (b,a,c) in the sum S, i.e. adding r twice in S, or is this considered as only
one triangle?

Sorry if that should have been obvious from the problem description but i didn't get it.
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hk
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Re: Problem 373

Post by hk »

(3,4,5) (4,3,5) (5,3,4) (5,4,3) (3,5,4) and (4,5,3) are congruent triangles. They count as one triangle.
If you insist on giving letters to the sides it seems best to require a≤b≤c.
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sbbone
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Re: Problem 373

Post by sbbone »

Thanks for clarifying. Of course i still don't know how to solve this :( but at least one question less to worry about...
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