Problem 353

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BostonBear
Posts: 17
Joined: Thu Apr 28, 2011 4:48 am
Location: Saugus, MA

Problem 353

I am a little confused about the new problem. This sphere has integer coordinates. Yet we are not given a specific size, so as far as I can tell , for a generic r, there are only 6 integer coordinates. (r,0,0), (0,r,0), (0,0,r) and its antipodals. Is there something I'm missing here? Does each trip to a station have to be a direct trip, or can you change directions before you get to the next station?

Edit, never mind, I figured out what I was not seeing right
Last edited by BostonBear on Sun Oct 09, 2011 5:36 am, edited 1 time in total.
TripleM
Posts: 382
Joined: Fri Sep 12, 2008 3:31 am

Re: Euler Problem 353

You are given a specific size - r is equal to 2^n - 1 for each value of n from 1 to 15.
cyrillic
Posts: 1
Joined: Wed Oct 12, 2011 11:25 am

Euler Problem 353

Could I check with someone the value of M(r) for n=15 I'm getting, to validate my code? Cheers.
thundre
Posts: 356
Joined: Sun Mar 27, 2011 10:01 am

Re: Euler Problem 353

cyrillic wrote:Could I check with someone the value of M(r) for n=15 I'm getting, to validate my code? Cheers.
PM it to me and I'll tell you whether it matches mine.
Arthelais
Posts: 3
Joined: Wed Oct 08, 2014 2:57 pm

Re: Problem 353

Could someone verify for me whether M(1000) is equal to 0.05174553024? I feel like I am facing a rounding error and I can't seem to figure it out.
MuthuVeerappanR
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Re: Problem 353

Is this problem prone to precision errors?? I tried a couple of methods and they all provide the same for value upto M(10). But somehow my answer is not getting accepted.

It is not knowledge, but the act of learning, not possession but the act of getting there, which grants the greatest enjoyment.
sjhillier
Posts: 561
Joined: Sun Aug 17, 2014 4:59 pm
Location: Birmingham, UK
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Re: Problem 353

Well you certainly have to be precise, but no more so than many other problems. There are other possible error modes, but it would be wrong to say more.
MuthuVeerappanR
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Re: Problem 353

Thanks sjhillier. I solved it yesterday. Turns out one of my assumptions only fails for the k = 8 case.

It is not knowledge, but the act of learning, not possession but the act of getting there, which grants the greatest enjoyment.
sjhillier
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Re: Problem 353

MuthuVeerappanR wrote: Thu Jun 08, 2017 2:34 pm Thanks sjhillier. I solved it yesterday. Turns out one of my assumptions only fails for the k = 8 case.
I'm sure I made the same mistake too.
bleh0.5
Posts: 1
Joined: Tue Aug 14, 2018 5:54 pm

Re: Problem 353

Are we supposed to calculate every single integer set of coordinates or are the only coordinates going to be (0,r,r) and stuff like that
RobertStanforth
Posts: 1690
Joined: Mon Dec 30, 2013 11:25 pm

Re: Problem 353

bleh0.5 wrote: Tue Aug 14, 2018 5:57 pm Are we supposed to calculate every single integer set of coordinates or are the only coordinates going to be (0,r,r) and stuff like that
The admissible paths (from which you are asked to find the one with the lowest risk) are permitted to use any of the integer-coordinate points on the surface of $C$. The problem statement does not confine you to a special subset of them.
ChicoTobi
Posts: 1
Joined: Thu May 27, 2021 9:04 pm

Re: Problem 353

Uhm, so I hope i don't spoil anything... so if it's too much please feel free to delete...

So I started doing this problem, because I love those fancy geometry problems! And also I was thinking, that a lot of symmetry is going on here. After reproducing the solution for the small number example from the question (N=3, r=7), I tried scaling it up. And boy, was I wrong! Both the generation of the pythagorean tuples as well as the graph-matrix as well as the dijkstra algo only perform until N=11, N=12 within around a minute. But there's no way that I will reach N=15 within a minute.

I produced hashes for the smaller numbers. Can somebode check these plz? It's a simple md5 of the the utf-8 string in the form as in the problem "0.1784943998".

Hash for r= 1 : d310cb367d993fb6fb584b198a2fd72c
Hash for r= 3 : fca3319e19aa9137881ea13ea824edce
Hash for r= 7 : d0dfb92eb56f02b897654fe4652b2f1f
Hash for r= 15 : 5f9e89ed8d08fd895368256943196c5a
Hash for r= 31 : eb166b2e0288c93cccd7a4d47229b7a7
Hash for r= 63 : 0ba2dd95494dc35f356eb185dfcd5a9a
Hash for r= 127 : f7049ff1b5cb032b8c1fcee6acc1ffdc
Hash for r= 255 : 02aa0151cd3f7eb54469ab56dcf65f1f
Hash for r= 511 : d58bbfcfa5f91cd8b530109f45743000
Hash for r= 1023 : 6ccc1e655cdbf556ea1f9f63e152053b
Hash for r= 2047 : 846004309705e1a24389b67e2f298f7e

Then I would know if my implementation is wrong, or if it's only too slow/inefficient. I attached part of my shortest path for r=255.

Edit: Nevermind, I let it run, and it was correct in the end But still took like 30 minutes, sadly.
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