Problem 317

A place to air possible concerns or difficulties in understanding ProjectEuler problems. This forum is not meant to publish solutions. This forum is NOT meant to discuss solution methods or giving hints how a problem can be solved.
Forum rules
As your posts will be visible to the general public you
are requested to be thoughtful in not posting anything
that might explicitly give away how to solve a particular problem.

This forum is NOT meant to discuss solution methods for a problem.

In particular don't post any code fragments or results.

Don't start begging others to give partial answers to problems

Don't ask for hints how to solve a problem

Don't start a new topic for a problem if there already exists one


See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
golfguy37
Posts: 4
Joined: Tue Jul 23, 2013 11:45 pm

Re: Problem 317

Post by golfguy37 »

Okay thanks
Image
muffoosta
Posts: 3
Joined: Fri Dec 06, 2013 9:41 pm

Re: Problem 317

Post by muffoosta »

Could someone please confirm or deny that the answer lies in this range: <snipped by hk> ?
muffoosta
Posts: 3
Joined: Fri Dec 06, 2013 9:41 pm

Re: Problem 317

Post by muffoosta »

So sorry for posting the number, I saw others post incorrect results here so I though it would be ok to ask if I was near the answer at all. But since it was quickly removed by hk, I figured I was on the right track ;)
oern
Posts: 1
Joined: Mon Dec 16, 2013 3:09 pm

Re: Problem 317

Post by oern »

Does it actually make sense to give g rounded to two decimal places but require the result with four decimal places? I learned in Physics that one should never have more precision in the result than in the given quantities.

EDIT: And the problem obviously does not mean a "velocity" of 20m/s, but a "speed" (velocity would be a vector, and the vectors of the speeds are definitely not the same).
thundre
Posts: 356
Joined: Sun Mar 27, 2011 10:01 am

Re: Problem 317

Post by thundre »

oern wrote:Does it actually make sense to give g rounded to two decimal places but require the result with four decimal places? I learned in Physics that one should never have more precision in the result than in the given quantities.
Constants given in math problems are not measured. They are assumed to be exactly true, with perfect precision. Using short ones saves typing. If a constant is not exact, the problem will explicitly say so.
Image
Alex-82w4
Posts: 20
Joined: Fri Jan 13, 2017 4:37 pm

Re: Problem 317

Post by Alex-82w4 »

What approximate value of $\pi$ should be used?
Image
The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2
v6ph1
Posts: 131
Joined: Mon Aug 25, 2014 7:14 pm

Re: Problem 317

Post by v6ph1 »

Alex-82w4 wrote: Fri Jul 14, 2017 9:46 pm What approximate value of $\pi$ should be used?
You need 4 decimals in your solution. - So if you append the next digit of pi, the solution should not change at all.

Hint:
If you calculate using float / double: Use always the maximum possible in this format.
Image
Alex-82w4
Posts: 20
Joined: Fri Jan 13, 2017 4:37 pm

Re: Problem 317

Post by Alex-82w4 »

I submit many answers but they were not accepted.
So I did approximation of covered area as cylinder with height $h=Ymax=120m$ and radius $r=X_{max}=100m$
$Volume_{cylinder}=\pi*r^2*h$ Volume of cylinder. So the answer can not exceed $3,769,911.1843 m^3$

Could someone PM me to verify my approach to solve the problem?

Thanks,
Alex.
Image
The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2
v6ph1
Posts: 131
Joined: Mon Aug 25, 2014 7:14 pm

Re: Problem 317

Post by v6ph1 »

Your max. height is not correct (it is a little bit higher)
Image
Alex-82w4
Posts: 20
Joined: Fri Jan 13, 2017 4:37 pm

Re: Problem 317

Post by Alex-82w4 »

formula $h = Y_{max}=H+\frac{V_0^2}{2g}$

exact result is $120.3874m$
Image
The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2
v6ph1
Posts: 131
Joined: Mon Aug 25, 2014 7:14 pm

Re: Problem 317

Post by v6ph1 »

This is my max. height too.

But (I think this is not to much hint) the problem did not describe some kind of cylinder - It's much less than a cylinder
Image
Alex-82w4
Posts: 20
Joined: Fri Jan 13, 2017 4:37 pm

Re: Problem 317

Post by Alex-82w4 »

OK. Cylinder is very rough maximum approximation. The real volume is much less than the volume of the cylinder. I agree with you, v6ph1.

My results are still not accepted :(
What angle increment $1^o$ or $0.5^o$ should I use in my calculations?
This is the important question because an answer to it gives volume difference in $79.4676m^3$.
Image
The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2
Alex-82w4
Posts: 20
Joined: Fri Jan 13, 2017 4:37 pm

Re: Problem 317

Post by Alex-82w4 »

v6ph1 wrote: Fri Jul 14, 2017 11:54 pm
Alex-82w4 wrote: Fri Jul 14, 2017 9:46 pm What approximate value of $\pi$ should be used?
You need 4 decimals in your solution. - So if you append the next digit of pi, the solution should not change at all.
I solved the problem. :)

v6ph1, Your suggestion to use 5 decimals in calculations gives an incorrect result. The difference from the correct answer is $1.5681 m^3$. Rounding $\pi$ to 10 or more decimals and using $\pi$ without rounding gives the correct result.

Here's the proof. The solution range is $3,769,911.1843m^3$ $10^7$ (see this post) plus 4 decimal digits. Therefore, we have $10^{11}$ range of significant digits. $\pi$ range is $10^1$. So a result will be not be affected if we round to $\log_{10}{(10^{11}/10^1)} = \log_{10}{10^{10}} = 10$ digits.
Image
The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2
User avatar
jaap
Posts: 559
Joined: Tue Mar 25, 2008 3:57 pm
Contact:

Re: Problem 317

Post by jaap »

Alex-82w4 wrote: Tue Aug 01, 2017 9:48 pm
v6ph1 wrote: Fri Jul 14, 2017 11:54 pm
Alex-82w4 wrote: Fri Jul 14, 2017 9:46 pm What approximate value of $\pi$ should be used?
You need 4 decimals in your solution. - So if you append the next digit of pi, the solution should not change at all.
I solved the problem. :)

v6ph1, Your suggestion to use 5 decimals in calculations gives an incorrect result.
I think you misunderstood v6ph1, though the post was somewhat ambiguous. I think the intended suggestion was to use as many decimals in the value of pi as you need such that adding an extra decimal does not change the four decimals in the calculated answer.
Alex-82w4
Posts: 20
Joined: Fri Jan 13, 2017 4:37 pm

Re: Problem 317

Post by Alex-82w4 »

Hi japp,
My initial question in the forum "Clarifications on Project Euler Problems" was
What approximate value of $\pi$ should be used?
The answer must be a number, for example $3.14$. v6ph1's initial answer was
You need 4 decimals in your solution. - So if you append the next digit of pi, the solution should not change at all.
This is a general answer that can be applied to any constant, for example $e$.
jaap, here's your comment
I think the intended suggestion was to use as many decimals in the value of pi as you need such that adding an extra decimal does not change the four decimals in the calculated answer.
The 317 problem can be solved by using scientific calculator. It may not have enough decimals of $\pi$. It may lead to an incorrect result.

Therefore, the value of $\pi$ must be stated in the problem description the same way it stated for $g$.

Thank you for understanding.
Regards,
Alex.

P.S. I suggest to moderators that problem 317 wording has to be updated by adding value of $\pi$ constant.
Image
The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2
v6ph1
Posts: 131
Joined: Mon Aug 25, 2014 7:14 pm

Re: Problem 317

Post by v6ph1 »

My intention was to calculate with as many decimals as possible/needed.
And this could be checked if the next digit of pi would not change the result.

Suggesting a specific precise approximation will gave the others too much hints.

The wording should not change as constants like pi and e have an exactly defined value.
Image
DJohn
Posts: 67
Joined: Sat Oct 11, 2008 12:24 pm

Re: Problem 317

Post by DJohn »

Alex-82w4 wrote: Wed Aug 02, 2017 12:14 pm Therefore, the value of $\pi$ must be stated in the problem description the same way it stated for $g$.
The problem is asking for the value you'd get by calculating the volume exactly, then rounding it to four decimal places. It doesn't mention $\pi$ at all. Any method of finding that rounded value is acceptable. You might use the exact value of $\pi$, some approximation, or not use $\pi$ at all.

Whichever method you use, you'll need to satisfy yourself that it gives the same result as rounding the exact volume. If you decide to use an approximation to $\pi$, finding a suitable approximation is up to you.

$g$ is a different matter. In the real world, it's a physical value that must be measured. The value you get will depend on the time and place of measurement, the equipment used, and the care and skill of the person taking the measurement. To ensure that everyone solving this problem gets the same answer, we all need to be using the same value. So that has to be given in the problem.
Alex-82w4
Posts: 20
Joined: Fri Jan 13, 2017 4:37 pm

Re: Problem 317

Post by Alex-82w4 »

v6ph1 wrote: Wed Aug 02, 2017 8:01 pm Suggesting a specific precise approximation will gave the others too much hints.
v6ph1, I understand your answer now, not to give the others to much hints. :)
Image
The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2
User avatar
RishadanPort
Posts: 79
Joined: Mon Jun 10, 2013 7:31 am

Re: Problem 317

Post by RishadanPort »

Can someone confirm to me, that the Max X distance, is ~<removed by moderator>, at t = ~xxx,
with Theta = ~xxx radians
?
Image

Rishada is the gateway to free trade—but the key will cost you.
User avatar
gaufowl
Posts: 5
Joined: Tue Sep 22, 2020 10:36 pm
Location: MD,USA

Re: Problem 317

Post by gaufowl »

I'm sure I've done this correctly, especially after reading old comments on here. I think I have a precision problem, could someone help me troubleshoot by checking that the 6 decimals after the 4 decimals asked for in the problem statement are 170668? In other words, the solution looks something like X.abcd170668.

Edit: nevermind, I had a mistake copying a number towards the end from my calculator to paper. Solved.
Image
1691991_rIEOKCNEDBtm7EzRUeWtIZDvhFNxQVp1
Post Reply