Problem 075

[Nelly]

I'm having trouble with P75. I believe my program is generating correct values for a unique p, the first few being;
12 24 30 36 40 48 56 70 72 80 96 108 112 126 140 150 154 156 160 176 182 192 198 ......
and the last being 1999992.

However I'm entering the number of unique values and it says its wrong. I've re-read the problem text so many times I think I can't see the wood for the trees. Any pointers?

[jaap]

The first numbers are all ok. The last ones I get however are 1999982, 1999984, 1999994.

1999992 is not a solution as it is the perimeter of the following Pythagorean triangles:
499998² + 666664² = 833330²
505986² + 661320² = 832686²
995988² + 7984² = 996020²

[Nelly]

Yeah, I got it now, I just hadn't iterated over enough values of m.

[wolf_II]

Dear all,

I have 3 issues regarding Problem 75 (View Problem):

1.) Clarification. Should it not be mentioned somewhere in the problem text, that the wire is only allowed to be bent into integer-length portions? As it is not mentioned yet, I like to bend my 20cm wires like this: 4cm + 7.5cm + 8.5cm and the resulting triangle is right-angled. In my humble opinion a hint to integers would be fair. But as you might have guessed, I currently work on the basis that only integer portions of wire length are allowed to form those triangles. Am I on the right track, please? Is it an intended part of the problem to figure that out? If yes, am I giving away too much?


2.) Wording. English is not my first language. Please forgive me the following suggestion:
Change the wording of the first sentence from this:

It turns out that 12 cm is the smallest length of wire can be bent to form a right angle triangle ...

to this:

It turns out that 12 cm is the smallest length of wire that can be bent to form a right angle triangle ...

or this:

It turns out that 12 cm is the smallest length of wire which can be bent to form a right angle triangle ...

3.) Forum. Am I violating any forum policies when I start a new topic regarding a problem for which there is already a topic in existence. Like I have done now. See P75 for the other topic.
Also, is there a better algorithm than my current one for finding topics regarding a certain problem:
- manually click on all pages in the section "Suggestions, Help, and FAQ" (7 pages at the time of writing)
- optically scan the topic subjects for the desired ProblemID (made easy, because they appear in large bold fonts)
- manually click on all pages in the section "Number" (3 pages at the time of writing)
- optically scan the topic subjects for the desired ProblemID (made easy, because they appear in large bold fonts)
- manually click on all pages in the section "Number Theory" (2 pages at the time of writing)
- optically scan the topic subjects for the desired ProblemID (made easy, because they appear in large bold fonts)
and so on ...
I don't seem to be able to use the search function properly.


Thank you in advance
Wolf

[ed_r]

1) You're right; the problem should openly state that all sides must have integer length.

2) Good spot. "... that ..." is the correct form.

3) Pass. Someone else will know.

[wolf_II]

1+2) Thank you for the confirmation.

[Tommy137]

It really is hard to find existing threads to a problem.

Maybe, we can open a special thread (or annonuncement so it's visible in all subforums) listing all these threads.


Edit: I'd do it myself if it's too much work for the moderators

[wolf_II]

I was thinking along the lines of sorting the topics in a way similar to the closed forum. (Each problem links somehow to the open forum discussion).
For example a logged in user, who enters a wrong answer will have a link available on the "Wrong answer" page that takes him/her to your suggested list of threads.

But I realise that this involves workload, and I can't even imagine how much.

[hk]

I will make the necessary changes to the problem description.

Edit: Changes made, please check.

When I discovered Project Euler there wasn't such a Forum like this one. That was when there were about 100 problems.
I figured them out all by myself under the motto "Don't contact Project Euler if you can't solve a problem."
I've still mixed feelings about these "problem xxx clarifications."
In the past I've deleted several of these "clarifications" about a month after the last post was made.
I don't guarantee that I will not resume doing so when I have some spare time, because they served there purpose: the person posing the question was helped.
This forum was never meant to be a exhaustive lookup table for hints and tips for the PE problems.
I would certainly dislike an exhaustive lookup table for the entire problem set.
I still think the problem set should be solvable without "problem xxx clarifications".

I strongly feel that the proposed link to a page where all help is concentrated is counterproductive in the sense that it makes the temptation rather big to enter nonsense answers to get as much hints as possible and in this way makes people dependant on hints to solve their problems.

[wolf_II]

1+2) Thank you for the changes made. I don't see any problem in deleting this thread, as the reason for starting it in the first place is no longer there.

3) How about allowing a number to be searched for, when searching in subjects only, not message bodies?

[hk]

This is a ready made Bulletin Board utility, not a self built one, so that's not a question of altering the code somewhat.
Take it as is.
One suggestion though: search for "clarification 75" or so.

[btilly]

This is a ready made Bulletin Board utility, not a self built one, so that's not a question of altering the code somewhat.
Take it as is.
One suggestion though: search for "clarification 75" or so.

One big problem with the search is that you can't search for a number. Try it. Enter 75 into search and it will reject the search because the word is too common.

Is there any way to turn that filter on searches off?

[Tommy137]

I strongly feel that the proposed link to a page where all help is concentrated is counterproductive in the sense that it makes the temptation rather big to enter nonsense answers to get as much hints as possible and in this way makes people dependant on hints to solve their problems.

Well, I don't think it'd be such a difference as members that do have question will ask it here anyway, either doing it in an existing thread or by creating a new one. But it would be much easier to monitor all given hints or clarifications and deleting them after some time.

Of course, it's the decision of the PE team

[MaJJ]

Hi there,
I'm having problem here - I'm generating the Pythagorean triplets (and I believe I have all unique ones <= 1500000).
The first 9 solutions my code outputs are:

Code: Select all

12 [3, 4, 5]
24 [6, 8, 10]
30 [5, 12, 13]
36 [9, 12, 15]
40 [8, 15, 17]
48 [12, 16, 20]
56 [7, 24, 25]
70 [20, 21, 29]
72 [18, 24, 30]

and the last 9 ones are:

Code: Select all

1499936 [236832, 609349, 653755]
1499946 [17892, 740919, 741135]
1499950 [188925, 641900, 669125]
1499952 [374988, 499984, 624980]
1499968 [421866, 456512, 621590]
1499970 [249995, 599988, 649987]
1499978 [157641, 661912, 680425]
1499984 [289416, 570688, 639880]
1499994 [5988, 746991, 747015]

I can PM the source code (Python). I don't get what's wrong with it
Thanks in advance!

[daniel.is.fischer]

461607 + 416576 + 621785 = 1499968.

[jsegrave]

I am a bit puzzled about this one. I worte a program, got an answer, refined it, tested it, checked it against a brute force all values up to 50,000 to see that it gave the same results. My result was still rejected.

I began googling and found 3 other solutions in Perl and C# (which I converted to perl). All use the same algorithm as I was using, they vary somewhat in what constraints they use to prune away PT parameters that need not be tried, And they all give the same answer as I get. I even found a page on google with numerical answers and it has the same answer as I have for this.

I don't wish to sound presumptious, as I'm hardly a mathematician, but I seriously wonder if, after the problem was apparently changed from 2,000,000 to 1,500,000 if the answer being sought was correctly updated.

I can provide the (commented) Python code and the answer I keep getting. For what it's worth, here are the first and last 10 lengths I found which give a single triangle, in the form length count always 1 (side, side, hypotenuse) ...

Code: Select all

      12:    1 (3 4 5)
      24:    1 (3 4 5)
      30:    1 (5 12 13)
      36:    1 (3 4 5)
      40:    1 (15 8 17)
      48:    1 (3 4 5)
      56:    1 (7 24 25)
      70:    1 (21 20 29)
      72:    1 (3 4 5)
      80:    1 (15 8 17)
...
 1499928:    1 (3 4 5)
 1499936:    1 (247 96 265)
 1499946:    1 (246973 5964 247045)
 1499950:    1 (7557 25676 26765)
 1499952:    1 (3 4 5)
 1499954:    1 (614473 229536 655945)
 1499970:    1 (5 12 13)
 1499978:    1 (9273 38936 40025)
 1499984:    1 (36177 71336 79985)
 1499994:    1 (248997 1996 249005)

[daniel.is.fischer]

I'll look.

Edit: correct answer, glitch during submission.

[EnDorphin]

Can someone verify these values.
If we let X(n) be the function which gives the number of solutions for the problem when L <= n,
then for L = 50, 100, 500 and 2000 I get these values:

X(50) = 6
X(100) = 17
X(500) = 144
X(2000) = 813

[Francky]

Can someone verify these values.
If we let X(n) be the function which gives the number of solutions for the problem when L <= n,
then for L = 50, 100, 500 and 2000 I get these values:

X(50) = 6
X(100) = 17
X(500) = 144
X(2000) = 813

I only agree with X(50)=6.

[EnDorphin]

Alright, I figured out what I did wrong for the one before and I fixed it, but still not correct.
So I wanted to check my new values.

If we let X(n) be the function which gives the number of solutions for the problem when L <= n,
then for L = 50, 100, 500 and 2000, I get these values:

X(50) = 6
X(100) = 11
X(500) = 56
X(2000) = 210

If these are still wrong, could I may be PM anyone the L values for when the max length is 2000?
If these values are all correct, then could someone take a look at my code? It's in Java