Project Euler's 500th problem(projecteuler.net)
projecteuler.net
Project Euler's 500th problem
https://projecteuler.net/problem=500
12 comments
Thanks for posting this.
https://www.hackerrank.com/ has a AI/Machine Learning section
I was asked this question at an interview at $COMPANY last week. Heh. I failed pretty miserably.
Screamed at JavaScript for failing at big integer multiplications. Got it after rewriting in python. Here is my poorly written explanation.
--------- MAJOR SPOILERS ---------
Let's multiply unique prime numbers and count their factors.
i.e. 2x3x5x...x500500th prime will have 2^500500 factors
The 500500th prime is 7376507 (thanks to wolframalpha), and it's not very big.
We can easily get the product of the above with 500500 multiplications while modding result of each step by 500500507 to get an answer.
Except that the product of first 500500 primes isn't the smallest number containing 2^500500 factors. This can be made smaller.
Observe that
Continuing with this, we see that
If we want continue doubling factor count by multiplying 2s, we will need to multiply by 2x2x2x2 next, and 2^8 after, which becomes inefficient quickly. Instead, why not use 3? Multiplying (2x3x5x...x500499th prime) by (3 x 3) also doubles the factor count.
So at this point, we think about doubling factor count and the ways we can do it. Out options are
<1>. Multiply by a prime that we have never used so far
<2>. Multiply by an existing prime k + 1 times, where k is the number of times it has been used
We repeat this 500500 times, using rule <1> and <2> (which can be generalized to one rule) and the result is the final answer.
--------- MAJOR SPOILERS ---------
Let's multiply unique prime numbers and count their factors.
n Number of factors
2 2 (1, 2)
2x3 4 (1, 2, 3, 6)
2x3x5 8
2x3x5x7 16
i.e. 2x3x5x...x500500th prime will have 2^500500 factors
The 500500th prime is 7376507 (thanks to wolframalpha), and it's not very big.
We can easily get the product of the above with 500500 multiplications while modding result of each step by 500500507 to get an answer.
Except that the product of first 500500 primes isn't the smallest number containing 2^500500 factors. This can be made smaller.
Observe that
(2x3x5x...x500499th prime)x2
2^500499 + 2^500498 factors.
Continuing with this, we see that
(2x3x5x...x500499th prime) x (2 x 2)
2^500499 + 2^500498 + 2^500498 = 2^500500 factors.
If we want continue doubling factor count by multiplying 2s, we will need to multiply by 2x2x2x2 next, and 2^8 after, which becomes inefficient quickly. Instead, why not use 3? Multiplying (2x3x5x...x500499th prime) by (3 x 3) also doubles the factor count.
So at this point, we think about doubling factor count and the ways we can do it. Out options are
<1>. Multiply by a prime that we have never used so far
<2>. Multiply by an existing prime k + 1 times, where k is the number of times it has been used
We repeat this 500500 times, using rule <1> and <2> (which can be generalized to one rule) and the result is the final answer.
factor count n
2 2
4 2*3 <1>
8 2*3*2*2 <2>
16 2*3*2*2*5 <1>
32 2*3*2*2*5*7 <1>
64 2*3*2*2*5*7*3*3 <2>
128 2*3*2*2*5*7*3*3*11 <1>
This relies heavily on requiring that the number of factors be exactly 2^500500. Can we find the smallest number with at least 2^500500 divisors?
I absolutely love project Euler. The math on the harder problems is way over my head but it is still my go-to site whenever I want to learn a new language. By the time you've done a bunch of them you'll be more than underway in understanding the territory and in a much better position to understand tutorials (which inevitably are written by experts in the language and usually make a ton of assumptions).
Alternating between project Euler, a book (or two), back to project Euler for a bit is a very fast school, and you determine how fast you can pace yourself.
Alternating between project Euler, a book (or two), back to project Euler for a bit is a very fast school, and you determine how fast you can pace yourself.
I'm hardly in the position to be telling you how you learn a language best, but I always felt programming type puzzles of the oj.leetcode.com variety were better for getting me to learn a programming language, where as projecteuler.net was better at developing mathematical/algorithmic reasoning skills. A non-trivial amount of the early puzzles are do-able by hand.
Anyone else feel similarly?
Anyone else feel similarly?
I agree. I have used the following subreddit to learn and practice new languages: http://www.reddit.com/r/dailyprogrammer
I will check that out, thank you for the pointer.
Project Euler is like sex or diabetes (both of which I have or have had), what it means to you and how it fits into your life is very much an individual thing.
Yes, a lot of the problems can be done by hand. You could treat the problems as "a math problem to solve", or as "an exercise in solving a problem with software."
When you've solved a problem, you have either "obtained the correct answer," or you have "figured out how to obtain the correct answer with software." Or both.
For me, even solving the early trivial ones, that could be done by hand, are more an exercise in software than in math.
Most of the time, I try to write my solutions in a general way, so that I can change the parameters of the problem, run it through software I've already written, and get the right results. The challenge then is defining the boundaries and parameters, and imagining the possible and reasonable limits of those boundaries and parameters. Just like we do every day with the rest of our software challenges.
The fact that these are math problems make it more fun for me, because I am not at all mathematically inclined. So I have to learn a little more math, and then decide if I've learned it sufficiently that my solution is reasonable.
Reading other people's posted solutions is almost as fun as solving the puzzle. I've learned a lot, or at least been impressed, from the various languages and hand solutions.
As for learning a new language, I think Euler is a reasonable tool in the total box. I find that I don't really know a new language until I've struggled with something "real" that matters, to me or my employer. And even then, if I come back to code a year or two later, I sometimes cringe, since I've likely learned more about using the language's strengths in that time, beyond just writing C in a new language.
Yes, a lot of the problems can be done by hand. You could treat the problems as "a math problem to solve", or as "an exercise in solving a problem with software."
When you've solved a problem, you have either "obtained the correct answer," or you have "figured out how to obtain the correct answer with software." Or both.
For me, even solving the early trivial ones, that could be done by hand, are more an exercise in software than in math.
Most of the time, I try to write my solutions in a general way, so that I can change the parameters of the problem, run it through software I've already written, and get the right results. The challenge then is defining the boundaries and parameters, and imagining the possible and reasonable limits of those boundaries and parameters. Just like we do every day with the rest of our software challenges.
The fact that these are math problems make it more fun for me, because I am not at all mathematically inclined. So I have to learn a little more math, and then decide if I've learned it sufficiently that my solution is reasonable.
Reading other people's posted solutions is almost as fun as solving the puzzle. I've learned a lot, or at least been impressed, from the various languages and hand solutions.
As for learning a new language, I think Euler is a reasonable tool in the total box. I find that I don't really know a new language until I've struggled with something "real" that matters, to me or my employer. And even then, if I come back to code a year or two later, I sometimes cringe, since I've likely learned more about using the language's strengths in that time, beyond just writing C in a new language.
I hope someday I have a project that someone understands well enough to compare it to sex or diabetes.
They can be done by hand but doing them algorithmically usually involves working around machine limitations or figuring out a non trivial algorithm to reduce the runtime complexity to something that will run in seconds rather than days.
> it is still my go-to site whenever I want to learn a new language
The great thing about Project Euler for this is that the correct answer isn't a piece of code or an algorithm in some language, but just a number.
Project Euler does not really care how that number was derived. Using a mathematical formula, plain pencil and paper, a program in some language, or cheating via Google-fu? Euler doesn't care.
Because of this, it is very inviting to code a solution in a different language: 1) to learn that language, 2) to use a concept of the language with a better fit with the problem, 3) to see which language yields the quickest answer, measured in development time or actual run time.
The drawback of Project Euler is that it's quite addictive.
The great thing about Project Euler for this is that the correct answer isn't a piece of code or an algorithm in some language, but just a number.
Project Euler does not really care how that number was derived. Using a mathematical formula, plain pencil and paper, a program in some language, or cheating via Google-fu? Euler doesn't care.
Because of this, it is very inviting to code a solution in a different language: 1) to learn that language, 2) to use a concept of the language with a better fit with the problem, 3) to see which language yields the quickest answer, measured in development time or actual run time.
The drawback of Project Euler is that it's quite addictive.
Unless you're switching from e.g. Java to e.g. Prolog or Erlang (or Haskell...), is it really useful for learning a new language? I mean, when learning a new language, the basic syntax is the easiest part anyway, but the ecosystem/batteries/best_practices is where the main bottleneck is. And I have an impression that Project Euler problems will help you only with the very basics of the language.
Yes, I do exactly the same thing! The first 20 problems in particular are excellent for learning a new language. It covers basic math, list operations, file IO, string parsing; in other words all the basics in learning a new language. Plus several of the problems are strung together and encourage you to build up a structured library of tools to help you solve any number of problems.
One of my favorite parts of Project Euler was getting something where the math seemed beyond me, but I could use a tool like Google to get further.
Not to cheat using Google, but to discover the name of the branch of mathematics I had no clue about, and then to learn enough to keep going.
I still remember the afternoon I spent doing chromatic polynomials by hand, each graph I drew having more mental exclamations on the end, because I was goddamn getting it.
For me, the lift from earning a right answer is palpable.
Not to cheat using Google, but to discover the name of the branch of mathematics I had no clue about, and then to learn enough to keep going.
I still remember the afternoon I spent doing chromatic polynomials by hand, each graph I drew having more mental exclamations on the end, because I was goddamn getting it.
For me, the lift from earning a right answer is palpable.
Yes! I had that same feeling with this one:
https://projecteuler.net/problem=307
I couldn't solve it analytically for the life of me. So I wrote a small simulator, computed the answer to within 7 decimal places or so and then tried to figure out why that was the right answer. Then the lightbulb went on and I realized what the real answer should be and it worked.
As I wrote above, I am not very good at math, but to be able to use the computer to make your math understanding better than what it was before you started solving a problem was a huge up for me.
https://projecteuler.net/problem=307
I couldn't solve it analytically for the life of me. So I wrote a small simulator, computed the answer to within 7 decimal places or so and then tried to figure out why that was the right answer. Then the lightbulb went on and I realized what the real answer should be and it worked.
As I wrote above, I am not very good at math, but to be able to use the computer to make your math understanding better than what it was before you started solving a problem was a huge up for me.
Can you solve that problem analytically? The straightforward answer is to set up a recurrence relation for p(k,n) and use dynamic programming, but even then it's far too laborious to compute p(20000,1000000) without a computer.
Now I can, but initially I had no idea. But the simulation can be coded up in 15 minutes or so and you'll have your first 3 decimal places in a few minutes after that. To get to 9 decimal places takes a while longer.
By analytically, do you mean by hand, or just getting the exact answer by computer? Can project euler problems often be solved by hand? For example this 500th problem doesn't seem very hard with a computer, but doing it by hand seems like it will take a long long time.
There is an elegant way to solve these types of problems analytically using a technique called Poissonization. The aspect that makes computing the correct answer difficult is the dependence among counts for the various chips. Instead imagine the the number of defects is now Poisson distributed with mean lambda. A convenient property of the Poisson distribution is its "binning" property whereby the number of defects hitting each chip as now Poisson(lambda/n) and, most important, the counts are mutually independent. From here you can compute the correct answer by a power series argument. I'll spare you the details but see http://bulletin.imstat.org/2014/02/student-puzzle-corner/ for a worked example.
Interesting idea, but I don't quite get how that lets us compute the answer. If I understand it correctly then we need to get the coefficient of l^(10^6) of (e^-(l/20000) + (l/20000)e^-(l/20000) + (l/20000)^2/2e^-(l/20000))^20000, and I do not know how to do that. In fact that is equivalent to finding the coefficient of x^(10^6) of (1+x+x^2)^20000, and it's easy too see that directly from the original problem, so I'm also not sure what Poissonization does for us here. Perhaps I'm completely misunderstanding the technique?
You have flipped the roles of 20000 and 10^6 in your formulation.
Yes I have, thanks.
No, it's still by computer. But some of them I can do by hand but that's only a few so far.
[deleted]
I agree 100%! I first learned to program in Python using Project Euler and they taught me a ton. Now I've solved 111 of them!
I'm currently trying to learn Rust, and since it's so new everything is geared towards people with experience. So I started Rust For Beginners:
http://www.rustforbeginners.com
I'm trying to make it a guide to the language, and programming in general, for true beginners. The very first post of actual coding (after installing, command line basics, and compiling/running with cargo) is a walk through of the first PE problem.
I know how you feel about most beginners guides not actually being suited for beginners. Since I'm a beginner, I'm hoping I'll be able to write one.
Feel free to check it out, and constructive criticism is encouraged!
I'm currently trying to learn Rust, and since it's so new everything is geared towards people with experience. So I started Rust For Beginners:
http://www.rustforbeginners.com
I'm trying to make it a guide to the language, and programming in general, for true beginners. The very first post of actual coding (after installing, command line basics, and compiling/running with cargo) is a walk through of the first PE problem.
I know how you feel about most beginners guides not actually being suited for beginners. Since I'm a beginner, I'm hoping I'll be able to write one.
Feel free to check it out, and constructive criticism is encouraged!
[deleted]
Happy 500th!
If you've never tried Project Euler, it's a great way to learn new mathematical concepts, sequences, dynamic programming and other algorithms.
https://projecteuler.net/problem=368 was one of my favorites, because it involved turning an algorithm described in a paper into code.
I don't consider this a spoiler for problem #500, but if you're counting the number of divisors, you'll need this function: http://oeis.org/wiki/Number_of_divisors_function#Formulae_fo...
As an example, 120 = 2^3 * 3^1 * 5^1
number of divisors of 120 => (3+1)(1+1)(1+1)
Even if you don't solve it, exploring the problem is a fun afternoon for a person with a computer.
If you've never tried Project Euler, it's a great way to learn new mathematical concepts, sequences, dynamic programming and other algorithms.
https://projecteuler.net/problem=368 was one of my favorites, because it involved turning an algorithm described in a paper into code.
I don't consider this a spoiler for problem #500, but if you're counting the number of divisors, you'll need this function: http://oeis.org/wiki/Number_of_divisors_function#Formulae_fo...
As an example, 120 = 2^3 * 3^1 * 5^1
number of divisors of 120 => (3+1)(1+1)(1+1)
Even if you don't solve it, exploring the problem is a fun afternoon for a person with a computer.
I love the way the harmonic series seems to juuuuuuust baaaaarely diverge....
One of the things that's freaky to me is that the harmonic series up to n terms is proportional to log(n)...
but the sum of the reciprocals of the primes is roughly proportional to log(log(n)). That's a crazy slow divergence.
but the sum of the reciprocals of the primes is roughly proportional to log(log(n)). That's a crazy slow divergence.
For even more crazy, consider the following series:
H(n) = 1/1 + 1/2 + ... + 1/n
H_2(n) = 1 / (H(1)) + 1 / (2 H(2)) + 1 / (3 H(3)) + ... + 1 / (n H(n))
H_3(n) = 1 / (H(1) H_2(1)) + 1 / (2 H(2) H_2(2)) + 1 / (3 H(3) H_2(3)) + ... + 1 / (n H(n) H_2(n))
...
H_k(n) = 1 / (H(1) H_2(1) ... H_{k-1}(1)) + 1 / (2 H(2) H_2(2) ... H_{k-1}(2)) + ... + 1 / (n H(n) H_2(n) ... H_{k-1}(n))
The series H_k(n) diverges roughly proportional to log(log(... k logs ... (log(n)) ...)).
H(n) = 1/1 + 1/2 + ... + 1/n
H_2(n) = 1 / (H(1)) + 1 / (2 H(2)) + 1 / (3 H(3)) + ... + 1 / (n H(n))
H_3(n) = 1 / (H(1) H_2(1)) + 1 / (2 H(2) H_2(2)) + 1 / (3 H(3) H_2(3)) + ... + 1 / (n H(n) H_2(n))
...
H_k(n) = 1 / (H(1) H_2(1) ... H_{k-1}(1)) + 1 / (2 H(2) H_2(2) ... H_{k-1}(2)) + ... + 1 / (n H(n) H_2(n) ... H_{k-1}(n))
The series H_k(n) diverges roughly proportional to log(log(... k logs ... (log(n)) ...)).
The n'th prime looks a bit like n log n. (This is basically the same fact as the "prime number theorem": the number of primes up to n is roughly n/log(n).)
We have:
sum (up to n) of 1/x ~= log n [harmonic series]
sum (up to n) of 1/(x log x) ~= log log n [reciprocals of primes, roughly]
sum (up to n) of 1/(x log x log log x) ~= log log log n
and so on. All these series diverge, but slower and slower and slower. And clearly these series "only just" diverge. In fact, iIf you take, e.g.,
sum of 1/(x log x log log x (log log log x)^1.1)
then it converges. If you have just the right kind of warped mind, the question that will immediately occur to you on contemplating this is: what if you define L(x) := x log x log log x log log log x ... continuing the product up to, but excluding, the first factor that's < 1, and ask about the sum of 1/L(x)?
Answer: The series diverges really slowly. The points at which L(x) starts to use one extra logarithm are e, e^e, e^e^e, e^e^e^e, etc. The sums between each adjacent pair of these are comparable in size. So the sum up to n is roughly the number of times you need to take logs, starting with n, before you get down to 1.
[EDITED to add: The first bit of the above is closely related to cperciva's comment that's a sibling of this one.]
We have:
sum (up to n) of 1/x ~= log n [harmonic series]
sum (up to n) of 1/(x log x) ~= log log n [reciprocals of primes, roughly]
sum (up to n) of 1/(x log x log log x) ~= log log log n
and so on. All these series diverge, but slower and slower and slower. And clearly these series "only just" diverge. In fact, iIf you take, e.g.,
sum of 1/(x log x log log x (log log log x)^1.1)
then it converges. If you have just the right kind of warped mind, the question that will immediately occur to you on contemplating this is: what if you define L(x) := x log x log log x log log log x ... continuing the product up to, but excluding, the first factor that's < 1, and ask about the sum of 1/L(x)?
Answer: The series diverges really slowly. The points at which L(x) starts to use one extra logarithm are e, e^e, e^e^e, e^e^e^e, etc. The sums between each adjacent pair of these are comparable in size. So the sum up to n is roughly the number of times you need to take logs, starting with n, before you get down to 1.
[EDITED to add: The first bit of the above is closely related to cperciva's comment that's a sibling of this one.]
[deleted]
I was only recently introduced to Project Euler but already I'm addicted. I've got a folder where I've named each solution as pXXX.ext and try solving the problems in multiple different languages. I refuse to move on to a new problem until I've solved the previous one (I don't solve every problem in multiple languages. Instead I've been using it to learn/play with languages here and there I don't really know).
Congrats to Project Euler on their 500th problem!
Congrats to Project Euler on their 500th problem!
Another way to go about all this is to sort them by how many people solved them and go from easiest to most difficult. This way you maximize your learning/time!
I would do this if I could handle going out of order :)
[deleted]
How far along are you?
Solved up to problem 25 so far, https://projecteuler.net/progress=joshstrange
Only friends' progress can be viewed, sharing this picture[0] would do the job ;)
0.https://projecteuler.net/profile/joshstrange.png
0.https://projecteuler.net/profile/joshstrange.png
Oh, I believed him. I'm actually at 24 (in Erlang).
I was just pointing it out :)
I solved 50 of them, I usually use Java, or Mathematica which feels...cheaty but sometimes all it takes is a paper and a pencil.
I solved 50 of them, I usually use Java, or Mathematica which feels...cheaty but sometimes all it takes is a paper and a pencil.
What is interesting is that some of the problems casually generate solutions that some languages can't easily represent. For those languages solving the problem suddenly gets an extra dimension and is a nice reminder that not all languages are created with equal expressive power and/or innate abilities.
Oh, yeah, I wasn't thinking. Thanks for providing the link!
Aw, they don't allow seeing progress unless you're contacts on the site.
Every couple of years I get addicted for a few weeks. Love the effort these guys put into the problems.
Same here - each time I learn a new programming language I like to solve a bunch.
[deleted]
For Project Euler enthusiasts: here's a similar site focused on bioinformatics that I've found very very good: http://rosalind.info/
Also suitable for learning new languages (and bioinformatics).
The implementation of the site is awesome.
Also suitable for learning new languages (and bioinformatics).
The implementation of the site is awesome.
Rosalind is a great way to pick up some bio along with your algorithms.
The instructors also created a course on Coursera called "Bioinformatics Algorithms" that I highly recommend. It's basically an ordered set of rosalind problems but accompanied by a textbook and lectures. It was one of the most interesting things I did last year.
The instructors also created a course on Coursera called "Bioinformatics Algorithms" that I highly recommend. It's basically an ordered set of rosalind problems but accompanied by a textbook and lectures. It was one of the most interesting things I did last year.
Agreed - it's one of the most enjoyable Coursera courses I've taken. Part Two starts February 16th. [1]
To expand a bit, what I liked about the course was the textbook, algorithms and their practical application in a non-CS field with a good set of programming problems that are automatically assessed.
[1] https://www.coursera.org/course/bioinformatics2
To expand a bit, what I liked about the course was the textbook, algorithms and their practical application in a non-CS field with a good set of programming problems that are automatically assessed.
[1] https://www.coursera.org/course/bioinformatics2
After you solve a problem Project Euler lets you browse through a giant list of solutions by those who have gone before. It's really fun to see and compare all the different approaches.
If I could do one thing to augment the project I would make it possible to filter previous solutions by language and/or author.
For example, if I'm learning rust, I would love to see all the rust solutions right there together.
If I could do one thing to augment the project I would make it possible to filter previous solutions by language and/or author.
For example, if I'm learning rust, I would love to see all the rust solutions right there together.
"Saturday 31st of January Project Euler will reach a major milestone with the publication of our 500th problem. To celebrate this the team has composed a special problem in which the number 500 occurs repeatedly. In addition we're going to publish an ordinary problem simultaneously as problem 501. Both problems are published at 13:00 London time.
We invite all members to participate in our problem solving party. As celebrator of this jubilee we're allowed to ask the participants for a present. Well, here's our wish: please don't share any information about our party and the dishes we're serving you outside the fora dedicated to our dishes after your meal. This to allow those that are late to the party to consume our dishes undisturbed by premature revelations about what we are serving."
https://projecteuler.net/news
We invite all members to participate in our problem solving party. As celebrator of this jubilee we're allowed to ask the participants for a present. Well, here's our wish: please don't share any information about our party and the dishes we're serving you outside the fora dedicated to our dishes after your meal. This to allow those that are late to the party to consume our dishes undisturbed by premature revelations about what we are serving."
https://projecteuler.net/news
If you're interesting in the birth, inspiration and pedagogy behind Project Euler, I recommend this article [1].
And it's an effective teacher because those problems are arranged like the programs in the ORIC-1's manual, in what Hughes calls an "inductive chain":
The problems range in difficulty and for many the experience is inductive chain learning. That is, by solving one problem it will expose you to a new concept that allows you to undertake a previously inaccessible problem. So the determined participant will slowly but surely work his/her way through every problem.
[1]http://www.theatlantic.com/technology/print/2011/06/how-i-fa...
And it's an effective teacher because those problems are arranged like the programs in the ORIC-1's manual, in what Hughes calls an "inductive chain":
The problems range in difficulty and for many the experience is inductive chain learning. That is, by solving one problem it will expose you to a new concept that allows you to undertake a previously inaccessible problem. So the determined participant will slowly but surely work his/her way through every problem.
[1]http://www.theatlantic.com/technology/print/2011/06/how-i-fa...
This is a nice little problem, I had fun solving it.
For those of us who are familiar with TopCoder, CodeForces or, perhaps, ACM ICPC, this should be daily bread and butter on prime numbers and data structures :)
Here is my solution if you are curious: https://gist.github.com/Slava/74276f5b7889a1d68a71 - computes the result in 400ms
Edit: hah, looking at the problems discussion board I took a very bruteforce approach, other people solved utilizing math, I used priority_queue for no particular reason :)
For those of us who are familiar with TopCoder, CodeForces or, perhaps, ACM ICPC, this should be daily bread and butter on prime numbers and data structures :)
Here is my solution if you are curious: https://gist.github.com/Slava/74276f5b7889a1d68a71 - computes the result in 400ms
Edit: hah, looking at the problems discussion board I took a very bruteforce approach, other people solved utilizing math, I used priority_queue for no particular reason :)
Any chance you guys know of a similar project for Computer Vision or general AI topics?