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Project Euler

projecteuler.net

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Re: Project Euler

#21

Post by jackweirdy »

delibes wrote: post #9

Ah, memories. Back in 2010 I whinged to colleagues who were using their fancy functional languages to solve problem 1 inefficiently in 2 lines of code... " Project Euler problem 1 :- It helps to know the sum of (1 .. x) is 0.5 * n * (n+1). Multiples of 3 less than 1000 are (3 .. 999) == (1 .. 333) * 3 = 0.5 * 333 * 334 * 3 = 166833 Multiples of 5 less than 1000 are (5 .. 995) == (1 .. 199) * 5 = 0.5 * 199 * 200 * 5 =…

> It helps to know the sum of (1 .. x) is 0.5 * n * (n+1)

This is cool. As someone out of full time education, and who dropped maths relatively early, where’s a good place to start learning some of this stuff? By stuff I mean things related to algebra and whatever the black magic I quoted is

Re: Project Euler

#22

Post by mulmen »

dhbradshaw wrote: post #5

My favorite part of Project Euler is the solutions forum that you can access after entering the correct answer to a problem. Having different solutions to the same problem to browse though is fantastic for learning new idioms and tricks. Sadly, only a small portion of the solutions posted are retained, so if you're not looking for Python or C you'll probably only see a couple of different variations in your language…

I like this idea but project Euler seems to focus much more on math than programming languages so it makes sense they don’t retain all answers.

I would really like to see something similar for actual programming challenges though. Some real-world examples, maybe with predefined test cases that you then solve and optimize for CPU/Memory/Time/lines/whatever.

Re: Project Euler

#23

Post by jklehm »

jackweirdy wrote: post #21
delibes wrote: post #9

Ah, memories. Back in 2010 I whinged to colleagues who were using their fancy functional languages to solve problem 1 inefficiently in 2 lines of code... " Project Euler problem 1 :- It helps to know the sum of (1 .. x) is 0.5 * n * (n+1). Multiples of 3 less than 1000 are (3 .. 999) == (1 .. 333) * 3 = 0.5 * 333 * 334 * 3 = 166833 Multiples of 5 less than 1000 are (5 .. 995) == (1 .. 199) * 5 = 0.5 * 199 * 200 * 5 =…

> It helps to know the sum of (1 .. x) is 0.5 * n * (n+1) This is cool. As someone out of full time education, and who dropped maths relatively early, where’s a good place to start learning some of this stuff? By stuff I mean things related to algebra and whatever the black magic I quoted is

Discrete mathematics is the branch of math you're after.

This looks like it has some ok examples: https://www.tutorialspoint.com/discrete_mathematics/discrete...

Re: Project Euler

#24

Post by pkaye »

jackweirdy wrote: post #21
delibes wrote: post #9

Ah, memories. Back in 2010 I whinged to colleagues who were using their fancy functional languages to solve problem 1 inefficiently in 2 lines of code... " Project Euler problem 1 :- It helps to know the sum of (1 .. x) is 0.5 * n * (n+1). Multiples of 3 less than 1000 are (3 .. 999) == (1 .. 333) * 3 = 0.5 * 333 * 334 * 3 = 166833 Multiples of 5 less than 1000 are (5 .. 995) == (1 .. 199) * 5 = 0.5 * 199 * 200 * 5 =…

> It helps to know the sum of (1 .. x) is 0.5 * n * (n+1) This is cool. As someone out of full time education, and who dropped maths relatively early, where’s a good place to start learning some of this stuff? By stuff I mean things related to algebra and whatever the black magic I quoted is

This would be a finite series. I think it is usually covered in calculus courses but some of the basics are discussed in earlier algebra courses. Here is a link to a brief summary on finite series. http://calculus.nipissingu.ca/tutorials/finiteseries.html

You can also start your way up from the basics in Khan's Academy.

Re: Project Euler

#25

Post by cantrip »

luckyt wrote: post #8

Solutions here: https://github.com/luckytoilet/projecteuler-solutions

Yeah why? Why would you publish strictly an answer key, with no solutions or discussion when the site itself, a well known and cherished part of the software development community has explicitly requested that you don't?

Re: Project Euler

#26

Post by metahost »

jackweirdy wrote: post #21
delibes wrote: post #9

Ah, memories. Back in 2010 I whinged to colleagues who were using their fancy functional languages to solve problem 1 inefficiently in 2 lines of code... " Project Euler problem 1 :- It helps to know the sum of (1 .. x) is 0.5 * n * (n+1). Multiples of 3 less than 1000 are (3 .. 999) == (1 .. 333) * 3 = 0.5 * 333 * 334 * 3 = 166833 Multiples of 5 less than 1000 are (5 .. 995) == (1 .. 199) * 5 = 0.5 * 199 * 200 * 5 =…

> It helps to know the sum of (1 .. x) is 0.5 * n * (n+1) This is cool. As someone out of full time education, and who dropped maths relatively early, where’s a good place to start learning some of this stuff? By stuff I mean things related to algebra and whatever the black magic I quoted is

Any arithmetic book on Sequence and Series will provide you with the basic toolset to derive the above.

Re: Project Euler

#27

Post by pkaye »

delibes wrote: post #9

Ah, memories. Back in 2010 I whinged to colleagues who were using their fancy functional languages to solve problem 1 inefficiently in 2 lines of code... " Project Euler problem 1 :- It helps to know the sum of (1 .. x) is 0.5 * n * (n+1). Multiples of 3 less than 1000 are (3 .. 999) == (1 .. 333) * 3 = 0.5 * 333 * 334 * 3 = 166833 Multiples of 5 less than 1000 are (5 .. 995) == (1 .. 199) * 5 = 0.5 * 199 * 200 * 5 =…

If you read the first paragraph of the website, "Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems."

Re: Project Euler

#28

Post by feikname »

jackweirdy wrote: post #21
delibes wrote: post #9

Ah, memories. Back in 2010 I whinged to colleagues who were using their fancy functional languages to solve problem 1 inefficiently in 2 lines of code... " Project Euler problem 1 :- It helps to know the sum of (1 .. x) is 0.5 * n * (n+1). Multiples of 3 less than 1000 are (3 .. 999) == (1 .. 333) * 3 = 0.5 * 333 * 334 * 3 = 166833 Multiples of 5 less than 1000 are (5 .. 995) == (1 .. 199) * 5 = 0.5 * 199 * 200 * 5 =…

> It helps to know the sum of (1 .. x) is 0.5 * n * (n+1) This is cool. As someone out of full time education, and who dropped maths relatively early, where’s a good place to start learning some of this stuff? By stuff I mean things related to algebra and whatever the black magic I quoted is

I do not have an formal math education, but I'm fairly sure what you quote is part of the Arithmetic Progression concept (the sum formula more specifically), so taking a look at the arithmetic branch of math might be a good idea.

Wikipedia[0], MathWords[1] and Brilliant[2] might be some good resources for you.

[0] https://en.wikipedia.org/wiki/Lists_of_mathematics_topics [1] http://www.mathwords.com/ [2] https://brilliant.org/

Re: Project Euler

#29

Post by pacuna »

DonbunEf7 wrote: post #18

The best tool for solving these problems is a whiteboard; writing code should be seen as a last resort, and brute-force solutions should be shunned, as they provide no insight.

good luck with that

Re: Project Euler

#30

Post by racer-v »

bunderbunder wrote: post #10
delibes wrote: post #9

Ah, memories. Back in 2010 I whinged to colleagues who were using their fancy functional languages to solve problem 1 inefficiently in 2 lines of code... " Project Euler problem 1 :- It helps to know the sum of (1 .. x) is 0.5 * n * (n+1). Multiples of 3 less than 1000 are (3 .. 999) == (1 .. 333) * 3 = 0.5 * 333 * 334 * 3 = 166833 Multiples of 5 less than 1000 are (5 .. 995) == (1 .. 199) * 5 = 0.5 * 199 * 200 * 5 =…

The best point to be made from this isn't that it's more computationally efficient. It's that, compared to something like sum [x | x which is really just bluntly translating the question into a programming language and letting it take care of the rest, your approach involves some actual insight into the underlying math.

Right. We don't actually know whether this Haskell code is less efficient - that depends entirely on the implementation.
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