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old Re: The Ultimate Lua Challenge

Infinite Rain
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Lee, theres is another way to earn Lua Programmer title?
IF YOU ONLY 13 YEARS OLD!?
I cant solve probs becouse my math knowledge is only on 6'th class level
so.. to be lua programmer you must end school?

old Re: The Ultimate Lua Challenge

DannyDeth
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You don't have to finish school, lol. You just need to know some math for most of this stuff, which you can learn on the internet.

old Re: The Ultimate Lua Challenge

Yates
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I might be able to help.
IMG:https://mozey.files.wordpress.com/2007/02/findx.gif

More >

old Re: The Ultimate Lua Challenge

Yates
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user palomino has written
user Yates has written
No.

3² = 9
4² = 12

9+12=21

The squared root of 21 is in between 4 and 5.
To be exact it's 4.58257569

*facepalm*
sqrt(3^2 + 4^2) = 5

Oh fuck yea. I knew I did something wrong, 4x4=16 xP
Damn 3 and 4 always get me mixed up.

old Re: The Ultimate Lua Challenge

palomino
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user Yates has written
user palomino has written
user Yates has written
No.

3² = 9
4² = 12

9+12=21

The squared root of 21 is in between 4 and 5.
To be exact it's 4.58257569

*facepalm*
sqrt(3^2 + 4^2) = 5

Oh fuck yea. I knew I did something wrong, 4x4=16 xP
Damn 3 and 4 always get me mixed up.

Happens to me all the time.

old Re: The Ultimate Lua Challenge

DoP3
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Quote
Haha lol Yates

BTW to add to that, it's also a particular configuration, forgot the name, but when you get 3 and 4 as the ( french "catets" don't know the term in English sorry and might have made a spelling mistake there too lol...) you allways get the 5 as the ("hypotenus" ... again )

Lua.. If only i had time for you

old Re: The Ultimate Lua Challenge

Lee
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user Yates has written
Oh fuck yea. I knew I did something wrong, 4x4=16 xP
Damn 3 and 4 always get me mixed up.


On my diff eq final, we had this easy question asking for eigenvalues of a forcing function superimposed over some wave equation, long story short, 2*3 accidentally 5 and I had to drop out of college

old Re: The Ultimate Lua Challenge

FlooD
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user Yates has written
I might be able to help.
IMG:https://mozey.files.wordpress.com/2007/02/findx.gif

More >

user Yates has written
No.

3² = 9
4² = 12

9+12=21

The squared root of 21 is in between 4 and 5.
To be exact it's 4.58257569

not sure if troll

or just stupid

old Re: The Ultimate Lua Challenge

Alistaire
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4² is 16 >/)<
kids these days

----

3² = 9
4² = 16

9+16=25
squareroot 25 = 5

So, how the hell are you going to explain anything with this sum?

old Re: The Ultimate Lua Challenge

palomino
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user Lee has written
user Yates has written
Oh fuck yea. I knew I did something wrong, 4x4=16 xP
Damn 3 and 4 always get me mixed up.


On my diff eq final, we had this easy question asking for eigenvalues of a forcing function superimposed over some wave equation, long story short, 2*3 accidentally 5 and I had to drop out of college

You mean they actually kicked out out because of THAT?

@2Fast4You That's what I just said.

old Re: The Ultimate Lua Challenge

Apache uwu
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user DannyDeth has written
You don't have to finish school, lol. You just need to know some math for most of this stuff, which you can learn on the internet.


Very bad advice. Programming/Scripting is very easy actually, anyone can spend a few weeks working on the syntax and understand the language. However concept is not so easy. Suppose you were creating a business application or inventory application you would need those concepts to even get started.

So no, don't stop school, and anyways you need your university or collage degree to get anyone to hire you as a programmer anyways. Extremely bad advice.

old Re: The Ultimate Lua Challenge

RAVENOUS
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Drop school for a "Lua Programmer"-title on a webforum. #winning

Actually, you can learn everything on web. With trying over studying it's even more easy.

old Re: The Ultimate Lua Challenge

IWhoopedPythagoras
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--11
function is_prime(n)

	if n % 2 == 0 and n ~= 2 then return false end
	ss = floor(sqrt(n))
	if ss * ss  == n then return false end
	local i  = 3
	while i <= ss do
		if n % i == 0 then return false end
		i = i + 2
	end
	return true
end
--12
function gcd(a,b)
	local c = a % b
	if c  == 0 then	return b else return gcd(b, c) end
end

--13
function relative_prime(a,b)
	return gcd(a,b) == 1
end

--14
function phi(n)

	local c = 0
	if is_prime(n) then return n-1
	else
		for i = 1, n do
			if relative_prime(i,n) then
				c = c + 1
			end
		end
	end
	return c
end

function get_primes(n)
	local primes = {}
	ss = math.floor(math.sqrt(n))
	for i =2, n do
		primes[i] = true
	end
	for i = 2, n do
		if primes[i] then
			for j = i*i, n, i do
				primes[j] = nil
			end
		end
	end
	return primes
end

--15
function prime_factors(n)

	local primes = get_primes(n)
	local factors = {}

	local function factor(n)
		if primes[n] then
			table.insert(factors,n)
			return
		else
			for i,v in pairs(primes) do
				if n % i  == 0 then
					table.insert(factors,i)
					return factor(n/i)
				end
			end
		end
	end

	factor(n)

	return factors
end

--16
function unique_prime_factors(n)
	local count = 0
	local factors = prime_factors(n)
	local uniques = {}
	local duplicates = {}
	local dup = 1

	for i = 1, #factors do
		local f = factors[i]
		local next = factors[i+1]

		if next == f then
			dup = dup + 1
		else
			table.insert(uniques,f)
			table.insert(duplicates,dup)
			dup = 1
		end
	end

	return uniques,duplicates
end

--17
function phi_2(n)
	local pow = math.pow
	p,k = unique_prime_factors(n)
	local sum = 1
	for i=1,#p do
		sum = sum * (p[i] - 1)*pow(p[i],(k[i] - 1))
	end
	return sum
end

print(phi(10090))
print(phi_2(10090))


--18
function gen(n)
	return get_primes(n)
end
--19
function goldbach(n)
	primes = get_primes(n)

	for i,v in pairs(primes) do
		if primes[i] and primes[n-i] then
			return i,n-i
		end
	end
	return false
end
for i=4, 1000,2 do
	if not goldbach(i) then print("not able "..i) end
end

old Re: The Ultimate Lua Challenge

Lee
Moderator Off Offline

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31. Christmas Challenge Problem:

Given f(n) = 12*f(n-2) + 4*f(n-1) + 5 such that f(0) = 0 and f(1) = 1, find f(100).

Note: You don't have to produce a program that gives the result, just something that can help you solve the problem.

Hint: start with http://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form and figure out a way to quickly square 3x3 matrices

Hint 2: the answer is 163329655875017726524172566789514455134285927618344355435787010709165677303125 (78 digits)

32. Challenge Problem part 2.

Show that for sufficiently large n, we can approximate the first 1/3 significant digits of f(n) with the function (6^100)/4. (you only need to come up with a weak proof for this, you don't have to find the strict lower bound, which is actually around 1.585/2.585)

33.

If we're given the following function:
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function f(a, x, x0)
	if x == 0 then
		return x0
	end
	return f(a,x-1)/4 * (5 - a*f(a,x-1,x0)^3)
end

find f(a, infinity, x0) for all a and x0.

Hint: the recursive function x[k+1] = x[k]/4*(5-a*x[k]^3), x[0] = x0 converges finitely and not at 0.

Hint: Converging at n implies that x[n+1] = x[n]
edited 5×, last 05.02.12 02:44:14 am
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