nayuki / project-euler-solutions Goto Github PK

Hi!

I'm solving the Project Euler problems with ReasonML, it's interesting to you if I open a PR with the solutions in this language into this repository?

Hi, I'm trying to understand what kind of crazy magic you're using for the twenty-sixth problem, specifically this little black box of black magic:

def reciprocal_cycle_len(n):
    seen = {}
    x = 1

    for i in itertools.count():
        if x in seen:
            return i - seen[x]
        else:
            seen[x] = i
            x = x * 10 % n

It's the simplest thing I've ever seen that I just can't figure out. I guess I'm just not good with numerology. Is there a spatial relationship that would make sense?

thought it would be neat :)

I've noticed that this repo posts most if not all of the Project Euler answers. That appears to go against the spirit of Project Euler. There is a write up near the bottom of the Project Euler FAQs that talks about this.

Hello there, I think I found a theoretical vital time improvement to question 27 "Quadratic primes"
It relies on the following 2 lemmas:
1.If b is composite OR smaller then 0 then x^2+ax+b returns a non-prime number for x=0
Meaning that the prime sequence for those b already ends at x=0 [Note that in the examples 41 and 1601 are both bigger then 0 and prime]
This can remove both the negative numbers AND the composite numbers from the b search.
2.If prime p divides a and b then x^2+ax+b returns a composite number for x=p
[Trivial proof since p divides p^2+xp*p+yp]
for those ab pairs, the sequence end at most in x=p
That means we don't have to check the ab pairs which share a prime factor smaller then the minimum required consecutive sequence (which in our range is at least 39)

https://github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p010.hs

I did like your code. But it's too slow to get the answer.

I was getting a syntax error on line:
NONPRIME_LAST_DIGITS = {0, 2, 4, 5, 6, 8}

Which I was able to "correct" by changing to an array notation [0,2...] on line:
if i == 1 or arr[-1] not in [0,2,4,5,6,8]:

I ran the code for Problem 21 for the first ten numbers.

def compute():
	# Compute sum of proper divisors for each number
	divisorsum = [0] * 10
	for i in range(1, len(divisorsum)):
		for j in range(i * 2, len(divisorsum), i):
			divisorsum[j] += i
		# Find all amicable pairs within range
	ans = 0
	for i in range(1, len(divisorsum)):
		j = divisorsum[i]
		if j != i and j < len(divisorsum) and divisorsum[j] == i:
			ans += i
	return str(ans)

and it doesn't seem to return the right answer.

For example, for the first 10 numbers the pair (n, d(n)) (where d(n) = sum of proper divisors of n)
is:
[(2, 1), (3, 1), (5, 1), (7, 1), (4, 3), (9, 4), (6, 6), (8, 7), (10, 8)]
and the answer should be 17.

The code outputs: [0, 0, 1, 1, 3, 1, 6, 1, 7, 4] for divisorsum and
0 for the ans.

Hi! Thank you for this repository. I use it frequently to check, whether my solutions are completely off the charts, in the right ballpark, or sane.

I believe however, that #026 might have a wrong solution. 1/171 has a recurring decimals length or 18, but 1/983 has a much longer sequence of recurring decimals: 982 digits long:

0 0 1 0 1 7 2 9 3 9 9 7 9 6 5 4 1 2 0 0 4 0 6 9 1 7 5 9 9 1 8 6 1 6 4 8 0 1 6 2 7 6 7 0 3 9 6 7 4 4 6 5 9 2 0 6 5 1 0 6 8 1 5 8 6 9 7 8 6 3 6 8 2 6 0 4 2 7 2 6 3 4 7 9 1 4 5 4 7 3 0 4 1 7 0 9 0 5 3 9 1 6 5 8 1 8 9 2 1 6 6 8 3 6 2 1 5 6 6 6 3 2 7 5 6 8 6 6 7 3 4 4 8 6 2 6 6 5 3 1 0 2 7 4 6 6 9 3 7 9 4 5 0 6 6 1 2 4 1 0 9 8 6 7 7 5 1 7 8 0 2 6 4 4 9 6 4 3 9 4 7 1 0 0 7 1 2 1 0 5 7 9 8 5 7 5 7 8 8 4 0 2 8 4 8 4 2 3 1 9 4 3 0 3 1 5 3 6 1 1 3 9 3 6 9 2 7 7 7 2 1 2 6 1 4 4 4 5 5 7 4 7 7 1 1 0 8 8 5 0 4 5 7 7 8 2 2 9 9 0 8 4 4 3 5 4 0 1 8 3 1 1 2 9 1 9 6 3 3 7 7 4 1 6 0 7 3 2 4 5 1 6 7 8 5 3 5 0 9 6 6 4 2 9 2 9 8 0 6 7 1 4 1 4 0 3 8 6 5 7 1 7 1 9 2 2 6 8 5 6 5 6 1 5 4 6 2 8 6 8 7 6 9 0 7 4 2 6 2 4 6 1 8 5 1 4 7 5 0 7 6 2 9 7 0 4 9 8 4 7 4 0 5 9 0 0 3 0 5 1 8 8 1 9 9 3 8 9 6 2 3 6 0 1 2 2 0 7 5 2 7 9 7 5 5 8 4 9 4 4 0 4 8 8 3 0 1 1 1 9 0 2 3 3 9 7 7 6 1 9 5 3 2 0 4 4 7 6 0 9 3 5 9 1 0 4 7 8 1 2 8 1 7 9 0 4 3 7 4 3 6 4 1 9 1 2 5 1 2 7 1 6 1 7 4 9 7 4 5 6 7 6 5 0 0 5 0 8 6 4 6 9 9 8 9 8 2 7 0 6 0 0 2 0 3 4 5 8 7 9 9 5 9 3 0 8 2 4 0 0 8 1 3 8 3 5 1 9 8 3 7 2 3 2 9 6 0 3 2 5 5 3 4 0 7 9 3 4 8 9 3 1 8 4 1 3 0 2 1 3 6 3 1 7 3 9 5 7 2 7 3 6 5 2 0 8 5 4 5 2 6 9 5 8 2 9 0 9 4 6 0 8 3 4 1 8 1 0 7 8 3 3 1 6 3 7 8 4 3 3 3 6 7 2 4 3 1 3 3 2 6 5 5 1 3 7 3 3 4 6 8 9 7 2 5 3 3 0 6 2 0 5 4 9 3 3 8 7 5 8 9 0 1 3 2 2 4 8 2 1 9 7 3 5 5 0 3 5 6 0 5 2 8 9 9 2 8 7 8 9 4 2 0 1 4 2 4 2 1 1 5 9 7 1 5 1 5 7 6 8 0 5 6 9 6 8 4 6 3 8 8 6 0 6 3 0 7 2 2 2 7 8 7 3 8 5 5 5 4 4 2 5 2 2 8 8 9 1 1 4 9 5 4 2 2 1 7 7 0 0 9 1 5 5 6 4 5 9 8 1 6 8 8 7 0 8 0 3 6 6 2 2 5 8 3 9 2 6 7 5 4 8 3 2 1 4 6 4 9 0 3 3 5 7 0 7 0 1 9 3 2 8 5 8 5 9 6 1 3 4 2 8 2 8 0 7 7 3 1 4 3 4 3 8 4 5 3 7 1 3 1 2 3 0 9 2 5 7 3 7 5 3 8 1 4 8 5 2 4 9 2 3 7 0 2 9 5 0 1 5 2 5 9 4 0 9 9 6 9 4 8 1 1 8 0 0 6 1 0 3 7 6 3 9 8 7 7 9 2 4 7 2 0 2 4 4 1 5 0 5 5 9 5 1 1 6 9 8 8 8 0 9 7 6 6 0 2 2 3 8 0 4 6 7 9 5 5 2 3 9 0 6 4 0 8 9 5 2 1 8 7 1 8 2 0 9 5 6 2 5 6 3 5 8 0 8 7 4 8 7 2 8 3 8 2 5 0 2 5 4 3 2 3 4 9 9 4 9 1 3 5 3

If we name that sequence s for "sequence", then 1/983 is:

0.sssss...

testing123

Are we allowed to include the Answers.txt file as-is in our own repos, thereby using it as reference material for testing self-written code?

Incidentally, the solution to Problem 6 can be further simplified as the product of linear terms and a constant:

n(n-1)(n+1)(3n+2)/12

I saw you calculating sqrt of some number but I didn't get how you are calculating floor of that number in this code

def sqrt(x):
	assert x >= 0
	i = 1
	while i * i <= x:
		i *= 2
	y = 0
	while i > 0:
		if (y + i)**2 <= x:
			y += i
		i //= 2
	return y

Import eulerlib module was not work