Voith voith

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Building on top of Ethereum

Trying to hone my rust skills.

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voith / sum_amicable.py

Created November 5, 2019 16:40

sum of all the amicable numbers under 10000

	from math import sqrt

	def _sum_of_proper_divisors(n, sum_of_proper_divisors):
	_sum = 1
	sqrt_n = int(sqrt(n))
	for i in range(2, sqrt_n+1):
	if n % i == 0:
	_sum += i
	if n // i != i:
	_sum += n // i

voith / sum_of_2_non_abundant.py

Created November 10, 2019 00:49

sum of all the positive integers which cannot be written as the sum of two abundant numbers.

	from math import sqrt


	def sum_of_divisors(number):
	_sum = 1
	nsqrt = int(sqrt(number))
	for i in range(2, nsqrt + 1):
	if number % i == 0:
	_sum += i
	if (number / i != nsqrt):

voith / nth_lexographic_permutation.py

Last active November 12, 2019 04:20

millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9

	from math import factorial


	numbers = list(range(10))
	nth_perm = 1000000
	total_numbers = len(numbers)
	num = ''


	def get_factorial_divisor(factorial, dividend):

voith / reciprocal_cycles.py

Created November 12, 2019 03:59

value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction par

	largest_num = 1000
	largest_divisor = 3
	longest_cycle = 1
	rdigits = ''

	def get_divide_result(i):
	seen_remainders = [False] * i
	remainder = 10 % i
	cycle_count = 0
	while remainder!= 0 and seen_remainders[remainder] is False:

voith / quadratic_primes.py

Created November 16, 2019 19:38

Find the product of the coefficients, a and b, for the quadratic expression (n**2+n+41) that produces the maximum number of primes for consecutive values of n, starting with n=0.

	from math import sqrt


	MAX = 999999
	sieve = [True] * (MAX + 1)
	sieve[0], sieve[1] = False, False

	for i in range(2, int(sqrt(MAX)) + 1):
	if sieve[i] is True:
	for j in range(2 * i, MAX, i):

voith / sum_spiral_diagonals.py

Created November 28, 2019 02:42

sum of the numbers on the diagonals in a 1001 by 1001 spiral

	grid_size = 1001
	num = grid_size ** 2
	sum_diagonals = num

	for i in range(grid_size - 1, 0, -2):
	for _ in range(4):
	num -= i
	sum_diagonals += num
	print(sum_diagonals)

voith / coin_change.py

Last active November 30, 2019 18:52

Findall possible ways €2 can be split using coins 1p, 2p, 5p, 10p, 20p, 50p, €1 (100p) and €2 (200p)

	final_target = 200
	possibilities = [0] * (final_target + 1)
	possibilities[0] = 1
	# keeping the coins sorted is the key here
	# we want to compute possibilities for lower coins first.
	# This helps us to fix one particular coin and
	# then find the other possibilites of the change needed
	coins = [1, 2, 5, 10, 20, 50, 100, 200]
	for coin in coins:
	for current_target in range(coin, final_target + 1):

voith / grid_dfs.py

Created December 13, 2019 18:22

total number of connected regions in a grid

	class Solution(object):

	def __init__(self, matrix, row, col):
	self.matrix = matrix
	self.row = row
	self.col = col

	def is_safe(self, i, j, visited):
	return (
	0 <= i < self.row and

voith / substrings.py

Created December 14, 2019 10:41

	def get_all_substrings(arr, index, subarr, record):
	if index == len(arr):
	if len(subarr) != 0:
	record.append(''.join(subarr))
	else:
	get_all_substrings(arr, index + 1, subarr, record)
	get_all_substrings(arr, index + 1, subarr + [arr[index]], record)

	return record

voith / Comptroller_ABIi.json

Created June 1, 2020 21:43

	[{
	"inputs": [],
	"payable": false,
	"stateMutability": "nonpayable",
	"type": "constructor"
	}, {
	"anonymous": false,
	"inputs": [{
	"indexed": false,
	"internalType": "string",

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