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euler.py
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2698 lines (2121 loc) · 74.9 KB
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#! /usr/bin/python3
# vim: set tabstop=4 shiftwidth=4 expandtab :
# Generated by letscode
"""Project Euler solutions"""
import string
import math
import itertools
import collections
import heapq
from prime import prime_factors_list, sieve, nb_divisors, yield_primes
from prime import primes_up_to, nb_prime_divisors, totient, divisors_sieve
from prime import is_prime, prime_divisors_sieve, mult
from functions import ceil, gcd, modinv
from functions import fibo, lcmm, yield_pythagorean_triples_of_peri
from functions import Tn, Pn, Hn, isPn, champernowne_digit
from poker import Hand
import os
from timeit import default_timer as timer
resource_folder = ''
def euler1(lim=1000):
"""Solution for problem 1."""
# could use sum formula here
return sum(i for i in range(lim) if i % 3 == 0 or i % 5 == 0)
def euler2(lim=4000000):
"""Solution for problem 2."""
s = 0
for f in fibo(1, 2):
if f > lim:
return s
if f % 2 == 0:
s += f
def euler3(n=600851475143):
"""Solution for problem 3."""
return prime_factors_list(n)[-1]
def euler4(l=3):
"""Solution for problem 4."""
# simple optimisation would be an early break
return max(n for n in (i * j for i, j in itertools.combinations(range(10 ** (l - 1), 10 ** l), 2)) if str(n) == str(n)[::-1])
def euler5(lim=20):
"""Solution for problem 5."""
return lcmm(*range(1, lim + 1))
def euler6(lim=100):
"""Solution for problem 6."""
# could use sum formula here
numbers = range(1, lim + 1)
sum_ = sum(numbers)
return sum_ * sum_ - sum(i * i for i in numbers)
def nth(iterable, n, default=None):
"""Returns the nth item or a default value.
From http://stackoverflow.com/questions/12007820/better-ways-to-get-nth-element-from-an-unsubscriptable-iterable ."""
return next(itertools.islice(iterable, n, None), default)
def euler7(n=10001):
"""Solution for problem 7."""
return nth(yield_primes(), n - 1)
def euler8(l=13):
"""Solution for problem 8."""
# Too lazy for optimisations
n = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'
return max(mult(int(c) for c in s) for s in (n[i:i + l] for i in range(len(n))))
def euler9(p=1000):
"""Solution for problem 9."""
for a, b, c in yield_pythagorean_triples_of_peri(p):
return a * b * c
def euler10(lim=2000000):
"""Solution for problem 10."""
return sum(primes_up_to(lim))
def euler11_():
"""Solution for problem 11."""
pass
def euler12(nb_div=500):
"""Solution for problem 12."""
# T(n) = n(n+1)/2
# T(2k + 0) = 2k*(2k+1) / 2 = k*(2k+1)
# and gcd(k, 2*k+1) = gcd(k, 1) = 1
# Thus, nb_div(T(2k)) = nb_div(k) * nb_div(2k+1)
# T(2k + 1) = (2k+1)*(2k+2) / 2 = (2k+1)(k+1)
# and gcd(2k+1, k+1) = gcd(k, k+1) = gcd(k, 1) = 1
# Thus, nb_div(T(2k + 1)) = nb_div(k+1) * nb_div(2k+1)
nb_div_k, nb_div_kp1 = None, 1
for k in itertools.count(1):
nb_div_k, nb_div_kp1 = nb_div_kp1, nb_divisors(k + 1)
c = 2 * k + 1
nb_div_c = nb_divisors(c)
if nb_div_k * nb_div_c >= nb_div:
return c * k
if nb_div_kp1 * nb_div_c >= nb_div:
return c * (k + 1)
def euler13_():
"""Solution for problem 13."""
pass
def euler14(lim=1000000):
"""Solution for problem 14."""
collatz = {1: 1}
for i in range(2, lim):
if i not in collatz:
chain = []
while i not in collatz:
chain.append(i)
i = (3 * i + 1) if i % 2 else (i // 2)
stop = collatz[i] + 1
for idx, val in enumerate(reversed(chain)):
collatz[val] = stop + idx
assert i in collatz
return max((v, k) for k, v in collatz.items() if k <= lim)[1]
def euler15(col=20, row=20):
"""Solution for problem 15."""
nb_routes = [[0] * (row + 1) for i in range(col + 1)]
nb_routes[0][0] = 1
for i in range(col + 1):
for j in range(row + 1):
if i:
nb_routes[i][j] += nb_routes[i - 1][j]
if j:
nb_routes[i][j] += nb_routes[i][j - 1]
return nb_routes[-1][-1]
def sum_digit(n):
return sum(int(c) for c in str(n))
def euler16(n=1000):
"""Solution for problem 16."""
return sum_digit(2 ** n)
def euler17(lim=1000):
"""Solution for problem 17."""
# Optimisation could be done by grouping similar number
# prefixes/suffixes and use multiplications instead of
# repeated sums but this is fast enough
small_numbers_str = {
1: "one", 2: "two", 3: "three", 4: "four", 5: "five", 6: "six",
7: "seven", 8: "eight", 9: "nine", 10: "ten", 11: "eleven",
12: "twelve", 13: "thirteen", 14: "fourteen", 15: "fifteen",
16: "sixteen", 17: "seventeen", 18: "eighteen", 19: "nineteen"
}
ten_multiples_str = {
2: "twenty", 3: "thirty", 4: "forty", 5: "fifty",
6: "sixty", 7: "seventy", 8: "eighty", 9: "ninety"
}
small_numbers_len = {k: len(v) for k, v in small_numbers_str.items()}
ten_multiples_len = {k: len(v) for k, v in ten_multiples_str.items()}
and_len = len("and")
hundred_len = len("hundred")
thousand_len = len("thousand")
def get_len_for_number(n):
n, units = divmod(n, 10)
n, tens = divmod(n, 10)
thousands, hundreds = divmod(n, 10)
total = 0
if thousands:
total += small_numbers_len[thousands] + thousand_len
if hundreds:
total += small_numbers_len[hundreds] + hundred_len
if tens or units:
total += and_len
if tens > 1:
total += ten_multiples_len[tens]
else:
units += 10 * tens # move tens in units
if units:
total += small_numbers_len[units]
return total
return sum(get_len_for_number(i) for i in range(1, lim + 1))
def euler18_():
"""Solution for problem 18."""
pass
def euler19():
"""Solution for problem 19."""
day = 2 # Tuesday 1 Jan 1901
nb_days = 7
freq = [0] * nb_days
months = [nb % nb_days for nb in (31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)]
for year in range(1901, 2001):
for i, month in enumerate(months):
freq[day] += 1
day = (day + month + (1 if (i == 1 and (year % 4 == 0 and (year % 100 != 0 or year % 400 == 0))) else 0)) % nb_days
return freq[0]
def euler20(n=100):
"""Solution for problem 20."""
return sum_digit(math.factorial(n))
def euler21(lim=10000):
"""Solution for problem 21."""
sum_div = [sum(l) for l in divisors_sieve(lim)]
return sum(
s + i
for i, s in enumerate(sum_div)
if s < i and sum_div[s] == i)
def euler22(f='p022_names.txt'):
"""Solution for problem 22."""
with open(os.path.join(resource_folder, f)) as file_:
return sum((i + 1) * sum(1 + ord(c) - ord('A') for c in name)
for i, name in enumerate(sorted(''.join(file_.readlines()).replace('"', '').split(','))))
def euler23(lim=28123):
"""Solution for problem 23."""
abun = [i > 0 and sum(l) > i for i, l in enumerate(divisors_sieve(lim))]
return sum(
n for n in range(lim)
if not any(i for i in range(1 + n // 2) if abun[i] and abun[n - i]))
def euler24():
"""Solution for problem 24."""
return int(''.join(nth(itertools.permutations(string.digits), 1000000 - 1)))
def euler25(nb_digits=1000):
"""Solution for problem 25."""
lim = 10 ** (nb_digits - 1)
for i, f in enumerate(fibo()):
if f > lim:
return 1 + i
def length_recur_cycle(a, b, base=10):
remainders = {}
for i in itertools.count():
if a == 0:
return 0
if a in remainders:
return i - remainders[a]
remainders[a] = i
a = (a * base) % b
def euler26(lim=1000):
"""Solution for problem 26."""
return max(range(2, lim), key=lambda i: length_recur_cycle(1, i))
def euler27(lim=1000):
"""Solution for problem 27."""
# P(0) = b must be prime
# P(1) = 1 + a + b must be prime so 1+a+b >= 2 => a => 1-b
maxa, maxb, maxn = None, None, 0
for b in primes_up_to(lim):
for a in range(1 - b, lim + 1):
assert -lim <= a <= lim
n = 0
while is_prime(n * n + a * n + b):
n += 1
if n > maxn:
maxa, maxb, maxn = a, b, n
return maxa * maxb
def euler28(n=1001):
"""Solution for problem 28."""
# For a level of size s (s>1), corners are :
# s*s, s*s-s+1, s*s-2s+2, s*s-3s+3
# Their sum is 4*s*s - 6*s + 6 (3 <= s <= n)
# Writing s = 2*j + 3, sum of corners is
# 16*j*j + 36j + 24
# This sum could be computed in constant time but this is
# fast enough.
return 1 + sum(16 * j * j + 36 * j + 24 for j in range(n // 2))
def euler29(lima=100, limb=100):
"""Solution for problem 29."""
n = set()
for a in range(2, lima + 1):
p = a
for _ in range(2, limb + 1):
p *= a
n.add(p)
return len(n)
def euler30():
"""Solution for problem 30."""
# sum(fifth(digits)) = n
# nb_digits * fifth(0) < sum(fifth(digits)) <= nb_digits * fifth(base - 1)
# base ^ (nb_digits - 1) <= n < base ^ nb_digits
# => base ^ (nb_digits - 1) <= nb_digits * fifth(base - 1)
# with base = 10, nb_digits <= 6
fifth = {c: int(c) ** 5 for c in string.digits}
dict_sum = {}
for nb_dig in range(2, 6 + 1):
for l in itertools.combinations_with_replacement('123456789', nb_dig):
dict_sum.setdefault(sum(fifth[c] for c in l), []).append(list(l))
return sum(n for n, l in dict_sum.items() if [c for c in sorted(str(n)) if c != '0'] in l)
def euler31(obj=200, coins=None):
"""Solution for problem 31."""
if coins is None:
coins = [1, 2, 5, 10, 20, 50, 100, 200]
nb_ways = [1] + [0] * obj
for c in coins:
for v in range(obj + 1 - c):
nb_ways[v + c] += nb_ways[v]
return nb_ways[-1]
def euler32():
"""Solution for problem 32."""
# nb_dig(a*b) = nb_dig(a) + nb_dig(b) or nb_dig(a) + nb_dig(b) - 1
# we want : a * b = c with 0 < a < b < c
# and we must have : nb_dig(a) + nb_dig(b) + nb_dig(c) = 9
# i + j + k = 9 and (i + j = k or i + j - 1 = k)
# => 2i + 2j = 9 (impossible) or 2i + 2j = 10
# => i + j = 10 => (1, 4, 4), (2, 3, 4).
return sum({a * b
for a in range(2, 98)
for b in range(1234 if a <= 10 else 123, 10000 // a + 1)
if ''.join(sorted(list(str(a) + str(b) + str(a * b)))) == '123456789'
})
def euler33():
"""Solution for problem 33."""
# Different options reducing to b/c (with 0 < b < c < 10 and 0 <= a < 10) are :
# ab / ac (possible only if a=0 or b=c : not interesting)
# ba / ca (possible only if a=0 or b=c : not interesting)
# ab / ca (possible iif 9bc = a (10c - b))
# ba / ac (possible iif 9bc = a (10b - c))
# Also, top / bottom = b/c <=> top * c = bottom * b
t, b = [mult(lst)
for lst in zip(*[(b, c)
for b, c in itertools.combinations(range(1, 10), 2)
for a in range(1, 10)
for top, bot in [(10 * a + b, 10 * c + a), (10 * b + a, 10 * a + c)]
if top * c == bot * b])]
return b // gcd(t, b)
def euler34():
"""Solution for problem 34."""
# sum(fact(digits)) = n
# nb_digits * fact(0) < sum(fact(digits)) <= nb_digits * fact(base - 1)
# base ^ (nb_digits - 1) <= n < base ^ nb_digits
# => base ^ (nb_digits - 1) <= nb_digits * fact(base - 1)
# with base = 10, we have nb_digits <= 7
fact = {c: math.factorial(int(c)) for c in string.digits}
dict_sum = {}
for nb_dig in range(2, 7 + 1):
for l in itertools.combinations_with_replacement(string.digits, nb_dig):
dict_sum.setdefault(sum(fact[c] for c in l), []).append(list(l))
return sum(n for n, l in dict_sum.items() if sorted(str(n)) in l)
def euler35(nb_dig_max=6):
# permutations of 2 digits or more must contain only 1, 3, 7, 9
count = 4 # counting 2, 3, 5 and 7
final_numbers = {'1', '3', '7', '9'}
for l in range(2, nb_dig_max + 1):
for p in itertools.product(final_numbers, repeat=l):
p_int = int(''.join(p))
perm = {int(''.join(p[i:] + p[:i])) for i in range(len(p))}
if p_int == min(perm) and all(is_prime(n) for n in perm):
count += len(perm)
return count
def euler36():
"""Solution for problem 36."""
# 999999
sol = []
for i in range(1000):
s = str(i)
for beg in (-1, -2):
n = int(s + s[beg::-1])
s2 = bin(n)[2:]
if s2 == s2[::-1]:
sol.append(n)
return sum(sol)
def euler37():
"""Solution for problem 37."""
left, sol = {0}, []
while left:
left = {n for n in (10 * l + i for i in range(10) for l in left) if is_prime(n)}
sol.extend(n for n in left if n > 10 and all(is_prime(n % (10 ** pow)) for pow in range(1, len(str(n)))))
return sum(sol)
def euler38():
"""Solution for problem 38."""
# '123456789' will be decomposed in at least two elements,
# the smallest being 4 at most characters long
sol = 0
digits = {str(d) for d in range(1, 10)}
for n in range(10000):
s = ""
for i in itertools.count(1):
s += str(n * i)
if len(s) >= len(digits):
if len(s) == len(digits) and set(s) == digits:
sol = max(sol, int(s))
break
return sol
def euler39(lim=1000):
"""Solution for problem 39."""
return max(range(1, lim + 1), key=lambda p: len(list(yield_pythagorean_triples_of_peri(p))))
def euler40():
"""Solution for problem 40."""
return mult(champernowne_digit(pow(10, i) - 1) for i in range(6))
def euler41():
"""Solution for problem 41."""
# sum(i, i=1..n) is is not divisible by 3 only if n = 1, 4, 7 or bigger than 9
for nb_dig in (7, 4, 1):
for l in itertools.permutations(str(d) for d in range(nb_dig, 0, -1)):
n = int(''.join(l))
if is_prime(n):
return n
def euler42_():
"""Solution for problem 42."""
pass
def euler43():
"""Solution for problem 43."""
# Could be optimised by using a constructive solution starting from the end
div = [(17, 7), (13, 6), (11, 5), (7, 4), (5, 3), (3, 2), (2, 1)]
return sum(int(''.join(p))
for p in itertools.permutations(string.digits)
if all(int(''.join(p[i:i + 3])) % d == 0 for d, i in div))
def euler44():
"""Solution for problem 44."""
# We look for minimal value of P(m) = P(k) - P(j)
# Such that P(k) + P(j) is pentagonal.
# We must have j < m < k.
for m in itertools.count():
pm = Pn(m)
for j in range(1, m):
pj = Pn(j)
pk = pm + pj
if isPn(pk) and isPn(pj + pk):
return pm
def euler45(nb_fact=4):
"""Solution for problem 45."""
t_gen = (Tn(n) for n in itertools.count())
p_gen = (Pn(n) for n in itertools.count())
h_gen = (Hn(n) for n in itertools.count())
t, p, h = next(t_gen), next(p_gen), next(h_gen)
while True:
if p > t:
t = next(t_gen)
elif t > p or h > p:
p = next(p_gen)
elif p > h:
h = next(h_gen)
elif t in [0, 1, 40755]:
t, p, h = next(t_gen), next(p_gen), next(h_gen)
else:
return t
def euler46():
"""Solution for problem 46."""
for i in itertools.count(9, 2):
if not is_prime(i) and not any(is_prime(i - 2 * n * n) for n in range(1, ceil(math.sqrt(i // 2)))):
return i
def euler47(nb_fact=4):
"""Solution for problem 47."""
cand = []
for i in itertools.count(2):
if nb_prime_divisors(i) == nb_fact:
cand.append(i)
if len(cand) == nb_fact:
return cand[0]
else:
cand = []
def euler48(n=1000, nb_dig=10):
"""Solution for problem 48."""
mod = 10 ** nb_dig
return sum(pow(i, i, mod) for i in range(1, n + 1)) % mod
def sorted_number(n):
# Reversed to keep 0 instead of` discarding them
return int(''.join(sorted(str(n), reverse=True)))
def euler49(nb_digit=4):
"""Solution for problem 49."""
low = 10 ** (nb_digit - 1)
high = 10 ** nb_digit - 1
prime = sieve(high)
prime_perm = {}
for i in range(low, high + 1):
if prime[i]:
prime_perm.setdefault(sorted_number(i), []).append(i)
for perms in prime_perm.values():
# could iterate only on a,b but not a bottleneck
for a, b, c in itertools.combinations(perms, 3):
assert c > b > a
if b - a == c - b and a != 1487:
return int(str(a) + str(b) + str(c))
def euler50(lim=1000000):
"""Solution for problem 50."""
primes = sieve(lim)
list_primes = [i for i, p in enumerate(primes) if p]
max_len, max_sum = 0, 0
for i in range(len(list_primes)):
for j in range(i + max_len + 1, len(list_primes)):
s = sum(list_primes[i:j]) # could use sum array here
if s > lim:
break
elif primes[s]:
assert j - i > max_len
max_len, max_sum = j - i, s
return max_sum
def euler51(nb=8):
"""Solution for problem 51."""
# Iterating on batches of number of same length to be able to use a sieve.
# For each prime, looking for digits one can change, then the possible
# combinations of these digits. For each combination, we compute the value
# of the corresponding mask and check primality.
base = 10 # Not so much of a variable due to various optimisations
lim = 1 + base - nb # replaced digit must be smaller than lim
# optimisation : precomputing powers of 10
pows = []
# optimisation : the mask (and so its length) must be divisible by 3 for
# big families, otherwise, we'd have too many multiple of 3.
len_mask_step = 3 if nb >= 8 else 1
# optimisation : last number can't be changed for families of 4 or more
last_nb_can_chg = nb <= 4
low = 1
while True:
high = base * low
primes = sieve(high)
pows.append(low)
for n in range(low, high):
if primes[n]:
n_str = [int(c) for c in reversed(str(n))]
for val_replaced in range(lim):
pos = [i for i, c in enumerate(n_str)
if c == val_replaced and (i or last_nb_can_chg)]
for len_mask in range(len_mask_step, 1 + len(pos), len_mask_step):
for pos_mask in itertools.combinations(pos, len_mask):
val_mask = sum(pows[i] for i in pos_mask)
if 1 + sum(1 for i in range(1, base - val_replaced) if primes[n + i * val_mask]) >= nb:
return n
low = high
def euler52(lim=6):
"""Solution for problem 52."""
for x in itertools.count(1):
digits = sorted_number(x)
if all(digits == sorted_number(i * x) for i in range(2, lim + 1)):
return x
def euler53(n_max=100, lim=1000000):
"""Solution for problem 53."""
# C(n, r) = n! / (r! (n-r)!) for r <= n
# = prod(n + 1 - i / i, i =1..r)
def C(n, r):
r = min(r, n - r) # symmetry
return mult(n - i for i in range(r)) // mult(i + 1 for i in range(r))
s = 0
for n in range(n_max + 1):
for r in range(n // 2 + 1): # stop halfway - use symmetry
if C(n, r) > lim:
s += 1 if 2 * r == n else 2
return s
def euler54(f='p054_poker.txt'):
"""Solution for problem 54."""
ret = 0
with open(os.path.join(resource_folder, f)) as file_:
for l in file_:
hand1, hand2 = Hand.from_string(l)
ret += hand1 > hand2
return ret
def euler55_():
"""Solution for problem 55."""
pass
def euler56(lim=100):
"""Solution for problem 56."""
int_ = {str(d): d for d in range(10)}
maxi = 1
for a in range(2, lim + 1):
p = a
for b in range(2, lim + 1):
p *= a
maxi = max(maxi, sum(int_[d] for d in str(p)))
return maxi
def euler57(nb_exp=1000):
"""Solution for problem 57."""
# If expansion at level n is a/b
# analysis shows that at level n+1, we have:
# (2*b+a) / (a+b)
# Also, gcd(2*b+a, a+b) = gcd(b, a+b) = gcd(a, b)
# Thus, gcd is conserved : if we start with a reduced
# fraction, all fractions will be reduced
nb = 0
a = b = 1
for i in range(nb_exp + 1):
assert gcd(a, b) == 1
if len(str(a)) > len(str(b)):
nb += 1
a, b = 2 * b + a, a + b
return nb
def euler58(ratio=0.1):
"""Solution for problem 58."""
# First analysis in euler28
# Corners are s*s-3s+3, s*s-2s+2, s*s-s+1, s*s
# s*s is likely not to be prime
nb_prime = 0
for s in itertools.count(3, 2):
for i in range(1, 4):
if is_prime(s * s - i * s + i):
nb_prime += 1
if nb_prime < ratio * (2 * s - 1):
return s
def euler59_():
"""Solution for problem 59."""
pass
def euler60_():
"""Solution for problem 60."""
# awful C solution on PE https://projecteuler.net/thread=60;page=3#20807
pass
def euler61_():
"""Solution for problem 61."""
pass
def euler62(nb_perm=5):
"""Solution for problem 62."""
cube_perm, l = {}, None
for i in itertools.count():
c = i * i * i
new_l = len(str(c))
if l != new_l:
cand = [numbers[0]
for c, numbers in cube_perm.items() if len(numbers) == nb_perm]
if cand:
return min(cand) ** 3
cube_perm, l = {}, new_l
cube_perm.setdefault(sorted_number(c), []).append(i)
def euler63():
"""Solution for problem 63."""
# 10^(n-1) <= x^n < 10^n
# (n-1) * log(10) <= n * log(x) < n * log(10)
# (n-1) * log(10) / n <= log(x) < log(10)
# exp((n-1) * log(10) / n) <= x < 10
# and LHS becomes bigger than 9 at n = 22
return sum(
10 - ceil(math.exp((n - 1) * math.log(10) / n))
for n in range(1, 22))
def euler64_():
"""Solution for problem 64."""
pass
def euler65_():
"""Solution for problem 65."""
pass
def euler66_():
"""Solution for problem 66."""
# http://en.wikipedia.org/wiki/Pell's_equation
pass
def euler67_():
"""Solution for problem 67."""
pass
def euler68_():
"""Solution for problem 68."""
# awful C solution on PE https://projecteuler.net/thread=68;page=3#21386
pass
def euler69(lim=1000000):
"""Solution for problem 69."""
return max((float(i) / t, i) for i, t in enumerate(totient(lim)) if i)[1]
def euler70(lim=10000000):
"""Solution for problem 70."""
n, val = lim, lim
for i, t in enumerate(totient(lim)):
if i > 1:
new_val = float(i) / t
if new_val < val and sorted_number(i) == sorted_number(t):
n, val = i, new_val
return n
def euler71_():
"""Solution for problem 71."""
pass
def euler72(lim=1000000):
"""Solution for problem 72."""
return sum(totient(lim))
def euler73_():
"""Solution for problem 73."""
pass
def euler74_():
"""Solution for problem 74."""
pass
def euler75(lim=1500000):
"""Solution for problem 75."""
# http://en.wikipedia.org/wiki/Pythagorean_triple
# a = (m*m-n*n)*d
# b = 2*m*n*d
# c = (m*m+n*n)*d
# p = 2*m*(m+n)*d
# with : pgcd(m,n)=1 and m < n < 2*m.
diff_c = [0] * (1 + lim) # 0: no solution / -1: more than 1 solution
for m in range(1 + int(math.sqrt(lim / 2))):
m2 = m * m
for k in range(m + 1, 2 * m): # k = n + m
if gcd(m, k) == 1:
n = k - m
n2 = n * n
mk2 = 2 * m * k
for d in range(1, 1 + lim // mk2):
p = mk2 * d
c = (m2 + n2) * d
old_res = diff_c[p]
if old_res == 0:
diff_c[p] = c
elif old_res != c:
diff_c[p] = -1
return sum(1 for d in diff_c if d > 0)
def euler76_():
"""Solution for problem 76."""
pass
def euler77_():
"""Solution for problem 77."""
pass
def euler78_():
"""Solution for problem 78."""
pass
def euler79_():
"""Solution for problem 79."""
pass
def euler80_():
"""Solution for problem 80."""
pass
def euler81_():
"""Solution for problem 81."""
pass
def euler82_():
"""Solution for problem 82."""
pass
def euler83_():
"""Solution for problem 83."""
pass
def euler84_():
"""Solution for problem 84."""
pass
def euler85(target=2000000):
"""Solution for problem 85."""
# nb_rect(i, 1) = 1 (of size i) + 2 (of size i-1) + ... + i (of size 1) = i*(i+1)/2
# nb_rect(i, j) = nb_rect(i, 1) (of height j) + 2 * nb_rect(i, 1) (of height j-1) + ... + j * nb_rect(i, 1) (of height 1)
# = nb_rect(i, 1) * nb_rect(j, 1) = i*(i+1)*j*(j+1)/4
# Solutions to nb_rect(i, j) = t for a given i are, with delta = 1 + 16 t / [i * (i+1)]
# j1 = (-1 - sqrt(delta))/2 < 0 and j2 = (-1 + sqrt(delta))/2 > 0
# Best approximations are ceil and floor of j2.
sol_dist, sol_area = target, 0
for i in itertools.count(1):
tmp_i = i * (i + 1)
delta = 1 + target * 16 / tmp_i
j2 = (-1 + math.sqrt(delta)) / 2
js = set([math.floor(j2), math.ceil(j2)])
for j in js:
val = tmp_i * j * (j + 1) / 4
dist = math.fabs(target - val)
if dist < sol_dist:
sol_dist, sol_area = dist, i * j
if i > max(js):
return sol_area
def euler86_():
"""Solution for problem 86."""
pass
def euler87(lim=50000000):
"""Solution for problem 87."""
# the biggest prime needed p is such that
# lim >= p**2 + 2**3 + 2**4
# p <= sqrt(lim - 24)
primes = list(primes_up_to(int(math.sqrt(lim - 24))))
sol = set()
for a in primes:
sum_a = a ** 4
if sum_a > lim:
break
for b in primes:
sum_b = sum_a + b ** 3
if sum_b > lim:
break
for c in primes:
sum_c = sum_b + c ** 2
if sum_c > lim:
break
sol.add(sum_c)
return len(sol)
def euler88_():
"""Solution for problem 88."""
pass
def euler89_():
"""Solution for problem 89."""
pass
def euler90():
"""Solution for problem 90."""
base = 10
nb_face = 6
def substitute(n):
assert 0 <= n < base
return 6 if n == 9 else n
# The values we want to display with 9 replaced by 6
squares = [tuple(substitute(n) for n in divmod(i * i, base)) for i in range(1, base)]
# Possible dice arrangements with 9 replaced by 6 to make the logic easier
permutations = (set(substitute(n) for n in p)
for p in itertools.combinations(range(base), nb_face))
# Filtering arrangements as an optimisation
candidates = (p for p in permutations
if all((a in p or b in p) for a, b in squares))
# Counting the matching arrangements
return sum(1 for (c1, c2) in itertools.combinations(candidates, 2)
if all(((a in c1 and b in c2) or (b in c1 and a in c2)) for a, b in squares))
def euler91_bruteforce(size=50):
"""Solution for problem 91.
Bruteforce is fast enough to get a solution."""
i = 0
for p1, q1, p2, q2 in itertools.product(range(size + 1), repeat=4):
if (p1, p2) < (q1, q2): # not to count them twice
low, med, high = sorted(x ** 2 + y ** 2 for x, y in [(p1, p2), (q1, q2), (p1 - q1, p2 - q2)])
if low and low + med == high:
i += 1
return i
def euler91(size=50):
"""Solution for problem 91."""
# We have 2 kinds of right angles :
# - angle in 0 (hypothenus is PQ) : size*size of them
# - without any loss of generality, angle in P (p1, p2):
# OP is the hypothenus and with H (h1, h2) middle of OP, we have :
# |OP| = 2*|OH| = 2*|HP| = 2*|HQ|
# p1^2 + p2^2 = 4 * ((q1-h1)^2 + (q2-h2)^2)
# = (2*q1 - p1)^2 + (2*q2 - p2)^2
# - If Q is on axis : 2*size*size obvious solutions
# - If Q is not on axis, we solve for all p1,p2,q1.
# A few optimisations :
# - we can assume p1 >= p2 and count solution twice if relevant
# - when looking for q1, we can stop early (h1 + |OH|)
i = 3 * size * size
for p1, p2 in itertools.combinations_with_replacement(range(size + 1), 2):
dp = p1 ** 2 + p2 ** 2
if dp:
for q1 in range(1, min(size, int((p1 + math.sqrt(dp)) / 2)) + 1):
square = dp - (2 * q1 - p1) ** 2
if square >= 0:
root = int(math.sqrt(square))
if root * root == square:
assert p2 % 2 == root % 2
for s in set([-root, +root]):
q2 = (p2 + s) // 2
if 1 <= q2 <= size and (q1, q2) != (p1, p2):
i += 1 if p1 == p2 else 2
return i
def euler92_():
"""Solution for problem 92."""
pass