def main(max_ = 30000):
# 各素数に対応する点の次数を管理して、次数が5未満の点は切り捨てたほうが良いかも.
ans = []
primes = intlib.primes(max_)
adj = defaultdict(set)
for p in primes:
for q in (q for q in primes if q > p):
if is_pair(p, q):
adj[p].add(q)
adj[q].add(p)
adj = {p: s for p, s in adj.items() if len(adj[p]) > 4} #次数4以下の点を削除
adj = {p: {q for q in ls if q in adj} for p, ls in adj.items()} #枝を更新
print(len(adj), sum((len(s) for s in adj.values()))/len(adj))
keys = sorted(adj.keys())
ans = float('inf')
for p in keys:
if p > ans: break # pが暫定解を超えたら終了
adj_p = sorted([x for x in adj[p] if x > p])
for q in adj_p:
if len(adj[p] & adj[q]) < 3: continue #pとqの共通隣接点は3個以上必要
common_adj = sorted([x for x in adj[p] & adj[q] if x > q])
for r, s, t in itertools.combinations(common_adj, 3):
tpl = (p, q, r, s, t)
if is_clique(tpl, adj):
print(tpl, sum(tpl))
ans = min((ans, sum(tpl)))
if ans < max_: return ans #答えが確定する場合に返す
def main(max_=30000):
# 各素数に対応する点の次数を管理して、次数が5未満の点は切り捨てたほうが良いかも.
ans = []
primes = intlib.primes(max_)
adj = defaultdict(set)
for p in primes:
for q in (q for q in primes if q > p):
if is_pair(p, q):
adj[p].add(q)
adj[q].add(p)
adj = {p: s for p, s in adj.items() if len(adj[p]) > 4} #次数4以下の点を削除
adj = {p: {q for q in ls if q in adj} for p, ls in adj.items()} #枝を更新
print(len(adj), sum((len(s) for s in adj.values())) / len(adj))
keys = sorted(adj.keys())
ans = float('inf')
for p in keys:
if p > ans: break # pが暫定解を超えたら終了
adj_p = sorted([x for x in adj[p] if x > p])
for q in adj_p:
if len(adj[p] & adj[q]) < 3: continue #pとqの共通隣接点は3個以上必要
common_adj = sorted([x for x in adj[p] & adj[q] if x > q])
for r, s, t in itertools.combinations(common_adj, 3):
tpl = (p, q, r, s, t)
if is_clique(tpl, adj):
print(tpl, sum(tpl))
ans = min((ans, sum(tpl)))
if ans < max_: return ans #答えが確定する場合に返す
def main():
TARGET = 50000000
primes = intlib.primes(int(TARGET ** 0.5))
p_squares = [p ** 2 for p in primes if p ** 2 <= TARGET]
p_cubes = [p ** 3 for p in primes if p ** 3 <= TARGET]
p_fths = [p ** 4 for p in primes if p ** 4 <= TARGET]
ans = set()
for a in p_squares:
for b in p_cubes:
for c in p_fths:
if a + b + c <= TARGET:
ans.add(a + b + c)
return len(ans)
def pattern(n):
pre_table = [0] * (n + 1)
for x in range(0, n + 1, 2):
pre_table[x] = 1
primes = intlib.primes(n)
for k in range(1, len(primes)):
table = [0] * (n + 1)
for i in range(n + 1):
table[i] = sum(pre_table[i - j * primes[k]] for j in
range(0, i//primes[k] + 1))
pre_table = table
return pre_table[n]
def main():
_max = 10**4
odd_iter = (2*n + 1 for n in itertools.count(1))
primes = intlib.primes(_max)
primes_set = set(primes) #membership判定用set
for n in odd_iter:
if n > _max: return 'Not found.'
if n in primes_set: continue
flag = True
for p in primes:
if p > n: break
if intlib.is_square((n - p)//2):
flag = False
break
if flag:
return n
'''
The arithmetic sequence, 1487, 4817, 8147, in which each of
the terms increases by 3330, is unusual in two ways:
(i) each of the three terms are prime, and,
(ii) each of the 4-digit numbers are permutations of one another.
There are no arithmetic sequences made up of three 1-, 2-, or
3-digit primes, exhibiting this property, but there is one other
4-digit increasing sequence.
What 12-digit number do you form by concatenating the three
terms in this sequence?
'''
import intlib
import itertools
if __name__ == '__main__':
four_digit_primes = [p for p in intlib.primes(10000) if 1000 < p < 10000]
for a, b, c in itertools.combinations(four_digit_primes, 3):
if (c-b == b-a and sorted(str(a)) == sorted(str(b)) == sorted(str(c))):
print(a, b, c)
