def solve():
sieve = Sieve()
sieve.compute_upto(100000) # prime the sieve
# we're starting with 23 to make life easy
found_soln = False
primes = (i for i in count(23) if sieve.is_prime(i))
digits = ''.join(str(d) for d in range(0,10)) # 0-9
for p in primes:
p_str = str(p)
for i in xrange(1,len(p_str)):
# combinations of indices into p_str of length i where every value index in
# p_str has the same character
idx_combs = (j for j in combinations(xrange(0,len(p_str)), i)
if all(imap(lambda x: p_str[x] == p_str[j[0]], j)) and len(j) > 0)
for comb in idx_combs:
group_members = set([])
# we've got all the possible index combinations, so let's try swapping them
for new_char in digits:
test_str = set_val_at_idxs(p_str, comb, new_char)
val = int(test_str)
if not test_str.startswith('0') and sieve.is_prime(val):
group_members.add(val)
if len(group_members) >= 7:
print sorted(group_members)
if len(group_members) == 8:
found_soln = True
break;
if found_soln:
break
if found_soln:
break
def problem46():
s = Sieve()
odd_composites = (i for i in counter(2) if not s.is_prime(i) and i % 2 == 1)
for c in odd_composites:
print "c = %d" % c
two_times_squares = (2 * j**2 for j in counter(1))
for tts in two_times_squares:
if s.is_prime(c - tts):
break
elif tts > c:
# this should mean that we couldn't find a sum of a prime and twice a square
print "{c = %d, tts = %d, c-tts = %d}" % (c, tts, c-tts)
return c
def problem49():
s = Sieve()
four_digit_primes = (i for i in xrange(1000, 9999+1) if s.is_prime(i))
grouped_primes = {}
for i in four_digit_primes:
# the key into the map will be a sorted representation of
# the digits in 'i'
key = ''.join(sorted(str(i)))
if not key in grouped_primes:
# create a list at that key
grouped_primes[key] = []
grouped_primes[key].append(i)
prime_groups_bigger_than_three = (grouped_primes[k] for k in grouped_primes if len(grouped_primes[k]) >= 3)
for grp in prime_groups_bigger_than_three:
diff_counts = {}
pairwise_diffs = [(grp[i],grp[j],grp[j]-grp[i]) for i in xrange(0,len(grp)) for j in xrange(i+1,len(grp))]
for p1,p2,diff in pairwise_diffs:
if diff not in diff_counts:
diff_counts[diff] = 0
diff_counts[diff]+=1
solution_diffs = (k for k in diff_counts if diff_counts[k] == 2)
for s in solution_diffs:
# there should only be two of these! (one should be the example)
pairs = filter(lambda x: x[2] == s, pairwise_diffs)
if pairs[0][1] == pairs[1][0] and pairs[0][0] != 1487:
return ''.join(map(str, [pairs[0][0], pairs[0][1], pairs[1][1]]))
return None
#!/usr/bin/env python from Sieve import Sieve upto = 2000000 s = Sieve(upto=2000000) primes_to_2_mil = (i for i in xrange(2,upto) if s.is_prime(i)) print reduce(lambda x,y: x + y, primes_to_2_mil)
