def find(digit_set):
s = set()
for p in permut(digit_set):
for c in couple(ops, length=3, optimized=False):
l = list(p)
# add ops
for i,cc in enumerate(c):
l.insert(2*i+1, cc)
# for the first parentheses
for i in range(int(len(l)/2)):
for j in range(i+1, int(len(l)/2)+2):
# for the second parentheses
for ii in range(i):
for jj in range(i+1, int(len(l)/2)+2):
# make a copy
ll = l[:]
# add first parentheses
ll.insert(2*i, '(')
ll.insert(2*j, ')')
# add second parentheses
# NB: can be optimized but i don't know how... :/
ll.insert(2*ii, '(')
ll.insert(2*jj, ')')
#raw_input(ll)
try:
v = eval(' '.join(ll))
if v > 0 and int(v) == v:
s.add(int(v))
#print ' '.join(ll), eval(' '.join(ll))
except:
pass
return s
def brute_force():
is_cube = lambda x: x == math.pow(round(math.pow(x,1./3)),3)
n = 5
cube_roots = []
for i in xrange(342,10000):
if i in cube_roots:
print "already checked", i
continue
l = []
cube = i**3
print "\n%d" % i
for p in permut(list(str(cube))):
p = int(p)
if p < 41063625: continue
if is_cube(p) and p not in l:
l.append(p)
cube_roots.append(round(p**(1./3)))
print p,
if len(l) == n:
return min(l)
return "[result]"
# math considerations
#
# 1+2+3+4+5+6+7+8+9 = 45 which is divisible by 3 so n <> 9
# 1+2+3+4+5+6+7+8 = 36 so n <> 8
# 1+2+3+4+5+6 = 21 so n <> 6
# 1+2+3+4+5 = 15 so n <> 5
# 1+2+3 = 6 so n <> 3
# 1+2 = 3 so n <> 2
import time
from permut import *
from prime import *
def is_pandigital(x):
return "".join(sorted(str(x))) == "123456789"
t = time.time()
m = 0
for n in [7, 4]:
for p in permut(range(n, 0, -1)):
if is_prime(int(p)):
m = max(m, p)
if m:
break
print m
print "time: %f s" % (time.time() - t)
# primes.append(int(p))
# if len(primes) == 12:
# print "eleven found! break for p =", p
# break
def fit(p):
p = str(p)
return all(is_prime(int(p[:-i])) and is_prime(int(p[i:])) for i in range(1,len(p)))
# test
#print len([c for c in couple(range(1,5), 4)])
primes,r = set(), [2]+range(1,10,2)
for n in range(2,10):
for c in couple(r, n):
for p in permut(c):
if p[0] in ['1','9'] or p[-1] in ['1','9']: continue
if is_prime(int(p)) and fit(int(p)):
primes.add(int(p))
print p, c, len(primes)
if len(primes) == 11:
break
if len(primes) == 11:
break
if len(primes) == 11:
print "11 found !!!"
break
print "sum:", sum(primes)
print primes
print "time: %f s" % (time.time() - t)
#
# 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 time
from collections import defaultdict
from prime import *
from permut import *
t = time.time()
for c in couple(range(10), length=4):
primes = set([int(p) for p in permut(c) if len(str(int(p))) == 4 and is_prime(int(p))])
if len(primes) <= 2:
continue
l = sorted(list(primes))
dist = list( ((p1,p2),p2-p1) for p1 in l for p2 in l if p2>p1 )
dist.sort(key=lambda x:x[0][0])
seqs = defaultdict(list)
for p1p2,d in dist:
if not seqs[d] or any(p1p2[0]==pp1pp2[1] for pp1pp2 in seqs[d]):
seqs[d].append(p1p2)
for k,v in seqs.items():
if len(v) <= 1:
del seqs[k]
v.sort(key=lambda x: x[0])
if seqs:
# Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:
#
# d2d3d4=406 is divisible by 2
# d3d4d5=063 is divisible by 3
# d4d5d6=635 is divisible by 5
# d5d6d7=357 is divisible by 7
# d6d7d8=572 is divisible by 11
# d7d8d9=728 is divisible by 13
# d8d9d10=289 is divisible by 17
# Find the sum of all 0 to 9 pandigital numbers with this property.
from permut import *
import time
def is_pandigital(x):
return "".join(sorted(str(x))) == "0123456789"
t = time.time()
s, r = 0, [2, 3, 5, 7, 11, 13, 17]
for p in permut(range(0, 10)):
if len(int(p)) != 10:
continue
if all(int(p[i + 1 : i + 1 + 3]) % d == 0 for i, d in enumerate(r)):
s += int(p)
print s
print "time: %f s" % (time.time() - t)
