def findpancat(pan):
rval = 0
ipan = array2int(pan)
# We only need to search for something < 5 digits
for ii in range(1, 6):
start = array2int(pan[0:ii])
cprod = makecatprod(start, 9)
if cprod == ipan:
rval = 1
return rval
def findlargestpandigital():
for n in range(9, 1, -1):
pan = makepandigital(n, True)
for p in pan:
pint = array2int(p)
# print "Testing: " + repr(pint)
if isprime(pint):
print "Found the largest pandigital prime: " + repr(pint)
return pint
def replaceprime(prime):
findall = lambda my_list: [i for i, x in enumerate(my_list) if x == '1']
ndig = len(int2array(prime))
pgroup = []
replacement_dig = make_palindrome_half(ndig-1)
# Ignore the last replacement rule, it is '0000'
for reprule in replacement_dig[:-1]:
# Run replacement for all integers
subgroup = []
newprime = int2array(prime)
for cint in range(10):
for ind in findall(reprule):
newprime[ind] = cint
# Check that the length is the same
if len(newprime) == len(int2array(array2int(newprime))):
subgroup.append(array2int(newprime))
pgroup.append(subgroup)
return pgroup
# print "List: " + repr(nums)
# print "Current sum: " + repr(cvalue)
# print "Additional terms: " + repr(naddperms)
# print "Target term: " + repr(lt)
num = 0
while num <= nleft:
# print repr(num) + " " + repr(nums[num-1]) +
# " " + repr(cvalue + (num)*naddperms)
if cvalue + (num)*naddperms == lt:
nt = nums.pop(num-1)
mlt.append(nt)
nums.reverse()
mlt.extend(nums)
nums = []
nleft = 0
break
elif cvalue + (num)*naddperms > lt:
cvalue += (num-1)*naddperms
nt = nums.pop(num-1)
mlt.append(nt)
break
elif num == nleft:
nt = nums.pop(num)
mlt.append(nt)
break
num += 1
# print mlt
nleft = len(nums)
print repr(array2int(mlt)) + " is the " + repr(lt) +
"th lexographic permutation of the set"
else:
if len(num) == maxlen:
rval = int(num)
return rval
# Takes in a pandigital number and determines if it can
# be formed by concatenating the product of an integer
# We do this by taking the first n digits, and multiplying
# them by the sequence (1,2,3,4,...) until we have a
# 9 digit number, then we can compare that
def findpancat(pan):
rval = 0
ipan = array2int(pan)
# We only need to search for something < 5 digits
for ii in range(1, 6):
start = array2int(pan[0:ii])
cprod = makecatprod(start, 9)
if cprod == ipan:
rval = 1
return rval
if __name__ == "__main__":
plist = makepandigital(9, True)
for pan in plist:
findpancat(pan)
if findpancat(pan):
print ("The largest pandigital that can be formed " +
"as a concatenated product is: %d" % (array2int(pan)))
break
def truncateRL(num):
arr = int2array(num)
arr.pop()
tnum = array2int(arr)
return tnum
# The first is; 1487, 4817, 8147 (difference is 3330)
from eulermath import (isprime, int2array, array2int,
find_primes_in_list, find_arithemetric_series)
from itertools import permutations, combinations
if __name__ == "__main__":
# Start by finding all 4-digit primes
print "Finding arithemetric series of primes between 1001 and 9999:"
exclude = []
for trial in xrange(1001, 9999, 2):
if trial not in exclude:
if isprime(trial):
# Find the permutations of this number
tp = permutations(int2array(trial))
# Get the unique values and sort them
tpl = list(set([array2int(x) for x in tp]))
tpp = find_primes_in_list(tpl)
tpn = []
for x in tpp:
if x > 1000:
tpn.append(x)
# Exclude the others
for p in tpn:
exclude.append(p)
# Find any arithemetric series of length 3 in the prime list
aseries = find_arithemetric_series(tpn, 3)
if len(aseries) > 0:
print aseries
for series in aseries:
print series
cat = ''.join(str(elem) for elem in series)
p12 = []
p23 = []
p1 = []
p2 = []
p3 = []
fnames = open('Euler79.dat')
for line in fnames:
val = int2array(int(line))
# All values
sub.append(val)
# Single values
p1.append(val[0])
p2.append(val[1])
p3.append(val[2])
# Pairs
p12.append(array2int(val[0:2]))
p23.append(array2int(val[1:3]))
p123.append(array2int(val))
# Flatten out the sublist and take all unique elements
fsl = list(set([x for arr in sub for x in arr]))
# this length is the shortest password size
v1, c1 = hist(p1)
v2, c2 = hist(p2)
v3, c3 = hist(p3)
v12, c12 = hist(p12)
v23, c23 = hist(p23)
v123, c123 = hist(p123)
print "We need to use all these:"
