Actual algorithm for Euler problem 37

Euler37Alg.py

from itertools import chain

def rtrunc_primes():
    """Generate a set of all right-truncatable primes"""
    rprimes = [[2, 3, 5, 7]]      # Seed with 1-digit primes
    rdigits = [1, 3, 7, 9]      # Only possible digits after first digit

    n = 2
    while rprimes[n-2]:         # Continue as long as we have found a prime in the previous n
        nprimes = [p for base in rprimes[n-2] for p in [10*base+d for d in rdigits ] if is_prime (p)]
        rprimes.append(nprimes)
        n += 1
    return {p for p in chain.from_iterable(rprimes)}

def ltrunc_primes():
    """Generate a set of all left-truncatable primes"""
    lprimes = [[2, 3, 5, 7]]      # Seed with 1-digit primes
    ldigits = range(1,10)      # Only possible digits after first digit

    n = 2
    factor = 1
    while lprimes[n-2]:         # Continue as long as we have found a prime in the previous n
        factor *= 10            # We need to stick the digits to the front, so mult with factor
        nprimes = [p for base in lprimes[n-2] for p in [base + l * factor for l in ldigits] if is_prime(p)]
        lprimes.append(nprimes)
        n += 1
    return {p for p in chain.from_iterable(lprimes)}


rtrunc_primes = rtrunc_primes()
ltrunc_primes = ltrunc_primes()

tprimes = sorted(list(rtrunc_primes & ltrunc_primes))
print "Truncatable primes: ", tprimes
print "Sum minus 1-digit primes: ", sum(tprimes[4:])
print "Number of truncatable primes: ", len(tprimes)

Guess list comprehensions could make these easier to read. Or shorter at least

Nprimes.append ( [ 10*base+p for p in rdigits if is_prime (base*10+r)])
or
Nprimes.append ( [ p for p in [10*base+d for d in rdigits ] if is_prime (p)])

Could this actually work?

nprimes = [p for base in rprimes[n-2] for p in [10*base+d for d in rdigits ] if is_prime (p)]