I've started working on some Project Euler problems, and have solved number 4 with a simple brute force solution:
def mprods(a,b):
c = range(a,b)
f = []
for d in c:
for e in c:
f.append(d*e)
return f
max([z for z in mprods(100,1000) if str(z)==(''.join([str(z)[-i] for i in range(1,len(str(z))+1)]))])
After solving, I tried to make it as compact as possible, and came up with that horrible bottom line!
Not to leave something half-done, I am trying to condense the mprods function into a list comprehension. So far, I've come up with these attempts:
[d*e for d,e in (range(a,b), range(a,b))]
Obviously completely on the wrong track. :-)[d*e for x in [e for e in range(1,5)] for d in range(1,5)]
This gives me [4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16], where I expect
[1, 2, 3, 4, 2, 4, 6, 8, 3, 6, 9, 12, 4, 8, 12, 16] or similar.
Any Pythonistas out there that can help? :)
c = range(a, b)
print [d * e for d in c for e in c]
from itertools import product
def palindrome(i):
return str(i) == str(i)[::-1]
x = xrange(900,1000)
max(a*b for (a,b) in (product(x,x)) if palindrome(a*b))
xrange(900,1000) is like range(900,1000) but instead of returning a list it returns an object that generates the numbers in the range on demand. For looping, this is slightly faster than range() and more memory efficient.
product(xrange(900,1000),xrange(900,1000)) gives the Cartesian product of the input iterables. It is equivalent to nested for-loops. For example, product(A, B) returns the same as: ((x,y) for x in A for y in B). The leftmost iterators are in the outermost for-loop, so the output tuples cycle in a manner similar to an odometer (with the rightmost element changing on every iteration).
product('ab', range(3)) --> ('a',0) ('a',1) ('a',2) ('b',0) ('b',1) ('b',2)
product((0,1), (0,1), (0,1)) --> (0,0,0) (0,0,1) (0,1,0) (0,1,1) (1,0,0) ...
str(i)[::-1] is list slicing shorthand to reverse a list.
Note how everything is wrapped in a generator expression, a high performance, memory efficient generalization of list comprehensions and generators.
Also note that the largest palindrome made from the product of two 2-digit numbers is made from the numbers 91 99, two numbers in the range(90,100). Extrapolating to 3-digit numbers you can use range(900,1000).
I think you'll like this one-liner (formatted for readability):
max(z for z in (d*e
for d in xrange(100, 1000)
for e in xrange(100, 1000))
if str(z) == str(z)[::-1])
Or slightly changed:
c = range(100, 1000)
max(z for z in (d*e for d in c for e in c) if str(z) == str(z)[::-1])
Wonder how many parens that would be in Lisp...
