This is my solution for the 5th problem in Project Euler. Is there any way to improve the while loop conditional instead of using the sum?
table = range(1,21)
result = 1
pf = 2
while sum(table) > len(table):
flag = False
for x,y in enumerate(table):
if y % pf == 0:
table[x] = y/pf
flag = True
if flag:
result *= pf
else:
pf += 1
print result
It's always clearer to write a program using a higher-order function instead of a loop because all of the extraneous variables that control the operation of the loop disappear. Here's a program that performs the same computation as yours, but the while and for loops disappear, as do the variables table, result, pf, and flag; variables x and y remain, but limited to a support role in an auxiliary function rather than as part of the main calculation:
>>> from fractions import gcd
>>> def lcm(x,y): return x*y/gcd(x,y)
...
>>> print reduce(lcm,range(1,21))
232792560
Use a constant. It's easier to understand when you want to go back and try a different value too.
MAX_DIVISOR = 20
table = range(1,MAX_DIVISOR + 1)
result = 1
pf = 2
while pf < MAX_DIVISOR + 1:
flag = False
for x,y in enumerate(table):
if y % pf == 0:
table[x] = y/pf
flag = True
if flag:
result *= pf
else:
pf += 1
print result
You could use any() to test the table, like so:
while any(n != 1 for n in table):
# do stuff
I think this is clearer than sum(table) > len(table).
Also, as @JoeClacks recommended, use a constant.
Complete revised solution:
MAX_DIVISOR = 20
table = list(range(1, MAX_DIVISOR+1))
result = 1
pf = 2
while any(n != 1 for n in table):
assert pf <= MAX_DIVISOR
flag = False
for i,n in enumerate(table):
if n % pf == 0:
table[i] = n//pf
flag = True
if flag:
result *= pf
else:
pf += 1
print(result)
I added an assert to make sure that pf only has legal values; this isn't needed as long as there is no bug in the code, but it can be a good idea to make code sanity-check itself.
I used i and n for the index and the number, rather than x and y.
I used Python's integer division operator // rather than /, so the code will work the same on Python 3.x as on Python 2.x. Also, the way I wrote the print statement it will work equally well in Python 3.x and Python 2.x.
I changed the indentation to steps of 4 spaces in accordance with PEP 8.
http://www.python.org/dev/peps/pep-0008/
Note: I like this algorithm for solving this problem. It's elegant. Did you invent this, get it from a book, or what?
EDIT: Actually, Project Euler problem 5 has been discussed before here on StackOverflow. Here is an answer that I just compared to the above answer. This one is almost exactly ten times faster than the one above. It's a bit trickier though!
