I decided to step up to Guff's challenge

[ncba93ivyase]

A solution to 25:

Code Select

x = []
y, z = 0, 2
while len(x) < 2:
x.append(1)
y += 1
while len(x) < 10:
x.append((x[y-2])+(x[y-1]))
y+=1
z+=1
if len(x) > 5:
del x[0]
y -= 1
if len(str(x[4])) > 999:
break
print z

[ncba93ivyase]

Last I checked, JMV only had 9 problems solved and I now have 10.

Code Select

x = []
y, z, i = 0, 2, 0
while len(x) < 2:
x.append(1)
y += 1
while len(x) < 6:
x.append((x[y-2])+(x[y-1]))
y+=1
z+=1
if x[y-1] %2 == 0: i+= x[y-1]
if len(x) > 5:
del x[0]
y -= 1
if x[4] > 4000000:
break
print iAnd there's my hideous solution to 2.

[guff]

hey those there fibonatchis numbers sure do show up a few times i reckon a feller such as yerself could write a function for them instead of resorting to while loops every cotton-pickin' time

also my solution for 24 is two lines long and takes ~0.06 seconds to run 
granted it can be solved by hand if you do the math but brute force is usually more fun

[Daddy]

Last I checked, JMV only had 9 problems solved and I now have 10.

Code Select

x = []
y, z, i = 0, 2, 0
while len(x) < 2:
x.append(1)
y += 1
while len(x) < 6:
x.append((x[y-2])+(x[y-1]))
y+=1
z+=1
if x[y-1] %2 == 0: i+= x[y-1]
if len(x) > 5:
del x[0]
y -= 1
if x[4] > 4000000:
break
print iAnd there's my hideous solution to 2.

Code Select

#!/usr/bin/python
# Filename : Euler_002.py
#Find the sum of all the even-valued terms in the sequence which do not exceed four million.
a,d,b = 0,0,1
while a <= 4000000:
c=a
a=b
b+=c
if a%2==0:
  d+= a
print d

[guff]

yeah i've got no idea what the hell lawlz is doing with his code

also god damn start using python3 already all of you 

[ncba93ivyase]

yeah i've got no idea what the hell lawlz is doing with his code

also god damn start using python3 already all of you 

i needed to create a list of fibonacci numbers for something and was like 'hey i can just recycle this'

although i see lots of problems with it and it really is terrible

And pygame doesn't support python 3.0, so I can't.

[guff]

i needed to create a list of fibonacci numbers for something and was like 'hey i can just recycle this'

although i see lots of problems with it and it really is terrible

And pygame doesn't support python 3.0, so I can't.

uh yeah that's what modules and functions are for 

uh yeah it is and it's extremely complicated and i'm still not sure what the hell is going on in it and why are there so many magic numbers in your code 

well then pygame SUCKS

[ncba93ivyase]

uh yeah that's what modules and functions are for 

uh yeah it is and it's extremely complicated and i'm still not sure what the hell is going on in it and why are there so many magic numbers in your code 

well then pygame SUCKS

for some reason, the crazier my code is the more i like it

short "simple" stuff wracks my brain for some reason, and really crazy methods of doing everything come smoother to me for some reason.

but i can really say this part of my code was completely pointless and stupid because i could've just started the list i wanted with the other thing with [1, 1] instead of this slop:

Code Select

while len(x) < 2:
  x.append(1)
  y += 1



My main problem in both writing and programming is I never review my work and I just go with what works the first time. I'm certain if I spent 30 more seconds doing things I could do everything better, but it's an impossible habit to break.

also pygame doesn't suck, it's just that it's updated like once every four score and seven years

and it's better than learning sepples

[ncba93ivyase]

a very, very slow solution to 5

Code Select

i = 0
while True:
if i != 0 and i % 11 == 0 and i % 13 == 0 and i % 14 == 0 and i % 15 == 0 and i % 16 == 0 and i % 17 == 0 and i % 18 == 0 and i % 19 == 0:
print i
break
i += 20

I'm certain guff has a better way of doing it, because this takes 5.63 seconds to run :'(

[guff]

I'm certain guff has a better way of doing it, because this takes 5.63 seconds to run :'(

uh well yeah look up euclid's algorithm
it's an easy peasy and pretty quick way to compute the gcd of two numbers, from which the lcm is easily to computed
then, to compute the gcd or lcm of a list, since both functions are associative, all you have to do is call reduce() with the function and that list (in this case range(1,21)) as arguments

and although i generally avoid recursion if you're too lazy to actually read the article the gcd function can be implemented as a one-liner:

Code Select

def gcd(a,b): return a if b==0 else gcd(b,a%b)
from there, you can write the lcm function as:

Code Select

def lcm(a,b): return a*b//gcd(a,b)
the rest is left as an exercise to the stupid, incompetent reader 

[ncba93ivyase]

Problem 55:

Code Select

import time
total = 0
for starter in range(0, 10000):
x = starter + int(str(starter)[::-1])
if x != int(str(x)[::-1]):
for i in range(1, 49):
x += int(str(x)[::-1])
if x == int(str(x)[::-1]):
break
if i == 48 and x != int(str(x)[::-1]):
total += 1
print "Answer is %d. Solved in %.3f" %(total, time.clock())
.18 seconds to solve, so not too bad.

Not even 6000 people solved it yet, making me feel even better about it.

[guff]

i already had 55 solved so i decided to go back and try and make my solution as short and obfuscated as possible a la lawlz except i did it on purpose:

Code Select

lychrel = lambda n,i=1: False if (n+int(str(n)[::-1]))==int(str(n+int(str(n)[::-1]))[::-1]) else True if i >= 50 else lychrel(n+int(str(n)[::-1]), i+1)
print([lychrel(i) for i in range(10000)].count(True))
takes about 0.46 seconds 

the solution i'm keeping for my records though is far more verbose and actually makes use of function definitions
also i could've done this in one line if python supported anonymous recursion but i don't think it does but i did learn that if you bind a name to a lambda you can recurse

[ncba93ivyase]

takes about 0.46 seconds 

so how fast is your real solution or is mine better

[guff]

so how fast is your real solution or is mine better

~0.26 (~0.36 if i use the recursive lyrchel function that's commented out instead which i think is prettier) so yours is slightly faster but god damnit i actually use functions so i've got overhead from that:

Code Select

def reverse_digits(n): return int(str(n)[::-1])

def is_palindrome(n): return n == reverse_digits(n)

#def lychrel(n, i=1):
# return False if is_palindrome(n + reverse_digits(n)) else True if i >= 50 else lychrel(n + reverse_digits(n), i+1)

def lychrel(n):
for i in range(1, 50):
n = n + reverse_digits(n)
if is_palindrome(n): return False
return True

if __name__ == '__main__':
from time import time

start = time()
print(len([True for i in range(10000) if lychrel(i)]), time() - start)




also open up an interpreter session and run:

Code Select

from test import pystone
pystone.full()
for comparison
my machine gets 49224.4 pystones per second

[ncba93ivyase]

~0.26 (~0.36 if i use the recursive lyrchel function that's commented out instead which i think is prettier) so yours is slightly faster but god damnit i actually use functions so i've got overhead from that:

Code Select

def reverse_digits(n): return int(str(n)[::-1])

def is_palindrome(n): return n == reverse_digits(n)

#def lychrel(n, i=1):
# return False if is_palindrome(n + reverse_digits(n)) else True if i >= 50 else lychrel(n + reverse_digits(n), i+1)

def lychrel(n):
for i in range(1, 50):
n = n + reverse_digits(n)
if is_palindrome(n): return False
return True

if __name__ == '__main__':
from time import time

start = time()
print(len([True for i in range(10000) if lychrel(i)]), time() - start)




also open up an interpreter session and run:

Code Select

from test import pystone
pystone.full()
for comparison
my machine gets 49224.4 pystones per second

>>> from test import pystone
>>> pystone.full()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'module' object has no attribute 'full'



also, functions don't really matter if i'm really only doing one little thing like this, and speed is more important

although i knew if i didn't make up some function to do something, you'd get grumpy but i was just like 'lol'