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General => Internet, Science, & Technology => Topic started by: ncba93ivyase on January 19, 2009, 11:18:13 PM

Title: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 19, 2009, 11:18:13 PM
And managed to solve the easiest Project Euler problem. badass

>>> x = 0
>>> for i in range(1, 1000):
...     if i % 3 == 0 or i % 5 == 0:
...             x+=i
...
>>> x
233168


I noticed wikipedia had another solution and compared my answer to theirs, and yes, it's right.

I SAY WE SHOULD ALL WORK TOWARDS SOLVING ALL OF THESE.

Also, I've been trying to think of a simple and efficient way to solve problem 5, and I keep getting answers that I know are wrong. :'(

and the website in case you're too stupid to google it: http://projecteuler.net/index.php?section=problems
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 20, 2009, 07:06:39 AM
better than jmv could do badass

also my naive solution for number one is sum([i for i in range(1000) if 0 in (i%3, i%5)]) which is much cooler than yours
the mathy way to do it would be to make use of the fact that the sum of the first n natural numbers is [tex]\frac{n(n+1)}{2}[/tex] so the sum of the multiples of 3 that are less than 1000 would be [tex]3*\frac{n(n+1)}{2}[/tex] where n is [tex]\lfloor \frac{1000}{3} \rfloor = 333[/tex]
so we do the same for five, and those two sums, then subtract all the multiples of 15
that's less work for the computer but also lots more work for the coder so yeah i'll just stick with the first one  baddood;

also my profile (http://projecteuler.net/index.php?section=profile&profile=Guff)
i haven't solved any new ones for a bit but for the last couple days i've been working my way through them again and actually keeping all my files and solutions organized this time for future reference
Title: Re: I decided to step up to Guff's challenge
Post by: Bolivian Army on January 20, 2009, 07:43:54 AM
guff is nerding up this place too i see  baddood;

i'll do one when i get home
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 20, 2009, 09:40:07 AM
okay i've redone the first ten now and have decent (i.e. less than two seconds to run) solutions for each

also number five is easy peasy if you use euclid's algorithm bassir;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 20, 2009, 10:50:11 PM
Quote from: andré the giant on January 20, 2009, 09:40:07 AM
okay i've redone the first ten now and have decent (i.e. less than two seconds to run) solutions for each

also number five is easy peasy if you use euclid's algorithm bassir;
ahhhhh i hate you :'(

I MUST TRY AGAIN
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 21, 2009, 04:13:14 AM
I think I've come up with a good solution for number 20. Here's what I typed up, but it's not going to give you the actual answer (of course):
Code Select
x = str(25373)
g = 0
n = 0
for items in x:
q = int(x[g:g+1])
n+=q
g+=1
print n
I tested it and it works perfectly fine. Didn't include the factorial part because if you can't do that (JMV baddood; ), you shouldn't even bother using a computer.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 21, 2009, 06:05:34 AM
Quote from: Pancake Persona on January 21, 2009, 04:13:14 AM
Code Select

q = int(x[g:g+1])

wat goonish

for strings, blah[foo : foo + 1] returns exactly the same thing that blah[foo] does

also you should probably learn how to use list comprehensions because then you could do:
Code Select
def sum_of_digits(numbar):
    return sum([int(digit) for digit in str(numbar)])

or at the very least slim down your for loop a bit by eliminating g:
Quote from: Pancake Persona on January 21, 2009, 04:13:14 AM
Code Select
x = str(25373)
total = 0
for digit in x:
total += int(digit)
print total


problem 20 is easy for languages with built in bignum support (e.g. python, ruby, and every other language that doesn't SUCK  madood;) but 100! is a pretty big number and it would definitely be a bit harder to do this in c

Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 21, 2009, 12:57:22 PM
Quote from: andré the giant on January 21, 2009, 06:05:34 AM
wat goonish

for strings, blah[foo : foo + 1] returns exactly the same thing that blah[foo] does
\

or at the very least slim down your for loop a bit by eliminating g:
what matters most is i got it done, albeit through unconventional methods

I just wonder if JMV could actually do any of them
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 21, 2009, 01:01:53 PM
Quote from: Pancake Persona on January 21, 2009, 12:57:22 PM
what matters most is i got it done, albeit through unconventional methods

I just wonder if JMV could actually do any of them
well yeah but given that you've been coding games you should probably have better style  baddood;

he did the first one but that was with a lot of help  akudood;
maybe he could do it in java n_u
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 21, 2009, 01:06:25 PM
Quote from: andré the giant on January 21, 2009, 01:01:53 PM
well yeah but given that you've been coding games you should probably have better style  baddood;

he did the first one but that was with a lot of help  akudood;
maybe he could do it in java n_u
games are just heaps and mounds of if statements and not much else, so they really don't help me learn and comprehend everything that i should know by now which i'll admit is shameful goonish

for example, a randomly selected chunk from my most recent project:
Code Select
if item.falling == False:
item.y += move_y
#If you reach 28 pixels above the starting point, fall
if item.y <item.y_init-30:
item.falling = True
#If you walk off the edge of a platform, fall
if item.y == item.y_init and my_color[0] > 0:
item.falling = True
#If falling... fall. If touching the ground, bring back ability to jump
if item.falling == True:
item.y +=1
if item.y >200:
item.y = 202
item.falling = False
item.y_init = item.y
#If you're on a platform, cease falling
if my_color[0] == 0 and my_color[2] == 0:
item.falling = False
item.y_init = item.y
while True:
for event in pygame.event.get():
if event.type == QUIT:
exit()
if event.type == KEYDOWN:
if event.key == K_w:
move_y = -2
# X controls are inverted because the level moves, not the character.
if event.key == K_d:
move_x = -1
if event.key == K_a:
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 21, 2009, 01:32:07 PM
Quote from: Pancake Persona on January 21, 2009, 01:06:25 PM
games are just heaps and mounds of if statements and not much else, so they really don't help me learn and comprehend everything that i should know by now which i'll admit is shameful goonish
okay so read the official docs  baddood;

Quote from: Pancake Persona on January 21, 2009, 01:06:25 PM

for example, a randomly selected chunk from my most recent project:[spoiler]
Code Select
if item.falling == False:
item.y += move_y
#If you reach 28 pixels above the starting point, fall
if item.y <item.y_init-30:
item.falling = True
#If you walk off the edge of a platform, fall
if item.y == item.y_init and my_color[0] > 0:
item.falling = True
#If falling... fall. If touching the ground, bring back ability to jump
if item.falling == True:
item.y +=1
if item.y >200:
item.y = 202
item.falling = False
item.y_init = item.y
#If you're on a platform, cease falling
if my_color[0] == 0 and my_color[2] == 0:
item.falling = False
item.y_init = item.y
while True:
for event in pygame.event.get():
if event.type == QUIT:
exit()
if event.type == KEYDOWN:
if event.key == K_w:
move_y = -2
# X controls are inverted because the level moves, not the character.
if event.key == K_d:
move_x = -1
if event.key == K_a:
[/spoiler]
okay well that mostly seems okay akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: FAMY2 on January 21, 2009, 03:41:09 PM
More please.    :3
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 21, 2009, 04:44:14 PM
Quote from: FAMY2 on January 21, 2009, 03:41:09 PM
More please.    :3
wat
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 22, 2009, 06:55:36 AM
okay how many you got now  baddood;

i haven't done any new ones yet but i did just redo number 12
my solution takes about 3.5 seconds to run though  saddood;
Title: Re: I decided to step up to Guff's challenge
Post by: FAMY2 on January 22, 2009, 07:41:37 AM
Quote from: andré the giant on January 21, 2009, 04:44:14 PM
wat

I like seeing how you guys discuss this stuff.  saddood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 22, 2009, 08:11:04 AM
Quote from: FAMY2 on January 22, 2009, 07:41:37 AM
I like seeing how you guys discuss this stuff.  saddood;
Learn Python <3
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 22, 2009, 08:17:30 AM
[spoiler]help> keywords

Here is a list of the Python keywords.  Enter any keyword to get more help.

and                 elif                if                  print
as                  else                import              raise
assert              except              in                  return
break               exec                is                  try
class               finally             lambda              while
continue            for                 not                 with
def                 from                or                  yield
del                 global              pass               

help> topics

Here is a list of available topics.  Enter any topic name to get more help.

ASSERTION           DEBUGGING           LITERALS            SEQUENCEMETHODS2
ASSIGNMENT          DELETION            LOOPING             SEQUENCES
ATTRIBUTEMETHODS    DICTIONARIES        MAPPINGMETHODS      SHIFTING
ATTRIBUTES          DICTIONARYLITERALS  MAPPINGS            SLICINGS
AUGMENTEDASSIGNMENT DYNAMICFEATURES     METHODS             SPECIALATTRIBUTES
BACKQUOTES          ELLIPSIS            MODULES             SPECIALIDENTIFIERS
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BITWISE             EXPRESSIONS         NUMBERMETHODS       STRINGS
BOOLEAN             FILES               NUMBERS             SUBSCRIPTS
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CALLS               FORMATTING          OPERATORS           TRUTHVALUE
CLASSES             FRAMEOBJECTS        PACKAGES            TUPLELITERALS
CODEOBJECTS         FRAMES              POWER               TUPLES
COERCIONS           FUNCTIONS           PRECEDENCE          TYPEOBJECTS
COMPARISON          IDENTIFIERS         PRINTING            TYPES
COMPLEX             IMPORTING           PRIVATENAMES        UNARY
CONDITIONAL         INTEGER             RETURNING           UNICODE
CONTEXTMANAGERS     LISTLITERALS        SCOPING             
CONVERSIONS         LISTS               SEQUENCEMETHODS1   

help> modules

Please wait a moment while I gather a list of all available modules...

Audio_mac           _hotshot            fnmatch             pydoc_topics
BaseHTTPServer      _json               formatter           pyexpat
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CGIHTTPServer       _lsprof             fractions           random
Canvas              _multibytecodec     ftplib              re
Carbon              _multiprocessing    functools           repr
CodeWarrior         _random             future_builtins     resource
ColorPicker         _sha256             gc                  rexec
ConfigParser        _sha512             genericpath         rfc822
Cookie              _socket             gensuitemodule      rlcompleter
Dialog              _sqlite3            gestalt             robotparser
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_functools          filecmp             py_compile         
_hashlib            fileinput           pyclbr             
_heapq              findertools         pydoc               

Enter any module name to get more help.  Or, type "modules spam" to search
for modules whose descriptions contain the word "spam".[/spoiler]

Uhm, so I have a question for the guy in red.  Why won't idle open a new window :(
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 22, 2009, 08:44:05 AM
multi3 = range(3, 1000, 3)
multi5 = range(5, 1000, 5)
multi3and5 = range(15, 1000, 15)
sum(multi3) +sum(multi5) - sum(multi3and5)
oh yay.
wait a minute, i can reduce that to one line why am i using variables.  baddood;

Code Select
sum(range(3, 1000, 3)) +sum(range(5, 1000, 5)) - sum(range(15, 1000, 15))


the first time I did that in december I didn't know about the sum function :(


My original solution for the problem.
[spoiler]a = 0
b = 0
guff = 0
for i in range(3, 1000, 3):
    a = a+ i
for i in range(5, 1000, 5):
    b = b+i
for i in range(15, 1000, 15):
    guff = guff+i;
print a+b-guff[/spoiler]


Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 22, 2009, 10:01:26 AM
Does anyone know why
Code Select
a = 0
b = 1
while a <= 100:
c=a
a=b
b=c+b
print b
Doesn't run directly in idle? It was pissing me off because I was all "wtf man, that code is correct" but it wouldn't run. I then saved it and did python ./Euler_2.py and it gave me.
QuoteValenteMac:Euler jamesvalente$ python ./Euler_2.py
1
2
3
5
8
13
21
34
55
89
144
233

Which isn't what the problem is asking for, nor what I coded. I just needed a working sequence first.
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 22, 2009, 10:08:08 AM
And there we go.  hocuspocus;
Code Select
#!/usr/bin/python
# Filename : Euler_2.py
#Find the sum of all the even-valued terms in the sequence which do not exceed four million.
a = 0
b = 1
d = 0
while a <= 4000000:
c=a
a=b
b+=c
if a%2==0:
  d+= a
print d

Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 22, 2009, 12:49:12 PM
Quote from: andré the giant on January 22, 2009, 06:55:36 AM
okay how many you got now  baddood;

i haven't done any new ones yet but i did just redo number 12
my solution takes about 3.5 seconds to run though  saddood;
no, i just reworked my solutions to the previous ones to run in ruby

But I'm going to try a couple after I wake up tonight.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 22, 2009, 12:58:13 PM
Quote from: Pancake Persona on January 22, 2009, 12:49:12 PM
no, i just reworked my solutions to the previous ones to run in ruby

But I'm going to try a couple after I wake up tonight.
i was into ruby for a while but then i gave up on it because it didn't have list comprehensions (yeah sure there are ways to do the same thing in ruby but they take a little more effort and more typing and aren't as pretty >:o)  akudood;

but seriously learn about list comprehensions and generator expressions and iterators and blah and blah etc. baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 22, 2009, 04:02:00 PM
also i rewrote my solution to problem 2:
Code Select
def fib(n):
fib1 = 0
fib2 = 1
for i in range(n - 1):
fib2, fib1 = fib2 + fib1, fib2
return fib2 if n > 0 else 0

import itertools
print(sum(itertools.takewhile(lambda x: x < 4e6, (fib(n) for n in itertools.count() if n % 3 == 0))))

fuck yes generator expressions  akudood;

also i'm using python3

also this guy (http://krenzel.info/?p=85) has a crazy fast implementation of a fibonacci function
i did a test of fib(2**n) for n from 0 to 19 using the one i wrote above and his version:
[spoiler]
Code Select
0: fib1 time: 1.09673e-05, fib2time: 2.14577e-06; ratio: 5.11111
1: fib1 time: 5.96046e-06, fib2time: 1.78814e-05; ratio: 0.333333
2: fib1 time: 5.96046e-06, fib2time: 9.05991e-06; ratio: 0.657895
3: fib1 time: 2.19345e-05, fib2time: 1.00136e-05; ratio: 2.19048
4: fib1 time: 8.10623e-06, fib2time: 2.7895e-05; ratio: 0.290598
5: fib1 time: 1.00136e-05, fib2time: 2.90871e-05; ratio: 0.344262
6: fib1 time: 1.50204e-05, fib2time: 5.22137e-05; ratio: 0.287671
7: fib1 time: 4.31538e-05, fib2time: 1.69277e-05; ratio: 2.5493
8: fib1 time: 6.60419e-05, fib2time: 2.09808e-05; ratio: 3.14773
9: fib1 time: 0.000129938, fib2time: 2.59876e-05; ratio: 5.0
10: fib1 time: 0.000296831, fib2time: 3.79086e-05; ratio: 7.83019
11: fib1 time: 0.000808954, fib2time: 6.81877e-05; ratio: 11.8636
12: fib1 time: 0.00269389, fib2time: 0.000169039; ratio: 15.9365
13: fib1 time: 0.00888014, fib2time: 0.000478983; ratio: 18.5396
14: fib1 time: 0.031538, fib2time: 0.00162387; ratio: 19.4215
15: fib1 time: 0.119242, fib2time: 0.00447392; ratio: 26.6527
16: fib1 time: 0.460428, fib2time: 0.0133159; ratio: 34.5773
17: fib1 time: 1.81237, fib2time: 0.0404069; ratio: 44.8529
18: fib1 time: 10.1864, fib2time: 0.123036; ratio: 82.7918
19: fib1 time: 43.4756, fib2time: 0.384916; ratio: 112.948[/spoiler]
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 23, 2009, 01:16:32 AM
my solution to number 6:
Code Select
g = []
x = 0
for i in range(1, 101):
g.append(i**2)
x += i
print x**2 - sum(g)
I'm certain guff will criticize my method of doing it, but I'm willing to learn.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 23, 2009, 04:45:40 AM
Quote from: Pancake Persona on January 23, 2009, 01:16:32 AM
my solution to number 6:
Code Select
g = []
x = 0
for i in range(1, 101):
g.append(i**2)
x += i
print x**2 - sum(g)
I'm certain guff will criticize my method of doing it, but I'm willing to learn.
it's not bad but as i've said before USE LIST COMPREHENSIONS
also it's kind of weird that for x you just use an int but for g you use a list
this would be more consistent:
Code Select

x, g = 0, 0
for i in range(1, 101):
g += i ** 2
x += i
print x**2 - g

but anyways what i did was something more like print(sum(i for i in range(101)) ** 2 - sum(i ** 2 for i in range(101)))
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 23, 2009, 07:16:31 AM
also if you want to do it all maths-like, then print(sum(i**3 - i**2 for i in range(101))) is prettier and also works akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 23, 2009, 01:04:18 PM
Quote from: andré the giant on January 23, 2009, 04:45:40 AM
it's not bad but as i've said before USE LIST COMPREHENSIONS
also it's kind of weird that for x you just use an int but for g you use a list
i don't know why i did it, but it's probably because i just slapped together something in 15 seconds and didn't bother thinking about alternative (and superior) methods
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 23, 2009, 01:39:27 PM
both of you still need to learn list comprehensions  baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 23, 2009, 01:40:32 PM
Quote from: andré the giant on January 23, 2009, 01:39:27 PM
both of you still need to learn list comprehensions  baddood;
help me understand better badass
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 23, 2009, 02:19:58 PM
Quote from: Pancake Persona on January 23, 2009, 01:40:32 PM
help me understand better badass
they're not all that complicated what don't you get  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Gladjaframpf on January 23, 2009, 05:55:00 PM
So I've been working with MIPS assembly recently and you've inspired me to try some of the PE problems with it. I solved the first two, but I've run into a bit of a problem in that the assembler I'm using doesn't actually have all the MIPS opcodes built into it and I'd really like to be able to use bitwise logic operations so I might have to find a new assembler.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 23, 2009, 05:57:33 PM
oh shit of course he has to show up

at least i can still beat lawlz and jmv  baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 23, 2009, 05:59:54 PM
so im on problem 3 and i think i did it right after some help on factorization and lists since i never used them.

now my code is taking a long time to run i hope it works ^___^
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 23, 2009, 06:04:21 PM
Quote from: JMV on January 23, 2009, 05:59:54 PM
...after some help...
AND I HAVE THE IM LOGS TO PROVE IT akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Gladjaframpf on January 23, 2009, 06:10:54 PM
Quote from: andré the giant on January 23, 2009, 05:57:33 PM
oh shit of course he has to show up

at least i can still beat lawlz and jmv  baddood;


You've solved like twice as many PE problems as I have. akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 23, 2009, 06:13:15 PM
Quote from: Gladjaframpf on January 23, 2009, 06:10:54 PM
You've solved like twice as many PE problems as I have. akudood;
yeah but i've had an account for a while now and i'm not using assembly language
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 23, 2009, 06:14:40 PM
QuoteOverflowError: long int too large to convert to int

akudood; akudood; akudood; akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 23, 2009, 06:24:24 PM
Code Select
#!/usr/bin/python
# Filename : Euler_003.py
# The prime factors of 13195 are 5, 7, 13 and 29.
# What is the largest prime factor of the number 600851475143?
EUL = 600851475143
#Function for finding primes
def isPrime(n):
import math
if n<2:
  return False #Negative Numbers, 0, and 1 are not prime.
elif n==2:
  return True #2 is Prime
elif n%2==0:
  return False #Even numbers, other than 2, are not prime.
else:
  for i in range(3, int(math.sqrt(n))+1,2):
   if n%i==0:
    return False #Number is composite
  else:
   return True  #All others are prime
#Find Factors
factor_list = []
primeFactors = []
i = 3
while True:
if EUL%i==0:
  while EUL % i == 0:
   EUL /= i
   factor_list.append(i)
if EUL == 1:
  break
i+=2 
#Take the list of factors and keep only the prime factors.
for factor in factor_list:
if isPrime(factor):
  primeFactors.append(factor)
print max(primeFactors) giggle;
Title: Re: I decided to step up to Guff's challenge
Post by: Gladjaframpf on January 23, 2009, 06:49:06 PM
Quote from: andré the giant on January 23, 2009, 06:13:15 PM
yeah but i've had an account for a while now and i'm not using assembly language


I actually did most of the problems I've done sometime last fall in Scheme.

Assembly isn't that bad for the easier ones, but it gets really messy when you need anything complicated. Function calls are horrible because you have to handle all the memory allocation manually.
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 24, 2009, 02:11:30 AM
Quote from: andré the giant on January 23, 2009, 05:57:33 PM
at least i can still beat jmv  baddood;
i think anyone can goonish
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 24, 2009, 04:55:54 AM
Quote from: Pancake Persona on January 24, 2009, 02:11:30 AM
i think anyone can goonish
because lack of actually doing the programs because I don't care enough = lack of ability. psyduck;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 24, 2009, 10:40:52 AM
Quote from: Pancake Persona on January 24, 2009, 02:11:30 AM
i think anyone can goonish
jmv is actually doing okay for a beginner but what's your excuse  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 24, 2009, 10:58:02 AM
Quote from: andré the giant on January 24, 2009, 10:40:52 AM
jmv is actually doing okay for a beginner but what's your excuse  akudood;
laziness and the vidya and a highly irregular sleep schedule  that keeps me from ever being able to focus and apply myself to anything

And if anything, JMV should be the "professional" here since he is (or was) studying to be a computer scientist. I just slowly picked up Python over the summer. :O
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 24, 2009, 11:04:23 AM
Just did problem 48  n_u
Code Select
#!/usr/bin/python
# Filename : Euler_048.py
#Find the last ten digits of the series, 1**1 + 2**2 + 3**3 + ... + 1000**1000
result = 0
for i in range(1,1001):
result+= i**i
print str(result)[-10:]


Quote from: Pancake Persona on January 24, 2009, 10:58:02 AMAnd if anything, JMV should be the "professional" here since he is (or was) studying to be a computer scientist. I just slowly picked up Python over the summer. :O
Because knowing one language makes you a professional in every programming language.
Write a compiler for Java so you don't have to use the JVM, since you know python and that makes you a professional in any language.
Also, as you said, my problem is:
Quote
laziness and the vidya and a highly irregular sleep schedule  that keeps me from ever being able to focus and apply myself to anything

Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 24, 2009, 11:10:13 AM
Quote from: JMV on January 24, 2009, 11:04:23 AM
Because knowing one language makes you a professional in every programming language.
when did i say we had to solve these in python
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 24, 2009, 11:18:57 AM
my solution for 48:
Code Select
x = str(sum(i**i for i in range(1, 1001)))
print x[-10:]

Looks like JMV got a whoopin', and I thought of guff when I worked towards this clean solution  baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 24, 2009, 11:28:26 AM
Quote from: Pancake Persona on January 24, 2009, 11:18:57 AM
my solution for 48:
Code Select
x = str(sum(i**i for i in range(1, 1001)))
print x[-10:]

Looks like JMV got a whoopin', and I thought of guff when I worked towards this clean solution  baddood;
yours is still silly
there's no need whatsoever to assign it to a variable and then have a separate line to print the answer:
Code Select
print str(sum(i**i for i in range(1, 1001)))[-10:]
would make more sense while still sticking to your method

but my solution is still more than ten times faster:
Code Select
print(sum(pow(i, i, 10**10) for i in range(1, 1001)) % 10 ** 10)

akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 24, 2009, 11:31:19 AM
Quote from: andré the giant on January 24, 2009, 11:28:26 AM
yours is still silly
there's no need whatsoever to assign it to a variable and then have a separate line to print the answer:
Code Select
print str(sum(i**i for i in range(1, 1001)))[-10:]
would make more sense while still sticking to your method

but my solution is still more than ten times faster:
Code Select
print(sum(pow(i, i, 10**10) for i in range(1, 1001)) % 10 ** 10)

akudood;
well then i learned some more today
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 24, 2009, 05:10:49 PM
oh god i just realized the link in my first post wasn't even to my profile but i fixed it now

also jamie solved problem 4 baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 24, 2009, 06:52:29 PM
Code Select
#!/usr/bin/python
# Filename : Euler_006.py
#Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
sumOfSquares, squareOfSums = 0, 0
for i in range(1,101):
sumOfSquares+= i**2
squareOfSums+=i
print squareOfSums**2 - sumOfSquares
Well, my original code was
Code Select
#!/usr/bin/python
# Filename : Euler_006.py
#Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
sumOfSquares=0
squareOfSums = 0
Sums = 0
for i in range(1,101):
sumOfSquares+= i**2
Sums+=i
squareOfSums=Sums**2
print squareOfSums - sumOfSquares
because I didn't know you could do math in the print statement, but other than that it was close to what guff had.

Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 24, 2009, 07:00:06 PM
Quote from: JMV on January 24, 2009, 06:52:29 PM
...but other than that it was close to what guff had.
uh no i had a one liner  akudood;

the long code i had wasn't anything i used it was just to show lawlz what he probably should have done instead  bassir;
Title: Re: I decided to step up to Guff's challenge
Post by: Gladjaframpf on January 24, 2009, 10:27:59 PM
A challenge: Do #79 on paper. I found it surprisingly easy. baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 24, 2009, 10:56:04 PM
Quote from: JMV on January 24, 2009, 06:52:29 PM
I didn't know you could do math in the print statement, but other than that it was close to what guff had.


That's the very first thing any python guide teaches you. goonish
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 25, 2009, 05:38:17 AM
Quote from: Pancake Persona on January 24, 2009, 10:56:04 PM
That's the very first thing any python guide teaches you. goonish
Well I guess me not reading any guides would be a factor.   n_u
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 25, 2009, 07:57:10 AM
Quote from: Gladjaframpf on January 24, 2009, 10:27:59 PM
A challenge: Do #79 on paper. I found it surprisingly easy. baddood;
that actually does seem doable  baddood;

i'll try it later maybe

also i think i'm going to try and do some of the problems i've already done in c akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 25, 2009, 09:48:31 AM
I did #79 in TextEdit only because I didn't want to waste paper.
[spoiler]73162890[/spoiler]

I had to register another account to test that so when I do it with actual code I won't feel like I cheated by submitting a problem with no programming.  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 25, 2009, 10:02:54 AM
Quote from: Pancake Persona on January 24, 2009, 11:23:55 PM
Solved 4:
Code Select
y = []
for i in range(100, 1000):
x = str(i*i)
if x == x[::-1]:
y.append(int(x))
print max(y)

Now I've just got to make a slimmer solution
uh that gives the wrong answer doodhuh;

you're assuming that the asnwer will be a perfect square (hint: it isn't)  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 25, 2009, 10:27:54 AM
Quote from: guff on January 25, 2009, 10:02:54 AM
uh that gives the wrong answer doodhuh;

you're assuming that the asnwer will be a perfect square (hint: it isn't)  akudood;
I just ran that too and the answer is way off.  doodhuh;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 25, 2009, 10:29:48 AM
lol jmv beat lawlz  n_u
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 25, 2009, 10:32:17 AM
Quote from: guff on January 25, 2009, 10:29:48 AM
lol jmv beat lawlz  n_u
what was that
Quote from: Pancake Persona on January 21, 2009, 12:57:22 PM
I just wonder if JMV could actually do any of them

baddood; baddood; baddood; baddood;

Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 25, 2009, 10:33:40 AM
Quote from: Pancake Persona on January 24, 2009, 02:11:30 AM
i think anyone can goonish
you can't lol
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 26, 2009, 12:05:22 AM
Quote from: guff on January 25, 2009, 10:02:54 AM
uh that gives the wrong answer doodhuh;

you're assuming that the asnwer will be a perfect square (hint: it isn't)  akudood;
oh goddamnit :'(

I'll work on the true solution
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 26, 2009, 12:11:54 AM
Code Select
"""A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91  99.

Find the largest palindrome made from the product of two 3-digit numbers."""

y = []
for i in range(100, 1000):
for o in range(100, 1000):
x = str(i*o)
if x == x[::-1]:
y.append(int(x))
print max(y)
That should be right... I hope.

But I really, really hate the way I did it and it's also really slow, so I'm going to rework it for a better solution.
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 26, 2009, 06:34:55 AM
Quote from: Pancake Persona on January 26, 2009, 12:05:22 AM
oh goddamnit :'(

I'll work on the true solution
You didn't check if you had the right answer before posting the code?  akudood;

and yeah that one works.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 26, 2009, 07:45:04 AM
QuoteJMV290
AIM
10:37
something is werid.
10:37
weird ;_;

Commodore Guff
AIM
10:37
how so

JMV290
AIM
10:37
/*Find the sum of all the multiples of 3 or 5 below 1000.*/
package euler_001;

/**
*
* @author jamesvalente
*/
public class Euler001 {

    public static void main(String[] args) {
        int sum = 0;
        for (int i = 1; i < 1000; i++) {
            if (3 % i == 0 || 5 % i == 0) {
                sum += i;
            }
        }
        System.out.println(sum);
    }
}
10:37
That should work
10:38
but it gives me 9
10:38
solution someone gave for the problem after i gave up and looked
10:38
public class mathschallenge
{
public static void main(String[] args)
{
int total = 0;
for(int i = 1; i < 1000; i ++)
{
if(i % 3 == 0 || i % 5 == 0)
total += i;
}
System.out.println(total);
}
}
10:38
same code except variable names
10:39
and the braces on the if statement don't change anything, they just make it more readable

Commodore Guff
AIM
10:40
dur
10:40
jamie
10:40
you have a serious problem with ordering things
10:41
the problem is in your if statement
10:41
3%i is the remainder you get when dividing 3 by i

JMV290
AIM
10:41
oh i forgot perentsitis

Commodore Guff
AIM
10:41
that's not what you want
10:41
NO

JMV290
AIM
10:41
lol

Commodore Guff
AIM
10:41
THAT'S NOT THE ISSUE
10:41
jamie
10:41
i%3 and i%5 are what you want
10:41
not 3%i and 5%i

JMV290
AIM
10:42
oh shit DYSLEXIA

Commodore Guff
AIM
10:42
:|

JMV290
AIM
10:42
forget this ever hapend THIS IS WHAT HAPPENS WHEN I EAT NO FOOD FOR HOURS

Commodore Guff
AIM
10:43
i save all my im logs

JMV290
AIM
10:43
IT NEVER AHPPENED GUFF


n_u
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 26, 2009, 07:46:24 AM
I'm hungry :(
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 26, 2009, 09:24:39 AM
PHP
Code Select
<?php /** Find the sum of all the multiples of 3 or 5 below 1000.*/
$sum = 0;
for($i=1;$i<1000;$i++){
if($i % 3 == 0 || $i % 5 == 0){
$sum+=$i;}}
echo $sum;
?>

Java
Code Select
/*Find the sum of all the multiples of 3 or 5 below 1000.*/
package euler_001;
/* @author jamesvalente*/
public class Euler001 {
    public static void main(String[] args) {
        int sum = 0;
        for (int i = 1; i < 1000; i++) {
            if (i % 3 == 0 || i % 5 == 0) {
                sum += i;
            }
        }
        System.out.println(sum);
    }
}

Python
Code Select
#!/usr/bin/python
# Filename : Euler_001.py
# Find the sum of all the multiples of 3 or 5 below 1000.
print sum(range(3, 1000, 3)) +sum(range(5, 1000, 5)) - sum(range(15, 1000, 15))
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 26, 2009, 09:39:47 AM
Code Select
/* Problem 1:
Find the sum of all the multiples of 3 or 5 below 1000.
*/

#include <stdio.h>

int main() {
int sum = 0;
int i;
for(i = 0; i < 1000; i++) {
if((i % 3 == 0) || (i % 5 == 0)) {
sum += i;
}
}
printf("%d\n", sum);
return(0);
}

i hate c i hope it dies  madood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 26, 2009, 09:42:44 AM
Quote from: guff on January 26, 2009, 09:39:47 AM
Code Select
/* Problem 1:
Find the sum of all the multiples of 3 or 5 below 1000.
*/

#include <stdio.h>

int main() {
int sum = 0;
int i;
for(i = 0; i < 1000; i++) {
if((i % 3 == 0) || (i % 5 == 0)) {
sum += i;
}
}
printf("%d\n", sum);
return(0);
}

i hate c i hope it dies  madood;
it looks like java.

speaking of java, brb removing multilne comments I don't need.  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 26, 2009, 10:14:11 AM
And, in Javascript.  baddood;
Code Select
<script type="text/javascript">
//Find the sum of all the multiples of 3 or 5 below 1000.
var sum=0;
for(var i=0;i<1000;i++){
if(i%3==0 || i%5==0){
sum+=i;
}
}
document.write(sum);
</script>
Title: Re: I decided to step up to Guff's challenge
Post by: Gladjaframpf on January 26, 2009, 01:16:36 PM
PLT Scheme:
Code Select

(display (foldr + 0 (build-list 1000 (lambda (x) (if (or (= (modulo x 3) 0) (= (modulo x 5) 0)) x 0)))))

Displayed on one line for maximum unreadability. akudood;


MIPS Assembly:
Code Select

lis $1
.word 0
lis $2
.word 1
lis $3
.word 1
lis $4
.word 3
lis $5
.word 5
lis $6
.word 1000

loop:
div $2, $4
mfhi $7
beq $7, $0, 3
div $2, $5
mfhi $7
bne $7, $0, 1
add $1, $1, $2

add $2, $2, $3
slt $7, $2, $6
bne $7, $0, loop

jr $31

This doesn't actually print the result, it just stores it in register $1, but I didn't want to stick the print routine in because it's big and ugly and my MIPS emulator automatically prints the contents of all the registers when it's done processing anyways.
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 26, 2009, 01:59:29 PM
I need to find out how to do loops in Clojure.  madood; madood; madood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on January 26, 2009, 02:11:23 PM
whoa, i didn't even notice guff did number 1 in c

Because that's what I just did:
Code Select
#include <stdio.h>
main(){
int i, x;
i = 0;
x = 0;
while (i<1000){
if (i % 5 == 0 || i % 3 == 0)
x+=i;
i++;
}
printf("%d \n\n", x);
}

i kept getting the wrong answer with this for a good twenty minutes and it was driving me nuts and then finally asked jmv. the second i hit my return key, i noticed i forgot to define x goonish
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 26, 2009, 02:39:00 PM
Code Select
sum [i|i<-[1..999],i`mod`3==0||i`mod`5==0]

haskell akudood;
i could probably shorten it a bit using the same trick i did for my python solution but this was easier to write in ten seconds because i didn't have to look anything up

Code Select
+/((-.*5|>:i.1000) +. (-.*3|>:i.1000))#>:i.1000

j  akudood;

i haven't looked at j for a while now and after seeing this code again i remember why
my code for this is horribly innefficient; ideally i should only have to call i.1000 once but using the stuff i knew this was the only way i could think of at the time baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 26, 2009, 04:10:54 PM
Quote from: guff on January 26, 2009, 02:39:00 PM
Code Select
sum [i|i<-[1..999],i`mod`3==0||i`mod`5==0]
okay shortened it to:
Code Select
sum [i|i<-[1..999],0`elem`[i`mod`3,i`mod`5]]

and by shorter i mean i made it a little longer but it would be shorter if shit damn haskell used shit damn % for modulo  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 26, 2009, 06:42:48 PM
okay did a ruby version too akudood;
Code Select

total = 0
for i in 1...1000 do total += i if [i % 3, i % 5].include? 0 end
puts total

gosh i gave up on ruby a while ago for python but on using it again i love its syntax so very much
it just needs list comprehensions

i also realized i ended up writing a sum function because i don't think ruby has one built in:
Code Select
module Sum
def sum
total = 0
for element in self do total += element end
total
end
end

class Array
include Sum
end


then i realized that my code didn't even use it but if it did it would have looked pretty n_u
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 27, 2009, 08:59:32 AM
okay jmv's supposedly been trying to learn clojure so i decided to try and make him look silly by solving problem one in it while having no idea how to do anything in it  baddood;

Code Select
((def sum 0)
(loop [i 0]
(when (< i 1000)
(if ((fn [x] (or (= 0 (rem x 3)) (= 0 (rem x 5)))) i) (def sum (+ sum i)))
(recur (inc i))))
(println sum))

it prints out the correct answer
but it also encounters some catastrophic errors or something at the end i dunno at least it gives the answer  akudood;

granted all i had to go on was skimming through this (http://www.moxleystratton.com/article/clojure/for-non-lisp-programmers) small tutorial so i had no idea how to do, say foo += bar all lisp-like so instead i had to do (def foo (+ foo bar)) and i'm pretty sure there's got to be a better way
i should also probably try and figure out why the loop explodes but there really doesn't seem to be that many tutorials for clojure floating around  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Gladjaframpf on January 27, 2009, 06:44:34 PM
People who use looping constructs in functional languages go to SUPER HELL. akudood;

I don't know much about Clojure specifically, but since it's a Lisp dialect I'd recommend making a recursive function that takes a parameter n and counts down from n to zero. Then you can just call the function with 999.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 27, 2009, 06:50:08 PM
Quote from: Gladjaframpf on January 27, 2009, 06:44:34 PM
People who use looping constructs in functional languages go to SUPER HELL. akudood;

I don't know much about Clojure specifically, but since it's a Lisp dialect I'd recommend making a recursive function that takes a parameter n and counts down from n to zero. Then you can just call the function with 999.
yeah probably but recall that this was hastily done in five minutes with no prior knowledge of clojure or any lisp for that matter baddood;

i'll just stick to haskell for my future functional programming needs screw you and your strict evaluation  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Gladjaframpf on January 27, 2009, 11:45:29 PM
Quote from: guff on January 27, 2009, 06:50:08 PM
yeah probably but recall that this was hastily done in five minutes with no prior knowledge of clojure or any lisp for that matter baddood;

i'll just stick to haskell for my future functional programming needs screw you and your strict evaluation  akudood;


Yeah I looked at the tutorial you were working from and it sucked so I don't blame you. I guess since it's designed for non-Lisp programmers it's trying to demonstrate things that someone used to imperative languages would recognize, but that's kind of silly because the whole point of learning languages with different paradigms is to explore new ways of doing things.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 28, 2009, 01:25:36 PM
okay after looking a little more into the language i was able to write this (also i didn't look at your scheme solution until after i posted this):
Code Select
(println (reduce + (filter (fn [i] (or (= (rem i 3) 0) (= (rem i 5) 0))) (range 1000))))

happy now  akudood;

if i am going to get into any lisp dialect i need an editor with automatic parentheses matching that isn't emacs or vim because i don't want a whole shit damn operating system  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Gladjaframpf on January 29, 2009, 05:09:52 AM
Yes that is much better.

I noticed something sort of odd when I was doing that problem. Simply looping through the numbers 1 to 999 should be faster than my solution, which first makes a list of the numbers and then adds them up so it has to go through all the numbers twice. However, tests show that my solution is faster. I guess the built-in functions for creating and folding lists are so heavily optimized that it makes up for it.
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 29, 2009, 08:45:13 PM
Yo dawg, I heard you like least common multiples.
Code Select
#!/usr/bin/python
# Filename : Euler_005.py
#What is the smallest number that is evenly divisible by all of the numbers from 1 to 20?
def gcd(a, b):
while b != 0:
  t =b
  b = a%b
  a = t
return a
def lcm(a,b):
d = a*b/gcd(a,b)
return d
print lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(1,2),3),4),5),6),7),8),9),10),11),12),13),14),15),16),17),18),19),20)

is there a way to get it so I don't need 1000000000000000 LCMs?
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 30, 2009, 06:16:25 AM
Quote from: Jacques Michel Valente on January 29, 2009, 08:45:13 PM
Code Select

print lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(1,2),3),4),5),6),7),8),9),10),11),12),13),14),15),16),17),18),19),20)
OH MY GOD  akudood;

there's more than one way to do it
you could use the reduce function, which takes a function and an iterable object as input so you could just do something like:
Code Select
reduce(lcm, range(1, 21))

or, you could just use a for loop like this:
Code Select
current_lcm = 1
for i in range(1, 21):
current_lcm = lcm(current_lcm, i)


the for loop might be a better idea if you intend to learn python3.0 because they moved reduce from the builtins to the functools module akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Feynman on January 30, 2009, 10:03:25 AM
lol computer nerds


Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 30, 2009, 11:57:55 AM
Quote from: Bassir on January 30, 2009, 10:03:25 AM
lol computer nerds



Can't copypaste these problems?  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Feynman on January 30, 2009, 12:52:35 PM
Quote from: Jacques Michel Valente on January 30, 2009, 11:57:55 AM
Can't copypaste these problems?  akudood;


i don't know python :'(

but i after seeing their neat logo i may consider it
Title: Re: I decided to step up to Guff's challenge
Post by: guff on January 30, 2009, 01:12:30 PM
Quote from: Bassir on January 30, 2009, 12:52:35 PM
i don't know python :'(
it takes all of five minutes to learn   baddood;

UNLESS YOU ARE JMV  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on January 30, 2009, 01:22:15 PM
Quote from: guff on January 30, 2009, 01:12:30 PM
it takes all of five minutes to learn   baddood;

UNLESS YOU ARE JMV  akudood;
then it takes java
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on February 06, 2009, 12:44:27 AM
Just decided to do problem 9:
Code Select
for a in range(1, 1000):
for b in range(1, 1000):
c = ((a**2)+(b**2))**.5
if a+b+c == 1000:
print a*b*c
exit()
it's an ugly way of doing it, but i got it right the first try

and i threw exit in there so it'd quit as soon as it found the solution instead of going for that tenth of a second more to solve it a second time
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on February 18, 2009, 09:57:38 PM
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
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on February 18, 2009, 10:22:11 PM
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.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 19, 2009, 06:06:35 AM
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 bassir;

also my solution for 24 is two lines long and takes ~0.06 seconds to run  n_u
granted it can be solved by hand if you do the math but brute force is usually more fun
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 19, 2009, 06:59:17 AM
Quote from: Pancake Persona on February 18, 2009, 10:22:11 PM
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
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 19, 2009, 07:09:18 AM
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  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on February 19, 2009, 12:43:42 PM
Quote from: guff on February 19, 2009, 07:09:18 AM
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  akudood;
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 goonish

And pygame doesn't support python 3.0, so I can't.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 19, 2009, 01:36:38 PM
Quote from: Pancake Persona on February 19, 2009, 12:43:42 PM
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 goonish

And pygame doesn't support python 3.0, so I can't.
uh yeah that's what modules and functions are for  baddood;

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  akudood;

well then pygame SUCKS akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on February 19, 2009, 02:01:56 PM
Quote from: guff on February 19, 2009, 01:36:38 PM
uh yeah that's what modules and functions are for  baddood;

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  akudood;

well then pygame SUCKS akudood;
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

goonish

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
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on February 20, 2009, 02:23:33 PM
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 :'(
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 20, 2009, 02:54:36 PM
Quote from: Pancake Persona on February 20, 2009, 02:23:33 PM
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  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on February 21, 2009, 09:41:18 AM
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.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 21, 2009, 01:24:47 PM
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  baddood;

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 akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on February 21, 2009, 01:29:05 PM
Quote from: guff on February 21, 2009, 01:24:47 PM
takes about 0.46 seconds  baddood;

so how fast is your real solution or is mine better spam;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 21, 2009, 01:39:32 PM
Quote from: Pancake Persona on February 21, 2009, 01:29:05 PM
so how fast is your real solution or is mine better spam;
~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)


maps;

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
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on February 21, 2009, 01:44:37 PM
Quote from: guff on February 21, 2009, 01:39:32 PM
~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)


maps;

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'

maps;

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' spam;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 21, 2009, 01:47:46 PM
Quote from: Pancake Persona on February 21, 2009, 01:44:37 PM
>>> from test import pystone
>>> pystone.full()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'module' object has no attribute 'full'
oh duh i'm an idiot it's pystone.main() not full()
Quote from: Pancake Persona on February 21, 2009, 01:44:37 PM
maps;

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' spam;
well it's just good practice  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: ncba93ivyase on February 21, 2009, 01:52:24 PM
Quote from: guff on February 21, 2009, 01:47:46 PM
oh duh i'm an idiot it's pystone.main() not full()well it's just good practice  akudood;
54945.1 pystones/second

and i just ran your solution and got like .201 s

Good practice doesn't necessarily mean good solutions. spam;

although i guess that's not something i should talk about maps;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 23, 2009, 09:49:15 AM
>>> from test import pystone
>>> pystone.main()
Pystone(1.1) time for 50000 passes = 0.99
This machine benchmarks at 50505.1 pystones/second
>>>

AND AGAIN

>>> from test import pystone
>>> pystone.main()
Pystone(1.1) time for 50000 passes = 0.97
This machine benchmarks at 51546.4 pystones/second
>>>


Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 23, 2009, 03:54:26 PM
Code Select
#!/usr/bin/python
# Filename : Euler_055.py
#How many Lychrel numbers are there below ten-thousand?
numLy=0
def isPalindrome(n): return str(n) == str(n)[::-1]
def rev(n): return int(str(n)[::-1])
def lychrel(n):
for i in range(50):
  n+=rev(n)
  if isPalindrome(n):
   return False
return True
for i in range(10000):
if lychrel(i):
  numLy+=1
print "Number of lychrel numbers is %d."%(numLy)   
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 23, 2009, 05:18:24 PM
Code Select
#!/usr/bin/python
# Filename : Euler_009.py
#There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.
#a<b<c && a**2 + b**2 = c**2
from time import time
start = time()
def equalsC(a,b,c):
    if c**2 == (a**2 + b**2):
        return True
def sizeCheck(a,b,c):
    if a<b<c:
        return True
def sumCheck(a,b,c):
    if a+b+c==1000:
        return True 
for a in range(500):
    for b in range(500):
        c=1000-a-b
        if equalsC(a,b,c) and sizeCheck(a,b,c) and sumCheck(a,b,c):
            print a*b*c
            print time() - start
            exit()
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 23, 2009, 09:19:30 PM
Code Select
#!/usr/bin/python
# Filename : Euler_008.py
#Find the greatest product of five consecutive digits in the 1000-digit number.
from time import time
start = time()
bigFuckingNumber="*number removed since it stretches the code box*"
prod = 0
for i in range(996):
dummy =int(bigFuckingNumber[i]) * int(bigFuckingNumber[i+1]) * int(bigFuckingNumber[i+2]) * int(bigFuckingNumber[i+3]) * int(bigFuckingNumber[i+4])
if dummy> prod:prod = dummy
print prod
print time() - start
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 23, 2009, 10:13:36 PM
Code Select
#!/usr/bin/python
# Filename : Euler_013.py
# Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
file = open('files/13.txt')
lineList = []
for line in file:lineList.append(int(line))
print str(sum(lineList))[:10]
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 24, 2009, 04:12:31 AM
okay so is jim or lawlz winning now  baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 25, 2009, 09:20:00 AM
solved problem 231, and my code runs in about 56 seconds  akudood;
Code Select
from euler_functions import sieve_of_erat

def factor_sum(factors_dict):
return sum(factor * exponent for factor, exponent in factors_dict.items())

def dict_add(dict_a, dict_b):
result_dict = {}
for key in set(dict_a.keys()) | set(dict_b.keys()):
result_dict[key] = dict_a.get(key, 0) + dict_b.get(key, 0)
return result_dict

def binomial_factors(n, k): # n!/(k!*(n-k)!) factorization = n! factors - k! factors - (n-k)! factors
primes = {n: sieve_of_erat(n), k: sieve_of_erat(k), n-k: sieve_of_erat(n-k)}
factors = {n: {}, k: {}, n-k: {}}

for number in (n, k, n - k):
for p in primes[number]:
p_count = 0
i = 1
while number // (p ** i) >= 1:
if number == n:
p_count += number // (p ** i)
else:
p_count -= number // (p ** i)
i += 1
factors[number][p] = p_count
return dict_add(factors[n], dict_add(factors[k], factors[n-k]))

doesn't include the code that actually generates the answer though just so no one gets tempted  baddood;

after looking at the forum, there's many different ways of doing it some of which are faster and some of which are shorter but this is just what i came up with first baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 25, 2009, 10:31:50 AM
GUFF GET ONLINE HELP ;_;

I did 14 and it runs in about 13 seconds(down from 49) if I check only odd numbers and start at 500,001(sicnce all of the odds will return an even which then gets cut in half, and all of the evens above 500k will be at least 1 more than all the evens below it)
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 25, 2009, 10:55:30 AM
Code Select
#!/usr/bin/python
# Filename : Euler_014.py
# Which starting number, under one million, produces the longest chain?
from time import time
start = time()
longestChain = 0
chainValue = 0
#Calculate and return the length of a chain
def chain(n):
lengthChain= 0
while n>1:
if n%2==0:
n/=2
else:
n= (3*n)+1
lengthChain+=1
return lengthChain
#Check up to 1 million.
for i in range(500001,1000000, 2):
if chain(i) > longestChain:
longestChain = chain(i)
chainValue=i
#Time to print shit
print chainValue
print chain(longestChain)
print time() - start

If I changed my function to
Code Select
def chain(n):
lengthChain= 0
while n>1:
if n%2:
n= (3*n)+1
else:
n/=2
lengthChain+=1
return lengthChain
It will run about one second faster, but it doesn't look as neat.
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 26, 2009, 12:20:42 PM
With Guffnigger's help I turned:
ValenteMac:Python jamesvalente$ python Euler_010.py
142913828922
29.1039888859
ValenteMac:Python jamesvalente$

into:

ValenteMac:Desktop jamesvalente$ python Euler_010.py
142913828922
3.5488049984
ValenteMac:Desktop jamesvalente$

Between 9 and 10 times faster.  baddood;

Further optimization:

ValenteMac:Python jamesvalente$ python Euler_010.py
142913828922
2.34858393669
ValenteMac:Python jamesvalente$
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 26, 2009, 02:16:51 PM
I think Guff and I are the only people who post in this thread. baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 26, 2009, 02:48:53 PM
Quote from: Raekewn on February 26, 2009, 02:16:51 PM
I think Guff and I are the only people who post in this thread. baddood;
those nerds  madood;
Title: Re: I decided to step up to Guff's challenge
Post by: Ezloﺕ on February 26, 2009, 02:51:29 PM
hy
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on February 26, 2009, 03:21:47 PM
Quote from: Ezlo on February 26, 2009, 02:51:29 PM
hy
what does this mean.

is it hi or or like how are you or what
Title: Re: I decided to step up to Guff's challenge
Post by: guff on February 28, 2009, 02:48:32 PM
QuoteCongratulations, the answer you gave to problem 46 is correct.

Nice work, Guff, you've just advanced another level.

baddood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on April 10, 2009, 10:35:41 AM
I REVIVE THIS THREAD. THE POWER OF GOD
Code Select
#Find the sum of all the multiples of 3 or 5 below 1000.
sum=0
for i in (1..999)
if i%3==0 or i%5==0
sum+=i
end
end
puts sum

Problem 1 in Ruby.
Title: Re: I decided to step up to Guff's challenge
Post by: guff on April 10, 2009, 05:18:54 PM
Quote from: ra-ˈkün on April 10, 2009, 10:35:41 AM
I REVIVE THIS THREAD. THE POWER OF GOD
Code Select
#Find the sum of all the multiples of 3 or 5 below 1000.
sum=0
for i in (1..999)
if i%3==0 or i%5==0
  sum+=i
end
end
puts sum

Problem 1 in Ruby.
already done punk  akudood;
Title: Re: I decided to step up to Guff's challenge
Post by: Daddy on April 10, 2009, 05:32:42 PM
you're the punk
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