project euler problem #1


If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.

  • try solving it as a python one-liner
  • try optimizing your solution for speed

highlight below for my solution:

sum(range(0,1000,3)) + sum(range(0,1000,5)) - sum(range(0,1000,15))

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project euler problem #4


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.

highlight below for my solution:

def is_palindrome(num):
    return str(num) == str(num)[::-1]

def fn(n):
    max_palindrome = 1
    for x in range(n,1,-1):
        if x*n < max_palindrome:
        for y in range(n,x-1,-1):
            if is_palindrome(x*y) and x*y > max_palindrome:
                max_palindrome = x*y
            elif x * y < max_palindrome:
    return max_palindrome

print fn(999)

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sudoku solution verifier – python


write a sudoku solution verifier in python.

that is, given a list of lists, representing a solution to an nxn sudoku puzzle, verify that the solution is a correct sudoku solution:

fn ([[1,2],[2,1]]) # True
fn ([[1,3],[2,1]]) # False

(question seen as a Udacity programming course homework question)

highlight to see a solution:

def fn(x):
    b = [(j + 1) for j in range(0,len(x))]
    for i in x:
        if sorted(i) != b: return False
    return True

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python counters

in my pythonic trolling of the internet, i found a character counter, written by ashoksk. given a string of text, return a count for each character in the string (a tool that might be used when attempting to crack an encrypted message)

so i wondered, could this be a python one liner? turns out that, from python 2.7 on, it can be (highlight to see the code below):

from collections import Counter
def fn(x): c = Counter(x); return {x: c[x] for x in c.keys()}

it’s a cheat, though. anyone have a real one-line implementation of this?

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recursive touch


i just moved a bunch of media files from one device, to the other, and the file creation dates for a small number of my directories (and their contents) moved a few years into the future.

this problem smelled like a python 1 liner; i came pretty close – highlight to see the code:

import os, sys, subprocess

for d,s,fs in os.walk(sys.argv[1]):
    for p in (s+fs):["touch",os.path.join(d,p)])

(source online, in case I improve it)

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riddle 2 (python)


Count the number of days between two days.

same as last time, highlight to see answer(s), in python, below.

def fn(*args): 
      x = '/'.join(str(i) for i in args[:3])
      y = '/'.join(str(i) for i in args[3:])
      return abs((datetime.strptime(x,"%Y/%m/%d")-datetime.strptime(y,"%Y/%m/%d")).days)


#depending on input format, could do a one-liner:

def fn(x,y): abs((datetime.strptime(x,"%Y/%m/%d")-datetime.strptime(y,"%Y/%m/%d")).days)

#it's probably possible to replace "%Y/%m/%d" with "%c"

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