andbroby

data Tree a = EmptyTree | Node a (Tree a) (Tree a) deriving (Show, Read, Eq)

singleton :: a -> Tree a
singleton x = Node x EmptyTree EmptyTree

treeInsert :: (Ord a) => a -> Tree a -> Tree a
treeInsert x EmptyTree = singleton x
treeInsert x (Node a left right)
    | x == a = Node x left right
    | x < a = Node a (treeInsert x left) right

andbroby

def shunting_yard(infix):
    precedence = {
        '^': 3,
        '*': 2,
        '/': 2,
        '+': 1,
        '-': 1
    }
    operators = precedence.keys()
    l_associative = ['*', '/', '+', '-']

andbroby

import string
import sys

def re_to_postfix(infix):
    precedence = {
        '|': 1,
        '.': 2,
        '*': 3,
        '+': 3,
        '?': 3

andbroby

def f(x):
  def g(y):
    def h(z):
      print(x,y,z)
    return h
  return g

andbroby

import java.util.regex.Pattern;

public class EmailPattern {
	private static Pattern pattern = Pattern.compile("(?:(?:\\r\\n)?[ \\t])*(?:(?:(?:[^()<>@,;:\\\\\".\\[\\] \\000-\\031]+(?:(?:(?:\\r\\n)?[ \\t])+|\\Z|(?=[\\[\"()<>@,;:\\\\\".\\[\\]]))|\"(?:[^\\\"\\r\\\\]|\\\\.|(?:(?:\\r\\n)?[ \\t]))*\"(?:(?:\\r\\n)?[ \\t])*)(?:\\.(?:(?:\\r\\n)?[ \\t])*(?:[^()<>@,;:\\\\\".\\[\\] \\000-\\031]+(?:(?:(?:\\r\\n)?[ \\t])+|\\Z|(?=[\\[\"()<>@,;:\\\\\".\\[\\]]))|\"(?:[^\\\"\\r\\\\]|\\\\.|(?:(?:\\r\\n)?[ \\t]))*\"(?:(?:\\r\\n)?[ \\t])*))*@(?:(?:\\r\\n)?[ \\t])*(?:[^()<>@,;:\\\\\".\\[\\] \\000-\\031]+(?:(?:(?:\\r\\n)?[ \\t])+|\\Z|(?=[\\[\"()<>@,;:\\\\\".\\[\\]]))|\\[([^\\[\\]\\r\\\\]|\\\\.)*\\](?:(?:\\r\\n)?[ \\t])*)(?:\\.(?:(?:\\r\\n)?[ \\t])*(?:[^()<>@,;:\\\\\".\\[\\] \\000-\\031]+(?:(?:(?:\\r\\n)?[ \\t])+|\\Z|(?=[\\[\"()<>@,;:\\\\\".\\[\\]]))|\\[([^\\[\\]\\r\\\\]|\\\\.)*\\](?:(?:\\r\\n)?[ \\t])*))*|(?:[^()<>@,;:\\\\\".\\[\\] \\000-\\031]+(?:(?:(?:\\r\\n)?[ \\t])+|\\Z|(?=[\\[\"()<>@,;:\\\\\".\\[\\]]))|\"(?:[^\\\"\\r\\\\]|\\\\

andbroby

### Keybase proof

I hereby claim:

  * I am andbroby on github.
  * I am andbroby (https://keybase.io/andbroby) on keybase.
  * I have a public key whose fingerprint is 1F9A A6E9 8E59 19F2 2D6D  2D83 3EB0 E4FE EE3C B2A1

To claim this, I am signing this object:

andbroby

cheeses={'cheddar':('A235-4','A236-1','A236-2','A236-3','A236-5','C31-1','C31-2'),'mozarella':('Q456-9','Q456-8','A234-5','Q457-1','Q457-2'),'gombost':('ZLAFS55-4','ZLAFS55-9','GOMBOS-7','A236-4'),'geitost':('SPAZ-1','SPAZ-3','EMACS45-0'),'portsalut':('B15-1','B15-2','B15-3','B15-4','B16-1','B16-2','B16-4'),'camembert':('A243-1','A234-2','A234-3','A234-4','A235-1','A235-2','A235-3'),'ridder':('GOMBOS-4','B16-3'),}


for item in cheeses.items():
    infected = ["A234","A235", "B13", "B14", "B15", "C31"]
    for shelves in item[1:]:
        if any(infected_shelf in shelf for infected_shelf in infected for shelf in shelves):
            print(item[0])
            break

andbroby

def warshall(matrix):
    n = len(matrix)
    for k in range(n):
        for i in range(n):
            for j in range(n):
                if matrix[i][k] and matrix[k][j]:
                    matrix[i][j] = 1
    return matrix
    
matrix = [[0, 0, 0, 1],

andbroby

listofMultiples factor1 factor2 l = 
    let p x =
        | x `mod` factor1 == 0 = True
        | x `mod` factor2 == 0 = True
        | otherwise = False
    in filter p [1..l]