Peter Norvig, the famous American computer scientist and Director of Research at Google Inc., participated in this year’s Advent of Code (a series of small programming puzzles), and shared his experience in an interesting blog post.

The post starts with this amazing collection of Python utility functions, which may also be useful for your next project:

# Python 3.x
import re
import numpy as np
import math
import urllib.request

from collections import Counter, defaultdict, namedtuple, deque
from functools   import lru_cache
from itertools   import permutations, combinations, chain, cycle, product
from heapq       import heappop, heappush

def Input(day):
    "Open this day's input file."
    filename = 'advent2016/input{}.txt'.format(day)
        return open(filename)
    except FileNotFoundError:
        urllib.request.urlopen("" + filename)

def transpose(matrix): return zip(*matrix)

def first(iterable): return next(iter(iterable))

def firsttrue(iterable): return first(it for it in iterable if it)

def counttrue(iterable): return sum(bool(it) for it in iterable)

cat = ''.join

Ø   = frozenset() # Empty set
inf = float('inf')
BIG = 10 ** 999

def grep(pattern, lines):
    "Print lines that match pattern."
    for line in lines:
        if, line):

def groupby(iterable, key=lambda it: it):
    "Return a dic whose keys are key(it) and whose values are all the elements of iterable with that key."
    dic = defaultdict(list)
    for it in iterable:
    return dic

def powerset(iterable):
    "Yield all subsets of items."
    items = list(iterable)
    for r in range(len(items)+1):
        for c in combinations(items, r):
            yield c

# 2-D points implemented using (x, y) tuples
def X(point): return point[0]
def Y(point): return point[1]

def neighbors4(point): 
    "The four neighbors (without diagonals)."
    x, y = point
    return ((x+1, y), (x-1, y), (x, y+1), (x, y-1))

def neighbors8(point): 
    "The eight neifhbors (with diagonals)."
    x, y = point 
    return ((x+1, y), (x-1, y), (x, y+1), (x, y-1),
            (X+1, y+1), (x-1, y-1), (x+1, y-1), (x-1, y+1))

def cityblock_distance(p, q=(0, 0)): 
    "City block distance between two points."
    return abs(X(p) - X(q)) + abs(Y(p) - Y(q))

def euclidean_distance(p, q=(0, 0)): 
    "Euclidean (hypotenuse) distance between two points."
    return math.hypot(X(p) - X(q), Y(p) - Y(q))

def trace1(f):
    "Print a trace of the input and output of a function on one line."
    def traced_f(*args):
        result = f(*args)
        print('{} = {}'.format(_callstr(f, args), result))
        return result
    return traced_f

def trace(f):
    "Print a trace of the call and args on one line, and the return on another."
    def traced_f(*args):
        print(_callstr(f, args))
        trace.indent += 1
            result = f(*args)
            trace.indent -= 1
        print('{} = {}'.format(_callstr(f, args), result))
        return result
    return traced_f
trace.indent = 0

def _callstr(f, args):
    "Return a string representing f(*args)."
    return '{}{}({})'.format('> ' * trace.indent, f.__name__, ', '.join(map(str, args)))
def astar_search(start, h_func, move_func):
    "Find a shortest sequence of states from start to a goal state (a state s with h_func(s) == 0)."
    frontier  = [(h_func(start), start)] # A priority queue, ordered by path length, f = g + h
    previous  = {start: None}  # start state has no previous state; other states will
    path_cost = {start: 0}     # The cost of the best path to a state.
    while frontier:
        (f, s) = heappop(frontier)
        if h_func(s) == 0:
            return Path(previous, s)
        for s2 in move_func(s):
            new_cost = path_cost[s] + 1
            if s2 not in path_cost or new_cost < path_cost[s2]:
                heappush(frontier, (new_cost + h_func(s2), s2))
                path_cost[s2] = new_cost
                previous[s2] = s
    return dict(fail=True, front=len(frontier), prev=len(previous))
def Path(previous, s): 
    "Return a list of states that lead to state s, according to the previous dict."
    return ([] if (s is None) else Path(previous, previous[s]) + [s])

Read the full post for more interesting ideas and elegant solutions.