fulmicoton

from collections import defaultdict
import heapq

ENGLISH = defaultdict(list)
for w in open("/usr/share/dict/american-english"):
    w_stripped = w.strip()
    ENGLISH[tuple(sorted(w_stripped))].append(w_stripped)

print heapq.nlargest(1, ENGLISH.itervalues(), key=len)[0]

fulmicoton

from collections import defaultdict
import heapq

ENGLISH = defaultdict(list)
for w in open("/usr/share/dict/american-english"):
    w_stripped = w.strip()
    ENGLISH[tuple(sorted(w_stripped))].append(w_stripped)

print heapq.nlargest(1, ENGLISH.itervalues(), key=len)[0]

fulmicoton

import numpy as np
import cProfile


def bits_to_int(bits,):
    n = 0
    for b in bits[::-1]:
        n *= 2
        n += b
    return n

fulmicoton

def iter_splits(l):
    if not l:
        yield ONE, ONE
    else:
        head, tail = l[0], l[1:]
        for (left, right) in iter_splits(tail):
            yield left + [head], right
            yield left, right + [head]

fulmicoton

def iter_splits(l):
    if not l:
        yield [], []
    else:
        head, tail = l[0], l[1:]
        for (left, right) in iter_splits(tail):
            yield left + [head], right
            yield left, right + [head]

fulmicoton

def bits_to_int(bits,):
    n = 0
    for b in bits[::-1]:
        n *= 2
        n += b
    return n


def zeros_matrix(N):
    return [[0] * N for i in range(N)]

fulmicoton

#include <iostream>

using namespace std;

int main()
{
  int size;
  cin >> size;
  
  // <- input

fulmicoton

Let's assume we found a solution with N not a power of 2.
The problem is N does not divide the number of configuration of the board...

Let's show that.
Since your pal is supposed to enter the room and point out the magic case, the solution
implies a mapping from the configuration of the board to one of the cases.

Let's call this mapping f and gives number to the cases. Call 
E = [0..N] and P=the set of configurations.
Its cardinal is 2^N

fulmicoton

import random

N = 100

def permutation(S=100000):
	l = range(N)
	for i in xrange(S):
		random.shuffle(l)
		yield l

fulmicoton

#!/usr/bin/env python

def quick_quantiles(data, ranks):
	# well there are better heuristics
	# than that but that's not the point of
	# this.
	pivot = (pivot_val, pivot_count) = data[0]
	if len(data) == 1:
		return [pivot_val]
	left = [it for it in data if it < pivot]