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"Id","City","State","Month","Year","Summary","Source","lon","lat"
1,"Bella Vista","Arkansas","September",2010,"The Bella Vista City Council met to consider banning the construction of mosques and the practice of Sharia Law within city limits.","http://nwahomepage.com/fulltext-news?nxd_id=188955 http://jonah-m-tebbetts.blogspot.com/2010/09/new-jerusalem-in-benton-cou...",-94.2732642,36.481464
2,"Phoenix","Arizona","December",2015,"Vandals shattered windows at the Islamic Cultural Center.","http://www.abc15.com/news/region-phoenix-metro/north-phoenix/vandalism-reported-at-islamic-community-center-in-phoenix-",-112.0740373,33.4483771
3,"Glendale","Arizona","August",2010,"Two men threw an acid bomb - a soda bottle filled with acid at a mosque - targeting mosque officials standing nearby.","http://www.reuters.com/article/idUSN0833871920070808",-112.1859866,33.5386523
4,"Phoenix","Arizona","August",2010,"Vandals threw paint on the floor, shot out windows, and wrote anti-Muslim slurs on the walls of a local mosque."

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# Find the probability distribution of the longest run of heads or tails in N tosses of a fair coin.

N = 10   # Number of coin tosses

from itertools import product, imap, groupby
from collections import Counter

# Adapted from http://stackoverflow.com/a/22214180/677398
def longest_run(sequence):
    return max(sum(1 for _ in group) for key, group in groupby(sequence))

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# -*- coding: utf-8 -*-
"""
Let a[n; c, r] = number of sequences of n coin flips
                 ending with a streak of c heads or tails
                 and having a longest streak of r heads or tails.

We can show that 
    a[n; 1, r] = sum (i=1 to r) a[n-1; i, r]
and
    a[n; r, r] = a[n-1; r-1, r-1] + a[n-1; r-1, r]

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# This R script generates a correspondence between ZIP codes and
# US Congressional Districts (113th Congress)
# by merging two data files from the US Census Bureau website.
#
# Note that the relation is many-to-many, since a ZIP code can cross
# several districts, and a district can cross several ZIP codes.

library(data.table)

district_zip <- fread(

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"Library","Phone","City","County","Zip","Latitude","Longitude","Street.Address"
"Ada Public Library","(218) 784-4480","Ada","Norman County","56510",47.3005818,-96.511009,"107 4th Avenue East"
"Adrian Branch Library","(507) 483-2541","Adrian","Nobles County","56110",43.6346305,-95.9331177,"214 Maine Avenue"
"Aitkin Public Library","(218) 927-2339","Aitkin","Aitkin County","56431",46.5326296,-93.7046646,"110 1st Avenue Northeast"
"Great River Regional Library - Albany","(320) 845-4843","Albany","Stearns County","56307",45.62953,-94.568801,"400 Railroad Avenue"
"Albert Lea Public Library","(507) 377-4350","Albert Lea","Freeborn County","56007",43.6501866,-93.3664103,"211 East Clark Street"
"Douglas County Library","(320) 762-3014","Alexandria","Douglas County","56308",45.8845467,-95.3792711,"720 Fillmore Street"
"Great River Regional Library - Annandale Branch","(320) 274-8448","Annandale","Wright County","55302",45.2638168,-94.1233636,"30 Cedar Street East"
"Rum River Library","(763) 576-4695","Anoka","Anoka Co

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import itertools
for a,e,f,h,i,l,t,n,o,w in itertools.permutations(range(10)):
    if h > 0 and t > 0 and f > 0 and w > 0:
        half = 1000*h + 100*a + 10*l + f
        fifth = 10100*f + i*1000 + t*10 + h
        tenth = 10010*t + 1000*e + n*100 + h
        whole = w*10000 + h*1000 + o*100 + l*10 + e
        if half + fifth + 3*tenth == whole:
            print "%s + %s + %s + %s + %s = %s" % (
                half, fifth, tenth, tenth, tenth, whole)

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function partition(arr, lo, hi) {
	var swap = function (i, j) {
		var tmp = arr[i];
		arr[i] = arr[j];
		arr[j] = tmp;
	}
	
	var pivot = arr[hi];
	var pivotLoc = lo;
	var i;