Heavy Hitters (The Britney Spears Algorithm)
June 2, 2015
We’ve looked at algorithms to process a stream of data in previous exercises (streaming median, streaming knapsack, reservoir sampling), and we also looked at an algorithm that finds the most frequent items (the “heavy hitters”) in a file. Today’s exercise combines the two to find the most frequent items in a stream; it’s called the “Britney Spears” algorithm because it is the algorithm used to cull the most frequently-seen items in Twitter feeds and other social media, and the pop starlet always seems to be on that list.
There are two similar algorithms used to find the n most frequently-appearing items in a stream. The first, due to Jayadev Misra and David Gries, uses an initially-empty associative array of items and their associated counts and processes each item in the input stream according to the following algorithm:
if the new item is already in the array
increment the associated count
else if there are less than n items in the array
add the new item to the array with a count of 1
else for each item in the array
decrement the associated count and
remove any items whose count becomes zero
The other algorithm, called the “space-saving” algorithm, is due to Ahmed Metwally, Divyakant Agrawal, and Amr El Abbadi. The first to parts of the algorithm are the same as the Misra-Gries algorithm. But, when a new item is found and there is no space to store it, the item already in the array that has the smallest associated count is replaced by the new item, and its count is incremented; thus, the counts are never reset, only the item is replaced.
With either algorithm, the array can be queried at any time to see the current list of most-frequent items in the array.
Your task is to implement the two heavy hitter algorithms. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
In Python3.
from collections import Counter from string import ascii_lowercase, punctuation from pprint import pprint TABLE =str.maketrans(ascii_lowercase, ascii_lowercase, ".,!?#)(*&^%$<>~") def misra_gries(stream, n): counts = Counter() for e in stream: if e in counts or len(counts) < n: counts[e] += 1 else: for c in list(counts.keys()): counts[c] -= 1 if counts[c] == 0: del counts[c] return counts.most_common(n) def metwally_agrawal_abaddi(stream, n): counts = Counter() for e in stream: if e in counts or len(counts) < n: counts[e] += 1 else: k, v = counts.most_common(n)[-1] del counts[k] counts[e] = v + 1 return counts.most_common(n) def wordgen(fname): with open(fname) as f: for line in f: words = line.lower().split() for w in words: yield w.translate(TABLE) def wordcounter(stream, n): return Counter(stream).most_common(n) N = 100 fname = "huckleberry_finn.txt" pprint(misra_gries(wordgen(fname), N)) pprint(metwally_agrawal_abaddi(wordgen(fname), N)) pprint(wordcounter(wordgen(fname), N))Also Python Counter. I grabbed some text from a web page I happened to have open and tested with random character frequencies: choose a character from the text 1000 times, track 6 counts.
from collections import Counter from random import choice text = ' '.join(''' if the new item is already in the array increment the associated count else if there are less than n items in the array add the new item to the array with a count of 1 else for each item in the array decrement the associated count and remove any items whose count becomes zero'''.split()) def first(n, stream): def dec(): for item in bag: bag[item] -= 1 return + bag bag = Counter() for item in stream: if item in bag: bag[item] += 1 elif len(bag) < n: bag[item] = 1 else: bag = dec() return set(bag) def other(n, stream): def rep(new): old, _ = bag.most_common()[-1] bag[new] = bag[old] + 1 del bag[old] bag = Counter() for item in stream: if item in bag: bag[item] += 1 elif len(bag) < n: bag[item] = 1 else: rep(item) return set(bag) for k in '---': print(other(6, (choice(text) for r in range(1000))), first(6, (choice(text) for r in range(1000)))) print() print(Counter(text).most_common(10))The frequency distribution goes rather flat after the two greatest hits, so there is variation, but those two, space and ‘e’, are in every sample. (The first algorithm results to the right, because their size varies and the other does not.)
{'c', 's', ' ', 'a', 'e', 'r'} {'e', 'o', ' ', 'n', 'i'} {' ', 'd', 'o', 'i', 'e', 'r'} {'s', ' ', 'u', 'y', 'e', 't'} {'c', 's', ' ', 'i', 'e', 't'} {'e', 't', 'r', ' '} [(' ', 54), ('e', 38), ('t', 25), ('a', 21), ('r', 16), ('n', 16), ('i', 15), ('s', 13), ('h', 13), ('o', 13)]Here’s a stab at it using Julia (just because).
function bscounter(iterable, size::Int, adjust::Function) counter = Dict{eltype(iterable), Int}() for ch in iterable if haskey(counter, ch) || length(counter) < size counter[ch] = get(counter, ch, 0) + 1 else adjust(counter, ch) end end return counter end function mg_rule(counter, elem) for k in keys(counter) counter[k] -= 1 if counter[k] == zero(k) delete!(counter, k) end end end function maa_rule(counter, elem) mink,minv = first(counter) for kv in counter k,v = kv if k < mink mink = k minv = v end end delete!(counter, mink) counter[elem] = minv + 1 end ### test text = "7675767476757636564525434132" # There are seven 7's, six 6's, five 5's, ... sz = 5 println("Using MG-rule: ", bscounter(text, sz, mg_rule)) println("Using MAA-rule: ", bscounter(text, sz, maa_rule)) #prints Using MG-rule: ['6'=>3,'7'=>4,'5'=>2,'4'=>1] Using MAA-rule: ['6'=>6,'7'=>7,'2'=>6,'5'=>5,'4'=>4]It looks like neither the MG-rule or the MAA-rule are certain to provide the correct results.
It’s easy to see that the associative array in question is functioning as a bounded multiset or bag, one that can contain only a given number of unique items. We can abstract the algorithms in these terms by saying that both algorithms start with:
If the item is not in the bag and the bag is not full, put the item in the bag.
Then the MG algorithm says:
Otherwise, remove one instance of every unique item in the bag.
The MAA algorithm, on the other hand, says:
Otherwise, remove all instances of the least common item in the bag and add the item to the bag.
The MG algorithm seems peculiar in that the new item is not in fact added to the bag, and indeed there is no guarantee that the next new item will be added either, since the bag may still be full; on the other hand, given an input where all items are distinct, a bag with maximum size n will be emptied completely every n items. The MAA algorithm always does add the new item to the bag, but is underspecified: it doesn’t say what to do if there is more than one “least common item”.
A solution in (Guile) scheme.