# Prime Product Compression

## Bigger isn't always better

Posted by Ashwin Madavan on February 8, 2017

# Introduction

This project was inspired by the famous Lockers Riddle. Suppose that there are 1000 lockers in a school with 1000 students. If the first student opens each locker, the second student closes every second locker, the third student opens every third locker if it’s closed and closes it if it’s open, and so on and so forth; which lockers will be closed once every student has had their turn? The answer to this puzzle is relatively straightforward - every perfect-square numbered locker will be closed and all others will be open. However, we may generalize this notion of lockers and students to compression of arbitrary sequences of binary data.

We may think of these 1000 lockers as a sequence of bits, or bit string, in which the value of each bit indicates whether or not the corresponding locker is open or closed. Furthermore, we may think of these students as applying periodic bit flips to this bit string - the first student flips every bit, the second second flips every second bit, etc. We may then represent any bit-string as the sequence of periodic bit flips required to generate it. For example, the bit string 0101 may be generated by flipping every second bit of the zero bit string. Therefore, we may represent the four bit string using just the two bits required to encode the number 2.

# Algorithm

Given some bit string $$B = \{0, 1\}^N$$, allocate some $$B' = \{0\}^N$$. For each $$i = 1, \ldots, N$$, if $$B_i \neq B'_i$$ then for all $$\alpha i \leq N$$ for $$\alpha \in \mathbb{Z}^{+}$$ set $$B'_{\alpha i} = \neg B'_{\alpha i}$$ and append $$i$$ to some sequence $$X$$. This procedure corresponds to $$i^{th}$$ student opening every $$i^{th}$$ locker if it is closed and closing it if it is open. Clearly, the original bit string $$B$$ may be recovered by applying the periodic bit flips in $$X$$ to the zero bit string.

## Correctness

Theorem: Upon termination, $$B = B'$$. Proof: If $$B_i \neq B'_i$$, the algorithm flips every $$i^{th}$$ bit. Therefore, after the $$i^{th}$$ iteration of the algorithm $$B_i = B'_i$$. Because the algorithm iterates over increasing $$i$$, future iterations are guaranteed to preserve previous results. Clearly, if $$j > i$$ and $$B_j \neq B'_j$$ then $$\alpha j > i$$ because $$j > i$$ and $$\alpha \geq 1$$ so the $$i^{th}$$ bit is unaffected by the $$j^{th}$$ iteration. Therefore, upon termination, for each $$i$$, $$B_i = B'_i$$. This implies that $$B = B'$$ upon termination.

## Complexity

The $$i^{th}$$ iteration of the algorithm may require up to $$\frac{N}{i}$$ bit flips. Therefore, in the worst case, all $$N$$ iterations may require up to $$N \sum_{i=1}^{N} \frac{1}{i}$$ bit flips. Clearly, $$\sum_{i=1}^{N} \frac{1}{i}$$ is the $$N^{th}$$ partial sum of a harmonic series, which is approximately equal to O(ln N). Therefore, the algorithm is $$O(N ln N)$$.

## Implementation

The algorithm encodes a binary string as a sequence of periodic bit flips and is correctly able to decode this sequence of bit flips back into the original binary string. We may encode this sequence of periodic bit flips as a single prime number $$p = \prod prime(X_i)$$ where $$prime(n)$$ returns the $$n^{th}$$ prime number. Clearly, this number can be decomposed into its prime factors to recover the sequence of periodic bit flips. The implementation of the algorithm and the prime product representation is including below.

from itertools import izip
import random
import math

# List of the first 168 prime numbers (http://primos.mat.br/indexen.html).
primes = [
-1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79,
83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173,
179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269,
271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373,
379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467,
479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593,
599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691,
701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821,
823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937,
941, 947, 953, 967, 971, 977, 983, 991, 997
]

def simulate(n, length):
"""
Perform a simulation of n trials of bit strings of the specified length. Prints the
average compression ratio and the percentage of bit strings that were compressable.

:param n: Number of trials.
:param length: Length of bit string.
"""
def step():
# Construct and encode a random sequence of bits.
r = [bool(random.getrandbits(1)) for i in range(length)]
e = encode(r)

# Ensure that the decoding is consistent with the original.
assert r == decode(*e)

# Calculate the size of the encoding.
return math.ceil(math.log(abs(e) + 1, 2)) + 1

s = [step() for i in range(n+1)]
print("Average bits: " + str(sum(s) / len(s)))
print("Compressable: " + str(sum(map(lambda x: 1.0 if x < length else 0.0, s)) / len(s)))

def encode(bits):
"""
Encodes the specified bits as the product of the prime numbers corresponding to the
indices in a bit string of all zeroes or all ones (depending on which is smaller),
that need to be flipped to generate the bits.

:param bits: Bits to encode.
:return: Compressed bits.
"""
copy = [False] * len(bits)
flip = []

# Determine the necessary bit flips required to copy the bit string.
for i, (b, c) in enumerate(izip(bits, copy)):
if b != c:
flip.append(i)
for j in range(i, len(copy), i + 1):
copy[j] = not copy[j]

if not flip:
# If there are no bit flips, then return 0.
return (len(bits), 0)
else:
# Otherwise, return the product of the prime numbers associated with each index.
product = reduce(lambda x, y: x * y, map(lambda x: primes[x], flip))
return (len(bits), product)

def decode(length, product):
"""
Decode the prime product by finding its prime factorization and performing the
required bit flips on a bit string of all zeroes or all ones (depending on the
sign of the product).

:param length: Length of bit string.
:param product: Prime product.
:return: Decoded bits.
"""
# Find the prime factorization of the product to determine the bit flips.
if product == 0:
flip = []
else:
flip =  if product < 0 else []
flip.extend([i for i in range(1, length) if abs(product) % primes[i] == 0])

# Perform the bit flips in order on the bit string.
bits = [False] * length
for f in flip:
for i in range(f, length, f + 1):
bits[i] = not bits[i]
return bits


# Results

The algorithm proved to have relatively poor compression ratios for randomly generated bit strings. However, it is possible that for certain kinds of files this compression strategy may prove to be performant. More rigorous validation is definitely required, but initial results are not promising.

Cover photograph by Tech Crunch.