def lempel_ziv_complexity(sequence):
    """Lempel-Ziv complexity for a binary sequence, in simple Python code."""
    sub_strings = set()
    n = len(sequence)
    ind = 0
    inc = 1
    # this while loop runs at most n times
    while True:
        if ind + inc > len(sequence):
            break
        # this can take some time, takes O(inc)
        sub_str = sequence[ind : ind + inc]
        # and this also, takes a O(log |size set|) in worst case
        # max value for inc = n / size set at the end
        # so worst case is that the set contains sub strings of the same size
        # and the worst loop takes a O(n / |S| * log(|S|))
        # ==> so if n/|S| is constant, it gives O(n log(n)) at the end
        # but if n/|S| = O(n) then it gives O(n^2)
        if sub_str in sub_strings:
            inc += 1
        else:
            sub_strings.add(sub_str)
            ind += inc
            inc = 1
    return len(sub_strings)

"""
Explanation: Table of Contents
<p><div class="lev1 toc-item"><a href="#Short-study-of-the-Lempel-Ziv-complexity" data-toc-modified-id="Short-study-of-the-Lempel-Ziv-complexity-1"><span class="toc-item-num">1&nbsp;&nbsp;</span>Short study of the Lempel-Ziv complexity</a></div><div class="lev2 toc-item"><a href="#Short-definition" data-toc-modified-id="Short-definition-11"><span class="toc-item-num">1.1&nbsp;&nbsp;</span>Short definition</a></div><div class="lev2 toc-item"><a href="#Python-implementation" data-toc-modified-id="Python-implementation-12"><span class="toc-item-num">1.2&nbsp;&nbsp;</span>Python implementation</a></div><div class="lev2 toc-item"><a href="#Tests-(1/2)" data-toc-modified-id="Tests-(1/2)-13"><span class="toc-item-num">1.3&nbsp;&nbsp;</span>Tests (1/2)</a></div><div class="lev2 toc-item"><a href="#Cython-implementation" data-toc-modified-id="Cython-implementation-14"><span class="toc-item-num">1.4&nbsp;&nbsp;</span>Cython implementation</a></div><div class="lev2 toc-item"><a href="#Numba-implementation" data-toc-modified-id="Numba-implementation-15"><span class="toc-item-num">1.5&nbsp;&nbsp;</span>Numba implementation</a></div><div class="lev2 toc-item"><a href="#Tests-(2/2)" data-toc-modified-id="Tests-(2/2)-16"><span class="toc-item-num">1.6&nbsp;&nbsp;</span>Tests (2/2)</a></div><div class="lev2 toc-item"><a href="#Benchmarks" data-toc-modified-id="Benchmarks-17"><span class="toc-item-num">1.7&nbsp;&nbsp;</span>Benchmarks</a></div><div class="lev2 toc-item"><a href="#Complexity-?" data-toc-modified-id="Complexity-?-18"><span class="toc-item-num">1.8&nbsp;&nbsp;</span>Complexity ?</a></div><div class="lev2 toc-item"><a href="#Conclusion" data-toc-modified-id="Conclusion-19"><span class="toc-item-num">1.9&nbsp;&nbsp;</span>Conclusion</a></div><div class="lev2 toc-item"><a href="#(Experimental)-Julia-implementation" data-toc-modified-id="(Experimental)-Julia-implementation-110"><span class="toc-item-num">1.10&nbsp;&nbsp;</span>(Experimental) <a href="http://julialang.org" target="_blank">Julia</a> implementation</a></div><div class="lev2 toc-item"><a href="#Ending-notes" data-toc-modified-id="Ending-notes-111"><span class="toc-item-num">1.11&nbsp;&nbsp;</span>Ending notes</a></div>

# Short study of the Lempel-Ziv complexity

In this short [Jupyter notebook](https://www.Jupyter.org/) aims at defining and explaining the [Lempel-Ziv complexity](https://en.wikipedia.org/wiki/Lempel-Ziv_complexity).

[I](http://perso.crans.org/besson/) will give examples, and benchmarks of different implementations.

- **Reference:** Abraham Lempel and Jacob Ziv, *« On the Complexity of Finite Sequences »*, IEEE Trans. on Information Theory, January 1976, p. 75–81, vol. 22, n°1.

----
## Short definition
The Lempel-Ziv complexity is defined as the number of different substrings encountered as the stream is viewed from begining to the end.

As an example:

```python
>>> s = '1001111011000010'
>>> lempel_ziv_complexity(s)  # 1 / 0 / 01 / 11 / 10 / 110 / 00 / 010
8
```

Marking in the different substrings, this sequence $s$ has complexity $\mathrm{Lempel}$-$\mathrm{Ziv}(s) = 6$ because $s = 1001111011000010 = 1 / 0 / 01 / 11 / 10 / 110 / 00 / 010$.

- See the page https://en.wikipedia.org/wiki/Lempel-Ziv_complexity for more details.

Other examples:

```python
>>> lempel_ziv_complexity('1010101010101010')  # 1, 0, 10, 101, 01, 010, 1010
7
>>> lempel_ziv_complexity('1001111011000010000010')  # 1, 0, 01, 11, 10, 110, 00, 010, 000
9
>>> lempel_ziv_complexity('100111101100001000001010')  # 1, 0, 01, 11, 10, 110, 00, 010, 000, 0101
10
```

----
## Python implementation
End of explanation
"""


s = '1001111011000010'
lempel_ziv_complexity(s)  # 1 / 0 / 01 / 11 / 10 / 110 / 00 / 010

%timeit lempel_ziv_complexity(s)

lempel_ziv_complexity('1010101010101010')  # 1, 0, 10, 101, 01, 010, 1010

lempel_ziv_complexity('1001111011000010000010')  # 1, 0, 01, 11, 10, 110, 00, 010, 000

lempel_ziv_complexity('100111101100001000001010')  # 1, 0, 01, 11, 10, 110, 00, 010, 000, 0101

%timeit lempel_ziv_complexity('100111101100001000001010')

import random

def random_string(size, alphabet="ABCDEFGHIJKLMNOPQRSTUVWXYZ"):
    return "".join(random.choices(alphabet, k=size))

def random_binary_sequence(size):
    return random_string(size, alphabet="01")

random_string(100)
random_binary_sequence(100)

for (r, name) in zip(
    [random_string, random_binary_sequence],
    ["random strings in A..Z", "random binary sequences"]
    ):
    print("\nFor {}...".format(name))
    for n in [10, 100, 1000, 10000, 100000]:
        print("  of sizes {}, Lempel-Ziv complexity runs in:".format(n))
        %timeit lempel_ziv_complexity(r(n))

"""
Explanation: Tests (1/2)
End of explanation
"""


%load_ext cython

%%cython
import cython

ctypedef unsigned int DTYPE_t

@cython.boundscheck(False) # turn off bounds-checking for entire function, quicker but less safe
def lempel_ziv_complexity_cython(str sequence not None):
    """Lempel-Ziv complexity for a string, in simple Cython code (C extension)."""
    
    cdef set sub_strings = set()
    cdef str sub_str = ""
    cdef DTYPE_t n = len(sequence)
    cdef DTYPE_t ind = 0
    cdef DTYPE_t inc = 1
    while True:
        if ind + inc > len(sequence):
            break
        sub_str = sequence[ind : ind + inc]
        if sub_str in sub_strings:
            inc += 1
        else:
            sub_strings.add(sub_str)
            ind += inc
            inc = 1
    return len(sub_strings)

"""
Explanation: We can start to see that the time complexity of this function seems to grow linearly as the size grows.

Cython implementation
As this blog post explains it, we can easily try to use Cython in a notebook cell.

See the Cython documentation for more information.
End of explanation
"""


s = '1001111011000010'
lempel_ziv_complexity_cython(s)  # 1 / 0 / 01 / 11 / 10 / 110 / 00 / 010

%timeit lempel_ziv_complexity(s)
%timeit lempel_ziv_complexity_cython(s)

lempel_ziv_complexity_cython('1010101010101010')  # 1, 0, 10, 101, 01, 010, 1010

lempel_ziv_complexity_cython('1001111011000010000010')  # 1, 0, 01, 11, 10, 110, 00, 010, 000

lempel_ziv_complexity_cython('100111101100001000001010')  # 1, 0, 01, 11, 10, 110, 00, 010, 000, 0101

"""
Explanation: Let try it!
End of explanation
"""


for (r, name) in zip(
    [random_string, random_binary_sequence],
    ["random strings in A..Z", "random binary sequences"]
    ):
    print("\nFor {}...".format(name))
    for n in [10, 100, 1000, 10000, 100000]:
        print("  of sizes {}, Lempel-Ziv complexity in Cython runs in:".format(n))
        %timeit lempel_ziv_complexity_cython(r(n))

"""
Explanation: Now for a test of the speed?
End of explanation
"""


from numba import jit

@jit
def lempel_ziv_complexity_numba(sequence : str) -> int:
    """Lempel-Ziv complexity for a sequence, in Python code using numba.jit() for automatic speedup (hopefully)."""

    sub_strings = set()
    n : int= len(sequence)

    ind : int = 0
    inc : int = 1
    while True:
        if ind + inc > len(sequence):
            break
        sub_str : str = sequence[ind : ind + inc]
        if sub_str in sub_strings:
            inc += 1
        else:
            sub_strings.add(sub_str)
            ind += inc
            inc = 1
    return len(sub_strings)

"""
Explanation: $\implies$ Yay! It seems faster indeed! but only x2 times faster...


Numba implementation
As this blog post explains it, we can also try to use Numba in a notebook cell.
End of explanation
"""


s = '1001111011000010'
lempel_ziv_complexity_numba(s)  # 1 / 0 / 01 / 1110 / 1100 / 0010

%timeit lempel_ziv_complexity_numba(s)

lempel_ziv_complexity_numba('1010101010101010')  # 1, 0, 10, 101, 01, 010, 1010

lempel_ziv_complexity_numba('1001111011000010000010')  # 1, 0, 01, 11, 10, 110, 00, 010, 000
    9

lempel_ziv_complexity_numba('100111101100001000001010')  # 1, 0, 01, 11, 10, 110, 00, 010, 000, 0101

%timeit lempel_ziv_complexity_numba('100111101100001000001010')

"""
Explanation: Let try it!
End of explanation
"""


from numpy.random import binomial

def bernoulli(p, size=1):
    """One or more samples from a Bernoulli of probability p."""
    return binomial(1, p, size)

bernoulli(0.5, 20)

"""
Explanation: $\implies$ Well... It doesn't seem that much faster from the naive Python code.
We specified the signature when calling @numba.jit, and used the more appropriate data structure (string is probably the smaller, numpy array are probably faster).
But even these tricks didn't help that much.
I tested, and without specifying the signature, the fastest approach is using string, compared to using lists or numpy arrays.
Note that the @jit-powered function is compiled at runtime when first being called, so the signature used for the first call is determining the signature used by the compile function


Tests (2/2)
To test more robustly, let us generate some (uniformly) random binary sequences.
End of explanation
"""


''.join(str(i) for i in bernoulli(0.5, 20))

def random_binary_sequence(n, p=0.5):
    """Uniform random binary sequence of size n, with rate of 0/1 being p."""
    return ''.join(str(i) for i in bernoulli(p, n))

random_binary_sequence(50)
random_binary_sequence(50, p=0.1)
random_binary_sequence(50, p=0.25)
random_binary_sequence(50, p=0.5)
random_binary_sequence(50, p=0.75)
random_binary_sequence(50, p=0.9)

"""
Explanation: That's probably not optimal, but we can generate a string with:
End of explanation
"""


def tests_3_functions(n, p=0.5, debug=True):
    s = random_binary_sequence(n, p=p)
    c1 = lempel_ziv_complexity(s)
    if debug:
        print("Sequence s = {} ==> complexity C = {}".format(s, c1))
    c2 = lempel_ziv_complexity_cython(s)
    c3 = lempel_ziv_complexity_numba(s)
    assert c1 == c2 == c3, "Error: the sequence {} gave different values of the Lempel-Ziv complexity from 3 functions ({}, {}, {})...".format(s, c1, c2, c3)
    return c1

tests_3_functions(5)

tests_3_functions(20)

tests_3_functions(50)

tests_3_functions(500)

tests_3_functions(5000)

"""
Explanation: And so, this function can test to check that the three implementations (naive, Cython-powered, Numba-powered) always give the same result.
End of explanation
"""


%timeit lempel_ziv_complexity('100111101100001000001010')
%timeit lempel_ziv_complexity_cython('100111101100001000001010')
%timeit lempel_ziv_complexity_numba('100111101100001000001010')

%timeit lempel_ziv_complexity('10011110110000100000101000100100101010010111111011001111111110101001010110101010')
%timeit lempel_ziv_complexity_cython('10011110110000100000101000100100101010010111111011001111111110101001010110101010')
%timeit lempel_ziv_complexity_numba('10011110110000100000101000100100101010010111111011001111111110101001010110101010')

"""
Explanation: Benchmarks
On two example of strings (binary sequences), we can compare our three implementation.
End of explanation
"""


%timeit tests_3_functions(10, debug=False)
%timeit tests_3_functions(20, debug=False)
%timeit tests_3_functions(40, debug=False)
%timeit tests_3_functions(80, debug=False)
%timeit tests_3_functions(160, debug=False)
%timeit tests_3_functions(320, debug=False)

def test_cython(n):
    s = random_binary_sequence(n)
    c = lempel_ziv_complexity_cython(s)
    return c

%timeit test_cython(10)
%timeit test_cython(20)
%timeit test_cython(40)
%timeit test_cython(80)
%timeit test_cython(160)
%timeit test_cython(320)

%timeit test_cython(640)
%timeit test_cython(1280)
%timeit test_cython(2560)
%timeit test_cython(5120)

%timeit test_cython(10240)
%timeit test_cython(20480)

"""
Explanation: Let check the time used by all the three functions, for longer and longer sequences:
End of explanation
"""


import matplotlib.pyplot as plt
import seaborn as sns
%matplotlib inline
sns.set(context="notebook", style="darkgrid", palette="hls", font="sans-serif", font_scale=1.4)

import numpy as np
import timeit

sizes = np.array(np.trunc(np.logspace(1, 6, 30)), dtype=int)

times = np.array([
    timeit.timeit(
        stmt="lempel_ziv_complexity_cython(random_string({}))".format(n),
        globals=globals(),
        number=10,
    )
    for n in sizes
])

plt.figure(figsize=(15, 10))
plt.plot(sizes, times, 'o-')
plt.xlabel("Length $n$ of the binary sequence $S$")
plt.ylabel(r"Time in $\mu\;\mathrm{s}$")
plt.title("Time complexity of Lempel-Ziv complexity")
plt.show()

plt.figure(figsize=(15, 10))
plt.loglog(sizes, times, 'o-')
plt.xlabel("Length $n$ of the binary sequence $S$")
plt.ylabel(r"Time in $\mu\;\mathrm{s}$")
plt.title("Time complexity of Lempel-Ziv complexity, loglog scale")
plt.show()

"""
Explanation: Complexity ?
$\implies$ The function lempel_ziv_complexity_cython seems to be indeed (almost) linear in $n$, the length of the binary sequence $S$.
But let check more precisely, as it could also have a complexity of $\mathcal{O}(n \log n)$.
End of explanation
"""


%%time
%%script julia

"""Lempel-Ziv complexity for a sequence, in simple Julia code."""
function lempel_ziv_complexity(sequence)
    sub_strings = Set()
    n = length(sequence)

    ind = 1
    inc = 1
    while true
        if ind + inc > n
            break
        end
        sub_str = sequence[ind : ind + inc]
        if sub_str in sub_strings
            inc += 1
        else
            push!(sub_strings, sub_str)
            ind += inc
            inc = 1
        end
    end
    return length(sub_strings)
end

s = "1001111011000010"
lempel_ziv_complexity(s)  # 1 / 0 / 01 / 1110 / 1100 / 0010

M = 1000;
N = 10000;
for _ in 1:M
    s = join(rand(0:1, N));
    lempel_ziv_complexity(s);
end
lempel_ziv_complexity(s)  # 1 / 0 / 01 / 1110 / 1100 / 0010

"""
Explanation: It is linear in $\log\log$ scale, so indeed the algorithm seems to have a linear complexity.
To sum-up, for a sequence $S$ of length $n$, it takes $\mathcal{O}(n)$ basic operations to compute its Lempel-Ziv complexity $\mathrm{Lempel}-\mathrm{Ziv}(S)$.

Conclusion


The Lempel-Ziv complexity is not too hard to implement, and it indeed represents a certain complexity of a binary sequence, capturing the regularity and reproducibility of the sequence.


Using the Cython was quite useful to have a $\simeq \times 100$ speed up on our manual naive implementation !


The algorithm is not easy to analyze, we have a trivial $\mathcal{O}(n^2)$ bound but experiments showed it is more likely to be $\mathcal{O}(n \log n)$ in the worst case, and $\mathcal{O}(n)$ in practice for "not too complicated sequences" (or in average, for random sequences).



(Experimental) Julia implementation
I want to (quickly) try to see if I can use Julia to write a faster version of this function.
See issue #1.
End of explanation
"""


%%time
%%pypy

def lempel_ziv_complexity(sequence):
    """Lempel-Ziv complexity for a binary sequence, in simple Python code."""
    sub_strings = set()
    n = len(sequence)

    ind = 0
    inc = 1
    while True:
        if ind + inc > len(sequence):
            break
        sub_str = sequence[ind : ind + inc]
        if sub_str in sub_strings:
            inc += 1
        else:
            sub_strings.add(sub_str)
            ind += inc
            inc = 1
    return len(sub_strings)

s = "1001111011000010"
lempel_ziv_complexity(s) # 1 / 0 / 01 / 11 / 10 / 110 / 00 / 010

from random import random

M = 1000
N = 10000
for _ in range(M):
    s = ''.join(str(int(random() < 0.5)) for _ in range(N))
    lempel_ziv_complexity(s)

"""
Explanation: And to compare it fairly, let us use Pypy for comparison.
End of explanation
"""


