def lempel_ziv_complexity(binary_sequence):
    """Lempel-Ziv complexity for a binary sequence, in simple Python code."""
    u, v, w = 0, 1, 1
    v_max = 1
    length = len(binary_sequence)
    complexity = 1
    while True:
        if binary_sequence[u + v - 1] == binary_sequence[w + v - 1]:
            v += 1
            if w + v >= length:
                complexity += 1
                break
        else:
            if v > v_max:
                v_max = v
            u += 1
            if u == w:
                complexity += 1
                w += v_max
                if w > length:
                    break
                else:
                    u = 0
                    v = 1
                    v_max = 1
            else:
                v = 1
    return complexity

"""
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 / 1110 / 1100 / 0010
6
```

Marking in the different substrings, this sequence $s$ has complexity $\mathrm{Lempel}$-$\mathrm{Ziv}(s) = 6$ because $s = 1001111011000010 = 1 / 0 / 01 / 1110 / 1100 / 0010$.

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

Other examples:

```python
>>> lempel_ziv_complexity('1010101010101010')  # 1 / 0 / 10
3
>>> lempel_ziv_complexity('1001111011000010000010')  # 1 / 0 / 01 / 1110 / 1100 / 0010 / 000 / 010
7
>>> lempel_ziv_complexity('100111101100001000001010')  # 1 / 0 / 01 / 1110 / 1100 / 0010 / 000 / 010 / 10
8
```

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


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

%timeit lempel_ziv_complexity(s)

lempel_ziv_complexity('1010101010101010')  # 1 / 0 / 10

lempel_ziv_complexity('1001111011000010000010')  # 1 / 0 / 01 / 1110

lempel_ziv_complexity('100111101100001000001010')  # 1 / 0 / 01 / 1110 / 1100 / 0010 / 000 / 010 / 10

%timeit lempel_ziv_complexity('100111101100001000001010')

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


%load_ext cython

%%cython
from __future__ import division
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 binary_sequence not None):
    """Lempel-Ziv complexity for a binary sequence, in simple Cython code (C extension)."""
    cdef DTYPE_t u = 0
    cdef DTYPE_t v = 1
    cdef DTYPE_t w = 1
    cdef DTYPE_t v_max = 1
    cdef DTYPE_t length = len(binary_sequence)
    cdef DTYPE_t complexity = 1
    # that was the only needed part, typing statically all the variables
    while True:
        if binary_sequence[u + v - 1] == binary_sequence[w + v - 1]:
            v += 1
            if w + v >= length:
                complexity += 1
                break
        else:
            if v > v_max:
                v_max = v
            u += 1
            if u == w:
                complexity += 1
                w += v_max
                if w > length:
                    break
                else:
                    u = 0
                    v = 1
                    v_max = 1
            else:
                v = 1
    return complexity

"""
Explanation: We can start to see that the time complexity of this function seems to grow exponentially as the complexity 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 / 1110 / 1100 / 0010

%timeit lempel_ziv_complexity_cython(s)

lempel_ziv_complexity_cython('1010101010101010')  # 1 / 0 / 10

lempel_ziv_complexity_cython('1001111011000010000010')  # 1 / 0 / 01 / 1110

lempel_ziv_complexity_cython('100111101100001000001010')  # 1 / 0 / 01 / 1110 / 1100 / 0010 / 000 / 010 / 10

%timeit lempel_ziv_complexity_cython('100111101100001000001010')

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


from numba import jit

@jit("int32(boolean[:])")
def lempel_ziv_complexity_numba_x(binary_sequence):
    """Lempel-Ziv complexity for a binary sequence, in Python code using numba.jit() for automatic speedup (hopefully)."""
    u, v, w = 0, 1, 1
    v_max = 1
    length = len(binary_sequence)
    complexity = 1
    while True:
        if binary_sequence[u + v - 1] == binary_sequence[w + v - 1]:
            v += 1
            if w + v >= length:
                complexity += 1
                break
        else:
            if v > v_max:
                v_max = v
            u += 1
            if u == w:
                complexity += 1
                w += v_max
                if w > length:
                    break
                else:
                    u = 0
                    v = 1
                    v_max = 1
            else:
                v = 1
    return complexity

def str_to_numpy(s):
    """str to np.array of bool"""
    return np.array([int(i) for i in s], dtype=np.bool)

def lempel_ziv_complexity_numba(s):
    return lempel_ziv_complexity_numba_x(str_to_numpy(s))

"""
Explanation: $\implies$ Yay! It seems faster indeed!


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


str_to_numpy(s)

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

lempel_ziv_complexity_numba('1001111011000010000010')  # 1 / 0 / 01 / 1110

lempel_ziv_complexity_numba('100111101100001000001010')  # 1 / 0 / 01 / 1110 / 1100 / 0010 / 000 / 010 / 10

%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)

"""
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
"""


x = [10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480]
y = [18, 30, 55, 107, 205, 471, 977, 2270, 5970, 17300, 56600, 185000]

plt.figure()
plt.plot(x, y, '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()
plt.semilogx(x, y, '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, semilogx scale")
plt.show()

plt.figure()
plt.semilogy(x, y, '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, semilogy scale")
plt.show()

plt.figure()
plt.loglog(x, y, '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: It's durty, but let us capture manually the times given by the experiments above.
End of explanation
"""


%%time
%%script julia

"""Lempel-Ziv complexity for a binary sequence, in simple Julia code."""
function lempel_ziv_complexity(binary_sequence)
    u, v, w = 0, 1, 1
    v_max = 1
    size = length(binary_sequence)
    complexity = 1
    while true
        if binary_sequence[u + v] == binary_sequence[w + v]
            v += 1
            if w + v >= size
                complexity += 1
                break
            end
        else
            if v > v_max
                v_max = v
            end
            u += 1
            if u == w
                complexity += 1
                w += v_max
                if w > size
                    break
                else
                    u = 0
                    v = 1
                    v_max = 1
                end
            else
                v = 1
            end
        end
    end
    return complexity
end

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

M = 100;
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.
Disclaimer: I am still learning Julia!
End of explanation
"""


%%time
%%pypy

def lempel_ziv_complexity(binary_sequence):
    """Lempel-Ziv complexity for a binary sequence, in simple Python code."""
    u, v, w = 0, 1, 1
    v_max = 1
    length = len(binary_sequence)
    complexity = 1
    while True:
        if binary_sequence[u + v - 1] == binary_sequence[w + v - 1]:
            v += 1
            if w + v >= length:
                complexity += 1
                break
        else:
            if v > v_max:
                v_max = v
            u += 1
            if u == w:
                complexity += 1
                w += v_max
                if w > length:
                    break
                else:
                    u = 0
                    v = 1
                    v_max = 1
            else:
                v = 1
    return complexity

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

from random import random

M = 100
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
"""


