\section{Application in Python}
\label{sec:barcode_python}

In this section, we will implement barcode generation in Python. We will first program a "Code-39" encoder, then an EAN-8 and finally an EAN-13.

\subsection{Code-39}
\label{ssec:code39_py}

This type of code being just a matter of translating each character to a particular group of wide and narrow stripes, the implementation is quite simple.

We first create a dictionary holding the codes for each character.

\begin{minted}[frame=single]{python}
code39_dict = {
    "A": "100001001", "B": "001001001",
    "C": "101001000", "D": "000011001",
    "E": "100011000", "F": "001011000",
    "G": "000001101", "H": "100001100",
    "I": "001001100", "J": "000011100",
    "K": "100000011", "L": "001000011",
    "M": "101000010", "N": "000010011",
    "O": "100010010", "P": "001010010",
    "Q": "000000111", "R": "100000110",
    "S": "001000110", "T": "000010110",
    "U": "110000001", "V": "011000001",
    "W": "111000000", "X": "010010001",
    "Y": "110010000", "Z": "011010000",
    "0": "000110100", "1": "100100001",
    "2": "001100001", "3": "101100000",
    "4": "000110001", "5": "100110000",
    "6": "001110000", "7": "000100101",
    "8": "100100100", "9": "001100100",
    " ": "011000100", "-": "010000101",
    "$": "010101000", "%": "000101010",
    ".": "110000100", "/": "010100010",
    "+": "010001010", "*": "010010100"
}
\end{minted}

To convert a string, we map each character to its corresponding binary representation and join the resulting codes with "0" in between.

\begin{minted}[frame=single]{python}
def code39(text):
    text = text.upper()
    text = map(lambda c: code39_dict[c], text)
    return "0".join(text)
\end{minted}

We will also need a function to render the barcode. For this, we will use the Pygame module.

\begin{minted}[frame=single]{python}
def draw_barcode(barcode, win):
    barcode = list(map(int, barcode))
    width = win.get_width()*0.8
    height = win.get_height()*0.5
    thicks = sum(barcode)
    thins = len(barcode)-thicks
    bar_w = width/(thicks*2+thins)

    win.fill((255,255,255))
    x = win.get_width()*0.1
    y = win.get_height()*0.25

    for i, c in enumerate(barcode):
        w = 2*bar_w if c else bar_w
        if i%2 == 0:
            pygame.draw.rect(win, (0,0,0), [x, y, w, height])

        x += w
\end{minted}

The full python script can be found in appendix \ref{app:code39_py} or on \hreffn{https://github.com/LordBaryhobal/5D\_Heredero\_Louis\_TM2022/blob/main/python/code39.py}{GitHub}.

\subsection{EAN-8}
\label{ssec:ean8_py}

The first step to create an EAN-8 barcode is to compute the check digit with Luhn's formula.
To make this function also usable for EAN-13, we need to redefine the formula as such:

\begin{enumerate}
  \item Multiply each digit by the alternating factors 1 and 3 starting with 3 \textbf{from the end}.

  \item Add them together then take the modulo ten and subtract the result from 10.

  \item If the result is equal to 10, change it to 0.
\end{enumerate}

%Since this function will also be used for EAN-13, with only the factors changing, we will make it general enough.

In python, the function multiplies the i\textsuperscript{th} to last digit by: \[
  \text{factor} = 3 - (i\ mod\ 2) * 2
\]
which basicly is step 1 above.

\def\arraystretch{1.5}
\begin{table}[H]
  \centering
  \begin{tabu}{|[2pt]c|[2pt]c|c|c|c|c|c|[2pt]}
    \tabucline[2pt]{-}
    i & ... & 4 & 3 & 2 & 1 & 0 \\
    \hline
    factor & ... & 3 & 1 & 3 & 1 & 3 \\
    \tabucline[2pt]{-}
  \end{tabu}
  \caption{Python Luhn formula example}
  \label{tab:luhn_py_ex}
\end{table}
\def\arraystretch{1}

%\begin{minipage}{\linewidth}
\begin{minted}[frame=single]{python}
def luhn(digits):
    checksum = sum([
        digits[-i-1]*(3-i%2*2)
        for i in range(len(digits))
    ])
    ctrl_key = 10 - checksum%10
    if ctrl_key == 10:
        ctrl_key = 0

    return ctrl_key
\end{minted}
%\end{minipage}

Both code types also need the table of elements:

\begin{minted}[frame=single]{python}
A = [
    0b0001101,
    0b0011001,
    0b0010011,
    0b0111101,
    0b0100011,
    0b0110001,
    0b0101111,
    0b0111011,
    0b0110111,
    0b0001011
]

# XOR 0b1111111
C = list(map(lambda a: a^127, A))

# Reverse bit order
B = list(map(lambda c: int(f"{c:07b}"[::-1], 2), C))
\end{minted}

The following function converts a number to the list of its bits:
\begin{minted}[frame=single]{python}
def bin_list(n):
    return list(map(int, f"{n:07b}"))
\end{minted}

Finally, the encoding function:
\begin{minted}[frame=single]{python}
def ean8(digits):
    digits.append(luhn(digits))
    elmts = []

    elmts += [1,0,1] #delimiter
    for digit in digits[:4]:
        elmts += bin_list(A[digit])

    elmts += [0,1,0,1,0] #middle delimiter
    for digit in digits[4:]:
        elmts += bin_list(C[digit])

    elmts += [1,0,1] #delimiter
    return elmts
\end{minted}

We will use a similar function as in \autoref{ssec:code39_py} to render the barcode

\begin{minted}[frame=single]{python}
def draw_barcode(barcode, win):
    width = win.get_width()*0.8
    height = win.get_height()*0.5
    bar_w = width/len(barcode)

    win.fill((255,255,255))
    x = win.get_width()*0.1
    y = win.get_height()*0.25

    for c in barcode:
        if c:
            pygame.draw.rect(win, (0,0,0), [x, y, bar_w, height])

        x += bar_w
\end{minted}

The full python script can be found in appendix \ref{app:ean_py} or on \hreffn{https://github.com/LordBaryhobal/5D\_Heredero\_Louis\_TM2022/blob/main/python/ean.py}{GitHub}

\subsection{EAN-13}
\label{ssec:ean13_py}

The main difference with EAN-8 is the encoding of the first digit, using an A/B pattern. We will create a list of these patterns:

\begin{minted}[frame=single]{python}
ean13_patterns = [
    "AAAAAA",
    "AABABB",
    "AABBAB",
    "AABBBA",
    "ABAABB",
    "ABBAAB",
    "ABBBAA",
    "ABABAB",
    "ABABBA",
    "ABBABA"
]
\end{minted}

And the appropriate encoding function:

\begin{minted}[frame=single]{python}
def ean13(digits):
    pattern = ean13_patterns[digits[0]]
    digits.append(luhn(digits))
    elmts = []

    elmts += [1,0,1] #delimiter
    for d in range(1,7):
        _ = A if pattern[d-1] == "A" else B
        digit = digits[d]
        elmts += bin_list(_[digit])

    elmts += [0,1,0,1,0] #middle delimiter
    for digit in digits[7:]:
        elmts += bin_list(C[digit])

    elmts += [1,0,1] #delimiter
    return elmts
\end{minted}

The full python script can be found in appendix \ref{app:ean_py} or on \hreffn{https://github.com/LordBaryhobal/5D\_Heredero\_Louis\_TM2022/blob/main/python/ean.py}{GitHub}