Scrambler

`init()`

Scramble the vectors \(v_t^{(0)}\) and \(v_t^{(1)}\), returning the vectors \(X[n]\) and \(Y[n]\) scrambled. The scrambling process is given by the expression below.

\[\begin{equation} X[n] = \begin{cases} A, & \text{if } n = 0 \pmod{3} \\ B, & \text{if } n = 1 \pmod{3} \\ C, & \text{if } n = 2 \pmod{3} \end{cases} \quad Y[n] = \begin{cases} A, & \text{if } n = 0 \pmod{3} \\ B, & \text{if } n = 1 \pmod{3} \\ C, & \text{if } n = 2 \pmod{3} \end{cases} \end{equation}\]

Where

\(X[n]\) and \(Y[n]\): Scrambled output vectors.
\(A\), \(B\) and \(C\): Combination of bits of the input vectors \(v_t^{(0)}\) and \(v_t^{(1)}\).
\(n\): Index of the bit to be scrambled.

The scrambling process is illustrated by the block diagram below.

pageplot

Examples:

>>> import argos3
>>> import numpy as np
>>> 
>>> vt0 = np.random.randint(0, 2, 15)
>>> vt1 = np.random.randint(0, 2, 15)
>>> 
>>> idx_vt0 = [f"X{i+1}" for i in range(len(vt0))]
>>> idx_vt1 = [f"Y{i+1}" for i in range(len(vt1))]
>>> 
>>> scrambler = argos3.Scrambler()
>>> Xn, Yn = scrambler.scramble(vt0, vt1)
>>> idx_Xn, idx_Yn = scrambler.scramble(idx_vt0, idx_vt1)
>>> 
>>> print(vt0)
[0 1 1 1 1 0 0 0 1 0 0 0 0 0 1]
>>> print(vt1)
[1 0 0 1 1 1 0 1 0 1 0 0 0 1 1]
>>> print(idx_vt0)
['X1', 'X2', 'X3', 'X4', 'X5', 'X6', 'X7', 'X8', 'X9', 'X10', 'X11', 'X12', 'X13', 'X14', 'X15']
>>> print(idx_vt1)
['Y1', 'Y2', 'Y3', 'Y4', 'Y5', 'Y6', 'Y7', 'Y8', 'Y9', 'Y10', 'Y11', 'Y12', 'Y13', 'Y14', 'Y15']
>>> 
>>> print(idx_Xn)
['Y1', 'X2', 'Y2', 'Y4', 'X5', 'Y5', 'Y7', 'X8', 'Y8', 'Y10', 'X11', 'Y11', 'Y13', 'X14', 'Y14']
>>> print(idx_Yn)
['X1', 'X3', 'Y3', 'X4', 'X6', 'Y6', 'X7', 'X9', 'Y9', 'X10', 'X12', 'Y12', 'X13', 'X15', 'Y15']
>>>
>>> vt0_prime, vt1_prime = scrambler.descramble(Xn,Yn)
>>> idx_vt0_prime, idx_vt1_prime = scrambler.descramble(idx_Xn, idx_Yn)
>>> 
>>> print(idx_vt0_prime)
['X1', 'X2', 'X3', 'X4', 'X5', 'X6', 'X7', 'X8', 'X9', 'X10', 'X11', 'X12', 'X13', 'X14', 'X15']
>>> print(idx_vt1_prime)
['Y1', 'Y2', 'Y3', 'Y4', 'Y5', 'Y6', 'Y7', 'Y8', 'Y9', 'Y10', 'Y11', 'Y12', 'Y13', 'Y14', 'Y15']

Bitstream Plot Example:

Referência:
AS3-SP-516-274-CNES (seção 3.1.4.5)

`scramble(vt0, vt1)`

Receive the vectors \(v_t^{(0)}\) and \(v_t^{(1)}\) as input and return the vectors \(X[n]\) and \(Y[n]\) scrambled.

Parameters:

Name	Type	Description	Default
`vt0`	`ndarray`	Input vector \(v_t^{(0)}\).	required
`vt1`	`ndarray`	Input vector \(v_t^{(1)}\).	required

Returns:

Name	Type	Description
`X_scrambled`	`ndarray`	Scrambled vector \(X[n]\).
`Y_scrambled`	`ndarray`	Scrambled vector \(Y[n]\).

Raises:

Type	Description
`AssertionError`	If the vectors X and Y do not have the same length.

`unscramble(X_prime, Y_prime)`

Receive the vectors \(X'[n]\) and \(Y'[n]\) scrambled and return the vectors \(v_t^{(0)'}\) and \(v_t^{(1)'}\) restored.

\[\begin{equation} v_t^{(0)'} = \begin{cases} A, & \text{if } n = 0 \pmod{3} \\ B, & \text{if } n = 1 \pmod{3} \\ C, & \text{if } n = 2 \pmod{3} \end{cases}, \quad v_t^{(1)'} = \begin{cases} A, & \text{if } n = 0 \pmod{3} \\ B, & \text{if } n = 1 \pmod{3} \\ C, & \text{if } n = 2 \pmod{3} \end{cases} \text{ .} \label{eq:desembaralhador_Y} \end{equation}\]

Where

\(v_t^{(0)'}\) and \(v_t^{(1)'}\): Output vectors restored.
\(A\), \(B\) and \(C\): Combination of bits of the input vectors \(X'[n]\) and \(Y'[n]\) scrambled.
\(n\): Index of the bit to be scrambled.

The descrambling process is illustrated by the block diagram below.