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I have to find the corresponding Boolean polynomial to the binary string $1101$ $ 1001$.

To start, I think that I have to use a truth table to find the values for $x_1$, $x_2$, and $x_3$.

We get:

$0=0000$
$x_1=00001111$
$x_2=00110011$
$x_3=01010101$
and so:
$x_1+x_2=00111100$
$x_1+x_3=01011010$
$x_2+x_3=01100110$
$1=11111111$
$1+x_1=11110000$
$1+x_2=11001100$
$1+x_3=10101010$
$1+x_1+x_2=11000011$
$1+x_1+x_3=10100101$
$1+x_2+x_3=10011001$
$1+x_1+x_2+x_3=10010110$

I think that I have to find a linear combination of the above in order to get the desired binary string, however I'm not sure that this is the correct process. Any hints would be greatly appreciated!

Also does the space between the binary string I have to convert change the polynomials I should be using?

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\begin{eqnarray*} 1 &=& 11111111 \\ x_2 &=& 00110011 \\ x_2x_3 &=& 00010001 \\ x_1 x_3 &=& 00000101 \\ x_1 x_2 x_3 &=& 00000001 \\ \end{eqnarray*}

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  • $\begingroup$ Thank you for answering! Could you explain how you found these? For example, what would be the process to find $x_1x_2$? $\endgroup$ – Silvia Rossi Apr 13 '18 at 0:51
  • $\begingroup$ Assuming $x_1x_2$ is $x_1$ AND $x_2$, this will give the string $00000011$ and $+$ is EXOR. Using the linear combinations that you have at the moment will only generate strings with $4$ $0$'s and $4$ $1$'s. $\endgroup$ – Donald Splutterwit Apr 13 '18 at 0:57
  • $\begingroup$ I've noticed! What am I missing to generate strings with five 1's? $\endgroup$ – Silvia Rossi Apr 13 '18 at 1:15

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