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Suppose that we want to implement the following equations in binary finite field.

$$ \begin{array}{l}\tag{1} y_{{1}}=x_{{1}}\oplus x_{{2}}\oplus x_{{7}}\oplus x_{{8}},\\ y_{{2}}=x_{{1}}\oplus x_{{2}}\oplus x_{{6}}\oplus x_{{8}},\\ y_{{3}}=x_{{1}}\oplus x_{{2}}\oplus x_{{7}},\\ y_{{4}}=x_{{2}}\oplus x_{{8}}. \end{array} $$

My question: Is it possible to implement the given equations in $(1)$ with five $\operatorname{XOR}$?

My try results in six $\operatorname{XOR}$ for implementation cost. For example;

$$ \begin{array}{l}\tag{2} u_1=x_1\oplus x_2,\\ u_2=x_2\oplus x_8=y_4\\ u_3=x_6\oplus x_8,\\ u_4=u_1\oplus x_7=y_3\\ u_5=u_4\oplus x_8=y_1\\ u_6=u_1\oplus u_3=y_2\\ \end{array} $$

Thanks for any suggestions.

Edit: As Dear fgrieu said with greedy thoughts, the answer is as follows: $$ \begin{array}{l}\tag{3} u_1=x_2\oplus x_8=y_4\\ u_2=x_1\oplus u_1\\ u_3=u_2\oplus x_7=y_1 \\ u_4=u_2\oplus x_6=y_2\\ u_5=u_3\oplus x_8=y_3 \end{array} $$

Thanks to greedy thoughts.

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  • $\begingroup$ This might better suits for CS? At the and this has graph based solutions? See MDS matrixes solved with A* algorithm. $\endgroup$ – kelalaka Jan 25 at 13:13
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    $\begingroup$ Instead of editing your question to include the answer, it would be better to post it as an answer of your own. $\endgroup$ – Ilmari Karonen Jan 25 at 21:59
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Yes. I have a solution with 5 $\oplus$.

Hint 1: That would not work with $*$ (ordinary multiplication, or even multiplication in $\Bbb Z^*_p$) where there is $\oplus$, because we must use a property of $\oplus$ that $*$ does not have.

Hint 2: The right path is greedy.

Hint 3: It's key to compute $y_3$ late, with just one extra step.

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  • $\begingroup$ @fgrieu Is there an algorithm for this kind of problem ? $\endgroup$ – Bissi Jan 25 at 15:48
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    $\begingroup$ @Bissi: the one I know is exploring a tree of possibilities, with trimming when the current node can't lead to a solution because too many remain to be found (e.g. when we want $n=5$ operations at most, and need to produce $m=4$ different quantities not in the givens, it is obvious that we can afford only $n-m=1$ operation not leading to a quantity to be produced). Note: greedy is the common term for exploration strategies that explore first those branches leading to a partial desired result. In many but not all cases, that's optimum. $\endgroup$ – fgrieu Jan 25 at 17:40

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