Reducing the XOR's in Implementation Cost 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.
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.