marked then $y_i = x_i$. Otherwise, $y_i = {\sqcup}$.

For example, consider the string $x = 01100110$. At the first iteration, the algorithm may mark the third and fourth bits to obtain $01\hat{1}\hat{0}0110$. Then, the second and fifth bits are marked 01100110. Lastly, the rightmost two bits are marked, and we obtain the marked string ${{\sf mark}}(x) = 0\hat{1}\hat{1}\hat{0}\hat{0}1\hat{1}\hat{0}$. Hence $y = {\sqcup}1100 {\sqcup}10$ and $C_y = \{ {\underline}{0}1100{\underline}{0}10, {\underline}{0}1100{\underline}{1}10, {\underline}{1}11100{\underline}{1}10 \}$.

(-5,1.7) node [$\hat{1}\hat{0}00$]{}; (-8+28/5,3.4) node [$\hat{1}\hat{0}01$]{}; (-5,5.1) node [$\hat{1}\hat{0}11$]{};

(-5+0.5,1.9) – (-8+28/5-0.3,3.1); (-8+28/5-0.3,3.7) – (-5+0.4,4.8);

(-5+8/3,1.7) node [$0\hat{1}\hat{0}0$]{}; (-8+42/5,3.4) node [$0\hat{1}\hat{0}1$]{}; (-5+8/3,5.1) node [$1\hat{1}\hat{0}1$]{};

(-5+8/3+0.5,1.9) – (-8+42/5-0.3,3.1); (-8+42/5-0.3,3.7) – (-5+8/3+0.4,4.8);

(-5+16/3,1.7) node [$00\hat{1}\hat{0}$]{}; (-8+56/5,3.4) node [$01\hat{1}\hat{0}$]{}; (-5+16/3,5.1) node [$11\hat{1}\hat{0}$]{};

(-5+16/3+0.5,1.9) – (-8+56/5-0.3,3.1); (-8+56/5-0.3,3.7) – (-5+16/3+0.4,4.8);

(-1,0) node [$0000$]{}; (3,1.7) node [$0001$]{}; (6,3.4) node [$0011$]{}; (3,5.1) node [$0111$]{}; (-1,6.8) node [$1111$]{};

(-0.5,0.1) – (2.4,1.4); (3.5,1.9) – (5.8,3.1); (5.8,3.7) – (3.6,4.8); (2.5,5.4) – (-0.5,6.7);

(-8+14/5,3.4) node [$\hat{1}\hat{0}\hat{1}\hat{0}$]{}; (-8,3.4) node [$\hat{1}\hat{1}\hat{0}\hat{0}$]{};

Readily, the algorithm induces a partition of ${\{0,1\}^{n}}$ to monotone symmetric chains. We stress that although the algorithm has some degree of freedom when choosing a $10$ pair out of, possibly, many pairs in a given iteration, the output of the algorithm, $y$, is independent of the specific choices that 