Lemma 3.

A sequence of rules is bijective and efficient, if s ð Þt <sup>k</sup> ¼ 1 ðcell ck active in step tÞ implies that.

$$\mathbf{i}.\qquad s\_{k-1}^{(t)} = \mathbf{0} \text{ and } s\_{k+1}^{(t)} = \mathbf{0} \text{ (both inactive)}.$$

ii. s ð Þ t�1 <sup>k</sup> ¼ 0 ðinactiveÞ, and.

iii. at least one of s ð Þ t�1 <sup>k</sup>�<sup>1</sup> and <sup>s</sup> ð Þ t�1 <sup>k</sup>þ<sup>1</sup> equals 1, active.
