$$ \newcommand \FilterTimeout {\mathrm{FilterTimeout}} \newcommand \Cert {\mathit{cert}} \newcommand \Soft {\mathit{soft}} \newcommand \Vote {\mathrm{Vote}} \newcommand \Bundle {\mathrm{Bundle}} $$

Filtering

On observing a timeout event of \( \FilterTimeout(p) \) (where \( \mu = (H, H’, l, p_\mu) = \mu(S, r, p) \)),

• if \( \mu \neq \bot \) and if

  • \( p_\mu = p \) or
  • there exists some \( s > \Cert \) such that \( \Bundle(r, p-1, s, \mu) \) was observed then the player broadcasts \( \Vote(I, r, p, \Soft, \mu) \).

• if there exists some \( s_0 > \Cert \) such that \( \Bundle(r, p-1, s_0, \bar{v}) \) was observed and there exists no \( s_1 > \Cert \) such that \( \Bundle(r, p-1, s_1, \bot) \) was observed, then the player broadcasts* \( \Vote(I, r, p, \Soft, \bar{v}) \).

• otherwise, the player does nothing.

For a detailed overview of how the filtering step may be implemented, refer to the Algorand ABFT non-normative section.

In other words, in the first case above,

$$ N(S, L, t(\FilterTimeout(p), p)) = (S, L, \Vote(I, r, p, \Soft, \mu)); $$

while in the second case above,

$$ N(S, L, t(\FilterTimeout(p), p)) = (S, L, \Vote(I, r, p, \Soft, \bar{v})); $$

and if neither case is true,

$$ N(S, L, t(\FilterTimeout(p), p)) = (S, L, \epsilon). $$