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Heuristic Strategy for the Practical Case

In previous sections an $\Omega(n^2)$ lower bound on the number of experiments even for the case of in-degree 2 was shown. Since O(n2) experiments are impossible for yeast (n=6200), this suggests that we cannot expect a single strategy identification method for the gene regulatory network of yeast and that the methods from the previous section should be employed only for determining the local network structure. This leads us to develop a strategy by which we can identify as many parts as possible using O(n) experiments. In such a case, we should find a set of edges E' such that $E' \subseteq E$. That is , we should find a set of edges not including false positive edges. Let InfFrom(x) denote the set of nodes influenced by x (excluding x itself). Note that InfFrom(x) is not necessarily determined uniquely because of inconsistent nodes. Only such nodes x that InfFrom(x) is determined uniquely will be considered.
  
Figure: These cases satisfy $InfFrom(b) \cup \{b\} = InfFrom(a)$ but do include cycles passing through node b.
\scalebox{0.85}{\includegraphics {lec14_fig/lect14_5.ps}}

Proposition   If $InfFrom(b) \cup \{b\} = InfFrom(a)$ holds and there is no cycle including node b, then the edge (a,b) appears in the gene regulatory network.


\begin{proof}Suppose that both conditions hold but the edge $(a,b)$\space does n...
..., $x \not \in
InfFrom(b)$\space holds, which is a contradiction.
\end{proof}
Note that the condition that there is no cycle including b can not be removed from the above proposition. For example, $InfFrom(b) \cup \{b\} = InfFrom(a)$ holds in both networks Fig. 14.12, but in each case, there not necessarily exists an edge (a,b). Note, that three nodes satisfy $InfFrom(b) \cup \{b\} = InfFrom(a)$ in case (i) while only one node satisfies this condition in case (ii). Although testing the existence of a cycle may require an exponential number of experiments as in Prop. 14.1, it is expected that such cases as in Fig. 14.12 seldom occur. Therefore, if only one node b satisfies $InfFrom(b) \cup \{b\} = InfFrom(a)$, we may conclude edge (a,b) appears in the network. Moreover, if we can identify the set of edges incoming to b (by such a method as above), we can identify the boolean function assigned to b by examining assignments only to incoming nodes.
next up previous
Next: Related Problems: Consistency and Up: Identification of Gene Regulatory Previous: More Efficient Strategies for
Peer Itsik
2001-03-04