[Cover] [Contents] [Index] Previous page Next Section

Page 215
9.37 Definition Let  be a nonempty finite multiset of grammars. Then the grammar majority(Â) is defined to be a grammar such that 0215-001.gif.
Clearly, majority(Â) can be defined using the s-m-n theorem (Theorem 2.1). Intuitively, majority(Â) is a grammar for a language that consists of all such elements that are enumerated by a majority of grammars in Â. Below, whenever we use a set as an argument to majority we assume the argument to be a multiset.
Proof (of Proposition 9.36): Let j and k be as given in the hypothesis of the proposition. Let 0215-002.gif by a team consisting of scientists M1, M2, . . . , Mk. We then describe a scientist M that 0215-003.gif.
For a finite sequence  s  and a scientist M', define
0215-004.gif
Intuitively, conv(M',  s ) is the length of the longest subsequence of a where M' conjectures a hypothesis distinct from the one it conjectures on  s .
Let 0215-005.gif, 0215-006.gif, . . ., 0215-007.gif be a permutation of l, 2, . . . , k, such that, for 0215-008.gif,
0215-009.gif
Note: ''<" here refers to ordering on pairs: (k0, l0) < (k1, l1) just in case either k0 < k1 or else k0 = kl and l0 < l1.
For each  s , define 0215-010.gif.
To complete the proof it must be shown that if the team consisting of scientists M1, M2, . . . , Mk 0215-011.gif, then M TxtEx-identifies L. This is the subject of Exercise 9-8.
Proposition 9.38 below says that the collections of languages that can be identified by a team with success ratio0215-012.gif (that is, at least two-thirds of the members in the team are required to be successful) are the same as those collections of languages that can be identified by a team of 3 scientists, at least 2 of which are required to be successful.
9.38 Proposition For each j > 0, 0215-013.gif.
Proof: Let j be as given in the hypothesis of the proposition. Suppose a team consisting of scientists M1, . . . , M3j 0215-014.gif. We then describe scientists 0215-015.gif, 0215-016.gif, and 0215-017.gif such that the team consisting of 0215-018.gif, 0215-019.gif, and 0215-020.gif witnesses 0215-021.gif.

 
[Cover] [Contents] [Index] Previous page Next Section