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Abstract:
Now observe that: c(T) = yp, q, a graphic answer to p, b graphic answer to p) c((p, 1), (p, 2), (p, a, q, b)) + Yp, q, a graphic answer to p, b graphic answer to p) c((1, q), (2, q), (p, a, q, b)) = y r-(p, q) = 2 (). ACCv,(A,B) P,q It follows that max.c(T) _('*)ACCv. Conversely, if (A, B) is a pair of prover strategies for which ACCv(x) = ACCv,(A,B)(X), then we can order X as suggested above such that for each p, ap = A(p), and for each q, bq = B(q). For the corresponding hypertree T c(T) = 2 (). ACCv,(,B) = 2 (). ACCv (x) In conclusion, raax:rc(T) = 2 (). ACCv(x). []
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