On Bounded Rationality And Computational Complexity

31 again, because if there are h more cooperating steps, the honest master receives payoo at least hp i ? 3w (the ?3w term is because the last 3 steps of the last block may be missing), whereas the deviating master receives no more than 2dw + hv i , which is smaller since h > 2dw== and p i ? v i > 4. In the nal cooperation section, if the master deviates in the last segment containing his most prootable pair it is clearly best to do it in the very last step; if he deviates earlier with h 0 steps remaining (note, h 0 c j ? 3 7w=(p i ? v i) ? 3), then he receives payoo w + h 0 v i whereas the honest master receives at least h 0 p i ? 3w (because the pairs are ordered within a block in increasing payoo to the master). Since h 0 c j ? 3 7w=(p i ? v i) ? 3, it follows that h 0 p i ? 3w > w + h 0 v i , so the deviation of the master from the prescribed play results again in a loss. The proof that the server cannot improve on his strategy is essentially the same as in part (a). We deene again for an impostor automaton A 0 the sets of cards Y 1 , Y 2 and the set of strong companies exactly the same way. The proof in (a) was based on using the normal paths to distinguish between the nodes. With the above choices for the strategies, we can deene normal paths for the server, and use the same arguments as in part (a) to show that A 0 can only save nodes used by weak companies, thus at most (n=3)Pr((0)=(=3) < 9nPr(Y 0 1)== states (every company uses at most 2d(L+1)+M +1+`n=3 states and is the maximum probability of a company). For the same reasons as before (and as for the master), it is not prootable for A 0 to deviate before the cooperation section of the last loop execution, and it needs n=2 extra states to do this, which must be distinct for cards of diierent companies. The best that A 0 can do is deviate in the last block of the game and the gain is certainly at most Mw. Thus, the total gain …