Interfacing sensors to the IBM® PC by Willis J Tompkins; John G Webster

By Willis J Tompkins; John G Webster

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Theorem 9: Let P be order decomposable. Then P(5), 5 = T1 can be com- puted in time 50rt(151) + T(151) where 50rt(n) is the time required to sort a set of n elements according to T( Ln/2j) + T( rn/2") + O(C(n)) for n <, and T(n) > 1 and T(1)= c for some constant c. Proof: The proof is a straightforvTard application of divide and conquer. We first sort 5 in time 50rt(151) and store 5 in sorted order in an array. This will allow us to split 5 in constant time. Next we either compute P(5) directly in constant time if 151 = 1 or we split 5 into sets A and B of size Ln/2 j and rn/2" respectively in constant time (if 5 = {a 1 < a 2 < ...

We will next describe how to insert into and delete from an augmented tree. We describe insertion in detail and leave deletion for the reader, deletion being very similar to insertion. Let a be a new point which we want to insert in S. Let D be an augmented tree for S. We first use D as a search tree. This will outline a path p down tree D. Let 23 p = v o ,v 1 , ... ,vk with Vo being the root. We walk down this path and reconstruct the P(S(v i )) IS as we walk down. More precisely, we start in root Vo with P(S(vo )) in our hands and use the sequence of actions a (vo ) stored in va and the leftover pieces p*(S(v 1 )) and p*(S(brother(v 1 ))) stored in v 1 and its brother to reconstruct P(S(v 1 )) and P(S(brother(v 1 ))) by running a(v o ) backwards.

This suggests that we can use binary search to find line L 2 . We first compute in time O(n 2 ) lines K1 , ... ,K k and points P 1 , ... ,P k . Next we find the median point of P 1 , ... ,P k in time O(n 2 ). Then we are either done or can restrict the search to k/2 points. This decision takes time O(n). Thus line L2 can be f~und in O(log n 2 ) iterations and the cost of the i-th iteration is O(k/2 l + n). Total cost is thus O(n 2 ). D Lemma 2 and the preceding discussion lead to: Definition: a) A 4-way polygon tree T for set S =R , 2 lSI = n is de- fined as follows: If set S is collinear then T is an ordinary one-dimensional search tree for S.

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