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14.7.1 Number of RDPs in P Set P of RDPs consists of all RDPs resulting from pruning PJ , the uniform partition of the unit square into n squares of sidelength 1/ n. We need to determine how many RDPs there are in P or, more speci cally, we need to know how many partitions there are with exactly squares/leafs. Since the RDP is based on recursive splits into four, the number of leafs in every partition in P is of the form = 3m + 1, for some integer 0 m (n 1)/3. The integer m corresponds to the number of recursive splits. For each RDP having 3m + 1 leafs there is a corresponding partially ordered sequence of m split points (at dyadic positions in the plane). In general, there are n m n! (n m)!m!
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possible selections of m points from n (n corresponding to the vertices of the nest resolution partition, PJ ). This number is an upper bound on the number of partitions in P with = 3m + 1 leafs (since RDPs can only have dyadic split points). 14.7.2 Kraft inequality Let n denote the set of all possible models of the eld. This set contains piecewise constant models (constant on the dyadic squares corresponding to one of the partitions in Pn ). The constant values are in a prescribed range [ R,R], and are quantized to k bits. The range corresponds to the upper and lower limits of the amplitude range of the sensors. The set n consists of a nite number of models derived in the previous section. Here we show that with the number of bits k employed per transmission and p(n) properly calibrated, we have e p(n)| | 1
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where for simplicity notation N (P) = | | is used. If (m) denotes the subset of n of models based on = 3m + 1 leaf partitions, then we have
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e p(n)| | =
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If A m log n + (3m + 1) log(2k ) (3m + 1) p(n) < 1 (then e A < e 1 ), then we have e p(n)| | 1/e
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To guarantee A < 1, we must have p(n) growing at least like log n. Therefore, set p(n) = log n, for some > 0. Also, as we will see later in the next section, to guarantee that the quantization of our models is suf ciently ne to contribute a negligible amount to the
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overall error we must select 2k : n 1/4 . With these calibrations we have A = [(7/4 3 ) m + (1/4 )] log n. In order to guarantee that the MSE converges to zero, we will see in the next section that m must be a monotonically increasing function of n. Therefore, for n suf ciently large, the term involving ( 1 ) is negligible, and the condition A < 1 is 4 satis ed by > 7/12. In References [119 125] = 2/3 is used. 14.7.3 Upper bounds on achievable accuracy Assume that p(n) satis es the condition de ned by Equation (14.4) where again | | denotes the number of squares (alternatively we shall call this the number of leafs in the pruned tree description of the boundary) in the partition . It is shown in the above section that p(n) log n satis es Equation (14.4). Let n denote the solution to n = arg min