Actual Actual Facts Of Our Icotinib Accomplishments

Clearly, the time complexity of FPC is O(nk) by maintaining a nearest neighbor table that records the nearest neighbor in R of each instance p �� X ? R and the corresponding distance between p and its nearest neighbor in R. The space complexity is O(n). So, the time complexity and the space complexity are check details both linear with n for a fixed k. Using a more complicated approach, the FPC algorithm can be implemented in O(nlog?k), but the implementation was exponentially dependent on the dimension d [3]. Now, a natural problem arises: if RSDopt(X) �� 1, how does the FPC algorithm perform? Although, in this paper, we cannot give performance guarantee of the FPC algorithm for the max-RSD problem and the max-min split problem if RSDopt(X) �� 1, Gonzalez [1] proved the following theorem (see also [2, 3]). Theorem 8 (see [1]). �� The FPC selleck chemical is a 2-approximation algorithm for the unsupervised min-max diameter problem with the triangle inequality satisfied for any k. Furthermore, for k �� 3, the (2 ? ��)-approximation of the unsupervised min-max diameter problem with the triangle inequality satisfied is NP-complete for any �� > 0. So as far as the approximation ratio is concerned, the FPC algorithm is the best for the unsupervised min-max diameter problem unless P = NP. 3.2. Semi-Supervised Learning For semisupervised learning, we present a nearest neighbor-based clustering (NNC) algorithm as shown in Algorithm 2. The algorithm is self-explanatory, and we do not give a further explanation. Algorithm 2 The NNC clustering algorithm for semisupervised learning. Theorem 9 . �� For semiunsupervised learning, if RSDopt(X) > 1, then the partition P returned by NNC is simultaneously the optimal solution of the semisupervised max-RSD problem, the semisupervised max-min split problem, and the semisupervised min-max diameter problem. Proof �� The proof of max-RSD(P) problem: let P�� = C1��, C2��,��, Ck�� be the optimal partition of the semisupervised max-RSD problem. Since P�� respects the supervision, we can replace Si by a super-instance ��i for i = 1,2,��, k; then each cluster Ci�� contains exactly one super-instance ��i for i = 1,2,��, k (without loss of generality, here we assume that ��i is in the cluster Ci�� for i = 1,2,��, k). Let P = C1, C2,��, Ck; then according YES1 to the algorithm NNC, each cluster also contains exactly one super-instance, and without loss of generality, we also assume that ��i is in the cluster Ci for i = 1,2,��, k. For each unlabeled instance p �� Cr�� for r = 1,2,��, k, since RSDopt(X) > 1, we have d(p, ��r) = dMax?(p, Sr)