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| + | = Informal Comparison of KNN and NN in case of k = 1 = | ||
This is an informal discussion about the Kn-Nearest-Neighbor (KNN) and Nearest Neighbor (NN) estimations in k = 1 case. My conclusion is that KNN and NN are essentially the same when k = 1. | This is an informal discussion about the Kn-Nearest-Neighbor (KNN) and Nearest Neighbor (NN) estimations in k = 1 case. My conclusion is that KNN and NN are essentially the same when k = 1. | ||
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| + | *Answer/comment here. | ||
| + | *Answer/comment here. | ||
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| + | [[ECE662|Back to ECE662]] | ||
Latest revision as of 10:47, 22 March 2012
Informal Comparison of KNN and NN in case of k = 1
This is an informal discussion about the Kn-Nearest-Neighbor (KNN) and Nearest Neighbor (NN) estimations in k = 1 case. My conclusion is that KNN and NN are essentially the same when k = 1.
In any testing point, let us assume that this testing point is category-k dominant using NN (trainning point from category k will be firstly captured). Then, if we switch to use KNN instead, when enlarge the volume surrounding this testing point, the volume in category k should be the smallest (with largest discriminant value), so this point is still considered as category-k dominant using KNN.
Based on the above discussion, I conclude that the classification using KNN and NN when k = 1 should be the same.
zge
- Answer/comment here.
- Answer/comment here.
