1 votes 1 votes amount of time it would take to send 500 packets using 6-packet size window and without sliding window will be, when the time to send in one direction is 5 ms 31000, 5000 4600, 21600 5000, 30000 30000, 5000.......... pls explain the logic behind? Himani Srivastava asked Oct 16, 2015 Himani Srivastava 7.9k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes No. of BST = 2nCn/n+1 = 6C3/4 = 5 No. of Binary tree = (2nCn/(n+1))n! = (6C3/4)*3!= 30 Umang Raman answered Oct 16, 2015 Umang Raman comment Share Follow See all 3 Comments See all 3 3 Comments reply Himani Srivastava commented Oct 16, 2015 reply Follow Share BUT answer for this is 12 ..What does ordered trees means here?? 0 votes 0 votes cse23 commented Jun 12, 2016 reply Follow Share an ordered tree is one in which node(root) is connected by disjoint set of subtree. In ordered tree (a "tree" of the first definition) has the subtrees in a particular order, and the subtree sequence cannot contain the same tree twice, because the subtrees must be disjoint. A rooted tree does not have these restrictions; by this definition a root might have a subtree twice, in a structure that resembles a cycle. I am getting 10 as answer. 0 votes 0 votes Start Living commented Sep 28, 2016 reply Follow Share It will 12. Case-1: Given the root node has two children,(A(root)->B(left)->C(right) and A(root)->B(right)->C(left))So there are 3*2 = 6 possible ordered trees Case-2: GIven the root node has one child(A(root)->B(left)->C(left) and A(root)->B(left)->C(leftt)), so there are a total of 3*2 = 6 possible ordered trees with three nodes given that the root node has only one child. In case-2,A(root)->B(left)->C(left) and (root)->B(right)->C(right) are considered as same as subsequence are not disjoint. 0 votes 0 votes Please log in or register to add a comment.