What is a loop statement that never resolves to a false value? an infinite loop

a continuous loop

a single loop

a statement loop

For each of the following claims, determine whether they are true or false. justify your determination (show your work). if the claim is false, state the correct asymptotic relationship as o, θ, or ω. unless otherwise specified, lg is log2.(a) (b) (c) (d) (e) (f) (g) (h) (i) (j)n+1 =22n =2n =1 =ln2 n =n2 +2n−4 =33n = 2n+1 =√n = 10100 =o(n4) o(2n)θ(2n+7 ) o(1/n)θ(lg2 n) ω(n2 )θ(9n ) θ(2n lg n )o(lg n) θ(1)

Some of your friends have gotten into the burgeoning field of time-series data mining, in which one looks for patterns in sequences of events that occur over time. purchases at stock exchanges"what's being bought" are one source of data with a natural ordering in time. given a long sequence s of such events, your friends want an efficient way to detect certain "patterns" in them"for example, they may want to know if the four events buy yahoo, buy ebay, buy yahoo, buy oracle occur in this sequence s, in order but not necessarily consecutively. they begin with a collection of possible events (e. g., the possible transactions) and a sequence s of n of these events. a given event may occur multiple times in s (e. g., yahoo stock may be bought many times in a single sequence s). we will say that a sequence s is a subsequence of s if there is a way to delete certain of the events from s so that the remaining events, in order, are equal to the sequence s . so, for example, the sequence of four events above is a subsequence of the sequence buy amazon, buy yahoo, buy ebay, buy yahoo, buy yahoo, buy oracle their goal is to be able to dream up short sequences and quickly detect whether they are subsequences of s. so this is the problem they pose to you: give an algorithm that takes two sequences of events"s of length m and s of length n, each possibly containing an event more than once"and decides in time o(m + n) whether s is a subsequence of s.

What percentage of companies is projected to use social media to locate new employees in 2012