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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.

)a grad student comes up with the following algorithm to sort an array a[1..n] that works by first sorting the first 2/3rds of the array, then sorting the last 2/3rds of the (resulting) array, and finally sorting the first 2/3rds of the new array. 1: function g-sort(a, n) . takes as input an array of n numbers, a[1..n] 2: g-sort-recurse(a, 1, n) 3: end function 4: function g-sort-recurse(a, `, u) 5: if u ⒠` ≤ 0 then 6: return . 1 or fewer elements already sorted 7: else if u ⒠` = 1 then . 2 elements 8: if a[u] < a[`] then . swap values 9: temp ↠a[u] 10: a[u] ↠a[`] 11: a[`] ↠temp 12: end if 13: else . 3 or more elements 14: size ↠u ⒠` + 1 15: twothirds ↠d(2 ◠size)/3e 16: g-sort-recurse(a, `, ` + twothirds ⒠1) 17: g-sort-recurse(a, u ⒠twothirds + 1, u) 18: g-sort-recurse(a, `, ` + twothirds ⒠1) 19: end if 20: end function first (5 pts), prove that the algorithm correctly sorts the numbers in the array (in increasing order). after showing that it correctly sorts 1 and 2 element intervals, you may make the (incorrect) assumption that the number of elements being passed to g-sort-recurse is always a multiple of 3 to simplify the notation (and drop the floors/ceilings).

How to print: number is equal to: 1 and it is odd number number is equal to: 2 and it is even number number is equal to: 3 and it is odd number number is equal to: 4 and it is even number in the console using java using 1 if statement, 1 while loop, 1 else loop also using % to check odds and evens