1. Analyze the word finder with wild card characters scenario to form the algorithm and compute the time complexity. You are given a list of characters with size n, m. You need to read character data provided as input into this list. There is also a list of p words that you need to find in the given problem. You can only search in left, right, up and down direction from a position. There are two wild characters * and 4,if there is any instance of wild card characters for example '+' appears you can match whatever character in your word that you are searching at this position (only one character). Similarly, if you got wildcard *' you can skip any number of characters (A sequence of characters not necessary same in between the characters) but the last character should be matched in the given data. This means you are not allowed to match a first and last character with "** wildcard. Input: The first line of the input will give you two integers n and m. From the next line you will get the characters list from word-finder. The next line read an integer p that represents number of the words to be searched. From the next line you will read word per line. Output: In output you need to write the starting position ( x andy space separated) of the word that you find and direction of the match (Either L,R,U and D one character only). As you only need to allow Left, Right, Up and Down so your valid match will be for any such direction. If the word is not presented, you just need to write N (not find)

To compute the average of the squares of the elements of a vector V, aka the mean square, we (Note: use randn to create a random vector) Select one: a. We use polyfit which does least squares and skip the minimization part   b.we use sum(V.^2)./length(V*V)   c.we use sum(V.*V)/length(V)   d.we use mean(V)   e.we use sum(V^2)/length(V)

Give a recursive definition for the set of all strings of a’s and b’s that begins with an a and ends in a b. Say, S = { ab, aab, abb, aaab, aabb, abbb, abab..} Let S be the set of all strings of a’s and b’s that begins with a and ends in a b. The recursive definition is as follows – Base:... Recursion: If u ∈ S, then... Restriction: There are no elements of S other than those obtained from the base and recursion of S.

Define a function shuffle that takes a sequence with an even number of elements (cards) and creates a new list that interleaves the elements of the first half with the elements of the second half. Let's better understand what it means to shuffle a list in this way. Let's say there is some list [1, 2, 3, 4, 5, 6, 7, 8]. To interleave the first half [1, 2, 3, 4] with the second half [5, 6, 7, 8] means that your final list should contain the first element from the first half, then the first element from the second half, then the second element of the first half, then the second element of the second half and so on. So the interleaved version of [1, 2, 3, 4, 5, 6, 7, 8] would be [1, 5, 2, 6, 3, 7, 4, 8]. def card (n): """Return the playing card numeral as a string for a positive n 0 and n >> shuffle (range(6)) [0, 3, 1, 4, 2, 5] >>> suits = ['H', 'D', 'S', 'C'] >>> cards = [card (n) + suit for n in range (1,14) for suit in suits] >>> cards [:12] ['AH', 'AD', 'AS', 'AC', '2H', '2D', '25',…

Define a function shuffle that takes a sequence with an even number of elements (cards) and creates a new list that interleaves the elements of the first half with the elements of the second half. Let's better understand what it means to shuffle a list in this way. Let's say there is some list [1, 2, 3, 4, 5, 6, 7, 8]. To interleave the first half [1, 2, 3, 4] with the second half [5, 6, 7, 8] means that your final list should contain the first element from the first half, then the first element from the second half, then the second element of the first half, then the second element of the second half and so on. So the interleaved version of [1, 2, 3, 4, 5, 6, 7, 8] would be [1, 5, 2, 6, 3, 7, 4, 8]. def card (n): """Return the playing card numeral as a string for a positive n 0 and n >> shuffle (range(6)) [0, 3, 1, 4, 2, 5] >>> suits = ['H', 'D', 'S', 'C'] >>> cards = [card (n) + suit for n in range (1,14) for suit in suits] >>> cards [:12] ['AH', 'AD', 'AS', 'AC', '2H', '2D', '25',…

4. Write the pseudocode for an algorithm that takes as input a list of numbers that are sorted in nondecreasing order, and finds the location(s) of the most frequently occurring element(s) in the list. If there are more than one element that is the most frequently occurring, then return the locations of all of them. Analyze the worst-case time complexity of this algorithm and give the O() estimate. (A list is in nondecreasing order if each number in the list is greater than or equal to the number preceding it.)

1. Write an algorithm to determine whether a given element x belongs to a set S := {s1, . . . , sn}.

Let ∑={a,b} and T be the set of words in ∑* that have no consecutive a’s (i.e. there cannot be 2 or more a’s in a row). (a) Give a recursive definition for the set T. (b) Use your recursive definition to show that the string “abbaba” is in T. (c) Is your recursive definition uniquely determined? Explain why or why not.

For part A, how what would the pseudo-code visually be when trying to visualize an efficient algorithm? As for Part B, what would determine the space and complexity?

With the Binary Search algorithm, the algorithm can be developed by the loop-based form as well as a recursive form. Which of the following is not true? O If the item is not found, the loop version returns when the range bounds reach, but the recursive version finishes when the recursive depth is more than half the initial search range. The search range starts with the whole array, and only the recursive version can work on a subrange passed through arguments. O Both forms of the program divide the search range repeatedly in half. O If the item is found, the loop version returns from the entire method, whereas the recursive version returns from one level of recursion.