1. Consider the knapsack problem with the capacity C= 8 and 5 items with weights 3, 6, 2, 5, 3. a) Find which items will exactly fill the knapsack using dynamic programming solution introduce in class. Show all your work. b) Give all the possible lists that can be added in the knapsack of question 1
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Q: 1. Consider the knapsack problem with the capacity C= 8 and 5 items with weights 3, 6, 2, 5, 3. a)…
A: The answer given as below:
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- 5. Consider the Josephus problem: in class, we looked at n elements in a circle and eliminated every second element until only one was left. The last element surviving this process was called the Josephus number. Instead of finding the last survivor, let I(n) be the element that survives second to last. To give a few small values, I(2) = 1, I(3) = 1, I(4) = 3, and I(5) = 5. Give a closed form expression for I(n) for any n ≥ 2.1. A library has n books that must be stored in alphabetical order on adjustable height shelves. Each book has a height and a thickness. The width of the shelf is fixed at W, and the sum of the thicknesses of books on a single shelf must be at most W. The next shelf will be placed on top, at a height equal to the maximum height of a book in the shelf. You are given the list of books in alphabetical order, b, = (h, t.), where h, is the height and t, is the thickness, and must organize the books in that order. Give an efficient dynamic programming algorithm that minimizes the total height of shelves used to store all the books.Consider the set A = {blue, yellow, green}. Prove or disprove the statements below. a) Vx E A, x contains the letter I. b) 3x E A such that x has more than 6 letters. c) Vx E A, x has more than 5 letters → x contains the letter w.
- Let’s assume M is a set defined as M ⊆ Z^+ × Z^+. Assume that M is defined using the concept of recursion taught in class as: i) (1, 2) ∈ M, ii) If (x, y) ∈ M, then (3x − y, 2x) ∈ M. Perform 5 recursive steps to obtain the updated M.The wolf-goat-cabbage ProblemDescription of the problem: There is a farmer who wishes to cross a river but he is not alone. He also has a goat, a wolf, and a cabbage along with him. There is only one boat available which can support the farmer and either of the goat, wolf or the cabbage. So at a time, the boat can have only two objects (farmer and one other). But the problem is, if the goat and wolf are left alone (either in the boat or onshore), the wolf will eat the goat. Similarly, if the goat and cabbage are left alone, then goat will eat the cabbage. The farmer wants to cross the river with all three of his belongings: goat, wolf, and cabbage.Complete the state space of this problem. The green state is valid state (you should expand it until to reach the goal state) and orange state is invalid state (you should not expand it).• w: wolf• g: goat• c: cabbage• f: framer• ||: rive.The following are the different operations that can be done using a doubly linked list. The corresponding algorithm and simulation/s are attached in the course material for doubly linked list. Choose only one operation and create the Java program for the chosen operation using its corresponding algorithm. The rubric below will be used to assess your output. Insertion in the beginning of the list Insertion after a node Insertion before a node Deletion of the first node Deletion of the last node Deletion of a given node
- Please help me Josephus Problem is a theoretical problem related to a certain counting-out game. On thiscase, people are standing in a circle waiting to be executed. After a specified number ofpeople are skipped, the next person is executed. The procedure is repeated with theremaining people, starting with the next person, going in the same direction and skippingthe same number of people, until one person remains, and is freed.Arrange the numbers 1 , 2, 3 , ... consecutively (say, clockwise) in a circle. Now removenumber 2 and proceed clockwise by removing every other number, among those thatremain, until one number is left. (a) Let denote the final number which remains. Find formula for .(b) If there are 70 people, what is the safe number (the number that remains)?Must be new solution and run on GNU Common Lisp! Using Lisp, write a program that solves the Missionaries and Cannibals problem that uses a DFS( depth first search). It should use (mac start end). Start is the current state (which can be (3 3 l) and End is the goal state (which can be (0 0 r). This should output the sequences of moves needed to reach the end state from the start state. This should print nil if there is no solution. For example, the call should be something like this! Call: (mac '(3 3 l) '(0 0 r)) Output: ((3 3 l) (2 2 r) (3 2 l) (3 0 r) (3 1 l) (1 1 r) (2 2 l) (0 2 r) (0 3 l) (0 1 r) (1 1 l) (0 0 r))An argument is expressed in English below. The domain is the set of people attending a party. Every person who was late to the party left the party early. There is a person who was late to the party and was not happy. .. There is a person who left the party early and was not happy. The form of the argument is: x (P(x) → Q(x)) 3X (P(x) ^ ¬R(x)) .: 3x (Q(x) ^ ¬R(x)) Select the definitions for predicates P, Q, and R. P(x): Pick Q(x): Pick R(x): Pick
- On a chess board of r rows and c columns there is a lone white rook surrounded by a group of opponent's black knights. Each knight attacks 8 squares as in a typical chess game, which are shown in the figure - the knight on the red square attacks the 8 squares with a red dot. The rook can move horizontally and vertically by any number of squares. The rook can safely pass through an empty square that is attacked by a knight, but it must move to a square that is not attacked by any knight. The rook cannot jump over a knight while moving. If the rook moves to a square that contains a knight, it may capture it and remove it from the board. The black knights. never move. Can the rook eventually safely move to the designated target square? The figure illustrates how the white rook can move to the blue target square at the top-right corner in the first sample case. The rook captures one black knight at the bottom-right of the board on its way. Rok nd kight lcoes by Chunen Input The first line…Please do this in JAVA PROGRAMMING. Given: List L of pairs of charactersString SOutput: TRUE if S is a valid string, FALSE otherwise. A string is considered valid if each character in the string can be paired with another character in the string, where the pair belongs to the input list L. Furthermore, two pairs cannot cross each other. In other words, a pair most completely enclose another, or be completely separate. Design and implement an efficient dynamic programming solution to this problem. Examples: Input L: (a b) (b c) (c d) (a a) Input S: aaba Output: True (pairs shown color-coded: aaba, "aa" fully encloses "ab") Input L: (a b) (b c) (c d) (a a) Input S: abcaad Output: True (pairs shown color-coded: abcaad, "ab" is separate from the other pairs) Input L: (a b) (b c) (c d) (a a) Input S: acbd Output: False (acbd is not valid because the pairs cross each other.) Input L: (a b) (b c) (c d) (a a) Input S: aaac Output: FalseThe Knapsack Problem is a famous computer science problem that is defined as follows: imagine you are carrying a knapsack with capacity to hold a total of weight C. You are selecting among n items with values A={a_1, a_2, ... , a_n} and associated weights W={w_1, w_2, ... , w_n}. Here the weights and values are all positive (but not necessarily unique). You wish to maximize the total value of the items you select not exceeding the given weight capacity, i.e. maximize sum_{a in A} such that sum_{w in W} <= C. Please note that you can only select your items once. a) We can reformulate this as a 2D bottom-up dynamic programming problem as follows. Define T_{i,j} as the highest possible value sum considering items 1 through i and total weight capacity j (j <= C). What is the base case i.e. T_{0,j} for all j and T_{i,0} for all i?, and What is the loop statement?