clrs

问题 Suppose that we wish to keep track of a point of maximum overlap in a set of intervals—a point that has the largest number of intervals in the database overlapping it. a. Show that there will always be a point of maximum overlap which is an endpoint of one of the segments. b. Design a data structure that efficiently supports the operations INTERVAL-INSERT, INTERVAL-DELETE, and FIND-POM, which returns a point of maximum overlap. (Hint: Keep a red-black tree of all the endpoints. Associate a

问题 Is it not true that depending on the language that way the parameters/arguments are passed will have a different time complexity. Then why is this not factored in or considered in the algorithms or programs that books measure time complexity of? CLRS or Data Structures and Algorithm Analysis by Mark Allen Weiss never would add the time complexity of how the arguments are being passed for the total runtime of the program? Am I misunderstanding something? I know CLRS was pseudocode but the

问题 Is it not true that depending on the language that way the parameters/arguments are passed will have a different time complexity. Then why is this not factored in or considered in the algorithms or programs that books measure time complexity of? CLRS or Data Structures and Algorithm Analysis by Mark Allen Weiss never would add the time complexity of how the arguments are being passed for the total runtime of the program? Am I misunderstanding something? I know CLRS was pseudocode but the

问题 CLRS says that We must show three things about a loop invariant: Initialization: It is true prior to the first iteration of the loop. Maintenance: If it is true before an iteration of the loop, it remains true before the next iteration. Termination: When the loop terminates, the invariant gives us a useful property that helps show that the algorithm is correct. My question is can I edit the steps and make them these instead: Initialization: It is true after the first iteration of the loop.

问题 CLRS says that We must show three things about a loop invariant: Initialization: It is true prior to the first iteration of the loop. Maintenance: If it is true before an iteration of the loop, it remains true before the next iteration. Termination: When the loop terminates, the invariant gives us a useful property that helps show that the algorithm is correct. My question is can I edit the steps and make them these instead: Initialization: It is true after the first iteration of the loop.

问题 Here is the question of exercise CLRS 10.3-4 I am trying to solve It is often desirable to keep all elements of a doubly linked list compact in storage, using, for example, the first m index locations in the multiple-array representation. (This is the case in a paged, virtual-memory computing environment.) Explain how to implement the procedures ALLOCATE OBJECT and FREE OBJECT so that the representation is compact. Assume that there are no pointers to elements of the linked list outside the

问题 In CLRS, third Edition, on page 155, it is given that in MAX-HEAPIFY, "the worst case occurs when the bottom level of the tree is exactly half full" I guess the reason is that in this case, Max-Heapify has to "float down" through the left subtree. But the thing I couldn't get is "why half full" ? Max-Heapify can also float down if left subtree has only one leaf. So why not consider this as the worst case ? 回答1: Read the entire context: The children's subtrees each have size at most 2n/3 - the

问题 In CLRS, third Edition, on page 155, it is given that in MAX-HEAPIFY, "the worst case occurs when the bottom level of the tree is exactly half full" I guess the reason is that in this case, Max-Heapify has to "float down" through the left subtree. But the thing I couldn't get is "why half full" ? Max-Heapify can also float down if left subtree has only one leaf. So why not consider this as the worst case ? 回答1: Read the entire context: The children's subtrees each have size at most 2n/3 - the

问题 I am reading Introduction to Algorithms by CLRS. Book shows pseudocode for simple divide and conquer matrix multiplication: n = A.rows let c be a new n x n matrix if n == 1 c11 = a11 * b11 else partition A, B, and C C11 = SquareMatrixMultiplyRecursive(A11, B11) + SquareMatrixMultiplyRecursive(A12, B21) //... return C Where for example, A11 is submatrix of A of size n/2 x n/2. Author also hints that I should use index calculations instead of creating new matrices to represent submatrices, so I

问题 I'm trying to do this exercise in Introduction to Algorithms by Cormen et al that has to do with the Disjoin Set data structure: Suppose that we wish to add the operation PRINT-SET(x) , which is given a node x and prints all the members of x 's set, in any order. Show how we can add just a single attribute to each node in a disjoint-set forest so that PRINT-SET(x) takes time linear in the number of members of x 's set , and the asymptotic running times of the other operations are unchanged.