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This website uses cookies to improve your experience. Well assume youre ok with this, but you can opt-out if you wish. Accept Read More This website uses cookies to improve your experience. Well assume youre ok with this, but you can opt-out if you wish. Accept Read More Tavu pieprasījumu nevarēja apstrādātIr radusies problēma saistībā ar šo pieprasījumu. Mēs strādājam, lai to pēc iespējas ātrāk atrisinātu.Lai turpinātu, ienāc vai reģistrējies Facebook.PieslēgtiesPievienotiesCOMPUTER SCIENCEAn AVL tree is a binary search tree that is height balanced: for each node x, the heights of the left and right subtrees of x differ by at most 1. To implement an AVL tree, we maintain an extra attribute in each node: x.h is the height of node x. As for any other binary search tree T , we assume that T:root points to the root node. a. Prove that an AVL tree with n nodes has height O(lg n). (Hint Prove that an AVL tree of height h has at least $$F_h$$ nodes, where $$F_h$$ is the hth Fibonacci number.) b. To insert into an AVL tree, we first place a node into the appropriate place in binary search tree order. Afterward, the tree might no longer be height balanced. Specifically, the heights of the left and right children of some node might differ by 2. Describe a procedure BALANCE (x), which takes a subtree rooted at x whose left and right children are height balanced and have heights that differ by at most 2, i.e., │x.right. h - x.left.h│≤ 2, and alters the subtree rooted at x to be height balanced. (Hint: Use rotations.) c. Show that AVL-INSERT, run on an n-node AVL tree, takes O(lg n) time and performs O(1) rotations.COMPUTER SCIENCEThis exercise concerns a variation of the Euclidean algorithm more suited to computer implementation. The original GCD algorithm relies on repeated integer divisions to compute remainders. The following variation, called the binary GCD algorithm, also uses divisions, but only by 2, plus subtraction and testing for parity (oddness or evenness). Given that the numbers are stored in the computer in binary form, these operations are simple and often done using built-in circuits. Testing for evenodd can be done using bitwise conjunction of N & 1, which results in 1 if and only if N is odd. Given an even N, division by 2 is easily accomplished by a 1-bit right shift operation, where all bits are shifted one place to the right (the rightmost bit disappears), and the leftmost bit is set to 0. (Multiplication by 2 is a 1-bit left shift.) Subtraction involves the 2s complement. As a result, the binary gcd algorithm, which avoids regular division, runs faster than the Euclidean algorithm, even though it does more (but simpler) steps. The binary GCD algorithm relies on three facts: 1. If both a and b are even, then $operatornamegcd(a, b)=2 operatornamegcd(a 2, b 2)$ 2. If a is even and b is odd, then $operatornamegcd(a, b)=operatornamegcd(a 2, b)$ 3. If a and b are both odd, and $a geq b,$ then gcd(a, b) = gcd((a − b)2, b). To prove that if both a and b are even, then gcd(a, b) = 2gcd(a2, b2), let a and b be even integers. Then 2 is a common factor of both a and b, so 2 is a factor of gcd(a, b). Let 2c = gcd(a, b). Then a = n(2c) and b = m(2c); a2 = nc and b2 = mc, so $c | a 2$ and $c | b 2.$ Finish this proof by showing that c = gcd(a2, b2).5th EditionJack T. Marchewka7th EditionRamez Elmasri, Shamkant B. Navathe5th EditionJack T. Marchewka8th EditionGodfrey Muganda, Judy Walters, Tony Gaddis0 ratings0% found this document useful (0 votes)6K views1 page, active
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