Example 9.22Solve following recurrence relations using backward substitution.а) T(п) %3D 2T(n/4) + 1b) T(п) 3D 2T(п — 2) + 3for n > 1,T(1) = 1(solve for n =4*)%3Dfor n > 2,T(0) = 0

Solution for a) Step 1: T(n) = 2T(n/4) + 1 T(n/4) = 2T(n/42) + 1 --------put n = n/4…

Example 9.22 Solve following recurrence relations using backward substitution. 4k) а) Т(п) 3D b) T(п) — 27 (п — 2) + 3 2T (п/4) + 1 for n > 1, T(1) = 1 (solve for n = for n > 2, T(0) = 0 - %3D

Example 9.22 Solve following recurrence relations using backward substitution. 4k) а) Т(п) 3D b) T(п) — 27 (п — 2) + 3 2T (п/4) + 1 for n > 1, T(1) = 1 (solve for n = for n > 2, T(0) = 0 - %3D

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

Example 9.22
Solve following recurrence relations using backward substitution.
а) T(п) %3D 2T(n/4) + 1
b) T(п) 3D 2T(п — 2) + 3
for n > 1,
T(1) = 1
(solve for n =
4*)
%3D
for n > 2,
T(0) = 0

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