Data Structures and Algorithms in Java
Data Structures and Algorithms in Java
6th Edition
ISBN: 9781119278023
Author: Michael T. Goodrich; Roberto Tamassia; Michael H. Goldwasser
Publisher: Wiley Global Education US
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Chapter 4, Problem 40C

Explanation of Solution

Total number of grains requested by the inventor of chess:

If a chessboard is placed with square then one grain is placed on first square, two on second, and four on third. On each subsequent square, the numbers of grains are doubled.

Thus, the exponential increase on each square contains 2n1 grains of rice.

Since, there are 64 squares on the chessboard, apply the value of 64 in the formula 2n1 to fill the final square on the chessboard. Substitute “n” as 64.

  Fillingofriceonfinalsquareonchessboard=2641=263=9223372036854775808=9.22×1018

To determine the total number of grains of rice let us use the exponential sum using the formula 2n 1

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Given a relation schema R = (A, B, C, D, E,G) with a set of functional dependencies F {ABCD BC → DE B→ D D→ A}. (a) Show that R is not in BCNF using the functional dependency A → BCD. (b) Show that AG is a superkey for R (c) Compute a canonical cover Fc for the set of functional dependencies F. Show your work. (d) Give a 3NF decomposition of R based on the canonical cover found in (c). Show your work. (e) Give a BCNF decomposition of R using F. Show your work.
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