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The standard algorithm for multiplying two n-bit binary integers x and y costs Theta(n^2). A naive divide-and-conquer algorithm is to let x = 2^n/2 a + b and y = 2^n/2 c + d, then xy = (2^n/2 a + b)(2^n/2 c + d) = 2^n ac + 2^n/2(ad + bc) + bd The complexity is T(n) = 4T(n/2) + Theta(n) = Theta(n^2). There is no improvement to the standard algorithm. By observing that ad + bc = (a + b)(c + d) - (ac + bd), we can use only three multiplications. Describe this divide-and-conquer algorithm in a pseudocode. What is the complexity of the algorithm. Illustrate the algorithm for multiplying integers x = 10011011 and y = 10111010 (just show one level of the recursion)