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Which algorithm inverts a complete resolution strategy?
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1. Consider the following algorithm: for ( i = 1 ; i <= 1 . 5 n ; i++) cout << i ; for ( i = n ; i >= 1 ; i - - ) cout << i ; (a) What is the output when n = 2, n = 4, and n = 6? (b) What is the time complexity T(n)? You may assume that the input n is divisible by 2.
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