Which algorithm in ‘unification and lifting’ takes two sentences and returns a unifier?
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Consider the following algorithm: j = 1 ; while ( j <= n/2) { i = 1 ; while ( i <= j ) { cout << j << i ; i++; } j++; } (a) What is the output when n = 6, n = 8, and n = 10? (b) What is the time complexity T(n)? You may assume that the input n is divisible by 2.
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Which algorithm in ‘unification and lifting’ takes two sentences and returns a unifier?
Consider the following algorithm: for ( i = 2 ; i <= n ; i++) { for ( j = 0 ; j <= n) { cout << i << j ; j = j + floor(n/4) ; } } (a) What is the output when n = 4 (b) What is the time complexity T(n). You may assume that n is divisible 4.
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