GATE | GATE-CS-2014-(Set-3) | Question 21
The minimum number of arithmetic operations required to evaluate the polynomial P(X) = X5 + 4X3 + 6X + 5 for a given value of X using only one temporary variable.
P(X) = x5 + 4x3 + 6x + 5 =x ( x4 + 4x2 + 6 ) +5 =x ( x ( x3 + 4x ) + 6 ) + 5 =x ( x ( x ( x2 + 4 ) ) + 6 ) + 5 =x ( x ( x (x (x) + 4 ) ) + 6 ) + 5 Let T be a temporary variable to store intermediate results. 1. T = (x) * (x) 2. T = T + 4 3. T = (x) * (T) 4. T = (x) * (T) 5. T = T + 6 6. T = (x) * T 7. T = T + 5 Thus, we need 7 operations if we are to use only one temporary variable.
Please comment below if you find anything wrong in the above post.
Quiz of this Question
Attention reader! Don’t stop learning now. Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.
Learn all GATE CS concepts with Free Live Classes on our youtube channel.