I'm reading Algorithm Design and Applications, by Michael T. Goodrich and Roberto Tamassia, published by Wiley. They teach the concept of primitive operations and how to count then in a given algorithm. Everything was clear to me until the moment they showed a recursive function (a simple recursive way to calculate the maximum value of an array) and its primitive operation count.

The function (in pseudo-code) is this:

Algorithm recursiveMax(A, n):
    Input: An array A storing n ≥ 1 integers.
    Output: The maximum element in A.
    if n = 1 then
        return A[0]
    return max{recursiveMax(A, n − 1), A[n − 1]}

where A is an array and n its length. The author states what follows concerning how we calculate the number of primitive operations this function has:

As with this example, recursive algorithms are often quite elegant. Analyzing the running time of a recursive algorithm takes a bit of additional work, however. In particular, to analyze such a running time, we use a recurrence equation, which defines mathematical statements that the running time of a recursive algorithm must satisfy. We introduce a function T (n) that denotes the running time of the algorithm on an input of size n, and we write equations that T (n) must satisfy. For example, we can characterize the running time, T (n), of the recursiveMax algorithm as T(n) = 3 if n = 1 or T(n - 1) + 7 otherwise, assuming that we count each comparison, array reference, recursive call, max calculation, or return as a single primitive operation. Ideally, we would like to characterize a recurrence equation like that above in closed form, where no references to the function T appear on the righthand side. For the recursiveMax algorithm, it isn’t too hard to see that a closed form would be T (n) = 7(n − 1) + 3 = 7n − 4.

I can clearly understand that in the case of a single item array, our T(n) would be just 3 (only 3 primitive operations will occur, i.e. the comparision n = 1, the array index A[0] and the return operation), but I cannot understand why in the case where n is not 1 we have T(n-1) + 7. Why + 7? From where did we get this constant?

Also, I cannot comprehend this closed form: how did he get that T(n) = 7(n - 1) + 3?

I appreciate any help.

  • 1
    $\begingroup$ Honestly, this constant doesn't matter. We almost never care for constants when we compute the running time. $\endgroup$
    – nir shahar
    Jun 20, 2021 at 23:52

1 Answer 1


For $n\geq 2$, I believe that the $T(n) = T(n-1) + 7$ because of the comaparison operation $n = 1$, returning $\max$, computing $\max$, calling recursiveMax, computing recursiveMax, i.e., $T(n-1)$ term, computing $n-1$ twice, accesing $A[n-1]$.

However, the constant should not matter, since we are interested in asymptotic running time analysis, as nir shahar is saying.

Now, we have $T(n) = T(n-1) + 7$ for $n\geq 2$. You can expand this term, as follows:

\begin{align} T(n) &= T(n-2) + 2\cdot7 \\ &= T(n-3) + 3 \cdot 7 \\ &= T(n-4) + 4 \cdot 7 \\ &=... \\ &=T(1) + 7 \cdot (n-1) \\ &= 3 + 7 \cdot (n-1) \end{align}

This answers your second question.


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