Counting Primitive Operations

Question

This is the solution I've been given for counting primitive Operation in an algorithm. I think I have my head around how all the operations are found, for instance the 2(n-1), the 2 is the primitive operations on that line and the n-1 is the amount of times it will be ran. I'm confused about how the final result was calculated 7n -2 and the use of brackets.

Algorithm arrayMax(A, n)                        # operations
Find Max element in an array A of n elements
currentMax <- A[0]                              2
for i <- 1 to n - 1 do                          2n
if A[i] > currentMax then                       2(n - 1)
currentMax <- A[i]                              2(n - 1)
{ increment counter i }                         n - 1
 return currentMax                              1
                                   Total        7n - 2

how can incremeting a counter be two primitive operations though? — mLstudent33, Apr 22 '20 at 22:30

5xum · Accepted Answer · 2015-03-16T13:45:06.160

1

$$2+2n+2(n-1)+2(n-1)+n-1+1 = \\=2+2n+2n-2+2n-2+n-1+1=\\=2n+2n+2n+n+2-2-2-1+1=7n-2$$

Is anything still unclear?

edited Mar 16 '15 at 13:45

answered Mar 16 '15 at 13:39

5xum

123,496
6
128
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No, that makes sense now, thank you. – JTK Mar 16 '15 at 13:55

Counting Primitive Operations

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