For the algorithm below, compute the number of multiplications performed, as well as the final value of s.
p ← 1
s ← 0
for
i is in {1, 2, 3, 4}
do

p ← p · 3

s ← s + p
number of multiplications
final value of s s =

Question

For the algorithm below, compute the number of multiplications performed, as well as the final value of s.
p ← 1
s ← 0
for
i is in {1, 2, 3, 4}
do
     	
p ← p · 3
	
s ← s + p
number of multiplications     		
final value of s     	s =

anonymous · Answer

@oldrin.bataku

anonymous · Answer

The for loop construct will go through $i=1,2,3,4$ so our loop code will occur $4$ times, meaning we multiply $4$ times since each iteration does $p\leftarrow p\cdot3$

anonymous · Answer

First, consider that iteration, $p$ is raised to a higher power of the base $3$. At initialization time, we have $p=1$. In the first iteration, we get $p=3^1$, in the second we get $p=3^2$, etc. giving us $p=3^4$ in our fourth iteration. These values of $p$ each iteration are accumulated in $s$ so at the end we just have the sum of four powers of $3$:$$s=3^1+3^2+3^3+3^4=3+9+27+81=120$$