Consider the following algorithm.
x ← 1
for i is in {1, 2, 3, 4} do
for j is in {1, 2, 3, 4, 5} do

x ← x + x

for k is in {1, 2, 3} do

x ← x + 1
x ← x + 5
Count the number of + operations done by this algorithm.

Question

Consider the following algorithm.
x ← 1
for i is in {1, 2, 3, 4} do
     for j is in {1, 2, 3, 4, 5} do
	     	     	

      x ← x + x

for k is in {1, 2, 3} do
			
       x ← x + 1
       x ← x + 5
Count the number of + operations done by this algorithm.

badhi · Answer

for each value i from 1 to 4, j has to go 1 to 5. Therefore $x \gets x+x $ happens $4\times 5=20$ times.

for each value k from 1 to 3, $x \gets x+1$ and $x\gets x+5$ happens. Therefore in that loop summation occurs $3\times 2=6$ times.

Therefore all together there are 20+6=26 summations

anonymous · Answer

hmm. I tried 26 and it was incorrect.. any ideas?

anonymous · Answer

@BAdhi

badhi · Answer

are you sure that the loop which uses k is not inside the loop which uses i and 
are you sure that they are not asking for the final value of x

anonymous · Answer

attachment

badhi · Answer

yeah now its a whole different story.(this is why I told you to use the indentation correctly)
consider inner loops for one value of i,

$x\gets x+x$ happens 5 times and  each $x\gets x+1$ and $x\gets x+5$ happens 3 times so all together $5+3\times2=11$ times. This is just for one value of i. 

For each value of i from 1 to 4, above scenario repeats for 4 times, which gives us $11\times 4=44$

anonymous · Answer

ok got it! thanks Badhi!

badhi · Answer

ur welcome:)

anonymous · Answer

I don't get it!! where did you get the 2?

badhi · Answer

where did you get this question?? :P its a year old one

badhi · Answer

you mean $5+3\times \color{red} 2$