Question

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

Answer 1

any ideas ?

Answer 2

you may wrap your code between ``` and ``` for highlighting the syntax

Answer 3

No very well familiar with this subject

Answer 4

for each value of i, you do the j loop. the j loop does 4 executions. the i loop does all of that 3 times. so 3*4= 12 times for both.
you do 1 add inside the i, j loops, so 12 adds for that part.

Answer 5

the indenting matters (a lot). According to the indenting, the k loop is independent of the i,j loops. So you do that one separately. How many adds for the k loop?
finally, the last statement is not in any loop. You do it once (so it does 1 last add)

Answer 6

15 for k loop

Answer 7

you would get 15 for k loop only if the k loop was "inside" the i loop. But according to the indentation, the k loop is not inside the i loop.

if the k loop statement was indented 2 spaces, it would be inside the i loop.

if it was indented a total of 4 spaces, the k would be inside both the i and j loops.

Answer 8

so what loop is it in

Answer 9

so i guess it is in the k and i loop

Answer 10

the k loop is at the same level as the i loop
the logic is:

do the i loop (which involves also doing the j loop)
when that is done, do the k loop
when that is done, do the last statement

Answer 11

Can anyone explain this problem to me

Answer 12

x <-- 1
for i is in {1, 2, 3} do

for j is in {1, 2, 3, 4} do
x <-- x + x
end j

for k is in {1, 2, 3, 4, 5} do
x<-- x + 1
x <-- x + 5
end k
end i

I added "end statements" to show where the control of each loop ends

the i loop is executed 3 times. that means everything "inside" it is executed 3 times

the j loop is executed 4 times. that means the statement x<-- x+x is performed 4 times, and then the j loop ends. But because we are inside the i loop, we will do the j loop 3 times, and the x<-- x+x will be executed 3*4= 12 times. (so 12 adds)

after the j look ends, the k loop is started. It goes 5 times. That means the 2 statements "inside" the k loop are executed 5 times. Because we are inside the i loop, the statements inside the k loop will be done 3*5=15 times total. There are 2 statements , and 2 adds, so 30 adds will be performed.

total adds will be 30+12= 42

Answer 13

the flow will go:
for i=1 do
j loop (which means for j=1 , up to j=4, do x<-- x+x)
(we do 4 adds)
then do the k loop (we do x<-- x + 1, x <-- x + 5 (2 adds), and we do them 5 times (for each value of k from k=1 to k=5). we do 10 adds

we get to the end of the i loop, having done 14 adds
we now do the i loop, with i=2. we end up doing another 14 adds
finally, we do the i loop with i=3, and another 14 adds

after i=3, the i loop is complete and we are done
total adds: 14+14+14= 42 adds