anonymous
  • anonymous
Consider the following algorithm. x ← 1 for i is in {1, 2, 3, 4} do for j is in {1, 2, 3} do x ← x + x for k is in {1, 2, 3, 4, 5, 6} do x ← x + 1 x ← x + 5 Count the number of + operations done by this algorithm.
Discrete Math
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
not my style of question. haha
amistre64
  • amistre64
it looks like some type of for/next looping but the syntax is hard to parse
anonymous
  • anonymous
Its suppose to be discrete mathematics. I am not a math person at all.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
its countable ... but the syntax/notation is hard to read thru. without knowing what the format is trying to convey its difficult to understand the process that is being described
amistre64
  • amistre64
was this a write up or did you copy/paste it?
anonymous
  • anonymous
I will do a screenshot and paste it here hang on
anonymous
  • anonymous
Here it is
1 Attachment
amistre64
  • amistre64
hmm, im guessing it follows as such: x = 1 for i = 1 to 4 for j = 1 to 3; x=x+x for k = 1 to 6; x=x+1, x=x+5 4(3(1)+6(2)) operations is what it looks like to me
amistre64
  • amistre64
x=n [x = n+n x = (n+n)+(n+n) x = (n+n)+(n+n)+(n+n)+(n+n) = 8(n) ] [x = 8(n)+6] x = 1 x'= 8(1)+6 x'= 8(8(14)+6)+6 x'= 8(8(8(14)+6)+6)+6 x'= 8(8(8(8(14)+6)+6)+6)+6 x = 60854
amistre64
  • amistre64
pfft, i left the 14 in there ... should be a 1 :)
amistre64
  • amistre64
x = 7606 if its 1 to start with
anonymous
  • anonymous
Is that the answer for "Count the number of + operations done by this algorithm."
amistre64
  • amistre64
of course not, i already surmised the results for that .... this was just an attempt to actually work the algorithm to see what the output would give us ...

Looking for something else?

Not the answer you are looking for? Search for more explanations.