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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
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not my style of question. haha
it looks like some type of for/next looping but the syntax is hard to parse
Its suppose to be discrete mathematics. I am not a math person at all.

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Other answers:

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
was this a write up or did you copy/paste it?
I will do a screenshot and paste it here hang on
Here it is
1 Attachment
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
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
pfft, i left the 14 in there ... should be a 1 :)
x = 7606 if its 1 to start with
Is that the answer for "Count the number of + operations done by this algorithm."
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 ...

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