From Wisdom90@21:1/5 to All on Fri Jan 10 08:53:19 2020
Hello,

I have just read the following webpage of a PhD Computer Scientist and researcher from Montreal Quebec Canada where i am living now from year
1989, here is the webpage and read it carefully:

Better computational complexity does not imply better speed

https://lemire.me/blog/2019/11/26/better-computational-complexity-does-not-imply-better-speed/

And here is his resume:

https://lemire.me/pdf/resume/resumelemire.pdf

So as you are noticing on the webpage above he is saying the following

"But it gets worse: these are not scientific models. A scientific model
would predict the running time of the algorithm given some
implementation, within some error margin. However, these models do
nothing of the sort. They are purely mathematical. They are not
falsifiable. As long as they are mathematically correct, then they are
always true. To be fair, some researchers like Knuth came up with models
that closely mimic reasonable computers, but that’s not what people
pursuing computational complexity bounds do, typically."

But i don't agree with him because i think he is not understanding
the goal of computational complexity, because when we say that
an algorithm has a time complexity of n*log(n), you have to
understand that it is by logical analogy like saying in physics what is
the material resistance, because the n*log(n) means how well the
algorithm is "amortizing"(it means reducing) the "time" that it takes
taking as a reference of measure the average computational complexity of
the giving algorithm, but we can easily notice that an algorithm is
becoming faster, in front of the data that is giving to him, by going
from an exponential time complexity towards a logarithmic time
complexity, so we are noticing that the computational complexity is
"universal" and it measures how well the algorithm amortizes(that means
it reduces) the time that it takes taking as a reference of measure the
average computational complexity of the giving algorithm,
so this is is why computational complexity is scientific and also it is engineering and it gives us information on the physical world.

So to give an interesting example of science in computing, we can ask of
what is the time complexity of a binary search algorithm, and here
is my mathematical calculations of its time complexity:

Recurrence relation of a binary search algorithm is: T(n)=T(n/2)+1

Because the "1" is like a comparison that we do in each step of
the divide and conquer method of the binary search algorithm.

So the calculation of the recurrence equation gives:

1st step=> T(n)=T(n/2) + 1

2nd step=> T(n/2)=T(n/4) + 1 ……[ T(n/4)= T(n/2^2) ]

3rd step=> T(n/4)=T(n/8) + 1 ……[ T(n/8)= T(n/2^3) ]

.

.

kth step=> T(n/2^k-1)=T(n/2^k) + 1*(k times)

Adding all the equations we get, T(n) = T(n/2^k) + k times 1

This is the final equation.

So how many times we need to divide by 2 until we have only one element
left?

So it must be:

n/2^k= 1

This gives: n=2^k

this give: log n=k [taken log(base 2) on both sides ]

Put k= log n in the final equation above and it gives:

T(n) = T(1) + log n

T(n) = 1 + log n [we know that T(1) = 1 , because it’s a base condition
as we are left with only one element in the array and that is the
element to be searched so we return 1]

So it gives:

T(n) = O(log n) [taking dominant polynomial, which is n here)

This is how we got “log n” time complexity for binary search.

Thank you,
Amine Moulay Ramdane.

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