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big o notation for reading data in blocks

Hi,

If there is a file of p bytes (characters), you would take O(p) time to read the file. But, if you read the file in blocks of size o, would the big o notation for the time complexity be O((p/o)*p)?

Thanks a lot
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zizi21
Asked:
zizi21
3 Solutions
 
ozoCommented:
Unlikely, but it may be conceivable that some implementation could have been written to behave that way.
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zizi21Author Commented:
I did a program that read the first o characters and then the second o characters. But, I am not sure of the complexity. To read the entire file, you would use p/o blocks. So, the complexity is (p/o)*p. However, I am not sure if this is correct for this implementation. Thanks a lot.
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ozoCommented:
Do you read the entire file each time you read a block?
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zizi21Author Commented:
Thanks for replying.

I only read a block each time (and not the entire file).
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ozoCommented:
So what is the time to read the first o characters?
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zizi21Author Commented:
The time to read the first o characters is O(o).
The time to read the second o characters is O(o)
The time to read the last o characters is O(o).

At first, I thought, it should be added. O(o)+O(o) which gives O(p). But, sometimes, I see people put the number of blocks or the number of rounds that is required to read the entire O(p). So, p/o gives you the number of blocks or rounds. So, you need p/o rounds or blocks to read the entire file. I see, you don't times with p but with o. That is, (p/o)*o. So, it is O(p) I think.

I understand that constant is ignored. p/o looks like a constant but it would given different values for different data. So, it can be considered a variable. That is the reason, I am multipying with  p/o...
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ozoCommented:
That is, (p/o)*o. So, it is O(p/o^2)
Look more carefully.
[Edit: sorry, I see you already did]
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zizi21Author Commented:
Sorry ozo, that is what I wrote and then, I realise that it is o^2.  

This means that the big o notation is very simple. That is, O(p). It doesn't matter whether you read the files in blocks or at once.

Sorry for sounding stupid. p/o is a variable (and not a constant) because it gives a different value for different files. But, it would give similar value for the same input file. Did I understand this wrong ?
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käµfm³d 👽Commented:
What if you try using actual values for demonstration?

Let's say you have a file of length 10 bytes (p). If you read 1 byte at a time, then you have 10 rounds, and that would be O(p). If you now break that up into 2 byte chunks (o), you now have 5 rounds. Since it takes 5 rounds to read the file, should the overall time not be O(p/o)--the overall size of the file divided by the size of the chunk. (I am assuming you're considering the time taken to read a chunk to be constant time, and not factoring in the reading of each byte into the complexity of reading a chunk.) I am not sure why you would want to multiply p or o to the division.
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zizi21Author Commented:
Hi kaufmed,

I am a newbie in big o notations (I am trying to understand myself). This is what I understand (I may be wrong).

Say o is 5 and p is 10
It takes O(5) to read the first 5 characters
Another O(5) to read the second 5 characters

In total, it takes O(5+5) = O(10)
Now, we can get the same number if take the number of rounds which is 2 here and multply it with O(5) which gives O(10).

The confusion is that you need to ignore the constants but you keep the variables..
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ozoCommented:
O(5) = O(10) = O(1)

Ignoring constants, what happens when p doubles?  What happens when o doubles?
What happens when p changes by a factor of 10?  100?  1000000?
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d-glitchCommented:
If you read a p-byte file one byte at a time, it will take p reads.
If you read a p-byte file o bytes at a time, it will take p/o reads.

But in either case, if the file is 2*p or 10*p it will take 2 or 10 times longer.

Reading is inherently a linear operation, so O(p) would be correct.
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zizi21Author Commented:
Thanks very much for the replies..
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zizi21Author Commented:
I think, i understand now. When p doubles, the time doubles but the doubling is never reflected in the big oh notation.
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ozoCommented:
I would say that the doubling is reflected in the big oh notation in the sense that
O(p) implies that when p doubles, the time doubles
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