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Help on big oh analysis
Hi,
I am trying to understand big oh notations. Please correct me.
Say there is a program that accesses a file on disk randomly and say the size of the file n.
So, the big oh notation for this program is O(n) this is because it may access the file randomly at n positions. However, there is a possibility that the program assess the data for the file from the cache but lets ignore this for now.
If the program accesses the file sequantially (one after the other), the big oh notation is O(1).
Is this right ?
Thank you very much.
I am trying to understand big oh notations. Please correct me.
Say there is a program that accesses a file on disk randomly and say the size of the file n.
So, the big oh notation for this program is O(n) this is because it may access the file randomly at n positions. However, there is a possibility that the program assess the data for the file from the cache but lets ignore this for now.
If the program accesses the file sequantially (one after the other), the big oh notation is O(1).
Is this right ?
Thank you very much.
AlgorithmsC
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phoffric
ASKER
Doesn't making random accesses to disk does not make a difference to the big oh notation ?
ASKER
Thanks ozo. I thought random accesses influeces big oh notations ?
It can.
What disk operations does the program require?
What disk operations does the program require?
ASKER
Thanks ozo.
The program accesses the data on file on random positions...But, how do I write that in big oh notation?
The program accesses the data on file on random positions...But, how do I write that in big oh notation?
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Thank you very much for the reply.
Does this mean that if seeking gets slower in large files, it is always sqrt(n)? But, why sqrt(n) and not only other number ? Seeking would get slower in larger files because the files would not be cached in RAM. But, if the file is small, it is cached in RAM and accessing it is fast.
For example, if the size of the file is larger than the available RAM, we can say the file is not cached into RAM. And, in this situation, making random accesses is expensive.
Does this mean that if seeking gets slower in large files, it is always sqrt(n)? But, why sqrt(n) and not only other number ? Seeking would get slower in larger files because the files would not be cached in RAM. But, if the file is small, it is cached in RAM and accessing it is fast.
For example, if the size of the file is larger than the available RAM, we can say the file is not cached into RAM. And, in this situation, making random accesses is expensive.
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Thank you very much for all the replies. I am studying it now.
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"If you read the data sequentially, it will take twice as long to read the entire file if the data doubles, so sequential reading is also an O(n) operation. "
Thank you very much for your reply. If I understand this correctly, accessing each record of data takes O(1) time (for sequential access) and O(n) for the entire file.
Thank you very much for your reply. If I understand this correctly, accessing each record of data takes O(1) time (for sequential access) and O(n) for the entire file.
Under that scenario, reading 1 record is O(1), reading n records is O(n)
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Thank you very much....
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Thank you .
ASKER
I spent days trying to understand and all of you helped lots...Thank you verymuch.
For very large files, I was saying that because additional cylinders will result in the average seek time in accessing a single record will increase linearly as the number of cylinders increase, then reading one record is O(n), not O(1).
Suppose a file is small, and that the additional records are placed in the same cylinder. In that case the reading of one records is O(1).
Suppose a file is small, and that the additional records are placed in the same cylinder. In that case the reading of one records is O(1).
O(n) time to access all of the file.