5 ms·

Let's say that our file consists of nothing but the new line character (1 byte). At 2^64 lines, that's 18.45 EB. Sure, for files larger than 18.45 EB at the m

Let's say that our file consists of nothing but the new line character (1 byte). At 2^64 lines, that's 18.45 EB. Sure, for files larger than 18.45 EB at the minimum, computing the new line count will take O(log N) time. I'm all for nitpicking the issues with Big O notation, but this instance doesn't strike me as one.

deleted 7y ago
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OskarS 7y ago
OP said “to be really precise, O(log n)”, to avoid nitpickers that might have added corrections. And it is absolutely true: to store the number N, you need O(log N) bits. That is how Big O notation works. The lesson to take away here is that O(log N) is REALLY REALLY small, for all sensible values of N. If an O(log N) algorithm is slow, it’s not because of the O(log N) part, it’s because of the hidden factors in O notation.
- lorenzhs 7y ago
  > That is how Big O notation works. No it is not. "Big O" is a way of expressing asymptotic growth of some quantity, most often running time. But you need a machine model to even express running time. And sure, you can define your machine model to require O(log n) time for adding two numbers, but that's not particularly useful. In particular, the word RAM model—which is what is normally used—assumes a word size of θ(log n), and that basic operations on those words take a constant amount of time. We use models because otherwise even the simplest things would be impossible to analyse. Or do you want to include a three-level cache hierarchy, micro-op decoding, the paging system, branch mispredictions and memory timings in your complexity analysis? Why is it always the "bUt AdDinG tWo nuMbERs tAkeS O(log n) tImE" hill that HN commenters are willing to die on?
  - nothrabannosir 7y ago
    To be fair he did say O(1), and qualified the O(log N) with both parentheses and “to be really precise”. If you can’t be precise about the definition of O on HN, then where? (which, “to be really precise”, is a polynomial time tape based Turing machine, not a modern RAM machine with log N words. It’s pointless, but fun to remember :) )
    - lorenzhs 7y ago
      No no no! Big O has nothing to do with Turing machines at all! You can, of course, express the number of steps a single-tape deterministic Turing machine needs to solve some problem using Big O notation. But you can just as well express the number of Qubits that a quantum computer needs to solve some other problem in Big O notation. It has nothing to do with the quantity being measured. Are you perhaps mixing up the definition of complexity classes such as P and NP with Big O notation? First you need to settle on a model. Then you can express time, space, or some other quantity in that model using Big O notation. Arguably, I've found that HN is one of the worst places to be precise about Big O notation. Everyone here seems to think they know what it is, only to lay bare their misconceptions about it a second later.
  - OskarS 7y ago
    You're simply wrong here. The algorithm is O(log n), whether you agree to it or not. First of all, there's absolutely nothing that says that we have to assume that the number of lines is less than 2^64. Big O notation does not care what happens below that number, it only cares what happens when you go to infinity, which is a lot larger than 2^64. At that point you need arbitrarily sized integers, which are not fixed-space. Hence O(log n). Word size or RAM model does not matter one whit. This is how Big O notation works, it has nothing to do with practicality. As another example of a similar thing, consider Sorting Networks. The AKS sorting networks has the best "asymptotic depth" of any known sorting networks (O(n log n) depth, i think), but the hidden constants are so massive that there are no situations where the AKS networks are actually practical, for any reasonable input they are always beaten by other (asymptotically worse) networks. That doesn't mean that they "aren't O(n log n)" or whatever. Second: if we were to actually write this algorithm in the way which is most conservative with space, we wouldn't use a 64-bit integer in the first place. If we only care about memory, we'd start the count with a single-byte datatype (i.e. unsigned char) to keep the count. The algorithm now uses 1 byte of space. When that overflows (i.e. reaches 256 lines), we would change to a bigger data type, i.e. unsigned short. The algorithm now takes 2 bytes of data. When THAT overflows, we again promote the datatype to an unsigned int. Now we use 4 bytes of data. When that overflows, we finally promote it to a uint64_t. Now we're on 8 bytes of data. When that overflows... (etc.) See how the memory usage is growing? And how it's growing exactly with the logarithm of the number of lines? You're just wrong about this. Yes, you can absolutely say "assuming storage of all numbers is O(1), then the algorithm is O(1)" (essentially what OP said) but that is not a default assumption and it's not accurate to reality (since actually storing numbers requires O(log n) of space). And yes, branch prediction misses and cache misses are far more important factors here, but that doesn't change the fact that it uses O(log n) space. That's just simply true. Also: we're not dying on a hill. OP made an entirely accurate and true statement, which was met with an avalanche of people criticizing him using incorrect arguments and misunderstandings of Big O notation. We're just saying that he was correct all along.
    - lorenzhs 7y ago
      By your logic, memory access takes time Ω(n^(1/3)) because memory cells need to occupy physical space and we live in a three-dimensional reality, so the time to access a memory cell must necessarily increase with the physical distance to that memory cell. But using a model that made such assumptions would be needlessly complex, so nobody does it. (Incidentally, there was a post on HN that argued this a few years ago, and the comments were full of people who felt the need to ignore the reasoning behind the assumptions backing commonly-used machine models, much like it is happening in this thread.) Arbitrarily large numbers, represented in binary, certainly take time that is linear in the length of their representation to add. Nobody's arguing with that. But using a model that assumes bitwise operations would be needlessly complex, so outside of some really specialized contexts, nobody does it. > You're just wrong about this. Yes, you can absolutely say "assuming storage of all numbers is O(1), then the algorithm is O(1)" (essentially what OP said) but that is not a default assumption I don't know the basis on which you argue that the word RAM model isn't the default assumption, but I feel confident in claiming that outside of some niche communities, it most certainly is. > and it's not accurate to reality (since actually storing numbers requires O(log n) of space). That's why it's called a model. We use models because reality is too messy to work with, and we usually don't care about the exact details. Just ask a physicist, they make simplifying assumptions all the time because without them, they'd be getting absolutely nowhere. It's pretty much the same in algorithmics.
  - mk89 7y ago
    For God's sake, for a second I thought I had forgotten everything about the O notation. Thanks for confirming what I thought is the correct definition of big O.

deleted 7y ago

[deleted]

OskarS 7y ago

OP said “to be really precise, O(log n)”, to avoid nitpickers that might have added corrections. And it is absolutely true: to store the number N, you need O(log N) bits. That is how Big O notation works. The lesson to take away here is that O(log N) is REALLY REALLY small, for all sensible values of N. If an O(log N) algorithm is slow, it’s not because of the O(log N) part, it’s because of the hidden factors in O notation.

lorenzhs 7y ago
> That is how Big O notation works. No it is not. "Big O" is a way of expressing asymptotic growth of some quantity, most often running time. But you need a machine model to even express running time. And sure, you can define your machine model to require O(log n) time for adding two numbers, but that's not particularly useful. In particular, the word RAM model—which is what is normally used—assumes a word size of θ(log n), and that basic operations on those words take a constant amount of time. We use models because otherwise even the simplest things would be impossible to analyse. Or do you want to include a three-level cache hierarchy, micro-op decoding, the paging system, branch mispredictions and memory timings in your complexity analysis? Why is it always the "bUt AdDinG tWo nuMbERs tAkeS O(log n) tImE" hill that HN commenters are willing to die on?
- nothrabannosir 7y ago
  To be fair he did say O(1), and qualified the O(log N) with both parentheses and “to be really precise”. If you can’t be precise about the definition of O on HN, then where? (which, “to be really precise”, is a polynomial time tape based Turing machine, not a modern RAM machine with log N words. It’s pointless, but fun to remember :) )
  - lorenzhs 7y ago
    No no no! Big O has nothing to do with Turing machines at all! You can, of course, express the number of steps a single-tape deterministic Turing machine needs to solve some problem using Big O notation. But you can just as well express the number of Qubits that a quantum computer needs to solve some other problem in Big O notation. It has nothing to do with the quantity being measured. Are you perhaps mixing up the definition of complexity classes such as P and NP with Big O notation? First you need to settle on a model. Then you can express time, space, or some other quantity in that model using Big O notation. Arguably, I've found that HN is one of the worst places to be precise about Big O notation. Everyone here seems to think they know what it is, only to lay bare their misconceptions about it a second later.
- OskarS 7y ago
  You're simply wrong here. The algorithm is O(log n), whether you agree to it or not. First of all, there's absolutely nothing that says that we have to assume that the number of lines is less than 2^64. Big O notation does not care what happens below that number, it only cares what happens when you go to infinity, which is a lot larger than 2^64. At that point you need arbitrarily sized integers, which are not fixed-space. Hence O(log n). Word size or RAM model does not matter one whit. This is how Big O notation works, it has nothing to do with practicality. As another example of a similar thing, consider Sorting Networks. The AKS sorting networks has the best "asymptotic depth" of any known sorting networks (O(n log n) depth, i think), but the hidden constants are so massive that there are no situations where the AKS networks are actually practical, for any reasonable input they are always beaten by other (asymptotically worse) networks. That doesn't mean that they "aren't O(n log n)" or whatever. Second: if we were to actually write this algorithm in the way which is most conservative with space, we wouldn't use a 64-bit integer in the first place. If we only care about memory, we'd start the count with a single-byte datatype (i.e. unsigned char) to keep the count. The algorithm now uses 1 byte of space. When that overflows (i.e. reaches 256 lines), we would change to a bigger data type, i.e. unsigned short. The algorithm now takes 2 bytes of data. When THAT overflows, we again promote the datatype to an unsigned int. Now we use 4 bytes of data. When that overflows, we finally promote it to a uint64_t. Now we're on 8 bytes of data. When that overflows... (etc.) See how the memory usage is growing? And how it's growing exactly with the logarithm of the number of lines? You're just wrong about this. Yes, you can absolutely say "assuming storage of all numbers is O(1), then the algorithm is O(1)" (essentially what OP said) but that is not a default assumption and it's not accurate to reality (since actually storing numbers requires O(log n) of space). And yes, branch prediction misses and cache misses are far more important factors here, but that doesn't change the fact that it uses O(log n) space. That's just simply true. Also: we're not dying on a hill. OP made an entirely accurate and true statement, which was met with an avalanche of people criticizing him using incorrect arguments and misunderstandings of Big O notation. We're just saying that he was correct all along.
  - lorenzhs 7y ago
    By your logic, memory access takes time Ω(n^(1/3)) because memory cells need to occupy physical space and we live in a three-dimensional reality, so the time to access a memory cell must necessarily increase with the physical distance to that memory cell. But using a model that made such assumptions would be needlessly complex, so nobody does it. (Incidentally, there was a post on HN that argued this a few years ago, and the comments were full of people who felt the need to ignore the reasoning behind the assumptions backing commonly-used machine models, much like it is happening in this thread.) Arbitrarily large numbers, represented in binary, certainly take time that is linear in the length of their representation to add. Nobody's arguing with that. But using a model that assumes bitwise operations would be needlessly complex, so outside of some really specialized contexts, nobody does it. > You're just wrong about this. Yes, you can absolutely say "assuming storage of all numbers is O(1), then the algorithm is O(1)" (essentially what OP said) but that is not a default assumption I don't know the basis on which you argue that the word RAM model isn't the default assumption, but I feel confident in claiming that outside of some niche communities, it most certainly is. > and it's not accurate to reality (since actually storing numbers requires O(log n) of space). That's why it's called a model. We use models because reality is too messy to work with, and we usually don't care about the exact details. Just ask a physicist, they make simplifying assumptions all the time because without them, they'd be getting absolutely nowhere. It's pretty much the same in algorithmics.
- mk89 7y ago
  For God's sake, for a second I thought I had forgotten everything about the O notation. Thanks for confirming what I thought is the correct definition of big O.