Big O Notation - Orders of Magnitude - algorithm

I am currently trying to work out what are the main quantitative differences between a quadratic algorithm, a cubic algorithm and one which is exponential.
What I don't understand is what it means by "Quantitative" and what it really is asking? I tried searching for information on this but no luck.
Thanks.

When you use big-O notation to estimate the computational complexity of an algorithm,
the goal is to provide a qualitative insight as to how changes in N affect the algorithmic
performance as N becomes large.
If you eliminate any term whose contribution to the total ceases to be significant as N becomes large and eliminate any constant factors, I guess you could say you are left with the main quantitative difference.

Quantitative differences just means differences in quantity -- i.e. what are the size differences between those different kinds of algorithms? It would be good idea to give numeric examples, e.g. show the running time of quadratic, cubic, and exponential algorithms for some example problem sizes.

Related

Big Theta of Factorial Multiplied by a Coefficient

For a function with a run time of (cn)! where c is a coefficient >= 0 and c != n, would the tight bound of the run be Θ(n!) or Θ((cn)!)? Right now, I believe it would be Θ((cn)!) since they would differ by a coefficent >= n since cn != n.
Thanks!
Edit: A more specific example to clarify what I'm asking:
Will (7n)!, (5n/16)! and n! all be Θ(n!)?
You can use Stirling's approximation to get that if c>1 then (cn)! is asymptotically larger than pow(c,n)*n!, which is not O(n!) since the quotient diverges. As a more elementary approach consider this example for c=2: (2n)!=(2n)(2n-1)...(n+1)n!>n!n! and (n!n!)/n!=n! diverges, so (2n)! is NOT O(n!).
Will (7n)!, (5n/16)! and n! all be Θ(n!)?
I think there are two answers to your question.
The shorter one is from the purely theoretical point of view. Of those 3 only the n! lies in the class of Θ(n!). The second lies in the O(n!) (note big-O instead of big-Theta) and (7n)! is slower than Θ(n!), it lies in Θ((7n)!)
There is also a longer but more practical answer. And to get to it we first need to understand what is the big deal with this whole big-O and big-Theta business in the first place?
The thing is that for many practical tasks there are many algorithms and not all of them are equally or even similarly efficient. So the practical question is: can we somehow capture this difference in performance in an easy to understand and compare way? And this is the problem that big-O/big-Theta are trying to solve. The idea behind this method is that if we look at some algorithm with some complicated real formula for the exact time, there is only 1 term that grows faster than all others and thus dominates the time as the problem gets bigger. So let's compress this big formula to that dominant term. Then we can compare those terms and if they are different, we can easily say which is the better algorithm (7*n^2 is clearly better than 2*n^3).
Another idea is that the term "operation" is usually not that well defined at the level people usually think about algorithms. Which "operation" actually maps to a single CPU instruction and which to a few depends on many factors such as particular hardware. Also the instructions themselves can take different time to execute. Moreover sometimes the algorithm's working time is dominated by memory access than CPU instructions and those components are not easily additive. The morale of this story is that if two algorithms are different only in a scalar coefficient, you can't really compare those algorithms just theoretically. You need to compare some implementations in some particular environment. This is why algorithms complexity measure typically boils down to something like O(n^k) where k is a constant.
There is one more consideration: practicality. If the algorithm is some polynomial, there is a huge practical difference between cases a=3 and a=4 in O(n^a). But if it is something like O(2^(n^a)), then there is not much difference what exactly the a as along as a>1. This is because 2^n grows fast enough to make it impractical for almost any realistic n irrespective of a. So in practical terms it is often good enough approximation to put all such algorithms into a single "exponential algorithms" bucket and say they are all impractical even despite the fact there is a huge difference between them. This is where some mathematically unconventional notations like 2^O(n) come from.
From this last practical perspective the difference between Θ(n!) and Θ((7n)!) is also very little: both are totally impractical because both lie beyond even the exponential bucket of 2^O(n) (see Stirling's formula that shows that n! grows a bit faster than (n/e)^n). So it makes sense to put all such algorithms in another bucket of "factorial complexity" and mark them as impractical as well.

Right way to discuss computational complexity for small n

When discussing computational complexity, it seems everyone generally goes straight to Big O.
Lets say for example I have a hybrid algorithm such as merge sort which uses insertion sort for smaller subarrays (I believe this is called tiled merge sort). It's still ultimately merge sort with O(n log n), but I want to discuss the behaviour/characteristics of the algorithm for small n, in cases where no merging actually takes place.
For all intents and purposes the tiled merge sort is insertion sort, executing exactly the same instructions for the domain of my small n. However, Big O deals with the large and asymptotic cases and discussing Big O for small n is pretty much an oxymoron. People have yelled at me for even thinking the words "behaves like an O(n^2) algorithm in such cases". What is the correct way to describe the algorithm's behaviour in cases of small n within the context of formal theoretical computational analysis? To clarify, not just in the case where n is small, but in the case where n is never big.
One might say that for such small n it doesn't matter but I'm interested in the cases where it does, for example with a large constant such as being executed many times, and where in practice it would show a clear trend and be the dominant factor. For example the initial quadratic growth seen in the graph below. I'm not dismissing Big O, more asking for a way to properly tell both sides of the story.
[EDIT]
If for "small n", constants can easily remove all trace of a growth rate then either
only the asymptotic case is discussed, in which case there is less relevance to any practical application, or
there must be a threshold at which we agree n is no longer "small".
What about the cases where n is not "small" (n is sufficiently big that the growth rate will not to affected significantly by any practical constant), but not yet big enough to show the final asymptotic growth rate so only sub growth rates are visible (for example the shape in the image above)?
Are there no practical algorithms that exhibit this behaviour? Even if there aren't, theoretical discussion should still be possible. Do we measure instead of discussing the theory purely because that's "what one should do"? If some behaviour is observed in all practical cases, why can't there be theory that's meaningful?
Let me turn the question around the other way. I have a graph that shows segmented super-linear steps. It sounds like many people would say "this is a pure coincidence, it could be any shape imaginable" (at the extreme of course) and wouldn't bat an eyelid if it were a sine wave instead. I know in many cases the shape could be hidden by constants, but here it's quite obvious. How can I give a formal explanation of why the graph produces this shape?
I particularly like #Sneftel's words "imprecise but useful guidance".
I know Big O and asymptotic analysis isn't applicable. What is? How far can I take it?
Discuss in chat
For small n, computation complexity - how things change as n increases towards infinity - isn't meaningful as other effects dominate.
Papers I've seen which discuss behaviour for small values of n do so by measuring the algorithms on real systems, and discuss how the algorithms perform in practice rather than from a theoretical viewpoint. For example, for the graph you've added to your post I would say 'this graph demonstrates an O(N) asymptotic behaviour overall, but the growth within each tile is bounded quadratic'.
I don't know of a situation where a discussion of such behaviour from a theoretical viewpoint would be meaningful - it is well known that for small n the practical effects outweigh the effects of scaling.
It's important to remember that asymptotic analysis is an analytic simplification, not a mandate for analyzing algorithms. Take selection sort, for instance. Yes, it executes in O(n^2) time. But it is also true that it performs precisely n*(n-1)/2 comparisons, and n-1-k swaps, where k is the number of elements (other than the maximum) which start in the correct position. Asymptotic analysis is a tool for simplifying the (otherwise generally impractical) task of performance analysis, and one we can put aside if we're not interested in the "really big n" segment.
And you can choose how you express your bounds. Say a function requires precisely n + floor(0.01*2^n) operations. That's exponential time, of course. But one can also say "for data sizes up to 10 elements, this algorithm requires between n and 2*n operations." The latter describes not the shape of the curve, but an envelope around that curve, giving imprecise but useful guidance about the practicalities of the algorithm within a particular range of use cases.
You are right.
For small n, i.e. when only insertion sort is performed, the asymptotic behavior is quadratic O(n^2).
And for larger n, when tiled merge sort enters into play, the behavior switches to O(n.Log(n)).
There is no contradiction if you remember that every behavior has its domain of validity, before the switching threshold, let N, and after it.
In practice there will be a smooth blend between the curves around N. But in practice too, that value of N is so small that the quadratic behavior does not have enough "room" to manifest itself.
Another way to deal with this analysis is to say that N being a constant, the insertion sorts take constant time. But I would disagree to say that this is a must.
Let's unpack things a bit. Big-O is a tool for describing the growth rate of a function. One of the functions to which it is commonly applied is the worst-case running time of an algorithm on inputs of length n, executing on a particular abstract machine. We often forget about the last part because there is a large class of machines with random-access memory that can emulate one another with only constant-factor slowdown, and the class of problems solvable within a particular big-O running-time bound is equivalent across these machines.
If you want to talk about complexity on small inputs, then you need to measure constant factors. One way is to measure running times of actual implementations. People usually do this on physical machines, but if you're hardcore like Knuth, you invent your own assembly language complete with instruction timings. Another way is to measure something that's readily identifiable but also a useful proxy for the other work performed. For comparison sorts, this could be comparisons. For numerical algorithms, this is often floating-point operations.
Complexity is not about execution time for one n on one machine, so there is no need to consider it even if constant is large. Complexity tells you how the size of the input affects execution time. For small n, you can treat execution time as constant. This is the one side.
From the second side you are saying that:
You have a hybrid algorithm working in O(n log n) for n larger than some k and O(n^2) for n smaller than k.
The constant k is so large that algorithm works slowly.
There is no sense in such algorithm, because you could easily improve it.
Lets take Red-black tree. Operations on this tree are performed in O(n log n) time complexity, but there is a large constant. So, on normal machines, it could work slowly (i.e. slower than simpler solutions) in some cases. There is no need to consider it in analyzing complexity. You need to consider it when you are implementing it in your system: you need to check if it's the best choice considering the actual machine(s) on which it will be working and what problems it will be solving.
Read Knuth's "The Art of Computer Programming series", starting with "Volume 1. Fundamental Algorithms", section "1.2.10: Analysis of an Algorithm". There he shows (and in all the rest of his seminal work) how exact analysis can be conducted for arbitrary problem sizes, using a suitable reference machine, by taking a detailed census of every processor instruction.
Such analyses have to take into account not only the problem size, but also any relevant aspect of the input data distribution which will influence the running time. For simplification, the analysis are often limited to the study of the worst case, the expected case or the output-sensitive case, rather than a general statistical characterization. And for further simplification, asymptotic analysis is used.
Not counting the fact that except for trivial algorithms the detailed approach is mathematically highly challenging, it has become unrealistic on modern machines. Indeed, it relies on a processor behavior similar to the so-called RAM model, which assumes constant time per instruction and per memory access (http://en.wikipedia.org/wiki/Random-access_machine). Except maybe for special hardware combined to a hard real-time OS, these assumptions are nowadays completely wrong.
When you have an algorithm with a time complexity say O(n^2).And you also have an another algorithm with a time complexity, say O(n).Then from these two time complexity you can't conclude that the latter algorithm is faster than the former one for all input values.You can only say latter algorithm is asymptotically(means for sufficiently large input values)faster than the former one.Here you have to keep in mind the fact that in case of asymptotic notations constant factors are generally ignored to increase the understand-ability of the time complexity of the algorithm.As example: marge sort runs in O(nlogn) worst-case time and insertion sort runs in O(n^2) worst case time.But as the hidden constant factors in insertion sort is smaller than that of marge sort, in practice insertion sort can be faster than marge sort for small problem sizes on many machines.
Asymptotic notation does not describe the actual running-time of an algorithm.Actual running time is dependent on machine as different machine has different architecture and different Instruction Cycle Execution time.Asymptotic notation just describes asymptotically how fast an algorithm is with respect to other algorithms.But it does not describe the behavior of the algorithm in case of small input values(n<=no).The value of no (threshold) is dependent on the hidden constant factors and lower order terms.And hidden constant factors are dependent on the machine on which it will be executed.

How to translate algorithm complexity to time necessary for computation

If I know complexity of an algorithm, can I predict how long it will compute in real life?
A bit more context:
I have been trying to solve university assignment which has to find the best possible result in a game from given position. I have written an algorithm and it works, however very slow. The complexity is O(n)=5^n . For 24 elements it computes a few minutes. I'm not sure if it's because my implementation is wrong, or if this algorithm is simply very slow. Is there a way for me to approximate how much time any algorithm should take?
Worst case you can base on extrapolation. So having time on N=1,2,3,4 elements (the more the better) and O-notation estimation for algorithm complexity you can estimate time for any finite number. Another question this estimation precision goes lower and lower as N increases.
What you can do with it? Search for error estimation algorithms for such approaches. In practice it usually gives good enough result.
Also please don't forget about model adequateness checks. So having results for N=1..10 and O-notation complexity you should check 'how good' your results correlate with your O-model (if you can select numbers for O-notation formula that meets your results). If you cannot get numbers, you need either more numbers to get wider picture or ... OK, you can have wrong complexity estimation :-).
Useful links:
Brief review on algorithm complexity.
Time complexity catalogue
Really good point to start - look for examples based on code as input.
You cannot predict running time based on time complexity alone. There are many factors involved: hardware speed, programming language, implementation details, etc. The only thing you can predict using the complexity is expected time increase when the size of the input increases.
For example, personally, I've seen O(N^2) algorithms take longer than O(N^3) ones, especially on small values of N, such as it is in your case. And by, the way, 5^24 is a huge number (5.9e16). I wouldn't be surprised if that took a few hours on a supercomputer, let alone on some mid-range personal pc, which most of us are using.

What is the purpose of Big-O notation in computer science if it doesn't give all the information needed?

What is the use of Big-O notation in computer science if it doesn't give all the information needed?
For example, if one algorithm runs at 1000n and one at n, it is true that they are both O(n). But I still may make a foolish choice based on this information, since one algorithm takes 1000 times as long as the other for any given input.
I still need to know all the parts of the equation, including the constant, to make an informed choice, so what is the importance of this "intermediate" comparison? I end up loosing important information when it gets reduced to this form, and what do I gain?
What does that constant factor represent? You can't say with certainty, for example, that an algorithm that is O(1000n) will be slower than an algorithm that's O(5n). It might be that the 1000n algorithm loads all data into memory and makes 1,000 passes over that data, and the 5n algorithm makes five passes over a file that's stored on a slow I/O device. The 1000n algorithm will run faster even though its "constant" is much larger.
In addition, some computers perform some operations more quickly than other computers do. It's quite common, given two O(n) algorithms (call them A and B), for A to execute faster on one computer and B to execute faster on the other computer. Or two different implementations of the same algorithm can have widely varying runtimes on the same computer.
Asymptotic analysis, as others have said, gives you an indication of how an algorithm's running time varies with the size of the input. It's useful for giving you a good starting place in algorithm selection. Quick reference will tell you that a particular algorithm is O(n) or O(n log n) or whatever, but it's very easy to find more detailed information on most common algorithms. Still, that more detailed analysis will only give you a constant number without saying how that number relates to real running time.
In the end, the only way you can determine which algorithm is right for you is to study it yourself and then test it against your expected data.
In short, I think you're expecting too much from asymptotic analysis. It's a useful "first line" filter. But when you get beyond that you have to look for more information.
As you correctly noted, it does not give you information on the exact running time of an algorithm. It is mainly used to indicate the complexity of an algorithm, to indicate if it is linear in the input size, quadratic, exponential, etc. This is important when choosing between algorithms if you know that your input size is large, since even a 1000n algorithm well beat a 1.23 exp(n) algorithm for large enough n.
In real world algorithms, the hidden 'scaling factor' is of course important. It is therefore not uncommon to use an algorithm with a 'worse' complexity if it has a lower scaling factor. Many practical implementations of sorting algorithms are for example 'hybrid' and will resort to some 'bad' algorithm like insertion sort (which is O(n^2) but very simple to implement) for n < 10, while changing to quicksort (which is O(n log(n)) but more complex) for n >= 10.
Big-O tells you how the runtime or memory consumption of a process changes as the size of its input changes. O(n) and O(1000n) are both still O(n) -- if you double the size of the input, then for all practical purposes the runtime doubles too.
Now, we can have an O(n) algorithm and an O(n2) algorithm where the coefficient of n is 1000000 and the coefficient of n2 is 1, in which case the O(n2) algorithm would outperform the O(n) for smaller n values. This doesn't change the fact, however, that the second algorithm's runtime grows more rapidly than the first's, and this is the information that big-O tells us. There will be some input size at which the O(n) algorithm begins to outperform the O(n2) algorithm.
In addition to the hidden impact of the constant term, complexity notation also only considers the worst case instance of a problem.
Case in point, the simplex method (linear programming) has exponential complexity for all known implementations. However, the simplex method works much faster in practice than the provably polynomial-time interior point methods.
Complexity notation has much value for theoretical problem classification. If you want some more information on practical consequences check out "Smoothed Analysis" by Spielman: http://www.cs.yale.edu/homes/spielman
This is what you are looking for.
It's main purpose is for rough comparisons of logic. The difference of O(n) and O(1000n) is large for n ~ 1000 (n roughly equal to 1000) and n < 1000, but when you compare it to values where n >> 1000 (n much larger than 1000) the difference is miniscule.
You are right in saying they both scale linearly and knowing the coefficient helps in a detailed analysis but generally in computing the difference between linear (O(cn)) and exponential (O(cn^x)) performance is more important to note than the difference between two linear times. There is a larger value in the comparisons of runtime of higher orders such as and Where the performance difference scales exponentially.
The overall purpose of Big O notation is to give a sense of relative performance time in order to compare and further optimize algorithms.
You're right that it doesn't give you all information, but there's no single metric in any field that does that.
Big-O notation tells you how quickly the performance gets worse, as your dataset gets larger. In other words, it describes the type of performance curve, but not the absolute performance.
Generally, Big-O notation is useful to express an algorithm's scaling performance as it falls into one of three basic categories:
Linear
Logarithmic (or "linearithmic")
Exponential
It is possible to do deep analysis of an algorithm for very accurate performance measurements, but it is time consuming and not really necessary to get a broad indication of performance.

Why to ignore the constants in computing the running time complexity of an Algorithm

Can someone explain the reason behind ignoring the constants in computing the running time complexity of an Algorithm please?
Thanks
When analysing time complexity constants are difficult and irrelevant to calculate.
On some architecture addition might take twice as long as multiplication, so now we have to go through the algorithm and calculate the number of additions we make, and the number of multiplications we make in order to get an accurate runtime analysis. Not pretty!
More importantly, even if that is true now, at some point in the future, or on another slightly different architecture, that constant may be different, so the runtime would vary between architectures. So now my algorithm has more than one runtime? One for now on this architecture, and another for another architecture, and each of those may change in the future... Again, not pretty.
In a world where the capabilities of computing change all the time, twice the CPU capabilities tomorrow, four times the memory in a week, etc. there is little relevance to constant factors. This isn't the case if we need to quantify real-life runtime, but when we are analysing the complexity of an algorithm in general, not the complexity of a solution in a specific environment it is the case.
Also, and probably most importantly, the constant factors are not a good measure of the complexity of the problem, and at the end of the day we are trying to measure complexity. An algorithm that is of a specific complexity class, behaves (or more accurately, is bounded) a certain way for all size inputs. So when I try and measure the general complexity of two solutions, I do so in as general a way as possible, and therefore try and consider all values of input size (i.e. n->infinity).
Finally, It allows theoreticians to group algorithms into the same class, irrespective of certain constant factors that may or may not change and may or may not be improved. This is helpful for, among other reasons, optimality proofs; finding that a problem is omega(f(n)) is only useful if we consider algorithms in the complexity class of O(f(n)), irrespective of the constants.
because while comparing very large values, (let's call it n)... n^2 will be higher then k*n for any k, where n->infinity.
this is of course true for any power/exponent of n.
note that in real life projects, you do make these optimizations and try to minimize the constants, but usually they are less siginificant then the degree of the polynom
You only ignore constants when doing rough estimates. Most of the time, it's a valid simplification: when comparing two algorithms for large input dimensions, such as sorting an array, O(n log n) will eventually be faster or smaller than O(n²) no matter what.
However, when two algorithms have the same complexity, or when the expected dataset is so small that the asymptotic behaviour is not a valid real world scenario, then the constant is definitely important. For instance, a bubble sort implementation is likely to perform faster on an array of 3 or 4 values than quicksort.
As the value of n grows in equation 6n^2 + 100n + 300 the constant becomes of less relevance and we tend to ignore that. See the following image
Picture taken from Asymptotic Notation section at Khan Academy.

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