I'm fairly new to the Big-O stuff and I'm wondering what's the complexity of the algorithm.
I understand that every addition, if statement and variable initialization is O(1).
From my understanding first 'i' loop will run 'n' times and the second 'j' loop will run 'n^2' times. Now, the third 'k' loop is where I'm having issues.
Is it running '(n^3)/2' times since the average value of 'j' will be half of 'n'?
Does it mean the Big-O is O((n^3)/2)?
We can use Sigma notation to calculate the number of iterations of the inner-most basic operation of you algorithm, where we consider the sum = sum + A[k] to be a basic operation.
Now, how do we infer that T(n) is in O(n^3) in the last step, you ask?
Let's loosely define what we mean by Big-O notation:
f(n) = O(g(n)) means c · g(n) is an upper bound on f(n). Thus
there exists some constant c such that f(n) is always ≤ c · g(n),
for sufficiently large n (i.e. , n ≥ n0 for some constant n0).
I.e., we want to find some (non-unique) set of positive constants c and n0 such that the following holds
|f(n)| ≤ c · |g(n)|, for some constant c>0 (+)
for n sufficiently large (say, n>n0)
for some function g(n), which will show that f(n) is in O(g(n)).
Now, in our case, f(n) = T(n) = (n^3 - n^2) / 2, and we have:
f(n) = 0.5·n^3 - 0.5·n^2
{ n > 0 } => f(n) = 0.5·n^3 - 0.5·n^2 ≤ 0.5·n^3 ≤ n^3
=> f(n) ≤ 1·n^3 (++)
Now (++) is exactly (+) with c=1 (and choose n0 as, say, 1, n>n0=1), and hence, we have shown that f(n) = T(n) is in O(n^3).
From the somewhat formal derivation above it's apparent that any constants in function g(n) can just be extracted and included in the constant c in (+), hence you'll never (at least should not) see time complexity described as e.g. O((n^3)/2). When using Big-O notation, we're describing an upper bound on the asymptotic behaviour of the algorithm, hence only the dominant term is of interest (however not how this is scaled with constants).
Related
My professor recently brushed over the formal definition of Big O:
To be completely honest even after him explaining it to a few different students we all seem to still not understand it at its core. The problems in comprehension mostly occurred with the following examples we went through:
So far my reasoning is as follows:
When you multiply a function's highest term by a constant, you get a new function that eventually surpasses the initial function at a given n. He called this n a "witness" to the function O(g(n))
How is this c term created/found? He mentioned bounds a couple of times but didn't really specify what bounds signify or how to find them/use them.
I think I just need a more solid foundation of the formal definition and how these examples back up the definition.
I think that the way this definition is typically presented in terms of c values and n0's is needlessly confusing. What f(n) being O(g(n)) really means is that when you disregard constant and lower order terms, g(n) is an asymptotic upper bound for f(n) (for a function to g to asymptotically upper bound f just means that past a certain point g is always greater than or equal to f). Put another way, f(n) grows no faster than g(n) as n goes to infinity.
Big O itself is a little confusing, because f(n) = O(g(n)) doesn't mean that g(n) grows strictly faster than f(n). It means when you disregard constant and lower order terms, g(n) grows faster than f(n), or it grows at the same rate (strictly faster would be "little o"). A simple, formal way to put this concept is to say:
That is, for this limit to hold true, the highest order term of f(n) can be at most a constant multiple of the highest order term of g(n). f(n) is O(g(n)) iff it grows no faster than g(n).
For example, f(n) = n is in O(g(n) = n^2), because past a certain point n^2 is always bigger than n. The limit of n^2 over n is positive, so n is in O(n^2)
As another example, f(n) = 5n^2 + 2n is in O(g(n) = n^2), because in the limit, f(n) can only be about 5 times larger than g(n). It's not infinitely bigger: they grow at the same rate. To be precise, the limit of n^2 over 5n^2 + 3n is 1/5, which is more than zero, so 5n^2 + 3n is in O(n^2). Hopefully this limit based definition provides some intuition, as it is completely equivalent mathematically to the provided definition.
Finding a particular constant value c and x value n0 for which the provided inequality holds true is just a particular way of showing that in the limit as n goes to infinity, g(n) grows at least as fast as f(n): that f(n) is in O(g(n)). That is, if you've found a value past which c*g(n) is always greater than f(n), you've shown that f(n) grows no more than a constant multiple (c times) faster than g(n) (if f grew faster than g by more than a constant multiple, finding such a c and n0 would be impossible).
There's no real art to finding a particular c and n0 value to demonstrate f(n) = O(g(n)). They can be literally whatever positive values you need them to be to make the inequality true. In fact, if it is true that f(n) = O(g(n)) then you can pick any value you want for c and there will be some sufficiently large n0 value that makes the inequality true, or, similarly you could pick any n0 value you want, and if you make c big enough the inequality will become true (obeying the restrictions that c and n0 are both positive). That's why I don't really like this formalization of big O: it's needlessly particular and proofs involving it are somewhat arbitrary, distracting away from the main concept which is the behavior of f and g as n goes to infinity.
So, as for how to handle this in practice, using one of the example questions: why is n^2 + 3n in O(n^2)?
Answer: because the limit as n goes to infinity of n^2 / n^2 + 3n is 1, which is greater than 0.
Or, if you're wanting/needing to do it the other way, pick any positive value you want for n0, and evaluate f at that value. f(1) will always be easy enough:
f(1) = 1^2 + 3*1 = 4
Then find the constant you could multiply g(1) by to get the same value as f(1) (or, if not using n0 = 1 use whatever n0 for g that you used for f).
c*g(1) = 4
c*1^2 = 4
c = 4
Then, you just combine the statements into an assertion to show that there exists a positive n0 and a constant c such that cg(n) <= f(n) for all n >= n0.
n^2 + 3n <= (4)n^2 for all n >= 1, implying n^2 + 3n is in O(n^2)
If you're using this method of proof, the above statement you use to demonstrate the inequality should ideally be immediately obvious. If it's not, maybe you want to change your n0 so that the final statement is more clearly true. I think that showing the limit of the ratio g(n)/f(n) is positive is much clearer and more direct if that route is available to you, but it is up to you.
Moving to a negative example, it's quite easy with the limit method to show that f(n) is not in O(g(n)). To do so, you just show that the limit of g(n) / f(n) = 0. Using the third example question: is nlog(n) + 2n in O(n)?
To demonstrate it the other way, you actually have to show that there exists no positive pair of numbers n0, c such that for all n >= n0 f(n) <= cg(n).
Unfortunately showing that f(n) = nlogn + 2n is in O(nlogn) by using c=2, n0=8 demonstrates nothing about whether f(n) is in O(n) (showing a function is in a higher complexity class implies nothing about it not being a lower complexity class).
To see why this is the case, we could also show a(n) = n is in g(n) = nlogn using those same c and n0 values (n <= 2(nlog(n) for all n >= 8, implying n is in O(nlogn))`), and yet a(n)=n clearly is in O(n). That is to say, to show f(n)=nlogn + 2n is not in O(n) with this method, you can't just show that it is in O(nlogn). You would have to show that no matter what n0 you pick, you can never find a c value large enough such that f(n) >= c(n) for all n >= n0. Showing that such a pair of numbers does not exist is not impossible, but relatively speaking it's a tricky thing to do (and would probably itself involve limit equations, or a proof by contradiction).
To sum things up, f(n) is in O(g(n)) if the limit of g(n) over f(n) is positive, which means f(n) doesn't grow any faster than g(n). Similarly, finding a constant c and x value n0 beyond which cg(n) >= f(n) shows that f(n) cannot grow asymptotically faster than g(n), implying that when discarding constants and lower order terms, g(n) is a valid upper bound for f(n).
I was given the function 5n^3+2n+8 to prove for big-O and big-Omega. I finished big-O, but for big-Omega I end up with a single-term function. I canceled out 2n and 8 because they're positive and make my function larger, so I just end up with 5n^3. How do I choose C and n_0? or is it simply trivial in this case?
From Big-Ω (Big-Omega) notation (slightly modified):
If a running time of some function f(n) is Ω(g(n)), then for large
enough n, say n > n_0 > 0, the running time of f(n) is at least
C⋅g(n), for some constant C > 0.
Hence, if f(n) is in Ω(g(n)), then there exists some positive constants C and n_0 such at the following holds
f(n) ≥ C⋅g(n), for all n > n_0 (+)
Now, the choice of C and n_0 is not unique, it suffices that you can show one such set of constants (such that (+) holds) to be able to describe the running time using the Big-Omega notation, as posted above.
Hence, you are indeed almost there
f(n) = 5n^3+2n+8 > 5n^3 holds for all n larger than say, 1
=> f(n) ≥ 5⋅n^3 for all n > n_0 = 1 (++)
Finally, (++) is just (+) for g(n) = n^3 and C=5, and hence, by (+), f(n) is in Ω(n^3).
In big O notation, we always say that we should ignore constant factors for most cases. That is, rather than writing,
3n^2-100n+6
we are almost always satisfied with
n^2
since that term is the fastest growing term in the equation.
But I found many algorithm courses starts comparing functions with many terms
2n^2+120n+5 = big O of n^2
then finding c and n0 for those long functions, before recommending to ignore low order terms in the end.
My question is what would I get from trying to understand and annalising these kinds of functions with many terms? Before this month I am comfortable with understanding what O(1), O(n), O(LOG(n)), O(N^3) mean. But am I missing some important concepts if I just rely on this typically used functions? What will I miss if I skipped analysing those long functions?
Let's first of all describe what we mean when we say that f(n) is in O(g(n)):
... we can say that f(n) is O(g(n)) if we can find a constant c such
that f(n) is less than c·g(n) or all n larger than n0, i.e., for all
n>n0.
In equation for: we need to find one set of constants (c, n0) that fulfils
f(n) < c · g(n), for all n > n0, (+)
Now, the result that f(n) is in O(g(n)) is sometimes presented in difference forms, e.g. as f(n) = O(g(n)) or f(n) ∈ O(g(n)), but the statement is the same. Hence, from your question, the statement 2n^2+120n+5 = big O of n^2 is just:
f(n) = 2n^2 + 120n + 5
a result after some analysis: f(n) is in O(g(n)), where
g(n) = n^2
Ok, with this out of the way, we look at the constant term in the functions we want to analyse asymptotically, and let's look at it educationally, using however, your example.
As the result of any big-O analysis is the asymptotic behaviour of a function, in all but some very unusual cases, the constant term has no effect whatsoever on this behaviour. The constant factor can, however, affect how to choose the constant pair (c, n0) used to show that f(n) is in O(g(n)) for some functions f(n) and g(n), i.e., the none-unique constant pair (c, n0) used to show that (+) holds. We can say that the constant term will have no effect of our result of the analysis, but it can affect our derivation of this result.
Lets look at your function as well as another related function
f(n) = 2n^2 + 120n + 5 (x)
h(n) = 2n^2 + 120n + 22500 (xx)
Using a similar approach as in this thread, for f(n), we can show:
linear term:
120n < n^2 for n > 120 (verify: 120n = n^2 at n = 120) (i)
constant term:
5 < n^2 for e.g. n > 3 (verify: 3^2 = 9 > 5) (ii)
This means that if we replace both 120n as well as 5 in (x) by n^2 we can state the following inequality result:
Given that n > 120, we have:
2n^2 + n^2 + n^2 = 4n^2 > {by (ii)} > 2n^2 + 120n + 5 = f(n) (iii)
From (iii), we can choose (c, n0) = (4, 120), and (iii) then shows that these constants fulfil (+) for f(n) with g(n) = n^2, and hence
result: f(n) is in O(n^2)
Now, for for h(n), we analogously have:
linear term (same as for f(n))
120n < n^2 for n > 120 (verify: 120n = n^2 at n = 120) (I)
constant term:
22500 < n^2 for e.g. n > 150 (verify: 150^2 = 22500) (II)
In this case, we replace 120n as well as 22500 in (xx) by n^2, but we need a larger less than constraint on n for these to hold, namely n > 150. Hence, we the following holds:
Given that n > 150, we have:
2n^2 + n^2 + n^2 = 4n^2 > {by (ii)} > 2n^2 + 120n + 5 = h(n) (III)
In same way as for f(n), we can, here, choose (c, n0) = (4, 150), and (III) then shows that these constants fulfil (+) for h(n), with g(n) = n^2, and hence
result: h(n) is in O(n^2)
Hence, we have the same result for both functions f(n) and h(n), but we had to use different constants (c,n0) to show these (i.e., somewhat different derivation). Note finally that:
Naturally the constants (c,n0) = (4,150) (used for h(n) analysis) are also valid to show that f(n) is in O(n^2), i.e., that (+) holds for f(n) with g(n)=n^2.
However, not the reverse: (c,n0) = (4,120) cannot be used to show that (+) holds for h(n) (with g(n)=n^2).
The core of this discussion is that:
As long as you look at sufficiently large values of n, you will be able to describe the constant terms in relations as constant < dominantTerm(n), where, in our example, we look at the relation with regard to dominant term n^2.
The asymptotic behaviour of a function will not (in all but some very unusual cases) depend on the constant terms, so we might as well skip looking at them at all. However, for a rigorous proof of the asymptotic behaviour of some function, we need to take into account also the constant terms.
Ever have intermediate steps in your work? That is what this likely is as when you are computing a big O, chances are you don't already know for sure what the highest order term is and thus you keep track of them all and then determine which complexity class makes sense in the end. There is also something to be said for understanding why the lower order terms can be ignored.
Take some graph algorithms like a minimum spanning tree or shortest path. Now, can just looking at an algorithm you know what the highest term will be? I know I wouldn't and so I'd trace through the algorithm and collect a bunch of terms.
If you want another example, consider Sorting Algorithms and whether you want to memorize all the complexities or not. Bubble Sort, Shell Sort, Merge Sort, Quick Sort, Radix Sort and Heap Sort are a few of the more common algorithms out there. You could either memorize both the algorithm and complexity or just the algorithm and derive the complexity from the pseudo code if you know how to trace them.
If I have already known f(n) is O(g(n)). From the definition of little-oh, how to prove that f(n) is o(n * g(n))?
Given: f(n) is in O(g(n)).
Using the definition of big-O notation, we can write this as:
f(n) is in O(g(n))
=> |f(n)| ≤ k*|g(n)|, for some constant k>0 (+)
for n sufficiently large (say, n>N)
For the definition of big-O used as above, see e.g.
https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-o-notation
Prove: Given (+), then f(n) is in o(n*g(n)).
Lets first state what little-o notation means:
Formally, f(n) = o(g(n)) (or f(n) ∈ o(g(n))) as n → ∞ means that
for every positive constant ε there exists a constant N such that
|f(n)| ≤ ε*|g(n)|, for all n > N (++)
From https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation.
Now, using (+), we can write
|f(n)| ≤ k*|g(n)|, som k>0, n sufficiently large
<=> { n > 0 } <=> n*|f(n)| ≤ k*n*|g(n)|
<=> n*|f(n)| ≤ k*|n*g(n)|
<=> |f(n)| ≤ (k/n)*|n*g(n)| (+++)
Return to the definition of little-o, specifically (++), and let, without loss of generality, k be fixed. Now, every positive constant ε can be described as
ε = k/C, for some constant C>0 (with k fixed, k>0) (*)
Now, assume, without loss of generality, that n is larger than this C, i.e., n>C. Then, (*) and (+++) yields
|f(n)| ≤ (k/n)*|n*g(n)| < (k/C)*|n*g(n)| = ε*|n*g(n)| (**)
^ ^
| |
since `n>C` (*)
Since we're studying asymptotic behaviour, we can choose to to assign a lower bound to n to any value larger than C (in fact, that's in the definition of both big-O and little-o, "n sufficiently large"), and hence---by the definition of little-oh above---, we have:
- As shown above, (+) implies (**)
- By the definition of little-o, (**) shows that f(n) is in o(n*g(n))
- Subsequently, we've shown that, given (+), then: f(n) is in o(n*g(n))
Result: If f(n) is in O(g(n)), then f(n) is in o(n*g(n)), where these two relations refer big-O and litte-O asymptotic bounds, respectively.
Comment: The result is, in fact, quite trivial. The big-O and little-o notation differ only in one of the two constants used in proving the upper bounds, i.e., we can write the definitions of big-O and little-O as:
f(n) is said to be in O(g(n)) if we can find a set of positive constants (k, N), such that f(n) < k*g(n) holds for all n>N.
f(n) is said to be in o(g(n)) if we can find a positive constant N, such that f(n) < ε*g(n) holds for all n>N, and for every positive constant ε.
The latter is obvious a stricter constraint, but if we can make use of one extra power of n in the left-hand-side of f(n) < ε*g(n) (i.e., f(n) < ε*n*g(n)), then even for infinitesimal values of ε, we can always choose the other constant N freely to be sufficiently large for ε*n to provide us any constant k that can be used to show that f(n) is in O(g(n)) (as, recall, n>N).
Can some one provide me a real time example for how to calculate big theta.
Is big theta some thing like average case, (min-max)/2?
I mean (minimum time - big O)/2
Please correct me if I am wrong, thanks
Big-theta notation represents the following rule:
For any two functions f(n), g(n), if f(n)/g(n) and g(n)/f(n) are both bounded as n grows to infinity, then f = Θ(g) and g = Θ(f). In that case, g is both an upper bound and a lower bound on the growth of f.
Here's an example algorithm:
def find-minimum(List)
min = +∞
foreach value in List
min = value if min > value
return min
We wish to evaluate the cost function c(n) where n is the size of the input list. This algorithm will perform one comparison for every item in the list, so c(n) = n.
c(n)/n = 1 which remains bounded as n goes to infinity, so c(n) grows no faster than n. This is what is meant by big-O notation c(n) = O(n). Conversely, n/C(n) = 1 also remains bounded, so c(n) grows no slower than n. Since it grows neither slower nor faster, it must grow at the same speed. This is what is meant by theta notation c(n) = Θ(n).
Note that c(n)/n² is also bounded, so c(n) = O(n²) as well — big-O notation is merely an upper bound on the complexity, so any O(n) function is also O(n²), O(n³)...
However, since n²/c(n) = n is not bounded, then c(n) ≠ Θ(n²). This is the interesting property of big-theta notation: it's both an upper bound and a lower bound on the complexity.
Big theta is a tight bound, for a function T(n): if: Omega(f(n))<=T(n)<=O(f(n)), then Theta(f(n)) is the tight bound for T(n).
In other words Theta(f(n)) 'describes' a function T(n), if both O [big O] and Omega, 'describe' the same T, with the same f.
for example, a quicksort [with correct median choices], always takes at most O(nlogn), at at least Omega(nlogn), so quicksort [with good median choices] is Theta(nlogn)
EDIT:
added discussion in comments:
Searching an array is still Theta(n). the Theta function does not indicate worst/best case, but the behavior of the desired case. i.e, searching for an array, T(n)=number of ops for worst case. in here, obviously T(n)<=O(n), but also T(n)>=n/2, because at worst case you need to iterate the whole array, so T(n)>=Omega(n) and therefore Theta(n) is asymptotic bound.
From http://en.wikipedia.org/wiki/Big_O_notation#Related_asymptotic_notations, we learn that "Big O" denotes an upper bound, whereas "Big Theta" denotes an upper and lower bound, i.e. in the limit as n goes to infinity:
f(n) = O(g(n)) --> |f(n)| < k.g(n)
f(n) = Theta(g(n)) --> k1.g(n) < f(n) < k2.g(n)
So you cannot infer Big Theta from Big O.
ig-Theta (Θ) notation provides an asymptotic upper and lower bound on the growth rate of an algorithm's running time. To calculate the big-Theta notation of a function, you need to find two non-negative functions, f(n) and g(n), such that:
There exist positive constants c1, c2 and n0 such that 0 <= c1 * g(n) <= f(n) <= c2 * g(n) for all n >= n0.
f(n) and g(n) have the same asymptotic growth rate.
The big-Theta notation for the function f(n) is then written as Θ(g(n)). The purpose of this notation is to provide a rough estimate of the running time, ignoring lower order terms and constant factors.
For example, consider the function f(n) = 2n^2 + 3n + 1. To calculate its big-Theta notation, we can choose g(n) = n^2. Then, we can find c1 and c2 such that 0 <= c1 * n^2 <= 2n^2 + 3n + 1 <= c2 * n^2 for all n >= n0. For example, c1 = 1/2 and c2 = 2. So, f(n) = Θ(n^2).