I've come across in CLRS (Introduction to Algorithms) a sentence which states
"Distinguishing asymptotic Upper Bounds from asymptotically tight bounds is standard in the algorithms literature"
While I understand the essence of what the text wants to convey, It would be better understood if I get an example illustrating the difference.
O-notation gives us asymptotic upper bound.
Consider a function f(x),
We can define a function g(x), such that f(x) = O(g(x)).
Here g(x) is the asymptotic upper bound of f(x), meaning for all values of x >= c, f(x) grows at the same rate or slower than g(x) as x increases.
Another thing to be noticed is that if h(x) is the asymptotic upper bound of g(x) then it could easily be concluded that h(x) is also the asymptotic upper bound of f(x). After all, if f(x) can only grow at an equal or smaller rate than g(x), it is bound to grow at an equal of smaller rate than h(x) as g(x) cannot grow at any faster than h(x).
Eg. if f(x) = 10x + 2,
g(x) = 12x + 1 and h(x) = 2x^2.
We can safely say that f(x) = O(g(x)), g(x) = O(h(x)) and f(x) = O(h(x)).
Here g(x) is said to be asymptotic tight upper bound and h(x) is said to be the asymptotic upper bound of f(x).
Related
Sometimes I see Θ(n) with the strange Θ symbol with something in the middle of it, and sometimes just O(n). Is it just laziness of typing because nobody knows how to type this symbol, or does it mean something different?
Short explanation:
If an algorithm is of Θ(g(n)), it means that the running time of the algorithm as n (input size) gets larger is proportional to g(n).
If an algorithm is of O(g(n)), it means that the running time of the algorithm as n gets larger is at most proportional to g(n).
Normally, even when people talk about O(g(n)) they actually mean Θ(g(n)) but technically, there is a difference.
More technically:
O(n) represents upper bound. Θ(n) means tight bound.
Ω(n) represents lower bound.
f(x) = Θ(g(x)) iff f(x) =
O(g(x)) and f(x) = Ω(g(x))
Basically when we say an algorithm is of O(n), it's also O(n2), O(n1000000), O(2n), ... but a Θ(n) algorithm is not Θ(n2).
In fact, since f(n) = Θ(g(n)) means for sufficiently large values of n, f(n) can be bound within c1g(n) and c2g(n) for some values of c1 and c2, i.e. the growth rate of f is asymptotically equal to g: g can be a lower bound and and an upper bound of f. This directly implies f can be a lower bound and an upper bound of g as well. Consequently,
f(x) = Θ(g(x)) iff g(x) = Θ(f(x))
Similarly, to show f(n) = Θ(g(n)), it's enough to show g is an upper bound of f (i.e. f(n) = O(g(n))) and f is a lower bound of g (i.e. f(n) = Ω(g(n)) which is the exact same thing as g(n) = O(f(n))). Concisely,
f(x) = Θ(g(x)) iff f(x) = O(g(x)) and g(x) = O(f(x))
There are also little-oh and little-omega (ω) notations representing loose upper and loose lower bounds of a function.
To summarize:
f(x) = O(g(x)) (big-oh) means that
the growth rate of f(x) is
asymptotically less than or equal
to to the growth rate of g(x).
f(x) = Ω(g(x)) (big-omega) means
that the growth rate of f(x) is
asymptotically greater than or
equal to the growth rate of g(x)
f(x) = o(g(x)) (little-oh) means that
the growth rate of f(x) is
asymptotically less than the
growth rate of g(x).
f(x) = ω(g(x)) (little-omega) means
that the growth rate of f(x) is
asymptotically greater than the
growth rate of g(x)
f(x) = Θ(g(x)) (theta) means that
the growth rate of f(x) is
asymptotically equal to the
growth rate of g(x)
For a more detailed discussion, you can read the definition on Wikipedia or consult a classic textbook like Introduction to Algorithms by Cormen et al.
There's a simple way (a trick, I guess) to remember which notation means what.
All of the Big-O notations can be considered to have a bar.
When looking at a Ω, the bar is at the bottom, so it is an (asymptotic) lower bound.
When looking at a Θ, the bar is obviously in the middle. So it is an (asymptotic) tight bound.
When handwriting O, you usually finish at the top, and draw a squiggle. Therefore O(n) is the upper bound of the function. To be fair, this one doesn't work with most fonts, but it is the original justification of the names.
one is Big "O"
one is Big Theta
http://en.wikipedia.org/wiki/Big_O_notation
Big O means your algorithm will execute in no more steps than in given expression(n^2)
Big Omega means your algorithm will execute in no fewer steps than in the given expression(n^2)
When both condition are true for the same expression, you can use the big theta notation....
Rather than provide a theoretical definition, which are beautifully summarized here already, I'll give a simple example:
Assume the run time of f(i) is O(1). Below is a code fragment whose asymptotic runtime is Θ(n). It always calls the function f(...) n times. Both the lower and the upper bound is n.
for(int i=0; i<n; i++){
f(i);
}
The second code fragment below has the asymptotic runtime of O(n). It calls the function f(...) at most n times. The upper bound is n, but the lower bound could be Ω(1) or Ω(log(n)), depending on what happens inside f2(i).
for(int i=0; i<n; i++){
if( f2(i) ) break;
f(i);
}
Theta is a shorthand way of referring to a special situtation where
the big O and Omega are the same.
Thus, if one claims The Theta is expression q, then they are also necessarily claiming that Big O is expression q and Omega is expression q.
Rough analogy:
If:
Theta claims, "That animal has 5 legs."
then it follows that:
Big O is true ("That animal has less than or equal to 5 legs.")
and
Omega is true("That animal has more than or equal to 5 legs.")
It's only a rough analogy because the expressions aren't necessarily specific numbers, but instead functions of varying orders of magnitude such as log(n), n, n^2, (etc.).
A chart could make the previous answers easier to understand:
Θ-Notation - Same order | O-Notation - Upper bound
In English,
On the left, note that there is an upper bound and a lower bound that are both of the same order of magnitude (i.e. g(n) ). Ignore the constants, and if the upper bound and lower bound have the same order of magnitude, one can validly say f(n) = Θ(g(n)) or f(n) is in big theta of g(n).
Starting with the right, the simpler example, it is saying the upper bound g(n) is simply the order of magnitude and ignores the constant c (just as all big O notation does).
f(n) belongs to O(n) if exists positive k as f(n)<=k*n
f(n) belongs to Θ(n) if exists positive k1, k2 as k1*n<=f(n)<=k2*n
Wikipedia article on Big O Notation
Using limits
Let's consider f(n) > 0 and g(n) > 0 for all n. It's ok to consider this, because the fastest real algorithm has at least one operation and completes its execution after the start. This will simplify the calculus, because we can use the value (f(n)) instead of the absolute value (|f(n)|).
f(n) = O(g(n))
General:
f(n)
0 ≤ lim ──────── < ∞
n➜∞ g(n)
For g(n) = n:
f(n)
0 ≤ lim ──────── < ∞
n➜∞ n
Examples:
Expression Value of the limit
------------------------------------------------
n = O(n) 1
1/2*n = O(n) 1/2
2*n = O(n) 2
n+log(n) = O(n) 1
n = O(n*log(n)) 0
n = O(n²) 0
n = O(nⁿ) 0
Counterexamples:
Expression Value of the limit
-------------------------------------------------
n ≠ O(log(n)) ∞
1/2*n ≠ O(sqrt(n)) ∞
2*n ≠ O(1) ∞
n+log(n) ≠ O(log(n)) ∞
f(n) = Θ(g(n))
General:
f(n)
0 < lim ──────── < ∞
n➜∞ g(n)
For g(n) = n:
f(n)
0 < lim ──────── < ∞
n➜∞ n
Examples:
Expression Value of the limit
------------------------------------------------
n = Θ(n) 1
1/2*n = Θ(n) 1/2
2*n = Θ(n) 2
n+log(n) = Θ(n) 1
Counterexamples:
Expression Value of the limit
-------------------------------------------------
n ≠ Θ(log(n)) ∞
1/2*n ≠ Θ(sqrt(n)) ∞
2*n ≠ Θ(1) ∞
n+log(n) ≠ Θ(log(n)) ∞
n ≠ Θ(n*log(n)) 0
n ≠ Θ(n²) 0
n ≠ Θ(nⁿ) 0
Conclusion: we regard big O, big θ and big Ω as the same thing.
Why? I will tell the reason below:
Firstly, I will clarify one wrong statement, some people think that we just care the worst time complexity, so we always use big O instead of big θ. I will say this man is bullshitting. Upper and lower bound are used to describe one function, not used to describe the time complexity. The worst time function has its upper and lower bound; the best time function has its upper and lower bound too.
In order to explain clearly the relation between big O and big θ, I will explain the relation between big O and small o first. From the definition, we can easily know that small o is a subset of big O. For example：
T(n)= n^2 + n, we can say T(n)=O(n^2), T(n)=O(n^3), T(n)=O(n^4). But for small o, T(n)=o(n^2) does not meet the definition of small o. So just T(n)=o(n^3), T(n)=o(n^4) are correct for small o. The redundant T(n)=O(n^2) is what? It's big θ!
Generally, we say big O is O(n^2), hardly to say T(n)=O(n^3), T(n)=O(n^4). Why? Because we regard big O as big θ subconsciously.
Similarly, we also regard big Ω as big θ subconsciously.
In one word, big O, big θ and big Ω are not the same thing from the definitions, but they are the same thing in our mouth and brain.
If I have Omega(1) , theta(1) and Big O(1) are these asymptotically same ,if not then whats the difference among all these ?
f(x) = O(g(x)) (big-oh) means that the growth rate of f(x) is asymptotically less than or equal to to the growth rate of g(x).
f(x) = Ω(g(x)) (big-omega) means that the growth rate of f(x) is asymptotically greater than or equal to the growth rate of g(x)
f(x) = o(g(x)) (small-oh) means that the growth rate of f(x) is asymptotically less than the growth rate of g(x).
f(x) = ω(g(x)) (small-omega) means that the growth rate of f(x) is asymptotically greater than the growth rate of g(x)
f(x) = Θ(g(x)) (theta) means that the growth rate of f(x) is asymptotically equal to the growth rate of g(x)
Sometimes I see Θ(n) with the strange Θ symbol with something in the middle of it, and sometimes just O(n). Is it just laziness of typing because nobody knows how to type this symbol, or does it mean something different?
Short explanation:
If an algorithm is of Θ(g(n)), it means that the running time of the algorithm as n (input size) gets larger is proportional to g(n).
If an algorithm is of O(g(n)), it means that the running time of the algorithm as n gets larger is at most proportional to g(n).
Normally, even when people talk about O(g(n)) they actually mean Θ(g(n)) but technically, there is a difference.
More technically:
O(n) represents upper bound. Θ(n) means tight bound.
Ω(n) represents lower bound.
f(x) = Θ(g(x)) iff f(x) =
O(g(x)) and f(x) = Ω(g(x))
Basically when we say an algorithm is of O(n), it's also O(n2), O(n1000000), O(2n), ... but a Θ(n) algorithm is not Θ(n2).
In fact, since f(n) = Θ(g(n)) means for sufficiently large values of n, f(n) can be bound within c1g(n) and c2g(n) for some values of c1 and c2, i.e. the growth rate of f is asymptotically equal to g: g can be a lower bound and and an upper bound of f. This directly implies f can be a lower bound and an upper bound of g as well. Consequently,
f(x) = Θ(g(x)) iff g(x) = Θ(f(x))
Similarly, to show f(n) = Θ(g(n)), it's enough to show g is an upper bound of f (i.e. f(n) = O(g(n))) and f is a lower bound of g (i.e. f(n) = Ω(g(n)) which is the exact same thing as g(n) = O(f(n))). Concisely,
f(x) = Θ(g(x)) iff f(x) = O(g(x)) and g(x) = O(f(x))
There are also little-oh and little-omega (ω) notations representing loose upper and loose lower bounds of a function.
To summarize:
f(x) = O(g(x)) (big-oh) means that
the growth rate of f(x) is
asymptotically less than or equal
to to the growth rate of g(x).
f(x) = Ω(g(x)) (big-omega) means
that the growth rate of f(x) is
asymptotically greater than or
equal to the growth rate of g(x)
f(x) = o(g(x)) (little-oh) means that
the growth rate of f(x) is
asymptotically less than the
growth rate of g(x).
f(x) = ω(g(x)) (little-omega) means
that the growth rate of f(x) is
asymptotically greater than the
growth rate of g(x)
f(x) = Θ(g(x)) (theta) means that
the growth rate of f(x) is
asymptotically equal to the
growth rate of g(x)
For a more detailed discussion, you can read the definition on Wikipedia or consult a classic textbook like Introduction to Algorithms by Cormen et al.
There's a simple way (a trick, I guess) to remember which notation means what.
All of the Big-O notations can be considered to have a bar.
When looking at a Ω, the bar is at the bottom, so it is an (asymptotic) lower bound.
When looking at a Θ, the bar is obviously in the middle. So it is an (asymptotic) tight bound.
When handwriting O, you usually finish at the top, and draw a squiggle. Therefore O(n) is the upper bound of the function. To be fair, this one doesn't work with most fonts, but it is the original justification of the names.
one is Big "O"
one is Big Theta
http://en.wikipedia.org/wiki/Big_O_notation
Big O means your algorithm will execute in no more steps than in given expression(n^2)
Big Omega means your algorithm will execute in no fewer steps than in the given expression(n^2)
When both condition are true for the same expression, you can use the big theta notation....
Rather than provide a theoretical definition, which are beautifully summarized here already, I'll give a simple example:
Assume the run time of f(i) is O(1). Below is a code fragment whose asymptotic runtime is Θ(n). It always calls the function f(...) n times. Both the lower and the upper bound is n.
for(int i=0; i<n; i++){
f(i);
}
The second code fragment below has the asymptotic runtime of O(n). It calls the function f(...) at most n times. The upper bound is n, but the lower bound could be Ω(1) or Ω(log(n)), depending on what happens inside f2(i).
for(int i=0; i<n; i++){
if( f2(i) ) break;
f(i);
}
Theta is a shorthand way of referring to a special situtation where
the big O and Omega are the same.
Thus, if one claims The Theta is expression q, then they are also necessarily claiming that Big O is expression q and Omega is expression q.
Rough analogy:
If:
Theta claims, "That animal has 5 legs."
then it follows that:
Big O is true ("That animal has less than or equal to 5 legs.")
and
Omega is true("That animal has more than or equal to 5 legs.")
It's only a rough analogy because the expressions aren't necessarily specific numbers, but instead functions of varying orders of magnitude such as log(n), n, n^2, (etc.).
A chart could make the previous answers easier to understand:
Θ-Notation - Same order | O-Notation - Upper bound
In English,
On the left, note that there is an upper bound and a lower bound that are both of the same order of magnitude (i.e. g(n) ). Ignore the constants, and if the upper bound and lower bound have the same order of magnitude, one can validly say f(n) = Θ(g(n)) or f(n) is in big theta of g(n).
Starting with the right, the simpler example, it is saying the upper bound g(n) is simply the order of magnitude and ignores the constant c (just as all big O notation does).
f(n) belongs to O(n) if exists positive k as f(n)<=k*n
f(n) belongs to Θ(n) if exists positive k1, k2 as k1*n<=f(n)<=k2*n
Wikipedia article on Big O Notation
Using limits
Let's consider f(n) > 0 and g(n) > 0 for all n. It's ok to consider this, because the fastest real algorithm has at least one operation and completes its execution after the start. This will simplify the calculus, because we can use the value (f(n)) instead of the absolute value (|f(n)|).
f(n) = O(g(n))
General:
f(n)
0 ≤ lim ──────── < ∞
n➜∞ g(n)
For g(n) = n:
f(n)
0 ≤ lim ──────── < ∞
n➜∞ n
Examples:
Expression Value of the limit
------------------------------------------------
n = O(n) 1
1/2*n = O(n) 1/2
2*n = O(n) 2
n+log(n) = O(n) 1
n = O(n*log(n)) 0
n = O(n²) 0
n = O(nⁿ) 0
Counterexamples:
Expression Value of the limit
-------------------------------------------------
n ≠ O(log(n)) ∞
1/2*n ≠ O(sqrt(n)) ∞
2*n ≠ O(1) ∞
n+log(n) ≠ O(log(n)) ∞
f(n) = Θ(g(n))
General:
f(n)
0 < lim ──────── < ∞
n➜∞ g(n)
For g(n) = n:
f(n)
0 < lim ──────── < ∞
n➜∞ n
Examples:
Expression Value of the limit
------------------------------------------------
n = Θ(n) 1
1/2*n = Θ(n) 1/2
2*n = Θ(n) 2
n+log(n) = Θ(n) 1
Counterexamples:
Expression Value of the limit
-------------------------------------------------
n ≠ Θ(log(n)) ∞
1/2*n ≠ Θ(sqrt(n)) ∞
2*n ≠ Θ(1) ∞
n+log(n) ≠ Θ(log(n)) ∞
n ≠ Θ(n*log(n)) 0
n ≠ Θ(n²) 0
n ≠ Θ(nⁿ) 0
Conclusion: we regard big O, big θ and big Ω as the same thing.
Why? I will tell the reason below:
Firstly, I will clarify one wrong statement, some people think that we just care the worst time complexity, so we always use big O instead of big θ. I will say this man is bullshitting. Upper and lower bound are used to describe one function, not used to describe the time complexity. The worst time function has its upper and lower bound; the best time function has its upper and lower bound too.
In order to explain clearly the relation between big O and big θ, I will explain the relation between big O and small o first. From the definition, we can easily know that small o is a subset of big O. For example：
T(n)= n^2 + n, we can say T(n)=O(n^2), T(n)=O(n^3), T(n)=O(n^4). But for small o, T(n)=o(n^2) does not meet the definition of small o. So just T(n)=o(n^3), T(n)=o(n^4) are correct for small o. The redundant T(n)=O(n^2) is what? It's big θ!
Generally, we say big O is O(n^2), hardly to say T(n)=O(n^3), T(n)=O(n^4). Why? Because we regard big O as big θ subconsciously.
Similarly, we also regard big Ω as big θ subconsciously.
In one word, big O, big θ and big Ω are not the same thing from the definitions, but they are the same thing in our mouth and brain.
I'm having an issue solving for big theta notation. I understand that big O notation denotes the worst case and upperbound while Omega notation denotes the best case and lower bound.
If I'm given an algorithm that runs in O(nlogn) time and Omega(n), how would I infer what Theta equals? I'm beginning to assume that there exists a theta notation if and only if O and Omega are equal, is this true?
Assume your algorithm runs in f(x).
f(x) = O(n*log(n)) means that for x high enough there is some constant c1 > 0 so that f(x) will always be smaller than c1*n*log(n).
f(x) = Omega(n) means that for x high enough there is some constant c2 > 0 so that f(x) will be bigger than c2*n
So all you know now is that from a certain point onward (x big enough) your algorithm will run faster than c2*n and slower than c1*n*log(n).
f(x) = Theta(g(x)) means that for x big enough there is some c3 > 0 and c4 > 0 so that c3*g(x) <= f(x) <= c4*g(x), this means f(x) will only run a constant factor faster or slower than g(x). So indeed, f(x) = O(g(x)) and f(x) = Omega(g(x)).
Conclusion: With only O and Omega given, if they are not the same you cannot conclude what Theta is. If you have the algorithm you could try and see if O was maybe chosen too high, or Omega might have been chosen too low.
If f(x) = O(g(x)) as x -> infinity, then
A. g is the upper bound of f
B. f is the upper bound of g.
C. g is the lower bound of f.
D. f is the lower bound of g.
Can someone please tell me when they think it is and why?
The real answer is that none of these is correct.
The definition of big-O notation is that:
|f(x)| <= k|g(x)|
for all x > x0, for some x0 and k.
In specific cases, |k| might be less than or equal to 1, in which case it would be correct to say that "|g| is the upper bound of |f|". But in general, that's not true.
Answer
g is the upper bound of f
When x goes towards infinity, worst case scenario is O(g(x)). That means actual exec time can be lower than g(x), but never worse than g(x).
EDIT:
As Oli Charlesworth pointed out, that is only true with arbitrary constant k <= 1 and not in general. Please look at his answer for the general case.
The question checks your understanding of the basics of asymptotic algebra, or big-oh notation. In
f(x) = O(g(x)) as x approaches infinity
the question says that when you feed the function f a value x, the value which f computes from x is then in the order of that returned from another function, g(x). As an example, suppose
f(x) = 2x
g(x) = x
then the value g(x) returns when fed x is of the same order as that f(x) returns for x. Specifically, the two functions return a value that is in the order of x; the functions are both linear. It doesn't matter whether f(x) is 2x or ½x; for any constant factor at all f(x) will return a value that is in the order of x. This is because big-oh notation is about ignoring constant factors. Constant factors don't grow as x grows and so we assume they don't matter nearly as much as x does.
We restrict g(x) to a specific set of functions. g(x) can be x, or ln(x), or log(x) and so on and so forth. It may look as if when
f(x) = 2x
g(x) = x
f(x) yields values higher than g(x) and therefore is the upper bound of g(x). But once again, we ignore the constant factor, and we say that the order-of upper bound, which is what big-oh is all about, is that of g(x).