Sometimes I see T(n) with the strange T 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?
This question is related to
big-o
time-complexity
notation
big-theta
The answer is
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 T(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 O(1) or O(log(n)), depending on what happens inside f2(i).
for(int i=0; i<n; i++){
if( f2(i) ) break;
f(i);
}
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 O, the bar is at the bottom, so it is an (asymptotic) lower bound.
When looking at a T, 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....
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....
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 -------- < 8 n?8 g(n)For
g(n) = n:f(n) 0 = lim -------- < 8 n?8 nExamples:
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) 0Counterexamples:
Expression Value of the limit ------------------------------------------------- n ? O(log(n)) 8 1/2*n ? O(sqrt(n)) 8 2*n ? O(1) 8 n+log(n) ? O(log(n)) 8f(n) = T(g(n))General:
f(n) 0 < lim -------- < 8 n?8 g(n)For
g(n) = n:f(n) 0 < lim -------- < 8 n?8 nExamples:
Expression Value of the limit ------------------------------------------------ n = T(n) 1 1/2*n = T(n) 1/2 2*n = T(n) 2 n+log(n) = T(n) 1Counterexamples:
Expression Value of the limit ------------------------------------------------- n ? T(log(n)) 8 1/2*n ? T(sqrt(n)) 8 2*n ? T(1) 8 n+log(n) ? T(log(n)) 8 n ? T(n*log(n)) 0 n ? T(n²) 0 n ? T(nn) 0
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 O, the bar is at the bottom, so it is an (asymptotic) lower bound.
When looking at a T, 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.
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 T(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 O(1) or O(log(n)), depending on what happens inside f2(i).
for(int i=0; i<n; i++){
if( f2(i) ) break;
f(i);
}
f(n) belongs to O(n) if exists positive k as f(n)<=k*n
f(n) belongs to T(n) if exists positive k1, k2 as k1*n<=f(n)<=k2*n
A chart could make the previous answers easier to understand:
T-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) = T(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).
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 -------- < 8 n?8 g(n)For
g(n) = n:f(n) 0 = lim -------- < 8 n?8 nExamples:
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) 0Counterexamples:
Expression Value of the limit ------------------------------------------------- n ? O(log(n)) 8 1/2*n ? O(sqrt(n)) 8 2*n ? O(1) 8 n+log(n) ? O(log(n)) 8f(n) = T(g(n))General:
f(n) 0 < lim -------- < 8 n?8 g(n)For
g(n) = n:f(n) 0 < lim -------- < 8 n?8 nExamples:
Expression Value of the limit ------------------------------------------------ n = T(n) 1 1/2*n = T(n) 1/2 2*n = T(n) 2 n+log(n) = T(n) 1Counterexamples:
Expression Value of the limit ------------------------------------------------- n ? T(log(n)) 8 1/2*n ? T(sqrt(n)) 8 2*n ? T(1) 8 n+log(n) ? T(log(n)) 8 n ? T(n*log(n)) 0 n ? T(n²) 0 n ? T(nn) 0
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....
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 O, the bar is at the bottom, so it is an (asymptotic) lower bound.
When looking at a T, 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.
A chart could make the previous answers easier to understand:
T-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) = T(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).
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....
f(n) belongs to O(n) if exists positive k as f(n)<=k*n
f(n) belongs to T(n) if exists positive k1, k2 as k1*n<=f(n)<=k2*n
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.).
f(n) belongs to O(n) if exists positive k as f(n)<=k*n
f(n) belongs to T(n) if exists positive k1, k2 as k1*n<=f(n)<=k2*n
Conclusion: we regard big O, big ? and big O 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 ?. 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 ?, 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 ?!
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 ? subconsciously.
Similarly, we also regard big O as big ? subconsciously.
In one word, big O, big ? and big O are not the same thing from the definitions, but they are the same thing in our mouth and brain.
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.).
Conclusion: we regard big O, big ? and big O 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 ?. 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 ?, 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 ?!
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 ? subconsciously.
Similarly, we also regard big O as big ? subconsciously.
In one word, big O, big ? and big O are not the same thing from the definitions, but they are the same thing in our mouth and brain.
Source: Stackoverflow.com