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On page 16 of this algorithms book, it states:

For example, suppose we are choosing between two algorithms for a particular computational task. One takes $f_1(n) = n^2$ steps, while the other takes $f_2(n) = 2n + 20$ steps (Figure 0.2).

He then goes on to say:

This superiority ... (of $f_2$ over $f_1$) ... is captured by the big-O notation: $f_2 = O(f_1)$, because ...

Now my problem is that in the original quote, he said that $f_1(n) = n^2$ steps and $f_2(n) = 2n+20$ steps, so thus $f_1 = O(n^2)$ and $f_2 = O(n)$ (big-O is defined in Section 0.3). But the second quote above states $f_2 = O(f_1)$, which means $f_2 = O(n^2)$ and contradicts his definition of big-O notation. What have I missed?

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The $=$ is a huge abuse of notation. I wonder if it wasn't chosen over $\in$ mostly for the convenience of typing it. –  Joe Nov 16 '12 at 3:43
    
@Joe: likely, though there is a closely related usage along the lines of $f = a + bx + cx^2 + O(x^3)$ for Taylor series. If you consider the special case $a,b,c= 0$, it would be quite tempting to simply write $f = O(x^3)$; but it feels really awkward (though it is more correct) to write $f \in a + bx + cx^2 + O(x^3)$. The fact that physicists (and other mathematicians) routinely write this sort of thing on black/white boards more or less suggests that the difficulty of writing $\in$ isn't the entire issue either. It's a place where good usage and idiom are at odds. –  Niel de Beaudrap Nov 16 '12 at 16:30
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2 Answers

up vote 8 down vote accepted

The key thing to keep in mind here is that the equals sign in $f = O(g)$ does not behave as an equals sign usually would. The reason is that $O(g)$ is not actually a function but a set of functions. For example, you can think of $O(n^2)$ as being a set which, among many other functions, contains the functions $n$, $5 \sqrt n$, $\log^5 n$, etc. So as the appropriate way to think of $f = O(g)$ is not as equality but instead as being an element of a set: $f \in O(g)$.

So the set $O(n)$ is a subset of $O(n^2)$ since any function that is asymptotically bounded above by a multiple of $n$ is also bounded asymptotically above by a multiple of $n^2$.

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The big-O denotes an upper bound. Since $O(n)\subset O(n^2) $, we have for any function $h$ $$h=O(n)\; \Rightarrow \; h= O(n^2).$$

In other words if $g_1(n)=n$ is an asymptotic upper bound for $f_2(n)$ then $g_2(n)=n^2$ is of course also an asymptotic upper bound for $f_2(n)$. The statement does not contradict the definition of the big-O definition.

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Thankyou for your reply; what confuses me is what is the purpose in saying that $h = O(n) \Rightarrow h = O(n^2)$. Because we could say that $h = O(n^{2382})$ which is also an upper bound. What's the point in talking about anything but the least upper bound? –  njp Nov 15 '12 at 13:46
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Well, usually you want to have good upper bounds. However, technically there is nothing wrong with it, when you say $h=O(n^{2382})$. It's just not a very strong statement. Notice that there is also the notion of $\Theta()$ which denotes a tight bound. The problem is, that it is often challenging to derive a $\Theta()$ bound, and that is why you stick to the big-O notation. –  A.Schulz Nov 15 '12 at 13:52
    
Sometimes we simply don't know the least upper bound. We can do matrix multiplication in $O(n^3)$, and even in $O(n^{2.38})$. It is a huge open question whether we can do it in $O(n^2)$. –  Pål GD Nov 15 '12 at 13:52
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Right, so in the case described in my original question, all we're really saying is that $f_2$ has equal or less time complexity than $f_1$, but on the other hand $f_1$'s complexity is not contained in $O(f_2)$ because $f_2 = O(n)$? I think the notation is odd, maybe it should be $f_2 \in O(f_1)$. –  njp Nov 15 '12 at 13:57
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@DesmondWolf: Indeed, the equality sign is slightly misleading as it is not symmetric. Membership is more correct, but when we speak about it, we really say "$f_2$ is $O(n)$", and then the equality sign seems more natural. –  Pål GD Nov 15 '12 at 18:23
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