Generalizing $n cdot m = O(n^2) iff m = O(n)$ to any two-variable function
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I came across the following statement in my Computer Science high-school textbook (I translated it to English).
There was some for loop that runs $m$ times nested inside a for loop that runs $n$ times, and it said:
"The function is $n cdot m$, and since $n$ and $m$ are as large as we want, we can set $m$ equal to $n$ and say that the complexity is $O(n^2).$"
Obviously I was suspicious, because $m$ in this case didn't depend on $n$.
So I started thinking about the problem with the big-$O$ definition.
Using the definition of for multiple variables, I proved the statement is false by contradiction.
I also thought about using the definition for a single variable, with $m$ as a function of $n$. I was able to prove that $n cdot m = O(n^2)$ if and only if $m=O(n)$.
My question is, is it possible to somehow generalize this thought and say that $f(n,m) = O(f(n,n))$ if some properties hold? Is this even legal, with Big-$O$ notation?
I'm also not sure if my original proofs are correct, so here they are:
Using the multiple variable definition, prove $n cdot m neq O(n^2)$.
Suppose $n cdot m = O(n^2)$, then there exist $c,n_0,m_0 in mathbb{R}_{>0}$ so that $n geq n_0, m geq m_0$ implies $n cdot m leq cn^2$.
But if we look at $n=n_0,m=max(m_0,cn_0+1)$, we have
$$n cdot m = n_0 cdot max(m_0,cn_0+1) geq n_0 cdot (cn_0+1) = cn_0^2+n_0 > cn_0^2$$
This is a contradiction, because $n=n_0 geq n_0$ and $m=max(m_0,cn_0+1) geq m_0$ and so we should have $n cdot m leq cn^2 = cn_0^2$.
Using the single variable definition, prove $n cdot m = O(n^2) iff m=O(n)$.
Suppose $n cdot m = O(n^2)$, then there exist $c,n_0 in mathbb{R}_{>0}$ so that $n geq n_0$ implies $n cdot m leq cn^2$. Divide both sides by $n$, and we have $m leq cn$ for $n geq n_0$, which directly means (by definition) that $m=O(n)$.
Suppose $m = O(n)$, then there exist $c,n_0 in mathbb{R}_{>0}$ so that $n geq n_0$ implies $m leq cn$. Now let's prove $n cdot m = O(n^2)$.
For every $n$ and specifically for every $n geq n_0$,
$$n cdot m leq n cdot cn = cn^2$$
Which directly means (by definition) that $n cdot m = O(n^2)$.
asymptotics computer-science computational-complexity
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add a comment |
$begingroup$
I came across the following statement in my Computer Science high-school textbook (I translated it to English).
There was some for loop that runs $m$ times nested inside a for loop that runs $n$ times, and it said:
"The function is $n cdot m$, and since $n$ and $m$ are as large as we want, we can set $m$ equal to $n$ and say that the complexity is $O(n^2).$"
Obviously I was suspicious, because $m$ in this case didn't depend on $n$.
So I started thinking about the problem with the big-$O$ definition.
Using the definition of for multiple variables, I proved the statement is false by contradiction.
I also thought about using the definition for a single variable, with $m$ as a function of $n$. I was able to prove that $n cdot m = O(n^2)$ if and only if $m=O(n)$.
My question is, is it possible to somehow generalize this thought and say that $f(n,m) = O(f(n,n))$ if some properties hold? Is this even legal, with Big-$O$ notation?
I'm also not sure if my original proofs are correct, so here they are:
Using the multiple variable definition, prove $n cdot m neq O(n^2)$.
Suppose $n cdot m = O(n^2)$, then there exist $c,n_0,m_0 in mathbb{R}_{>0}$ so that $n geq n_0, m geq m_0$ implies $n cdot m leq cn^2$.
But if we look at $n=n_0,m=max(m_0,cn_0+1)$, we have
$$n cdot m = n_0 cdot max(m_0,cn_0+1) geq n_0 cdot (cn_0+1) = cn_0^2+n_0 > cn_0^2$$
This is a contradiction, because $n=n_0 geq n_0$ and $m=max(m_0,cn_0+1) geq m_0$ and so we should have $n cdot m leq cn^2 = cn_0^2$.
Using the single variable definition, prove $n cdot m = O(n^2) iff m=O(n)$.
Suppose $n cdot m = O(n^2)$, then there exist $c,n_0 in mathbb{R}_{>0}$ so that $n geq n_0$ implies $n cdot m leq cn^2$. Divide both sides by $n$, and we have $m leq cn$ for $n geq n_0$, which directly means (by definition) that $m=O(n)$.
Suppose $m = O(n)$, then there exist $c,n_0 in mathbb{R}_{>0}$ so that $n geq n_0$ implies $m leq cn$. Now let's prove $n cdot m = O(n^2)$.
For every $n$ and specifically for every $n geq n_0$,
$$n cdot m leq n cdot cn = cn^2$$
Which directly means (by definition) that $n cdot m = O(n^2)$.
asymptotics computer-science computational-complexity
$endgroup$
$begingroup$
$m$ must depend on $n$ in some way, otherwise $mn=O(n)ne O(n^2)$
$endgroup$
– Exodd
Nov 30 '18 at 9:34
$begingroup$
Yes. As I wrote, we think of m as a function of n.
$endgroup$
– user554564
Nov 30 '18 at 10:18
add a comment |
$begingroup$
I came across the following statement in my Computer Science high-school textbook (I translated it to English).
There was some for loop that runs $m$ times nested inside a for loop that runs $n$ times, and it said:
"The function is $n cdot m$, and since $n$ and $m$ are as large as we want, we can set $m$ equal to $n$ and say that the complexity is $O(n^2).$"
Obviously I was suspicious, because $m$ in this case didn't depend on $n$.
So I started thinking about the problem with the big-$O$ definition.
Using the definition of for multiple variables, I proved the statement is false by contradiction.
I also thought about using the definition for a single variable, with $m$ as a function of $n$. I was able to prove that $n cdot m = O(n^2)$ if and only if $m=O(n)$.
My question is, is it possible to somehow generalize this thought and say that $f(n,m) = O(f(n,n))$ if some properties hold? Is this even legal, with Big-$O$ notation?
I'm also not sure if my original proofs are correct, so here they are:
Using the multiple variable definition, prove $n cdot m neq O(n^2)$.
Suppose $n cdot m = O(n^2)$, then there exist $c,n_0,m_0 in mathbb{R}_{>0}$ so that $n geq n_0, m geq m_0$ implies $n cdot m leq cn^2$.
But if we look at $n=n_0,m=max(m_0,cn_0+1)$, we have
$$n cdot m = n_0 cdot max(m_0,cn_0+1) geq n_0 cdot (cn_0+1) = cn_0^2+n_0 > cn_0^2$$
This is a contradiction, because $n=n_0 geq n_0$ and $m=max(m_0,cn_0+1) geq m_0$ and so we should have $n cdot m leq cn^2 = cn_0^2$.
Using the single variable definition, prove $n cdot m = O(n^2) iff m=O(n)$.
Suppose $n cdot m = O(n^2)$, then there exist $c,n_0 in mathbb{R}_{>0}$ so that $n geq n_0$ implies $n cdot m leq cn^2$. Divide both sides by $n$, and we have $m leq cn$ for $n geq n_0$, which directly means (by definition) that $m=O(n)$.
Suppose $m = O(n)$, then there exist $c,n_0 in mathbb{R}_{>0}$ so that $n geq n_0$ implies $m leq cn$. Now let's prove $n cdot m = O(n^2)$.
For every $n$ and specifically for every $n geq n_0$,
$$n cdot m leq n cdot cn = cn^2$$
Which directly means (by definition) that $n cdot m = O(n^2)$.
asymptotics computer-science computational-complexity
$endgroup$
I came across the following statement in my Computer Science high-school textbook (I translated it to English).
There was some for loop that runs $m$ times nested inside a for loop that runs $n$ times, and it said:
"The function is $n cdot m$, and since $n$ and $m$ are as large as we want, we can set $m$ equal to $n$ and say that the complexity is $O(n^2).$"
Obviously I was suspicious, because $m$ in this case didn't depend on $n$.
So I started thinking about the problem with the big-$O$ definition.
Using the definition of for multiple variables, I proved the statement is false by contradiction.
I also thought about using the definition for a single variable, with $m$ as a function of $n$. I was able to prove that $n cdot m = O(n^2)$ if and only if $m=O(n)$.
My question is, is it possible to somehow generalize this thought and say that $f(n,m) = O(f(n,n))$ if some properties hold? Is this even legal, with Big-$O$ notation?
I'm also not sure if my original proofs are correct, so here they are:
Using the multiple variable definition, prove $n cdot m neq O(n^2)$.
Suppose $n cdot m = O(n^2)$, then there exist $c,n_0,m_0 in mathbb{R}_{>0}$ so that $n geq n_0, m geq m_0$ implies $n cdot m leq cn^2$.
But if we look at $n=n_0,m=max(m_0,cn_0+1)$, we have
$$n cdot m = n_0 cdot max(m_0,cn_0+1) geq n_0 cdot (cn_0+1) = cn_0^2+n_0 > cn_0^2$$
This is a contradiction, because $n=n_0 geq n_0$ and $m=max(m_0,cn_0+1) geq m_0$ and so we should have $n cdot m leq cn^2 = cn_0^2$.
Using the single variable definition, prove $n cdot m = O(n^2) iff m=O(n)$.
Suppose $n cdot m = O(n^2)$, then there exist $c,n_0 in mathbb{R}_{>0}$ so that $n geq n_0$ implies $n cdot m leq cn^2$. Divide both sides by $n$, and we have $m leq cn$ for $n geq n_0$, which directly means (by definition) that $m=O(n)$.
Suppose $m = O(n)$, then there exist $c,n_0 in mathbb{R}_{>0}$ so that $n geq n_0$ implies $m leq cn$. Now let's prove $n cdot m = O(n^2)$.
For every $n$ and specifically for every $n geq n_0$,
$$n cdot m leq n cdot cn = cn^2$$
Which directly means (by definition) that $n cdot m = O(n^2)$.
asymptotics computer-science computational-complexity
asymptotics computer-science computational-complexity
asked Nov 30 '18 at 9:14
user554564
$begingroup$
$m$ must depend on $n$ in some way, otherwise $mn=O(n)ne O(n^2)$
$endgroup$
– Exodd
Nov 30 '18 at 9:34
$begingroup$
Yes. As I wrote, we think of m as a function of n.
$endgroup$
– user554564
Nov 30 '18 at 10:18
add a comment |
$begingroup$
$m$ must depend on $n$ in some way, otherwise $mn=O(n)ne O(n^2)$
$endgroup$
– Exodd
Nov 30 '18 at 9:34
$begingroup$
Yes. As I wrote, we think of m as a function of n.
$endgroup$
– user554564
Nov 30 '18 at 10:18
$begingroup$
$m$ must depend on $n$ in some way, otherwise $mn=O(n)ne O(n^2)$
$endgroup$
– Exodd
Nov 30 '18 at 9:34
$begingroup$
$m$ must depend on $n$ in some way, otherwise $mn=O(n)ne O(n^2)$
$endgroup$
– Exodd
Nov 30 '18 at 9:34
$begingroup$
Yes. As I wrote, we think of m as a function of n.
$endgroup$
– user554564
Nov 30 '18 at 10:18
$begingroup$
Yes. As I wrote, we think of m as a function of n.
$endgroup$
– user554564
Nov 30 '18 at 10:18
add a comment |
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$begingroup$
$m$ must depend on $n$ in some way, otherwise $mn=O(n)ne O(n^2)$
$endgroup$
– Exodd
Nov 30 '18 at 9:34
$begingroup$
Yes. As I wrote, we think of m as a function of n.
$endgroup$
– user554564
Nov 30 '18 at 10:18