From wikipedia

In computational complexity theory, a numeric algorithm runs in pseudo-polynomial time if its running time is a polynomial in the length of the input (the number of bits required to represent it) and the numeric value of the input (the largest integer present in the input). In general, the numeric value of the input is exponential in the input length, which is why a pseudo-polynomial time algorithm does not necessarily run in polynomial time with respect to the input length.

From this SO post that was very helpful.

An algorithm runs in pseudo-polynomial time if the runtime is some polynomial in the numeric value of the input, rather than in the number of bits required to represent it.

And this MIT OCW lecture on the knapsack DP running time and definition of pseudo-polynomial time.

They are all talking about the same thing in that a running time is pseudo-polynomial when it is a multivariate polynomial of $(x, k)$ where $x$ denotes the size of the input and $k$ is some integer(numeric value) involved. Since $k$ is often exponential in $x$, pseudo-polynomial time is not necessarily polynomial time in $x$, the number of bits needed to represent the input. Mathematically, $k = \Theta(2^{x})$, 2 is replaceable by a constant.

This works absolutely fine when we have a numeric value $k$ as the input with running time $\Theta(k)$. Then $x = \Theta(\log k)$ and we have running time $\Theta(2^{x})$. Everything is clean.

And then I start to see more complicated issues in other algorithms. For example, in the knapsack case, let $N$ denote the number of things or objects in the input. As described in the MIT video, the input is $x = \Theta(NlogS)$ and the running time is $\Theta(NS)$. The instructor describes that since $S$ is exponential in the input size $\log S$ (the number of bits needed to represent the numeric value), we can see how knapsack running time is not polynomial time. $\log S$ is only a part of $x$, so I don't see how $\Theta(N\cdot 2^{\log S})$ is exponential in $\Theta(N\log S)$. How is running time expressed in the form $O(c^{x})$ where $c$ is a constant?

Something similar happens for counting sort running time. The SO post gives running time $O(N + k)$. It gives the same logic as above to say that the running time is exponential in the input size $\log k$. But to me it seems like $x = \Theta(N\log K)$ since the input is a list of $N$ things with the maximum value $k$. I don't really see how $O(N + k)$ can be exponential in $x$ just because $k$ is exponential in the number of bits needed to represent it, since $\log k$ is definitely not $x$, only a part of it. Same thing: running time doesn't seem exponential in $x$, rather only in $\log S$.

So either I am still not getting what it means to have exponential "length of the input" is or there is some conflict with how people define length of the input. MIT video defines input size similar to how I think of $x$ while many other sources have a different opinion it seems. I see it as the total number of bits needed to represent the whole input, but some analysis just sees it as the number of bits needed to represent a numeric value that is only a part of the whole input. Is running time exponential in the input size if it is exponential in the size of a part of the input or the whole thing?? Please clarify!!

  • 1
    $\begingroup$ As in my answer to your previous query, saying the algorithm is exponential doesn't mean it always runs in exponential time. If $N=S$, the running time of the naive knapsack algorithm is actually polynomial. $\endgroup$
    – Ariel
    Jan 13, 2018 at 18:59

1 Answer 1


First, the size of the input is the number of bits required to represent the input. When talking about complexity, we often assume that our input is represented as some bit-string.

For the knapsack case, when we have $x= \Theta(N\log S)$ and running time $\Theta(N\cdot2^{\log S})$, we say that the algorithm is exponential in the size of the input, as $\Theta(N\cdot2^{\log S}) = \Theta(N\cdot2^{(N\log S)/N}) = \Theta(N\cdot 2^{x/N})$. For fixed $N$, the latter formula is exponential in $x$. However, when we fix $S$ instead, the formula is linear in $x$ (which then is $\Theta(N)$). Since we often do not assume bounds on $S$ in the general knapsack problem, we can consider the formula exponential.

A similar analysis can be done for the case of counting sort.

So, in summary, although the length of the input is indeed the full encoding of the input (usually as a bit-string), this doesn't change the fact that if you encode a number $N$ in your input $x$ with $\log N$ bits alongside perhaps other things, $\Theta(N)$ will be exponential in $x$.

Finally, let me make some comments on pseudo-polynomial time. Personally, I think that both the definition on Wikipedia and in the SO post lack generality and clarity. I prefer to take the definition alluded to in this answer: An algorithm is pseudo-polynomial if it is polynomial only in a specific encoding, such as unary. This implies the definition given on Wikipedia, as the unary encoding on the number $N$ has size $N$, but the binary encoding has size $\log N$.

I prefer the definition above, as it shows a bit more what is going on: it matters what sort of input encoding we use whether our algorithm is polynomial, which explains the name. The fact that this occurs precisely when we have a number in the input is a nice corollary, but hides a bit too much and is probably a reason for your confusion.

  • $\begingroup$ $2^x\cdot 2^{1/N}\neq 2^{x/N}$. You will not be able the write the running time in the form of $2^{\Omega(x)}$, since there exists an infinite series of inputs on which the algorithm is polynomial. $\endgroup$
    – Ariel
    Jan 13, 2018 at 22:08
  • $\begingroup$ @Ariel Oops, my bad. I fixed my math and changed the interpretation of the result. Indeed, the algorithm cannot be 'always' exponential in one parameter. $\endgroup$
    – Discrete lizard
    Jan 13, 2018 at 22:25
  • $\begingroup$ but isn't fixing $N$ a very special case? not something than happens in general? the number of things in the input doesn't seem like something that will be constant in most of cases. $\endgroup$
    – namesake22
    Jan 15, 2018 at 8:08
  • $\begingroup$ @namesake22 We only 'fix' $N$ to look at the growth of the function w.r.t. the size of $x$. If we don't do this, then $x=\Theta(N)$ and the running time will be $\Theta(x)$. So, it turns out that fixing $N$ gives the worst case running time w.r.t. $x$. $\endgroup$
    – Discrete lizard
    Jan 15, 2018 at 8:21

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