6

I just stumbled upon this question here Why is the clique problem NP-complete? and I am confused by the given answer. The question about about whether $k$-clique is NP-hard for a fixed $k$ and the answer is no. However, we know that clique in general is NP-hard. Also we know that $0 \leq k \leq n$ must hold (for a graph of size $n$).

What is wrong about the following reasoning: Assume I have some algorithm $A_i$ that solves $i$-clique for some fixed $i$ in polynomial time. Now given some general clique problem without a specific $k$ (these problems are supposedly NP-hard), I simply run $A_i$ for $i=1$ to $n$. Now since $n$ is poly and every $A_i$ runs in polynomial time $p_i$, the resulting algorithm also runs in polynomial time $\sum p_i$, but this cannot be the case, since clique is NP-hard.

What is wrong about my reasoning?

Community
  • 1
Cryptonaut
  • 289
  • 2
  • 9

1 Answers1

6

The polynomial depends on the parameter $k$. In particular, the algorithm that people have in mind runs in time $O(n^k)$ (better algorithms might exist, but I believe that we don't know a running time better than $O(n^{O(k)})$. The running time of your proposed algorithm is then big O of $$ \sum_{k=1}^n n^k, $$ which is no longer polynomial.

Yuval Filmus
  • 276,994
  • 27
  • 311
  • 503
  • Why would that not be polynomial. If I add a polynomial amount of polynomials, then I end up with a polynomial not something exponential, no? Basically, in your big-O notation, the addition would only change the hidden constant. – Cryptonaut Sep 22 '16 at 18:53
  • 3
    No. The function $n^n$ is not a polynomial. – Yuval Filmus Sep 22 '16 at 18:54
  • 1
    Ah I think I understand. So it's tractable, if I fix k independent of the graph size, because then it's just a "constant". If I ask: Is there a clique of size n-1 in a graph of size n, then it's still hard, because k depends on n? Basically I need to make sure that the exponent has some dependency of the size of the graph or what would be a simple way to look at it? – Cryptonaut Sep 22 '16 at 18:57
  • 2
    Actually the trivial algorithm (which goes over all potential cliques) runs in time proportional to $\binom{n}{k}$, so the hardest setting is $k = n/2$; when $k = n-1$, the problem becomes easy again. – Yuval Filmus Sep 22 '16 at 18:59
  • 4
    If you're interested in pursuing this direction further, you should read on parameterized complexity. – Yuval Filmus Sep 22 '16 at 19:00
  • thanks for the pointer. That seems very close to what I need! – Cryptonaut Sep 22 '16 at 19:01
  • "better than $O(n^{O(k)})$" -- that doesn't make any sense. You probably mean $\Theta$s, but nested Landau terms are hard to make sense of either way. Can you express more clearly what you mean to say? – Raphael Sep 22 '16 at 20:03
  • @Raphael Perhaps $O(n^{ck})$ algorithms exist for constant $c$ (e.g. for $c=1/2$), but nothing better is known. – Yuval Filmus Sep 22 '16 at 20:19