# Relationship between complexity classes XP and W[1]?

I am reading the introductory chapter in Parameterized Algorithms by Cygan et al. and I am having some problems with the distinction between complexity classes $$\mathsf{W[1]}$$ and $$\mathsf{XP}$$.

They explicitly define complexity class $$\mathsf{XP}$$ on p. 13 as the those problems that, given an instance $$(x, k)$$ of a parameterized problem, can be correctly decided by an algorithm in $$O(f(k)\cdot|(x,k)|^{g(k)})$$ time for functions $$g, k$$. I believe I understand this.

Complexity class $$\mathsf{W[1]}$$ is not explicitly defined. However, in the discussion of $$k$$-cliques on p. 9, they give an $$O(n^k)$$ algorithm for deciding whether a $$k$$-clique exists in a graph. As I understand it, therefore, the $$k$$-clique problem must be in $$\mathsf{XP}$$. However, the authors discuss it as a $$\mathsf{W[1]}$$ problem in a way that I find confusing.

My question, then, is twofold: (1) Am I right in thinking that $$\mathsf{W[1]} \subseteq \mathsf{XP}$$? I haven't been able to settle this by looking ahead in the book or by searching online. (2) Intuitively, what is the difference between the two classes? I have found defitions online, e.g. on Wikipedia, but not in language that I have found accessible. I would appreciate, therefore, if someone could provide more of an overview.

1) Yes, you are right: W[1] is a subset of XP. As you noticed, $$k$$-clique is in XP. The class W[1] is (usually by definition) the closure of that problem under FPT-reductions. XP is closed under these reductions. Hence the subsetship.
Gaining an intuition for W[1] is not exactly made easy (in the past I tried in this answer). As said, it is usually defined by completeness of a problem (like $$k$$-clique). Definitions using machine models exist (e.g. Chen, Flum, Grohe: Machine-based methods in parameterized complexity theory. TCS 339(2-3), 2005). But that does not imply they are suitable for entry-level understanding of the class. I suggest you just take W[1] as the sum of the facts you know about it.