SIAM Journal on Discrete Mathematics, Volume 34, Issue 3, Page 1587-1601, January 2020.
An undirected graph $G$ is $d$-degenerate if every subgraph of $G$ has a vertex of degree at most $d$. By the classical theorem of Erdös and Gallai from 1959, every graph of degeneracy $d>1$ contains a cycle of length at least $d+1$. The proof of Erdös and Gallai is constructive and can be turned into a polynomial time algorithm constructing a cycle of length at least $d+1$. But can we decide in polynomial time whether a graph contains a cycle of length at least $d+2$? An easy reduction from Hamiltonian Cycle provides a negative answer to this question: Deciding whether a graph has a cycle of length at least $d+2$ is NP-complete. Surprisingly, the complexity of the problem changes drastically when the input graph is 2-connected. In this case we prove that deciding whether $G$ contains a cycle of length at least $d+k$ can be done in time $2^{\mathcal{O}(k)}\cdot|V(G)|^{\mathcal{O}(1)}$. In other words, deciding whether a 2-connected $n$-vertex $G$ contains a cycle of length at least $d+\log{n}$ can be done in polynomial time. Similar algorithmic results hold for long paths in graphs. We observe that deciding whether a graph has a path of length at least $d+1$ is NP-complete. However, we prove that if graph $G$ is connected, then deciding whether $G$ contains a path of length at least $d+k$ can be done in time $2^{\mathcal{O}(k)}\cdot n^{\mathcal{O}(1)}$. We complement these results by showing that the choice of degeneracy as the “above guarantee parameterization” is optimal in the following sense: For any $\varepsilon>0$ it is NP-complete to decide whether a connected (2-connected) graph of degeneracy $d$ has a path (cycle) of length at least $(1+\varepsilon)d$.

中文翻译：

---

远离堕落

SIAM离散数学杂志，第34卷，第3期，第1587-1601页，2020年1月。
如果$ G $的每个子图最多具有一个度为$ d $的度数顶点，则无向图$ G $会退化。根据1959年的Erdös和Gallai的经典定理，退化的每个图$ d> 1 $都包含一个长度至少为$ d + 1 $的循环。Erdös和Gallai的证明是建设性的，可以转化为构造长度至少为$ d + 1 $的循环的多项式时间算法。但是我们可以在多项式时间内确定图是否包含长度至少为$ d + 2 $的循环吗？通过汉密尔顿循环的简单归纳法可以解决这个问题：判断一个图的循环长度至少为$ d + 2 $是否是NP完全的。令人惊讶的是，当输入图形为2连接时，问题的复杂性发生了巨大变化。在这种情况下，我们证明了确定$ G $是否包含长度至少为$ d + k $的循环可以在$ 2 ^ {\ mathcal {O}（k）} \ cdot | V（G）| ^ { \ mathcal {O}（1）} $。换句话说，可以在多项式时间内确定2个连接的$ n $-顶点$ G $是否包含长度至少为$ d + \ log {n} $的循环。类似的算法结果适用于图中的长路径。我们观察到确定图是否具有长度至少为$ d + 1 $的路径是NP完全的。但是，我们证明如果连接了图$ G $，则可以在$ 2 ^ {\ mathcal {O}（k）} \ cdot的时间确定$ G $是否包含长度至少为$ d + k $的路径。 n ^ {\ mathcal {O}（1）} $。我们通过证明在以下意义上将简并性选择为“以上保证参数化”是最佳的，可以对这些结果进行补充：对于任何$ \ varepsilon>