將每個員工視為一個頂點，並將每對仇人之間連線，從而將問題轉化到圖 $(V, E)$ 上。考慮用若干顏色去塗這些點的塗色法；一個塗法被稱為”合法的”，若且唯若任兩邊的兩端點的顏色不同。則原題等價於：若最少色的合法塗法為 $k$ 色，則圖 $(V, E)$ 上至少存在 $2^{k-1} - k$ 個環。

對於任何 $k$ 色塗法, 以 $\hat{V} = (V_1, V_2, \dots, V_k)$ 來表示之, 其中 $V_i$ 為被塗成第 $i$ 色的點所成集合。讓我們用以下方式將塗法排序：對於兩種塗法 $\hat{V} = (V_1, \dots, V_k)$ 和 $\hat{U} = (U_1, \dots, U_k)$, 若 $|V_1| < |U_1|$ 則 $\hat{V} < \hat{U}$。若 $|V_1| = |U_1|$, 則再比較 $|V_2|$ 和 $|U_2|$, 若 $|V_2| < |U_2|$ 則 $\hat{V} < \hat{U}$。若再相等則比較 $|V_3| < |U_3|$, 依此類推。換言之, 若 $k^* = \inf\{n : |V_n| \neq |U_n|\}$, 則 $\hat{V} < \hat{U}$ 若且唯若 $|V_{k^*}| < |U_{k^*}|$。

We will consider some minimal coloring among all valid $k$-colorings. Call a cycle a "rainbow cycle" if and only if the vertices on it include all colors. We will prove that a minimal coloring has the following property:

Property 1. If $k$ is odd, then under a minimal valid coloring, there exists a rainbow cycle with an odd number of vertices.

Proof of Property 1. Clearly $V_1$ is not empty, so there exists $v \in V_1$. We will prove that there exists a rainbow cycle on which there is only one point of $V_1$, namely $v$.

Construct a subgraph: first, mark $v$. Next, mark the points of $V_2$ that are connected to $v$. Next, mark all the points of $V_3$ that are connected to some marked point of $V_2$. Continue in this manner, each time marking all the points of $V_{i+1}$ that are connected to some marked point of $V_i$, where $V_{k+1} = V_2, V_{k+2} = V_3$, and so on (in other words, after $v$ we no longer consider points of $V_1$). Repeat this process until no more points can be marked (since there are finitely many points, this process must terminate in finitely many steps).

Let the set of all marked points be denoted by $M$. Clearly, points in $V_i \cap M$ cannot be connected to points in $V_{i+1} \cap M^c$. Moreover, $v$ can reach any point of $M$ by a finite path of edges.

Now, consider a new coloring $W = (W_1, \cdots, W_k)$, where:
$$- W_1 = V_1.$$
$$- \forall 3 \le i \le k, W_i = (V_i \cap M^c) \cup (V_{i-1} \cap M).$$
$$- W_2 = (V_2 \cap M^c) \cup (V_k \cap M).$$
In other words, $W$ is obtained by first coloring all points using $V$, and then recoloring the marked points of $V_2$ to color 3, the marked points of $V_3$ to color 4, and so on, and finally the marked points of $V_k$ to color 2. By the construction above, $W$ must also be valid (since it can be viewed as a relabeling of colors on the entire connected subgraph). Moreover, $v$ must be connected to some $w \in W_2$; otherwise, we could recolor $v$ with color 2, and this coloring would still be valid and smaller (since color 1 would have one fewer point), contradicting the assumption that the original coloring is minimal. Also note that if $w \notin M$, then this means $w \in V_2$; but if so, since $v$ and $w$ are connected, $w$ should have been marked in the first step and would be sent to $W_3$ when recolored to $W$, a contradiction! Therefore $w \in M$ and $w \in V_k$.

Now, since $w \in M$, there must exist a finite path of edges connecting $v$ and $w$, which passes through $V_2$ up to $V_k$. At the same time, from the discussion above, $v$ and $w$ are connected by an edge. These two facts together form a rainbow cycle with an odd number of vertices, as desired!

Proof of the original problem: Consider a minimal valid coloring $\hat{V}$. Suppose $C \subset \{1, 2, \cdots, k\}$ and $|C|$ is an odd number greater than 1. We will prove that there exists a cycle in the original graph whose vertices have exactly the colors corresponding to $C$, and no others; if so, since there are $2^{k-1} - k$ choices for $C$ (subsets with an odd number of elements, excluding singleton subsets), the original graph has $2^{k-1} - k$ distinct cycles, and the original claim follows.

Let $V_c$ be the set of all vertices of color $c$ in $V$, let $V_C = \bigcup_{c \in C} V_c$, and let $G_C$ be the subgraph induced by $V_C$. Note that the valid coloring remains valid when restricted to this subgraph. Moreover, it is still minimal, because if it were not, then there would exist a smaller coloring $\hat{U}$ on the subgraph, and we could then consider a coloring of the original graph that colors all points of $V_C$ according to $\hat{U}$ and all points of $V \setminus V_C$ according to $\hat{V}$; this coloring would be smaller than $\hat{V}$, a contradiction. Hence, the original coloring restricted to the subgraph remains a minimal valid coloring, so by Property 1, there must exist a rainbow cycle in the subgraph whose vertices have exactly the colors corresponding to $C$, and no others. This completes the proof.