Some computers of a computer room have a following network. Each computers are connected by three cable to three computers. Two arbitrary computers can exchange data directly or indirectly (through other computers). Now let's remove $K$ computers so that there are two computers, which can not exchange data, or there is one computer left. Let $k$ be the minimum value of $K$. Let's remove $L$ cable from original network so that there are two computers, which can not exchange data. Let $l$ be the minimum value of $L$. Show that $k=l$.
1. **Establishing the inequality \( k \le \ell \):**
   - Whenever we can remove \( \ell \) edges to disconnect a graph, we can also disconnect the graph by removing an endpoint of each of the \( \ell \) edges. This means that we can remove at most \( \ell \) vertices to achieve the same disconnection.
   - Therefore, \( k \le \ell \).

2. **Establishing the inequality \( \ell \le k \):**
   - We need to show that the minimum number of edges to remove to disconnect the graph is at most the minimum number of vertices to remove to disconnect the graph.

3. **Considering the case \( k \ge 3 \):**
   - If \( k \ge 3 \), then we can remove 3 vertices to disconnect the graph. Since each vertex is connected by 3 edges, removing these vertices will also remove at least 3 edges.
   - Therefore, \( k \ge 3 \ge \ell \), which implies \( \ell \le k \).

4. **Considering the case \( k = 1 \):**
   - If \( k = 1 \), removing one vertex disconnects the graph. This means the graph is not connected initially, which contradicts the problem statement that any two computers can exchange data directly or indirectly.
   - Therefore, \( k = 1 \) is not possible.

5. **Considering the case \( k = 2 \):**
   - If \( k = 2 \), the graph is 2-connected, meaning each pair of vertices is connected by at least two disjoint paths.
   - Assume we remove vertices \( x \) and \( y \), and let \( u \) and \( v \) be two vertices that become disconnected.
   - Let \( x_1, x_2, x_3 \) be the neighbors of \( x \), and \( y_1, y_2, y_3 \) be the neighbors of \( y \).
   - Since \( u \) and \( v \) cannot both be linked to two of the \( x_i \)'s (through paths not containing \( x \) or \( y \)), assume \( u \) is only linked to \( x_1 \).
   - Similarly, assume \( u \) is only linked to \( y_1 \).
   - Removing the edges \( xx_1 \) and \( yy_1 \) will disconnect the graph, so \( \ell \le 2 = k \).

6. **Conclusion:**
   - Since we have shown \( k \le \ell \) and \( \ell \le k \), it follows that \( k = \ell \).

The final answer is \( \boxed{ k = \ell } \).