Let $ n $ be the number of switches, $ m $ the number of lamps, and $ A \in \{0,1\}^{n \times m} $ the incidence matrix where $ A_{i,j} = 1 $ if switch $ i $ controls lamp $ j $, and $ 0 $ otherwise.
Let $ S_i \subseteq \{1, 2, \dots, m\} $ denote the set of lamps controlled by switch $ i $, i.e., $ S_i = \{ j \mid A_{i,j} = 1 \} $.
Define the union of all switches:
$$
U = \bigcup_{i=1}^n S_i = \{1, 2, \dots, m\}
$$
(given: all lamps are turned on when all switches are pressed).
For each switch $ i $, define the union of all other switches:
$$
U_{-i} = \bigcup_{\substack{j=1 \\ j \ne i}}^n S_j
$$
We are to determine:
Does there exist an index $ i \in \{1, 2, \dots, n\} $ such that $ U_{-i} = \{1, 2, \dots, m\} $?
Equivalently:
Is there a switch $ i $ such that for every lamp $ j \in \{1, 2, \dots, m\} $, there exists at least one switch $ k \ne i $ with $ A_{k,j} = 1 $?
That is, for each lamp $ j $, define the set of switches that control it:
$$
T_j = \{ i \in \{1, \dots, n\} \mid A_{i,j} = 1 \}
$$
Then the condition holds if and only if there exists an $ i $ such that for all $ j \in \{1, \dots, m\} $, $ |T_j| \ge 2 $ or $ i \notin T_j $.
In other words:
There exists an $ i \in \{1, \dots, n\} $ such that for every $ j \in \{1, \dots, m\} $, if $ T_j = \{i\} $, then the condition fails. So we require:
$$
\forall j \in \{1, \dots, m\}, \quad |T_j| \ge 2 \quad \text{or} \quad i \notin T_j
$$
But since we are to find *some* $ i $ such that removing it still covers all lamps, the condition is:
$$
\exists i \in \{1, \dots, n\} \text{ such that } \bigcup_{k \ne i} S_k = \{1, \dots, m\}
$$
Which is equivalent to:
$$
\exists i \in \{1, \dots, n\} \text{ such that } \forall j \in \{1, \dots, m\}, \quad T_j \not\subseteq \{i\}
$$
That is, no lamp is **uniquely** controlled by switch $ i $ — but more precisely, for the chosen $ i $, **no lamp** has its *only* controlling switch being $ i $.
Thus, define for each lamp $ j $:
- If $ |T_j| = 1 $, then the unique switch in $ T_j $ is **critical** for lamp $ j $.
Let $ C = \{ i \in \{1, \dots, n\} \mid \exists j \text{ such that } T_j = \{i\} \} $ — the set of critical switches.
Then:
There exists a switch $ i $ to ignore such that all lamps remain on **if and only if** $ i \notin C $.
That is, if $ C \ne \{1, \dots, n\} $, then any switch not in $ C $ can be ignored.
But note: It is guaranteed that $ \bigcup_{i=1}^n S_i = \{1, \dots, m\} $, so every lamp is covered.
Therefore:
> **Answer is "YES" if and only if there exists at least one switch $ i $ that is not the unique controller of any lamp.**
> Equivalently: **"YES" if and only if $ C \ne \{1, \dots, n\} $** — i.e., not every switch is critical for at least one lamp.
Formally:
Let $ \text{critical}(i) = \exists j \in \{1, \dots, m\} \text{ such that } T_j = \{i\} $
Then:
$$
\boxed{
\text{Answer} =
\begin{cases}
\text{YES} & \text{if } \exists i \in \{1,\dots,n\} \text{ such that } \neg \text{critical}(i) \\
\text{NO} & \text{otherwise}
\end{cases}
}
$$