4. A mysterious machine contains a sequence of 2017 integers, $x_{1}, x_{2}, \ldots, x_{2016}$, where all but one of the numbers are equal. If a sequence of 2017 integers $y_{1}, y_{2}, \ldots, y_{2017}$ is input into the machine, it will output the value $\sum_{i=1}^{2017} x_{i} \cdot y_{i}$. After the first input, a second, third, and so on can be made. Question: In the following scenarios, how many times at least must an input be made to determine the sequence of numbers in the machine? (1) The number in the machine that is different from the others is 0; (2) It is unknown what the number in the machine that is different from the others is.
Four, Analysis: Let $u$ be the number that appears 2016 times, $d$ be the number that appears once, and in the sequence $x_{t}=d$.
(1) When the number in the machine that is different from the others is 0, the first input of 2017 ones results in an output of $2016 u$; the second input of $1,2,3, \ldots, 2017$ results in an output of $\left(C_{2018}^{2}-t\right) \cdot u$, which allows us to find the value of $t$.

To prove that one input is insufficient to determine the result, we need to show that for every different input sequence, there exist two different number combinations in the machine that produce the same output.
When the input is $y_{1}, y_{2}, \ldots, y_{2017}$, (i) if there exists $y_{i}=y_{j}(i, j \in\{1,2, \ldots, 2017\})$, swapping $x_{i}$ and $x_{j}$ in the machine results in the same output; (ii) if there does not exist $y_{i}=y_{j}(i, j \in\{1,2, \ldots, 2017\})$, let $S=\sum_{i=1}^{2016} y_{i}$. For $i \neq j, i \neq k$, let $x_{i}=0, x_{k}=S-y_{j}$, the output is $\left(S-y_{j}\right) \cdot S-y_{i} \cdot\left(S-y_{j}\right)=\left(S-y_{i}\right) \cdot\left(S-y_{j}\right)$. If we swap $x_{i}$ and $x_{j}$, let $x_{j}=0, x_{k}=S-y_{i}$, the output is $\left(S-y_{i}\right) \cdot S-y_{j} \cdot\left(S-y_{i}\right)=\left(S-y_{i}\right) \cdot\left(S-y_{j}\right)$, which is the same as before. Therefore, for every different input, there exist two different number combinations in the machine that produce the same output. Hence, at least two inputs are required.
(2) From (1), we know that at least two inputs are required. We now prove that two inputs are sufficient.
The following construction is given: The first input: $1,-1,1,-1, \ldots$, let the output be $R$, and let $p=u-d$, then $R=(-1)^{t} \cdot p$.

The second input should satisfy: (i) $R$ and $S=\sum_{i=1}^{2016} y_{i}$ are coprime; (ii) no two $y_{i} \cdot(-1)^{i}$ are congruent modulo $S$; then the output $R^{\prime}=\left(S-y_{t}\right) \cdot u+y_{t} \cdot d=S u-y_{t} \cdot p=S u-y_{t} \cdot(-1)^{t} \cdot R$. Since $S$ and $R$ are coprime, $R^{\prime}$ is uniquely determined by the remainder of $y_{i} \cdot(-1)^{i}$ modulo $S$, which allows us to find the value of $t$.