100 dot product riddle I was given this riddle by one of my professors, and was wondering if anyone could give some hints on this problem. Say I have 100 vectors, $x_1, x_2, ... x_{100}$. I compute every dot product pair, excluding self-pairs, so a vector is not dotted with itself, i.e $x_1 \cdot x_2$, $x_1 \cdot x_3$, ... $x_1 \cdot x_{100}$, $x_2 \cdot x_3$, $x_2 \cdot x_4$, ... $x_2 \cdot x_{100}$ ... $x_{99} \cdot x_{100}$, and tabulate the result in a list $L_1$. I then create another list $L_2 = -L_1$, $L_2$ is just $L_1$ with the signs flipped. If I only gave you the two lists, could you tell which one is $L_1$ and which one is $L_2$, i.e which list is the true dot product pairs, and which list had it's sign flipped?

Consider a set of vectors $x_k\in \mathbb{R}^{100}:$
$$
\Big\{ x_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \\0 \end{bmatrix}, \; x_2 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ \vdots \\ 0 \\0\end{bmatrix}\; x_3 = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots \\ 0\\0 \end{bmatrix}, \; \dots,\;  x_{99} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots\\1 \\ 0\end{bmatrix},\;x_{100} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots  \\ 1\\1\end{bmatrix}\Big\}
$$
Then $L_1=\big\{(1)_{\times 99},(2)_{\times 98},\dots,(99)_{\times 1}\big\}, \; L_2=\big\{(-1)_{\times 99},(-2)_{\times 98},\dots,(-99)_{\times 1}\big\}$.
Now consider a set of vectors $x_k\in \mathbb{R}^{100}:$
$$
\Big\{ x_1 = \begin{bmatrix} -1 \\ 0 \\ 0 \\ \vdots \\ 0 \\0 \end{bmatrix}, \; x_2 = \begin{bmatrix} 1 \\ -3 \\ 0 \\ \vdots \\ 0 \\0\end{bmatrix}\; x_3 = \begin{bmatrix} 1 \\ 1 \\ -5 \\ \vdots \\ 0\\0 \end{bmatrix}, \; \dots,\;  x_{99} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots\\-197 \\ 0\end{bmatrix},\;x_{100} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots  \\ 1\\1\end{bmatrix}\Big\}
$$
Then $L_1=\big\{(-1)_{\times 99},(-2)_{\times 98},\dots,(-99)_{\times 1}\big\},\;  L_2=\big\{(1)_{\times 99},(2)_{\times 98},\dots,(99)_{\times 1}\big\}$.
