168 cm tall. Try to find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for all $x, y \in \mathbf{R}$, we have
$$
f(f(x+f(y))-1)=f(x)+f(x+y)-x .
$$
First, we prove that $f$ is injective.
Let $y_{1}, y_{2} \in \mathbf{R}, f\left(y_{1}\right)=f\left(y_{2}\right)$.
We need to prove that $y_{1}=y_{2}$.
From $f\left(y_{1}\right)=f\left(y_{2}\right)$, we get
$f\left(x+y_{1}\right)=f\left(x+y_{2}\right)$ (for any $\left.x \in \mathbf{R}\right)$.
Thus, $f\left(x+y_{1}-y_{2}\right)=f(x)$.
Let $T=y_{1}-y_{2}$, then
$f(x+T)=f(x)$ (for any $x \in \mathbf{R}$ ).
Substituting $x+T$ for $x$ in equation (1) and using equation (2), we get
$$
f(f(x+f(y))-1)=f(x)+f(x+y)-x-T \text {. }
$$

From equations (1) and (3), we get $T=0$, i.e., $y_{1}=y_{2}$, so $f$ must be injective.
Substituting $y=0$ into equation (1), we get
$$
f(f(x+f(0))-1)=2 f(x)-x \text {. }
$$

Substituting $x=f(y)-f(0)$ into equation (4), we get
$$
\begin{array}{l}
f(f(f(y))-1) \\
=2 f(f(y)-f(0))-f(y)+f(0) .
\end{array}
$$

Substituting $x=0$ into equation (1), we get
$$
f(f(f(y))-1)=f(0)+f(y) \text {. }
$$

From equations (5) and (6), we get
$$
f(f(y)-f(0))=f(y) \text {. }
$$

From equation (7) and the fact that $f$ is injective, we get $f(y)-f(0)=y$, i.e.,
$$
f(y)=y+f(0) \text {. }
$$

From equations (1) and (8), we get
$$
x+y+3 f(0)-1=x+y+2 f(0) \text {, }
$$

i.e., $f(0)=1$.
Thus, $f(y)=y+1$.
Upon verification, $f(x)=x+1$ satisfies equation (1).
In conclusion, the solution to this problem is $f(x)=x+1$ (for any $x$ $\in \mathbf{R})$.