Existence of terminal $3$-fold flips Does there exist a terminal $3$-fold $X$ with a curve $C\subset X$ such that $K_X\cdot C < 0$ admitting a Mori flip $X\dashrightarrow Y$, flipping $C$ to a curve $C'\subset Y$, where the singular locus of $Y$ consists of a single point of type $\frac{1}{2}(1,1,1)$, $C'$ passes through the singular point of $Y$, and $-K_Y\cdot C' < -\frac{1}{2}$?
Is there any classification result of $3$-fold terminal flips that could answer to this question?
Thank you.
 A: Yes - there are very many such examples, and you can cook up examples by a procedure called 'Mori's algorithm'.
A k2A flipping neighbourhood is a 3-fold flipping contraction $f\colon(C\subset X)\to (P\in Y)$, where the exceptional curve $C$ is irreducible and $X$ has two $cA_{d_i-1}/\tfrac1{r_i}(1,-1,a_i,0)$ hyperquotient singularities along $C$. You can construct such neighbourhoods as a 1-parameter $\mathbb Q$-Gorenstein smoothing of the contraction of a rational curve in a surface $f\colon (C\subset H_X)\to (P\in H_Y)$, where $H_X$ has two $T$-singularities of type $\tfrac{1}{r_i^2d_i}(1,r_id_ia_i-1)$ for $i=1,2$ along $C$.
Mori's algorithm says that if $\delta=r_1a_2+r_2a_1-r_1r_2>0$ (which is the integer such that $K_XC=-\frac{\delta}{r_1r_2}$), then set $d_1=d_3=d_5=\ldots$ and $d_2=d_4=d_6=\ldots$ and write down a table
$$ \begin{array}{ccccccc}
d_1\delta & d_2\delta & d_1\delta & d_2\delta & \ldots & d_n\delta & d_{n+1}\delta \\ \hline
r_1 & r_2 & r_3 & r_4 & \cdots & r_n & r_{n+1}  \\
r_1-a_1 & a_2 & a_3 & a_4 & \cdots & a_n & a_{n+1} 
\end{array} $$
where $r_{i+1} = \delta d_ir_i - r_{i-1}$ and $a_{i+1} = \delta d_ia_i - a_{i-1}$. If $r_1>r_2$ then eventually the sign of $r_i$ will become negative. Let $n$ be the first $n$ such that $r_n>0$ and $r_{n+1}<0$. Then the flip $(C^+\subset X^+)$ will have singularities of type $cA_{d_i-1}/\tfrac1{r_i}(1,-1,a_i,0)$ along $C^+$ for $i=n,n+1$. Moreover $K_{X^+}C^+=\frac{\delta}{r_n(-r_{n+1})}>0$.
Note that if $d=1$ then a $cA_{d-1}/\tfrac1{r}(1,-1,a,0)$ singularity is simply a $\tfrac1{r}(1,-1,a)$ quotient singularity, and if $r=1$ then this quotient singularity is actually just a smooth point. Therefore, to give a fairly arbitrary example which address your question, we can take $(r_1,a_1,d_1)=(34,25,1)$ and $(r_2,a_2,d_2)=(7,2,1)$ with $\delta=5$ and write down the table
$$ \begin{array}{cccc}
5 & 5 & 5 & 5 \\ \hline
34 & 7 & 1 & -2 \\
9 & 2 & 1 & 3 
\end{array} $$
which tells me that $(C\subset X)$ has $\tfrac1{34}(1,9,25)$ and $\tfrac17(1,2,5)$ singularities with $K_XC=-\frac{5}{238}$ and the flip $(C^+\subset X^+)$ has one $\tfrac12(1,1,1)$ singularity along $C^+$ and $K_{X^+}C^+=\frac{5}{2}$. If you look at the resolution of the surface singularity $(P\in H_Y)$ downstairs, you will find that it is a $\tfrac1{15}(1,4)$ singularity whose resolution is a chain of two rational curves of self-intersection $-4$. The flip $X^+$ is obtained by resolving one of the $-4$ curves and taking a $\mathbb Q$-Gorenstein smoothing of the $\tfrac14(1,1)$ point on the resulting surface.
You make arbitrarily complicated examples by cooking up the two rightmost columns of the table correctly so that $\delta>1$, and then extending the table to the left as far as you want.
