\node at (0.4,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (1,2.5) -- (2.5,0);
\draw (2,0) -- (3.5,2.5);
\node at (2.4,1.25) {\scriptsize $E_{l}$};

\draw (3,2.1) -- (6,2.1);
\node at (4.5,2.35) {\scriptsize $E_{l+1}[-m]$};

\draw (5.5,2.5) -- (7,0);
\node at (5.8,1.25){\scriptsize $E_{l+m}$};
\draw (6.5,0) -- (8,2.5);
\node at (8.05,1.25){\scriptsize $E_{l+m-1}$};
\draw[dashed] (7.5,2.5) -- (9,0);
\node at (9.75,1.25){\scriptsize $E_{l+2}$};
\draw (8.5,0) -- (10,2.5);

\draw (6.7,1) -- (5.5,-1);
\node at (6.75,-0.25){\scriptsize $E_{l+m+1}$};
\draw[very thick] (6,-1) -- (4.8,1);
\node at (5.75,0.7){\scriptsize $E_{l+m+2}$};

\draw[very thick] (0.1,0.8) to [out=-40, in=210] (5.6,0.1);
\node at (2.7,-0.7) {\scriptsize $C_{l+m+2}$};

\draw[<-] (9.4,0.7) -- (10.2,0.7);

\draw (10.5,0) -- (12,2.5);
\node at (10.9,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (11.5,2.5) -- (13,0);
\draw (12.5,0) -- (14,2.5);
\node at (12.9,1.25) {\scriptsize $E_{l}$};

\draw (13.5,2.1) -- (16.5,2.1);
\node at (15,2.35) {\scriptsize $E_{l+1}[-m]$};

\draw (16,2.5) -- (17.5,0);
\node at (16.3,1.25){\scriptsize $E_{l+m}$};
\draw (17,0) -- (18.5,2.5);
\node at (18.55,1.25){\scriptsize $E_{l+m-1}$};
\draw[dashed] (18,2.5) -- (19.5,0);
\node at (20.25,1.25){\scriptsize $E_{l+2}$};
\draw (19,0) -- (20.5,2.5);

\draw (17.2,1) -- (16,-1);
\node at (17.6,-0.25){\scriptsize $E_{l+m+1}[-3]$};
\draw[dashed] (14.7,-0.8) -- (16.3,-0.8);
\draw (15.1,-1)--(13.9,1);

\node at (15.5,0.65){\scriptsize $E_{l+m+2}[\leq-3]$};

\draw[very thick] (10.6,0.8) to [out=-40, in=210] (14.8,0.1);
\node at (13,-0.6) {\scriptsize $C_n$};

\end{tikzpicture}
\caption{Case $k=1$, $\delta = 1$.}\label{fig:i5}
\end{figure}

Suppose that this resolution is $\pi$. Then $\eta$ contracts $E_{l+m}$ before the $(-2)$-curves $E_{l+m-1}$ and $E_{l+m+1}$, but this is not possible. Hence $\pi$ has another base-point, which must be the intersection point between $E_{l+m+1}$ and $E_{l+m+2}$, otherwise there would be a loop in the resolution in $X$. Now in $X_{l+m+3}$, the curve $C_{l+m+3}$ intersects the $(-2)$-curves $E_1$ and $E_{l+m+2}$. Thus there is another base-point of  $\pi$, which is the intersection point between $E_{l+m+2}$ and $E_{l+m+3}$. But this implies that $E_{l+m+1}$ has self-intersection $-3$ in $X$ (see the right part of Figure~$\ref{fig:i5}$). We know that $\eta$ contracts $E_{l+m}$ before $E_{l+m-1}$ and $E_{l+m+1}$. After contracting $E_{l+m}$, the self-intersections of the images of $E_{l+m-1}$ and $E_{l+m+1}$ are $-1$ and $-2$ respectively. But then $E_{l+m-1}$ intersects no other $(-2)$-curve, so we have $E_{l+m-1} = E_{l+2}$ and hence $m=3$. The multiplicity sequence of $C$ is thus of the form $(3,2_{(l)})$. Using $\delta = 1$ and the genus degree formula, we obtain
\begin{align*}
0 &= d^2 - 4l - 10, \\
0 &= d^2 - 3d - 2l - 4.
\end{align*}
Subtracting these equations and rearranging terms, we obtain $l = \frac{3d-6}{2}$, which we can substitute in the first equation and get $d^2-6d+2=0$, which has no integer solution in $d$. Thus $\delta = 1$ is not possible.
\end{subproof}

\begin{claim*}[\mathrm{A.}5]\label{claim10}
If $\delta \geq 2$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(\mathrm{A}.5)$]
For $i=l+m,\ldots, l+m+\delta$, the base-point $p_{i+1}$ is then the unique intersection point between $C_i$ and $E_i$. As $\delta \geq 2$, this means that $E_{l+m+1}$ has self-intersection $-2$ in $X$ (see Figure~$\ref{fig:i6}$). But this leads to a contradiction, since $\eta$ contracts $E_{l+m}$ before the $(-2)$-curves $E_{l+m-1}$ and $E_{l+m+1}$, whose images both have self-intersection $-1$, after $E_{l+m}$ is contracted.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.75]
\draw (0,0) -- (1.5,2.5);
\node at (0.4,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (1,2.5) -- (2.5,0);
\draw (2,0) -- (3.5,2.5);
\node at (2.4,1.25) {\scriptsize $E_{l}$};

\draw (3,2.1) -- (6,2.1);
\node at (4.5,2.35) {\scriptsize $E_{l+1}[-m]$};

\draw (5.5,2.5) -- (7,0);
\draw (6.5,0) -- (8,2.5);
\draw[dashed] (7.5,2.5) -- (9,0);
\draw (8.5,0) -- (10,2.5);
\node at (5.75,1.25){\scriptsize $E_{l+m}$};
\node at (8.05,1.25){\scriptsize $E_{l+m-1}$};
\node at (9.75,1.25){\scriptsize $E_{l+2}$};
\node at (6.73,-0.26){\scriptsize $E_{l+m+1}$};

\draw (6.7,1) -- (5.5,-1);
\draw[dashed] (4.2,-0.8) -- (5.9,-0.8);
\draw[very thick] (4.5,-1) -- (3.3,1);

\node at (4.5,0.8) {\scriptsize $E_{l+m+\delta+1}$};

\draw[very thick] (0.1,0.8) to [out=-40, in=210] (4.3,0.1);
\node at (2.5,-0.55) {\scriptsize $C_{l+m+\delta+1}$};

\end{tikzpicture}
\caption{Case $k=1$, $\delta \geq 2$.}\label{fig:i6}
\end{figure}
\end{subproof}

This concludes the case $k=1$.

\bigskip

(B) Assume now that $l=1$, as shown in Figure~$\ref{fig:ii0}$. We can also assume that $k \geq 2$, since we have already considered the case $k=1$. If $C_{k+1}$ has self-intersection~$-1$, then by the symmetry of the configuration, there exists a morphism $X \to \p^2$ whose contracted locus is $E_1 \cup \ldots \cup E_{n-1} \cup C_n$.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.75]
\draw (0,0) -- (1.5,2.5);
\node at (0.4,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (1,2.5) --(2.5,0);
\draw (2,0) -- (3.5,2.5);
\node at (2.4,1.25) {\scriptsize $E_{k}$};

\draw[very thick] (3,2.1) -- (6,2.1);
\node at (5.5,2.35) {\scriptsize $E_{k+1}$};

\draw (2.2,0.8) to [out=-20, in=180] (4.5,2.4);
\draw (4.5,2.4) to [out=0, in=110] (5.5,1);

\node at (3.75,1.25) {\scriptsize $C_{k+1}$};

\end{tikzpicture}
\caption{Minimal resolution of singularities for $l=1$.}\label{fig:ii0}
\end{figure}

\noindent From $(C_{k+1})^2 = -1$ and the genus-degree formula we get the following two identities
\begin{align*}
0 &= d^2 - km^2 - (m-1)^2 + 1, \\
0 &= d^2 -3d + 2 - km(m-1) - (m-1)(m-1).
\end{align*}
Subtracting the second identity from the first yields $3d - 1 - km - (m-1) = 0$. We then substitute  the equality $km = 3d - 1 - (m-1)$ in the first equation and obtain $d^2 = m(3d-2)$. Let $p$ be a prime number that divides $3d-2$ and thus also $d$. But then $p=2$ and hence we can write $3d-2 = 2^r$ for some natural number $r$. It then follows that $m = \frac{(2^r+2)^2}{9\cdot 2^r}$, in particular $2^r$ divides $2^{2r}+4\cdot 2^r + 4$ and thus $r=1$ or $r=2$. If $r=1$, then $d = \frac{4}{3}$, which is absurd. If $r=2$, then $d=2$ and $m=1$, which is excluded by hypothesis.

We thus know that $(C_{k+1})^2 > -1$ and hence $\pi$ has a base-point on $E_{k+1}$ that also lies on $C_{k+1}$. Since $C$ is not unicuspidal, the curves $C_{k+1}$ and $E_{k+1}$ intersect in at least two points.

There are now two possibilities: either $C_{k+1}$ passes through the intersection point between $E_k$ and $E_{k+1}$, or it does not. We will look at these cases separately (parts (i) and (ii) below).

\bigskip

(i) We suppose that $C_{k+1}$ passes through the intersection point between $E_k$ and $E_{k+1}$. Then this point is the next base-point of $\pi$, since there can be no triple intersections in the tree of the curves $E_1,\ldots,E_{n-1},C_n$ in $X$. Moreover the intersection multiplicity between $C_{k+1}$ and $E_{k+1}$ at $p_{k+2}$ is $m-2$ as $C_n$ and $E_{k+1}$ intersect transversely in $X$, see the configuration on the left in Figure~$\ref{fig:triplecase1}$.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.75]
\draw (0,0) -- (1.5,2.5);
\node at (0.4,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (1,2.5) --(2.5,0);
\draw (2,0) -- (3.5,2.5);
\node at (3.1,1.25) {\scriptsize $E_{k}$};

\draw[very thick] (2.8,2.1) -- (6,2.1);
\node at (4.3,2.35) {\scriptsize $E_{k+1}$};

\draw (2.2,1) to [out=70, in=180] (3.3,2.1);
\draw (3.3,2.1) to [out=0, in=180] (4.4,1);
\draw (4.4,1) to [out=0, in=180] (5.4,2.5);

\node at (5.3,1.25) {\scriptsize $C_{k+1}$};

\draw[<-] (6,0.7) -- (6.8,0.7);

\draw (7,0) -- (8.5,2.5);
\node at (7.4,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (8,2.5) -- (9.5,0);
\draw (9,0) -- (10.4,2.5);
\node at (9.73,1.25) {\scriptsize $E_{k}[-3]$};
\draw (10,2.5) -- (11.5,0);
\draw[dashed] (11,0) -- (12.5,2.5);
\node at (11.3,1.25) {\scriptsize $E_{k+2}$};

\draw[very thick] (12,2.1) -- (15,2.1);
\node at (13.47,2.35) {\scriptsize $E_{k+m-1}$};

\node at (16.6,1.25) {\scriptsize $E_{k+1}[-m+1]$};

\draw (14.5,2.5) -- (16,0);

\draw (12.6,2.4) to [out=-90, in=200] (16,0.5);
\node at (14.12,1.25) {\scriptsize $C_{k+m-1}[\delta]$};

\end{tikzpicture}
\caption{Blow-up of $p_{k+2},\ldots,p_{k+m-1}$.}\label{fig:triplecase1}
\end{figure}

It follows that the base-point $p_{i+1}$ is the intersection point between $E_{k+1}$ and $E_i$ for $i=k+1,\ldots,k+m-2$. We then denote by $\delta$ the self-intersection of $C_{k+m-1}$ in $X_{k+m-1}$, see the configuration on the right in Figure~$\ref{fig:triplecase1}$. We have $\delta = d^2 - km^2 - (m-1)^2 - (m-2)$ and $\delta \geq -1$, since $\pi$ is a $(-1)$-tower resolution.

\begin{claim*}[\mathrm{B.i.}1]\label{claim11}
If $\delta = -1$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(\mathrm{B.i.}1)$]
From $\delta = -1$ and the genus-degree formula we obtain
\begin{align*}
0 &= d^2 - km^2 - (m-1)^2 - m + 3, \\
0 &= d^2 -3d + 2 - km(m-1) - (m-1)(m-2).
\end{align*}
Subtracting these identities yields $3d - km - 2m + 2 = 0$. Thus the greatest common divisor of $d$ and $m$ divides $2$. We then substitute $k = \frac{3d-2m+2}{m}$ in the first equation and obtain $d^2 - 3dm + m^2 - m +2 = 0$. Let $p$ be any prime number that divides $m$. Then $p$ divides $3d+2$ and also $d^2+2$. But then $p$ also divides $d^2-3d=d(d-3)$. Assume that $p$ does not divide $d$, then $p$ divides $d-3$. Then $p$ divides $3d+2 - 3(d-3) = 11$. On the other hand $p$ also divides $(d^2+2) - (d-3)^2 -3(d-3) = 2$ and thus we have a contradiction. It follows that $p$ divides $d$ and hence $p=2$. Dividing the equation above by $2$ yields
$$d\frac{d}{2} - 3d\frac{m}{2} + m\frac{m}{2} - \frac{m}{2} + 1 = 0.$$
We conclude that $\frac{m}{2}$ must be odd. Since $m$ is a power of $2$ it then follows that $m = 2$. We hence obtain the equation $d^2 - 6d + 4 = 0$, which has no integer solution in $d$. We conclude that $\delta = -1$ is not possible.
\end{subproof}

\begin{claim*}[\mathrm{B.i.}2]\label{claim12}
If $\delta = 0$, then $C$ has degree $13$ and multiplicity sequence $(5_{(6)},4)$.
\end{claim*}

\begin{subproof}[Proof of Claim~$(\mathrm{B.i.}2)$]
From $\delta = 0$ and the genus-degree formula we obtain
\begin{align*}
0 &= d^2 - km^2 - (m-1)^2 - m + 2, \\
0 &= d^2 -3d + 2 - km(m-1) - (m-1)(m-2).
\end{align*}
Subtracting these identities yields $3d - km - 2m + 1 = 0$. We thus see that $d$ and $m$ are coprime and that $m$ divides $3d+1$. We substitute $k = \frac{3d - 2m + 1}{m}$ in the first equation and obtain $d^2 - 3dm + m^2 + 1 = 0$. From this we see that $m$ divides $d^2 + 1$. But then $m$ also divides $(d^2+1)-(3d+1)=d(d-3)$. Since $d$ and $m$ are coprime, $m$ divides $d-3$. On the other hand, $m$ also divides $(d^2+1)+(3d+1)=(d+1)(d+2)$. Let $p$ be a prime number that divides $m$. Then $p$ divides $d-3$ and either $d+1$ or $d+2$, but not both since they are coprime. Thus $p$ must be either $2$ or $5$. Assume moreover that $p^2$ divides $m$. Then $p^2$ also divides $d^2+1$ and $3d+1$. Since $p$ divides $d-3$, it follows that $p^2$ divides $(d-3)^2 = d^2 - 6d + 9 = d^2+1 - 2(3d+1) + 10$. But then $p^2$ divides $10$, which is not possible. We conclude that $m \in \{5,10\}$ (since $m \geq 3$). We then check for integer solutions for $d$ in the equation $d^2 - 3dm + m^2 + 1 = 0$ for these values of $m$ and find $(d,m) = (13,5)$ as the only possibility. For a diagram of a resolution of such an isomorphism see Remark~$\ref{remark:5555554}$. We assume from now on that we are not in this case.
\end{subproof}

\begin{claim*}[\mathrm{B.i.}3]\label{claim13}
If $\delta = 1$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(\mathrm{B.i.}3)$]
From $\delta = 1$ and the genus-degree formula we get the equations
\begin{align*}
0 &= d^2 - km^2 - (m-1)^2 - m + 1, \\
0 &= d^2 -3d + 2 - km(m-1) - (m-1)(m-1).
\end{align*}
Subtracting these identities yields $3d - km - 2m = 0$. We then substitute $k = \frac{3d - 2m}{m}$ in the first equation and obtain $d^2 = m(3d+m+1)$. Let $p$ be any prime number that divides $m$. But then $p$ divides $d^2$ and thus also $d$. It then follows that $p$ divides $1$ and we have a contradiction.
\end{subproof}

\begin{claim*}[\mathrm{B.i.}4]\label{claim14}
If $\delta \geq 2$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(\mathrm{B.i.4})$]
Since $\pi$ is a $(-1)$-tower resolution of $C$, the base-point $p_{i+1}$ is the unique intersection point between $C_i$ and $E_i$, for $i = k+m-1,\ldots,k+m+\delta-1$. The configuration after these blow-ups is shown in Figure~$\ref{fig:triplecase2}$.
\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.75]
\draw (7,0) -- (8.5,2.5);
\node at (7.4,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (8,2.5) -- (9.5,0);
\draw (9,0) -- (10.4,2.5);
\node at (9.73,1.25) {\scriptsize $E_{k}[-3]$};
\draw[dashed] (10,2.5) -- (11.5,0);
\draw (11,0) -- (12.5,2.5);
\node at (11.72,1.25) {\scriptsize $E_{k+m-2}$};

\draw (12,2.1) -- (15,2.1);
\node at (13.35,2.37) {\scriptsize $E_{k+m-1}$};

\node at (16.6,1.25) {\scriptsize $E_{k+1}[-m+1]$};

\draw (14.5,2.5) -- (16,0);

\draw (13.25,2.25)--(14,1);
\node at (13.2,1.5) {\scriptsize $E_{k+m}$};
\draw (14,1.25)--(13.25,0);
\node at (12.86,0.7) {\scriptsize $E_{k+m+1}$};
\draw[dashed] (13.25,0.25)--(14,-1);
\draw[very thick] (13.6,-0.8)--(15.5,-0.8);
\node at (14.9,-1.11) {\scriptsize $E_{k+m+\delta}$};

\draw[very thick] (15,-1.1) to [out=90, in=195] (16,0.7);
\node at (15.95,-0.25) {\scriptsize $C_{k+m+\delta}$};

\end{tikzpicture}
\caption{Case $l=1$, $\delta \geq 2$.}\label{fig:triplecase2}
\end{figure}\\
Since no more base-point of $\pi$ can lie on $E_{k+m}$, its strict transform in $X$ has self-intersection $-2$. If $m>3$, then $E_{k+m-1}$ intersects the two $(-2)$-curves $E_{k+m-2}$ and $E_{k+m}$ in $X$. But $\eta$ contracts $E_{k+m-1}$ before these two curves and thus this situation is not possible and we have $m=3$. Since $d < 3m = 9$ by Lemma~$\ref{Lem:inequalities2}$, the multiplicity sequence of $C$ is in Table~$\ref{table:sequences}$ and can only be $(3_{(3)},2)$ in degree $6$. In this case $\delta = 5$. But this implies that $E_{k+m+1}$ is also a $(-2)$-curve in $X$. We hence get a contradiction after $\eta$ contracts $E_{k+m-1}$. Then the image of $E_{k+m}$ intersects the $(-2)$-curves $E_k$ and $E_{k+m+1}$.
\end{subproof}

This concludes (i) of part (B).

\bigskip

(ii) Suppose now that $C_{k+1}$ does not pass through the intersection point between $E_k$ and $E_{k+1}$. Then $C_{k+1}$ intersects $E_{k+1}$ in one point with intersection multiplicity $m-1$, otherwise there would be a loop in the configuration of the curves $E_1,\ldots,E_{n-1},C_n$. The configuration of curves in $X_{k+1}$ is shown in the left part of Figure~$\ref{fig:ii1}$. Since $C_n$ and $E_{k+1}$ do not intersect in $X$, it follows that the base-point $p_{i+1}$ for $i=k+1,\ldots,k+m-1$ is the unique intersection point between $C_i$ and $E_i$, which also lies on $E_{k+1}$. The configuration of curves in $X_{k+m}$ is shown in the right part of Figure~$\ref{fig:ii1}$. We denote the self-intersection of $C_{k+m}$ by $\delta$ and this number is equal to $d^2 - km^2 - (m-1)^2 - (m-1)$. Since $\pi$ is a $(-1)$-tower resolution of $C$, it follows that $\delta \geq -1$.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.75]
\draw (0,0) -- (1.5,2.5);
\node at (0.4,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (1,2.5) --(2.5,0);
\draw (2,0) -- (3.5,2.5);
\node at (2.4,1.25) {\scriptsize $E_{k}$};

\draw[very thick] (3,2.1) -- (6,2.1);
\node at (4.5,2.35) {\scriptsize $E_{k+1}$};

\draw (2.2,0.8) to [out=-20, in=180] (4.5,2.1);
\draw (4.5,2.1) to [out=0, in=110] (5.3,1);

\node at (3.9,1.25) {\scriptsize $C_{k+1}$};

\draw[<-] (6,0.7) -- (6.8,0.7);

\draw (7,0) -- (8.5,2.5);
\node at (7.4,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (8,2.5) -- (9.5,0);
\draw (9,0) -- (10.5,2.5);
\node at (9.4,1.25) {\scriptsize $E_{k}$};

\draw (10,2.1) -- (13,2.1);
\node at (11.5,2.35) {\scriptsize $E_{k+1}[-m]$};

\draw[very thick] (12.5,2.5) -- (14,0);
\draw[dashed] (13.5,0) -- (15,2.5);
\draw (14.5,2.5) -- (16,0);
\node at (13.88,1.25) {\scriptsize $E_{k+m}$};
\node at (15.8,1.25) {\scriptsize $E_{k+2}$};

\draw (9.2,1) to [out=-40, in=220] (13.8,1);
\node at (11.8,-0.1) {\scriptsize $C_{k+m}[\delta]$};

\end{tikzpicture}
\caption{Blow-up of $p_{k+2},\ldots,p_{k+m}$.}\label{fig:ii1}
\end{figure}

In the surface $X$, let $E \neq E_k$ in $\{E_1,\ldots,E_n\}$ be a curve that intersects $C_n$. We know that the map $\eta$ first contracts $C_n$ and then the chain $E_k,\ldots,E_1$. Since $k\geq 2$, it follows that the image of $E$ is tangent to $E_{k+1}$, after these contractions. This implies that $E$ is not contracted by $\eta$ and thus $E=E_n$ is the last exceptional curve in the $(-1)$-tower resolution $\pi$. We now discuss what happens for different values of $\delta$.

\begin{claim*}[\mathrm{B.ii.}1]\label{claim15}
If $\delta = -1$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(\mathrm{B.ii.}1)$]
In this case we already have a $(-1)$-tower resolution of $C$. This resolution must be $\pi$, since there is no more base-point on $C_{k+m}$ and $C_n$ intersects $E_n$. But we observe that the curves $E_1,\ldots,E_{k+m-1},C_{k+m}$ are not connected and thus cannot be the contracted locus of $\eta$. Hence $\delta = -1$ is not possible.
\end{subproof}

\begin{claim*}[\mathrm{B.ii.}2]\label{claim16}
If $\delta = 0$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(\mathrm{B.ii.}2)$]
The base-point $p_{k+m+1}$ is the unique intersection point between $C_{k+m}$ and $E_{k+m}$. After this blow-up, we have a $(-1)$-tower resolution of $C$, which must be $\pi$, for the same reason as in the case $\delta = -1$. The configuration of curves is shown in Figure~$\ref{fig:ii2}$.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.75]
\draw (0,0) --(1.5,2.5);
\node at (0.4,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (1,2.5) --(2.5,0);
\draw (2,0) --(3.5,2.5);
\node at (2.4,1.25) {\scriptsize $E_{k}$};

\draw (3,2.1) --(6,2.1);
\node at (4.5,2.35) {\scriptsize $E_{k+1}[-m]$};

\draw (5.5,2.5) --(7,0);
\node at (6.88,1.25){\scriptsize $E_{k+m}$};
\draw[dashed] (6.5,0) --(8,2.5);
\draw (7.5,2.5) --(9,0);
\node at (8.8,1.25){\scriptsize $E_{k+2}$};

\draw[very thick] (6.7,1) --(5.5,-1);
\node at (6.8,-0.25){\scriptsize $E_{k+m+1}$};

\draw[very thick] (2.2,1) to [out=-50, in=200] (6.1,-0.6);
\node at (4.55,-0.25) {\scriptsize $C_{k+m+1}$};

\end{tikzpicture}
\caption{Case $l=1$, $\delta = 0$.}\label{fig:ii2}
\end{figure}

The map $\eta$ contracts first $C_{k+m+1}$ and then the chain $E_k,\ldots,E_1$. After these contractions the self-intersection of the image of $E_{k+1}$ is $-m+k$, but must also be $-1$ and hence $k=m-1$. From $\delta = 0$ we then obtain the equation $d^2 = m(m^2-1)$. Since $m$ and $m^2-1$ are coprime, they are both squares, as $d>0$. But if $m \geq 2$ is a square, then $m^2-1$ is not a square. Hence the only integer solutions to the equation are $(d,m) = (0,-1), (0,0), (0,1)$, and thus $\delta = 0$ is also not possible.
\end{subproof}

\begin{claim*}[\mathrm{B.ii.}3]
If $\delta \geq 1$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(\mathrm{B.ii.}3)$]
For $i=l+m,\ldots, l+m+\delta$, the base-point $p_{i+1}$ is the unique intersection point between $C_i$ and $E_i$. After these blow-ups we have a $(-1)$-tower resolution of $C$, which has to be $\pi$ for the same reason as in the previous cases. The configuration of curves is shown in Figure~$\ref{fig:ii5}$.
\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.75]
\draw (0,0) --(1.5,2.5);
\node at (0.4,1.25) {\scriptsize $E_{1}$};
\draw[dashed] (1,2.5) --(2.5,0);
\draw (2,0) --(3.5,2.5);
\node at (2.4,1.25) {\scriptsize $E_{k}$};

\draw (3,2.1) --(6,2.1);
\node at (4.5,2.35) {\scriptsize $E_{k+1}[-m]$};

\draw (5.5,2.5) --(7,0);
\node at (5.75,1.25){\scriptsize $E_{k+m}$};
\draw (6.5,0) --(8,2.5);
\node at (8.05,1.25){\scriptsize $E_{k+m-1}$};
\draw[dashed] (7.5,2.5) --(9,0);
\draw (8.5,0) --(10,2.5);
\node at (9.8,1.25){\scriptsize $E_{k+2}$};

\draw (6.7,1) --(5.5,-1);
\node at (6.8,-0.25){\scriptsize $E_{k+m+1}$};
\draw[dashed] (6,-1) --(4.8,1);
\draw[very thick] (5.3,1) --(4.1,-1);
\node at (4.17,0.85){\scriptsize $E_{k+m+\delta+1}$};

\draw[very thick] (2.2,1) to [out=-60, in=200] (4.9,-0.5);
\node at (3.78,0.3) {\scriptsize $C_{k+m+\delta+1}$};

\end{tikzpicture}
\caption{Case $l=1$, $\delta \geq 1$.}\label{fig:ii5}
\end{figure}

Since $\delta \geq 1$, the curve $E_{l+m+1}$ has self-intersection $-2$. But we know that $\eta$ contracts $E_{k+m}$ before the $(-2)$-curves $E_{l+m-1}$ and $E_{l+m+1}$, which leads to a contradiction.
\end{subproof}

This concludes (ii) of part (B) and hence finishes the proof of Proposition~$\ref{Prop:1jump}$.
\end{proof}

\begin{remark}\label{remark:5555554}
{\rm Below we see the configuration of exceptional curves of a resolution of a non-extendable isomorphism between two curves of degree $13$ with multiplicity sequence $(5_{(6)},4)$. All the unlabeled curves have self-intersection $-2$. Starting with either of the $(-1)$-curves, one can successively contract all curves in this configuration, except the other $(-1)$-curve. The image of this curve in $\p^2$, denoted $C$, then has self-intersection $169 = 13^2$. It remains to be verified whether such curves exist and whether new counterexamples to Conjecture~$\ref{conjecture:yoshihara}$ may arise in this way. We remark that $C \setminus \Sing(C) \simeq \A^1 \setminus \{0\}$ and thus $C$ is different from the unicuspidal examples of degree $13$ constructed in [7].

\begin{center}
\begin{tikzpicture}[scale=0.75]
\draw (-2,0)--(-0.5,2.5);
\draw (-1,2.5) --(0.5,0);
\draw (0,0) -- (1.5,2.5);
\draw (1,2.5) --(2.5,0);
\draw (2,0) -- (3.5,2.5);
\draw (3,2.5) -- (4.5,0);
\node at (3.52,1) {\scriptsize $-3$};
\draw (4,0) -- (5.5,2.5);
\draw (5,2.5) -- (6.5,0);

\draw (6.1,0.2) -- (9.7,0.2);
\node at (8.5,-0.05) {\scriptsize $-5$};

\draw (9.25,0) -- (10,1.25);
\node at (7.27,0.6) {\scriptsize $-4$};
\draw[very thick] (10,1)--(9.25,2.25);
\node at (7.27,1.7) {\scriptsize $-1$};
\draw (9.25,2)--(10,3.25);
\draw (10,3)--(9.25,4.25);

\draw (8,0)--(7.25,1.25);
\node at (9.92,0.6) {\scriptsize $-4$};
\draw[very thick] (7.25,1)--(8,2.25);
\node at (9.92,1.7) {\scriptsize $-1$};
\draw (8,2)--(7.25,3.25);
\draw (7.25,3)--(8,4.25);

\draw (7.75,4.1)--(9.5,4.1);
\node at (8.5,4.3) {\scriptsize $-3$};

\end{tikzpicture}
\end{center}
\end{remark}

\begin{corollary}\label{cor:1jumpcase}
Let $C \subset \p^2$ be an irreducible curve with one of the multiplicity sequences $(3,2_{(3)})$, $(3_{(2)},2_{(4)})$, $(3_{(3)},2)$, $(3_{(4)},2_{(3)})$, $(4,3_{(3)})$, $(4,3_{(5)})$, $(4_{(2)},3_{(3)})$, or $(4_{(3)},3)$. Then either $C$ is unicuspidal or any open embedding $\p^2 \setminus C \hookrightarrow \p^2$ extends to an automorphism of $\p^2$.
\end{corollary}

\begin{proof}
This is a direct consequence of Proposition~$\ref{Prop:1jump}$.
\end{proof}

\begin{remark}
{\rm Note that in Corollary~$\ref{cor:1jumpcase}$, only curves with the multiplicity sequences $(3_{(3)},2)$ and $(4_{(3)},3)$ can be unicuspidal.}
\end{remark}

\begin{proposition}\label{Prop:alleven}
Let $C \subset \p^2$ be a rational curve of degree $d$ and multiplicity sequence $(m_1,\ldots,m_k)$ such that all multiplicities are even and there exists $l<k$ such that $m_{l+1} = \ldots = m_k = 2$ and $m_{j} < m_{j+1}+\ldots+m_k$ for all $j \leq l$. Let $\varphi \colon \p^2 \setminus C \hookrightarrow \p^2$ be an open embedding that does not extend to an automorphism of $\p^2$. Then $C$ is unicuspidal.
\end{proposition}

\begin{proof}
Suppose that $C$ is not unicuspidal. By Proposition~$\ref{Prop:nojump}$, we can assume that the multiplicity sequence of $C$ is non-constant. By Lemma~$\ref{Lem:tower}$, there exists $\pi \colon X = X_n \xrightarrow{\pi_n} \ldots \xrightarrow{\pi_2} X_1 \xrightarrow{\pi_1} X_0 = \p^2$ a $(-1)$-tower resolution  of $C$ with base-points $p_1,\ldots,p_n$ and exceptional curves $E_1,\ldots,E_n$, and a $(-1)$-tower resolution $\eta \colon X \to \p^2$ of some curve $D \subset \p^2$ such that $\varphi \circ \pi = \eta$. Then $E_1 \cup \ldots \cup E_{n-1} \cup C_n$ is the exceptional locus of $\eta$, being the support of an SNC-divisor that has a tree structure. The composition $\pi_1 \circ \ldots \circ \pi_k$ is the minimal resolution of singularities of $C$. For $i = 1,\ldots,k$, we obtain the following intersection numbers, by Lemma~$\ref{Lem:proximate mult}$:
$$C_k \cdot E_i = m_i - \sum_{p_j \succ p_i} m_j.$$
In particular, $C_k \cdot E_k = 2$. Since $m_{j} < m_{j+1}+\ldots+m_k$ for all $j \leq l$, it follows that, for $i=1,\ldots,l$, the curves $E_i$ and $E_k$ do not intersect in $X_k$ and hence also not in $X$. Since all $m_i$ are even, it follows that the intersection numbers $C_k \cdot E_i$ are even. It follows moreover that the intersection numbers $C_n \cdot E_i$ are also even for $i=1,\ldots,l$, since $E_k$ and $E_i$ do not intersect in $X_k$. The curve $E_1\cup\ldots \cup E_{n-1} \cup C_n$ is SNC and therefore $E_i$ and $C_n$ do not intersect at all, for $i=1,\ldots,l$. Since the multiplicities $m_{l+1},\ldots,m_k$ are all equal to $2$, it follows that $C_k$ does not intersect any of the curves $E_1,\ldots,E_{k-1}$, but only $E_k$. Since $C$ is not unicuspidal, the curves $C_k$ and $E_k$ intersect in two distinct points. We denote by $\delta$ the self-intersection of $C_k$, which is given by $\delta = d^2 - \sum_{i=1}^k m_i^2$. Since $C$ has a $(-1)$-tower resolution, we have $\delta \geq -1$.

\begin{claim*}[1]
If $\delta = -1$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(1)$]
We already have a $(-1)$-tower resolution of $C$ (see Figure~$\ref{fig:even1}$). Since $C_k$ and $E_k$ intersect in two points and there is no more base-point on $C_k$, there is no more base-point at all. But we observe that $C_k$ and $E_1 \cup \ldots \cup E_{k-1}$ are not connected. This is not possible and hence $\delta$ must be $\geq 0$.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.75]
\draw[dashed] (2,0) -- (3.5,2.5);

\draw[very thick] (3,2.1) -- (6,2.1);
\node at (5.65,2.35) {\scriptsize $E_{k}$};

\draw[very thick] (3.2,0.5) to [out=70, in=180] (4.5,2.6);
\draw[very thick] (4.5,2.6) to [out=0, in=110] (5.8,0.5);

\node at (3.75,1.25) {\scriptsize $C_{k}$};

\end{tikzpicture}
\caption{Case $\delta = -1$.}\label{fig:even1}
\end{figure}
\end{subproof}

\begin{claim*}[2]
If $\delta = 0$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(2)$]
The genus-degree formula yields
$$d^2 - 3d + 2 = \sum_{i=1}^k m_i(m_i-1).$$
Using $\delta = 0$, we get $3d-2 = \sum_{i=1}^k m_i$. This identity implies that $d$ is even. We can thus find the equations
\begin{align*}
\left(\frac{d}{2}\right)^2 & = \sum_{i=1}^k \left(\frac{m_i}{2}\right)^2 ,\\
3\left(\frac{d}{2}\right)+1 &= \sum_{i=1}^k \frac{m_i}{2}.
\end{align*}
Adding these identities yields
\begin{align*}
\frac{d}{2}\left(\frac{d}{2}+3\right) + 1 &= \sum_{i=1}^k \frac{m_i}{2}\left(\frac{m_i}{2}+1\right).
\end{align*}
The left-hand side of this equation is odd, whereas the right-hand side is even. This is a contradiction and thus $\delta = 0$ is not possible.
\end{subproof}

\begin{claim*}[3]
If $\delta = 1$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(3)$]
The base-point $p_{k+1}$ is one of the intersection points between $C_k$ and $E_k$. The curve $C_{k+1}$ has then self-intersection $0$ in $X_{k+1}$ and thus the base-point $p_{k+2}$ is the unique intersection point between $C_{k+1}$ and $E_{k+1}$. The configuration of curves in $X_{k+2}$ is shown in Figure~$\ref{fig:even2}$. In the surface $X$, the curve $E_k$ has self-intersection $-2$. This implies that $\eta$ first contracts $C_n$ and then $E_{k},\ldots,E_1$, in this order. By assumption, the multiplicity sequence of $C$ is non-constant. This implies that there exists a curve $E_j$ with $j<k$ that intersects $3$ other exceptional curves. But this implies that the image of $E_{k+1}$, after contracting $C_n,E_k,\ldots,E_1$, is singular and hence cannot be contracted. We thus reach a contradiction and conclude that $\delta \neq 1$.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.75]
\draw (0,0) -- (1.5,2.5);
\draw (1,0) -- (2.5,2.5);
\draw (1,2.5) -- (2.5,0);
\draw[dashed] (2,0) -- (3.5,2.5);
\node at (2.25,0.9) {\scriptsize $E_{j}$};

\draw (3,2.1) -- (6,2.1);
\node at (4.5,2.35) {\scriptsize $E_{k}$};

\draw[very thick] (5.5,2.5) -- (7,0);
\node at (5.75,1.25) {\scriptsize $E_{k+1}$};
\draw (6,-0.83) -- (8,2.5);
\node at (7.8,1.25){\scriptsize $E_{k+2}$};

\draw[very thick] (4,2.3) to [out=-90, in=180] (6.6,-0.3);
\node at (4.8,-0.02) {\scriptsize $C_{k+2}$};

\end{tikzpicture}
\caption{Case $\delta = 1$.}\label{fig:even2}
\end{figure}
\end{subproof}

\begin{claim*}[4]
If $\delta \geq 2$, we reach a contradiction.
\end{claim*}

\begin{subproof}[Proof of Claim~$(4)$]
Again, the base-point $p_{k+1}$ is one of the intersection points between $C_k$ and $E_k$. Since $\pi$ is a $(-1)$-tower resolution of $C$, it follows that for $i=k+1,\ldots,k+\delta$, the base-point $p_{i+1}$ is the unique intersection point between $C_k$ and $E_k$ (see Figure~$\ref{fig:even3}$). This implies that in $X$, the curve $E_{k+1}$ has self-intersection $-2$. We observe that $E_k$ also intersects the $(-2)$-curve $E_{k-1}$ in $X$. Since $\eta$ contracts $E_k$ before $E_{k-1}$ and $E_{k+1}$, this leads to a contradiction.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=0.75]
\draw[dashed] (1,2.5) -- (2.5,0);
\draw (2,0) --(3.5,2.5);
\node at (2.25,1.25){\scriptsize $E_{k-1}$};

\draw (3,2.1) --(6,2.1);
\node at (4,2.35) {\scriptsize $E_{k}$};

\draw (5.5,2.5) --(7,0);
\node at (6.8,1.25){\scriptsize $E_{k+1}$};
\draw[dashed] (6.5,0) --(8,2.5);
\draw[very thick] (7.5,2.5) --(9,0);
\node at (9.05,1.25){\scriptsize $E_{k+\delta+1}$};

\draw[very thick] (4.5,2.3) to [out=-100, in=180] (7,-0.3);
\draw[very thick] (7,-0.3) to [out=0, in=240] (9,1);
\node at (6.55,-0.5) {\scriptsize $C_{k+\delta+1}$};

\end{tikzpicture}
\caption{Case $\delta \geq 2$.}\label{fig:even3}
\end{figure}
\end{subproof}
This concludes the proof of Proposition~$\ref{Prop:alleven}$.
\end{proof}

\begin{corollary}\label{cor:proximatecase}
Let $C \subset \p^2$ be an irreducible curve with one of the multiplicity sequences $(4,2_{(4)})$, $(4_{(3)},2_{(3)})$, or $(6,2_{(6)})$. If $C$ is not unicuspidal, then any open embedding $\p^2 \setminus C \hookrightarrow \p^2$ extends to an automorphism of $\p^2$.
\end{corollary}

\begin{proof}
This is a direct consequence of Proposition~$\ref{Prop:alleven}$.
\end{proof}

\subsection{A special sextic curve and the proof of Theorem~$\ref{Thm:degree8}$}

\begin{proposition}\label{prop:deg6equivalent}
Let $C \subset \p^2$ be a curve of degree $6$ and multiplicity sequence $(3,2_{(7)})$ such that there exists $\varphi \colon \p^2 \setminus C \to \p^2 \setminus D$ an isomorphism, where $D \subset \p^2$ is a curve. Then $C$ and $D$ are projectively equivalent.
\end{proposition}

\begin{proof}
If $\varphi$ extends to an automorphism of $\p^2$ the claim is trivial, so we assume this is not the case. Then by Lemma~$\ref{Lem:tower}$, there exists a $(-1)$-tower resolution $\pi \colon X \to \p^2$ of $C$ and a $(-1)$-tower resolution $\eta \colon X \to \p^2$ of $D$ such that $\eta = \varphi \circ \pi$. The curve $C$ has $8$ singular points $p_1,\ldots,p_8$, where $p_{i+1}$ lies in the first neighborhood of $p_i$ for $i=1,\ldots,7$. The map $\pi$ is a $(-1)$-tower resolution of $C$ and thus blows up the points $p_1,\ldots,p_8$. We denote by $E_i$ the exceptional curve of the blow-up of $p_i$, for $i=1,\ldots,8$. After blowing up these $8$ points, the strict transform $\hat{C}$ of $C$ has self-intersection $6^2 - 3^2 - 7 \cdot 2^2 = -1$. We observe that $\hat{C}$ and $E_8$ intersect with multiplicity $2$. Since no other base-point of $\pi$ lies on $\hat{C}$, it follows that also the strict transforms of $\hat{C}$ and $E_8$ intersect with multiplicity $2$ in $X$. But this means that $E_8$ is not contracted by $\eta$. It follows that $E_8$ is the last exceptional curve of $\pi$ and $\eta(E_8) = D$.

By B\'ezout's theorem the points $p_1, p_2, p_3$ are not collinear and hence there exists a conic $Q_1 \subset \p^2$ that passes through $p_1,\ldots,p_5$. Again by B\'ezout's theorem, it follows that $C$ and $Q_1$ intersect transversely in some proper point of $\p^2$ that is different from $p_1$. It then follows that the strict transform $\hat{Q}_1$ of $Q_1$ in $X$ transversely intersects $E_5$ and $\hat{C}$. By symmetry there also exists a conic $Q_2 \subset \p^2$ whose strict transform $\hat{Q}_2$ by $\eta$ intersects $E_3$ and $\hat{D}$ transversely. The configuration of curves in $X$ is shown below.

\bigskip

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw (0,0)--(1.5,2.5);
\draw (1,2.5) -- (2.5,0);
\draw (2,0) --(3.5,2.5);
\node at (2.45,1.3){\scriptsize $E_3$};
\node at (0.45,1.3){\scriptsize $E_1$};
\node at (1.45,1.3){\scriptsize $E_2$};

\draw (3,2.1) --(6,2.1);
\node at (4.5,2.35) {\scriptsize $E_4$};

\draw (5.5,2.5) --(7,0);
\node at (6.55,1.3){\scriptsize $E_5$};
\draw (6.5,0) --(8,2.5);
\draw (7.5,2.5) --(9,0);
\node at (8.55,1.3){\scriptsize $E_7$};
\node at (7.55,1.3){\scriptsize $E_6$};

\node at (3.05,0.7){\scriptsize $\hat{Q}_2$};
\node at (6,0.7){\scriptsize $\hat{Q}_1$};

\draw[very thick] (2.7,1.7)--(3.9,-0.3);
\draw[very thick] (6.3,1.7) --(5.1,-0.3);

\draw[red] (3.5,0) to [out=90, in=180] (5.5,1.5);
\draw[red] (3.5,0) to [out=-90, in=180] (5.5,-1.5);
\draw[red] (5.5,-1.5) to [out=0, in=-90] (8.8,0.8);

\draw[blue] (5.5,0) to [out=90, in=0] (3.5,1.5);
\draw[blue] (5.5,0) to [out=-90, in=0] (3.5,-1.5);
\draw[blue] (3.5,-1.5) to [out=180, in=-90] (0.2,0.8);

\node at (7.5,-1.35) {\scriptsize $E_8$};
\node at (1.5,-1.35) {\scriptsize $\hat{C}$};

\end{tikzpicture}
\end{center}

\bigskip

To see that $\hat{Q}_1$ and $\hat{Q}_2$ do not intersect in $X$, we observe that $\pi$ sends $\hat{Q}_2$ to a rational quartic curve with multiplicity sequence $(2_{(3)})$ and singular points $p_1,p_2,p_3$. It then follows that
$$\hat{Q}_1\cdot \hat{Q}_2 =  Q_2 \cdot \pi(\hat{Q}_2)  - 2 - 2 - 2 - 1 - 1 = 0.$$
Moreover, the curves $\hat{Q}_1$ and $\hat{Q}_2$ both have self-intersection $-1$ in $X$. We can thus construct a morphism $\rho$ by contracting the curves $\hat{Q}_2, E_3, E_2, E_1$ and $\hat{Q}_1, E_5, E_6, E_7$. The rank of the Picard group of $X$ is $9$, and hence the rank of the Picard group of the image of $\rho$ is $1$. It thus follows that $\rho$ is a morphism $X \to \p^2$. The images of $\hat{C}, E_4$ and $E_8$ all have self-intersection $4$ and are thus smooth conics in $\p^2$. The curves $\rho(E_4)$ and $\rho(\hat{C})$ intersect in two distinct points $p,q \in \p^2$, with multiplicity $1$ in $p$ and multiplicity $3$ in $q$. The curves $\rho(E_4)$ and $\rho(E_8)$ also intersect in $p$ and $q$, but with multiplicity $3$ in $p$ and multiplicity $1$ in $q$. The configuration of the $3$ conics is shown below.

\bigskip

\begin{center}
\begin{tikzpicture}[scale=0.95]

\draw[red] (-2,0) to [out=-90, in=180] (0.3,-1.8);
\draw[red] (-2,0) to [out=90, in=120] (2,0);
\draw[red] (0.3,-1.8) to [out=0, in=-60] (2,0);

\draw[blue] (2,0) to [out=90, in=60] (-2,0);
\draw[blue] (-2,0) to [out=-120, in=180] (-0.3,-1.8);
\draw[blue] (-0.3,-1.8) to [out=0, in=-90] (2,0);

\draw (0,0) ellipse (2 and 1.5);

\fill (-2,0) circle (2pt);
\fill (2,0) circle (2pt);

\node at (0,-1.2){\scriptsize $\rho(E_4)$};
\node at (-2.4,-1){\scriptsize $\rho(\hat{C})$};
\node at (2.4,-1){\scriptsize $\rho(E_8)$};

\node at (-2.2,0){\scriptsize $p$};
\node at (2.2,0){\scriptsize $q$};

\end{tikzpicture}
\end{center}

\bigskip

\noindent Up to a linear change of coordinates, we can assume that the smooth conic $\rho(E_4)$ has equation $xz+y^2 = 0$ and the points $p$ and $q$ are $[1:0:0]$ and $[0:0:1]$ respectively. Conics that pass through the points $[1:0:0]$ and $[0:0:1]$ are of the form
$$ay^2 + bxy + cxz + dyz = 0$$
where $a,b,c,d \in \k$. A smooth conic with this equation intersects $xz+y^2=0$ with multiplicity $3$ in $[1:0:0]$ if and only if $a=c \neq 0$, $b=0$ and $d \neq 0$. Thus there exists some $\lambda \in \k^*$ such that $\rho(\hat{C})$ has equation $xz + y^2 + \lambda yz = 0$. Analogously, there exists $\mu \in \k^*$ such that $\rho(E_8)$ has equation $xz + y^2 + \mu yz = 0$.

We then find $\theta \in \Aut(\p^2)$ that sends a point $[x:y:z]$ to $[\frac{\lambda}{\mu}z:y:\frac{\mu}{\lambda}x]$. Thus $\theta$ preserves the conic $xz+y^2=0$ and exchanges $\rho(\hat{C})$ and $\rho(E_8)$. It follows that $\hat{\theta} \coloneqq \rho^{-1}\circ \theta \circ \rho$ is an automorphism of $X$ that exchanges $\hat{C}$ and $E_8$ and sends $E_i$ to $E_{8-i}$ for $i = 2,\ldots,7$. But then $\eta \circ \hat{\theta} \circ \pi^{-1}$ is an automorphism of $\p^2$ that sends $C$ to $D$, and hence $C$ and $D$ are projectively equivalent.
\end{proof}

Before we are able to prove Theorem~$\ref{Thm:degree8}$, we need to look at one more special case.

\begin{lemma}\label{lemma:522222}
Let $C \subset \p^2$ be a curve of degree $7$ and multiplicity sequence $(5,2_{(5)})$. Then every open embedding $\p^2 \setminus C \hookrightarrow \p^2$ extends to an automorphism of $\p^2$.
\end{lemma}

\begin{proof}
Suppose that there exists an open embedding $\varphi \colon \p^2 \setminus C \hookrightarrow \p^2$ that does not extend to an automorphism of $\p^2$. Then by Lemma~$\ref{Lem:tower}$, there exists $\pi \colon X = X_n \xrightarrow{\pi_n} \ldots \xrightarrow{\pi_2} X_1 \xrightarrow{\pi_1} X_0 = \p^2$ a $(-1)$-tower resolution of $C$ with base-points $p_1,\ldots,p_n$ and exceptional curves $E_1,\ldots,E_n$, and a $(-1)$-tower resolution $\eta \colon X \to \p^2$ of some curve $D \subset \p^2$ such that $\varphi \circ \pi = \eta$. Then $E_1 \cup \ldots \cup E_{n-1} \cup C_n$ is the exceptional locus of $\eta$, being the support of an SNC-divisor that has a tree structure. By Lemma~$\ref{Lem:proximate mult}$, we obtain the intersection number
$$C_n \cdot E_1 = m_i - \sum_{p_j \succ p_1} m_j.$$
Thus either $C_n \cdot E_1 =3$ or $C_n \cdot E_1 =1$. Since $C_n$ can intersect $E_1$ only transversely in at most one point, we conclude that $C_n \cdot E_1 =1$ and that $p_3$ is proximate to $p_1$. For the first $6$ blow-ups of $\pi$, we then obtain the configuration of curves illustrated below.

\begin{center}
\begin{tikzpicture}[scale=0.75]

\draw (2,0) --(3.5,2.5);
\node at (2.4,1.25) {\scriptsize $E_{1}$};
\draw (3,2.1) --(6,2.1);
\node at (4.8,2.35) {\scriptsize $E_3$};
\draw (5.5,2.5) --(7,0);
\node at (6.6,1.25) {\scriptsize $E_2$};

\draw (4.25,2.5) --(5,1.25);
\node at (4.45,1.7) {\scriptsize $E_4$};
\draw (5,1.5) --(4.25,0.25);
\node at (4.9,0.8) {\scriptsize $E_5$};
\draw[very thick] (4.25,0.5) --(5,-0.75);
\node at (4.45,-0.28) {\scriptsize $E_6$};

\draw[very thick] (2,0.45) to [out=-50, in=120] (4.8,-0.2);
\draw[very thick] (4.8,-0.2) to [out=300, in=20] (4.6,-0.8);
\node at (3.5,-0.07) {\scriptsize $C_6$};

\end{tikzpicture}
\end{center}
The curves $E_2$ and $E_4$ have self-intersection $-2$ in $X$ since the resolution $\pi$ is obtained by blowing up more points on $E_6$. Moreover, the map $\eta$ contracts $E_3$ before $E_2$ and $E_4$, but this leads to a contradiction.
\end{proof}

We are now ready to give the proof of the second main result.

\begin{proof}[Proof of Theorem~$\ref{Thm:degree8}$]
We assume that $C$ is not a line, conic, or a nodal cubic. We can also assume that $C$ is rational and has a unique proper singular point with one of the multiplicity sequences in Table~$\ref{table:sequences}$, by Corollary~$\ref{Cor:table}$. Otherwise, $\varphi$ extends to an automorphism of $\p^2$. If $C$ is unicuspidal, then $C$ and $D$ are projectively equivalent by Corollary~$\ref{cor:unicuspidal8}$. If $C$ is not unicuspidal, then $\varphi$ extends to an automorphism of $\p^2$ by Corollary~$\ref{cor:jumpcase}$, Corollary~$\ref{cor:equalcase}$, Corollary~$\ref{cor:1jumpcase}$, Corollary~$\ref{cor:proximatecase}$, and Lemma~$\ref{lemma:522222}$, except when $C$ is of degree $6$ with multiplicity sequence $(3,2_{(7)})$ or $C$ is of degree $8$ with multiplicity sequence $(3_{(7)})$. If $C$ has multiplicity sequence $(3,2_{(7)})$, the claim follows from Proposition~$\ref{prop:deg6equivalent}$. If $C$ has multiplicity sequence $(3_{(7)})$, then $C \setminus \Sing(C)$ is isomorphic to $\A^1 \setminus \{0\}$, by Proposition~$\ref{Prop:nojump}$.
\end{proof}

\begin{remark}
{\rm For all known examples of irreducible curves $C \subset \p^2$ that have non-extendable open embeddings $\p^2 \setminus C \hookrightarrow \p^2$, we have that $C \setminus \Sing(C) \simeq \p^1 \setminus \{p_1,\ldots,p_k\}$, where $k \in \{1,2,3,9\}$. There are only very few known non-unicuspidal examples. Do there exist examples for any $k \in \N$?}
\end{remark}

\subsection{A counterexample of degree 8}\label{sec:deg8}
It follows from Theorem~$\ref{Thm:degree8}$ that if two irreducible curves $C, D \subset \p^2$ of degree $\leq 8$ are counterexamples to Conjecture~$\ref{conjecture:yoshihara}$, then $C$ and $D$ are of degree $8$ and have multiplicity sequence $(3_{(7)})$. In this section, we show that such counterexamples do indeed exist. First we need the following auxiliary construction.

\begin{lemma}\label{lemma:conicconfiguration}
We denote the conic
$$\Lambda \colon xy + xz + yz = 0$$
and for $\lambda \in \k \setminus \{0,-1\}$ the conics
\begin{align*}
\Gamma_\lambda &\colon x^2 - (1+\lambda)xy - \lambda xz - (1+\lambda)yz = 0, \\
\Delta_\lambda &\colon z^2 - \left(1+\frac{1}{\lambda}\right)xy - \frac{1}{\lambda}xz - \left(1+\frac{1}{\lambda}\right)yz = 0.
\end{align*}
Then the curves $\Lambda$, $\Gamma_\lambda$ and $\Delta_\lambda$ intersect in $[0:1:0]$ with multiplicity $3$ for each pair. Moreover, the curves
\begin{itemize}
\item $\Lambda$ and $\Gamma_\lambda$ intersect in $[0:0:1]$,
\item $\Lambda$ and $\Delta_\lambda$ intersect in $[1:0:0]$,
\item $\Gamma_\lambda$ and $\Delta_\lambda$ intersect in $[\lambda:0:1]$,
\end{itemize}
and in no other point apart from $[0:1:0]$. The configuration of these conics is shown below.

\begin{center}
\begin{tikzpicture}[scale=0.9]

\draw[red] (-2,0) to [out=90, in=170] (1,1.3);
\draw[red] (-2,0) to [out=-90, in=170] (1,-2.5);
\draw[red] (1,1.3) [out=-10,in=90] to (4,-1.2);
\draw[red] (4,-1.2) [out=-90, in=-10] to (1,-2.5);

\draw[blue] (-2,0) to [out=90, in=110] (1,-1.3);
\draw[blue] (1,-1.3) to [out=-70, in=45] (1,-2.5);
\draw[blue] (-2,0) to [out=-90,in=-145] (1,-2.5);

\draw[green] (0,0) ellipse (2 and 1.5);

\fill (1,-2.5) circle (2pt);
\fill (1,-1.3) circle (2pt);
\fill (1,1.3) circle (2pt);
\fill (-2,0) circle (2pt);

\node at (1.9,-0.8){\scriptsize $\Lambda$};
\node at (3.65,-2.4){\scriptsize $\Delta_\lambda$};
\node at (-1.1,-2.4){\scriptsize $\Gamma_\lambda$};

\node at (-2.75,0){\scriptsize $[0:1:0]$};
\node at (1.75,1.5){\scriptsize $[1:0:0]$};
\node at (1.75,-1.4){\scriptsize $[0:0:1]$};
\node at (1.75,-2.75){\scriptsize $[\lambda:0:1]$};

\end{tikzpicture}
\end{center}

\noindent Furthermore, there exists an automorphsim of $\p^2$ that preserves $\Lambda$ and exchanges $\Gamma_\lambda$ and $\Delta_\lambda$ if and only if $\lambda = 1$.

\end{lemma}

\begin{proof}
The curves $\Lambda$, $\Gamma_\lambda$ and $\Delta_\lambda$ are given by explicit equations and it is a straightforward computation to determine the intersection points and multiplicities.

To prove the last claim, suppose that $\theta \in \Aut(\p^2) = \PGL_3(\k)$ preserves $\Lambda$ and exchanges $\Gamma_\lambda$ and $\Delta_\lambda$. Then $\theta$ fixes $[0:1:0]$ and exchanges $[1:0:0]$ and $[0:0:1]$. These conditions imply that $\theta$ is of the form $[x:y:z] \mapsto [\alpha z : y : \beta x]$, for some $\alpha, \beta \in \k^*$. The image of $\Lambda$ under $\theta$ then has equation $\beta xy + \alpha \beta xz + \alpha yz = 0$. Since $\Lambda$ is preserved, it follows that $\alpha = \beta = \alpha \beta$ and hence $\alpha = \beta = 1$. The map $\theta$ also fixes the intersection point $[\lambda : 0 : 1]$ between $\Gamma_\lambda$ and $\Delta_\lambda$. Since $\theta([\lambda:0:1]) = [1:0:\lambda]$, it follows that $\lambda = 1$. For the converse, suppose that $\lambda = 1$. Then the automorphism $[x:y:z] \mapsto [z:y:x]$ preserves $\Lambda$ and exchanges $\Gamma_1$ and $\Delta_1$.
\end{proof}

\begin{proof}[Proof of Theorem~$\ref{theorem:counterexample}$]
With the same notations as in Lemma~$\ref{lemma:conicconfiguration}$, we choose some $\lambda \in \k \setminus \{0,\pm 1\}$ and conics $\Lambda$, $\Gamma = \Gamma_\lambda$, $\Delta = \Delta_\lambda$. We denote moreover by $L_y$ the line $y=0$ and by $L_\lambda$ the line through $[0:1:0]$ and $[\lambda : 0 : 1]$. The line $L_\lambda$ has equation $x - \lambda z = 0$ and intersects $\Lambda$ in the points $[0:1:0]$ and $[1+\lambda:-1:1+\frac{1}{\lambda}]$. The configuration of these curves in shown below.

\begin{center}
\begin{tikzpicture}[scale=0.9]

\draw[red] (-2,0) to [out=90, in=170] (1,1.3);
\draw[red] (-2,0) to [out=-90, in=170] (1,-2.5);
\draw[red] (1,1.3) [out=-10,in=90] to (4,-1.2);
\draw[red] (4,-1.2) [out=-90, in=-10] to (1,-2.5);

\draw[blue] (-2,0) to [out=90, in=110] (1,-1.3);
\draw[blue] (1,-1.3) to [out=-70, in=45] (1,-2.5);
\draw[blue] (-2,0) to [out=-90,in=-145] (1,-2.5);

\draw[green] (0,0) ellipse (2 and 1.5);

\draw[orange] (1,1.8)--(1,-3);
\draw[orange] (-2.35,0.3)--(1.4,-2.85);

\fill (1,-2.5) circle (2pt);
\fill (1,-1.3) circle (2pt);
\fill (1,1.3) circle (2pt);
\fill (-2,0) circle (2pt);
\fill (-0.22,-1.48) circle (2pt);

\node at (1.9,-0.8){\scriptsize $\Lambda$};
\node at (3.5,-2.4){\scriptsize $\Delta$};
\node at (-1.1,-2.4){\scriptsize $\Gamma$};
\node at (0.7,0.2){\scriptsize $L_y$};
\node at (-0.6,-0.8){\scriptsize $L_\lambda$};

\node at (-2.75,0){\scriptsize $[0:1:0]$};
\node at (1.75,1.5){\scriptsize $[1:0:0]$};
\node at (1.75,-1.4){\scriptsize $[0:0:1]$};
\node at (1.75,-2.75){\scriptsize $[\lambda:0:1]$};
\node at (-1.8,-1.7){\scriptsize $[1+\lambda:-1:1+\frac{1}{\lambda}]$};

\end{tikzpicture}
\end{center}

We then blow up the points $[1:0:0]$, $[0:0:1]$ and $[\lambda:0:1]$, with exceptional curves $E_1$, $E_2$, and $E_3$ respectively. The configuration after these blow-ups is shown below. By abuse of notation, we use the same names for the strict transforms of all curves. Curves with self-intersection $-1$ are drawn with thick lines and all other self-intersection numbers are indicated, except if they are $-2$.

\begin{center}
\begin{tikzpicture}[scale=0.9]

\draw[orange] (1,2)--(1,-2);
\draw[green] (-0.5,0) to [out=90,in=200] (1,1.5);
\draw[green] (-0.5,0) to [out=-90,in=160] (1,-1.5);
\draw[green] (1,1.5) to [out=20,in=180] (2,1.7);
\draw[green] (1,-1.5) to [out=-20,in=180] (2,-1.7);
\node at (-0.15,0){\scriptsize $\Lambda[2]$};
\node at (1.45,-1){\scriptsize $L_\lambda[0]$};
\node at (6.3,-1.9){\scriptsize $L_y$};

\draw[orange] (3.8,-1.7)--(7,-1.7);

\draw[red] (0.3,0.8) to [out=80,in=200] (1,1.5);
\draw[red] (1,1.5) to [out=20,in=190] (6.5,0);
\node at (4.1,1){\scriptsize $\Delta[2]$};

\draw[blue] (0.5,0.5) to [out=80,in=200] (1,1.5);
\draw[blue] (1,1.5) to [out=20,in=190] (5.5,-0.5);
\node at (2.9,0.5){\scriptsize $\Gamma[2]$};

\draw[very thick] (0.7,0) to [out=0,in=180] (3.5,-2);
\draw[very thick] (3.5,-2) to [out=0,in=-100] (4.9,0.6);

\draw[very thick] (1.6,-1.4) to [out=-80,in=180] (4,-2.3);
\draw[very thick] (4,-2.3) to [out=0,in=-100] (5.3,-0.2);

\draw[very thick] (1.3,-1.3) to [out=-80,in=180] (4.5,-2.6);
\draw[very thick] (4.5,-2.6) to [out=0,in=-100] (6.2,0.4);

\node at (2.75,-1.4){\scriptsize $E_3$};
\node at (2.4,-1.95){\scriptsize $E_2$};
\node at (1.3,-1.95){\scriptsize $E_1$};

\node at (1.2,1.75){\scriptsize $p$};
\node at (1.2,0.15){\scriptsize $q$};

\fill (1,1.5) circle (2pt);
\fill (1,-0.04) circle (2pt);

\end{tikzpicture}
\end{center}

Next, we blow up the intersection point $q$ between $L_\lambda$ and $E_3$, with exceptional curve $E_4$. The curves $\Gamma$, $\Delta$ and $\Lambda$ each intersect with multiplicity $3$ in the point $p$. We then blow up $p$ and two points proximate to $p$ (with exceptional curves $E_5$, $E_6$, $E_7$) so that the strict transforms of $\Gamma$, $\Delta$ and $\Lambda$ are disjoint. We thus obtain the following configuration of curves.

\begin{center}
\begin{tikzpicture}[scale=0.9]

\draw[orange] (-1,2.5) --(0.5,0);
\draw (0,0) -- (1.5,2.5);
\draw (1,2.5) --(2.5,0);
\draw[very thick] (2,0) -- (3.5,2.5);
\draw (4,2.5) --(5.5,0);

\draw[very thick,red] (2.7,1.9)--(6.5,1.9);
\node at (5.2,2.1){\scriptsize $\Delta$};
\draw[very thick,blue] (2.3,1.2)--(6,1.2);
\node at (5.2,1.4){\scriptsize $\Gamma$};

\draw[orange] (4.5,0.4)--(7.3,0.4);
\node at (-0.45,1.25){\scriptsize $L_\lambda$};

\draw[very thick,green] (-1.5,0.3) to [out=80,in=110] (3.5,2);
\node at (0.2,2.55){\scriptsize $\Lambda$};
\node at (6.6,0.15){\scriptsize $L_y$};

\draw[very thick] (-1.5,1) to [out=-80,in=180] (4.4,-1);
\draw[very thick] (4.4,-1) to [out=0,in=-80] (6.3,2.1);

\draw[very thick] (-1.2,1.5) to [out=-80,in=180] (4.5,-0.6);
\draw[very thick] (4.5,-0.6) to [out=0,in=-70] (5.8,1.4);

\draw[very thick] (0.25,0.2) to [out=-20,in=180] (3.8,-0.2);
\draw[very thick] (3.8,-0.2) to [out=0,in=220] (5.6,0.3);

\node at (-0.8,-0.5){\scriptsize $E_1$};
\node at (0.55,-0.55){\scriptsize $E_2$};
\node at (4.75,0.75){\scriptsize $E_3$};
\node at (1.3,0.15){\scriptsize $E_4$};
\node at (0.4,1.25){\scriptsize $E_5$};
\node at (1.45,1.25){\scriptsize $E_6$};
\node at (2.75,0.75){\scriptsize $E_7$};

\node at (3.1,2.3){\scriptsize $r$};

\fill (3.36,2.27) circle (2pt);

\end{tikzpicture}
\end{center}

Finally, we blow up the intersection point $r$ between $\Lambda$ and $E_7$ and two points proximate to $r$, with exceptional curves $E_8$, $E_9$, $E_{10}$, and obtain the configuration shown below. We denote the surface obtained after these blow-ups by $X$ and denote the composition of all 10 blow-ups by $\rho \colon X \to \p^2$. The curves $E_1$, $E_2$, $E_4$, $E_{10}$ are dashed and unlabeled because they will not be used for what follows.

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw (-4,0)--(-2.5,2.5);
\draw (-3,2.5) --(-1.5,0);
\draw[green] (-2,0)--(-0.5,2.5);
\draw[orange] (-1,2.5) --(0.5,0);
\draw (0,0) -- (1.5,2.5);
\draw (1,2.5) --(2.5,0);
\draw (2,0) -- (3.5,2.5);
\draw (4,2.5) --(5.5,0);
\node at (-0.45,1.25){\scriptsize $L_\lambda$};
\node at (-1.5,1.25){\scriptsize $\Lambda$};
\node at (6.6,0.15){\scriptsize $L_y$};

\draw[very thick, red] (2.7,1.9)--(6.5,1.9);
\node at (5.2,2.1){\scriptsize $\Delta$};
\draw[very thick, blue] (2.3,1.2)--(6,1.2);
\node at (5.2,1.4){\scriptsize $\Gamma$};

\draw[orange] (4.5,0.4)--(7.3,0.4);

\draw[dashed] (-3.45,0.2) to [out=150,in=180] (-2.2,3.2);
\draw[dashed] (-2.2,3.2) to [out=0,in=145] (3.6,2.05);

\draw[dashed] (-1.5,1) to [out=-80,in=180] (4.4,-1);
\draw[dashed] (4.4,-1) to [out=0,in=-80] (6.3,2.1);

\draw[dashed] (-1.2,1.5) to [out=-80,in=180] (4.5,-0.6);
\draw[dashed] (4.5,-0.6) to [out=0,in=-70] (5.8,1.4);

\draw[dashed] (0.25,0.2) to [out=-20,in=180] (3.8,-0.2);
\draw[dashed] (3.8,-0.2) to [out=0,in=220] (5.6,0.3);

\node at (4.75,0.75){\scriptsize $E_3$};

\node at (0.4,1.25){\scriptsize $E_5$};
\node at (1.47,1.25){\scriptsize $E_6$};
\node at (3.05,0.75){\scriptsize $E_7[-4]$};
\node at (-2.5,1.25){\scriptsize $E_8$};
\node at (-3.55,1.25){\scriptsize $E_9$};

\end{tikzpicture}
\end{center}

The rank of the Picard group of $X$ is $11$, since this surface is obtained from $\p^2$ by $10$ blow-ups.
We can now find a morphism $\pi \colon X \to \p^2$, by contracting the $10$ curves $\Delta$, $E_3$, $L_y$, $E_7$, $E_6$, $E_5$, $L_\lambda$, $\Lambda$, $E_8$, $E_9$, in this order. The image $C \coloneqq \pi(\Gamma)$ is then a curve of degree $8$ in $\p^2$ with multiplicity sequence $(3_{(7)})$. Likewise, we find a morphism $\eta \colon X \to \p^2$, where we first contract $\Gamma$ instead of $\Delta$. The image $D \coloneqq \eta(\Delta)$ is then also a curve of degree $8$ with multiplicity sequence $(3_{(7)})$. The complements $\p^2 \setminus C$ and $\p^2 \setminus D$ are both isomorphic to the complement of the union of the curves $\Gamma$, $\Delta$, $E_3$, $L_y$, $E_7$, $E_6$, $E_5$, $L_\lambda$, $\Lambda$, $E_8$, $E_9$ in $X$.

Suppose now that $C$ and $D$ are projectively equivalent, i.e. there exists $\theta \in \PGL_3(\k)$ with $\theta(C) = D$. We observe that the base-points of $\pi$ are completely determined by $C$, since $\pi$ is the minimal SNC-resolution of $C$ followed by the blow-up of the unique intersection point between $E_3$ and $E_7$. Likewise, the base-points of $\eta$ are determined by $D$. It follows that $\hat{\theta} \coloneqq \eta^{-1}\circ \theta \circ \pi$ defines an automorphism of $X$ that exchanges $\Gamma$ and $\Delta$ and preserves the other exceptional curves. But then $\hat{\theta}$ induces an automorphism of $\p^2$ (via $\rho$) that exchanges the conics $\Gamma, \Delta \subset \p^2$ and preserves $\Lambda$, $L_y$ and $L_\lambda$. But this is not possible by Lemma~$\ref{lemma:conicconfiguration}$, since we have chosen $\lambda \neq 1$. We thus reach a contradiction and conclude that $C$ and $D$ are not projectively equivalent.
\end{proof}

\begin{remark}
{\rm The construction in the proof of Theorem~$\ref{theorem:counterexample}$ also works if the base-field $\k$ is not algebraically closed, except if $\k$ has only $2$ or $3$ elements. In these cases we cannot choose $\lambda \in \k \setminus \{0,\pm1\} = \varnothing$.}
\end{remark}

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\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\arXiv}[2][]{\href{https://arxiv.org/abs/#2}{arXiv:#1#2}}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
% \MRhref is called by the amsart/book/proc definition of \MR.
\providecommand{\MRhref}[2]{%
\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
\providecommand{\href}[2]{#2}
\begin{thebibliography}{XXXX}
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\end{thebibliography}

\end{document}