\section{Proofs by hand for the cells}\label{sec:hand}

Proposition~\ref{prop:exclusion} is a computation: it excludes $52{,}432$ cells by certified linear programs and $38{,}798$ by Gr\"obner bases. In this section we explain the exclusion of most cells by lemmas whose hypotheses are conditions on the trees and the argmin sets of a cell and whose proofs are by hand. Each lemma either gives a contradiction at every positive mass or gives a polynomial in the masses that vanishes wherever the cell occurs. The polynomial is nonzero by the statement of the lemma or, for Lemmas~\ref{lem:handW3}, \ref{lem:handEQ2} and~\ref{lem:handLIN}, by a check in each cell where the lemma is used. Proposition~\ref{prop:handcover} counts the cells to which the lemmas apply. The list of cells is that of Proposition~\ref{prop:cells}, and a program decides which lemma applies to which cell.

As in Section~\ref{sec:exclusion}, $m\in(\R_{>0})^6$ and $C\subseteq Y_m$ is a nontrivial curve. We fix a listed cell of a representative $D$, with trees $T_z$, $T_w$ and polyhedron $\Pi$. Recall from \S\ref{ss:celldef} that a place lies in the cell if its diagram is $D$ and its trees and its sets of terms of least exponent are those of the cell, and that its heights then lie in $\Pi$. A lemma \emph{excludes} the cell if it shows that no singular place of $C$ lies in it.

\subsection{Notation}\label{ss:handnot}
We write $c$ for one of the colours $z$, $w$ and $c'$ for the other one, $c_k$ for the position $z_k$ or $w_k$, $Z^c_{kl}$ for $Z_{kl}$ or $W_{kl}$, $T_c$ for the tree of colour $c$, and $X^c_{v,u}$ for its node coordinates ($X^z=X$, $X^w=Y$ in Lemma~\ref{lem:tree}). For a body $k$, $A^c_k$ is the argmin set of the $c$-equation $c_k=\sum_{l\ne k}m_lZ^c_{kl}$ of $k$. The \emph{position} of $k$ is \emph{minimal} if $c_k\in A^c_k$, and the \emph{force} of $l$ on $k$ is \emph{minimal} if $m_lZ^c_{kl}\in A^c_k$. If $T$ is a set of bodies and $k\in T$, then $k$ is \emph{balanced in colour $c$ with respect to $T$} if $A^c_k$ consists of the forces of the bodies of $T\setminus k$ on $k$.

For points $p,q\in B$, $\meet_c(p,q)$ is the lowest common ancestor of their leaves in $T_c$. For nodes $u$, $v$ we write $u\preceq v$ if $u$ lies on the path from the root to $v$. Then $h_u\le h_v$, and $h_u<h_v$ on all of $\Pi$ unless $u=v$. If $O$ lies below a node $v$, the \emph{$O$-child} of $v$ is the child below which $O$ lies. We choose it as the child $c_0$ of Lemma~\ref{lem:tree}, so that its node coordinate is $0$. We write $\hat f=\lc f{e(f)}$ for the leading coefficient of a function $f$, and $P_{kl}=\hat p_{kl}$. Lemma~\ref{lem:tree} gives the following. If $v=\meet_c(k,O)$ and $k$ lies below the child $u$ of $v$, then $e(c_k)=h_v$ and $\hat c_k=X^c_{v,u}\ne0$. If moreover $l$ is a body with $\meet_c(k,l)=v$, then
\begin{equation}\label{eq:factO}
\lc{c_k-c_l}{h_v}=\begin{cases}\hat c_k&\text{if $l$ lies below the $O$-child of $v$ or shares the leaf of $O$,}\\ \hat c_k-\hat c_l&\text{otherwise,}\end{cases}
\end{equation}
and in the second case $\hat c_l\ne0$. If $v$ is the root of $T_c$, the coefficient of exponent $h_v$ in $\sum_km_kc_k=0$ is
\begin{equation}\label{eq:rootcm}
\textstyle\sum_uM_uX^c_{v,u}=0,
\end{equation}
the sum over the children $u$ of $v$ other than the $O$-child, where $M_u$ is the sum of the masses below $u$ (proof of Lemma~\ref{lem:constraints}(b)).

\begin{lemma}\label{lem:Ssum}
Let $K$ be a set of bodies whose $c$-equations have the same least exponent, and suppose that every minimal force on a body of $K$ is the force of a body of $K$. Then $\sum m_k\hat c_k=0$, the sum over the bodies $k\in K$ whose position is minimal.
\end{lemma}

\begin{proof}
Let $e$ be the common least exponent. The initial form of the $c$-equation of $k\in K$ is $\delta_k\hat c_k-\sum_lm_l\lc{Z^c_{kl}}e$, where $\delta_k=1$ if the position of $k$ is minimal and $\delta_k=0$ otherwise, and the sum runs over the bodies $l$ whose force on $k$ is minimal. These $l$ lie in $K$. Since $e(Z^c_{lk})=e(Z^c_{kl})=e$, the force of $k$ on $l$ is then minimal as well, and $m_km_l\lc{Z^c_{kl}}e+m_lm_k\lc{Z^c_{lk}}e=0$. So the sum of the initial forms, weighted by the masses $m_k$, is $\sum_k\delta_km_k\hat c_k$.
\end{proof}

\subsection{Lemmas at every positive mass}\label{ss:handall}
The lemmas of this subsection exclude a cell at every positive mass. In the first three, an identity has on $\Pi$ a single term of least exponent, or terms of least exponent whose coefficients add up to a positive multiple of one leading coefficient.

Let $J$ be a nonempty proper subset of the bodies. The terms of the torque $T_J$ (\S\ref{ss:identities}) correspond to the ordered pairs $(p,q)$ of bodies with exactly one of $p$, $q$ in $J$, the \emph{cross pairs}. The term of $(p,q)$ is $\pm m_pm_qp_{pq}^3z_pw_q$, of exponent $(a_p-\frac32\zeta_{pq})+(b_q-\frac32\omega_{pq})$, and it is absent if $z_p$ or $w_q$ vanishes identically, i.e., if $p$ shares the leaf of $O$ in $T_z$ or $q$ does in $T_w$.

\begin{lemma}\label{lem:handT}
Let $(p,q)$ be a cross pair of $J$ whose term is present, and suppose:
\begin{enumerate}[label=\textup{(\Alph*)},leftmargin=*]
\item[\textup{(Z)}] $\meet_z(p,O)\preceq\meet_z(p,q)$, and $\meet_z(k,l)\preceq\meet_z(p,q)$ for every cross pair $(k,l)$;
\item[\textup{(W)}] $\meet_w(q,O)\preceq\meet_w(p,q)$, and $\meet_w(k,l)\preceq\meet_w(p,q)$ for every cross pair $(k,l)$;
\item[\textup{(S)}] no other cross pair $(k,l)$ whose term is present has the same nodes $\meet_z(k,l)$, $\meet_z(k,O)$, $\meet_w(k,l)$ and $\meet_w(l,O)$ as $(p,q)$.
\end{enumerate}
Then the term of $(p,q)$ is the only term of $T_J$ of least exponent on $\Pi$, and Lemma~\ref{lem:single} excludes the cell.
\end{lemma}

\begin{proof}
Let $(k,l)\ne(p,q)$ be a cross pair whose term is present, and let $t=\meet_z(p,q)$ and $t'=\meet_z(k,l)$, so that $t'\preceq t$ by (Z). If $O$ lies below $t'$, then $t'\preceq\meet_z(k,O)$, so $a_k\ge\zeta_{kl}$ and
\[
a_k-\tfrac32\zeta_{kl}\ \ge\ -\tfrac12\zeta_{kl}\ \ge\ -\tfrac12\zeta_{pq}\ \ge\ a_p-\tfrac32\zeta_{pq},
\]
where the last step is $a_p\le\zeta_{pq}$, which holds by (Z). If $O$ does not lie below $t'$, then $\meet_z(k,O)=\meet_z(p,O)$, since $k$ and $p$ both lie below $t'$. So $a_k=a_p$ and again $a_k-\frac32\zeta_{kl}\ge a_p-\frac32\zeta_{pq}$. Each inequality used is of the form $h_u\le h_v$ with $u\preceq v$, which is strict on $\Pi$ unless $u=v$. So equality at a point of $\Pi$ forces $t'=t$ and $\meet_z(k,O)=\meet_z(p,O)$. In the same way (W) gives $b_l-\frac32\omega_{kl}\ge b_q-\frac32\omega_{pq}$, with equality only if $\meet_w(k,l)=\meet_w(p,q)$ and $\meet_w(l,O)=\meet_w(q,O)$. By (S), the term of $(k,l)$ has a larger exponent than the term of $(p,q)$ at every point of $\Pi$.
\end{proof}

For a pair $kl$ we say that $x_{kl}>0$ is \emph{forced} if one of the following holds: (x1) $kl$ is a $zw$-edge of $D$; (x2) $kl$ is a $z$-stroke of $D$ and neither $k$ nor $l$ is $z$-circled, or the same with $w$; (x3) the term $m_km_lp_{kl}$ belongs to the argmin set of $U=E$ and the constant $E$ does not; (x4) the force of $l$ on $k$ is minimal in the $z$-equation of $k$, the position of $k$ is not, and $\meet_z(k,O)=\meet_z(k,l)$, or the same with $w$ in place of $z$ or with $k$ and $l$ exchanged. Each of these implies $x_{kl}>0$ on $\Pi$: (x1) by (O4); (x2) since $3x_{kl}-2=\zeta_{kl}\ge\min(a_k,a_l)>-2$ by (O3); (x3) since $-x_{kl}$ is then the least exponent of $U-E$, which is smaller than $e(E)=0$; (x4) since $-\frac12(\zeta_{kl}+3\omega_{kl})=e(Z_{kl})<a_k=\zeta_{kl}$.

\begin{lemma}\label{lem:handP}
Suppose that for every body $k$ with $z_k,w_k\not\equiv0$ there is a body $l$ such that $O$ lies below $\meet_z(k,l)$ and below $\meet_w(k,l)$ and $x_{kl}>0$ is forced. Then the constant $E$ is the only term of least exponent of the identity $\sum_km_kz_kw_k=E$ on $\Pi$, and Lemma~\ref{lem:single} excludes the cell.
\end{lemma}

\begin{proof}
If $O$ lies below $\meet_z(k,l)$, then $\meet_z(k,l)\preceq\meet_z(k,O)$, so $a_k\ge\zeta_{kl}$. Likewise $b_k\ge\omega_{kl}$. So $e(z_kw_k)=a_k+b_k\ge2x_{kl}>0=e(E)$ on $\Pi$.
\end{proof}

For a set $J$ of bodies, each term of the subset sum $\sum_{k\in J}m_kc_k=\sum_{k\in J,\,l\notin J}m_km_lZ^c_{kl}$ is $m_k$ times a term of the $c$-equation of one body $k\in J$: its position, or the force of a body outside $J$.

\begin{lemma}\label{lem:handC}
Let $J$ be a set of at least two bodies, and suppose:
\begin{enumerate}[label=\textup{(C\arabic*)},leftmargin=*]
\item the graph on $J$ in which $k$ and $l$ are joined if the forces of $l$ on $k$ and of $k$ on $l$ are minimal, or if the positions of $k$ and $l$ are minimal and $\meet_c(k,O)=\meet_c(l,O)$, is connected;
\item the set $X$ of the minimal terms of the $c$-equations of the bodies of $J$ that are positions or forces of bodies outside $J$ is not empty, and it consists of one term, or of the positions of the bodies of a set $I$ that lie below one child of a node $v$ with $\meet_c(k,O)=v$ for all $k\in I$.
\end{enumerate}
Then the cell is excluded.
\end{lemma}

\begin{proof}
Two bodies joined in (C1) have $c$-equations with the same least exponent, since $e(Z^c_{kl})=e(Z^c_{lk})$, and since positions that meet $O$ at the same node have the same exponent. So these least exponents have a common value $\mu$. Each term of the subset sum of $J$ has exponent at least $\mu$, with equality exactly for the elements of $X$. If $X$ has one element, Lemma~\ref{lem:single} applies. Otherwise all $\hat c_k$ with $k\in I$ equal the node coordinate $X^c_{v,u}\ne0$ of the child $u$ of $v$ below which they lie, and the initial form of the subset sum is $(\sum_{k\in I}m_k)X^c_{v,u}\ne0$, contradicting Lemma~\ref{lem:initial}.
\end{proof}

\begin{lemma}\label{lem:handB0}
Let $k\ne l$ be bodies, and suppose:
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=*]
\item $A^c_l$ consists of the position of $l$ and the force of $k$ on $l$, $\meet_c(l,k)=\meet_c(l,O)$, and $k$ lies below the $O$-child of this node or shares the leaf of $O$;
\item $A^{c'}_l$ consists of the position of $l$ and the force of $k$ on $l$, $\meet_{c'}(l,k)=\meet_{c'}(l,O)$, and $k$ lies below a child of this node other than the $O$-child.
\end{enumerate}
Then the cell is excluded.
\end{lemma}

\begin{proof}
Let $P=P_{kl}$. By (i) and \eqref{eq:factO}, the leading coefficient of $c_l-c_k$ is $\hat c_l$, so the initial form of the $c$-equation of $l$ is $\hat c_l(1-m_kP^3)=0$, so $m_kP^3=1$. By (ii) and \eqref{eq:factO}, the leading coefficient of $c'_l-c'_k$ is $\hat c'_l-\hat c'_k$ with $\hat c'_k\ne0$. So the initial form of the $c'$-equation of $l$ is $\hat c'_l-m_kP^3(\hat c'_l-\hat c'_k)=\hat c'_k$, which is not zero.
\end{proof}

\begin{lemma}\label{lem:handK}
Let $k\ne l$ be bodies, and suppose:
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=*]
\item $k$, $l$ and $O$ lie below three distinct children of one node of $T_c$; $A^c_k$ consists of the position of $k$ and the force of $l$ on $k$, and $A^c_l$ of the position of $l$ and the force of $k$ on $l$;
\item $k$, $l$ and $O$ lie below three distinct children of one node of $T_{c'}$; for a labelling $\{p,q\}=\{k,l\}$, $A^{c'}_p$ consists of the position of $p$ and the force of $q$ on $p$, and $A^{c'}_q$ of the position of $q$, the force of $p$ on $q$ and the force of one body $j\notin\{k,l\}$ on $q$.
\end{enumerate}
Then the cell is excluded.
\end{lemma}

\begin{proof}
Let $P=P_{kl}$. In colour $c$, Lemma~\ref{lem:Ssum} for $\{k,l\}$ gives $m_k\hat c_k+m_l\hat c_l=0$. With \eqref{eq:factO}, the initial form of the $c$-equation of $k$ becomes $\hat c_k-m_lP^3(\hat c_k-\hat c_l)=\hat c_k(1-(m_k+m_l)P^3)$. Hence $(m_k+m_l)P^3=1$. In colour $c'$, the initial form of the equation of $p$ is $\hat c'_p(1-m_qP^3)+m_qP^3\hat c'_q=P^3(m_p\hat c'_p+m_q\hat c'_q)$, so $m_p\hat c'_p+m_q\hat c'_q=0$. Then the initial form of the equation of $q$ is $P^3(m_q\hat c'_q+m_p\hat c'_p)-m_j\lc{Z^{c'}_{qj}}{}=-m_j\lc{Z^{c'}_{qj}}{}$, which is not zero.
\end{proof}

\begin{lemma}\label{lem:handL}
Let $T=\{a,b,s\}$ be three bodies, and suppose:
\begin{enumerate}[label=\textup{(L\arabic*)},leftmargin=*]
\item $a$, $b$, $s$ lie below three distinct children of one node $v$ of $T_c$; $O$ lies below $v$ but not below these children; and for each $k\in T$, $A^c_k$ consists of the position of $k$ and the forces of the other two bodies of $T$ on $k$;
\item $a$, $b$, $s$ lie below three distinct children of one node $v'$ of $T_{c'}$; $s$ lies below the $O$-child of $v'$ or shares the leaf of $O$; $A^{c'}_s$ consists of the forces of $a$ and $b$ on $s$; and for $\{k,l\}=\{a,b\}$, $A^{c'}_k$ consists of the position of $k$ and the forces of $s$ and $l$ on $k$.
\end{enumerate}
Then the cell is excluded.
\end{lemma}

\begin{proof}
Put $X_k=\hat c_k$ for $k\in T$ and $Y_k=\hat c'_k$ for $k=a,b$. By \eqref{eq:factO}, the leading coefficients of $c_k-c_l$, $c'_k-c'_s$ and $c'_a-c'_b$ are $X_k-X_l$, $Y_k$ and $Y_a-Y_b$. The three pairs of $T$ have the same $\zeta$ and the same $\omega$. So the forces between the bodies of $T$ have a common exponent in each colour, and it is the least exponent of each of the six equations in (L1) and (L2).

Lemma~\ref{lem:Ssum} in colour $c'$ gives $m_aY_a+m_bY_b=0$, and the initial form of the $c'$-equation of $s$ gives $m_aP_{sa}^3Y_a+m_bP_{sb}^3Y_b=0$. Hence $P_{sa}^3=P_{sb}^3=:Q$. Lemma~\ref{lem:Ssum} in colour $c$ gives $m_aX_a+m_bX_b+m_sX_s=0$. So the initial form of the $c$-equation of $s$ is $X_s-Q(m_a(X_s-X_a)+m_b(X_s-X_b))=X_s(1-QM_T)$, where $M_T=m_a+m_b+m_s$, and $Q=1/M_T$ since $X_s\ne0$. The initial form of the $c'$-equation of $a$ is $Y_a(1-m_sQ)-m_bP_{ab}^3(Y_a-Y_b)=(m_a+m_b)Y_a(1/M_T-P_{ab}^3)$, so $P_{ab}^3=1/M_T$ as well. Hence $P_{kl}^{-2}=\varepsilon_{kl}M_T^{2/3}$ with cube roots of unity $\varepsilon_{kl}$. The distance equations give $X_k-X_l=P_{kl}^{-2}\lc{c'_k-c'_l}{}^{-1}$. We insert this and $Y_b=-m_aY_a/m_b$ into $(X_a-X_b)+(X_b-X_s)+(X_s-X_a)=0$ and multiply by $m_a(m_a+m_b)Y_aM_T^{-2/3}$. This gives
\[
m_am_b\,\varepsilon_{ab}=m_b(m_a+m_b)\,\varepsilon_{bs}+m_a(m_a+m_b)\,\varepsilon_{sa}.
\]
If $\varepsilon_{bs}=\varepsilon_{sa}$, the right side has modulus $(m_a+m_b)^2$. Otherwise its squared modulus is $\beta^2+\gamma^2-\beta\gamma\ge\beta\gamma$, where $\beta$ and $\gamma$ are its two coefficients, so the modulus is at least $(m_a+m_b)\sqrt{m_am_b}$. Both bounds exceed $m_am_b$.
\end{proof}

\subsection{Triangles}\label{ss:handtri}
Let $K$ be a set of bodies that lie below distinct children of one node $v_z$ of $T_z$ and of one node $v_w$ of $T_w$. Then all pairs of $K$ have $\zeta=h_{v_z}$ and $\omega=h_{v_w}$, and so the same value $x$. Let $\xi_t=X_{v_z,u}$ and $\eta_t=Y_{v_w,u'}$ be the node coordinates of the children $u$ and $u'$ below which $t\in K$ lies, and $P_{tt'}=\lc{p_{tt'}}{-x}$. The differences $\xi_t-\xi_{t'}$ and $\eta_t-\eta_{t'}$ are the leading coefficients of $z_t-z_{t'}$ and $w_t-w_{t'}$, and they are not zero for $t\ne t'$. The distance equations give
\begin{equation}\label{eq:triD}
P_{tt'}^2(\xi_t-\xi_{t'})(\eta_t-\eta_{t'})=1.
\end{equation}
In each colour $c$ the forces between bodies of $K$ have a common exponent $e^c_K$. We put
\[
\Phi^z_t=\sum_{t'\in K\setminus t}m_{t'}P_{tt'}^3(\xi_t-\xi_{t'}),\qquad \Phi^w_t=\sum_{t'\in K\setminus t}m_{t'}P_{tt'}^3(\eta_t-\eta_{t'}).
\]
If $t$ is balanced in colour $c$ with respect to $K$, the initial form of its $c$-equation is $-\Phi^c_t$, so $\Phi^c_t=0$. When $K=\{a,b,s\}$ has three bodies, we call it a \emph{triangle}.

\begin{lemma}\label{lem:cliquesum}
Identically in $\xi$, $\eta$ and the symmetric $P$, $\sum_tm_t\Phi^z_t=\sum_tm_t\Phi^w_t=0$ and $\sum_tm_t(\xi_t\Phi^w_t-\eta_t\Phi^z_t)=0$.
\end{lemma}

\begin{proof}
The terms $m_tm_{t'}P_{tt'}^3(\xi_t-\xi_{t'})$ and $m_tm_{t'}P_{tt'}^3(\xi_t(\eta_t-\eta_{t'})-\eta_t(\xi_t-\xi_{t'}))=m_tm_{t'}P_{tt'}^3(\xi_{t'}\eta_t-\xi_t\eta_{t'})$ are antisymmetric in $(t,t')$, and so is the first one with $\eta$ in place of $\xi$.
\end{proof}

\begin{theorem}\label{thm:clique3}
Let $K=\{a,b,s\}$ be a triangle.
\begin{enumerate}[label=\textup{(\roman*)}]
\item If $\Phi^z=\Phi^w=0$, then one of $m_a^{-1/2}$, $m_b^{-1/2}$, $m_s^{-1/2}$ is the sum of the other two.
\item If $\Phi^w=0$ and $\Phi^z_a=\Phi^z_b\ne0$, then $m_s^2(m_a+m_b)^4=m_a^3m_b^3$.
\item If $\Phi^w=0$ and $\Phi^z_a=0$, then $\Phi^z=0$.
\end{enumerate}
The same holds with $z$ and $w$ exchanged.
\end{theorem}

\begin{proof}
Let $L$ be the $3\times3$ matrix with $L_{tt'}=m_{t'}P_{tt'}^3$ for $t\ne t'$ and zero row sums. Then $\Phi^z=-L\xi$, $\Phi^w=-L\eta$, $L\mathbf1=0$, and $L\,\mathrm{diag}(m)^{-1}$ is symmetric with nonzero entries off the diagonal. We write $\xi_{tt'}=\xi_t-\xi_{t'}$, $\eta_{tt'}=\eta_t-\eta_{t'}$ and $n_t=m_t^{-1/2}$.

(i) If $\mathbf1$, $\xi$, $\eta$ were linearly independent, $L$ would vanish. Since $\xi$ and $\eta$ are not constant, $\eta=\lambda\xi+\nu\mathbf1$ with $\lambda\ne0$. Then $(P_{tt'}\xi_{tt'})^2=1/\lambda$, so $P_{tt'}\xi_{tt'}=\sigma_{tt'}\psi$ with $\psi^2=1/\lambda$ and antisymmetric signs $\sigma_{tt'}$, and $\Phi^z_t=\psi^3\sum_{t'\ne t}\sigma_{tt'}m_{t'}\xi_{tt'}^{-2}$. The equations $\Phi^z_a=\Phi^z_b=0$ give $\xi_{as}^2/\xi_{ab}^2=\pm m_s/m_b$ and $\xi_{bs}^2/\xi_{ab}^2=\pm m_s/m_a$. Hence $\xi_{as}=\alpha\xi_{ab}n_b/n_s$ and $\xi_{bs}=\beta\xi_{ab}n_a/n_s$ with $\alpha,\beta\in\{\pm1,\pm i\}$, and $\xi_{as}=\xi_{ab}+\xi_{bs}$ gives $n_s+\beta n_a=\alpha n_b$. Since the $n_t$ are positive, $\alpha$ and $\beta$ are real, and one of $n_a$, $n_b$, $n_s$ is the sum of the other two.

(ii) Since $L\eta=L\mathbf1=0$ and $\eta$ is not constant, $L$ has rank $1$, so $L=uv^{\mathsf T}$. The symmetry of $L\,\mathrm{diag}(m)^{-1}$ gives $v=\lambda\,m\circ u$, so $P_{tt'}^3=\lambda u_tu_{t'}$ with all $u_t\ne0$, $\sum_tm_tu_t=0$ and $\Phi^z=-\lambda\langle m\circ u,\xi\rangle u$. As $\Phi^z_a=\Phi^z_b\ne0$, we get $u_a=u_b$ and $u_s=-u_a(m_a+m_b)/m_s$. So $P_{bs}=\mathrm j^kP_{as}$ for some $k\in\{0,1,2\}$, where $\mathrm j=e^{2\pi i/3}$, and $(P_{ab}/P_{as})^3=-m_s/(m_a+m_b)$. The equation $\Phi^w=0$ means $m_a\eta_{as}+m_b\eta_{bs}=0$, so $\eta_{as}=m_b\nu$, $\eta_{bs}=-m_a\nu$ and $\eta_{ab}=(m_a+m_b)\nu$. The distance equations of $as$ and $bs$ give $\xi_{bs}=-\xi_{as}m_b/(\mathrm j^{2k}m_a)$, so $\xi_{ab}=\xi_{as}(\mathrm j^{2k}m_a+m_b)/(\mathrm j^{2k}m_a)$, and the distance equation of $ab$ divided by that of $as$ gives $(P_{ab}/P_{as})^2=\mathrm j^{2k}m_am_b/((\mathrm j^{2k}m_a+m_b)(m_a+m_b))$. Comparing sixth powers, $m_a^3m_b^3=m_s^2(m_a+m_b)(\mathrm j^{2k}m_a+m_b)^3$. If $k\ne0$, the argument of $\mathrm j^{2k}m_a+m_b$ lies strictly between $0$ and $\pm2\pi/3$, so its cube is not positive. Hence $k=0$.

(iii) As in (ii), $\Phi^z=-\lambda\langle m\circ u,\xi\rangle u$ with $u_a\ne0$, and $\Phi^z_a=0$ forces $\Phi^z=0$.
\end{proof}

The two conditions are those that \cite[(16), (20)]{AK12} find for the isolated triangle and for the kite among the diagrams of four bodies. Here, however, both come from one three-body system. For a triangle $T=\{a,b,s\}$ let
\[
Q_T=m_a^2m_b^2+m_b^2m_s^2+m_s^2m_a^2-2m_am_bm_s(m_a+m_b+m_s)=m_a^2m_b^2m_s^2\textstyle\prod(n_a\pm n_b\pm n_s),
\]
the product over the four choices of signs. At positive masses $Q_T=0$ exactly when one of $n_a$, $n_b$, $n_s$ is the sum of the other two.

\begin{corollary}\label{cor:handE1}
Let $T$ be a triangle. If two bodies of $T$ are balanced in one colour and one body of $T$ is balanced in the other colour, with respect to $T$, then $Q_T(m)=0$ at every place in the cell.
\end{corollary}

\begin{proof}
Let $a$ and $b$ be balanced in colour $c$. Then $\Phi^c_a=\Phi^c_b=0$, and $\Phi^c_s=0$ by Lemma~\ref{lem:cliquesum}. If $t\in T$ is balanced in colour $c'$, then $\Phi^{c'}_t=0$, and Theorem~\ref{thm:clique3}(iii), with $t$ in the role of $a$, gives $\Phi^{c'}=0$. Now Theorem~\ref{thm:clique3}(i) applies.
\end{proof}

Suppose that three bodies of $D$ are pairwise joined by $zw$-edges, that none of them is circled, and that no other stroke meets them. At every singular place with diagram $D$ they lie below distinct children of one node in each tree, since $\zeta=\omega=1$ on their pairs. In their equations the forces between them have exponent $-2$ and all other terms have larger exponents. So each of the three is balanced in both colours, and Corollary~\ref{cor:handE1} applies in every cell of $D$. Seven of the $191$ classes have such a triangle, among them diagrams $7$ and $12$ of \cite{CC23}. Six of them have listed cells, $50{,}399$ in all.

A triangle is \emph{centred} in colour $c$ at $s$ if, at its node $v$ in $T_c$, the body $s$ lies below the $O$-child or shares the leaf of $O$, while $a$ and $b$ lie below two other children $A\ne B$ of $v$.

\begin{lemma}\label{lem:handR}
Let $T=\{a,b,s\}$ be a triangle centred in colour $c$ at $s$, with node $v$ and children $A$, $B$ as above, and suppose that $a$ and $b$ are balanced in colour $c$ with respect to $T$. Suppose further that one of the following holds, and let $M_A$, $M_B$ be the sums of masses that it names:
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=*]
\item $v$ is the root of $T_c$ and has exactly three children; $M_A$ and $M_B$ are the sums of the masses below $A$ and $B$;
\item $v$ is the root of $T_c$; every body below a child of $v$ other than $A$, $B$ and the $O$-child has a minimal position, and these bodies form a disjoint union of sets $K$ as in Lemma~\textup{\ref{lem:Ssum}}; $M_A$ and $M_B$ are as in \textup{(a)};
\item for some set $J\supseteq\{a,b\}$, the argmin set of the subset sum of $J$ in colour $c$ is constant on the cell and consists of the terms $m_ac_a$ and $m_bc_b$; $M_A=m_a$ and $M_B=m_b$.
\end{enumerate}
Then $N(m_a,m_b,m_s,M_A,M_B)=0$ at every place in the cell, where
\[
N=\Res_\kappa\big(M_Am_b\kappa^3-M_Bm_a,\ m_s^2(M_A+M_B)(m_a+m_b\kappa)^3-m_a^3m_b^2M_B\big),
\]
and $N$ is a nonzero polynomial in $m$. If $M_A=m_a$ and $M_B=m_b$, then $m_s^2(m_a+m_b)^4=m_a^3m_b^3$ at every place in the cell.
\end{lemma}

\begin{proof}
We write the proof for $c=w$. The case $c=z$ follows by exchanging the colours. Let $G_{kl}=P_{kl}^3(\eta_k-\eta_l)$. Since $a$ and $b$ are balanced, $\Phi^w_a=\Phi^w_b=0$, i.e., $m_bG_{ab}=m_sG_{sa}$ and $m_aG_{ab}=m_sG_{bs}$. So $(G_{ab},G_{bs},G_{sa})=\lambda(m_s,m_a,m_b)$ with $\lambda\ne0$. As $\eta_s=0$, the node coordinates of $A$ and $B$ are $Y_A=\eta_a=-\lambda m_b/P_{sa}^3$ and $Y_B=\eta_b=\lambda m_a/P_{bs}^3$. We claim that $M_AY_A+M_BY_B=0$. In case (a) this is \eqref{eq:rootcm}. In case (b) it is \eqref{eq:rootcm} minus the relations of Lemma~\ref{lem:Ssum} for the sets $K$, since each body $k$ of these sets has a minimal position and $\hat w_k$ is the coordinate of its child of $v$. In case (c) it is the initial form of the subset sum, as $\hat w_a=Y_A$ and $\hat w_b=Y_B$. Hence $\kappa=P_{bs}/P_{sa}$ satisfies $M_Am_b\kappa^3=M_Bm_a$. The relation $(\eta_a-\eta_b)+(\eta_b-\eta_s)+(\eta_s-\eta_a)=0$ reads $m_s/P_{ab}^3+m_a/P_{bs}^3+m_b/P_{sa}^3=0$, so $m_s(P_{sa}/P_{ab})^3=-m_b(M_A+M_B)/M_B$. By \eqref{eq:triD}, $\xi_k-\xi_l=P_{kl}/G_{kl}$, and the same relation for $\xi$ reads $P_{ab}/m_s+P_{bs}/m_a+P_{sa}/m_b=0$, so $P_{ab}/P_{sa}=-m_s(m_a+m_b\kappa)/(m_am_b)$. Comparing cubes gives $m_s^2(M_A+M_B)(m_a+m_b\kappa)^3=m_a^3m_b^2M_B$. So the two polynomials of the resultant have the common root $\kappa$, and $N=0$.

If the masses below $A$ and $B$ other than $m_a$ and $m_b$ are set to zero, $N$ becomes $(m_am_b)^3\prod_\kappa\big(m_s^2(m_a+m_b)(m_a+m_b\kappa)^3-m_a^3m_b^3\big)$, the product over the cube roots of unity $\kappa$. At $m_s=0$ this is $-m_a^{12}m_b^{12}\ne0$, so $N\ne0$. If $M_A=m_a$ and $M_B=m_b$, then $\kappa^3=1$ and $(m_a+m_b\kappa)^3=m_a^3m_b^3/(m_s^2(m_a+m_b))>0$. For $\kappa\ne1$ the argument of $m_a+m_b\kappa$ lies strictly between $0$ and $\pm2\pi/3$, so $\kappa=1$.
\end{proof}

\begin{lemma}\label{lem:handR6}
Let $T=\{a,b,s\}$ be a triangle whose node $v$ in $T_c$ is the root of $T_c$, with exactly three children: the leaves of $a$ and of $b$, and a child below which all other points lie. Suppose that $s$ is balanced in colour $c$ and $a$ in colour $c'$, with respect to $T$. Then at every place in the cell
\[
m_am_b^5=m_s^2(m_a+m_b)^3(m_a+\omega m_b)
\]
for a cube root of unity $\omega$. The product of the differences of the two sides over the three values of $\omega$ is a nonzero polynomial, and at positive masses $\omega=1$, that is, $m_s^2(m_a+m_b)^4=m_am_b^5$.
\end{lemma}

\begin{proof}
We write the proof for $c=w$. By \eqref{eq:rootcm}, $m_a\eta_a+m_b\eta_b=0$, and $\eta_s=0$. The initial form of the $w$-equation of $s$ gives $m_aP_{sa}^3\eta_a+m_bP_{sb}^3\eta_b=0$, so $P_{sb}=\omega P_{sa}$ with $\omega^3=1$. By \eqref{eq:triD}, $\xi_k-\xi_l=1/(P_{kl}^2(\eta_k-\eta_l))$. With $\eta_a-\eta_b=(m_a+m_b)\eta_a/m_b$, the relation $m_bP_{ab}^3(\xi_a-\xi_b)+m_sP_{as}^3(\xi_a-\xi_s)=0$, which is the initial form of the $z$-equation of $a$, becomes $m_b^2P_{ab}/(m_a+m_b)+m_sP_{as}=0$ after multiplication by $\eta_a$, so $P_{ab}^2/P_{sa}^2=m_s^2(m_a+m_b)^2/m_b^4$. The relation $(\xi_a-\xi_b)+(\xi_b-\xi_s)+(\xi_s-\xi_a)=0$, multiplied by $\eta_aP_{ab}^2$, reads
\[
\frac{m_b}{m_a+m_b}-\omega\,\frac{m_b}{m_a}\,\frac{P_{ab}^2}{P_{sa}^2}-\frac{P_{ab}^2}{P_{sa}^2}=0,
\]
which is the stated relation. At $m_s=0$ the product over $\omega$ is $(m_am_b^5)^3\ne0$. At positive masses the left side of the stated relation is positive, so $m_a+\omega m_b$ is real and $\omega=1$.
\end{proof}

\begin{lemma}\label{lem:handW3}
Let $T=\{a,b,s\}$ be a triangle with node $v$ in $T_c$, and suppose that $O$ lies below $v$ but not below the children of $v$ that contain $a$, $b$ and $s$, and that two bodies of $T$ are balanced in colour $c$ with respect to $T$. Suppose further that there are $e_a,e_b,e_s\in\Z[m]$ and a nonzero $g\in\Z[m]$ such that, at every place in the cell with $g(m)\ne0$, the node coordinates of these three children are proportional to $(e_a(m),e_b(m),e_s(m))\ne0$. Let $u_1=m_s^2(e_a-e_b)$, $u_2=m_a^2(e_b-e_s)$, $u_3=m_b^2(e_s-e_a)$ and
\[
H_W=(u_1u_2+u_2u_3+u_3u_1)^3-27(u_1u_2u_3)^2.
\]
Then $g(m)H_W(m)=0$ at every place in the cell.
\end{lemma}

\begin{proof}
Suppose that $g(m)\ne0$, and write the proof for $c=w$. Then $(\eta_a,\eta_b,\eta_s)=\tau(e_a,e_b,e_s)$ at $m$ with $\tau\ne0$, and the $e_k(m)$ are pairwise distinct. By Lemma~\ref{lem:cliquesum}, $\Phi^w$ vanishes at all three bodies, and as in the proof of Lemma~\ref{lem:handR}, $(G_{ab},G_{bs},G_{sa})=\lambda(m_s,m_a,m_b)$ with $\lambda\ne0$. So $P_{ab}^3=\lambda m_s/(\tau(e_a-e_b))$, and cyclically. By \eqref{eq:triD}, $\xi_k-\xi_l=P_{kl}/G_{kl}$, and $(\xi_a-\xi_b)+(\xi_b-\xi_s)+(\xi_s-\xi_a)=0$ becomes $\alpha+\beta+\gamma=0$ for $\alpha=P_{ab}/m_s$, $\beta=P_{bs}/m_a$ and $\gamma=P_{sa}/m_b$. Here $\alpha^3=\lambda/(\tau u_1)$, $\beta^3=\lambda/(\tau u_2)$ and $\gamma^3=\lambda/(\tau u_3)$. From $\alpha+\beta+\gamma=0$ we get $\alpha^3+\beta^3+\gamma^3=3\alpha\beta\gamma$, and cubing gives $(u_1^{-1}+u_2^{-1}+u_3^{-1})^3=27/(u_1u_2u_3)$, that is, $H_W=0$.
\end{proof}

In the cells where we use Lemma~\ref{lem:handW3}, the coordinates of the children of $v$ other than the $O$-child satisfy the linear relations \eqref{eq:rootcm} (if $v$ is the root) and those of Lemma~\ref{lem:Ssum} at $v$, with coefficients in $\Z[m]$. Over $\Q(m)$ their solutions, restricted to the three children of $T$, form a line. Linear algebra over $\Q(m)$ then gives $e$ and $g$ as in the lemma. The relations of Lemma~\ref{lem:Ssum} at $v$ are those of the following sets. Call a body $k$ a \emph{candidate} if $\meet_c(k,O)=v$ and the position of $k$ is minimal. For a candidate $k$, let $K$ be the set of candidates reached from $k$ by chains of minimal forces between candidates. If every minimal force on a body of $K$ is the force of a body of $K$, Lemma~\ref{lem:Ssum} gives $\sum_{k\in K}m_kX^c_{v,u(k)}=0$, where $k$ lies below the child $u(k)$ of $v$. The counts of Proposition~\ref{prop:handcover} use exactly these relations, and these sets in Lemma~\ref{lem:handTQ}(ii).

\begin{lemma}\label{lem:handEQ2}
Let $T=\{a,b,s\}$ be a triangle with nodes $v$ in $T_c$ and $v'$ in $T_{c'}$. Suppose that $s$ lies below the $O$-children of $v$ and of $v'$ or shares the leaf of $O$, and $a$ and $b$ do not; that $s$ is balanced in colour $c$ with respect to $T$; that $A^c_a$ and $A^{c'}_a$ consist of the position of $a$ and the forces of $s$ and $b$ on $a$; and that there are $M_A,M_B\in\Z[m]$ and a nonzero $g\in\Z[m]$ such that $M_A\hat c_a+M_B\hat c_b=0$ at every place in the cell with $g(m)\ne0$. Then $g(m)\big(m_a^2M_A^4-m_b^2M_B^4\big)=0$ at every place in the cell.
\end{lemma}

\begin{proof}
Suppose that $g(m)\ne0$, and put $Y_k=\hat c_k$ and $X_k=\hat c'_k$ for $k=a,b$. By \eqref{eq:factO}, the leading coefficients of $c_a-c_s$, $c_a-c_b$ and $c_s-c_b$ are $Y_a$, $Y_a-Y_b$ and $-Y_b$, and in colour $c'$ they are the same with $X$ in place of $Y$. The initial forms of the two equations of $a$ give $Y_a(1-m_sP_{sa}^3)=m_bP_{ab}^3(Y_a-Y_b)$ and $X_a(1-m_sP_{sa}^3)=m_bP_{ab}^3(X_a-X_b)$. Since $Y_a\ne Y_b$, the factor $1-m_sP_{sa}^3$ is not zero, and dividing gives $X_bY_a=X_aY_b$. If $M_B(m)=0$, then $M_A(m)Y_a=0$, so $M_A(m)=0$ and the claim holds. Otherwise $Y_b=-rY_a$ and $X_b=-rX_a$ with $r=M_A/M_B$. The initial form of the $c$-equation of $s$ is $m_aP_{sa}^3Y_a+m_bP_{sb}^3Y_b=0$, so $\rho=P_{sb}/P_{sa}$ satisfies $\rho^3=m_a/(m_br)$. The distance equations $P_{sa}^2X_aY_a=P_{sb}^2X_bY_b=1$ give $\rho^2=r^{-2}$. Hence $\rho=\rho^3/\rho^2=m_ar/m_b$, and $\rho^2=r^{-2}$ becomes $m_a^2r^4=m_b^2$.
\end{proof}

At positive masses $m_a^2M_A^4=m_b^2M_B^4$ says $m_aM_A^2=m_bM_B^2$. In the cells where we use Lemma~\ref{lem:handEQ2}, $M_A$, $M_B$ and $g$ come from \eqref{eq:rootcm} and Lemma~\ref{lem:Ssum} at $v$, as after Lemma~\ref{lem:handW3}.

The last lemma of this subsection concerns a body that sees a cluster of bodies as one point.

\begin{lemma}\label{lem:handQ}
Let $s$ be a body and $K$ a set of at least two bodies such that $A^c_s$ consists of the forces of the bodies of $K$ on $s$. Suppose that in each tree there is a node $v$ with $\meet(s,k)=v$ for all $k\in K$, and that the bodies of $K$ lie below one child of $v$. Then $\sum_{k\in K}\varepsilon_km_k=0$ at every place in the cell, for some signs $\varepsilon_k=\pm1$.
\end{lemma}

\begin{proof}
In each colour the leading coefficient of $c_s-c_k$ is the same for all $k\in K$, say $\gamma$ in colour $c$ and $\gamma'$ in colour $c'$, and so are the exponents. By the distance equations, $P_{sk}^2\gamma\gamma'=1$, so $P_{sk}=\varepsilon_kP$ for a common $P$ and signs $\varepsilon_k$. The initial form of the $c$-equation of $s$ is $-P^3\gamma\sum_k\varepsilon_km_k$.
\end{proof}

The product of the sums $\sum_k\varepsilon_km_k$ over all choices of signs is a nonzero polynomial. At positive masses the signs are not all equal. For example, for $K=\{a,b\}$ the lemma says $m_a=m_b$.

\subsection{Cliques}\label{ss:cliques}
A set $K$ of at least three bodies is a \emph{$zw$-clique} if all its pairs are $zw$-edges, and a \emph{$z$-clique} if all its pairs are $z$-strokes and none is a $zw$-edge, and likewise for $w$. For cliques the hypotheses of Theorem~\ref{thm:clique3} can be read off the argmin sets in another way, which also covers four bodies.

\begin{lemma}\label{lem:clique}
\begin{enumerate}[label=\textup{(\alph*)}]
\item On the pairs of a $zw$-clique, $\zeta=\omega=x=1$.
\item On the pairs of a $z$-clique $K$, $\zeta=3x_K-2$ and $\omega=2-x_K$ for a common $x_K\in[0,1)$, and $x_K>0$ if no body of $K$ is $z$-circled.
\end{enumerate}
In both cases the bodies of $K$ lie below distinct children of one internal node of $T_z$, and of one of $T_w$.
\end{lemma}

\begin{proof}
(a) is (O4) and (O1). (b) By (E3), applied to the triangles of $K$, the values of $\zeta$, $\omega$ and $x$ are the same on all pairs of $K$, and (O3), (O4) give $\zeta=3x_K-2$, $\omega=2-x_K$ with $0\le x_K<1$. If $x_K=0$, then $\zeta=-2$ on the pairs, and $\zeta_{kl}\ge\min(a_k,a_l)$ makes one body of each pair $z$-circled. The last statement holds because $\zeta$ and $\omega$ are constant on the pairs of $K$ (Lemma~\ref{lem:tree}).
\end{proof}

So a clique $K$ is a set as in \S\ref{ss:handtri}. Its internal order $e^c_K$ is $-2$ in both colours for a $zw$-clique, and $e^z_K=-2$, $e^w_K=2-4x_K$ for a $z$-clique. We call $K$ \emph{$c$-clean} if $e(Z^c_{tX})\ge e^c_K$ for all $t\in K$ and all bodies $X\notin K$. A body $t\in K$ is \emph{$c$-forced} if $e(Z^c_{tX})=e^c_K$ for some $X\notin K$. We put $f^c_t=\sum_Xm_X\lc{Z^c_{tX}}{e^c_K}$, the sum over these $X$. A $zw$-clique is clean in both colours, a $z$-clique is $z$-clean, and in these cases $t$ is $c$-forced if and only if a $c$-stroke joins $t$ to a body outside $K$.

\begin{lemma}\label{lem:cliquelead}
Let $K$ be a clique that is clean in both colours, with $x_K>0$ if it is a $z$- or $w$-clique. Then $e(c_t)\ge e^c_K$ for $t\in K$, the coefficient $\kappa^c=\lc{c_t}{e^c_K}$ does not depend on $t$, and
\[
\Phi^z_t=\kappa^z-f^z_t,\qquad \Phi^w_t=\kappa^w-f^w_t.
\]
\end{lemma}

\begin{proof}
By cleanliness, every term on the right of the $c$-equation of $t$ has exponent at least $e^c_K$, and the internal terms have exponent exactly $e^c_K$ with the coefficients of $\Phi^c_t$. If $e(c_t)<e^c_K$, the coefficient of exponent $e(c_t)$ would be nonzero on the left and zero on the right. So $e(c_t)\ge e^c_K$, and the coefficient of exponent $e^c_K$ gives $\lc{c_t}{e^c_K}=\Phi^c_t+f^c_t$. The separations inside $K$ have exponents larger than $e^c_K$, since $1>-2$, $3x_K-2>-2$ and $2-x_K>2-4x_K$, so $\lc{c_t}{e^c_K}$ does not depend on $t$.
\end{proof}

\begin{corollary}\label{cor:cliquecell}
Let $K$ be a $zw$-clique, or a $z$- or $w$-clique none of whose bodies is circled in the colour of its strokes, of three or four bodies, in a cell of a diagram. Suppose that for each colour $c$ and each $t\in K$ the terms $Z^c_{tt'}$, $t'\in K\setminus t$, belong to the argmin set of the $c$-equation of $t$. Let $L^c$ be the span of the vectors $e_t$ of the bodies $t\in K$ whose $c$-equation has a term $Z^c_{tX}$ with $X\notin K$ in its argmin set, together with the vector $\mathbf1$ if the position $c_t$ belongs to the argmin set of the $c$-equation of some $t\in K$. If $K$ has three bodies, $\dim L^z,\dim L^w\le2$ and $\dim L^z+\dim L^w\le3$, then at every place in the cell the masses of $K$ satisfy, for some labelling $K=\{a,b,s\}$, the conclusion of Theorem~\textup{\ref{thm:clique3}}(i) or (ii). If $K$ has four bodies, $\dim L^z,\dim L^w\le1$ and a place lies in the cell, then $m_K\in B_E$ \textup{(}Proposition~\textup{\ref{prop:E4}}\textup{)}.
\end{corollary}

\begin{proof}
At a place in the cell the hypotheses say that $K$ is clean in both colours, that the $c$-forced bodies are those of the vectors $e_t$ in $L^c$, and that $\kappa^c=0$ unless $\mathbf1\in L^c$. So $\Phi^c=\kappa^c\mathbf1-f^c\in L^c$, and $\sum_tm_t\Phi^c_t=0$ by Lemma~\ref{lem:cliquesum}. If $\dim L^c\le1$, this forces $\Phi^c=0$. If $K$ has four bodies, $\Phi^z=\Phi^w=0$ and \eqref{eq:triD} say that $(\xi,\eta,P)$ is a solution of $E_4(m_K)$. If $K$ has three bodies and $\dim L^z=2$, then $\Phi^w=0$, and either $L^z$ is spanned by $\mathbf1$ and $e_s$, so that $\Phi^z_a=\Phi^z_b$, or $L^z$ is spanned by $e_b$ and $e_s$, so that $\Phi^z_a=0$. In the first case Theorem~\ref{thm:clique3}(ii) applies if $\Phi^z_a\ne0$, and otherwise $\Phi^z$ is a multiple of $e_s$ and vanishes. In the second case $\Phi^z=0$ by Theorem~\ref{thm:clique3}(iii). So either Theorem~\ref{thm:clique3}(ii) applies or $\Phi^z=\Phi^w=0$. The same holds with the colours exchanged if $\dim L^w=2$, and $\Phi^z=\Phi^w=0$ if $\dim L^z,\dim L^w\le1$. Whenever $\Phi^z=\Phi^w=0$, Theorem~\ref{thm:clique3}(i) applies.
\end{proof}

\begin{proposition}\cert\label{prop:cliquecount}
Of the $91{,}273$ cells of Proposition~\ref{prop:cells}, $86{,}938$ have a clique as in Corollary~\ref{cor:cliquecell}. They include all cells of $18$ classes, among them all $44{,}247$ cells of diagram~$7$ and all $32{,}864$ cells of diagram~$11$ of \cite{CC23}.
\end{proposition}

The $zw$-cliques are read off the diagram, and the other cliques off the argmin sets of each cell. Of the $43$ cells of Proposition~\ref{prop:exclusion}, the $11$ cells of diagram~$64$ have a clique of four bodies as in Corollary~\ref{cor:cliquecell}, which gives the system $E_4$ as in Section~\ref{sec:k4}, and the other $32$ have no clique as in Corollary~\ref{cor:cliquecell}. Of the ten cells whose cores Macaulay2 does not confirm (Section~\ref{sec:verification}), nine have a clique as in Corollary~\ref{cor:cliquecell}, while the one of diagram~$16$ has none.

\subsection{Lemmas with further identities}\label{ss:handsec}
The hypotheses of Lemmas~\ref{lem:handP4} and~\ref{lem:handLIN} are relations among leading coefficients at every place in the cell. In the cells where we use them, the relations are initial forms of base identities and of identities of \S\ref{ss:identities} whose argmin sets are constant on the cell (Section~\ref{sec:exclusion}), certified by Lemma~\ref{lem:farkas}. Lemma~\ref{lem:handTQ} assumes such an argmin set for a torque.

\begin{lemma}\label{lem:handP4}
Let $a$, $b$, $s$ be bodies, and suppose that at every place in the cell the following hold, where $Y_k=\lc{c_k-c_s}{}$ for $k=a,b$, at a common exponent:
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=*]
\item $m_aY_a+m_bY_b=0$;
\item $m_aP_{sa}^3Y_a+m_bP_{sb}^3Y_b=0$;
\item $\lc{c'_s-c'_a}{}=\lc{c'_s-c'_b}{}$, at a common exponent.
\end{enumerate}
Then $m_a^3+m_b^3=0$ at every place in the cell, so the cell is excluded.
\end{lemma}

\begin{proof}
Let $\xi$ be the common value in (iii). By the distance equations, $P_{sk}^2=-1/(\xi Y_k)$ for $k=a,b$, so $P_{sk}^3Y_k=-P_{sk}/\xi$, and (ii) becomes $m_aP_{sa}+m_bP_{sb}=0$. The computation in the proof of Lemma~\ref{lem:binary} now gives $m_a^3+m_b^3=0$.
\end{proof}

In the cells where we use Lemma~\ref{lem:handP4}, (i) is the initial form of the centre of mass or of a subset sum, (ii) is that of the equation of $s$, of a torque or of a subset sum, and (iii) holds because $a$ and $b$ lie below one child of the node $\meet_{c'}(s,a)=\meet_{c'}(s,b)$.

\begin{lemma}\label{lem:handLIN}
Suppose that at every place in the cell, nonzero leading coefficients $y_1,\dots,y_n$ satisfy $r$ relations $\sum_{j=1}^nM_{ij}y_j=0$ with $M_{ij}\in\Z[m]$. Then every $n\times n$ minor of the matrix $(M_{ij})$ vanishes at $m$ if a place lies in the cell.
\end{lemma}

\begin{proof}
The vector $(y_j)$ is a nonzero vector in the kernel of $(M_{ij}(m))$.
\end{proof}

In the cells where we use Lemma~\ref{lem:handLIN}, the relations are initial forms of subset sums or of the centre of mass whose terms of least exponent are positions of one colour, $y_j$ is the common leading coefficient $X^c_{v,u}$ of the positions below one child $u$ of a node $v$, and $M_{ij}$ is the sum of their masses in the $i$-th relation. For example, for two relations with single masses as coefficients, a minor reads $m_am_b=m_cm_d$.

\begin{lemma}\label{lem:handTQ}
Let $l_1$, $l_2$, $k_1$, $k_2$ be bodies, and suppose:
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=*]
\item the root of $T_c$ has exactly three children: the leaves of $l_1$ and of $l_2$, and the $O$-child, below which $k_1$ and $k_2$ lie;
\item $k_1$, $k_2$ and $O$ lie below three distinct children of the node $v'=\meet_{c'}(k_1,k_2)$ of $T_{c'}$, $l_1$ and $l_2$ lie below the $O$-child of $v'$, and $\{k_1,k_2\}$ is a set as in Lemma~\textup{\ref{lem:Ssum}} for the colour $c'$, with both positions minimal;
\item the argmin set of the torque $T_{\{l_1,l_2\}}$ is constant on the cell and consists of the four terms $\pm m_{l_i}m_{k_j}p_{l_ik_j}^3c_{l_i}c'_{k_j}$ with $i,j\in\{1,2\}$.
\end{enumerate}
Then $m_{l_1}=m_{l_2}$ and $m_{k_1}=m_{k_2}$ at every place in the cell.
\end{lemma}

\begin{proof}
Put $Y_i=\hat c_{l_i}$, $X_j=\hat c'_{k_j}$ and $P_{ij}=P_{l_ik_j}$. By \eqref{eq:rootcm} and Lemma~\ref{lem:Ssum}, $m_{l_1}Y_1+m_{l_2}Y_2=0$ and $m_{k_1}X_1+m_{k_2}X_2=0$. By \eqref{eq:factO}, the leading coefficients of $c_{k_j}-c_{l_i}$ and of $c'_{k_j}-c'_{l_i}$ are $-Y_i$ and $X_j$. So the distance equations give $P_{ij}^2=-1/(X_jY_i)$, and the initial form of the torque is $\pm\sum_{i,j}m_{l_i}m_{k_j}P_{ij}^3Y_iX_j=\mp\sum_{i,j}m_{l_i}m_{k_j}P_{ij}$. With $u=-X_1Y_1$ we get $P_{11}^2=1/u$, $P_{12}^2=-m_{k_2}/(m_{k_1}u)$, $P_{21}^2=-m_{l_2}/(m_{l_1}u)$ and $P_{22}^2=m_{k_2}m_{l_2}/(m_{k_1}m_{l_1}u)$. Multiplied by $(m_{k_1}m_{l_1}u)^{1/2}$, the initial form becomes $\varepsilon_{11}A_{11}+i\varepsilon_{12}A_{12}+i\varepsilon_{21}A_{21}+\varepsilon_{22}A_{22}$ with signs $\varepsilon_{ij}$ and $A_{ij}=(m_{l_i}m_{k_j})^{3/2}>0$. Its real and imaginary parts vanish, so $A_{11}=A_{22}$ and $A_{12}=A_{21}$, i.e., $m_{l_1}m_{k_1}=m_{l_2}m_{k_2}$ and $m_{l_1}m_{k_2}=m_{l_2}m_{k_1}$. Hence $m_{l_1}=m_{l_2}$ and $m_{k_1}=m_{k_2}$.
\end{proof}

\subsection{Coverage}\label{ss:handcover}
\begin{proposition}\cert\label{prop:handcover}
Of the $91{,}273$ cells of Proposition~\textup{\ref{prop:cells}}:
\begin{enumerate}[label=\textup{(\alph*)}]
\item $70{,}406$ are excluded at every positive mass by the following lemmas: $70{,}233$ by Lemmas~\textup{\ref{lem:handT}}--\textup{\ref{lem:handL}}, $172$ by Lemmas~\textup{\ref{lem:handP4}} \textup{(}$171$ cells\textup{)} and~\textup{\ref{lem:handLIN}} \textup{(}$1$\textup{)}, and one by Lemma~\textup{\ref{lem:single}}, applied to a torque;
\item $90{,}943$ satisfy the hypotheses of one of Lemmas~\textup{\ref{lem:handT}}--\textup{\ref{lem:handL}}, Corollary~\textup{\ref{cor:handE1}}, Lemmas~\textup{\ref{lem:handR}}\textup{(a), (b)}, \textup{\ref{lem:handR6}}, \textup{\ref{lem:handW3}}, \textup{\ref{lem:handEQ2}} and~\textup{\ref{lem:handQ}}, and Corollary~\textup{\ref{cor:cliquecell}} for a clique of three bodies, whose hypotheses are conditions on the trees and the argmin sets of the cell; $287$ others satisfy those of Lemmas~\textup{\ref{lem:handR}}\textup{(c)} \textup{(}$1$ cell\textup{)}, \textup{\ref{lem:handTQ}} \textup{(}$1$\textup{)}, \textup{\ref{lem:handP4}} \textup{(}$169$\textup{)} or~\textup{\ref{lem:handLIN}} \textup{(}$116$\textup{)}; and the remaining $43$ belong to the diagrams $64$ \textup{(}$10$ cells\textup{)}, $66$ \textup{(}$11$\textup{)}, $88$ \textup{(}$9$\textup{)}, $94$ \textup{(}$9$\textup{)}, $93$ \textup{(}$2$\textup{)}, $89$ \textup{(}$1$\textup{)} and $98$ \textup{(}$1$\textup{)}.
\end{enumerate}
In each cell of the first two groups of~\textup{(b)} the lemma gives a contradiction or a nonzero polynomial in the masses that vanishes wherever the cell occurs. So every cell is excluded at all masses outside a proper algebraic subset, by this section or, for the last $43$, by Sections~\textup{\ref{sec:ak50}} and~\textup{\ref{sec:k4}}.
\end{proposition}

\begin{proof}
A program checks the hypotheses of the lemmas on each listed cell, reading its trees and argmin sets, and Proposition~\ref{prop:cliquecount} gives the cliques (Table~\ref{tab:verif4}). Table~\ref{tab:handuse} lists the results that apply most often. Without Corollary~\ref{cor:cliquecell}, this program and a second one, written independently, both find the split $90{,}919$, $311$ and $43$. Corollary~\ref{cor:cliquecell} adds $24$ cells, each with a clique of three bodies and covered by no other lemma of the first group of~(b): $22$ cells of diagram~$48$ and one cell each of diagrams $31$ and $95$. The argmin sets of the subset sums and torques in Lemmas~\ref{lem:handR}(c) and~\ref{lem:handTQ} and in the torque of~(a) are certified by Lemma~\ref{lem:farkas}, and so are those of the identities whose initial forms give the hypotheses of Lemmas~\ref{lem:handP4} and~\ref{lem:handLIN}. Lemmas~\ref{lem:handT}--\ref{lem:handL} and~\ref{lem:handP4} give a contradiction at every positive mass. The polynomials of Corollaries~\ref{cor:handE1} and~\ref{cor:cliquecell} and Lemmas~\ref{lem:handR}, \ref{lem:handR6}, \ref{lem:handQ} and~\ref{lem:handTQ} are nonzero by their statements. Those of Lemmas~\ref{lem:handW3} and~\ref{lem:handEQ2} depend on the cell. In each of the $74$ cells in which one of these two lemmas is the only one of the first group of~(b) that applies, $16$ for Lemma~\ref{lem:handW3} and $58$ for Lemma~\ref{lem:handEQ2}, the polynomial $H_W$, resp.\ $m_a^2M_A^4-m_b^2M_B^4$, is not zero at $m=(1,2,3,4,5,6)$. The minors of Lemma~\ref{lem:handLIN} are nonzero polynomials, and in the cell of~(a) the minor has positive coefficients. The $43$ cells of the last group are excluded by Theorems~\ref{thm:d89}, \ref{thm:d98}, \ref{thm:d93}, \ref{thm:d88}, \ref{thm:d64} and~\ref{thm:d66}. They include the $10$ cells of diagram~$64$ in which Corollary~\ref{cor:cliquecell} applies with a clique of four bodies, since its conclusion there rests on Proposition~\ref{prop:E4}.
\end{proof}

Unlike Corollary~\ref{cor:cliquecell}, Corollary~\ref{cor:handE1} needs no strokes: one node in each tree and three balance conditions suffice. The program applies Lemmas~\ref{lem:handP4} and~\ref{lem:handLIN} only to cells in which no lemma of the first group of~(b) applies, but these two lemmas also exclude further cells at every positive mass, so the count $70{,}406$ of~(a) is a lower bound. Lemma~\ref{lem:single} excludes at every positive mass the cells outside~(a) with an identity that has a single term of least exponent on $\Pi$ (Proposition~\ref{prop:exclusion}), and Theorem~\ref{thm:d89} and Proposition~\ref{prop:d82} exclude at every positive mass the cells of diagram~$89$ and further cells of diagram~$82$, none of them in~(a) or among the cells excluded by Lemma~\ref{lem:single}. Table~\ref{tab:handuse} counts these cells: Sections~\ref{sec:exclusion}, \ref{sec:ak50} and~\ref{sec:hand} exclude at least $71{,}235$ cells at every positive mass.

Proposition~\ref{prop:handcover} rests on four computations: the list of cells with their argmin sets (Proposition~\ref{prop:cells}), the programs that check the hypotheses of the lemmas on each cell (Table~\ref{tab:verif4}), the certified argmin sets of further identities for the $287$ cells and, in~(a), for three cells of diagram~$82$, and the computations of Sections~\ref{sec:ak50} and~\ref{sec:k4} for the $43$ cells (Tables~\ref{tab:verif2} and~\ref{tab:verif3}). These $43$ cells, however, differ from those of Proposition~\ref{prop:exclusion} in two cells: Lemma~\ref{lem:handC} excludes one cell of diagram~$64$ at every positive mass, and no lemma here was found to apply to one cell of diagram~$93$, which Proposition~\ref{prop:exclusion} excludes by a core.