Proof. Note that, for each $i\in [s]$ , $V_i$ has the same distribution as a uniformly chosen $n_i$ -element subset of $V(G)$ . For each vertex $v\in V(G)$ and index $i\in [s]$ , Lemma2.2 gives

\begin{equation*}\mathbb P\Big (d(v,V_i)=(1\pm \gamma )d(v)\cdot \frac {n_i}{n}\Big )\ge 1-n^{-2}.\end{equation*}

Therefore, by a union-bound over all vertices $v\in V(G)$ and indices $i\in [s]$ , we have

\begin{equation*}\mathbb P\left (\exists v\in V(G)\,\text{and } i\in [s] \text{ s.t. }\big |d(v,V_i)-d(v)\cdot \frac {n_i}{n}\big |\gt \gamma d(v)\cdot \frac {n_i}{n}\right )\le s\cdot n\cdot n^{-2}\le \frac {\log ^4 n}{n}.\end{equation*}

Proof. Let $S\subseteq V(G)$ be a vertex subset of $G$ with $|S| \leq (k-1)\lambda n / d$ . By Lemma2.4 (iii), the following inequality holds.

(2.1) \begin{equation} kD\lambda |S| \leq e(S, \Gamma _G(S)\cap X) \leq \frac {d}{n}|S||\Gamma (S)\cap X| + \lambda \sqrt {|S||\Gamma (S)\cap X|}. \end{equation}

Assume $|\Gamma (S)\cap X| \lt D|S|$ . Then (2.1) implies that we have

\begin{equation*} kD\lambda |S| \lt \frac {d D}{n}|S|^2 + \lambda \sqrt {D} |S|. \end{equation*}

Thus, we deduce that $|S| \gt (k-1)\lambda n/ d$ , a contradiction. Hence, we have $|N(S) \cap X| \geq |\Gamma (S)\cap X| - |S| \geq (D-1)|S|$ . This completes the proof.

Proof of Theorem 1.1. We introduce auxiliary constants satisfying

\begin{equation*}0 \lt 1/C\ll 1/d\ll \gamma \ll \varepsilon \ll c, p\ll 1\end{equation*}

so that $C\ge 100C_{3.2}/p$ . Let $V(G)=X_1\cup X_2\cup X_3$ be a random partition so that for each $v\in V(G)$ , $\mathbb P(v\in X_2)=\mathbb P(v\in X_3)=p$ and $\mathbb P(v\in X_1)=1-2p$ , and all choices are made independently. Then, Lemma2.1 implies that there exist pairwise disjoint sets $X_1,X_2,X_3$ such that $|X_1|=(1-2p\pm \gamma )n$ , $|X_2|,|X_3|=(p\pm \gamma )n$ and for each $v\in V(G)$ ,

\begin{align*}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (\! \star \!) \,\,\, d(v,X_1)=(1-2p \pm \gamma )d, d(v,X_2)=(p\pm \gamma )d, \textrm{and} \ d(v,X_3)=(p\pm \gamma )d \\[-20pt] \end{align*}

Our first step is to find the branch vertices of the $K_t$ -subdivision.

Proof. Let $\{S_1,\ldots , S_\ell \}$ be a maximal collection of vertex-disjoint stars with $t-1$ leaves in $G[X_1]$ . Suppose $\ell \lt t$ and let $Y = X_1\setminus \cup _{i\in [\ell ]}V(S_i)$ . Using that $|Y|\ge |X_1|-t^2\ge (1-4p)n$ , the expander mixing lemma (Lemma2.4 (iii)) gives

\begin{equation*}e(G[Y])\ge \frac {1}{2} \left (\frac {d}{n}|Y|^2-\lambda |Y|\right )\ge \frac {|Y|}{2}\cdot \left ( (1-4p)d-\lambda \right ) \ge \frac {|Y|}{2}\cdot (1-5p)d,\end{equation*}

which implies that there is a vertex in $Y$ with at least $(1-5p)d\ge t$ neighbours in $Y$ , contradicting the maximality of $S_1,\ldots , S_\ell$ . The factor of $1/2$ in the first inequality comes from the observation that $e(Y,Y)=2 \cdot e(G[Y]).$

For each $i\in [t]$ , let $s_i\in V(S_i)$ be the centre of $S_i$ which we identify as the $i$ -th branch vertex of the $K_t$ -subdivision and let $L_i = V(S_i) \setminus \{s_i\}$ be the leaves of $S_i$ . Set $Z=(X_1\cup X_2)\setminus \{s_1,\ldots , s_t\}$ and let $L\,:\!=\, I \left ( \cup _{i\in [t]}L_i\right ) \subset G[Z]$ be the empty graph. We now claim that $L$ is $(10,\frac {\lambda n}{d})$ -extendable in $G[Z]$ . To prove this, we will show that each set $U\subset Z$ with $1\le |U|\le 2 \lambda n/d$ satisfies $|N(U)\cap (X_1\cup X_2)\setminus \cup _{i\in [t]}V(S_i)|\ge 10|U|$ , from which the $(10, \lambda n/d)$ -extendability is a consequence of Proposition2.8. Note that (⋆) implies that every vertex $v\in V(G)$ satisfies

\begin{equation*}d_G(v,X_2)\ge (p-\gamma )d\ge \frac {pd}{2}\ge 40\lambda \end{equation*}

and thus, for any subset $U\subset Z$ of size $1\le |U|\le 2\lambda n/d$ , Lemma2.5 implies

\begin{equation*}|(N(U)\cap Z)\setminus V(L)|\ge |N(U)\cap X_2| \ge 10|U|.\end{equation*}

Our goal now is to find $\binom {t}{2}-1$ vertex-disjoint paths joining distinct branch vertices so that the size of the unused vertices in $X_1\cup X_2$ is at most $200\varepsilon n$ . Let $\mathcal Q$ be the set of pairs $(i,j)$ with $1\le i\lt j\le t$ such that $(i,j)\not =(1,t)$ and find a collection of positive integers $(\ell _{e})_{e\in \mathcal Q}$ such that

1. (i) $\ell _{e}\ge 100\log n$ for each $e\in \mathcal Q$ , and

2. (ii) $(1-p)n-200\varepsilon n \leq \sum _{e\in \mathcal Q}\ell _e\le (1-p)n-100\varepsilon n.$

This is clearly possible as $t^2\le c^2n/\log n$ and $c$ is small enough.

Proof. We will construct a collection of paths $(P'_e)_{e\in \mathcal Q}$ connecting distinct vertices in $L$ and using only vertices from $Z\setminus V(L)$ in the process. Say a vertex $v\in V(L)$ is saturated if it has already been used in one of the constructed paths. For each $e=(i,j)\in \mathcal Q$ in turn, find a path $P'_{e}$ of length $\ell _e-2$ in $G[Z]$ so that $P'_e$ is an $(L_i,L_j)$ -path connecting two unsaturated vertices and such that $L$ together with all the paths we have constructed so far is a $(10,\frac {\lambda n}{d})$ -extendable subgraph of $G[Z]$ . If this algorithm ends with $\binom {t}{2}-1$ paths $(P'_e)_{e\in \mathcal Q}$ as described above, we can extend each of these paths to their respective branch vertex to find the desired collection of paths. So suppose the algorithm stops with a collection of paths $P'_{e_1},\ldots , P'_{e_r}$ so that $r\lt \binom {t}{2}-1$ and $L+P'_{e_1}+\ldots +P'_{e_r}$ is $(10,\frac {\lambda n}{d})$ -extendable. Let $e=(i,j)\in \mathcal Q\setminus \{e_1,\ldots , e_r\}$ . Note that both $L_i$ and $L_j$ contain at least one unsaturated vertex, say $v_i\in L_i$ and $v_j\in L_j$ , as the path $P'_e$ is missing. As $\ell _e-2\ge 100\log n-2\ge 10\log _9(\frac {2\lambda n}{d})$ and

\begin{equation*}|L+P'_{e_1}+\ldots +P'_{e_r}|+(\ell _e-2)\le \sum _{e\in \mathcal Q}\ell _e\le (1-p)n-100\varepsilon n\le |Z|-100\cdot \frac {\lambda n}{d},\end{equation*}

we can use Lemma2.9 to find a $(v_i,v_j)$ -path of length $\ell _e-2$ in $G[Z]$ whose internal vertices avoid $V(L+P'_{e_1}+\ldots +P'_{e_r})$ and such that $(L+P'_{e_1}+\ldots +P'_{e_r})+P'_e$ is $(10,\frac {\lambda n}{d})$ -extendable in $G[Z]$ . This proves the algorithm ends with all the required paths, finishing the proof.

Note that up to this point, we have found all but one of the paths in the $K_t$ -subdivision. Thus, it only remains to find a $(s_1,s_t)$ path using all the remaining vertices in the graph at the same time. Let $Z'=(Z\cup X_3)\setminus \cup _{e\in \mathcal Q}V(P'_e)$ and let $u_1\in L_1\cap Z'$ and $u_t\in L_t\cap Z'$ be the remaining leaves after $\cup _{e\in \mathcal Q}P_e$ is constructed. We know $d/\lambda \geq C\geq 100C_{3.2}/p$ , so suppose $d/\lambda =100C'/p$ for some $C' \geq C_{3.2}$ .

Finally, use Theorem3.2 to find a Hamilton path in $G[Z']$ with endpoints $v_i$ and $v_j$ to complete the spanning $K_t$ -subdivision.

Proof of Theorem 1.2. We start by choosing constants

\begin{equation*}1/n_0\ll 1/C\ll \varepsilon \ll c\ll 1/K_{3.3}\ll 1/C_{3.3}\ll \alpha , 1/D\ll 1.\end{equation*}

Suppose that $n\ge n_0$ and let $G$ be an $(n,d,\lambda )$ -graph with $d/\lambda \ge C\log ^3 n$ . Let $t\le \min \{cd,c\sqrt {\frac {n}{\log ^3n}}\}$ be fixed and let $\ell =\lfloor C_{3.3}\log ^3 n\rfloor$ . Set $m=2c^2n/\log ^3n$ , $m'=m-\binom {t}{2}$ and $k=\lfloor (\log ^3n)/(20c)\rfloor$ $-1$ , and pick disjoint random subsets $V_0,V_1,\ldots , V_{k+3}\subset V(G)$ subject to $|V_{0}|=\alpha n$ , $|V_1|=\ldots =|V_k|=\varepsilon m$ , and $|V_{k+1}|=|V_{k+2}|=|V_{k+3}|=m$ . Then, by Lemma2.2, we have w.h.p. that

• • for every $v\in V(G)$ and $i\in [k+3]_0$ , $d(v,V_i)=(1\pm 0.1)d\cdot \frac {|V_i|}{n}$ .

Use Lemma3.3 to find a $(D,m)$ -extendable subgraph $S_{res}\subset G-\cup _{i\in [k+1]_0}V_i$ of size $|S_{res}|=$ $(\ell +1)\cdot m$ and such that $V_{k+2},V_{k+3}\subset V(S_{res})$ and

1. B for every bijection $\sigma \,:\,V_{k+2}\to V_{k+3}$ there is a $P_\ell$ -factor in $S_{res}$ such that each copy of $P_\ell$ has as its endpoints some $v\in V_{k+2}$ and $\sigma (v)\in V_{k+3}$ .

Let $W=V(S_{res})\setminus (V_{k+2}\cup V_{k+3})$ . The next claim finds a collection of vertex-disjoint stars, each of size $t$ , and we omit its proof as it is identical to the proof of Claim1 in Theorem1.1.

For each $i\in [t]$ , let $s_i$ be the centre of $S_i$ and let $L_i=\{u_{i,j}\,:\,j\in [t]\setminus \{i\}\}$ be the set of leaves of $S_i$ . We will use $s_1,\ldots , s_t$ as the branch vertices of the $K_t$ -subdivision. Let $V'=\cup _{i\in [k]}V_i\cup W\cup \{s_1,\ldots , s_t\}$ , $Z=\cup _{i\in [t]}L_i \cup V_{k+1}\cup V_{k+2}\cup V_{k+3}$ , and set $G'=G-V'$ .

We now briefly describe how we will connect the branch vertices $s_i$ and $s_j$ . Clearly, it suffices to do so for $u_{i,j} \in L_i$ and $u_{j,i}\in L_j$ . We only give an overview at this point and postpone exact definitions until later.

The numbers above the arrows indicate the lengths of the corresponding path segments. The sets $V'_i$ , $i \in \left [ k \right ]$ , will satisfy $V_i \subset V'_i$ and $|V'_i|=m$ , and the path lengths $q_{i,j}$ , $q_{j,i} \in \mathbb{N}$ will be appropriately chosen as to ensure that the final spanning subdivision is nearly balanced. The backward arrow indicates that we may visit some vertex parts multiple times (if $|J_{\{i,j\}}|$ is non-zero).

We let $\mathbb{M}=100\log n +\ell + k+1$ , and note that this is the length of a single loop segment and that $\mathbb{M} \leq \log ^3n/(10c)$ . Furthermore, observe that if the above sketch is true, then the length of the branching path from $u_{i,j}$ to $u_{j,i}$ will be equal to $q_{i,j}+q_{j,i}+ |J_{\{i,j\}}| \cdot \mathbb{M}+\ell +k +1$ .

We now connect the vertices from $\cup _{i\in [t]}L_i$ with $V_{k+1}$ and $V_{k+3}$ . To do so, we first pick subsets $A\subset V_{k+1}$ and $B\subset V_{k+3}$ of size $|A|=|B|=m'\,:\!=\,m-\binom {t}{2}$ . Letting $a_1,\ldots , a_{m'}$ and $b_1,\ldots , b_{m'}$ be a labelling of $A$ and $B$ , respectively, use Lemma2.9 to find a collection of paths $P_1,\ldots , P_{m'}$ such that, for each $i\in [m']$ ,

• • $P_i$ is an $(a_i,b_i)$ -path of length $100\log n$ ,

• • $\text{int}(P_i)$ is disjoint from $I(Z)+P_1+\ldots +P_{i-1}$ , and

• • $I(Z)+P_1+\ldots +P_{i}$ is $(D,m)$ -extendable.

This is clearly possible, as $m'\le m\le 2c^2n/\log ^3n$ implies

(3.1) \begin{equation} |I(Z)+P_1+\ldots +P_{m'}|\le t(t-1)+3m+m'\cdot 100\log n\le 10c^2n. \end{equation}

Let $[m']=\cup _{\{i, j\}\in \binom {[t]}{2}} J_{\{i,j\}}$ be a partition so that $||J_{\{i,j\}}|-|J_{\{i',j'\}}||\le 1$ for all $\{i,j\}, \{i',j'\}\in \binom {[t]}{2}$ . Pick a collection of positive integers $(q_{i,j})_{i\neq j\in [t]}$ such that for all $i\neq j\in [t]$ and $a\neq b \in [t]$ ,

1. C1 $100\log ^3 n\le q_{i,j},q_{j,i}\le 0.99n$ ,

2. C2 $\Big |\left (q_{i,j} +q_{j,i}+|J_{\{i,j\}}|\cdot \mathbb{M}\right )-\left (q_{a,b}+q_{b,a}+|J_{\{a, b\}}| \cdot \mathbb{M}\right )\Big |\le 1$ , and

3. C3 $n-|V'|-|Z|-m'(100\log n-1)- \sum _{1\leq i\lt j \leq t}\left ((q_{i,j}-1)+(q_{j,i}-1)\right )=(1-\varepsilon )km$ .

This is possible as the number of leftover vertices after the $m'$ paths of length $100\log n$ are embedded is more than $n-|V'|-|Z|-100m'\log n\ge 0.8n$ and we want to find a collection of $t(t-1)\le c^2n/\log ^3 n$ numbers, each of them greater than $100\log ^3 n$ . Moreover, note that only the sum $(q_{i,j}+q_{j,i})$ matters, as one could increase either at the cost of decreasing the other and the final branching path’s length will remain the same. Also, for each $i\neq j\in [t]$ , $|J_{\{i,j\}}|=x\,\, \textrm {or} \,\,x+1$ for some number $x$ and this follows from construction. Hence, we must have that, for each $i\neq j \in [t]$ , $(q_{i,j}+q_{j,i})=y\pm 1 \,\,\textrm {or}\,\,y+\mathbb{M} \pm 1$ for some number $y$ , in order to satisfy C2. From then on, we may ‘increase’ $y$ so that condition C3 is also satisfied.

Set $\mathcal Q=\emptyset$ and $F=I(Z)+P_1+\ldots +P_{m'}$ , and iteratively, while $\mathcal Q\not =\binom {[t]}{2}$ , do the following. Choose a pair $\{i, j\} \in \binom {[t]}{2}\setminus \mathcal Q$ and pick some unused vertices $u'_{i,j}\in V_{k+1}\setminus A$ and $u'_{j,i}\in V_{k+3}\setminus B$ , and, if possible, perform the following step.

1. D _ij Find vertex-disjoint paths $Q_{j,i}$ and $Q_{i,j}$ such that $Q_{j,i}$ is a $(u_{j,i},u'_{j,i})$ -path of length $q_{j,i}$ , $Q_{i,j}$ is a $(u_{i,j},u'_{i,j})$ -path of length $q_{i,j}$ , $Q_{i,j}$ and $Q_{j,i}$ are internally disjoint from $F$ , and $F+Q_{i,j}+ Q_{j,i}$ is $(D,\frac {\lambda n}{d})$ -extendable in $G'$ . Then we update the set $Q$ and the graph $F$ as $\mathcal Q\,:\!=\,\mathcal Q\cup \{ij\}$ and $F\,:\!=\,F+ Q_{i,j}+ Q_{j,i}$ .

We now show that it will be always possible to perform step D _ij . Indeed, by C3, we know that there are always at least $(1-\varepsilon )km\geq cn/20$ many unused vertices in $G'$ . Thus, by Lemma2.9 we may find the desired paths $Q_{i,j}$ and $Q_{j,i}$ to complete step D _ij .

Note that to finish the embedding of the $K_t$ -subdivision, we only need to connect, for each $\{i,j\}\in \binom {[t]}{2}$ , the vertices $u'_{ij}\in V_{k+1}$ and $u'_{ji}\in V_{k+3}$ . Let $U$ denote the set of vertices that we have not yet used in $G'$ . By C3, $|U|=(1-\varepsilon )km$ and so we may distribute these leftover vertices to the sets $V_1,V_2,\ldots ,V_k$ , to form pairwise disjoint sets $V_1',V_2',\ldots ,V_k'$ satisfying

• • $|V'_1|=\ldots =|V'_k|=m$ ,

• • $V_i\subset V'_i$ for each $i\in [k]$ ,

• • $\left (V'_1\cup \ldots \cup V_k'\right )\setminus \left (V_1,\ldots ,V_k\right )=V(G')\setminus V(F)$ , and

• • for every vertex $v\in V(G)$ , $d(v,V'_i)\ge 0.9d\cdot \frac {|V_i|}{n}\ge \frac {\varepsilon c^2d}{2\log ^3 n}$ .

Use Lemma2.6, iteratively, to find a collection of matchings $M_0, M_1,\ldots , M_{k+1}$ such that

• • $M_0$ is a perfect matching in $G[V_{k+1},V'_1]$ ,

• • for $i\in [k-1]$ , $M_i$ is a perfect matching in $G[V'_i,V'_{i+1}]$ , and

• • $M_{k+1}$ is a perfect matching in $G[V'_k,V_{k+2}]$ .

The union of these $k+2$ matchings gives a family of $m$ vertex-disjoint paths of length $k+1$ so that each such path has as endpoints some vertex from $V_{k+1}$ and some vertex from $V_{k+2}$ . It is thus left only to connect the endpoints of those paths in a suitable order to finish the embedding.

To do so, we first partition $B$ into sets $B=\cup _{(i,j)\in \binom {[t]}{2}}B_{(i,j)}$ , where $B_{(i,j)}=\{b^{(i,j)}_1,\ldots ,b^{(i,j)}_{|J_{\{i,j\}}|}\}$ . For each $(i,j)\in \binom {[t]}{2}$ , let $v'_{i,j}\in V_{k+2}$ be the other endpoint of the path starting at $u'_{i,j}\in V_{k+1}$ and using edges from $M_0,\ldots , M_{k+1}$ . For each $(i,j)\in \binom {[t]}{2}$ , and $1\le z\le |J_{\{i,j\}}|$ , let $c^{(i,j)}_z\in V_{k+2}$ be the other endpoint of the path which starts at $b^{(i,j)}_z$ , goes through the corresponding path $P_l$ , $l\in [m']$ , and then uses edges from the matchings $M_0, M_1,\ldots , M_{k+1}$ . Let $\sigma \,:\,V_{k+2}\to V_{k+3}$ be the permutation defined as follows: for each $(i,j)\in \binom {[t]}{2}$ , set $\sigma (v'_{i,j})=b^{(i,j)}_1$ , $\sigma (c_{z}^{(i,j)})=b^{(i,j)}_{z+1}$ for each $1\le z\lt |J_{\{i,j\}}|$ , and $\sigma (c_{|J_{\{i,j\}}|}^{(i,j)})=u'_{j,i}$ . Then, using property B, we can complete the embedding of the $K_t$ -subdivision.

Proof of Theorem 1.5. We start by choosing constants

\begin{equation*}0 \lt 1/n_0\ll c\ll \varepsilon \ll \theta \ll \alpha \ll \mu \lt 1.\end{equation*}

Suppose that $n\ge n_0$ and that $G$ is an $n$ -vertex graph which is $\varepsilon n$ -joined and has minimum degree $\delta (G)\geq \mu n$ . During the proof, we will repeatedly use the following observation. Let $U, Z\subset V(G)$ and let $v,v'$ be distinct vertices.

1. F If $d(v,U),d(v',U)\gt |Z|+4\varepsilon n$ , then there is a $v,v'$ -path of length 3 whose interior vertices belong to $U\setminus Z$ .

Indeed, if $d(v,U),d(v',U)\gt |Z|+4\varepsilon n$ , we can find disjoint sets $X\subset N(v,U)\setminus (Z\cup \{v'\})$ and $Y\subset N(v',U)\setminus (Z\cup \{v\})$ such that $|X|=|Y|=\varepsilon n$ and then find an edge between $X$ and $Y$ using that $G$ is $\varepsilon n$ -joined.

Step 1. Fixing the reservoirs. Let $1\le t\le c\sqrt {n}$ be fixed from now on, and let $r=\theta n$ and $k=4\cdot \binom {t}{2}+8\varepsilon n$ . We start by setting aside three disjoint random sets $R_1, R_2, R_3\subset V(G)$ with the following properties.

1. G1 $|R_1|=2r$ , $|R_2|=r +k$ , and $|R_3|=\alpha n$ .

2. G2 For all $v\in V(G)$ and $i\in [3]$ , $d(v,R_i)\ge \frac {\mu |R_i|}{2}$ .

3. G3 For all $v\in V(G)$ , $d(v,V(G)\setminus \cup _{i\in [3]}R_i)\ge \frac {\mu n}{2}$ .

Step 2. Constructing the absorbers. Our aim now is to construct a collection of absorbing paths that is capable of absorbing $R_1$ together with any set of $r$ vertices from $R_2$ . To do so, we first construct a collection of 100 vertex-disjoint absorbers for each vertex in $R_1\cup R_2$ .

Proof of Claim 1. We observe first that an absorber of size 2 for $v$ is just a triangle supported at $v$ . We will find the triangles greedily using G3 and the joinedness condition. Indeed, firstly note that for every vertex $v\in V(G)$ we have

(3.2) \begin{equation} d(v,V(G)\setminus \cup _{i\in [3]}R_i)\overset {G3}{\ge }\frac {\mu n}{2}\ge \frac {\mu n}{4}+200|R_1\cup R_2|, \end{equation}

as $1/n_0\ll \alpha ,\theta \ll \mu$ .

Secondly, for each vertex $v\in R_1\cup R_2$ , use that $G$ is $\varepsilon n$ -joined to find $100$ vertex-disjoint edges in the neighbourhood of $v$ , disjoint from all the other choices and disjoint from $\cup _{i\in [3]}R_i$ . This is possible as

\begin{equation*}d(v,V(G)\setminus \cup _{i\in [3]}R_i)\overset {(3.2)}{\ge } \frac {\mu n}{4} +200 |R_1 \cup R_2|\ge 2\varepsilon n+200|R_1\cup R_2|.\end{equation*}

So, for a vertex $v\in R_1\cup R_2$ , at each step we can pick disjoint subsets $X_v, Y_v\subset N(v)\setminus \cup _{i\in [3]}R_i$ of unused vertices such that $|X_v|=|Y_v|=\varepsilon n$ and then pick an edge between $X_v$ and $Y_v$ using that $G$ is $\varepsilon n$ -joined.

Our aim now is to suitably glue these absorbers to create robust absorbers.

Proof of Claim 2. Let $H$ be the bipartite graph from Lemma3.5 with $r$ in the place of $n$ , and let $R_1$ and $R_2$ play the role of the sets $Y$ and $Z$ , respectively, and let $X=[3r]$ . For every vertex $v\in R_1\cup R_2$ , let $f_v\,:\,N_H(v)\to [100]$ be an injective function listing the neighbours of $v$ in $H$ . Letting $Q_0$ be the empty graph, suppose that we have found $Q_0,\ldots , Q_{i-1}$ for some $i\in [3r]$ and let us show how to find $Q_i$ . Let $v_1,\ldots , v_q$ be a labelling of $N_H(i)$ and note that $q\le \Delta (H)\le 100$ . Using F, for each $1\le j\lt q$ , find a path $Q'_{ij}$ of length $3$ between $b_{v_j,f_{v_j}(i)}$ and $a_{v_{j+1},f_{v_{j+1}}(i)}$ , which is possible because

\begin{equation*}d(v,V(G)\setminus \cup _{i\in [3]}R_i)\overset {(3.2)}{\ge }\frac {\mu n}{4}+200|R_1\cup R_2|\ge \frac {\mu n}{8}+2\cdot 10^5|R_1\cup R_2|,\end{equation*}

provided $1/n_0\ll \alpha ,\theta \ll \mu$ . Then, repeatedly using F, find a path $Q'_{iq}$ starting from $b_{v_q,f_{v_q}(i)}$ and having suitable size so that $\cup _{j\in [q]}Q'_{i,j} \cup \{ a_{v_1,f_1(i)}b_{v_1,f_1(i)}\}$ has exactly 1000 vertices. Again, this is possible due to (3.2) and because we are looking for a path with $1000\gt 4q$ vertices.

Step 3. Finding an almost spanning $K_t$ -subdivision. Let $A= \cup _{i\in [3r]}V(Q_i)$ be the set of vertices covered by the absorbing paths we found in Claim2. Note that $|A|=3000r$ as each path has exactly $1000$ vertices. Let $V'=V(G)\setminus (R_1\cup R_2\cup R_3\cup A)$ and pick arbitrarily our branching vertices $u_1,\ldots ,u_t \in V'$ . To connect the branching vertices, we will find a collection of $\binom {t}{2}$ vertex-disjoint paths covering most of $G[V']$ . Indeed, as $G$ is $\varepsilon n$ -joined, we can use Lemma3.6 to find a path $P$ in $G'-\{u_1,\ldots , u_t\}$ such that

\begin{equation*}|P|\ge |V'|-t-2\varepsilon n.\end{equation*}

Divide $P$ into paths $(P'_{e})_{e\in \binom {[t]}{2}}$ so that $\big ||P'_{e}|-|P'_{e'}|\big |\le 1$ for all $e,e'\in \binom {[t]}{2}$ . Next, for each $e=ij\in \binom {[t]}{2}$ , form a $(u_i,u_j)$ -path $P_{ij}=P_e$ by using $P'_{ij}$ and connecting through $R_3$ , using F, the endpoints of $P'_{ij}$ with $u_i$ and $u_j$ . This is possible, as for all $v\in V(G)$ , G2 gives $d(v,R_3)\ge \mu \alpha n/2\ge c^2n+ 100\varepsilon n\ge t^2n+100\varepsilon n$ . Thus, we have found a $K_t$ -subdivision, say $S$ , such that

• • S is nearly balanced, and

• • $|V(S)\setminus \cup _{i\in [3]}R_i|\ge n-|R_1\cup R_2\cup R_3|-|A|-2\varepsilon n$ , and

• • $|V(S)\cap R_3|=4\cdot \binom {t}{2}$ .

Step 4. Finding a spanning $K_t$ -subdivision. The last step of the proof consists of using the absorbing paths we built in Step 2 to turn the $K_t$ -subdivision $S$ into a spanning nearly balanced subdivision. To do so, we first need to incorporate each of the absorbing paths $(Q_i)_{i\in [3r]}$ into $S$ .

For $1\le i\le 3r$ , pick a path $P_{e}$ of minimal length and take two consecutive vertices $x,y\in V(P_e)$ , and replace the edge $xy\in E(P_e)$ with the $(x,y)$ -path which consists of $Q_i$ together with two paths of length $3$ connecting the endpoints of $Q_i$ with $x$ and $y$ , respectively, and whose interior vertices belong to $R_3$ . This is done, again, using G2 and F, and it is possible as, for every $v\in V(G)$ ,

\begin{equation*}d(v,R_3\setminus V(S)){\ge } \frac {\mu |R_3|}{2}-4\cdot \binom {t}{2}\ge \frac {\mu \alpha n}{2}-2c^2n\ge \frac {\mu \alpha n}{4}+4\cdot 3r,\end{equation*}

as $1/n_0\ll c\ll \theta \ll \alpha \ll \mu$ . Thus we have found a $K_t$ -subdivision, say $S'$ , whose branching path lengths differ by at most $1005$ . Indeed, unless $3r$ divides ${\binom{t}{2}}$ , then some of the branching paths will contain one more absorbing path $Q_i$ than others. Hence, they may contain at most $1005$ more vertices, $1000$ for the path itself, $2$ to glue each side and one final vertex which is not in $S'$ yet but will be embedded right at the end of the proof. Moreover, $S'$ satisfies $|V(S')\setminus \cup _{i\in [3]}R_i|\ge n-|R_1\cup R_2\cup R_3|-2\varepsilon n$ , $|V(S')\cap R_3|=4\cdot \binom {t}{2}+4\cdot 3r$ , and

1. I each path $Q_i$ is a subpath of one of the paths from $S'$ .

The last step we need before using the absorbing paths is to include each vertex from $V(G)\setminus (V(S')\cup R_1\cup R_2)$ into the $K_t$ -subdivision. To do that, we will connect through $R_2$ now. Observe that the number of vertices used from $R_3$ thus far is exactly $4\cdot {\binom{t}{2}}+4\cdot 3r$ and that

(3.3) \begin{equation} |R_3|=\alpha n\geq 4\cdot {\binom{t}{2}}+4\cdot 3r +1005\cdot {\binom{t}{2}}+2\varepsilon n . \end{equation}

Therefore, by Lemma3.6 we may find a path $P''$ in $V(G)\setminus (V(S')\cup R_1\cup R_2)$ covering all but exactly $2\varepsilon n$ vertices. Furthermore, by 3.3 $|P''|\geq 1005 \cdot {\binom{t}{2}}$ .

Firstly, we embed into $S'$ those vertices in $V(G)\setminus (V(S')\cup R_1\cup R_2)$ which are not covered by $P''$ . This is done using F to make the connections through paths of length $3$ with internal vertices in $R_2$ . We do this by embedding one vertex at a time in a shortest branching path, ensuring that the lengths of the branching paths differ by at most $1005$ still. Since $|P''| \geq 1005\cdot {\binom{t}{2}}$ , we may subdivide the path into ${\binom{t}{2}}$ segments of appropriate lengths, taking into consideration the fact that some branching paths may contain one more absorbing path $Q_i$ than others, and embed them into the subdivision. This is once again done by using F and connecting by paths of length 3 with internal vertices in $R_2$ . In particular, embedding the $2\varepsilon n$ vertices as well as the ${\binom{t}{2}}$ segments of $P''$ is possible as, as for every every vertex $v\in V(G)$ we have

\begin{equation*}d(v,R_2)\overset {G2}{\ge } \frac {\mu |R_2|}{2}\ge \frac {\mu \theta n}{4}+4\cdot 2\varepsilon n+4\cdot \tbinom {t}{2},\end{equation*}

and at each step, we use exactly four new vertices from $R_2$ . Note that picking the segment lengths of $P''$ carefully ensures that at this point the branching path sizes differ by at most one. We are now left to embed the remainder of $R_1 \cup R_2$ .

Let $R_2'\subset R_2$ be the set of leftover vertices in $R_2$ and note that $|R_2'|=|R_2|-4\cdot {\binom{t}{2}}-8\varepsilon n=r$ . Now we are ready to use the property of the absorbing paths. By H2, there is perfect matching, say $M$ , between $[3r]$ and $R_1\cup R_2'$ . For each edge $iz_i\in E(M)$ , by H1 we know that $Q_i$ is an absorbing path for $z_i$ and thus, by I and E, we can include $z_i$ into the $K_t$ -subdivision. Since $M$ is a perfect matching, we have included each vertex from $R_1\cup R_2'$ and thus found a spanning nearly balanced $K_t$ -subdivision.