11institutetext: Advanced Computing and Microelectronics unit
Indian Statistical Institute, 203 B. T. Road, Kolkata 700108, India
Emails: 11email<EMAIL_ADDRESS><EMAIL_ADDRESS>

# Optimal $L(1,2)$-edge Labeling of Infinite Octagonal Grid

Subhasis Koley  Sasthi C. Ghosh

###### Abstract

For two given non-negative integers $h$ and $k$, an $L(h,k)$-edge labeling of
a graph $G=(V(G),E(G))$ is a function
$f^{\prime}:E(G)\xrightarrow{}\\{0,1,\cdots,n\\}$ such that $\forall
e_{1},e_{2}\in E(G)$, $|f^{\prime}(e_{1})-f^{\prime}(e_{2})|\geq h$ when
$d^{\prime}(e_{1},e_{2})=1$ and $|f^{\prime}(e_{1})-f^{\prime}(e_{2})|\geq k$
when $d^{\prime}(e_{1},e_{2})=2$ where $d^{\prime}(e_{1},e_{2})$ denotes the
distance between $e_{1}$ and $e_{2}$ in $G$. Here
$d^{\prime}(e_{1},e_{2})=k^{\prime}$ if there are at least $(k^{\prime}-1)$
number of edges in $E(G)$ to connect $e_{1}$ and $e_{2}$ in $G$. The objective
is to find span which is the minimum $n$ over all such $L(h,k)$-edge labeling
and is denoted as $\lambda^{\prime}_{h,k}(G)$. Motivated by the channel
assignment problem in wireless cellular network, $L(h,k)$-edge labeling
problem has been studied in various infinite regular grids. For infinite
regular octagonal grid $T_{8}$, it was proved that
$25\leq\lambda^{\prime}_{1,2}(T_{8})\leq 28$ [Tiziana Calamoneri,
International Journal of Foundations of Computer Science, Vol. 26, No. 04,
2015] with a gap between lower and upper bounds. In this paper we fill the gap
and prove that $\lambda^{\prime}_{1,2}(T_{8})=28$.

###### Keywords:

$L(h,k)$-edge labeling Regular octagonal grid Span Edge labeling

##  1 Introduction

In a wireless network, frequency channels are assigned to the transmitters for
communications. As the number of frequency channels are limited, the main
challenge of the Channel Assignment Problem ($CAP$) is to assign frequencies
to the transmitters effectively such that interference can not occur. Hale [8]
formulated $CAP$ as a classical vertex coloring problem. To incorporate the
effect of interference in multi hop distance into the formulation, distance
vertex labeling of graph was proposed in [11], [7]. A special case of distance
vertex labeling of a graph is $L(h,k)$-vertex labeling and is defined as
follows.

###### Definition 1

For two given non-negative integers $h$ and $k$, an $L(h,k)$-vertex labeling
of a graph $G=(V(G),E(G))$ is a function
$f:V(G)\xrightarrow{}\\{0,1,\cdots,n\\}$ such that $\forall v_{1},v_{2}\in
V(G)$, $|f(v_{1})-f(v_{2})|\geq h$ when $d(v_{1},v_{2})=1$ and
$|f(v_{1})-f(v_{2})|\geq k$ when $d(v_{1},v_{2})=2$ where $d(v_{1},v_{2})$
denotes the distance between $v_{1}$ and $v_{2}$. Here
$d(v_{1},v_{2})=k^{\prime}$ if there are at least $k^{\prime}$ number of edges
in $E(G)$ to connect $v_{1}$ and $v_{2}$ in $G$.

The span $\lambda_{h,k}(G)$ of $L(h,k)$-vertex labeling is the minimum $n$
such that $G$ admits an $L(h,k)$-vertex labeling. The $L(h,k)$-vertex labeling
problem on different types of graphs have been studied [12], [2], [6]. The
$L(h,k)$-edge labeling problem was introduced in [5] and is defined as
follows.

###### Definition 2

For two given non-negative integers $h$ and $k$, an $L(h,k)$-edge labeling of
a graph $G=(V(G),E(G))$ is a function
$f^{\prime}:E(G)\xrightarrow{}\\{0,1,\cdots,n\\}$ such that $\forall
e_{1},e_{2}\in E(G)$, $|f^{\prime}(e_{1})-f^{\prime}(e_{2})|\geq h$ when
$d^{\prime}(e_{1},e_{2})=1$ and $|f^{\prime}(e_{1})-f^{\prime}(e_{2})|\geq k$
when $d^{\prime}(e_{1},e_{2})=2$ where $d^{\prime}(e_{1},e_{2})$ denotes the
distance between $e_{1}$ and $e_{2}$. Here
$d^{\prime}(e_{1},e_{2})=k^{\prime}$ if there are at least $(k^{\prime}-1)$
number of edges in $E(G)$ to connect $e_{1}$ and $e_{2}$ in $G$.

The span $\lambda^{\prime}_{h,k}(G)$ of $L(h,k)$-edge labeling is the minimum
$n$ such that $G$ admits an $L(h,k)$-edge labeling. Various studies for
$L(h,k)$-edge labelling have been carried out for different types of graphs.
Among the various graph classes, many studies have been done on infinite
regular hexagonal grid, infinite square grid, infinite triangular grid and
infinite octagonal grid specially for $h=1,2$ and $k=1,2$ [4], [10], [9], [3]
[1]. In [3], infinite octagonal grid $T_{8}$ has been studied and the lower
and upper bound of $\lambda^{\prime}_{1,2}(T_{8})$ have been given. More
specifically, in [3], it is given that
$25\leq\lambda^{\prime}_{1,2}(T_{8})\leq 28$, where there is a gap between
lower and upper bounds. In this paper, we fill the gap and prove that
$\lambda^{\prime}_{1,2}(T_{8})\geq 28$. As in [3],
$\lambda^{\prime}_{1,2}(T_{8})\leq 28$, it is concluded that
$\lambda^{\prime}_{1,2}(T_{8})=28$. Throughout this paper, we use color and
label interchangeably. To prove the lower bound of
$\lambda^{\prime}_{1,2}(T_{8})$, we use the structural properties of $T_{8}$.
More specifically, we first identify a subgraph $G^{\prime}$ of $T_{8}$ where
no two edges can have the same color. After that we consider all edges where a
pair of consecutive colors $(c,c\pm 1)$ can be used in $G^{\prime}$. Based on
this, we identify the subgraphs in $T_{8}$ where $c$ and $c\pm 1$ can not be
used. Then using the structural properties of those subgraphs we conclude how
many additional colors other the colors used in $G^{\prime}$ must be required
to color $T_{8}$ by using the pigeon hole principle and accordingly derive the
span.

##  2 Results

Figure 1: An infinite regular octagonal grid $T_{8}$.

Fig. 1 shows a portion of infinite regular octagonal grid $T_{8}$. Observe
that there are four types of edges in $T_{8}$. The edges which are along or
parallel to the $X$ axis and the edges which are along or parallel to the $Y$
axis are said to be horizontal edges and vertical edges respectively. The
edges which are at $45^{\circ}$ to any of a horizontal edge and the edges
which are at $135^{\circ}$ to any of a horizontal edge are said to be right
slanting edges and left slanting edges respectively. Here the angular distance
between two adjacent edges $e_{1}$ and $e_{2}$ represents the smaller of the
two angles measured in anticlockwise from $e_{1}$ to $e_{2}$ and from $e_{2}$
to $e_{1}$. In Fig. 1, the edges $(p,q)$, $(p,r)$, $(p,s)$ and $(p,t)$ are a
horizontal, a right slanting, a vertical and a slanting edges respectively. By
slanting edges, we mean all the left and right slanting edges and by non
slanting edges, we mean all the horizontal and vertical edges.

Figure 2: The $G_{S}$ corresponding to the $K_{4}$ having vertex set
$S=\\{v_{1},v_{2},v_{3},v_{4}\\}$.

Consider any $K_{4}$ (complete graph of $4$ vertices) in $T_{8}$ and let
$S=V(K_{4})$ be the vertex set of the $K_{4}$. Let $v$ be a vertex in $T_{8}$
and $N_{v}$ be the set of vertices in $T_{8}$ which are adjacent to $v$. Let
$N(S)=\displaystyle\bigcup_{v\in S}N_{v}$ be the set of all vertices in
$T_{8}$ which are adjacent to at least one vertex in $S$. Let us define
$G_{S}$ as the subgraph of $T_{8}$ such that $V(G_{S})=S\cup N(S)$ and
$E(G_{S})$ is the set of all edges of $T_{8}$ which are incident to at least
one vertex in $S$. Fig. 2 shows the the $K_{4}$ with vertex set
$S=\\{v_{1},v_{2},v_{3},v_{4}\\}$ and the corresponding $G_{S}$. It is evident
that any two edges of $G_{S}$ are at distance at most $2$. Hence no two edges
of $G_{S}$ can be given the same color for $L(1,2)$-edge labeling of $G_{S}$.
As $|E(G_{S})|=26$ and $d^{\prime}(e_{1},e_{2})\leq 2$, $\forall
e_{1},e_{2}\in E(G_{S})$, $\lambda^{\prime}_{1,2}(G_{S})\geq 25$ for $L(1,2)$
edge labeling of $G_{S}$.

Figure 3: A subgraph $G$ of an infinite octagonal grid $T_{8}$.

Consider two edges $e_{1},e_{2}\in E(G_{S})$ such that
$d^{\prime}(e_{1},e_{2})=2$. Let the color $c$ be assigned to $e_{1}$.
Clearly, the colors $c\pm 1$ can not be used at $e_{2}$ for $L(1,2)$-edge
labeling. As there exists no pair of edges $e_{1}$ and $e_{2}$ such that
$d^{\prime}(e_{1},e_{2})\geq 3$ in $G_{S}$, both $c\pm 1$ must be used at the
adjacent edges of $e_{1}$ in $G_{S}$.

Consider the $K_{4}$ having vertex set $S=\\{v_{1},v_{2},v_{3},v_{4}\\}$ as
shown in Fig. 3. Now define $S^{\prime}=S\cup N(S)\cup N(N(S))$. Fig. 3 shows
the graph $G$ such that $V(G)=S^{\prime}\cup N(S^{\prime})$ and $E(G)$ is the
set of all edges in $T_{8}$ which are incident to at least one vertex in
$S^{\prime}$. Note that there are total $25$ distinct $K_{4}$s including the
$K_{4}$ having vertex set $S$ in $G$. Note that here $G$ is built over the
$G_{S}$. Consider a $G_{S}$ and the corresponding $G$. Let two consecutive
colors $c$ and $c+1$ be used in $G_{S}$. In the following lemma we identify
all the $K_{4}$s having vertex sets $S_{1},S_{2},\cdots,S_{m}$ such that $c$
and $c+1$ can not be used at $G_{S_{1}},G_{S_{2}},\cdots,G_{S_{m}}$ when $c$
and $c+1$ are used in $G_{S}$.

###### Lemma 1

For every pair of consecutive colors $c$ and $c+1$ used in two adjacent edges
$e_{1}$ and $e_{2}$ in $G_{S}$, except when $e_{1}$ and $e_{2}$ form an angle
of $45^{\circ}$ at their common incident vertex, there exists at least $4$
different $K_{4}$s having vertex sets $S_{1}$, $S_{2}$, $S_{3}$ and $S_{4}$
other than $S$ such that 1) $c$ can not be used at $G_{S_{1}}$ and $G_{S_{2}}$
and 2) $c+1$ can not be used at $G_{S_{3}}$ and $G_{S_{4}}$. When $e_{1}$ and
$e_{2}$ are at angle $45^{\circ}$, there exists $3$ different $K_{4}$s having
vertex sets $S_{1}$, $S_{2}$ and $S_{3}$ other than $S$ such that either 1)
$c$ can not be used at $G_{S_{1}}$ and $G_{S_{2}}$ and $c+1$ can not be used
at $G_{S_{3}}$ or 2) $c+1$ can not be used at $G_{S_{1}}$ and $G_{S_{2}}$ and
$c$ can not be used at $G_{S_{3}}$.

###### Proof

Note that the angular distance between any two adjacent edges in $T_{8}$ can
be any one of $45^{\circ}$, $90^{\circ}$, $135^{\circ}$, $180^{\circ}$,
$225^{\circ}$, $270^{\circ}$ and $315^{\circ}$. From symmetry, it is suffice
to consider the cases where the angular distance between two adjacent edges in
$T_{8}$ is $180^{\circ}$ or $135^{\circ}$ or $90^{\circ}$ or $45^{\circ}$.

  * •

When angular distance between two adjacent edges is $180^{\circ}$: Observe
that two adjacent horizontal edges or two adjacent vertical edges or two
adjacent left slanting edges or two adjacent right slanting edges can be at
$180^{\circ}$. Clearly the cases where two horizontal edges and two vertical
edges forming $180^{\circ}$ are symmetric. Similarly the cases where two left
slanting edges and two right slanting edges forming $180^{\circ}$ are also
symmetric. So we need to consider the following two cases.

Case $1$) When two adjacent horizontal edges $e_{1}=(v_{1},v_{2})$ and
$e_{2}=(v_{2},u_{6})$ form $180^{\circ}$ (Fig. 3). Let $f^{\prime}(e_{1})=c$
and $f^{\prime}(e_{2})=c+1$. Note that any edge $e$ incident to any of the
vertices in $\\{u_{5},u_{6},u_{7}\\}$ is at distance at most two from $e_{1}$.
As $f^{\prime}(e_{1})=c$, the color $c$ can not be used at any of those edges
for $L(1,2)$ edge labeling. Similarly observe that any edge $e$ incident to
any of the vertices in $\\{w_{8},w_{9},w_{10}\\}$ but not incident to any of
the vertices in $\\{u_{5},u_{6},u_{7}\\}$ is at distance two from $e_{2}$. As
$f^{\prime}(e_{2})=c+1$, the color $c$ can not be used at any of those edges
for $L(1,2)$ edge labeling. Hence there exists $2$ different $K_{4}$s having
vertex sets $S_{1}=\\{u_{5},u_{6},w_{8},w_{9}\\}$ and
$S_{2}=\\{u_{6},u_{7},w_{9},w_{10}\\}$ such that $c$ can not be used at
$G_{S_{1}}$ and $G_{S_{2}}$. With similar argument, it can be shown that there
exists $2$ different $K_{4}$s having vertex sets
$S_{3}=\\{v_{1},v_{4},u_{11},u_{12}\\}$ and
$S_{4}=\\{v_{1},u_{9},u_{10},u_{11}\\}$ such that $c+1$ can not be used at
$G_{S_{3}}$ and $G_{S_{4}}$.

Case $2$) When two adjacent right slanting edges $e_{1}=(v_{1},v_{3})$ and
$e_{2}=(v_{1},u_{10})$ form $180^{\circ}$ (Fig. 3). Let $f^{\prime}(e_{1})=c$
and $f^{\prime}(e_{2})=c+1$. In this case, there exists $3$ different $K_{4}$s
having vertex sets $S_{1}=\\{u_{10},w_{15},w_{16},w_{17}\\}$,
$S_{2}=\\{u_{9},u_{10},w_{14},w_{15}\\}$ and
$S_{3}=\\{u_{10},u_{11},w_{17},w_{18}\\}$ such that $c$ can not be used at
$G_{S_{1}}$, $G_{S_{2}}$ and $G_{S_{3}}$. Similarly, there exists $3$
different $K_{4}$s having vertex sets $S_{3}=\\{v_{3},v_{4},u_{2},u_{3}\\}$,
$S_{4}=\\{v_{3},u_{3},u_{4},,u_{5}\\}$ and
$S_{5}=\\{v_{2},v_{3},u_{5},u_{6}\\}$ such that $c+1$ can not be used at
$G_{S_{3}}$, $G_{S_{4}}$ and $G_{S_{5}}$.

  * •

When angular distance between two adjacent edges is $135^{\circ}$: Observe
that a horizontal edge and its adjacent left slanting edge or a horizontal
edge and its adjacent right slanting edge or a vertical edge and its adjacent
left slanting edge or a vertical edge and its adjacent right slanting edge may
be at $135^{\circ}$. Clearly all the cases are symmetric. So we need to
consider the following case only. Let us consider the horizontal edge
$e_{1}=(v_{1},v_{2})$ and the right slanting edge $e_{2}=(v_{2},u_{5})$. Let
$f^{\prime}(e_{1})=c$ and $f^{\prime}(e_{2})=c+1$. From similar discussion
stated in the previous case, there exists $3$ different $K_{4}$s having vertex
sets $S_{1}=\\{v_{3},u_{3},u_{4},u_{5}\\}$,
$S_{2}=\\{u_{4},u_{5},w_{7},w_{8}\\}$ and
$S_{3}=\\{u_{5},u_{6},w_{8},w_{9}\\}$ such that $c$ can not be used at
$G_{S_{1}}$, $G_{S_{2}}$ and $G_{S_{3}}$. Similarly, there exists $2$
different $K_{4}$s having vertex sets $S_{4}=\\{v_{1},v_{4},u_{11},u_{12}\\}$
and $S_{5}=\\{v_{1},u_{9},u_{10},u_{11}\\}$ such that $c+1$ can not be used at
$G_{S_{4}}$ and $G_{S_{5}}$.

  * •

When angular distance between two adjacent edges is $90^{\circ}$: Observe that
a horizontal edge and its adjacent vertical edge or a vertical edge and its
adjacent horizontal edge or a right slanting edge and its adjacent left
slanting edge may be at $90^{\circ}$. Clearly the first two cases are
symmetric. So we need to consider the following two cases.

Case $1$) Consider the horizontal edge $e_{1}=(v_{1},v_{2})$ and the vertical
edge $e_{2}=(v_{2},v_{3})$. Let $f^{\prime}(e_{1})=c$ and
$f^{\prime}(e_{2})=c+1$. As similar argument stated in the previous cases,
there exists $2$ different $K_{4}$s having vertex sets
$S_{1}=\\{v_{3},v_{4},u_{2},u_{3}\\}$ and
$S_{2}=\\{v_{3},u_{3},u_{4},u_{5}\\}$ such that $c$ can not be used at
$G_{S_{1}}$ and $G_{S_{2}}$. Similarly, there exists $2$ different $K_{4}$s
having vertex sets $S_{3}=\\{v_{1},v_{4},u_{11},u_{12}\\}$ and
$S_{4}=\\{v_{1},u_{9},u_{10},u_{11}\\}$ such that $c+1$ can not be used at
$G_{S_{3}}$ and $G_{S_{4}}$.

Case $2$) Consider the right slanting edge $e_{1}=(v_{1},v_{3})$ and the left
slanting edge $e_{2}=(v_{1},u_{12})$. Let $f^{\prime}(e_{1})=c$ and
$f^{\prime}(e_{2})=c+1$. As similar argument stated in the previous cases,
there exists $3$ different $K_{4}$s having vertex sets
$S_{1}=\\{u_{1},u_{12},w_{19},w_{20}\\}$,
$S_{2}=\\{u_{1},u_{2},u_{12},v_{4}\\}$ and
$S_{3}=\\{u_{11},u_{12},w_{18},w_{19}\\}$ such that $c$ can not be used at
$G_{S_{1}}$, $G_{S_{2}}$ and $G_{S_{3}}$. Similarly, there exists $3$
different $K_{4}$s having vertex sets $S_{4}=\\{v_{3},v_{4},u_{2},u_{3}\\}$,
$S_{5}=\\{v_{3},u_{3},u_{4},u_{5}\\}$ and
$S_{6}=\\{v_{2},v_{3},u_{5},u_{6}\\}$ such that $c+1$ can not be used at
$G_{S_{4}}$, $G_{S_{5}}$ and $G_{S_{6}}$.

  * •

When angular distance between two edges is $45^{\circ}$: Observe that a
horizontal edge and its adjacent left slanting edge or a horizontal edge and
its adjacent right slanting edge or a vertical edge and its adjacent left
slanting edge or a vertical edge and its adjacent right slanting edge may be
at $45^{\circ}$. Clearly all the cases are symmetric. So we need to consider
the following case only. Consider the horizontal edge $e_{1}=(v_{1},v_{2})$
and the left slanting edge $e_{2}=(v_{2},v_{4})$. Let $f^{\prime}(e_{1})=c$
and $f^{\prime}(e_{2})=c+1$. As similar argument stated in the previous cases,
there exists $2$ different $K_{4}$s having vertex sets
$S_{1}=\\{v_{4},u_{1},u_{2},u_{12}\\}$ and
$S_{2}=\\{v_{3},v_{4},u_{2},u_{3}\\}$ such that $c$ can not be used at
$G_{S_{1}}$ and $G_{S_{2}}$. Similarly, there exists a $K_{4}$s having vertex
set $S_{3}=\\{v_{1},u_{9},u_{10},u_{11}\\}$ such that $c+1$ can not be used at
$G_{S_{3}}$. If $f^{\prime}(e_{1})=c+1$ and $f^{\prime}(e_{2})=c$, then there
exists $2$ different $K_{4}$s having vertex sets
$S_{1}=\\{v_{4},u_{1},u_{2},u_{12}\\}$ and
$S_{2}=\\{v_{3},v_{4},u_{2},u_{3}\\}$ such that $c+1$ can not be used at
$G_{S_{1}}$ and $G_{S_{2}}$. Similarly, there exists a $K_{4}$s having vertex
set $S_{3}=\\{v_{1},u_{9},u_{10},u_{11}\\}$ such that $c$ can not be used at
$G_{S_{3}}$.

∎

Now we state and prove the following Theorems.

###### Theorem 2.1

$\lambda^{\prime}_{1,2}(T_{8})\geq 26$.

###### Proof

Let us consider the $K_{4}$ with vertex set $S=\\{v_{1},v_{2},v_{3},v_{4}\\}$
and the corresponding $G_{S}$ as shown in Fig. 2. Observe that there are $26$
edges in $G_{S}$. Note that there are no two edges at more than distance two
apart in $G_{S}$. Hence all the colors used in $G_{S}$ must be distinct for
$L(1,2)$-edge labeling. Hence $26$ consecutive colors $\\{0,1,\cdots,25\\}$
are to be used in $G_{S}$, otherwise $\lambda^{\prime}_{1,2}(G_{S})\geq 26$.
In that case, any two consecutive colors $c$ and $c+1$ must be used at two
adjacent edges in $G_{S}$. Therefore, from Lemma 1, there exists at least a
$K_{4}$ having vertex set $S_{1}$ such that $c$ can not be used at
$G_{S_{1}}$. But in $G_{S_{1}}$, $26$ distinct colors must be used. So if we
do not use $c$ in $G_{S_{1}}$, at least a new color which is not used in
$G_{S}$ must be used in $G_{S_{1}}$. So at least the color $26$ must be
introduced in $G_{S_{1}}$. Hence $\lambda^{\prime}_{1,2}(G_{S_{1}})\geq 26$
implying $\lambda^{\prime}_{1,2}(T_{8})\geq 26$.

∎

###### Theorem 2.2

$\lambda^{\prime}_{1,2}(T_{8})\geq 27$.

###### Proof

Consider $S=\\{v_{1},v_{2},v_{3},v_{4}\\}$ and the corresponding $G_{S}$ as
shown in Fig. 3. Note that $\lambda^{\prime}_{1,2}(G_{S})\leq 26$ only if 1)
$26$ consecutive colors $\\{0,1,\cdots,25\\}$ or $\\{1,2,\cdots,26\\}$ are
used in $G_{S}$, or 2) the colors
$\\{0,1,\cdots,26\\}\setminus\\{c^{\prime}\\}$ are used in $G_{S}$, where
$1\leq c^{\prime}\leq 25$. The cases of $\\{0,1,\cdots,25\\}$ and
$\\{1,2,\cdots,26\\}$ are clearly symmetric. Hence we need to consider only
the following two cases.

Consider the first case. Observe that the set of colors $\\{0,1,\cdots,25\\}$
can be partitioned into $13$ disjoint consecutive pairs of colors $(0,1)$,
$(2,3)$, $\cdots$, $(24,25)$. Consider a pair of consecutive colors $(c,c+1)$,
where $0\leq c\leq 24$. From Lemma 1, it follows that for the pair $(c,c+1)$,
there exists two distinct $S_{1}$ and $S_{2}$ other that $S$ such that either
$c$ or $c+1$ can not be used in $G_{S_{1}}$ and $G_{S_{2}}$. So for the $13$
disjoint pairs mentioned above, there must be $26$ such
$S_{1},S_{2},\cdots,S_{26}$ other than $S$. Note that there are total $25$
$S_{j}$s including $S$ in $G$. Therefore, there exists only $24$ such $S_{j}$s
other than $S$ in $G$. Hence from pigeon hole ($26$ pigeons and $24$ holes)
principle there must be at least one $G_{S_{j}}$ where two different colors
which are used in $G_{S}$ can not be used there and hence
$\lambda^{\prime}_{1,2}(G_{S_{j}})\geq 27$ implying
$\lambda^{\prime}_{1,2}(T_{8})\geq 27$.

Now we consider the second case. Consider that the colors
$\\{0,1,\cdots,26\\}\setminus\\{c^{\prime}\\}$ have been used in $G_{S}$,
where $1\leq c^{\prime}\leq 25$. First let us consider the case when
$c^{\prime}$ is even. In that case the set of colors
$\\{0,\cdots,c^{\prime}-1,c^{\prime}+1,\cdots 26\\}$ can be partitioned into
$13$ disjoint consecutive pairs of colors
$(0,1),\cdots,(c^{\prime}-2,c^{\prime}-1),(c^{\prime}+1,c^{\prime}+2),\cdots,(25,26)$.
Hence proceeding similarly as above case, from pigeon hole principle, we get
that $\lambda^{\prime}_{1,2}(T_{8})\geq 27$. Now consider the case when
$c^{\prime}$ is odd. Let us first consider $c^{\prime}\neq 1,25$. Note that
the set of colors $\\{0,\cdots,c^{\prime}-1,c^{\prime}+1,\cdots 26\\}$ can be
partitioned into $12$ disjoint consecutive pairs of colors
$(0,1),\cdots,(c^{\prime}-3,c^{\prime}-2),(c^{\prime}+1,c^{\prime}+2),\cdots,(24,25)$.
So there must be $24$ $K_{4}$s having vertex sets $S_{1},S_{2},\cdots,S_{24}$
other than $S$ in $G$. Now consider the pair of consecutive colors $(25,26)$.
Now, from Lemma 1, there must exists at least one $S_{25}$ other than $S$ such
that color $26$ can not be used in $G_{S_{25}}$. So we need $25$ such
$S_{1},S_{2},\cdots,S_{25}$ other than $S$. But there exists only $24$ such
distinct $S_{j}$s other than $S$ in $G$. Hence from pigeon hole ($25$ pigeons
and $24$ holes) principle there must be at least one $G_{S_{j}}$ where two
different colors used in $G_{S}$ can not be used there and hence
$\lambda^{\prime}_{1,2}(G_{S_{j}})\geq 27$ implying
$\lambda^{\prime}_{1,2}(T_{8})\geq 27$. When $c^{\prime}=1$ the set of colors
$\\{0,2,\cdots 26\\}$ can be partitioned into $12$ disjoint consecutive pairs
of colors $(2,3),\cdots,(24,25)$. Considering these $12$ pairs and the pair of
consecutive colors $(25,26)$, we get $\lambda^{\prime}_{1,2}(T_{8})\geq 27$ by
proceeding similarly as above. When $c^{\prime}=25$ the set of colors
$\\{0,\cdots 24,26\\}$ can be partitioned into $12$ disjoint consecutive pairs
of colors $(1,2,),\cdots,(23,24)$. Considering these $12$ pairs and the pair
of consecutive colors $(0,1)$, we get $\lambda^{\prime}_{1,2}(T_{8})\geq 27$
by proceeding similarly as above.

∎

Observe that a $K_{3}$ contains one horizontal, one vertical and one slanting
edge. Note that three consecutive colors $c-1,c,c+1$ can be used at three
edges with or without forming a $K_{3}$ (complete graph of $3$ vertices) in
$G_{S}$, where $c$ being used at a slanting or non slanting edge. Accordingly,
we now have the following two Lemmas.

Figure 4: The all possible cases where $c$ is used in a slanting edge $e$ and
both $c\pm 1$ are used at edges forming $45^{\circ}$ with $e$.

###### Lemma 2

When three consecutive colors $c-1$, $c$ and $c+1$ are used at three edges
forming a $K_{3}$ in $G_{S}$ with $c$ being used at the slanting edge, then
there exists a $K_{4}$ with vertex set $S_{1}$ in $G$ such that $c$ can not be
used in $G_{S_{1}}$. When three consecutive colors $c-1$, $c$ and $c+1$ are
used at three edges without forming a $K_{3}$ in $G_{S}$ with $c$ being used
at the slanting edge, then there exists at least $2$ different $K_{4}$s with
vertex sets $S_{1}$ and $S_{2}$ in $G$ such that $c$ can not be used at
$G_{S_{1}}$ and $G_{S_{2}}$.

###### Proof

Let us consider $c-1$, $c$ and $c+1$ are used at three edges which are forming
a $K_{3}$ with $c$ being used at the slanting edge. The different cases where
such scenarios occur are shown in Fig. 4 a. to Fig. 4 d. Note that all the
cases shown in Fig. 4 a. to Fig. 4 d. are symmetric. So we consider any one of
the cases. Consider the case where the colors $c-1$, $c$ and $c+1$ are used at
three edges $(v_{1},v_{2})$, $(v_{1},v_{3})$ and $(v_{2},v_{3})$ respectively
(Fig. 3). Now consider the $K_{4}$ with vertex set
$S_{1}=\\{v_{2},u_{6},u_{7},u_{8}\\}$ and observe that any edge incident to
any of the vertices in $S_{1}\setminus\\{v_{2}\\}$ is at distance two from
$(v_{1},v_{2})$. As $f^{\prime}(v_{1},v_{2})=c-1$, $c$ can not be used at
those edges for $L(1,2)$ edge labeling. As any edge incident to $v_{2}$ is at
distance at most $2$ from $(v_{1},v_{3})$ and $f^{\prime}(v_{1},v_{3})=c$, the
color $c$ can not be used there as well. So $c$ can not be used at
$G_{S_{1}}$.

Now we consider the case where $c-1$, $c$ and $c+1$ are used at three edges
which are not forming a $K_{3}$ with $c$ being used at the slanting edge. Note
that the colors $c-1$ and $c+1$ can be used at two edges both of which are at
$45^{\circ}$ with the edge having color $c$. The different cases where such
scenarios arrive are shown in Fig. 4 e. to Fig. 4 l. Note that the cases shown
in Fig. 4 e. to Fig. 4 h. are symmetric and the cases shown in Fig. 4 i. to
Fig. 4 l. are also symmetric. So here we need to consider only two cases.

  * •

Consider the case where the colors $c-1$, $c$ and $c+1$ are used at three
edges $(v_{1},v_{4})$, $(v_{1},v_{3})$ and $(v_{2},v_{3})$ respectively (Fig.
3). In this case, the said three edges form a structure isomorphic to Fig.4 e.
to Fig.4 h. Here note that there exists two $K_{4}$s with vertex sets
$S_{1}=\\{v_{4},u_{1},u_{2},u_{12}\\}$ and
$S_{2}=\\{v_{2},u_{6},u_{7},u_{8}\\}$ such that $c$ can not be used in
$G_{S_{1}}$ and $G_{S_{2}}$.

  * •

Consider the case where the colors $c-1$, $c$ and $c+1$ are used at three
edges $(v_{1},v_{4})$, $(v_{1},v_{3})$ and $(v_{1},v_{2})$ respectively (Fig.
3). In this case the said three edges form a structure isomorphic to Fig. 4 i.
to Fig. 4 l. Here also, there exists two $K_{4}$s with vertex sets
$S_{1}=\\{v_{4},u_{1},u_{2},u_{12}\\}$ and
$S_{2}=\\{v_{2},u_{6},u_{7},u_{8}\\}$ such that $c$ can not in $G_{S_{1}}$ and
$G_{S_{2}}$.

Now we consider the case where the edge having color $c+1$ (or $c-1$) is not
forming $45^{\circ}$ with the slanting edge having color $c$ and the other
edge having color $c-1$ (or $c+1$) forming $45^{\circ}$ with the slanting
edge. In that case, from Lemma 1, it follows that there exists at least $2$
distinct $K_{4}$s having vertex sets $S_{1}$ and $S_{2}$ other than $S$ such
that $c$ can not be used in $G_{S_{1}}$ and $G_{S_{2}}$. As the other color
$c-1$ (or $c+1$) is used at an edge forming $45^{\circ}$ with the slanting
edge, there also exists at least another $K_{4}$ having vertex set $S_{3}$
other than $S$ such that $c$ can not be used in $G_{S_{3}}$. So in this case
there are at least $3$ distinct $K_{4}$s.

We now consider the case where both the colors $c-1$ and $c+1$ both are used
at edges not forming $45^{\circ}$ with the slanting edge with color $c$. In
this case, from Lemma 1, there exists at least $4$ different $K_{4}$s having
vertex sets $S_{1}$, $S_{2}$, $S_{3}$ and $S_{4}$ other than $S$ such that $c$
can not be used in $G_{S_{1}}$, $G_{S_{2}}$, $G_{S_{3}}$ and $G_{S_{4}}$. So
in this case there are at least $4$ distinct $K_{4}$s.

∎

###### Lemma 3

If three consecutive colors $c-1$, $c$ and $c+1$ are used at three edges of
$G_{S}$ forming a $K_{3}$ with $c$ being used at a non slanting edge, then
there exists two $K_{4}$s with vertex sets $S_{1}$ and $S_{2}$ in $G$ such
that $c$ can not be used in $G_{S_{1}}$ and $G_{S_{2}}$. If three consecutive
colors $c-1$, $c$ and $c+1$ are used at three edges of $G_{S}$ without forming
a $K_{3}$ with $c$ being used at a non slanting edge, then there exists at
least $3$ $K_{4}$s with vertex sets $S_{1}$, $S_{2}$ and $S_{3}$ in $G$ such
that $c$ can not be used in $G_{S_{1}}$, $G_{S_{2}}$ and $G_{S_{2}}$.

###### Proof

  * •

Let us first consider the case where $c-1$, $c$ and $c+1$ are used at three
edges which are forming a $K_{3}$ with $c$ being used at the non slanting
edge. Consider the $K_{3}$ with vertex set $\\{v_{1},v_{2},v_{3}\\}$ in
$G_{S}$ as shown in Fig. 3. Suppose the colors $c-1$, $c$ and $c+1$ are used
at $(v_{1},v_{2})$, $(v_{2},v_{3})$ and $(v_{1},v_{3})$ respectively. In this
case, there exists two $K_{4}$s with vertex sets
$S_{1}=\\{v_{1},u_{9},u_{10},u_{11}\\}$ and
$S_{2}=\\{v_{1},v_{4},u_{11},u_{12}\\}$ such that $c$ can not be used at
$G_{S_{1}}$ and $G_{S_{2}}$.

  * •

Now we consider the case where $c-1$, $c$ and $c+1$ are used at three edges
which are not forming a $K_{3}$ with $c$ being used at the non slanting edge.
In the previous case, as the two edges with colors $c-1$ and $c+1$ have a
common vertex other than the end vertices of the edge with color $c$, we get
only two different $K_{4}$s. Note that if three edges do not form a $K_{3}$,
there does not exists a common vertex of the two edges with colors $c-1$ and
$c+1$, other than an end vertex of the edge with color $c$. In that case,
there exists at least $3$ different $K_{4}$s having vertex sets $S_{1}$,
$S_{2}$ and $S_{3}$ such that $c$ can not be used at $G_{S_{1}}$, $G_{S_{2}}$
and $G_{S_{3}}$.

∎

We now consider how many $K_{3}$S can be there in $G_{S}$ such that for each
such $K_{3}$ three consecutive colors can be used. For this, we have the
following Lemma.

###### Observation 1

Consider the $K_{4}$ having vertex set $S=\\{v_{1},v_{2},v_{3},v_{4}\\}$ and
the subgraph $G_{S}$. There can be at most $8$ $K_{3}$s in $G_{S}$ such that
in each of the $K_{3}$, three consecutive colors can be used.

###### Proof

Consider the edges $e_{1}=(v_{1},v_{2})$, $e_{2}=(v_{2},v_{3})$,
$e_{3}=(v_{3},v_{4})$ and $e_{4}=(v_{4},v_{1})$ of the $K_{4}$ having vertex
set $S=\\{v_{1},v_{2},v_{3},v_{4}\\}$ (Fig. 2). Observe that there are total
$12$ $K_{3}$s in $G_{S}$ and each $K_{3}$ has at least one edge in
$\\{e_{1},e_{2},e_{3},e_{4}\\}$. Each edge $e\in\\{e_{1},e_{2},e_{3},e_{4}\\}$
is a common edge of $4$ different $K_{3}$s. Let a color $c$ be used in an edge
$e\in\\{e_{1},e_{2},e_{3},e_{4}\\}$. In order to form a $K_{3}$ with three
consecutive colors with $e$, either $c+1$ or $c-1$ must be used in that
$K_{3}$. So, out of the $4$ $K_{3}$s that include $e$, at most two of then can
have $3$ consecutive colors. As there are $4$ edges in
$\\{e_{1},e_{2},e_{3},e_{4}\\}$, we can have at most $8$ $K_{3}$s where each
of them has $3$ consecutive colors.

∎

Now we state and prove the following Theorem.

###### Theorem 2.3

$\lambda^{\prime}_{1,2}(T_{8})\geq 28$.

###### Proof

Consider the graph $G$, the $K_{4}$ with vertex set
$S=\\{v_{1},v_{2},v_{3},v_{4}\\}$ and the subgraph $G_{S}$ as shown in Fig. 3.
Note that $26$ distinct colors from $\\{0,1,\cdots,27\\}$ must be used in
$G_{S}$. In other words, there must be $2$ colors in $\\{0,1,\cdots,27\\}$
which should remain unused. Let these two colors be $c_{1}$ and $c_{2}$ where
$c_{1}<c_{2}$. Let us now consider the $6$ colors $0$, $c_{1}-1$, $c_{1}+1$,
$c_{2}-1$, $c_{2}+1$ and $27$.

We first consider the case when all these $6$ colors are distinct and denote
$X=\\{0,c_{1}-1,c_{1}+1,c_{2}-1,c_{2}+127\\}$. In this case, for each color
$c\in X$, only one of $c\pm 1$ is used and the other is not used in $G_{S}$.
From Lemma 1, for each $c\in X$, there exists at least one $K_{4}$ having
vertex set $S_{1}$ other than $S$ in $G$ such that $c$ can not be used at
$G_{S_{1}}$. Moreover, $c$ must be used at a slanting edge in $G_{S}$ in this
case. Considering all $6$ colors in $X$ are used in $6$ slanting edges, we get
at least $6$ such $K_{4}$s. For each $c$ among the remaining $26-6=20$ colors,
both $c\pm 1$ are used in $G_{S}$. There are total $26$ edges in $G_{S}$ among
which $14$ are slanting edges and $12$ are non-slanting edges. Therefore, we
are yet to consider the colors used in the remaining $14-6=8$ slanting edges.
Note that for each such color $c$, both $c\pm 1$ are used in $G_{S}$. Assume
these $8$ colors, $x$ many colors
$c^{\prime}_{1},c^{\prime}_{2},\cdots,c^{\prime}_{x}$ are there such that for
each $c^{\prime}_{i}$, three consecutive colors $c^{\prime}_{i}-1$,
$c^{\prime}_{i}$, $c^{\prime}_{i}+1$ can be used in a $K_{3}$. From Lemma 2,
for each such color $c^{\prime}_{i}$, there exists at least one $K_{4}$ having
vertex set $S_{1}$ other than $S$ in $G$ such that $c^{\prime}_{i}$ can not be
used at $G_{S_{1}}$. Considering all these $x$ colors, we get at least $x$
such $K_{4}$s in $G$. Now consider the remaining $8-x$ colors
$c^{\prime}_{x+1},c^{\prime}_{x+2},\cdots,c^{\prime}_{8}$ used in slanting
edges. Note that for each such $c^{\prime}_{i}$, three consecutive colors
$c^{\prime}_{i}-1$, $c^{\prime}_{i}$ and $c^{\prime}_{i}+1$ can not be used in
a $K_{3}$. From Lemma 2, for each such $c^{\prime}_{i}$, there exists at least
$2$ different $K_{4}$s having vertex sets $S_{1}$ and $S_{2}$ other than $S$
in $G$ such that $c^{\prime}_{i}$ can not be used at $G_{S_{1}}$ and
$G_{S_{2}}$. Considering all these $8-x$ colors, we get at least $2(8-x)$
$K_{4}$s in $G$.

Now consider the $12$ non slanting edges in $G_{S}$. Assume that among them,
$y$ many colors
$c^{\prime\prime}_{1},c^{\prime\prime}_{2},\cdots,c^{\prime\prime}_{y}$ are
there such that for each $c^{\prime\prime}_{i}$, three consecutive colors
$c^{\prime\prime}_{i}-1$, $c^{\prime\prime}_{i}$, $c^{\prime\prime}_{i}+1$ can
be used in a $K_{3}$. Clearly all those $y$ many $K_{3}$s must be different
from those $x$ many $K_{3}$ considered for slanting edges. From Observation 1,
there exists at most $8$ $K_{3}$s in $G_{S}$ such that for each of them, three
consecutive colors can be used. As $x$ many $K_{3}$s have already been
considered for slanting edges, $y$ can be at most $8-x$. From Lemma3, for each
such $c^{\prime\prime}_{i}$, there exists at least $2$ different $K_{4}$s
having vertex sets $S_{1}$ and $S_{2}$ other than $S$ such that
$c^{\prime\prime}_{i}$ can not be used at $G_{S_{1}}$ and $G_{S_{2}}$.
Considering all these $8-x$ colors, we get at least $2(8-x)$ $K_{4}$s. We are
yet to consider the remaining $z=12-(8-x)=4+x$ non slanting edges. For each
such color $c^{\prime\prime}$, the colors $c^{\prime\prime}-1$,
$c^{\prime\prime}$ and $c^{\prime\prime}+1$ can not be used in a $K_{3}$ in
$G_{S}$. So from Lemma3, for each such $c^{\prime\prime}$, there exists at
least $3$ different $K_{4}$s having vertex sets $S_{1}$, $S_{2}$ and $S_{3}$
other than $S$ in $G$ such that $c^{\prime\prime}$ can not be used at
$G_{S_{1}}$, $G_{S_{2}}$ and $G_{S_{3}}$. Considering all these $4+x$ colors,
we get at least $3(4+x)$ $K_{4}$s. In total we get at least
$6+x+2(8-x)+2(8-x)+3(4+x)=50$ $K_{4}$s in $G$. But there are only $24$
distinct $K_{4}$s in $G$ other than $S$. From pigeon hole principle ($50$
pigeons and $24$ holes), there exists at least one $S_{i}$ in $G$ such that at
least $3$ colors which are used in $G_{S}$ can not be used in $G_{S_{i}}$ and
hence $\lambda^{\prime}_{1,2}(G_{S_{i}})\geq 28$ implying
$\lambda^{\prime}_{1,2}(T_{8})\geq 28$.

If the colors $0$, $c_{1}-1$, $c_{1}+1$, $c_{2}-1$, $c_{2}+1$ and $27$ are not
distinct, proceeding similarly, we can show that there are a need of more that
$50$ $K_{4}$s in $G$ and hence from pigeon hole principle (more than $50$
pigeons and $24$ holes) $\lambda^{\prime}_{1,2}(T_{8})\geq 28$.

∎

##  3 Conclusion

It was proved in [3] that $25\leq\lambda^{\prime}_{1,2}(T_{8})\leq 28$ with a
gap between the lower and upper bounds. In this paper, we filled the gap and
proved that $\lambda^{\prime}_{1,2}(T_{8})\geq 28$. This essentially implies
$\lambda^{\prime}_{1,2}(T_{8})=28$.

## References

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# Incorporating tunability of shear thickening transitions into a universal
scaling framework

††preprint: APS/123-QED

Incorporating tunability of shear thickening transitions into a universal
scaling framework

Supplementary Materials

##  S1 Sample Preparation and Experimental Protocol

The samples were prepared by weighing out the cornstarch (Argo) and glycerol
(Sigma-Aldrich). We used cornstarch suspensions as they are ubiquitous in the
study of shear thickening suspensions. Glycerol is used as a solvent as it is
less easily absorbed by the cornstarch and can be easily loaded into the small
gap in the double gap couette geometry. The volume fraction was calculated as

$\phi=\frac{\rho_{g}\phi_{M}}{\rho_{g}\phi_{M}+(1-\phi_{M})\rho_{s}}$ (S1)

where $\phi$ is the volume fraction, $\phi_{M}$ is the mass fraction of
cornstarch, $\rho_{g}$ is the density of glycerol, and $\rho_{s}=1.62$g/cm3 is
the density of the cornstarch. The suspension is mixed manually with a spatula
for 10 minutes and then used immediately.

An TA instruments, ARES-G2 strain controlled rheometer is used with a double
wall concentric cylinder geometry and the OSP flow. The sample was presheared
at a constant strain rate of 1$s^{-1}$ for five minutes. The sample was then
sheared with only a primary shear at shear rates ranging from $0.01s^{-1}$ to
$10s^{-1}$. For the higher volume fractions, the strain rate ramp was stopped
when the stress exceeded the maximum stress allowed by the rheometer.
Following the steady shear ramp, the orthogonal oscillatory shear is applied.
At each primary shear rate, orthogonal shear is applied with a constant strain
amplitude of $0.05\%$, and at angular frequencies from $0.05s^{-1}$ to 100
$s^{-1}$. The viscosity, measured as the stress along the axial direction
divided by the primary shear rate is reported in the paper.

##  S2  Estimation of $\phi_{0}$

Figure S1: Estimation of $\phi_{0}$. Plot of $\eta_{min}^{-1/2}$ as a function
of the volume fraction $\phi$, where $\eta_{min}$ is the viscosity prior to
shear thickening for the steady shear system. The red line is a linear fit to
the data and the x intercept is the estimated value of $\phi_{0}$.

We use the low strain rate viscosity with the primary shear rate alone across
all volume fractions $\eta_{{Min}}$ to estimate the value of $\phi_{0}$. It
has been shown previously that the low stress viscosity prior to shear
thickening diverges at $\phi_{0}$ as $\eta\sim 1/(\phi_{0}-\phi)^{2}$. Thus,
we plot $1/\sqrt{\eta_{Min}}$ as a function of $\phi$, fit it to a straight
line. The x intercept of the line is is used as $\phi_{0}$ (Fig. S1). We find
that this value of $\phi_{0}$ collapses the data well in the regime of small
values of the scaling variable $\tilde{x}$.

##  S3 Scaling collapse

Figure S2:  Scaling collapse of data binned at constant $\dot{\Gamma}$ (a)
Independent scaling collapse of data at each value of $\dot{\Gamma}$,
following the protocol in ramaswamy2021universal . The scaling variable,
$x_{1}(\sigma,\phi)=e^{-\sigma^{*}/\sigma}C(\phi)/(\phi_{0}-\phi)$, can be
used to get excellent scaling collapse of the data to generate the scaling
function, $\mathcal{F}=\eta(\phi_{0}-\phi)^{2}$ for all values of
$\dot{\Gamma}$. (b) The anisotropy factor $C(\phi)$ used for the scaling
collapse of the data. Figure S3:  Scaling collapse of data with an exponential
$f(\sigma)$ (a). Failed scaling of data across all volume fractions, $\phi$,
primary strain rates, $\dot{\gamma}_{P}$ and normalized strain rates
$\dot{\Gamma}$, using previously determined scaling variable
ramaswamy2021universal  that is independent of $\dot{\Gamma}$,
$x_{1}(\sigma,\phi)=e^{-\sigma^{*}/\sigma}C(\phi)/(\phi_{0}-\phi)$. The
scaling function is $\mathcal{F}=\eta(\phi_{0}-\phi)^{2}$. As expected, data
at a given stress and volume fraction now has the same value of the scaling
variable, irrespective of $\dot{\Gamma}$, and as such, the dethickening in the
viscosity appears to occur at a constant $x_{1}$, evident in the lack of
scaling collapse of the data. (b) Multiplicative function of $\dot{\Gamma}$,
$g(\dot{\Gamma})$, used to better collapse the data. The blue data is the
point by point data that best collapses the data. The red line is an
exponential function of the normalized strain rate
$e^{-\dot{\Gamma}/\dot{\Gamma}^{*}}$ that captures the transition in
$g(\dot{\Gamma})$ with increasing $\dot{\Gamma}$. (c) Attempted scaling
collapse with a new scaling variable,
$x_{2}(\sigma,\dot{\Gamma},\phi)=e^{-\sigma^{*}/\sigma}g(\dot{\Gamma})C(\phi)/(\phi_{0}-\phi)$,
that depends on $\dot{\Gamma}$. We find that while the scaling collapse is
better than in Fig S3(a), the collapse is not perfect. In particular, we note
that the data at different primary strain rates diverge at different values of
the scaling variable $x_{2}(\sigma,\dot{\Gamma},\phi)$ Figure S4: Scaling
collapse of viscosity data at each volume fraction The figures on the left
show the failed scaling where the scaling variable is
$x_{2}(\sigma,\phi,\dot{\Gamma})=C(\phi)e^{-\sigma^{*}/\sigma}e^{-\dot{\Gamma}/\dot{\Gamma}^{*}}/(\phi_{0}-\phi)$
and the scaling function $\mathcal{F}=\eta(\phi_{0}-\phi)^{2}$. We can clearly
see that the data at different primary stresses diverge at different points,
indicating that a stress dependent modification is required to the scaling
variable. The figures on the right show the excellent scaling collapse
obtained when $f(\sigma)$ was changed to a stretched exponential, where
$\tilde{x}(\sigma,\phi,\dot{\Gamma})=C(\phi)e^{-(\sigma^{*}/\sigma)^{0.75}}e^{-\dot{\Gamma}/\dot{\Gamma}^{*}}/(\phi_{0}-\phi)$
and the scaling function $\mathcal{F}=\eta(\phi_{0}-\phi)^{2}$. The different
colours indicate the different stresses, with blue as the largest primary
stress and green is the smallest primary stress.

We first fit for the global parameter $\sigma^{*}$ by fitting the raw
viscosity data to the Wyart and Cates model. Using this value of $\sigma^{*}$,
we can calculate $f(\sigma)=e^{-\sigma^{*}/\sigma}$ for all packing fractions
and stresses. Importantly, the stress in this expression is the stress in the
absence of any orthogonal flow. Thus the stress used in the scaling collapse
is $\sigma=\eta_{P}\dot{\gamma}_{P}$.

To collapse the data, we first attempt a scaling collapse for the steady state
data $\dot{\Gamma}=0$. A volume fraction dependent function, $C(\phi)$, (Fig.
S2(b)) is then determined that best collapses the data across different volume
fractions onto a single curve (Fig. S2(a) upper left panel). As expected, we
find excellent scaling collapse across several decades in the suspension
viscosity and the scaling variable. We then bin the data into bins of constant
$\dot{\Gamma}$. We find similarly good collapse at other OSP strain rates and
the scaled curves for $\dot{\Gamma}=1,4,$ and $10$ are shown in the remaining
panels of Fig. S2(a) using the same values of $C(\phi)$ and the same function
$f(\sigma)$.

Though similar, the curves for different values of $\dot{\Gamma}$ do not
collapse (Fig. S3(a)), indicating that a scaling variable that depends only on
$\sigma$ and $\phi$ is not sufficient for capturing the dethickening behavior.
We thus extend the scaling variable to account for the decrease in the
viscosity to OSP flow. To construct $g(\dot{\Gamma})$, we manually shift the
curves for each $\dot{\Gamma}$ onto the $\dot{\Gamma}=0$ curve. We plot the
shift, $g(\dot{\Gamma})$ versus $\dot{\Gamma}$ in Fig. S3(b) (blue points). We
find that the behaviour of $g(\dot{\Gamma})$ is well captured by the
functional form $g(\dot{\Gamma})=e^{-\dot{\Gamma}/16}$. Incorporating this
functional form into $x_{2}$,

$x_{2}(\sigma,\phi,\dot{\Gamma})=\frac{C(\phi)e^{-\sigma^{*}/\sigma}e^{-\dot{\Gamma}/\dot{\Gamma}^{*}}}{(\phi_{0}-\phi)}$
(S2)

we attempt a scaling collapse for the entire data set (Fig. S3(c)). While much
improved, we find that data at different primary stresses diverge at different
values of $x_{2}$. This deviation is evident especially when the scaling
collapse is attempted with each volume fraction independently (Fig. S4).

The systematic stress dependent deviation in the scaling collapse suggests
that our form for $f(\sigma)$ should be modified. To determine this modified
form, we shift the curves for each volume fraction and applied stress onto a
single curve. We find that there is no dependence on the volume fraction and
that the entire range is well described by a stretched exponential of the
form, $f(\sigma)=-e^{-(\sigma^{*}/\sigma)^{0.75}}$. This stretched exponential
form results in excellent scaling collapse for data at each volume fraction
(Fig. S4). Using this modified form for $f(\sigma)$, we determine again the
best value of $C(\phi)$ that collapses the data (see main text for the final
$C(\phi)$ and $f(\sigma)$). With this new form of $f(\sigma)$ and $C(\phi)$
and a scaling variable

$\tilde{x}(\sigma,\dot{\Gamma},\phi)=\frac{f(\sigma)g(\dot{\Gamma})C(\phi)}{(\phi_{0}-\phi)}\equiv\frac{\zeta(\sigma,\dot{\Gamma},\phi)}{(\phi_{0}-\phi)}$
(S3)

we find excellent scaling collapse across all the volume fractions, strain
rates and OSP flows.

##  S4  Determining the scaling exponent

Figure S5: Power law scaling of cornstarch viscosity data. The scaling
function $\mathcal{F}=\eta(\phi_{0}-\phi)^{2}$ as a function of the
$x_{c}-\tilde{x}$ where the scaling variable
$\tilde{x}=e^{-(\sigma^{*}/\sigma)^{0.75}}e^{-\dot{\Gamma}/\dot{\Gamma}_{0}}C(\phi)/(\phi_{0}-\phi)$.
We find that the data shows power law scaling with an exponent of $\sim-1.5$.
The black solid line is a line with power law -1.5.

From the scaling collapse of the data shown in Fig, 3b, the scaling function
$\mathcal{F}$ appears to diverge at $x_{c}$: $\mathcal{F}\sim
1/(x_{c}-\tilde{x})^{\delta}$. To determine the scaling exponent $\delta$, we
plot the scaling function $\mathcal{F}=\eta(\phi_{0}-\phi)^{2}$ as a function
of $x_{c}-\tilde{x}$ (Fig. S5). We find that the divergence is indeed a power
law, with $\delta\neq 2$. The best estimate for $\delta\approx-3/2$. However,
since the data is still reasonably far from $x_{c}$, this value of $\delta$
should not be taken as a fixed parameter.

##  S5  Shear Jamming Line Calculation

The shear jamming line is given by the divergence of the scaling function
$\mathcal{F}$ at $\tilde{x}=x_{c}$. The scaling variable depends on the
stress, volume fraction and the OSP flow rate:

$\tilde{x}=\frac{f(\sigma)C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}$ (S4)

where $f(\sigma)=e^{-(\sigma^{*}/\sigma)^{0.75}}$ and
$g(\dot{\Gamma})=e^{-\dot{\Gamma}/\dot{\Gamma}_{0}}$ Since $C(\phi)$ is fit
for each volume fraction, we estimate that value of the $C(\phi)$ at high
volume fraction by fitting the data points at the highest three volume
fractions to a straight line:

$C(\phi)=c_{1}\phi+c_{2}$ (S5)

We can then determine the shear jamming volume fraction for each $\sigma$ and
$\dot{\Gamma}$ as:

$\phi_{J,\sigma,\phi}=\frac{x_{c}\phi_{0}-c_{2}g(\dot{\Gamma})f(\sigma)}{c_{1}f(\sigma)g(\dot{\Gamma})+x_{c}}$
(S6)

For the given value of $C(\phi)$, $x_{c}=6.1$. The values of $c_{1}$ and
$c_{2}$ used to generate the phase diagram are listed in Table 1.

Figure S6:  Fits to the anisotropy $C(\phi)$. The anisotropy factor $C(\phi)$
and the linear fit at high packing fraction (red) used to generate the phase
diagram.

##  S6 Estimating the effect of the OSP flow on the DST region:

The DST line is calculated as

$\frac{d\log\eta}{d\log\sigma}=1$ (S7)

or,

$\frac{\sigma}{\eta}\frac{d\eta}{d\sigma}=1$ (S8)

.

Using the scaling form of the viscosity -

$\eta(\phi_{0}-\phi)^{2}=\mathcal{F}\left(\tilde{x}\right)$ (S9)

, where

$\tilde{x}=\frac{f(\sigma)g(\dot{\Gamma})C(\phi)}{\phi_{0}-\phi}$ (S10)

Taking derivatives,

$\frac{\sigma}{\eta}\frac{d\eta}{d\sigma}=\frac{\sigma}{\eta}\frac{1}{(\phi_{0}-\phi)^{2}}\frac{d\mathcal{F}}{d\tilde{x}}\frac{df(\sigma)}{d\sigma}\frac{C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}=1$
(S11)

or

$\frac{\sigma}{\frac{\mathcal{F}}{(\phi_{0}-\phi)^{2}}}\frac{1}{(\phi_{0}-\phi)^{2}}\frac{d\mathcal{F}}{d\tilde{x}}\frac{df(\sigma)}{d\sigma}\frac{C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}=1$
(S12)

$\frac{\sigma}{\mathcal{F}}\frac{d\mathcal{F}}{d\tilde{x}}\frac{df(\sigma)}{d\sigma}\frac{C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}=1$
(S13)

Fig S5 suggests that we can estimate
$\mathcal{F}(\tilde{x})=(x_{c}-\tilde{x})^{\delta}$. In the limit that
$g(\dot{\Gamma})\rightarrow 0$, the left hand side of the equation

$\sigma\frac{1}{x_{c}}\frac{df(\sigma)}{d\sigma}\frac{C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}\rightarrow
0$ (S14)

As such, we expect that as the OSP flow increases and $g(\dot{\Gamma})$
decreases, the DST region gets progressively smaller and vanishes. This
behaviour is further reinforced by the calculations performed with the
estimate for the crossover scaling function $\mathcal{H}$.

##  S7 Comparisons with previous scaling collapse

Figure S7:  Cardy Scaling with fit from previous scaling function. The scaling
function $\mathcal{H}=\eta(\zeta(\sigma,\dot{\Gamma},\phi))^{2}$ versus
$|1/x_{c}-1/\tilde{x}|$ where
$\zeta(\sigma,\dot{\Gamma},\phi)=f(\sigma)g(\dot{\Gamma})C(\phi)$ as in Fig.
3(c). The overlaid pink line is the scaling function from
ramaswamy2021universal , scaled by 1.7 to match the y axis and matching
$x_{c}$.

The scaling collapse in Fig. 3(c) demonstrates a change in the exponent,
however, we don’t have data at sufficiently high stresses and volume fractions
to fully determine the scaling collapse. However, we anticipate that the
scaling function to be universal suggesting that the we should be able to
overlay the scaling function obtained from the steady shear data in
ramaswamy2021universal . We overlay the crossover between two power laws fit
to $\mathcal{H}$ onto the Cardy scaling with the OSP flow. To overlay the two
curves, we match the values of $x_{c}$ by multiplying $C(\phi)$ by a constant.
In addition we find that we need to multiply the fit by a constant factor of
1.7 to best overlap the two functions. We find that the scaling functions
obtained with the scaling collapse of the OSP flow data is consistent with the
scaling function obtained in the steady shear data in ramaswamy2021universal .

We can use the previous functional form of $\mathcal{H}$ to explore in more
detail how the DST line changes with $g(\dot{\Gamma})$. We determine the
derivative the DST line as

$\frac{d\log\eta}{d\log\sigma}=1$ (S15)

or,

$\frac{\sigma}{\eta}\frac{d\eta}{d\sigma}=1$ (S16)

From the Cardy scaling collapse, we can write:

$\eta(C(\phi)f(\sigma)g(\dot{\Gamma}))^{2}=\mathcal{H}(1/\tilde{x}-1/x_{c})$
(S17)

Taking the derivative of the viscosity,

$\frac{\sigma}{\eta}\frac{d\eta}{d\sigma}=\frac{\sigma}{\eta}\frac{1}{(C(\phi)g(\dot{\Gamma}))^{2}}\left(\frac{-2}{f^{3}}\mathcal{H}\frac{df}{d\sigma}+\frac{1}{f^{2}}\frac{d\mathcal{H}}{d\sigma}\right)=1$
(S18)

Noting that $\mathcal{H}$ is a function of $\left(1/\tilde{x}-1/x_{c}\right)$,

$\frac{\sigma}{\eta}\frac{d\eta}{d\sigma}=\frac{\sigma}{\eta}\frac{1}{(C(\phi)g(\dot{\Gamma}))^{2}}\left(\frac{-2}{f^{3}}\mathcal{H}\frac{df}{d\sigma}+\frac{1}{f^{2}}\frac{d\mathcal{H}\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}{d\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}\frac{d\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}{d\sigma}\right)=1$
(S19)
$\frac{\sigma}{\eta}\frac{d\eta}{d\sigma}=\frac{\sigma}{\eta}\frac{1}{(C(\phi)g(\dot{\Gamma}))^{2}}\left(\frac{-2}{f^{3}}\mathcal{H}\frac{df}{d\sigma}+\frac{1}{f^{2}}\frac{d\mathcal{H}\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}{d\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}\frac{-1}{\tilde{x}^{2}}\frac{d\tilde{x}}{d\sigma}\right)=1$
(S20)
$\frac{\sigma}{\eta}\frac{d\eta}{d\sigma}=\frac{\sigma}{\eta}\frac{1}{(C(\phi)g(\dot{\Gamma}))^{2}}\left(\frac{-2}{f^{3}}\mathcal{H}\frac{df}{d\sigma}+\frac{1}{f^{2}}\frac{d\mathcal{H}\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}{d\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}\frac{-1}{\tilde{x}^{2}}\frac{C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}\frac{df}{d\sigma}\right)=1$
(S21)

$\frac{\sigma}{\eta}\frac{d\eta}{d\sigma}=\frac{\sigma}{\eta}\frac{1}{(C(\phi)g(\dot{\Gamma})f(\sigma))^{2}}\frac{df}{d\sigma}\left(\frac{-2}{f}\mathcal{H}+\frac{d\mathcal{H}\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}{d\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}\frac{-1}{\tilde{x}^{2}}\frac{C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}\right)=1$
(S22)

Figure S8:  Phase diagrams with the DST Lines as estimated by using the
scaling function obtained in ramaswamy2021universal . The DST lines are shown
with the dashed lines and the shear jamming lines are the solid lines for
$\dot{\Gamma}=0$ (blue), $\dot{\Gamma}=6.3$ (green), $\dot{\Gamma}=40$
(maroon), and $\dot{\Gamma}=1000$ (pink). We see that the DST region decreases
as the $\dot{\Gamma}$ increases.

Substituting for the viscosity,

$\frac{\sigma}{\frac{1}{(C(\phi)g(\dot{\Gamma})f(\sigma))^{2}}\mathcal{H}}\frac{1}{(C(\phi)g(\dot{\Gamma})f(\sigma))^{2}}\frac{df}{d\sigma}\left(\frac{-2}{f}\mathcal{H}+\frac{d\mathcal{H}\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}{d\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}\frac{-1}{\tilde{x}^{2}}\frac{C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}\right)=1$
(S23)

$\sigma\frac{df}{d\sigma}\left(\frac{-2}{f}+\frac{1}{\mathcal{H}}\frac{d\mathcal{H}\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}{d\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}\frac{-1}{\tilde{x}^{2}}\frac{C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}\right)=1$
(S24)

$\sigma\frac{df}{d\sigma}\left(\frac{-2}{f}+\frac{d\log\mathcal{H}\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}{d\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}\frac{-1}{\tilde{x}^{2}}\frac{C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}\right)=1$
(S25)

$\sigma\frac{df}{d\sigma}\left(\frac{-2}{f}-\frac{1}{\left(\frac{1}{\tilde{x}}-\frac{1}{x_{c}}\right)}\frac{d\log\mathcal{H}(t)}{d\log
t}\frac{1}{\tilde{x}^{2}}\frac{C(\phi)g(\dot{\Gamma})}{(\phi_{0}-\phi)}\right)=1$
(S26)

where $t=1/\tilde{x}-1/x_{c}$. Using the fact that
$f(\sigma)=e^{-(\sigma^{*}/\sigma)^{0.75}}$,

$\frac{df}{d\sigma}=e^{-(\sigma^{*}/\sigma)^{0.75}}(\sigma^{*})^{-0.75}(0.75)(\sigma)^{-1.75}$
(S27)

or

$\frac{df}{d\sigma}=(0.75)f\left(\frac{\sigma^{*}}{\sigma}\right)^{-0.75}\left(\frac{1}{\sigma}\right)$
(S28)

Substituting in Eq. S26,

$(0.75)\left(\frac{\sigma^{*}}{\sigma}\right)^{-0.75}\left(-2-\frac{x_{c}}{(x_{c}-\tilde{x})}\frac{d\log\mathcal{H}(t)}{d\log
t}\right)=1$ (S29)

We can use the fit from the steady state data and substitute for
$\mathcal{H}$:

$\log\mathcal{H}(y)=ay+b+\left(\frac{1+\tanh(e(y-f))}{2}\right)\left((c-a)y+d-b\right)$
(S30)

where $y=\log(1/\tilde{x}-1/x_{c})$, and $a$, $b$, $c$, $d$, $e$ and $f$ are
fitting parameters. We can take the derivative of this function:

$\frac{d\log\mathcal{H}}{dy}=a+((c-a)y+d-b)\frac{1}{2\cosh^{2}(e(y-f))}e+\left(\frac{1+\tanh(e(y-f))}{2}\right)(c-a)$
(S31)

From Eq. S31 and Eq. S29, and the fits for $a$, $b$, $c$, $d$, $e$ and $f$, we
can solve for the DST line at different values of $g(\dot{\Gamma})$ (Fig. S8).
We find that as the OSP flow increases, the DST area shrinks until it is
nonexistent at $\dot{\Gamma}\rightarrow\infty$. The exact parameters used to
generate the phase diagram in Fig. S8 is shown in Table 1.

Parameters | Values
---|---
$C(\phi)$ | -3.5$\phi$ \+ 2.8
$\sigma^{*}$ | 1.8827
$\phi_{0}$ | 0.688
$x_{c}$ | 6.1
Fit $\mathcal{H}$ \- $a$ | -1.5 $\pm$ 0.3
Fit $\mathcal{H}$ \- $b$ | -1 $\pm$ 1
Fit $\mathcal{H}$ \- $c$ | -2.00 $\pm$ 0.04
Fit $\mathcal{H}$ \- $d$ | -1.3 $\pm$ 0.3
Fit $\mathcal{H}$ \- $e$ | 0.3 $\pm$ 0.2
Fit $\mathcal{H}$ \- $f$ | -3 $\pm$ 2
Table 1: Parameters to determine the shear jamming and DST lines for the phase
diagram.

## References

  * (1) Ramaswamy, M. _et al._ Universal scaling of shear thickening transitions.  _arXiv preprint arXiv:2107.13338_ (2021).