# Disorder-induced currents as signatures of chiral superconductivity

Madhuparna Karmakar<EMAIL_ADDRESS>Department of Physics, Indian
Institute of Technology Madras, Chennai 600036, India  R. Ganesh
<EMAIL_ADDRESS>The Institute of Mathematical Sciences, HBNI, C I T Campus,
Chennai 600113, India

###### Abstract

Chiral superconductors are expected to carry a spontaneous, chiral and
perpetual current along the sample edge. However, despite the availability of
several candidate materials, such a current has not been observed in
experiments. In this article, we suggest an alternative probe in the form of
impurity-induced chiral currents. We first demonstrate that a single non-
magnetic impurity induces an encircling chiral current. Its direction depends
on the chirality of the order parameter and the sign of the impurity
potential. Building on this observation, we consider the case of multiple
impurities, e.g., realized as adatoms deposited on the surface of a candidate
chiral superconductor. We contrast the response that is obtained in two cases:
(a) when the impurities are all identical in sign and (b) when the impurities
have mixed positive and negative signs. The former leads to coherent currents
within the sample, arising from the fusion of individual current loops. The
latter produces loops of random chirality that lead to incoherent local
currents. These two scenarios can be distinguished by measuring the induced
magnetic field using recent probes such as diamond nitrogen-vacancy (NV)
centres. We argue that impurity-induced currents may be easier to observe than
edge currents, as they can be tuned by varying impurity strength,
concentration and correlations. We demonstrate these results using a toy model
for $p_{x}\pm ip_{y}$ superconductivity on a square lattice. We develop an
improved scheme for Bogoliubov deGennes (BdG) simulations where both the order
parameter as well as the magnetic field are determined self-consistently.

##  I Introduction

A superconductor is said to be chiral if it spontaneously breaks time reversal
symmetryKallin and Berlinsky (2016); Wysokiński (2019). Such states have long
been known, starting with liquid He3 which is technically a superfluid and not
a superconductorLeggett  _et al._ (2006). Interest in chiral superconductivity
has grown alongside studies on non-superconducting time-reversal-symmetry-
breaking states such as quantum Hall systems. In both classes of systems,
typical excitation spectra show non-trivial topology. This shared thread has
motivated extensive studies into edge currents. Despite several experimental
studies and theoretical refinements, such a current has hitherto not been
observed in a chiral superconductor. This situation can be contrasted with the
relatively quick detection of edge states in topological insulatorsNowack  _et
al._ (2013) and with the recent detection of an edge current in a topological
non-chiral superconductorWang  _et al._ (2020). In this article, we propose
measuring a different type of current – one induced by impurities. This
provides a direct, observable and tuneable manifestation of chirality.

Several materials have been proposed as candidates for chiral
superconductivity. The most prominent is Sr2RuO4, long thought to be a $p+ip$
superconductor. However, experimental studies have proved inconclusive. Recent
studies have revived this debateLeggett and Liu (2020). Chiral
superconductivity has also been proposed to occur in heavy fermion
materialsMaeno  _et al._ (2012), SrPtAsBiswas  _et al._ (2013), highly doped
grapheneNandkishore  _et al._ (2012); Black-Schaffer and Honerkamp (2014) and
transition metal dichalcogenidesHsu  _et al._ (2017); Ganesh  _et al._ (2014).
In these materials, any evidence for chirality comes from bulk measurements
such as $\mu$SR or the Kerr effect. In some cases, scanning tunnelling probes
have found in-gap states at step edgesJiao  _et al._ (2020). However, an
electrical current at an edge has not been observed so farKirtley  _et al._
(2007); Hicks  _et al._ (2010).

In this backdrop, recent technologies to detect microscopic currents can break
new ground. The most prominent of these uses nitrogen-vacancy (NV) centres –
atomic defects within a diamond probe. A spectroscopic measurement of
excitations at the defect conveys information about the local magnetic
fieldCasola  _et al._ (2018). As NV centres can operate over a wide
temperature range and at nanometre separations from the sample, they can
measure detailed field distributions. This has been used to map the spatial
distribution of a current within grapheneTetienne  _et al._ (2017); Ku  _et
al._ (2020). With regard to edge currents of chiral superconductors, NV
centres may succeed if they are sensitive enough to detect weak currents that
were unobservable with earlier probes. However, edge currents may be
unobservable due to deeper issues such as dissipationMarguerite  _et al._
(2019), disorderAshby and Kallin (2009); Lederer  _et al._ (2014) and
compensating magnetisation texturesImai  _et al._ (2012). Such hurdles cannot
be circumvented easily as edges are simple objects with no tuneable handles –
at least over coarse-grained length scales that are appropriate for mesoscopic
probes. In this article, we describe an alternative signature where the
magnitude and range of chiral currents can be tuned. We discuss signatures in
magnetic field distributions that can be easily measured with probes such as
NV centres.

##  II Toy model for chiral superconductivity

We consider a simple model with electrons hopping on a square lattice. The
Hamiltonian is given by

$\displaystyle H=-t\sum_{\sigma,\langle
ij\rangle}\left\\{e^{i\frac{e}{\hbar}\mathcal{A}_{ij}}c_{i,\sigma}^{\dagger}c_{j,\sigma}+h.c.\right\\}+H_{int},$
(1)

We restrict hoppings to nearest neighbours for simplicity. The hopping
amplitude, $t$, sets the energy scale for the problem. Each hopping element
carriers a Peierls’ phase that encodes the vector potential, with
$\mathcal{A}_{ij}=\int_{i}^{j}\vec{A}(\mathbf{r})\cdot\mathbf{dl}$. We have
denoted the charge of the electron as $e$. The vector potential here
originates from supercurrents in the sample and not from an externally imposed
magnetic field. Its precise form will be discussed below.

We choose an artificial interaction term designed to give rise to chiral
superconductivity,

$\displaystyle H_{int}=-g\sum_{\langle
ij\rangle}\mathcal{D}_{ij}^{\dagger}\mathcal{D}_{ij}.$ (2)

The sum runs over all nearest neighbour pairs of sites. The coupling constant,
$g$, is taken to be positive. We introduce a two-site Cooper pair annihilation
operator,

$\displaystyle\mathcal{D}_{ij}=c_{i,\mu}(\sigma_{3}i\sigma_{2})_{\mu,\mu^{\prime}}c_{j,\mu^{\prime}}=\\{c_{i,\uparrow}c_{j,\downarrow}+c_{i,\downarrow}c_{j,\uparrow}\\},$
(3)

where the $\sigma$’s are Pauli matrices. While a generic triplet Cooper pair
can contain three components with spin $m_{z}=-1,0,1$, we have only retained
the $m_{z}=0$ component in $\mathcal{D}_{ij}$. In order to gain energy from
the interaction term, the Cooper pair operator acquires a non-zero expectation
value with $\langle\mathcal{D}_{ij}\rangle=\Delta_{ij}$, a complex scalar.
However, this expectation value must satisfy certain constraints. From
fermionic anticommutation relations, we see that $\Delta_{ij}=-{\Delta}_{ji}$.
This signifies that the order parameter has $p$-wave symmetry, changing sign
under a $\pi$-rotation. For instance, upon a $\pi$ rotation about a bond
centre, the bond direction changes from $ij$ to $ji$ so that the order
parameter flips sign. Each site has four bonds emanating from it. In order to
lower energy on each of these bonds, the order parameter naturally develops
chiral $p_{x}\pm ip_{y}$ symmetry. Moving counterclockwise around a site, the
phase of the order parameter changes by $\pm\pi/2$ from one bond to the next.

A detailed study of this model can be found in Ref. Clara, 2017. In summary,
the system spontaneously chooses between one of two chiralities, i.e., between
$p_{x}+ip_{y}$ and $p_{x}-ip_{y}$ character. Ordering introduces a
quasiparticle gap in an infinite system (or in one with periodic boundaries),
with the same spectrum for either choice of chirality. However, in the
presence of an edge, we find a sharp signature that distinguishes the two
chiralities. The quasiparticle spectrum hosts a mode within the bulk gap that
propagates along the edge. The direction of propagation is determined by the
chirality of the bulk order parameter. It is this mode that carries the edge
current, providing a signature of chiral superconductivity. As a quasiparticle
mode in a superconductor, this mode is a mixture of particle and hole
components. As a consequence, it does not carry a quantized charge current –
unlike an edge current in a Chern insulator.

###  II.1 Role of impurities

We now introduce impurities into the problem in the form of potentials on a
certain fraction of sites. For simplicity, we assume that each impurity
generates a local potential with a fixed amplitude, denoted as $W$. As for the
sign of the impurity potential, we consider two schemes: (i) correlated
disorder – with every impurity carrying the same sign, and (ii) random
disorder – with each impurity having a random sign. We represent these as

$\displaystyle H_{corr.}$ $\displaystyle=$ $\displaystyle
W\sum_{\\{i\\}}\sum_{\sigma}c_{i,\sigma}^{\dagger}c_{i,\sigma},$ (4)
$\displaystyle H_{rand.}$ $\displaystyle=$ $\displaystyle
W\sum_{\\{i\\}}\delta_{i}\sum_{\sigma}c_{i,\sigma}^{\dagger}c_{i,\sigma}.$ (5)

In both cases, the sum over $i$ runs over a randomly selected subset of sites.
We consider the fraction of impurity sites to be a tuneable parameter, $\eta$.
In the random case, $\delta_{i}=\pm 1$ is an Ising-like random variable that
is independently chosen at each impurity site. These two impurity schemes can
be realized by placing a collection of adatoms on the surface of a chiral
superconductor. If all adatoms are of the same type, this corresponds to the
situation encoded in $H_{corr.}$. Here, the sign of the potential is decided
by the relative electronegativity of the adatom with respect to the substrate.
On the same lines, deposition of two species of adatoms can realize
$H_{rand.}$.

##  III Bogoliubov deGennes simulations

We consider the system on a $L\times L$ lattice with periodic boundary
conditions. We perform a mean-field decomposition of the interaction term in
Eq. 2, by defining bond variables
$\Delta_{ij}\equiv\langle\mathcal{D}_{ij}\rangle$. This allows us to write the
Hamiltonian as a quadratic form with

$\displaystyle\hat{H}=\Psi^{\dagger}M\Psi,$ (6)

where $\Psi=\\{\ldots c_{i,\uparrow}\ldots
c_{i,\downarrow}^{\dagger}\ldots\\}^{T}$. The $2L\times 2L$ Hamiltonian matrix
can be written as

$\displaystyle M=\left(\begin{array}[]{cc}K(t,\vec{A},W)&P(\Delta_{mn})\\\
P(\Delta_{mn})^{\dagger}&-K(t,\vec{A},W)^{T}\end{array}\right).$ (9)

The diagonal blocks, $K(t,\vec{A},W)$ encode hoppings with Peierls’ phases and
impurity potentials. The off-diagonal blocks contain the mean-fields, with
$\Delta_{mn}$ defined on the nearest-neighbour bond connecting sites $m$ and
$n$. In typical Bogoliubov deGennes (BdG) calculations, the superconducting
order parameters are determined using
$\Delta_{mn}\equiv\langle\mathcal{D}_{mn}\rangle$. At the same time, the
vector potential is either ignored or assumed to be fixed by a strong (non-
self-consistent) external field. In this article, we present an improved BdG
scheme where the vector potential is also determined self-consistently.

To determine the vector potential, we note that each bond carries a current.
By adapting the arguments in Ref. Scalapino  _et al._ , 1993, we define the
bond current operator as
${J}_{mn}=\langle(\partial{\hat{H}}/\partial{\mathcal{A}_{mn}})\rangle$, where
$\hat{H}$ is the Hamiltonian of Eq. 6 and $<\cdot>$ represents the expectation
value. Here, $(m,n)$ represents a pair of neighbouring sites. We obtain

$\displaystyle{J}_{mn}$ $\displaystyle\approx$
$\displaystyle\frac{e}{\hbar}\langle\hat{\mathcal{J}}_{mn}\rangle-\left(\frac{e}{\hbar}\right)^{2}\langle\hat{\mathcal{K}}_{mn}\rangle\mathcal{A}_{mn}+\mathcal{O}(\mathcal{A}_{mn}^{2}),$
(10)

where

$\displaystyle\hat{\mathcal{J}}_{mn}$ $\displaystyle=$
$\displaystyle-t\sum_{\sigma}\\{ic_{m,\sigma}^{\dagger}c_{n,\sigma}-ic_{n,\sigma}^{\dagger}c_{m,\sigma}\\},$
(11) $\displaystyle\hat{\mathcal{K}}_{mn}$ $\displaystyle=$
$\displaystyle-t\sum_{\sigma}\\{c_{m,\sigma}^{\dagger}c_{n,\sigma}+c_{n,\sigma}^{\dagger}c_{m,\sigma}\\}.$
(12)

In the absence of a vector potential, $\hat{\mathcal{J}}_{mn}$ and
$\hat{\mathcal{K}}_{mn}$ would encode the bond current and the bond energy
respectively. We neglect $\mathcal{O}(\mathcal{A}_{mn}^{2})$ terms here,
assuming that $\mathcal{A}_{mn}$ is small on every bond. This is justified in
the limit of weak disorder. Currents in our system are induced solely by
disorder. Weak disorder will give rise to weak currents and, in turn, to a
weak vector potential.

We next discuss the vector potential induced by the bond currents of Eq. 12.
Assuming the Coulomb gauge condition ($\vec{\nabla}\cdot\vec{A}=0$), a closed
form expression can be derived for the vector potential induced by a set of
static source currents in two dimensions,

$\displaystyle\vec{A}(\mathbf{r})=\frac{\mu_{0}}{4\pi}\int
d^{2}r^{\prime}\frac{\vec{J}(r^{\prime})}{|\mathbf{r}-\mathbf{r}^{\prime}|},$
(13)

where $\vec{J}(r^{\prime})$ is the current density, i.e., current per unit
cross-sectional length. This relation follows from the Biot-Savart law and the
integral solution to the Poisson’s equation. We adapt this equation to our
lattice problem using

$\displaystyle\mathcal{A}_{ij}=\frac{\mu_{0}}{4\pi}\ell^{2}\left[\sum_{\\{kl\\}\parallel\\{ij\\}}\frac{J_{kl}}{|\mathbf{R}_{kl}-\mathbf{R}_{ij}|}+\frac{J_{ij}}{\delta\ell}\right].$
(14)

Here, $\ell$ represents the lattice constant. The index ${ij}$ represents the
reference bond where the vector potential is to be determined. The position of
the centre of the bond is denoted as $\mathbf{R}_{ij}$. The sum runs over
${kl}$, all other bonds that are parallel to ${ij}$, i.e., if ${ij}$ is a bond
in the $x$-direction, $kl$ runs over all $x$-bonds except ${ij}$. The centre
of each of these bonds is denoted as $\mathbf{R}_{kl}$. We calculate inter-
bond distances ($|\mathbf{R}_{kl}-\mathbf{R}_{ij}|$), in units of $\ell$, the
lattice constant. The last term in the above equation represents the
contribution from the very same bond, $ij$. Naively this represents a singular
contribution as the source current and the target bond coincide, i.e.,
$|\mathbf{R}_{kl}-\mathbf{R}_{ij}|$ vanishes. However, this contribution will
be cut off by microscopic length scales which can be visualized as follows. We
may view each bond as being composed of a bundle of wires, with each wire
corresponding to an orbital overlap. If the wires are slightly displaced from
the bond vector, the singular contribution is regularized. We
phenomenologically model such effects using a cutoff length scale denoted as
$\delta\ell$.

We obtain the order parameters and the vector potential self-consistently.
Starting with random values of $\Delta_{ij}$’s and $\mathcal{A}_{ij}$’s, we
obtain a new set of values as described above. We iterate this process until
self-consistency is achieved. The calculations are carried out while
respecting periodic boundary conditions. Each distance
($\mathbf{R}_{ij}-\mathbf{R}_{kl}$) in Eq. 14 is chosen in accordance with the
periodicity of the lattice. For example, two bonds are placed in neighbouring
unit cells if that gives the lowest distance between them. We note that our
approach does not give consistent results with open boundary conditions. In
the presence of an edge, the vector potential outside the sample cannot be
taken into account. This prevents us from ensuring
$\vec{\nabla}\cdot\vec{A}=0$, a necessary condition for the use of Eq. 13.

We emphasize that our scheme is an improvement over standard BdG approaches.
As the vector potential is determined self-consistently, our approach allows
for ‘screening’ currents. These can be viewed as a manifestation of the
Meissner effect, with any magnetic field diminished by counter-currents that
naturally arise in a superconductor. In the context of Sr2RuO4, it has been
shown that edge currents are indeed strongly screenedMatsumoto and Sigrist
(1999); Furusaki  _et al._ (2001).

###  III.1 Parameters in the simulation

In the BdG scheme outlined above, the input parameters are the hopping
strength ($t$) and the coupling constant ($g$). The vector potentials on
bonds, $\mathcal{A}_{ij}$’s, are the desired output as they immediately yield
the magnetic flux through each plaquette.

The currents and the vector potentials require particular attention as they
are accompanied by dimension-carrying constants. They can be determined in a
consistent fashion as follows. We first assert that the bond currents in Eq.
12 are proportional to $(te/\hbar)$, a combination that carries dimensions of
current. Operationally, we determine the currents up to this factor.
Similarly, we calculate the vector potentials in Eq. 14 up to
$(\mu_{0}\ell/4\pi)(te/\hbar)$, a combination that carries dimensions of the
line integral of the vector potential ($\int\vec{A}\cdot\vec{dl}$). We have a
lower power of $\ell$ as compared to the prefactor in Eq. 14 as the lattice
distances ($|\mathbf{R}_{kl}-\mathbf{R}_{ij}|$) are also proportional to
$\ell$. The factor of $(te/\hbar)$ comes from the current term ($J_{kl}$) on
the right hand side of Eq. 14.

To determine currents and vector potentials, we require numerical values for
two dimensionless parameters:

  1. 1.

$\zeta\equiv(\mu_{0}\ell/4\pi)(te^{2}/\hbar^{2})$: this quantity appears in
two places. The first is in the Peierls’ phases of Eq. 1 where the vector
potential, $\mathcal{A}_{ij}$, is multiplied by $e/\hbar$. The second is the
current expression of Eq. 12, as a prefactor to the second term on the right
hand side. This quantity, $\zeta$, contains two material-dependent parameters
$t$ and $\ell$, apart from fundamental constants. To estimate its numerical
value, we take $\ell=10\AA$ and $t\sim 2.5eV$ so that $\zeta\sim 10^{-4}$.

  2. 2.

$\delta\ell/\ell$: this quantity encodes the regularization parameter in Eq.
14. In our calculations, we assume an ad hoc value for this parameter,
$\delta\ell/\ell=0.1$.

##  IV Response to a single impurity

We first consider the case of a single impurity in the chiral superconductor.
Before discussing our results, we review some basic ideas. Chirality allows
for a spontaneous local current that circulates around an impurity. This can
be understood as follows. We imagine a hole cut into the superconductor. As
the superconductor is chiral, it hosts a chiral boundary current that
propagates along the edge. The hole can be replaced by a plateau-shaped
potential which repels electrons from the interior region. We now imagine
shrinking the radius of the ‘hole’, with the boundary current smoothly
deforming to cling to the shrinking perimeter. When the radius approaches
zero, we are left with a point-like potential. In this limit, the boundary
current takes the form of a local current encircling the impurity. The current
is no longer quantized as we can now have scattering between opposite edges of
the hole. Nevertheless, we expect a non-zero current strength. Based on
similar arguments, a impurity-induced current has been studied in a Chern
insulatorJha  _et al._ (2017).

Before presenting results on impurity response, we discuss chirality in our
mean-field solutions. In clean as well as in disordered systems, we find two
mean-field solutions that correspond to a choice of chirality. They can be
understood as $\Delta_{x}\pm i\Delta_{y}$ configurations: the order parameters
on the $x$ and $y$ bonds differ by a phase of $\pm\pi/2$. The chirality of the
configuration is determined by the choice of initial values in our BdG scheme.
If the initial values are chosen to be unbiased towards either chirality, the
system picks one of the two chiralities at random. For the situations
considered in this article, we did not find ‘domain-wall’ solutions containing
regions of differing chirality. For concreteness, in the discussion below, we
restrict our attention to solutions of the form $\Delta_{x}+i\Delta_{y}$.

Our results for the response to a single impurity are shown in Fig. 1. The
panels on the left show the order parameter amplitude on $x$ bonds. The panels
in the second column show amplitudes on the $y$ bonds. Here,
$\Delta_{x/y}^{i}$ is the order parameter on the bond emanating from site $i$
in the positive $x/y$-direction. In these plots, we see that the impurity
creates a local distortion in order parameters. In addition, magnetic fields
are generated as shown in the third column. The fields (fluxes, to be more
precise) are defined on plaquettes, given by the sum of Peierls’ phases
(vector potentials) along four bonds. The field is strongest at the plaquettes
that are immediately adjacent to the impurity, weakening as we move further.
Crucially, the direction of magnetic flux on a given plaquette is determined
by the sign of the impurity potential. This field originates from a current
that circulates around the impurity, as shown in the panels on the right. The
current is only appreciable on bonds that are close to the impurity. For a
given background chirality, the direction of the current loop is set by the
sign of the impurity potential.

In Fig. 2, we show the profile generated by a single impurity. We show the
current induced on neighbouring bonds, the resulting vector potential and the
magnetic field. We see that the response is highly localized, decaying sharply
as we move away from the impurity. This figure is to be compared with well-
known results on edge currents, e.g., in Ref. Furusaki  _et al._ , 2001. Our
problem is fundamentally different as it lacks the translational symmetry of a
linear edge. Nevertheless, the profiles show some degree of similarity. This
comparison also brings out an advantage of impurities over edges. An edge is a
sharply defined feature with no additional tuning parameters (apart from
deforming the edge shape). In contrast, an impurity is inherently tuneable as
both the sign as well as the amplitude of the impurity potential can be
varied. This allows for tunability of the response. For example, the direction
of the current can be switched by simply changing the sign of the potential.

Figure 1: (Colour online) Response to a single impurity corresponding to a
positive (top panels) and a negative (bottom panels) on-site potential. The
data have been calculated for $g=4t$ with the impurity position indicated by
an ‘x’. The chirality in all panels is of the $\Delta_{x}+i\Delta_{y}$ type.
From left to right, the panels show the order parameter on the x-bonds, those
on the y-bonds, the magnetic field through each plaquette and the induced bond
currents.  Figure 2: (Colour online) Response to a single impurity. Profiles
of induced current, vector potential and magnetic field for $W=3.5t$ and
$g=4t$. In the enclosed cartoon, the impurity location is shown as a dark
square. This point is taken to be $(0,0)$. The current and vector potential
are shown along bonds parallel to the $x$-axis at various values of $y$. The
field shown is on plaquettes at various $y$ values.

##  V Response to multiple impurities

We next consider the response to multiple impurities.

###  V.1 Correlated disorder

We first take up the case of correlated impurities, as defined in Eq. 4. As
discussed in the previous section, each impurity induces a local current loop
whose chirality is determined by the sign of the impurity potential. As all
impurities are of the same sign, we have many current loops with the same
chirality. As a result, where impurities are close to one another, their
currents fuse to form a larger loop as depicted in Fig. 3(left).

Figure 3: (Colour online) Currents with proximate impurities for $W=0.5t$ and
$g=4t$. In (a), all impurities have the same sign. Their positions are
labelled with ‘x’. In (b), impurities have mixed sign, indicated using ‘x’ and
‘o’.

The fusing of current loops lead to observable macroscopic consequences. In
Fig. 4, we show physical properties for two disorder realizations with
concentrations $\eta=0.05$ and $\eta=0.2$. As the order parameters are complex
scalars, we write them as
$\Delta_{x/y}^{j}=|\Delta_{x/y}^{j}|e^{i\theta_{x/y}^{j}}$, where $j$ is a
site index and the $\theta_{x/y}^{j}$ is the order parameter phase. In the
panels on the left, we show the amplitudes of the order parameters,
$|\Delta_{x}|$ and $|\Delta_{y}|$. We see that impurities locally suppress the
amplitudes of both $\Delta_{x}$ and $\Delta_{y}$ components. This is a
consequence of the positive sign of the impurity potential. When the negative
sign is used, we see local enhancement of order parameters. In the third
panel, we plot $\sin(\theta_{x}-\theta_{y})$. Here, $\theta_{x}$ is the phase
of $\Delta_{x}$ while $\theta_{y}$ is that of $\Delta_{y}$. The sine of the
difference in angles is a measure of chirality. In the $\Delta_{x}\pm
i\Delta_{y}$ states, we expect $(\theta_{x}-\theta_{y})=\pm\pi/2$. The sine of
this quantity takes the value of $\pm 1$, serving as an indicator of
chirality. In the figure, we see that chirality is robust to impurities.
Within the parameter regime that we have explored, we do not see any domain
walls separating regions of differing chirality.

In the fourth column in the figure, we show the spatial distribution of the
magnetic field. At low impurity concentration, we see localized (negative)
magnetic fields around each impurity. At larger impurity concentrations, these
field regions come closer to one another. In the figure, for the two
concentrations shown, we see field distributions that are skewed towards
negative values. This is tied to the sign of the impurity potential. The
spread is set by $W$, the magnitude of each impurity potential, as well as the
impurity concentration, $\eta$. This field distribution is a measurable
signature of impurities that can be accessed by probes such as NV centres.

Figure 4: (Colour online) Response to correlated impurities with $W=0.5t$ and
$g=4t$ on a $32\times 32$ lattice. We show snapshots for two disorder
realizations with concentrations $\eta=0.05$ and $\eta=0.2$. From left to
right, we show order parameter amplitudes on $x$ bonds, those on $y$ bonds,
sine of the phase difference in the two components and finally, the magnetic
field.

###  V.2 Random disorder

We next consider disorder with random sign as given by Eq. 5. Here, impurity-
generated current loops have random chiralities. As shown in Fig. 3(right),
this leads to incoherent currents with no loop structure. We obtain strong
local currents in random directions. In Fig. 5, we show physical properties
that arise from random disorder. In the left two panels, we show the
amplitudes of the order parameters. These plots are to be compared with the
corresponding panels in Fig. 4 where order parameters were suppressed at
impurity sites. Here, we see suppression in some regions and enhancement at
others. The suppression or enhancement directly correlates with the sign of
the impurity potential. As the impurity potential changes sign from one site
to another, the response in the amplitude changes as well.

In the third panel, we plot $\sin(\theta_{x}-\theta_{y})$, an indicator of
chirality as described above. The chirality remains robust with no sign of
domain formation. This is similar to the case of correlated disorder, shown in
Fig. 4. We note that this robustness of chirality to disorder may only hold in
our specific model and regime of interest. In other systems, it is conceivable
that disorder (especially in the form of magnetic impurities) may lead to
domain formation.

The fourth column depicts the induced magnetic flux. Each impurity gives rise
to a local flux, with the flux direction fixed by the sign of the impurity.
Away from impurities, the flux is close to zero. Around a positive or negative
impurity, the flux takes a negative or positive value, respectively. This
leads to a broad distribution with roughly equal spread on either side of
zero. The width of the distribution is set by $W$, the strength of the
impurity potential, as well as $\eta$, the impurity concentration. The field
distribution here is visibly different from that induced by correlated
disorder, especially at low impurity concentrations. In the correlated case,
as seen in Fig. 4, impurities generate local flux peaks that have the same
sign. With random disorder, we see peaks of both signs.

Figure 5: (Colour online) Response to random impurities. The parameters and
the plotted quantities are the same as in Fig. 4.  Figure 6: (Colour online)
Magnetic field distributions generated by correlated and random disorder.
Histograms of field values are shown in (a) and (b), with the disorder
concentration fixed at 20%. (c) The variance of the distributions is shown as
a function of disorder concentration. (d) The skewness of the distributions
vs. disorder concentration. In all panels, the data corresponds to $g=4t$,
$W=2t$ and $L=32$, averaged over 100 disorder realizations.

###  V.3 Comparing field distributions

We focus on magnetic field distributions, as they are particularly suited for
measurements by $NV$ centre probes. In Fig. 6, we compare the field
distributions generated by correlated and random disorder. In panels (a) and
(b), we show histograms of field values. In both cases, the mean of the
distribution is pinned at zero, i.e., $\langle b_{z}\rangle=0$. This is a
consequence of periodic boundary conditions, whereby our lattice system forms
a torus. As this torus does not enclose a magnetic monopole charge, the net
magnetic flux must vanish and the mean of the field distribution must remain
zero. Despite having the same mean value, the two field distributions are
markedly different.

Correlated disorder creates a skewed distribution as each impurity induces a
current loop with the same chirality. The distribution has a longer tail on
one side as seen from Fig. 6(a). This reflects a bias towards one magnetic
field direction over the other. In contrast, random disorder leads to local
currents with both chiralities. This leads to a nearly symmetric field
distribution centred at zero. The distribution is wider than in the correlated
case. This is a consequence of having strong local currents when impurities
with opposite signs are in close proximity. These lead to localized regions
with strong fields. Such regions can correspond to either of the two field
directions.

We compare quantitative measures of these distributions in panels (c) and (d).
Random disorder leads to a higher variance for a given disorder concentration.
This reflects a wider spread in field values, as can be seen visually by
comparing the histograms in panels (b) and (a). In panel (c), we see that the
variance in the field distribution changes non-monotonically with disorder
concentration. This is because strong disorder suppresses order parameter
amplitudes and, in turn, the supercurrents in the sample.

We next compare the skewness of the distributions, defined as
$\tilde{\mu}_{3}=E[\\{(X-\mu)/\sigma\\}^{3}]$. This quantifies the asymmetry
of the distribution about the mean. As seen in Fig. 6(d), correlated disorder
leads to significantly higher skewness for any value of the impurity
concentration. In the random case, the skewness is small but non-zero. This
reflects an inherent bias in the system that arises from the chirality of the
ground state. In other words, there is a small preference for current loops of
one chirality over another. However, this is a much weaker effect than that
induced by correlations in disorder.

Measurements with $NV$ centres are particularly suited to finding statistical
properties of field distributions. As shown above, correlations in the field
distribution reflect underlying correlations of the disorder realization. This
makes the case for an additional tuning handle in the form of disorder-
correlations. This provides for an additional layer of measurable signatures
in a chiral superconductor. With adatoms on a candidate material, these
measurements can reveal the character of the underlying superconductor. In
particular, the contrast between random and correlated disorder serves as a
clear signature of chiral superconductivity.

##  VI Discussion

A fascinating consequence of chiral superconductivity is the presence of
persistent currents along the sample edge. Despite sustained efforts over
decades, such currents have not been observed in experiments. We discuss an
alternative in the form of disorder-induced current. Our results show that (a)
impurities generically induce local currents in a chiral superconductor and
(b) correlated disorder can lead to long ranged currents within the sample.
These properties offer significant advantages over an edge current setup. At a
simple edge, the bulk chirality serves as a biasing field that drives a
current. This bias is in-built and not tuneable. In contrast, a disorder
potential provides an ‘external’ biasing field that is both tuneable and
flippable. This allows for disorder-currents to be tuneable in orientation,
strength and spatial extent. As a general principle, currents driven by an
external biasing field are easier to detect. This is exemplified by the recent
detection of an edge current in MoTe2Wang  _et al._ (2020), where an external
current was used to bias the edge.

Our analysis is based on a toy model for chiral $p_{x}\pm ip_{y}$
superconductivity on a square lattice. However, our results are not model-
dependent. Rather, they reflect a generic symmetry-allowed phenomenon – an
impurity in a chiral (quasi-)two-dimensional system gives rise to a localized
circulating current. This idea can be traced back to studies on integer
quantum Hall systems with plateau-shaped potentialsFord  _et al._ (1994). More
recently, impurity-induced currents have been demonstrated in a Chern
insulator, a non-superconducting phase that breaks time-reversal symmetryJha
_et al._ (2017). Extrapolating this physics to a chiral superconductor
requires careful consideration for the following reasons: (a) The
superconducting order parameter can re-organize itself to generate ‘screening’
currents. This can be viewed as a local manifestation of the Meissner effect,
where the superconductor resists any magnetic flux generated by impurity-
currents. (b) The current in a chiral superconductor is carried by
‘quasiparticles’ - a mixture of electrons and holes. If the electron and hole
contributions are comparable, the net charge current may vanish despite there
being a robust quasiparticle current. In this light, the calculations
presented in this article clarify that (i) an impurity in a chiral
superconductor generates a localized current loop, and (ii) its net charge
current is non-zero, generating a magnetic field.

Impurity-generated currents have been previously reported in chiral
superconductors by Guo et. al.Guo  _et al._ (2017). However, their approach
was not fully self-consistent as the vector potential was ignored. Our study
addresses this issue by fully accounting for vector potentials and screening
currents. We have used an improved Bogoliubov-deGennes schemeZhu (2016), where
the order parameters and the vector potential are determined self-
consistently. As far as we are aware, this has only been achieved previously
in certain simple cases that are amenable to semi-analytic methods. A classic
example is the calculation of the profile of a single vortex in a
superconductorTinkham (2004). A second, more pertinent, example is edge
current in the context of Sr2RuO4Furusaki  _et al._ (2001). In both these
examples, symmetry reduces the vector potential and the order parameter into
functions of one variable. Both can be calculated self-consistently in this
effective one-dimensional problem. Such approaches are inherently limited and
cannot be extended beyond idealized cases. Our algorithm, based on Eq. 13, is
suitable for a wider array of problems. For example, it can be used to
determine field distributions in vortices and vortex lattices.

Our approach complements earlier studies on impurities in chiral
superconductors. In particular, Refs. Kaladzhyan  _et al._ , 2016 and Kreisel
_et al._ , 2021 present detailed studies of impurity-induced in-gap states.
They further discuss periodic arrangements of impurities, where bound states
hybridize to form bands that can have topological character. Here, we do not
focus on bound states or quasiparticle bands. Rather, we discuss currents
induced in the superconducting ground state and in turn, the magnetic fields
that they generate. In the discussion above, we have restricted our attention
to weak disorder and zero temperature, a situation in which impurity-induced
bound states do not play a role. In future studies on realistic systems, their
effects can be taken into account.

Going beyond isolated impurities, we have discussed currents generated by two
disorder schemes. We show that correlated disorder can lead to long-ranged
current loops within the sample. This is reflected in the magnetic field
distribution, providing more complex signatures of chirality. Our two disorder
schemes are inspired by studies of impuritiesBalatsky  _et al._ (2006) and
adatoms. In several contexts, adatoms have been shown to bring out interesting
properties. For instance, adatoms on graphene have been shown to give rise to
superconductivityProfeta  _et al._ (2012); Ludbrook  _et al._ (2015).

##  VII Acknowledgements

The authors acknowledge use of the Nandadevi computing cluster facility at The
Institute of Mathematical Sciences (IMSc), Chennai, India.

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# Partisan Combinatorial Game of Edge and Vertex Removal on Graphs

Nathan Shank  Moravian College 1200 Main Street Bethlehem, PA 18018
<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Devon Vukovich

(Date: August 2020)

###### Abstract.

We consider three variants of a partisan combinatorial game between two
players, Left and Right, played on an undirected simple graph. Left is able to
delete vertices (and incident edges) while Right is able to delete edges. This
natural extension of a similar impartial game gives a clear advantage to one
player by allowing them the ability to play on a small subgraph which the
other player can not. Our last variant removes this advantage by assuming a
move is valid for one player if and only if the other player has a valid move
on the same graph. In this case, we show that the ability to remove a vertex
is more advantageous compared to removing edges.

###### Key words and phrases:

Combinatorial Games, Graph Theory, Atomic Weight, All-small

###### 2010 Mathematics Subject Classification:

Primary. 94C15

##  1\. Introduction

Many games, both partial and impartial, can be played on graphs. Some games
generate graphs of larger size and/or order, for example Sprouts [4]. While
others remove edges and/or vertices to reduce the graph, for example the
impartial version of Hackenbush [2]. In either case, the game ends when a
particular end state of the graph is reached. For edge deletion games a turn
consists of removing an edge from the graph and possibly additional edges or
vertices depending on the structure of the game. For example, Gallant et. al.
[5] considered deleting edges from a graph so long as the graph does not
contain an isolated vertex. Similarly in a vertex deletion game, a turn
consists of removing a vertex which may also result in other vertices or edges
being removed. For example Adams et. al. [1] considered a game where a turn
consists of removing a vertex, all incident edges, and all vertices which
become isolated.

Often, the ability for a vertex or edge to be removed is dependent on some
properties of the object. Nowakowski and Ottaway [7] considered a vertex
removal game based on the parity of the vertex while Harding and Ottaway [6]
considered variants of an edge removal game based on the parity of the
incident vertices. In either case, the underlying principle of the game
clearly define what objects may or may not be removed.

In this paper we consider three partisan variants of the impartial edge
removal game presented in [5]. Two players, Left and Right, play on a finite
graph $G$. A turn for Left consists of removing a vertex and all incident
edges while a turn for Right consists of removing a single edge. The variants
differ based on the underlying principles which define what objects can be
removed. In section 3.1 we will analyze the Classic Variant which follows the
model introduced by Gallant et. al. [5] which implies the game continues so
long as the graph does not contain an isolated vertex. In section 3.2, the
Forbidden Leaf Variant will impose an additional restriction not allowing the
deletion of leaf vertices. Finally, in section 3.3, the Mutual Failures
Variant will impose additional restrictions which will allow Left the ability
to delete a vertex from a subgraph if and only if Right has the ability to
delete an edge from the same subgraph.

As in the original result of Gallant et. al. [5], we will concern ourselves
with finding which values of $n$ in a particular graph class can be won by a
particular player, rather than the value of the game or the size of the
advantage, although these would be interesting additions for future
consideration.

##  2\. Background and definitions

We will adopt standard notation from game theory literature. The theorems and
notation introduced in this section can be found in [2] and [3].

Since we are interested in a game with only two players, Left and Right, we
will denote a game position, $G$, as
$G=\left\\{{\mathcal{L}}~{}|~{}{\mathcal{R}}\right\\}$, where $\mathcal{L}$
and $\mathcal{R}$ represent the move sets available to Left and Right
respectively. When Left moves from $G$ to some option $L\in\mathcal{L}$, we
will note her move ${G}~{}\overrightarrow{{}_{L}}~{}{L}$. Similarly,
${G}~{}\overrightarrow{{}_{R}}~{}{R}$ is a move by Right from $G$ to some
$R\in\mathcal{R}$. The zero game will be denoted
$G=\left\\{{}~{}|~{}{}\right\\}$, star will be denoted
$\ast=\left\\{{0}~{}|~{}{0}\right\\}$, far star will be denoted $\bigstar$,
the game position $\left\\{{0}~{}|~{}{\ast}\right\\}$ is defined as up
$(\uparrow)$ and the negative of up, $\left\\{{\ast}~{}|~{}{0}\right\\}$ is
defined as down, $(\downarrow)$. We will write $G\|H$ if $G$ is incomparable
with $H$. If $G>H$ or $G\|H$ we will denote this by $G\rhd H$.

Since our analysis will be primarily focused on all-small games, we will
provide a review of some useful definitions and theorems related to atomic
weights.

###  2.1. Atomic Weights

When examining an all-small game, complicated infinitesimal positions are
frequently encountered. Many of the analysis techniques used to make sense of
a game as a whole are useless when applied to such positions. It then becomes
desirable to relate a complicated position to one that is understood. In this
way, the dominant features of a position can be equated to the features of a
known position provided the error can be accommodated. We use an equivalence
relation, far star, to do just this.

######  Definition 2.1.

We say a game position $G$ is far star equivalent to a game position $H$ if
$G+X+\bigstar$ and $H+X+\bigstar$ have the same outcome for all game positions
$X$. We denote this equivalence ${G}\sim_{\bigstar}{H}$.

The following establishes a useful way to determine far star equivalence.

######  Theorem 2.2.

For game positions $G$ and $H$, ${G}\sim_{\bigstar}{H}$ if and only if
${\downarrow}{\bigstar}<G-H<{\uparrow}{\bigstar}$.

Theorem 2.2 provides a scale based upon $\uparrow$ that we can use to measure
infinitesimals. Recall that despite being an infinitesimal and consequently
less than all positive numbers, $\uparrow>0$. This makes $\uparrow$ a Left
win, and similarly $\downarrow$ is a Right win. To be able to draw an
equivalence between an infinitesimal position and some integer multiple of
$\uparrow$ would allow us to measure how much an infinitesimal position favors
a player.

######  Definition 2.3.

For an integer $n$ and a game position $G$, the atomic weight of $G$ is $n$ if
${G}\sim_{\bigstar}{n\cdot\uparrow}$. We denote this AW$(G)=n$.

The above definition restricts the atomic weight to an integer value, but
there is no reason the atomic weight must be integer valued. We make this
restriction as our usage of atomic weights will yield integer values. In the
above definition, $n\cdot\uparrow$ represents the game sum of $|n|$ copies of
$\uparrow$ if $n$ is positive and $|n|$ copies of $\downarrow$ if $n$ is
negative. Atomic weights are well defined, and because far star equivalence
respects addition, atomic weights are also additive.

######  Theorem 2.4.

For game positions $G$ and $H$,

$\text{AW}(G)+\text{AW}(H)=\text{AW}(G+H).$

Clearly, this result allows for the atomic weight of a game sum to be easily
computed by summing the atomic weights of each summand. We now present the
result that makes atomic weights the most effective tool for solving
infinitesimal positions.

######  Theorem 2.5.

For a game position $G$,

$\displaystyle\text{if AW}(G)$ $\displaystyle\geq 2\text{, then }G>0,$
$\displaystyle\text{if AW}(G)$ $\displaystyle\geq 1\text{, then }G\rhd 0,$
$\displaystyle\text{if AW}(G)$ $\displaystyle\leq-1\text{, then }G\lhd 0,$
$\displaystyle\text{and if AW}(G)$ $\displaystyle\leq-2\text{, then }G<0.$

This theorem is known as the Two Ahead Rule. If the atomic weight is at least
two in favor of a player, that player then has sufficient advantage in the
position to win regardless which player moves first. While it is possible to
compute the atomic weight of a position $G$ using only the definition of
atomic weight, the following theorem, known as the Atomic Weight Calculus,
provides a method to compute the atomic weight using only the atomic weights
of the positions in $G$.

######  Theorem 2.6.

For a game position
$G=\left\\{{L_{1},L_{2},L_{3},\cdots}~{}|~{}{R_{1},R_{2},R_{3},\ldots}\right\\}$
if $L_{i}$ and $R_{i}$ have atomic weights for all $i$, then

$\displaystyle\text{AW}(G)=\left\\{{\text{AW}(L_{1})-2,\text{AW}(L_{2})-2,\ldots}~{}|~{}{\text{AW}(R_{1})+2,\text{AW}(R_{2})+2,\ldots}\right\\}$

unless this is an integer. If this is is an integer, let $x$ be the least
integer such that,

$\displaystyle\text{AW}(L_{1})-2\lhd x,\text{AW}(L_{2})-2\lhd x,\ldots$

and let $y$ be the greatest integer such that,

$\displaystyle\text{AW}(R_{1})+2\rhd y,\text{AW}(R_{2})+2\rhd y,\ldots.$

Consider $G$.

$\displaystyle\text{If }G~{}\|~{}\bigstar\text{, then AW}(G)=0.$
$\displaystyle\text{If }G>\bigstar\text{, then AW}(G)=y.$
$\displaystyle\text{If }G<\bigstar\text{, then AW}(G)=x.$

In most situations, the exception is invoked and it is necessary to compare
$G$ to $\bigstar$. We will see this later in the analysis of the Mutual
Failures Variant. There is one final useful result regarding the relation
between a position’s atomic weight and far star.

######  Theorem 2.7.

For a position, $G$, $G+\bigstar>0$ exactly when AW$(G)\geq 1$.

###  2.2. Graph Theory

We will adopt the graph theory notation and definitions from [8]. We will
denote a graph $\mathscr{G}:=(V,E)$ such that $V$ is a set of vertices, and
$E$ is a set of edges. We will assume throughout that $\mathscr{G}$ is a
simple undirected graph. For any vertex $v$, let $\delta(v)$ be the degree of
$v$ in $G$. A vertex with degree zero is called an isolated vertex and a
vertex of degree one is called a leaf. We will let $\mathscr{P}_{n}$ denote a
path graph on $n$ vertices and $\mathscr{K}_{n}$ be a complete graph on $n$
vertices.

Our games will require players to make deletions from a graph thus many of the
possible game positions are disconnected graphs. These disconnected positions
can be considered a game sum of connected positions. As such, we will abuse
notation and refer to the positions of a game sum as the components of the
game sum for they are in fact the components of the overall graph position.
Further, it is desirable to have a clear but related notation to distinguish
between graphs and game positions. Thus, we will notate a graph class as
$\mathscr{X}$ and the game positions played on that class as $X$.

##  3\. Results

###  3.1. The Mixed Deletion Game

######  Definition 3.1.

The Mixed Deletion Game (Classic Variant) is a combinatorial game defined by
the following rules:

  * •

The game is played on a finite graph.

  * •

There are two players, Left and Right, who alternate turns.

  * •

Left must delete one vertex and all incident edges from the graph.

  * •

Right must delete one edge.

  * •

The game ends when a deletion creates an isolated vertex.

  * •

The player whose deletion created an isolated vertex loses.

Notice in path graphs there is an inherent bias towards Left. Since any path
position will only be broken into smaller path position components, the
ability for only Left to be able to move on $P_{3}$ causes any
$\mathscr{P}_{3}$ component to act as a “free move” for Left in the game sum
of path position components. No such “free move” path exists for Right,
denying him the ability to counter the existence of a $P_{3}$ component. This
implies there is no path position $P_{n}$, with a negative integer or dyadic
rational value. To more formalize this argument, consider the following
proposition:

######  Proposition 3.2.

There is no graph that contains a move for Right and no move for Left in the
Classic Variant.

###### Proof.

Let $\mathscr{G}$ be a graph. Assume there exists a move for Right. This
implies the existence of an edge, $\\{a,b\\}$, such that
$\delta(a),\delta(b)>1$. Thus, $a$ must be adjacent to some vertex
$v^{\prime}\neq b$. Consider the degree of $v^{\prime}$. If
$\delta(v^{\prime})=1$, then Left can legally delete $v^{\prime}$. If
$\delta(v^{\prime})>1$, then Left can legally delete $a$. Therefore, if Right
has a legal move in a graph position, then Left must also have a legal move.

We can apply the logic of Proposition 3.2 to path graph positions and find
that it is impossible to have a path position with a negative numeric value.

######  Corollary 3.3.

There is no path position, $P_{n}$, with a negative integer or dyadic rational
value.

###### Proof.

On $\mathscr{P}_{{n\geq 3}}$, Left can delete a leaf vertex without creating
an isolated vertex, so the only path position without a move for Left is
$P_{2}$. However, Right also has no move in $P_{2}$. Thus, in $P_{n}$ if
$\mathcal{L}=\emptyset$, then $\mathcal{R}=\emptyset$. Thus, it is impossible
for $k\in\mathbb{Z}^{+}$ to exist such that,
$P_{n}=-k=\left\\{{}~{}|~{}{-(k-1)}\right\\}$. Therefore, there does not exist
any path position with an integer value less than zero.

Further, for $j>0$ and an odd $q\in\mathbb{Z}$, the game value
$\frac{q}{2^{j}}=\left\\{{\frac{q-1}{2^{j}}}~{}|~{}{\frac{q+1}{2^{j}}}\right\\}$.
So any negative dyadic rational for which $|q|<2^{j}$ implies the existence of
a position $-\frac{1}{2}=\left\\{{-1}~{}|~{}{0}\right\\}$, but as shown above
there is no path position valued $-1$. Notice in the case where $|q|\geq
2^{j}$, we can express $\frac{q}{2^{j}}$ as a mixed number requiring both a
negative integer value and a dyadic rational between $-1$ and $0$, both of
which have already been shown to be impossible values for a path position.
Therefore, there does not exist any path position with a negative dyadic
rational value.

Corollary 3.3 implies Right’s inability to counter Left’s “free moves.”
Because Left can move ${P_{3}}~{}\overrightarrow{{}_{L}}~{}{P_{2}}$ and Right
cannot move on $P_{3}$, every $P_{3}$ component is worth a game value of $1$.
If Left adopts a strategy of creating a $P_{3}$ component with every one of
her moves and then moving on the $P_{3}$ components only when no more can be
created, Right would need to create a component of equal and opposite game
value to counter Left’s strategy. Corollary 3.3 shows this is impossible.
Hence, Left will be the winner of $P_{n}$ for $n$ sufficiently large.

######  Theorem 3.4.

For an integer $n\geq 5$, Left will win $P_{n}$ in the Classic Variant
regardless of which player moves first.

###### Proof.

Let $n$ be a positive integer, and consider the path position $P_{n}$ in the
Classic Variant. Notice

$\displaystyle P_{2}$ $\displaystyle=\left\\{{}~{}|~{}{}\right\\},$
$\displaystyle P_{3}$ $\displaystyle=\left\\{{0}~{}|~{}{}\right\\},$
$\displaystyle P_{4}$ $\displaystyle=\left\\{{1}~{}|~{}{0}\right\\},\text{
and}$ $\displaystyle P_{5}$
$\displaystyle=\left\\{{P_{4},0}~{}|~{}{P_{3}}\right\\}.$

Since $P_{3}=1$, we know if Left moves first and $n\geq 6$, she can move to
$1+P_{n-4}$. However by Corollary 3.3, we know there is no
$k\in\mathbb{Z}^{+}$ such that $1+P_{k}\leq 0$. Thus, regardless of which
player wins the $P_{n-4}$ position, Left will always have the last legal move
on $P_{3}$ and win. We have not shown that no path position with value greater
than 1 exists, and if one did it would surely be more optimal for Left to win
by a larger margin. Yet, we have shown Left is able to win by a potentially
not optimal strategy.

If Right moves first, we know he will want to move $P_{n}$ to a game sum
without creating a $P_{3}$ position or path position long enough for Left to
create a $P_{3}$ on her following turn. This implies Right will move to
$P_{2}+P_{n-2}=P_{n-2}$ or $P_{5}+P_{n-5}$.

Let us consider the case where Right moves to $P_{n-2}$ first. From above, we
know Left wins playing first on $P_{m\geq 3}$. Hence, Left will win if Right
moves ${P_{n}}~{}\overrightarrow{{}_{R}}~{}{P_{n-2}}$ for $n\geq 5$.

Now consider the case where Right moves to $P_{5}+P_{n-5}$. We can assume
$3<n-5<6$. So if Right moved to $P_{5}+P_{4}$, Left will move
${P_{5}+P_{4}}~{}\overrightarrow{{}_{L}}~{}{P_{5}+P_{3}}$, and if Right moved
to $P_{5}+P_{5}$, Left will move
${P_{5}+P_{5}}~{}\overrightarrow{{}_{L}}~{}{P_{5}}$. Right’s move on $P_{5}$
must create $P_{3}$. Thus, Right has no winning strategy if he moves
${P_{n}}~{}\overrightarrow{{}_{R}}~{}{P_{5}+P_{n-5}}$.L

Therefore, for $n\geq 5$, Left is guaranteed to win $P_{n}$ regardless which
player moves first.

Solutions for more graph classes, including cycles and wheels, can be
extrapolated similarly if they can be reduced by a single move into a graph
class that has already been solved. Cycles follow immediately from Theorem 3.4
because the first move will move to $P_{n}$ or $P_{n-1}$.

######  Theorem 3.5.

For an integer $n\geq 3$ and $n\neq 5$, Left will win the Classic Variant
played on the cycle position, $C_{n}$, regardless of which player moves first.

For wheel graphs, Left will want to remove the center vertex first,
independent of Right’s move or who goes first. This will result in $P_{n}$
after Left’s first move. Thus we can apply Theorem 3.4 and we have the
following:

######  Theorem 3.6.

For an integer $n\geq 3$ and $n\neq 5$, Left will win the wheel position
$W_{n}$ regardless of which player moves first.

For a complete graph, $K_{n}$, Left also has the advantage by removing a
vertex incident to the edge removed by Right. Thus Right can always be forced
to play on a complete subgraph $K_{n-1}$. Eventually Right will be forced to
move ${K_{3}}~{}\overrightarrow{{}_{R}}~{}{P_{3}}$, and Left wins $P_{3}$.
Therefore, we have the following result for complete graphs:

######  Theorem 3.7.

For a positive integer $n\geq 3$, Left will win the complete graph position
$K_{n}$.

Other graph classes, including star graphs and complete bipartite graphs, can
be handled similarly. All of which show an advantage for Left.

The analysis of the Classic Variant begins to show the imbalance between an
edge deletion and a vertex deletion. Paths, cycles, wheels, stars, complete
bipartite, and complete graphs are biased towards Left winning, and this bias
largely resulted from Left’s ability to delete leaf vertices. It then becomes
natural to ask, “Who wins if Left cannot delete leaf vertices?”

###  3.2. The Forbidden Leaf Variant

In the interest of trying to remove the bias towards Left, we will add a rule
that denies her the ability to delete leaf vertices.

######  Definition 3.8.

The Forbidden Leaf Variant adopts all of the base rules and adds to it the
condition that Left may not delete leaf vertices.

It should be immediately clear that $P_{2}=P_{3}=0$. However, notice Left can
no longer move on $P_{4}$ but Right can. For path graphs, this creates an
advantage for Right because any $\mathscr{P}_{4}$ component acts as a “free
move” for Right and there are no “free moves” for Left. This implies there is
no path position $P_{n}$, with a positive integer or dyadic rational value. So
we see our additional condition is too strong and rather than eliminating the
bias, it has shifted it from Left to Right. To more formalize this argument,
consider the following proposition:

######  Proposition 3.9.

There is no graph that contains a move for Left and no move for Right in the
Forbidden Leaf Variant.

###### Proof.

Let $\mathscr{G}$ be a graph, and $G$ be the corresponding game position.
Assume $\mathcal{L}\neq\emptyset$. This implies the existence of a vertex,
$v$, such that $\delta(v)>1$, and for any vertex, $v^{\prime}$, that is
adjacent to $v$, $\delta(v^{\prime})>1$. However, deleting the edge
$\\{v,v^{\prime}\\}$ is a legal move for Right. Thus, if Left has a legal move
in $G$, then Right must also have a legal move in $G$. Therefore, there is no
graph that contains a move for Left and does not contain a move for Right.

Thus the advantage has now shifted to Right’s ability to play on
$\mathscr{P}_{4}$ whereas Left can not.

######  Corollary 3.10.

There is no path position $P_{n}$, with a positive integer or dyadic rational
value.

###### Proof.

We know from Proposition 3.9 that if $\mathcal{R}=\emptyset$ then
$\mathcal{L}=\emptyset$. Thus, it is impossible for $k\in\mathbb{Z}^{+}$ to
exist such that $P_{n}=k=\left\\{{(k-1)}~{}|~{}{}\right\\}$. Therefore, there
does not exist any path position with an integer value greater than zero.

Further, for $j>0$ and an odd $q\in\mathbb{Z}$, the game value
$\frac{q}{2^{j}}=\left\\{{\frac{q-1}{2^{j}}}~{}|~{}{\frac{q+1}{2^{j}}}\right\\}$.
So any positive dyadic rational for which $|q|<2^{j}$ implies the existence of
a position $\frac{1}{2}=\left\\{{0}~{}|~{}{1}\right\\}$, but as shown above
there is no path position valued $1$. Notice in the case where $|q|\geq
2^{j}$, we can express $\frac{q}{2^{j}}$ as a mixed number requiring both a
positive integer value and a dyadic rational between $0$ and $1$, both of
which have already been shown to be impossible values for a path position.
Therefore, there does not exist any path position with a positive dyadic
rational value.

We will now use this result to show Right wins for sufficiently large path
positions.

######  Theorem 3.11.

For an integer $n\geq 8$ and not equal to $11$, Right is guaranteed to win
$P_{n}$ in the Forbidden Leaf Variant regardless of which player moves first.

###### Proof.

Let $n$ be a positive integer, and consider the path position $P_{n}$ in the
Forbidden Leaf Variant. Notice

$\displaystyle P_{2}$ $\displaystyle=\left\\{{}~{}|~{}{}\right\\},$
$\displaystyle P_{3}$ $\displaystyle=\left\\{{}~{}|~{}{}\right\\},$
$\displaystyle P_{4}$ $\displaystyle=\left\\{{}~{}|~{}{0}\right\\}\text{,
and}$ $\displaystyle P_{5}$ $\displaystyle=\ast.$

Consider $n\geq 6$. Since $P_{4}=-1$, we know Right can adopt a strategy of
creating a $P_{4}$ component with each move and we know Left has no ability to
counter this strategy. If Right moves first, then he will be able to create a
$P_{4}$ component and win.

Hence, we may restrict our examination to the case where Left moves first.
Left will avoid creating any $P_{4}$ component or $P_{k}$ component where $k$
is sufficiently large to allow Right to create a $P_{4}$ component on his
subsequent move. Left has two options: move
${P_{n}}~{}\overrightarrow{{}_{L}}~{}{P_{n-4}}$ or move
${P_{n}}~{}\overrightarrow{{}_{L}}~{}{P_{5}+P_{n-6}}$. For the first strategy,
if $n\geq 8$, Right will be able to win. For the second strategy, Left can
only win if $n=11$ as $P_{5}+P_{5}=0$. Notice Left can only attempt the second
strategy if $n\geq 8$.

We can again extrapolate our result on paths to solve cycles since any edge or
vertex deletion in a cycle will result in a path. Wheel graphs do not follow
easily from paths because of the inability of Right to move a wheel to a cycle
and Left’s desire to avoid cycles.

######  Theorem 3.12.

For an integer $n\geq 8$ such that $n\neq 11$, Right is guaranteed to win
$C_{n}$ in the Forbidden Leaf Variant regardless of which player moves first.

###### Proof.

Let $n$ be a positive integer, and $C_{n}$ be the cycle position on $n$
vertices in the Forbidden Leaf Variant. If Right moves first, he must move
${C_{n}}~{}\overrightarrow{{}_{R}}~{}{P_{n}}$. Left will lose moving first on
$P_{k}$ if $k\geq 8$ and $k\neq 11$. If Left moves first, she must move
${C_{n}}~{}\overrightarrow{{}_{L}}~{}{P_{n-1}}$. Right will win moving first
on $P_{k}$ if $k\geq 4$. Therefore, Right will win $C_{n}$ if $n\geq 8$ and
$n\neq 11$ regardless of which player moves first.

We conclude the Forbidden Leaf Variant by considering complete graphs. As with
the Classic Variant, regardless of who goes first, Right can be forced to
always move on a complete subgraph. This can continue until Right must move on
$K_{4}$ to which Left will counter and move the graph position to $P_{3}$.
Thus Left will win on $K_{n}$ if $n\geq 5$ regardless of which player moves
first.

######  Theorem 3.13.

For an integer $n\geq 5$, Left is guaranteed to win $K_{n}$ in the Forbidden
Leaf Variant regardless of which player moves first.

###### Proof.

Let $n$ be a positive integer. Notice $K_{1}=K_{2}=0$ and $K_{3}=\ast$. If
$n=4$ and Left moves first, she must move
${K_{4}}~{}\overrightarrow{{}_{L}}~{}{K_{3}}$ and Right wins moving first on
$\ast$. If $n=4$ and Right moves first, then any edge he deletes results in
the same graph position due to symmetry. The corresponding graph has two
vertices of degree 2 and two vertices of degree 3. Left can win this position
by deleting a vertex of degree 3 which moves the graph position to $P_{3}$. If
$n\geq 5$, and Left moves first, any vertex she deletes moves
${K_{n}}~{}\overrightarrow{{}_{L}}~{}{K_{n-1}}$. Hence, it will suffice to
show that Left wins moving second. Again on account of symmetry, any single
edge deleted from $\mathscr{K}_{n}$ results in one distinct graph after
excluding isomorphisms, so Right only has one possible first move. If Left
counters by deleting one of the two vertices incident to the edge Right
deleted, she moves to $K_{n-1}$. Eventually, Right will be forced to move
first on $K_{4}$. From above, we know Left can win if this happens. Therefore,
Left will win $K_{n}$ if $n\geq 5$ regardless of which player moves first.

Left still has the ability to win complete graphs because complete graphs are
connected enough to overcome the bias towards Right. In this way, they are
different than the other graph classes because Left can prevent the graph from
breaking down into paths until the paths are small enough to be unplayable.

We saw early in the analysis of the Forbidden Leaf Variant that our additional
rule was too strong. The bias given to Left in the Classic Variant was given
to Right in the Forbidden Leaf Variant. Although we did see Left still wins
complete graphs, it is clear that for paths, the bias shifted to Right because
of his ability to play on $P_{4}$.

###  3.3. The Mutual Failures Variant

In both of the prior variants, we found that the game was inherently biased
towards one player or the other. Examining the Classic Variant and the
Forbidden Leaf Variant for similarities revealed an interesting relationship
in the failure states of the players. In the Classic Variant, the only path
graph that did not contain a move for Left was $P_{2}$, while Right could not
move on $P_{2}$, and, $P_{3}$. In the Forbidden Leaf Variant, Right could not
move on $P_{2}$, and $P_{3}$, while Left could not move on $P_{2}$, $P_{3}$,
and $P_{4}$. In both cases one players set of failure states was a proper
subset of the other player’s, thus we move to examining the Mutual Failures
Variant.

######  Definition 3.14.

The Mutual Failures Variant adopts all of the base rules, and adds to it the
condition that in any graph position, $\mathcal{R}=\emptyset$ if and only if
$\mathcal{L}=\emptyset$.

This added condition dramatically changes the nature of the Mixed Deletion
game and requires a different technique for analysis. The condition,
$\mathcal{R}=\emptyset$ if and only if $\mathcal{L}=\emptyset$, implies that
every position in the Mutual Failures Variant is all-small and infinitesimal.

We will need to utilize atomic weights to solve this variant. Although we can
calculate the atomic weight for small games based on the game tree, it is
impractical to do so for large games. For the benefit of the reader, we will
show the calculation of the atomic weight $P_{5}$ and $P_{6}$ using two
different methods.

######  Example 3.15.

Consider $P_{5}$ in the Mutual Failures Variant. Notice
$P_{4}=\left\\{{0}~{}|~{}{0}\right\\}$, so
$P_{5}=\left\\{{P_{4},0}~{}|~{}{0}\right\\}={\uparrow}{\ast}$. By Theorem 2.2,
to show ${P_{5}}\sim_{\bigstar}{\uparrow}$, we need to show that
${\downarrow}{\bigstar}<P_{5}-\uparrow<{\uparrow}{\bigstar}$. Thus, it will
suffice to show ${\uparrow}{\ast}+\bigstar>0$ and
${\downarrow}{\ast}+\bigstar<0$.

First, we can fix $\bigstar=\ast 2$ as any larger nim heap will be reduced to
$\ast 2$ by optimal players. If Left moves first on ${\uparrow}{\ast}+\ast 2$,
she will move ${{\uparrow}{\ast}+\ast
2}~{}\overrightarrow{{}_{L}}~{}{\uparrow}$ which she wins. If Right moves
first on ${\uparrow}{\ast}+\ast 2$, he can move to $\uparrow$, $\ast 2$, or
$\uparrow+\ast 2$. Left obviously wins the first two, and can win the third
moving ${\uparrow+\ast 2}~{}\overrightarrow{{}_{L}}~{}{\uparrow}$. Thus,
${\uparrow}{\ast}+\bigstar>0$. Since
${\downarrow}{\ast}+\bigstar=-({\uparrow}{\ast}+\bigstar)$, we also know
${\downarrow}{\ast}+\bigstar<0$. Therefore, AW$(P_{5})=1$.

Let us now consider
$P_{6}=\left\\{{P_{5},P_{2}+P_{3}}~{}|~{}{P_{2}+P_{4},P_{3}+P_{3}}\right\\}.$
Let

$\displaystyle g=$
$\displaystyle\\{\text{AW}(P_{5})-2,\text{AW}(P_{2}+P_{3})-2$
$\displaystyle|\text{AW}(P_{2}+P_{4})+2,\text{AW}(P_{3}+P_{3})+2\\}.$

After simplifying, $g=\left\\{{-1}~{}|~{}{2}\right\\}=0$ which is an integer
and thus invokes the exception of Theorem 2.6. Then $x=0$ is the least integer
such that $-1\lhd x$ and $y=1$ is the greatest integer such that $2\rhd y$. We
must now compare $P_{6}$ to $\bigstar$.

Consider $P_{6}+\bigstar$. Again we can fix $\bigstar=\ast 2$. If Right moves
first, he can move to $\ast 2$ or $\ast+\ast 2$. Left will win both. Since
Left wins moving second, we know $P_{6}+\bigstar\geq 0$, so $P_{6}>\bigstar$.
Therefore, AW$(P_{6})=y=1$.

To compute some additional base case atomic weights, we used CGSuite, a
program used to analyze combinatorial games:

Path Position | Atomic Weight
---|---
$P_{2}$ | 0
$P_{3}$ | 0
$P_{4}$ | 0
$P_{5}$ | 1
$P_{6}$ | 1
$P_{7}$ | 1
$P_{8}$ | 1
$P_{9}$ | 2
$P_{10}$ | 2
$P_{11}$ | 2
$P_{12}$ | 2
Table 1. Atomic Weight for $P_{2}$ through $P_{12}$.

As path positions move to sums of smaller path positions, and the smallest
paths have integer valued atomic weights the exception will be invoked to find
the atomic weight of every path position. For this reason, we will need to
know how a path position compares to far star. Additionally, as we show every
path position on 5 or more vertices is greater than far star, we can use
Theorem 2.7 to show a lower bound of the atomic weights of these positions.

######  Theorem 3.16.

For an integer $n\geq 5$, the following hold:

  * •

$P_{n}>\bigstar$, and

  * •

AW$(P_{n})\geq 1$.

###### Proof.

For an integer $n\geq 5$, let $P_{n}$ be the path position on $n$ vertices in
the Mutual Failures Variant. We will prove by strong induction that
$P_{n}>\bigstar$ and AW$(P_{n})\geq 1$. Notice $P_{n}>\bigstar$ holds if Left
wins $P_{n}+\bigstar$ regardless which player moves first, so that
$P_{n}+\bigstar>0$.

Base Cases: After we have shown a base case to be a Left win or a first player
win, we will assume that Right will not move to that position. We will also
assume Right will not move to $\bigstar$ as Left can move a single nim heap
directly to $0$. Thus, we will omit these base case positions from Right’s
move set. Further, for $k\geq 6$, Left can move
${P_{k+1}+\bigstar}~{}\overrightarrow{{}_{L}}~{}{P_{k}+\bigstar}$, where
$P_{k}+\bigstar$ is an earlier base case. For this reason we will not consider
Left’s move set for $P_{n}$ when $n\geq 6$.

Consider $P_{4}+\bigstar$. Both players will move $P_{4}+\bigstar$ to
$\ast+\ast=0$. So $P_{4}+\bigstar$ is a first player win.

Consider $P_{5}+\bigstar$. If Left moves first on $P_{5}+\bigstar$, she will
move to $P_{5}+\ast$. Since $P_{5}+\ast=\uparrow$, she will win moving first.
If Right moves first on $P_{5}+\bigstar$, his move set is
$\mathcal{R}=\\{P_{5},P_{5}+\ast\\}$. Left can move the first position
directly to $0$, and she moves
${P_{5}+\ast}~{}\overrightarrow{{}_{L}}~{}{P_{4}+\ast=0}$, so Left wins if
Right moves first, and $P_{5}+\bigstar>0$.

Consider $P_{6}+\bigstar$. Right’s move set is
$\mathcal{R}=\\{P_{6},\ast+\bigstar,P_{6}+\ast\\}$. Left can move the first
two positions directly to $0$, and the last position to $P_{5}+\ast=\uparrow$,
so Left wins if Right moves first, and $P_{6}+\bigstar>0$.

Consider $P_{7}+\bigstar$. Right’s move set is
$\mathcal{R}=\\{P_{7},P_{7}+\ast\\}$. Left wins $P_{7}=\uparrow$, and she will
move ${P_{7}+\ast}~{}\overrightarrow{{}_{L}}~{}{\ast+\ast=0}$, so Left wins if
Right moves first, and $P_{7}+\bigstar>0$.

Consider $P_{8}+\bigstar$. Right’s move set is
$\mathcal{R}=\\{P_{8},P_{8}+\ast,P_{4}+P_{4}+\bigstar\\}$. Notice
$P_{4}+P_{4}+\bigstar=\bigstar$, so Right will not make this move. Left moves
${P_{8}}~{}\overrightarrow{{}_{L}}~{}{P_{7}=\uparrow}$, and she will move
${P_{8}+\ast}~{}\overrightarrow{{}_{L}}~{}{P_{5}+\ast=\uparrow}$, so Left wins
if Right moves first, and $P_{8}+\bigstar>0$.

Consider $P_{9}+\bigstar$. Right’s move set is
$\mathcal{R}=\\{P_{9},P_{9}+\ast,P_{5}+P_{4}+\bigstar\\}$. Left moves
${P_{9}}~{}\overrightarrow{{}_{L}}~{}{P_{4}+P_{4}=0}$, and she will move
${P_{9}+\ast}~{}\overrightarrow{{}_{L}}~{}{P_{5}+\ast=\uparrow}$. Lastly, she
will move $P_{5}+P_{4}+\bigstar$ to $P_{5}+P_{4}=\uparrow$, so Left wins if
Right moves first, and $P_{9}+\bigstar>0$.

Consider $P_{10}+\bigstar$. Right’s move set is
$\mathcal{R}=\\{P_{10},P_{10}+\ast,P_{5}+P_{5}+\bigstar,P_{4}+P_{6}+\bigstar\\}$.
Left will move ${P_{10}}~{}\overrightarrow{{}_{L}}~{}{P_{4}+P_{5}=\uparrow}$,
and she will move ${P_{10}+\ast}~{}\overrightarrow{{}_{L}}~{}{P_{10}}$. Right
can only move $P_{10}$ to $P_{8}$, $P_{7}$, $P_{5}+P_{5}=\Uparrow$ and
$P_{6}+P_{4}$. We have shown already how Left wins the first three, and Left
will win $P_{6}+P_{4}$ by moving to $P_{5}+P_{4}=\uparrow$. As for the other
first moves Right can make, Left is able to move the smallest path component
to $0$ leaving a base case shown to be a Left win. Thus, Left wins if Right
moves first, and $P_{10}+\bigstar>0$.

We will now proceed with strong induction on $n>10$. Assume for all integers,
$k$, such that $5\leq k\leq n$, that $P_{k}+\bigstar>0$.

Consider $P_{n+1}+\bigstar$. If Left moves first, she will move
${P_{n+1}+\bigstar}~{}\overrightarrow{{}_{L}}~{}{P_{n}+\bigstar}$. By the
induction hypothesis, $P_{n}+\bigstar$ is a Left win regardless which player
moves first.

If Right moves first, his move set is
$\mathcal{R}=\\{P_{n+1},P_{n+1}+\ast,P_{n-1}+\bigstar,P_{n-2}+\bigstar,P_{n-3}+P_{4}+\bigstar,P_{a}+P_{b}+\bigstar\\}$
for some integers $a,b\geq 5$ such that $a+b=n+1$. Left will move
${P_{n+1}}~{}\overrightarrow{{}_{L}}~{}{P_{n-5}+P_{5}}$, and because $n-5\geq
5$, we can apply Theorem 2.7 to see AW$(P_{n-5}+P_{5})\geq 2$. Then by Theorem
2.5, Left will win. Similarly, Left will move
${P_{n+1}+\ast}~{}\overrightarrow{{}_{L}}~{}{P_{n-5}+P_{5}+\ast}$, and
AW$(P_{n-5}+P_{5}+\ast)\geq 2$. Application of the induction hypothesis shows
that $P_{n-1}+\bigstar$ and $P_{n-2}+\bigstar$ are both Left wins. Left will
move ${P_{n-3}+P_{4}+\bigstar}~{}\overrightarrow{{}_{L}}~{}{P_{n-3}+\bigstar}$
which is also a Left win by the induction hypothesis. Finally, because
$a,b\geq 5$, $P_{a}+\bigstar$ and $P_{b}+\bigstar$ are both Left wins, so by
Theorem 2.7 AW$(P_{a}+P_{b}+\bigstar)\geq 2$. Thus, Left will win
$P_{n+1}+\bigstar$ regardless which player moves first.

Therefore, $P_{n}>\bigstar$. Further, by Theorem 2.7, AW$(P_{n})\geq 1$.

Knowing that $P_{n}>\bigstar$, we can approach finding the atomic weight of
$P_{n}$ inductively. We will use the fact that atomic weights are additive to
determine the maximum atomic weight to which Left can move and the minimum
atomic weight to which Right can move. The game position composed of Left’s
maximum minus two and Right’s minimum plus two will be an integer requiring a
comparison of $P_{n}$ to $\bigstar$. However, Theorem 3.16 makes this a
trivial exercise. We then know AW$(P_{n})$ is the greatest integer such that
AW$(P_{n})$ is less than two more than Right’s minimum.

######  Theorem 3.17.

For an integer $n\geq 5$ and path position, $P_{n}$, the atomic weight of
$P_{n}$ is $\left\lceil\frac{n}{4}\right\rceil-1$.

###### Proof.

For an integer $n\geq 5$, let $P_{n}$ be the path position on $n$ vertices in
the Mutual Failures Variant. We will prove by strong induction that
AW$(P_{n})=\left\lceil\frac{n}{4}\right\rceil-1$.

Base Cases: See Example 3.15.

We will now proceed with strong induction on $n>10$. Assume for all integers,
$k$, such that $5\leq k\leq n$, that
AW$(P_{k})=\left\lceil\frac{n}{4}\right\rceil-1$.

Consider AW$(P_{n+1})$. We will apply Theorem 2.6 to find AW$(P_{n+1})$. We
know Left is going to want to maximize the atomic weight, so she will not move
to $P_{n-2}$, $P_{n-3}$, or $P_{n-4}+P_{4}$ as these moves potentially
decrease the atomic weight of the larger component while creating a smaller
component with atomic weight $0$. So then, Left will move to $P_{n}$ or
$P_{a}+P_{n-a}$ where $a$ is an integer and $5\leq a\leq n-5$. By the
induction hypothesis, AW$(P_{n})=\left\lceil\frac{n}{4}\right\rceil-1$. For
her other first move,

$\displaystyle\text{AW}(P_{a}+P_{n-a})$
$\displaystyle=\text{AW}(P_{a})+\text{AW}(P_{n-a})$
$\displaystyle=\left\lceil\frac{a}{4}\right\rceil-1+\left\lceil\frac{n-a}{4}\right\rceil-1$
$\displaystyle\geq\left\lceil\frac{n}{4}\right\rceil-2.$

However, since
$\left\lceil{x}\right\rceil+\left\lceil{y}\right\rceil\leq\left\lceil{x+y}\right\rceil+1$,
$\text{AW}(P_{a}+P_{n-a})\leq\left\lceil\frac{n}{4}\right\rceil-1$. So we will
assume Left will always move ${P_{n+1}}~{}\overrightarrow{{}_{L}}~{}{P_{n}}$
as this consistently yields the largest atomic weight.

Let us now consider Right’s first moves. We know Right is going to want to
minimize the atomic weight, so he will move to $P_{n-3}+P_{4}$ rather than
$P_{n-1}$ or $P_{n-2}$. He could also move
${P_{n+1}}~{}\overrightarrow{{}_{R}}~{}{P_{b}+P_{n+1-b}}$ for an integer $b$
such that $5\leq b\leq n-4$. Notice if $n~{}$mod$~{}4\neq 0$, then

$\left\lceil\frac{n}{4}\right\rceil=\left\lceil\frac{n+1}{4}\right\rceil\text{
and
}\left\lceil\frac{n-3}{4}\right\rceil=\left\lceil\frac{n+1}{4}\right\rceil-1.$

Additionally, if $n~{}$mod$~{}4=0$, then

$\left\lceil\frac{n}{4}\right\rceil=\left\lceil\frac{n-3}{4}\right\rceil\text{
and
}\left\lceil\frac{n}{4}\right\rceil=\left\lceil\frac{n+1}{4}\right\rceil-1.$

Then it follows that

$\displaystyle\text{AW}(P_{n-3}+P_{4})$ $\displaystyle=\text{AW}(P_{n-3})$
$\displaystyle=\left\lceil\frac{n-3}{4}\right\rceil-1$
$\displaystyle=\left\lceil\frac{n+1}{4}\right\rceil-2.$

For his other first move,

$\displaystyle\text{AW}(P_{b}+P_{n+1-b})$
$\displaystyle=\text{AW}(P_{b})+\text{AW}(P_{n+1-b})$
$\displaystyle=\left\lceil\frac{b}{4}\right\rceil-1+\left\lceil\frac{n+1-b}{4}\right\rceil-1$
$\displaystyle\geq\left\lceil\frac{n+1}{4}\right\rceil-2.$

Similarly to the atomic weight of Left’s move from $P_{n+1}$ to
$P_{a}+P_{n-a}$, this atomic weight is bounded above by
$\left\lceil\frac{n+1}{4}\right\rceil-1$. We will assume Right always moves
${P_{n+1}}~{}\overrightarrow{{}_{R}}~{}{P_{n-3}+P_{4}}$ as this consistently
yields the lowest atomic weight.

By applying Theorem 2.6, we see

$\displaystyle\text{AW}(P_{n+1})$
$\displaystyle=\left\\{{\text{AW}(P_{n})-2}~{}|~{}{\text{AW}(P_{n-3}+P_{4})+2}\right\\}$
$\displaystyle=\left\\{{\left\lceil\frac{n}{4}\right\rceil-3}~{}|~{}{\left\lceil\frac{n+1}{4}\right\rceil}\right\\}.$

This is clearly a number and the simplest number that can be assigned to this
game position is $\left\lceil\frac{n}{4}\right\rceil-2$ which is an integer.
Since the exception is invoked, we must compare $P_{n+1}$ to $\bigstar$. By
Theorem 3.16, we know $P_{j}>\bigstar$ for $j\geq 5$. So, AW$(P_{n+1})$ is the
greatest integer, $y$, such that $y\lhd\left\lceil\frac{n+1}{4}\right\rceil$.
Thus, AW$(P_{n+1})=\left\lceil\frac{n+1}{4}\right\rceil-1$.

Therefore, for an integer $n\geq 5$, the atomic weight of $P_{n}$ is
$\left\lceil\frac{n}{4}\right\rceil-1$.

We now have the atomic weight for any $P_{n}$, and this is the only
information needed to solve the Mutual Failures Variant on paths.

######  Theorem 3.18.

For an integer $n\geq 9$, Left will win the path position $P_{n}$ in the
Mutual Failures Variant regardless of which player moves first.

###### Proof.

Let $n\geq 9$ be a positive integer, and let $P_{n}$ be the path position on
$n$ vertices in the Mutual Failures Variant. By Theorem 2.5, we know Left wins
a game position, $G$, if AW$(G)\geq 2$, and by Theorem 3.17, we know
AW$(P_{n})=\left\lceil\frac{n}{4}\right\rceil-1$. Thus, AW$(P_{n})\geq 2$ when
$n\geq 9$. Therefore, for $n\geq 9$, Left will win the path position $P_{n}$
regardless of which player moves first.

As should be expected at this point, we will use the solution for paths to
solve cycles. Since any move on a cycle will produce a path, we see for an
integer $n\geq 10$, Left will win the cycle position $C_{n}$ regardless of
which player moves first. Thus we have the following immediate result:

######  Theorem 3.19.

For an integer $n\geq 10$, Left will win the cycle position $C_{n}$ in the
Mutual Failures Variant regardless of which player moves first.

In wheel graphs, Left will want to delete the hub vertex, moving to $P_{n}$ or
$C_{n}$, both of which are won by Left. Therefore, for an integer $n\geq 10$,
Left will win the wheel position $W_{n}$ regardless of which player moves
first. Thus we have the following immediate result:

######  Theorem 3.20.

For an integer $n\geq 10$, Left will win the wheel position $W_{n}$ in the
Mutual Failures Variant regardless of which player moves first.

Finally, complete graphs work exactly the same in the Mutual Failures Variant
as they do in the Forbidden Leaf Variant. Any complete graph position on more
than four vertices will reduce to the complete graph position on four vertices
in which Left will win moving second.

######  Theorem 3.21.

For an integer $n\geq 5$, Left is guaranteed to win $K_{n}$ in the Mutual
Failures Variant regardless of which player moves first.

The results obtained in the Mutual Failures Variant show that Left is favored
to win. We sought to remove the bias from the Mixed Deletion Game by forcing
Left and Right to have identical sets of unplayable positions, but ultimately
failed to do so. However, we can conclude that this bias does not result from
one player being able to stockpile “free moves.” The only difference between
players is what they are legally allowed to delete from a graph. This bias
must then result from the differences between an edge and a vertex deletion.

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