Theoremn $3\mathrm{.}4.$ For the ring Zpt$,$xPG$($Z$,$:$)=$p$-$p$+2^22=5=$p$+3,$Proof. Any two non$-$zero elements of PGi $($Z$,$:$)$ are adjacent if and only if both these elementsare divisible by p$^2$ and p$^2.$ There are p$^2-$ l elements divisible by p$^2$ and p and all are adjacent toeach other. So$,$ these vertices induces a complete subgraph k$-1$ and the vertices of this cliquecan be colored withp$^2-$l colors.There are p$\text{'}($p$-1)$ elements which are divisible byp but not and p$^2.$ These elements arenot adjacent to each other but are adjacent to the elements divisible by p and the elements ina set of units. So$,$ the vertices divisible by p can be colored with any one color assigned to thevertices divisible by p$\text{'}.$Theorem $3\mathrm{.}5.$ For the ring Z$\text{'},$ifp is odd prime.The remaining p$^2($p $-1)$ elements which are not divisible by p$,$ p$^2$ and p are adjacent to allthe elements in a set of non$-$zero zero divisors and form two complete subgraphs having Pelements in each subgraph. So$,$ the vertices in these two complete subgraphs can be properlycolored with colors except the colors assigned to the vertices of the clique ky$-1$:ifp $=2.$Also, the vertex $0$ is adjacent to all other vertices in PGi$($Z$,).$ So$,$ the vertex $0$ can becolored with a single color except the colors assigned to the vertices of the cliques cliques k$-]$:kpn$-.$ Thus, the graph PG$($Z$)$ can be properly colored with $($p $-1)+$ P$+$l colors.Therefore, xPGi $($Z$)=$ PP$,$ ifp is an odd prime.In case when p $=2,$ Zi$6=\{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}.$ Here, the ele$-$ments $\{0,4,8,12\}$ forms a complete subgraph k and the vertices of this clique can be coloredwith $4$ colors. The elements $\{2,6,10,14\}$ are not adjacent to each other and the elements $4$and $12.$ So$,$ these vertices can be colored with any one color assigned to the vertices $4$ and $12.$The elements $\{1,3,5,7,9,11,13,15\}$ are not adjacent to each other but are adjacent to all theelements in a set $\{0,2,4,6,8,10,12,14\}.$ So$,$ these vertices can be colored with a single colorexcept the colors assigned to the vertices of the clique k$4.$ Thus, the graph PGi$($Z$6)$ can beproperly colored by $5$ colors. Hence xPGi$($Z$6)=5=$p$+3.$XPG$($Z$,)=$P$-$p$\text{'}+2(+1)2=6=$p$+4,$ifp is odd prime.fp $=2.$Proof. Any two non$-$zero elements of PGi $($Zs$)$ are adjacent if and only if both these elementsare divisible by p and p$^2.$ There are p$^2-$l elements divisible by p and p$^2$ and all are adjacent toeach other. So$,$ these vertices induces a complete subgraph k$-$ and the vertices of this cliquecan be colored withp$-$ l colors.There are p$^2($p$-1)$ elements which are divisible by p$\text{'}.$ These elements are not adjacent toeach other and the elements divisible by p but are adjacent to the elements divisible by p andp'. So$,$ these vertices can be colored with a single color except the colors assigned to the verticesdivisible by p and p$.$There are p$($p$-1)$ elements which are divisible by p$.$ These elements are not adjacent toeach other and the elements divisible by p and p$\text{'},$ but are adjacent to every element from the setof units in a graph. So$,$ these vertices can be colored with any one color assigned to the verticesdivisible by p and p$.$The remaining p $($p$-1)$ elements which are not divisible by p$,$ p$^2,$ p and p are adjacent to allthe elements in a set of non$-$zero zero divisors and form two complete subgraphs having Pelements in each subgraph. So$,$ the vertices in these two complete subgraphs can be properlyColoring of Prime Graph PGi$($R$)$ and PG$2($R$)$ of a Ring$101$colored with colors except the colors assigned to the vertices of the clique k:$-)$ and theAlso, the vertex O is adjacent to all other vertices in PGi$($Z$).$ So$,$ the vertex $0$ can be coloredwith a single color except the colors assigned to the vertices of the cliques k$-|,$ kyp$-)$ and thevertices divisible by p$^2.$ Thus, the graph PG$1($Z$)$ can be properly colored with $(-1)+1+$p$\text{'}$p$-$D $+1.$ Therefore, xPG$,($Z$,)$ PP$2+$D $,$ifp is an odd prime.In case when p $=2,$ In PGi$($Z$32)$ the elements $\{0,8,16,24\}$ form a complete subgraph k$4.$So$,$ the vertices of this clique can be colored with $4$ colors. The elements $\{4,12,20,28\}$ arenot adjacent to each other but are adjacent to all the elements in a set $\{0,8,16,24\}.$ So$,$ thesevertices can be colored with a single color except the colors assigned to a clique k$4.$ The elements$\{2,6,10,14,18,22,26,30\}$ are not adjacent to each other and the elements divisible by $2$ and$22.$ So$,$ these vertices can be colored by any one color assigned to the vertices divisible by $2$ and$22.$ Also, The elements in a set of units are not adjacent to each other but are adjacent to all theelements in a set of zero$-$divisors. So$,$ these elements can be colored by a single color except thecolors assigned to the clique k$4$ and the elements divisible by $22.$ Thus, the graph PGi$($Z$2)$ canbe properly colored with $6$ colors. Hence xPGi $($Z$2)=6=$p$+4.$vertices divisible by $.$

question : From the code above, how do you color element $0$ in p$**4$ differently from the other elements in p$**4$ with p $=3??$ Is using greedy coloring correct??? please fix my code, I$\text{'}$m really stuck

import networkx as nx

import matplotlib.pyplot as plt

def is$\_$adjacent$\_$zp$4($v$,$ u$,$ p$)$:

if v $==$ u:

return False # No self$-$loops

if v $==0$ or u $==0$:

return True # Vertex $0$ is adjacent to all other vertices

if v $\%($p$*$p$)==0$ and u $\%($p$*$p$)==0$:

return True # divisible by p$^2$

if v $\%($p$*$p$*$p$)==0$ and u $\%($p$*$p$*$p$)==0$:

return True # divisible by p$^3$

if v $\%$ p $==0$ and u $\%$ p $==0$:

return False

if $($v $\%$ p $==0$ and v $\%($p$**2)!=0$ and v $\%($p$**3)!=0)$ and $($u $\%$ p $==0$ and u $\%($p$**2)!=0$ and u $\%($p$**3)!=0)$:

return False

#if $($v $\%$ p $==0$ and v $\%($p$**2)!=0$ and u $\%$ p $==0$ and u $\%($p$**2)!=0)$:

#return True

# Kondisi adjacency jika tidak habis dibagi p$,$ p$^2,$ atau p$^3$

if $($v $\%$ p $!=0$ and v $\%($p$**2)!=0$ and v $\%($p$**3)!=0)$ and $($u $\%$ p $!=0$ and u $\%($p$**2)!=0$ and u $\%($p$**3)!=0)$:

return $($v$+$u$)\%$ p $!=0$ and $($v$+$u$)\%($p$*$p$)!=0$ and $($v$+$u$)\%($p$*$p$*$p$)!=0$

return True

def generate$\_$graph$($vertices$,$ p$,$ is$\_$adjacent$)$:

edges $=[($v$,$ u$)$ for v in vertices for u in vertices if is$\_$adjacent$($v$,$ u$,$ p$)]$

# Create the graph

G $=$ nx$.$Graph$()$

G$.$add$\_$nodes$\_$from$($vertices$)$

G$.$add$\_$edges$\_$from$($edges$)$

return G

def analyze$\_$graph$($G$)$:

max$\_$clique $=$ max$($nx$.$find$\_$cliques$($G$),$ key$=$len$)$

max$\_$clique$\_$size $=$ len$($max$\_$clique$)$

color$\_$map $=$ nx$.$coloring.greedy$\_$color$($G$,$ strategy$=$"largest$\_$first"$)$

chromatic$\_$number $=$ max$($color$\_$map.values$())+1$

return chromatic$\_$number, max$\_$clique, max$\_$clique$\_$size

def draw$\_$graph$($G$,$ color$\_$map, title$)$:

plt$.$figure$($figsize$=(8,6))$

pos $=$ nx$.$spring$\_$layout$($G$)$ # Layout adjusted for better visualization

node$\_$colors $=[$color$\_$map$[$node$]$ for node in G$.$nodes$()]$

nx$.$draw$\_$networkx$\_$nodes$($G$,$ pos, node$\_$color$=$node$\_$colors, node$\_$size$=300,$ cmap$=$plt$.$cm$.$rainbow$)$

nx$.$draw$\_$networkx$\_$edges$($G$,$ pos, width$=1\mathrm{.}0,$ alpha$=0\mathrm{.}5)$

nx$.$draw$\_$networkx$\_$labels$($G$,$ pos, labels$=\{$node: str$($node$)$ for node in G$.$nodes$()\},$ font$\_$color$=$'black', font$\_$size$=10,$ font$\_$weight$=$'medium'$)$

plt$.$title$($title$)$

plt$.$axis$(\text{'}$off$\text{'})$

plt$.$show$()$

# Example usage for p$=3($adjust accordingly for p$=2$ or other odd primes$)$

p $=3$

vertices$\_$zp$4=$ list$($range$($p $*$ p $*$ p $*$ p$))$

# Generate graph for Zp$^4$

G$\_$zp$4=$ generate$\_$graph$($vertices$\_$zp$4,$ p$,$ is$\_$adjacent$\_$zp$4)$

# Analyze the graph

chromatic$\_$number$\_$zp$4,$ max$\_$clique$\_$zp$4,$ max$\_$clique$\_$size$\_$zp$4=$ analyze$\_$graph$($G$\_$zp$4)$

# Expected chromatic number

expected$\_$chromatic$\_$number$\_$zp$4=(($p $*$ p $*$ p$*$p$)-($p $*$ p$*$p$)+<$