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Consider an even order, balanced(both partitions have same vertices) bipartite regular graph of order greater than or equal to $12$ and degree atleast six and divisible by $3$. Then is the graph of Type 1(totally colorable by $\Delta+1$ colors where $\Delta$ is the maximum degree)?

By petersen theorem, the graph has 2-factor( in fact k-factors for $k\le2n$, where $2n$ be the total number of vertices). Again, it is a union of disjoint $1$ factors(perfect matchings).Could these be used to provide a total coloring?

Apart from this, since the degree is also divisible by $3$(i.e. divisible by $6$), then, we may be able to find cycles of order divisible by $3$($6$); and since cycles whose order is divisible by $3$ can be totally colored(each pair of adjacent or incident elements of graph receive different colors) with $3$ colors, therefore we can also totally color each disjoint cycle in the graph using just three colors. Can we use this fact to totally color the whole graph with ($\Delta+1$) colors?

Consider an even order, balanced(both partitions have same vertices) bipartite regular graph of order greater than or equal to $12$ and degree atleast six and divisible by $6$. Then is the graph of Type 1(totally colorable by $\Delta+1$ colors where $\Delta$ is the maximum degree)?

By petersen theorem, the graph has 2-factor( in fact k-factors for $k\le2n$, where $2n$ be the total number of vertices). Again, it is a union of disjoint $1$ factors(perfect matchings).Could these be used to provide a total coloring?

Apart from this, since the degree is also divisible by $3$(i.e. divisible by $6$), then, we may be able to find cycles of order divisible by $3$($6$); and since cycles whose order is divisible by $3$ can be totally colored(each pair of adjacent or incident elements of graph receive different colors) with $3$ colors, therefore we can also totally color each disjoint cycle in the graph using just three colors. Can we use this fact to totally color the whole graph with ($\Delta+1$) colors?

Consider an even order, balanced(both partitions have same vertices) bipartite regular graph of order greater than or equal to $12$ and degree atleast six and divisible by $3$. Then is the graph of Type 1(totally colorable by $\Delta+1$ colors where $\Delta$ is the maximum degree)?

By petersen theorem, the graph has 2-factor( in fact k-factors for $k\le2n$, where $2n$ be the total number of vertices). Again, it is a union of disjoint $1$ factors(perfect matchings). Now, if the degree be divisible by $3$(i.e. divisible by $6$), then, we may be able to find cycles of order divisible by $3$($6$); and since cycles whose order is divisible by $3$ can be totally colored(each pair of adjacent or incident elements of graph receive different colors) with $3$ colors, therefore we can also totally color each disjoint cycle in the graph using just three colors. Can we use this fact to totally color the whole graph with ($\Delta+1$) colors?

Consider an even order, balanced(both partitions have same vertices) bipartite regular graph of order greater than or equal to $12$ and degree atleast six and divisible by $3$. Then is the graph of Type 1(totally colorable by $\Delta+1$ colors where $\Delta$ is the maximum degree)?

By petersen theorem, the graph has 2-factor( in fact k-factors for $k\le2n$, where $2n$ be the total number of vertices). Again, it is a union of disjoint $1$ factors(perfect matchings).Could these be used to provide a total coloring?

Apart from this, since the degree is also divisible by $3$(i.e. divisible by $6$), then, we may be able to find cycles of order divisible by $3$($6$); and since cycles whose order is divisible by $3$ can be totally colored(each pair of adjacent or incident elements of graph receive different colors) with $3$ colors, therefore we can also totally color each disjoint cycle in the graph using just three colors. Can we use this fact to totally color the whole graph with ($\Delta+1$) colors?

Consider an even order, balanced(both partitions have same vertices) bipartite regular graph of order greater than or equal to $12$ and degree atleast six and divisible by $3$. Then is the graph of Type 1(totally colorable by $\Delta+1$ colors where $\Delta$ is the maximum degree)?

By petersen theorem, the graph has 2-factor( in fact k-factors for $k\le2n$, where $2n$ be the total number of vertices). Again, it is a union of disjoint $1$ factors(perfect matchings). Now, if the degree be divisible by $3$(i.e. divisible by $6$), then, since cycles whose order is divisible by $3$ can be totally colored(each pair of adjacent or incident elements of graph receive different colors) with $3$ colors, therefore we can also totally color each disjoint cycle in the graph using just three colors. Can we use this fact to totally color the whole graph with ($\Delta+1$) colors?

Consider an even order, balanced(both partitions have same vertices) bipartite regular graph of order greater than or equal to $12$ and degree atleast six and divisible by $3$. Then is the graph of Type 1(totally colorable by $\Delta+1$ colors where $\Delta$ is the maximum degree)?

By petersen theorem, the graph has 2-factor( in fact k-factors for $k\le2n$, where $2n$ be the total number of vertices). Again, it is a union of disjoint $1$ factors(perfect matchings). Now, if the degree be divisible by $3$(i.e. divisible by $6$), then, we may be able to find cycles of order divisible by $3$($6$); and since cycles whose order is divisible by $3$ can be totally colored(each pair of adjacent or incident elements of graph receive different colors) with $3$ colors, therefore we can also totally color each disjoint cycle in the graph using just three colors. Can we use this fact to totally color the whole graph with ($\Delta+1$) colors?