Equitable total coloring of \(C_m\square C_n\)
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Publication:1028450
DOI10.1016/j.dam.2008.08.030zbMath1173.05324OpenAlexW1570792098MaRDI QIDQ1028450
Chunling Tong, Lin Xiaohui, Li Zhihe, Yuansheng Yang
Publication date: 30 June 2009
Published in: Discrete Applied Mathematics (Search for Journal in Brave)
Full work available at URL: https://doi.org/10.1016/j.dam.2008.08.030
Related Items (12)
Equitable colorings of Cartesian products of square of cycles and paths with complete bipartite graphs ⋮ Determining equitable total chromatic number for infinite classes of complete \(r\)-partite graphs ⋮ Equitable total coloring of complete $r$-partite $p$-balanced graphs ⋮ Total colorings-a survey ⋮ On total coloring the direct product of cycles and bipartite direct product of graphs ⋮ Equitable total chromatic number of splitting graph ⋮ An Algorithmic Approach to Equitable Total Chromatic Number of Graphs ⋮ Total colorings of product graphs ⋮ Total equitable list coloring ⋮ Total coloring conjecture for certain classes of graphs ⋮ Equitable total coloring of corona of cubic graphs ⋮ On list equitable total colorings of the generalized theta graph
Cites Work
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- On the neighbour-distinguishing index of a graph
- A result on the total colouring of powers of cycles
- Determining the total colouring number is NP-hard
- Total chromatic numbers
- The total coloring of a multigraph with maximal degree 4
- Total colouring regular bipartite graphs is NP-hard
- Equitable total coloring of graphs with maximum degree 3
- The determination of the total chromatic number of series-parallel graphs with \((G) \geq 4\)
- Total chromatic number of one kind of join graphs
- A note on total colourings of digraphs
- Total chromatic number of planar graphs with maximum degree ten
- Behzad-Vizing conjecture and Cartesian product graphs
- Total colourings of Cartesian products
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