2 n 2 therefore 2 m n 2 n 2 k yielding k n 2 n 2 2 m

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2 = n 2 . Therefore, 2 m n 2 - n 2 k , yielding k n 2 n 2 - 2 m . Cartesian product Theorem 4 (Vizing 1963, Aberth 1964) . Let G and H be graphs. Then χ ( G 2 H ) = max { χ ( G ) , χ ( H ) } . Proof. Since G 2 H contains copies of G and H as subgraphs, we have χ ( G 2 H ) χ ( G ) and χ ( G 2 H ) χ ( H ). Hence χ ( G 2 H ) max { χ ( G ) , χ ( H ) } . Proof (continued). Denote k = max { χ ( G ) , χ ( H ) } . Let g be a proper colour- ing of G using χ ( G ) colours, say 1 , 2 , . . . , χ ( G ). Let h be a proper colouring of H using χ ( H ) colours, say 1 , 2 , . . . , χ ( H ). For each vertex ( u, v ) V ( G ) × V ( H ), define f ( u, v ) = g ( u ) + h ( v ) mod k, which is an integer in { 1 , 2 , . . . , k } . (We use { 1 , 2 , . . . , k } instead of the more commonly used { 0 , 1 , . . . , k - 1 } for the set of reminders.) We claim that f is a proper colouring of G 2 H . Indeed, if ( u, v ) and ( x, y ) are adjacent in G 2 H , then g ( u )+ h ( v ) and g ( x )+ h ( y ) agree in one of the summands and differ in the other by some integer between 1 and k . So g ( u ) + h ( v ) and g ( x ) + h ( y ) differ by some integer between 1 and k . Thus they are in different congruence classes modulo k . In other words, ( u, v ) and ( x, y ) receive different colours under f . Therefore, χ ( G 2 H ) k .
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3 A greedy heuristic A greedy colouring heuristic colours = { 1 , 2 , 3 , . . . } Algorithm GreedyColour Input : A graph G Output : A proper colouring of G 1. Arrange the vertices of G in a linear order: u 1 , u 2 , . . . , u n . 2. Colour the vertices one by one in this order, assigning to u i the smallest positive integer not yet assigned to one of its already-coloured neighbours. By the rule in step 2, a proper colouring of G is determined This is a heuristic; it does not necessarily compute χ (G) Sometimes it performs badly; often it performs reasonably well The number of colours required depends on the linear order of vertices Welsh-Powell Theorem 5. (Welsh-Powell 1967) Let G be a graph with degree sequence d 1 d 2 ≥ · · · ≥ d n . Then χ ( G ) 1 + max 1 i n min { d i , i - 1 } . Proof. Denote the vertices of G by u 1 , u 2 , . . . , u n such that deg( u i ) = d i , 1 i n . Apply GreedyColour to G using this order. At the time when we colour u i , it has at most min { d i , i - 1 } earlier neighbours (i.e. from { u 1 , . . . , u i - 1 } ), and so at most this many colours appear on its earlier neighbours. Hence the colour we assign to u i is at most 1 + min { d i , i - 1 } . Since this holds for every vertex u i , we need at most max 1 i n (1 + min { d i , i - 1 } ) = 1 + max 1 i n min { d i , i - 1 } colours. Therefore, χ ( G ) 1 + max 1 i n min { d i , i - 1 } . A greedy colouring heuristic colours = { 1 , 2 , 3 , . . . } Algorithm ColourGraph ( G ) 1. let n := | V ( G ) | 2. for i = 1 , 2 , . . . , n do let G i := G - { v 1 , . . . , v i - 1 } [ G 1 = G ] choose v i to be a vertex with minimum degree in G i end-for
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3. for i = n, n - 1 , . . . , 1 do colour v i by the smallest colour not used by any neighbour of v i that is already coloured (i.e., with a higher subscript) end-for Analysis of ColourGraph Theorem 6. Let G be a graph such that every subgraph of it has minimum degree at most k . Then ColourGraph ( G ) produces a ( k + 1)-colouring of G .
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