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Example 1. \( {y}^{\prime } = x - 1/y\;\left( {y > 0}\right) \) .
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In the first example, the special solution \( \phi \) is the only bounded global solution. The solutions above \( \phi \) tend to \( \infty \) as \( x \rightarrow \infty \), while every positive solution beneath \( \phi \) exists only in a finite interval \( \lbrack 0, b) \) and tends to 0 as \( x \rightarrow b - \) .
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No
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Example 1. Here the global solutions \( \phi ,\psi \) are bounded; that is, there exists \( L > 0 \) such that \( 0 < \phi < \psi < L \) holds in \( \lbrack 0,\infty ) \) and hence \( {f}_{y}\left( {x, y}\right) = 1/{y}^{2} > \) \( 1/{L}^{2} \mathrel{\text{:=}} \alpha \) .
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\[ {u}^{\prime } \geq {\alpha u},\;\text{ which implies that }\;u\left( x\right) \geq u\left( 0\right) {\mathrm{e}}^{\alpha x}. \] But \( u \) is bounded by assumption. This contradiction proves the assertion made at the beginning that there is only one bounded global solution. The estimate \( {v}_{1} < \phi < 1/x \) (see XIII) shows that \( \phi \) behaves like \( 1/x \) as \( x \rightarrow \infty \) and that \[ \frac{1}{x} - \frac{1}{{x}^{4}} < \phi \left( x\right) < \frac{1}{x} \]
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Yes
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Let \( y \) be a solution and \( y\left( a\right) \geq - a \) for some \( a \geq 0 \). It is easy to see from the differential equation that there exists \( b > a \) with \( y\left( b\right) > 0 \). Since the solution of the initial value problem \( {v}^{\prime } = {v}^{3}, v\left( b\right) = y\left( b\right) \) is a lower solution to \( y \) and since there exists \( c < \infty \) such that \( v\left( x\right) \rightarrow \infty \) as \( x \rightarrow c \), it follows that \( y\left( x\right) \) is not a global solution.
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If \( \phi \) and \( \psi \) are global solutions with \( \phi < \psi \), then accordingly, \( \psi \left( x\right) < - x \). Thus in (14) we have \( {f}_{y}\left( {x,{y}^{ * }}\right) = 3{y}^{*2} > 3{x}^{2} \), and hence \( u = \psi - \phi \geq \delta \exp \left( {x}^{3}\right) \), where \( \delta = u\left( 0\right) > 0 \). In particular, \( z = - \phi \geq \delta \exp \left( {x}^{3}\right) \). Since \( z \) satisfies \( {z}^{\prime } = {z}^{3} - {x}^{3} \), we have \( {z}^{\prime } > \frac{1}{2}{z}^{3} \) for large \( x \). This implies in a manner similar to the case above that \( z = - \phi \) exists only in a finite interval \( \lbrack a, b) \) and tends to \( \infty \) as \( x \rightarrow b \). Therefore, there exists only one global solution. This proves the assertion made at the beginning for the second example.
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Yes
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Corollary 1. If the sets \( A \) and \( B \) are homeomorphic and if \( A \) has the fixed point property, then \( B \) also has the fixed point property.
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The proof is very simple. Let \( h : A \rightarrow B \) be a homeomorphism and \( f \) : \( B \rightarrow B \) a continuous mapping. Then \( F = {h}^{-1} \circ f \circ h \) is a continuous mapping of \( A \) to itself. If \( x \) is a fixed point of \( F \), then the image point \( \xi = h\left( x\right) \) is a fixed point of \( f \), as one easily verifies.
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Yes
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Corollary 2. Let the set \( A \subset {\mathbb{R}}^{n} \) be compact, and let there exist a continuous mapping \( P : {\mathbb{R}}^{n} \rightarrow A \) with \( {\left. P\right| }_{A} = {\operatorname{id}}_{A} \), i.e., \( P\left( x\right) = x \) for \( x \in A \) . Then \( A \) has the fixed point property.
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For the proof let \( B \supset A \) be a closed ball and \( f : A \rightarrow A \) continuous. Then \( F = f \circ P \) is a continuous mapping of \( B \) into itself. By the Brouwer fixed point theorem, \( F \) has a fixed point \( \xi \), and because \( F\left( B\right) \subset A \), this fixed point belongs to \( A \), whence \( \xi = f\left( \xi \right) \), i.e., \( \xi \) is a fixed point of \( f \) .
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Yes
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Corollary 3. A nonempty, convex, and compact set \( A \subset {\mathbb{R}}^{n} \) has the fixed point property.
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Proof. For every \( x \in {\mathbb{R}}^{n} \) there exists, since \( A \) is convex and compact, exactly one \
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No
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Theorem 13. Given a separable Banach space \( E \) such that every sequence of elements of \( E \) that is bounded in norm has a subsequence weakly convergent to an element of \( E \), the space \( E \) is isometrically isomorphic to the space \( {E}^{* * } \) (the dual of \( {E}^{ * } \) ).
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Though some of these terms have yet to be defined here, their meanings are not important at the moment. An examination of Banach's proof of this theorem shows that the isometric isomorphism he had in mind is the natural map from \( E \) into \( {E}^{* * } \), so the conclusion of Banach’s theorem is that \( E \) must be reflexive. In a note on that theorem given on page 243 of Banach's book, he made the following statement.
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No
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Theorem 1 The edge set of a graph can be partitioned into cycles if, and only if, every vertex has even degree.
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Proof. The condition is clearly necessary, since if a graph is the union of some edge disjoint cycles and isolated vertices, then a vertex contained in \( k \) cycles has degree \( {2k} \) .\n\nSuppose that every vertex of a graph \( G \) has even degree and \( e\left( G\right) > 0 \) . How can we find a single cycle in \( G \) ? Let \( {x}_{0}{x}_{1}\cdots {x}_{\ell } \) be a path of maximal length \( \ell \) in \( G \) . Since \( {x}_{0}{x}_{1} \in E\left( G\right) \), we have \( d\left( {x}_{0}\right) \geq 2 \) . But then \( {x}_{0} \) has another neighbour \( y \) in addition to \( {x}_{1} \) ; furthermore, we must have \( y = {x}_{i} \) for some \( i,2 \leq i \leq \ell \), since otherwise \( y{x}_{0}{x}_{1}\cdots {x}_{\ell } \) would be a path of length \( \ell + 1 \) . Therefore, we have found our cycle: \( {x}_{0}{x}_{1}\cdots {x}_{i} \) .\n\nHaving found one cycle, \( {C}_{1} \), say, all we have to do is to repeat the procedure over and over again. To formalize this, set \( {G}_{1} = G \), so that \( {C}_{1} \) is a cycle in \( {G}_{1} \) , and define \( {G}_{2} = {G}_{1} - E\left( {C}_{1}\right) \) . Every vertex of \( {G}_{2} \) has even degree, so either\n\n\( E\left( {G}_{2}\right) = \varnothing \) or else \( {G}_{2} \) contains a cycle \( {C}_{2} \) . Continuing in this way, we find vertex disjoint cycles \( {C}_{1},{C}_{2},\ldots ,{C}_{s} \) such that \( E\left( G\right) = \mathop{\bigcup }\limits_{{i = 1}}^{s}E\left( {C}_{i}\right) \) .
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Yes
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Theorem 2 Every graph of order \( n \) and size greater than \( \left\lfloor {{n}^{2}/4}\right\rfloor \) contains a triangle.
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Proof. Let \( G \) be a triangle-free graph of order \( n \) . Then \( \Gamma \left( x\right) \cap \Gamma \left( y\right) = \varnothing \) for every edge \( {xy} \in E\left( G\right) \), so\n\n\[ d\left( x\right) + d\left( y\right) \leq n. \]\n\nSumming these inequalities for all \( e\left( G\right) \) edges \( {xy} \), we find that\n\n\[ \mathop{\sum }\limits_{{x \in G}}d{\left( x\right) }^{2} \leq {ne}\left( G\right) \]\n\n(3)\n\nNow by (1) and Cauchy's inequality,\n\n\[ {\left( 2e\left( G\right) \right) }^{2} = {\left( \mathop{\sum }\limits_{{x \in G}}d\left( x\right) \right) }^{2} \leq n\left( {\mathop{\sum }\limits_{{x \in G}}d{\left( x\right) }^{2}}\right) . \]\n\nHence, by (3),\n\n\[ {\left( 2e\left( G\right) \right) }^{2} \leq {n}^{2}e\left( G\right) \]\n\nimplying that \( e\left( G\right) \leq {n}^{2}/4 \) .
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Yes
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Theorem 4 A graph is bipartite iff it does not contain an odd cycle.
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Proof. Suppose \( G \) is bipartite with vertex classes \( {V}_{1} \) and \( {V}_{2} \) . Let \( {x}_{1}{x}_{2}\cdots {x}_{l} \) be a cycle in \( G \) . We may assume that \( {x}_{1} \in {V}_{1} \) . Then \( {x}_{2} \in {V}_{2},{x}_{3} \in {V}_{1} \), and so on: \( {x}_{i} \in {V}_{1} \) iff \( i \) is odd. Since \( {x}_{l} \in {V}_{2} \), we find that \( l \) is even.\n\nSuppose now that \( G \) does not contain an odd cycle. Since a graph is bipartite iff each component of it is, we may assume that \( G \) is connected. Pick a vertex \( x \in V\left( G\right) \) and put \( {V}_{1} = \{ y : d\left( {x, y}\right) \) is odd \( \} ,{V}_{2} = V \smallsetminus {V}_{1} \) . There is no edge joining two vertices of the same class \( {V}_{i} \), since otherwise \( G \) would contain an odd cycle. Hence \( G \) is bipartite.
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Yes
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Theorem 5 A graph is a forest iff for every pair \( \{ x, y\} \) of distinct vertices it contains at most one \( x - y \) path.
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Proof. If \( {x}_{1}{x}_{2}\cdots {x}_{l} \) is a cycle in a graph \( G \), then \( {x}_{1}{x}_{2}\cdots {x}_{l} \) and \( {x}_{1}{x}_{l} \) are two \( {x}_{1} - {x}_{l} \) paths in \( G \) .\n\nConversely, let \( {P}_{1} = {x}_{0}{x}_{1}\cdots {x}_{l} \) and \( {P}_{2} = {x}_{0}{y}_{1}{y}_{2}\cdots {y}_{k}{x}_{l} \) be two distinct \( {x}_{0} - {x}_{l} \) paths in a graph \( G \) . Let \( i + 1 \) be the minimal index for which \( {x}_{i + 1} \neq {y}_{i + 1} \) and let \( j \) be the minimal index for which \( j \geq i \) and \( {y}_{j + 1} \) is a vertex of \( {P}_{1} \), say \( {y}_{j + 1} = {x}_{h} \) . Then \( {x}_{i}{x}_{i + 1}\cdots {x}_{h}{y}_{j}{y}_{j - 1}\cdots {y}_{i + 1} \) is a cycle in \( G \) .
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Yes
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Theorem 6 The following assertions are equivalent for a graph \( G \) .\ni. \( G \) is a tree.\nii. \( G \) is a minimal connected graph, that is, \( G \) is connected and if \( {xy} \in E\left( G\right) \), then \( G - {xy} \) is disconnected. [In other words, \( G \) is connected and every edge is a bridge.]\niii. \( G \) is a maximal acyclic graph; that is, \( G \) is acyclic and if \( x \) and \( y \) are nonadjacent vertices of \( G \), then \( G + {xy} \) contains a cycle.
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Proof. Suppose \( G \) is a tree. For an edge \( {xy} \in E\left( G\right) \), the graph \( G - {xy} \) cannot contain an \( x - y \) path \( x{z}_{1}{z}_{2}\cdots {z}_{k}y \), since otherwise \( G \) contains the cycle \( x{z}_{1}{z}_{2}\cdots {z}_{k}y \) . Hence \( G - {xy} \) is disconnected; and so \( G \) is a minimal connected graph. Similarly, if \( x \) and \( y \) are nonadjacent vertices of the tree \( G \) then \( G \), contains a path \( x{z}_{1}{z}_{2}\cdots {z}_{k}y \), and so \( G + {xy} \) contains the cycle \( x{z}_{1}{z}_{2}\cdots {z}_{k}y \) . Hence \( G + {xy} \) contains a cycle, and so \( G \) is a maximal acyclic graph.\n\nSuppose next that \( G \) is a minimal connected graph. If \( G \) contains a cycle \( x{z}_{1}{z}_{2}\cdots {z}_{k}y \), then \( G - {xy} \) is still connected, since in any \( u - v \) walk in \( G \) the edge \( {xy} \) can be replaced by the path \( x{z}_{1}{z}_{2}\cdots {z}_{k}y \) . As this contradicts the minimality of \( G \), we conclude that \( G \) is acyclic and so it is a tree.\n\nSuppose, finally, that \( G \) is a maximal acyclic graph. Is \( G \) connected? Yes, since if \( x \) and \( y \) belong to different components, the addition of \( {xy} \) to \( G \) cannot create a cycle \( x{z}_{1}{z}_{2}\cdots {z}_{k}y \), since otherwise the path \( x{z}_{1}{z}_{2}\cdots {z}_{k}y \) is in \( G \) . Thus \( G \) is a tree.
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Yes
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Corollary 7 Every connected graph contains a spanning tree, that is, a tree containing every vertex of the graph.
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## Proof. Take a minimal connected spanning subgraph.
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No
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Corollary 9 A tree of order at least 2 contains at least 2 vertices of degree 1.
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Proof. Let \( {d}_{1} \leq {d}_{2} \leq \cdots \leq {d}_{n} \) be the degree sequence of a tree \( T \) of order \( n \geq 2 \) . Since \( T \) is connected, \( \delta \left( T\right) = {d}_{1} \geq 1 \) . Hence if \( T \) had at most one vertex of degree 1, by (1) and Corollary 8 we would have\n\n\[ \n{2n} - 2 = {2e}\left( T\right) = \mathop{\sum }\limits_{1}^{n}{d}_{i} \geq 1 + 2\left( {n - 1}\right) .\n\]
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Yes
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Theorem 10 Each of the four methods described above produces an economical spanning tree. If no two edges have the same cost, then there is a unique economical spanning tree.
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Proof. Choose an economical spanning tree \( T \) of \( G \) that has as many edges in common with \( {T}_{1} \) as possible, where \( {T}_{1} \) is a spanning tree constructed by the first method.\n\nSuppose that \( E\left( {T}_{1}\right) \neq E\left( T\right) \) . The edges of \( {T}_{1} \) have been selected one by one: let \( {xy} \) be the first edge of \( {T}_{1} \) that is not an edge of \( T \) . Then \( T \) contains a unique \( x - y \) path, say \( P \) . This path \( P \) has at least one edge, say \( {uv} \), that does not belong to \( {T}_{1} \), since otherwise \( {T}_{1} \) would contain a cycle. When \( {xy} \) was selected as an edge of \( {T}_{1} \), the edge \( {uv} \) was also a candidate. As \( {xy} \) was chosen and not \( {uv} \), the edge \( {xy} \) cannot be costlier then \( {uv} \) ; that is, \( f\left( {xy}\right) \leq f\left( {uv}\right) \) . Then \( {T}^{\prime } = T - {uv} + {xy} \) is a spanning tree, and since \( f\left( {T}^{\prime }\right) = f\left( T\right) - f\left( {uv}\right) + f\left( {xy}\right) \leq f\left( T\right) \), the new tree \( {T}^{\prime } \) is an economical spanning tree of \( G \) . (Of course, this inequality implies that \( f\left( {T}^{\prime }\right) = f\left( T\right) \) and \( f\left( {xy}\right) = f\left( {uv}\right) \) .) This tree \( {T}^{\prime } \) has more edges in common with \( {T}_{1} \) than \( T \), contradicting the choice of \( T \) . Hence \( T = {T}_{1} \), so \( {T}_{1} \) is indeed an economical spanning tree.\n\nSlight variants of the proof above show that the spanning trees \( {T}_{2} \) and \( {T}_{3} \) , constructed by the second and third methods, are also economical. We invite the reader to furnish the details (Exercise 44).\n\nSuppose now that no two edges have the same cost; that is, \( f\left( {xy}\right) \neq f\left( {uv}\right) \) whenever \( {xy} \neq {uv} \) . Let \( {T}_{4} \) be the spanning tree constructed by the fourth method and let \( T \) be an economical spanning tree. Suppose that \( T \neq {T}_{4} \), and let \( {xy} \) be the first edge not in \( T \) that we select for \( {T}_{4} \) . The edge \( {xy} \) was selected, since it is the least costly edge of \( G \) joining a vertex of a subtree \( F \) of \( {T}_{4} \) to a vertex outside \( F \) . The \( x - y \) path in \( T \) has an edge \( {uv} \) joining a vertex of \( F \) to a vertex outside \( F \) so \( f\left( {xy}\right) < f\left( {uv}\right) \) . However, this is impossible, since \( {T}^{\prime } = T - {uv} + {xy} \) is a spanning tree of \( G \) and \( f\left( {T}^{\prime }\right) < f\left( T\right) \) . Hence \( T = {T}_{4} \) . This shows that \( {T}_{4} \) is indeed an economical spanning tree. Furthermore, since the spanning tree constructed by the fourth method is unique, the economical spanning tree is unique if no two edges have the same cost.
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No
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Theorem 12 A non-trivial connected graph has an Euler circuit iff each vertex has even degree.
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Proof. The conditions are clearly necessary. For example, if \( G \) has an Euler circuit \( {x}_{1}{x}_{2}\cdots {x}_{m} \), and \( x \) occurs \( k \) times in the sequence \( {x}_{1},{x}_{2},\ldots ,{x}_{m} \), then \( d\left( x\right) = {2k} \) .\n\nWe prove the sufficiency of the first condition by induction on the number of edges. If there are no edges, there is nothing to prove, so we proceed to the induction step.\n\nLet \( G \) be a non-trivial connected graph in which each vertex has even degree. Since \( e\left( G\right) \geq 1 \), we find that \( \delta \left( G\right) \geq 2 \), so by Corollary \( 9, G \) contains a cycle. Let \( C \) be a circuit in \( G \) with the maximal number of edges. Suppose \( C \) is not Eulerian. As \( G \) is connected, \( C \) contains a vertex \( x \) that is in a non-trivial component \( H \) of \( G - E\left( C\right) \) . Every vertex of \( H \) has even degree in \( H \), so by the induction hypothesis, \( H \) contains an Euler circuit \( D \) . The circuits \( C \) and \( D \) (see Fig. I.16) are edge-disjoint and have a vertex in common, so they can be concatenated to form a circuit with more edges than \( C \) . As this contradicts the maximality of \( e\left( C\right) \), the circuit \( C \) is Eulerian.\n\nSuppose now that \( G \) is connected and \( x \) and \( y \) are the only vertices of odd degree. Let \( {G}^{ * } \) be obtained from \( G \) by adding to it a vertex \( u \) together with the edges \( {ux} \) and \( {uy} \) . Then, by the first part, \( {G}^{ * } \) has an Euler circuit \( {C}^{ * } \) . Clearly, \( {C}^{ * } - u \) is an Euler trail from \( x \) to \( y \) .
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Yes
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Theorem 15 If a connected plane graph \( G \) has \( n \) vertices, \( m \) edges, and \( f \) faces, then\n\n\[ n - m + f = 2\text{.} \]
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Proof. Let us apply induction on the number of faces. If \( f = 1 \), then \( G \) does not contain a cycle, so it is a tree, and the result holds by Corollary 8.\n\nSuppose now that \( f > 1 \) and the result holds for smaller values of \( f \) . Let \( {ab} \) be an edge in a cycle of \( G \) . Since a cycle separates the plane, the edge \( {ab} \) is in the boundary of two faces, say \( S \) and \( T \) . Omitting \( {ab} \), in the new plane graph \( {G}^{\prime } \) the faces \( S \) and \( T \) join up to form a new face, while all other faces of \( G \) remain unchanged. Thus if \( {n}^{\prime },{m}^{\prime } \) and \( {f}^{\prime } \) are the parameters of \( {G}^{\prime } \), then \( {n}^{\prime } = n,{m}^{\prime } = m - 1 \) , and \( {f}^{\prime } = f - 1 \) . Hence \( n - m + f = {n}^{\prime } - {m}^{\prime } + {f}^{\prime } = 2 \) .
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Yes
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Theorem 16 A planar graph of order \( n \geq 3 \) has at most \( {3n} - 6 \) edges. Furthermore, a planar graph of order \( n \) and girth at least \( g,3 \leq g < \infty \), has size at most\n\n\[ \max \left\{ {\frac{g}{g - 2}\left( {n - 2}\right), n - 1}\right\} . \]
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Proof. The first assertion is the case \( g = 3 \) of the second, so it suffices to prove the second assertion. Let \( G \) be a planar graph of order \( n \), size \( m \), and girth at least \( g \) . If \( n \leq g - 1 \), then \( G \) is acyclic, so \( m \leq n - 1 \) . Assume now that \( n \geq g \) and the assertion holds for smaller values of \( n \) . We may assume without loss of generality that \( G \) is connected. If \( {ab} \) is a bridge then \( G - {ab} \) is the union of two vertex disjoint subgraphs, say \( {G}_{1} \) and \( {G}_{2} \) . Putting \( {n}_{i} = \left| {G}_{i}\right| ,{m}_{i} = e\left( {G}_{i}\right), i = 1,2 \), by induction we find that\n\n\[ m = {m}_{1} + {m}_{2} + 1 \leq \max \left\{ {\frac{g}{g - 2}\left( {{n}_{1} - 2}\right) ,{n}_{1} - 1}\right\} \n\n+ \max \left\{ {\frac{g}{g - 2}\left( {{n}_{2} - 2}\right) ,{n}_{2} - 1}\right\} + 1 \n\n\leq \max \left\{ {\frac{g}{g - 2}\left( {n - 2}\right), n - 1}\right\} . \n\nOn the other hand, if \( G \) is bridgeless,(4) and (5) imply that\n\n\[ {2m} = \mathop{\sum }\limits_{i}i{f}_{i} = \mathop{\sum }\limits_{{i \geq g}}i{f}_{i} \geq g\mathop{\sum }\limits_{i}{f}_{i} = {gf}. \n\nHence, by Euler's formula,\n\n\[ m + 2 = n + f \leq n + \frac{2}{g}m \n\nand so\n\n\[ m \leq \frac{g}{g - 2}\left( {n - 2}\right) \]
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Yes
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Theorem 19 Let \( R \) be a commutative ring and let the matrices \( {A}_{1},{A}_{2},\ldots ,{A}_{2k} \) be in \( {M}_{k}\left( R\right) \) . Then \( \left\lbrack {{A}_{1},{A}_{2},\ldots ,{A}_{2k}}\right\rbrack = 0 \) .
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Proof. We shall deduce the result from a lemma about Euler trails in directed multigraphs. Let \( \overrightarrow{G} \) be a directed multigraph of order \( n \) with edges \( {e}_{1},{e}_{2},\ldots ,{e}_{m} \) . Thus to each edge \( {e}_{i} \) we associate an ordered pair of not necessarily distinct vertices: the initial vertex of \( {e}_{i} \) and the terminal vertex of \( {e}_{i} \) . Every (directed) Euler trail \( P \) is readily identified with a permutation of \( \{ 1,2,\ldots, m\} \) ; define \( \varepsilon \left( P\right) \) to be the sign of this permutation. Given not necessarily distinct vertices \( x, y \) of \( \overrightarrow{G} \), put \( \varepsilon \left( {\overrightarrow{G};x, y}\right) = \mathop{\sum }\limits_{P}\varepsilon \left( P\right) \), where the summation is over all Euler trails from \( x \) to \( y \) .\n\nLemma 20 If \( m \geq {2n} \) then \( \varepsilon \left( {\overrightarrow{G};x, y}\right) = 0 \) .\n\nBefore proving this lemma, let us see how it implies Theorem 19. Write \( {E}_{ij} \in \) \( {M}_{n}\left( R\right) \) for the matrix whose only non-zero entry is a 1 in the \( i \) th row and \( j \) th column. Since \( \left\lbrack {{A}_{1},{A}_{2},\ldots ,{A}_{2n}}\right\rbrack \) is \( R \) -linear in each variable and \( \left\{ {{E}_{ij} : 1 \leq i, j \leq }\right. \) \( n\} \) is a basis of \( {M}_{n}\left( R\right) \) as an \( R \) -module, it suffices to prove Theorem 19 when \( {A}_{k} = \) \( {E}_{{i}_{k}{j}_{k}} \) for each \( k \) . Assuming that this is the case, let \( \overrightarrow{G} \) be the directed multigraph with vertex set \( \{ 1,2,\ldots, n\} \) whose set of directed edges is \( \left\{ {{i}_{1}{j}_{1},{i}_{2}{j}_{2},\ldots ,{i}_{2n}{j}_{2n}}\right\} \) . By the definition of matrix multiplication, a product \( {A}_{▟}{A}_{\blacksquare }\cdots {A}_{\blacksquare n} \) is \( {E}_{ij} \) if the corresponding sequence of edges is a (directed) Euler trail from \( i \) to \( j \) and otherwise it is 0 . Hence \( \left\lbrack {{A}_{1},{A}_{2},\ldots ,{A}_{2n}}\right\rbrack = \mathop{\sum }\limits_{{i, j}}\varepsilon \left( {\overrightarrow{G};i, j}\right) {E}_{ij} \) . By Lemma 20 each summand is 0 , so the sum is also 0 , and Theorem 19 is proved.
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Yes
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Lemma 20 If \( m \geq {2n} \) then \( \varepsilon \left( {\overrightarrow{G};x, y}\right) = 0 \) .
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Proof of Lemma 20. We may clearly assume that \( \overrightarrow{G} \) has no isolated vertices. Let \( {\overrightarrow{G}}^{\prime } \) be obtained from \( \overrightarrow{G} \) by adding to it a vertex \( {x}^{\prime } \), a path of length \( m + 1 - {2n} \) from \( {x}^{\prime } \) to \( x \), and an edge from \( y \) to \( {x}^{\prime } \) (see Fig. I.22). Then \( {\overrightarrow{G}}^{\prime } \) has order \( n + \left( {m + 1 - {2n}}\right) = \) \( m + 1 - n \) and size \( m + m + 1 - {2n} + 1 = 2\left( {m + 1 - n}\right) \) . Furthermore, it is easily checked that \( \left| {\varepsilon \left( {\overrightarrow{G};x, y}\right) }\right| = \left| {\varepsilon \left( {{\overrightarrow{G}}^{\prime };{x}^{\prime },{x}^{\prime }}\right) }\right| \) . Hence it suffices to prove the theorem when \( m = {2n} \) and \( x = y \) .\n\nGiven a vertex \( z \), recall that \( {d}^{ + }\left( z\right) \) is the number of edges starting at \( z \) and recall that \( {d}^{ - }\left( z\right) \) is the number of edges ending at \( z \) . Call \( d\left( z\right) = {d}^{ + }\left( z\right) + {d}^{ - }\left( z\right) \) the degree of \( z \) and \( f\left( z\right) = {d}^{ + }\left( z\right) - {d}^{ - }\left( z\right) \) the flux at \( z \) . We may assume that \( \overrightarrow{G} \) contains an Euler circuit (an Euler trail from \( x \) to \( x \) ; otherwise, there is nothing to prove. In this case, each vertex has 0 flux, even degree, and the degree is at least 2. Furthermore, we may assume that there is no double edge (and so no double loop), for otherwise the assertion is trivial.\n\nIn order to prove the theorem in the case \( m = {2n} \) and \( x = y \) we apply induction on \( n \) . The case \( n = 1 \) being trivial, we turn to the induction step. We shall distinguish three cases.\n\n(i) There is a vertex \( b \neq x \) of degree 2; say \( {e}_{m - 1} = {ab} \) ends at \( b \) and \( {e}_{m} = {bc} \) starts at \( b \) . If \( a = c \), the assertion follows by applying the induction hypothesis to \( \overrightarrow{G} - b \) . If \( a \neq c \), then without loss of generality \( x \neq c \) . Let \( {e}_{1} = c{c}_{1} \) , \( {e}_{2} = c{c}_{2},\ldots ,{e}_{t} = c{c}_{t} \) be the edges starting at \( c \) . For each \( i,1 \leq i \leq t \), construct a graph \( {\overrightarrow{G}}_{i} \) from \( \overrightarrow{G} - b \) by omitting \( {e}_{i} \) and adding \( {e}_{i}^{\prime } = a{c}_{i} \)
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Yes
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Theorem 5 A rectangle has a perfect squaring if, and only if, the ratio of two neighbouring sides is rational.
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The result can be proved by putting together appropriate perfect rectangles; for the proof we refer the reader to the original paper of Sprague.
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No
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Theorem 6 Let a rectangle \( T \) be tiled with rectangles \( {T}_{1},\ldots ,{T}_{\ell } \) . If each \( {T}_{i} \) has an integer side then so does \( T \) .
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First Proof. Construct a bipartite graph \( G \), with vertex classes \( L \) and \( R \), as follows. Let \( L \) (for ’left’ or ’lattice points’) be the set of integer lattice points in the tiled rectangle: \( L = \left\{ {\left( {x, y}\right) \in {\mathbb{Z}}^{2} : 0 \leq x \leq a,0 \leq y \leq b}\right\} \), and let \( R \) (for ’right’ or ’rectangles’) be the set of tiling rectangles \( {T}_{1},\ldots ,{T}_{\ell } \) . Our graph \( G \) has vertex set \( L \cup R \), and \( \left( {x, y}\right) \in L \) is joined to \( {T}_{i} \in R \) if \( \left( {x, y}\right) \) is a vertex (’corner’) of \( {T}_{i} \) . Then, since each \( {T}_{i} \) has an integer side, each \( {T}_{i} \) has degree 0,2 or 4, so \( e\left( G\right) \) is even.\n\nAlso, every vertex in \( L \), other than the corners of \( T \), has degree 0,2 or 4, but the corner \( \left( {0,0}\right) \in L \) has degree 1 . Hence \( G \) has at least one edge incident with another corner: in particular, at least one other corner belongs to \( L \), and we are done.
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No
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Theorem 7 Let a box \( B \) in \( {\mathbb{R}}^{n} \) be tiled with boxes \( {B}_{1},\ldots ,{B}_{\ell } \) . If each \( {B}_{i} \) has at least \( k \) integer sides, then \( B \) itself has at least \( k \) integer sides.
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Proof. The fourth proof above carries over, mutatis mutandis. Defining \( {\phi }_{\varepsilon } \) as before, with \( {\phi }_{\varepsilon }\left( \mathbf{z}\right) = \left( {{\phi }_{\varepsilon }\left( {z}_{1}\right) ,\ldots ,{\phi }_{\varepsilon }\left( {z}_{n}\right) }\right) \), we find that, if \( \varepsilon > 0 \) is small enough, each \( \left| {{\phi }_{\varepsilon }\left( {B}_{i}\right) }\right| \) is a polynomial of degree at most \( n - k \) . Also, if \( B \) has precisely \( h \) integer sides then \( \left| {{\phi }_{\varepsilon }\left( B\right) }\right| \) has degree \( n - h \) .
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Yes
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Theorem 8 Let \( {a}_{1},\ldots ,{a}_{n} \) be natural numbers with \( {a}_{1}\left| {{a}_{2},\ldots ,{a}_{n - 1}}\right| {a}_{n} \), and let \( B \) be an \( {A}_{1} \times \cdots \times {A}_{n} \) box filled with \( {a}_{1} \times \cdots \times {a}_{n} \) bricks standing in any position. Then B can also be filled with these bricks positioned the same way. Equivalently, there is a permutation \( \pi \) of \( \{ 1,\ldots, n\} \) such that \( {a}_{i} \) divides \( {A}_{\pi \left( i\right) } \) .
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Proof. By Theorem 6, we know that \( {a}_{n} \) divides an \( {A}_{i} \) : let \( \pi \left( n\right) \) be such that \( {a}_{n} \) divides \( {A}_{\pi \left( n\right) } \) . Next, we know by Theorem 7 that \( {a}_{n - 1} \) divides at least two \( {A}_{i} \) : let \( \pi \left( {n - 1}\right) \neq \pi \left( n\right) \) be such that \( {a}_{n - 1} \) divides \( {A}_{\pi \left( {n - 1}\right) } \) . Continuing in this way, we get a permutation as desired.
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Yes
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Theorem 10 Let \( D = \left( {D}_{ij}\right) \) be the \( n \times n \) diagonal matrix with \( {D}_{ii} = d\left( {v}_{i}\right) \), the degree of \( {v}_{i} \) in \( G \) . Then \[ B{B}^{t} = D - A\text{. } \]
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Proof. What is \( {\left( B{B}^{t}\right) }_{ij} \) ? It is \( \mathop{\sum }\limits_{{l = 1}}^{m}{b}_{il}{b}_{jl} \), which is \( d\left( {v}_{i}\right) \) if \( i = j, - 1 \) if \( {v}_{i}{v}_{j} \) is an edge (if \( {e}_{l} = {v}_{i}{v}_{j} \) is directed from \( {v}_{i} \) to \( {v}_{j} \), then \( {b}_{il}{b}_{jl} = 1\left( {-1}\right) = - 1 \) and all other products are 0 ), and 0 if \( {v}_{i}{v}_{j} \) is not an edge and \( i \neq j \) .
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Yes
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Theorem 11 The electric current \( \mathbf{w} \) satisfying \( \mathbf{p} = R\mathbf{w} + \mathbf{g} \) is given by \( \mathbf{w} = \) \( - C{\left( {C}^{t}RC\right) }^{-1}{C}^{t}\mathbf{g} \) .
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Proof. Equation (1) implies that \( {B}_{T}{\mathbf{w}}_{T} + {B}_{N}{\mathbf{w}}_{N} = \mathbf{0} \), so \( {\mathbf{w}}_{T} = - {B}_{T}^{-1}{B}_{N}{\mathbf{w}}_{N} = \) \( {C}_{T}{\mathbf{w}}_{N} \) . Hence \( \mathbf{w} = C{\mathbf{w}}_{N} \) . Combining (2) and (3) we find that \( {C}^{t}R\mathbf{w} + {C}^{t}\mathbf{g} = \mathbf{0} \) and so \( \left( {{C}^{t}{RC}}\right) {\mathbf{w}}_{N} = - {C}^{t}\mathbf{g} \) . As \( {C}^{t}{RC} \) is easily shown to be invertible, the result follows.
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Yes
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Theorem 12 With the notation above, \( {N}^{ * }\left( G\right) = {K}^{ * }\left( G\right) \) .
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Proof. We may assume that \( G \) is connected, since otherwise \( {N}^{ * }\left( G\right) = {K}^{ * }\left( G\right) = \) 0 . Also, the result is trivial for \( n = 1 \) since then \( {N}^{ * }\left( G\right) = {K}^{ * }\left( G\right) = 1 \) .\n\nLet us apply induction on the number of edges of \( G \) . As the result holds for no edges, we turn to the proof of the induction step. Suppose then that \( n > 1, G \) is connected, and the assertion holds for networks with fewer edges. Assuming, as we may, that \( {v}_{1} \) and \( {v}_{2} \) are adjacent, let \( G - {v}_{1}{v}_{2} \) be obtained from \( G \) by cutting (deleting) the edge \( {v}_{1}{v}_{2} \), and let \( G/{v}_{1}{v}_{2} \) be obtained from \( G \) by fusing (contracting) the edge \( {v}_{1}{v}_{2} \) . Thus in \( G/{v}_{1}{v}_{2} \) the vertices \( {v}_{1} \) and \( {v}_{2} \) are replaced by a new vertex, \( {v}_{12} \), say, which is joined to a vertex \( {v}_{i}, i > 2 \), by an edge of conductance \( {c}_{1i} + {c}_{2i} \), provided \( {c}_{1i} + {c}_{2i} > 0 \) .\n\nThe crunch of the proof is that \( {N}^{ * } \) and \( {K}^{ * } \) satisfy the same cut-and-fuse relation:\n\n\[ \n{N}^{ * }\left( G\right) = {N}^{ * }\left( {G - {v}_{1}{v}_{2}}\right) + {c}_{12}{N}^{ * }\left( {G/{v}_{1}{v}_{2}}\right) , \n\]\n\n(4)\n\nand\n\n\[ \n{K}^{ * }\left( G\right) = {K}^{ * }\left( {G - {v}_{1}{v}_{2}}\right) + {c}_{12}{K}^{ * }\left( {G/{v}_{1}{v}_{2}}\right) . \n\]\n\n(5)\n\nIndeed, \( {N}^{ * }\left( {G - {v}_{1}{v}_{2}}\right) \) ’counts’ the spanning trees not containing \( {v}_{1}{v}_{2} \), and \( {c}_{12}{N}^{ * }\left( {G/{v}_{1}{v}_{2}}\right) \) ’counts’ the remaining spanning trees. To see (5), simply consider the cofactors belonging to \( {v}_{1} \) and \( {v}_{12} \).\n\nThis is all: by the induction hypothesis, the right-hand sides of (4) and (5) are equal.
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Yes
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Theorem 14 Let \( G \) be a directed multigraph with vertex set \( V\left( G\right) = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) . For \( 1 \leq i \leq n \), denote by \( {t}_{i}\left( G\right) \) the number of spanning trees oriented towards \( {v}_{i} \) . Also, let \( L = \left( {\ell }_{ij}\right) \) be the combinatorial Laplacian of \( G \) : for \( i \neq j, - {\ell }_{ij} \) is the number of edges from \( {v}_{i} \) to \( {v}_{j} \), and \( {\ell }_{ii} = - \mathop{\sum }\limits_{{j \neq i}}{\ell }_{ij} \) . Then \( {t}_{i}\left( G\right) \) is precisely the first cofactor of \( L \) belonging to \( {\ell }_{ii} \) .
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The proof is entirely along the lines of the proof of Theorem 12: when considering \( {t}_{1}\left( G\right) \), say, all we have to take care is to contract all edges from \( {v}_{i} \) to \( {v}_{1} \) for some \( i > 1 \) . Note that this result contains Corollary 13: given a multigraph, replace each edge by two edges, oriented in either direction, and apply Theorem 14.
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No
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Theorem 14 Let \( G \) be a directed multigraph with vertex set \( V\left( G\right) = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) . For \( 1 \leq i \leq n \), denote by \( {t}_{i}\left( G\right) \) the number of spanning trees oriented towards \( {v}_{i} \) . Also, let \( L = \left( {\ell }_{ij}\right) \) be the combinatorial Laplacian of \( G \) : for \( i \neq j, - {\ell }_{ij} \) is the number of edges from \( {v}_{i} \) to \( {v}_{j} \), and \( {\ell }_{ii} = - \mathop{\sum }\limits_{{j \neq i}}{\ell }_{ij} \) . Then \( {t}_{i}\left( G\right) \) is precisely the first cofactor of \( L \) belonging to \( {\ell }_{ii} \).
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The proof is entirely along the lines of the proof of Theorem 12: when considering \( {t}_{1}\left( G\right) \), say, all we have to take care is to contract all edges from \( {v}_{i} \) to \( {v}_{1} \) for some \( i > 1 \) . Note that this result contains Corollary 13: given a multigraph, replace each edge by two edges, oriented in either direction, and apply Theorem 14.
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No
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Theorem 1 (Max-Flow Min-Cut Theorem.) The maximal flow value from s to t is equal to the minimum of the capacities of cuts separating \( s \) from \( t \) .
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Proof. We have remarked already that there is a flow \( f \) with maximal value, say \( v \), and the capacity of every cut is at least \( v \) . Thus, in order to prove the theorem we have to show that there is a cut with capacity \( v \) . We shall, in fact, do considerably more than this: we shall give a very simple procedure for constructing such a cut from a flow \( f \) with maximal value.\n\nDefine a subset \( S \subset V \) recursively as follows. Let \( s \in S \) . If \( x \in S \), and\n\n\[ c\left( {x, y}\right) > f\left( {x, y}\right) \]\n\nor\n\n\[ f\left( {y, x}\right) > 0 \]\n\nthen let \( y \in S \) .\n\nWe claim that \( \overrightarrow{E}\left( {S,\bar{S}}\right) \) is a cut separating \( s \) from \( t \) with capacity \( v = v\left( f\right) \) . Let us see first why \( t \) cannot belong to \( S \) . If \( t \) belongs to \( S \), we can find vertices \( {x}_{0} = s,{x}_{1},\ldots ,{x}_{\ell } = t \) such that\n\n\[ {\varepsilon }_{i} = \max \left\{ {c\left( {{x}_{i},{x}_{i + 1}}\right) - f\left( {{x}_{i},{x}_{i + 1}}\right), f\left( {{x}_{i + 1},{x}_{i}}\right) }\right\} > 0 \]\n\nfor every \( i,0 \leq i \leq l - 1 \) . Put \( \varepsilon = \mathop{\min }\limits_{i}{\varepsilon }_{i} \) . Then \( f \) can be augmented to a flow \( {f}^{ * } \) in the following way: if \( {\varepsilon }_{i} > f\left( {{x}_{i + 1},{x}_{i}}\right) \) then increase the flow in \( {\overrightarrow{{x}_{i}{x}_{i}}}_{+1} \) by \( \varepsilon \) ; otherwise, decrease the flow in \( \overrightarrow{{x}_{i + 1}{x}_{i}} \) by \( \varepsilon \) . Clearly, \( {f}^{ * } \) is a flow and its value is \( v\left( {f}^{ * }\right) = v\left( f\right) + \varepsilon \), contradicting the maximality of \( f \) . This shows that \( t \notin S \) so \( \overrightarrow{E}\left( {S,\bar{S}}\right) \) is a cut separating \( s \) from \( t \) .\n\nNow, \( v\left( f\right) \) is equal to the value of the flow from \( S \) to \( \bar{S} \) defined in the obvious way:\n\n\[ \mathop{\sum }\limits_{{x \in S, y \in \bar{S}}}f\left( {x, y}\right) - \mathop{\sum }\limits_{{x \in \bar{S}, y \in S}}f\left( {x, y}\right) . \]\nBy the definition of \( S \) the first sum is exactly\n\n\[ \mathop{\sum }\limits_{{x \in S, y \in \bar{S}}}c\left( {x, y}\right) = c\left( {S,\bar{S}}\right) \]\n\nand each summand in the second sum is zero. Hence \( c\left( {S,\bar{S}}\right) = v\left( f\right) \), as required.
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Yes
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Theorem 5 (i) Let \( s \) and \( t \) be distinct nonadjacent vertices of a graph \( G \) . Then the minimal number of vertices separating \( s \) from \( t \) is equal to the maximal number of independent \( s - t \) paths.
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Proof. (i) Replace each edge \( {xy} \) of \( G \) by two directed edges, \( \overrightarrow{xy} \) and \( \overrightarrow{yx} \), and give each vertex other than \( s \) and \( t \) capacity 1 . Then by Theorem 4 the maximal flow value from \( s \) to \( t \) is equal to the minimum of the capacity of a cut separating \( s \) from \( t \) . By the integrality theorem (Theorem 2) there is a maximal flow with current 1 or 0 in each edge. Therefore, the maximal flow value from \( s \) to \( t \) is equal to the maximal number of independent \( s - t \) paths. The minimum of the cut capacity is clearly the minimal number of vertices separating \( s \) from \( t \) .
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Yes
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Theorem 7 A bipartite graph \( G \) with vertex sets \( {V}_{1} \) and \( {V}_{2} \) contains a complete matching from \( {V}_{1} \) to \( {V}_{2} \) iff\n\n\[ \left| {\Gamma \left( S\right) }\right| \geq \left| S\right| \text{ for every }S \subset {V}_{1}. \]
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We have already seen that the condition is necessary so we have to prove only the sufficiency.\n\nFirst Proof. Both Menger’s theorem (applied to the sets \( {V}_{1} \) and \( {V}_{2} \) as at the end of Section 2) and the max-flow min-cut theorem (applied to the directed graph obtained from \( G \) by sending each edge from \( {V}_{1} \) to \( {V}_{2} \), and giving each vertex capacity 1) imply the following. If \( G \) does not contain a complete matching from \( {V}_{1} \) to \( {V}_{2} \) then there are \( {T}_{1} \subset {V}_{1} \) and \( {T}_{2} \subset {V}_{2} \) such that \( \left| {T}_{1}\right| + \left| {T}_{2}\right| < \left| {V}_{1}\right| \) and there is no edge from \( {V}_{1} - {T}_{1} \) to \( {V}_{2} - {T}_{2} \) . Then \( \Gamma \left( {{V}_{1} - {T}_{1}}\right) \subset {T}_{2} \) so\n\n\[ \left| {\Gamma \left( {{V}_{1} - {T}_{1}}\right) }\right| \leq \left| {T}_{2}\right| < \left| {V}_{1}\right| - \left| {T}_{1}\right| = \left| {{V}_{1} - {T}_{1}}\right| . \]\n\nThis shows the sufficiency of the condition.
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Yes
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Corollary 9 Suppose that a bipartite graph \( G = {G}_{2}\left( {m, n}\right) \), with vertex sets \( {V}_{1},{V}_{2} \), satisfies the following condition:\n\n\[ \left| {\Gamma \left( S\right) }\right| \geq \left| S\right| - d \]\n\nfor every \( {S}_{1} \subset {V}_{1} \) . Then \( G \) contains \( m - d \) independent edges.
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Proof. Add \( d \) vertices to \( {V}_{2} \) and join them to each vertex in \( {V}_{1} \) . The new graph \( G * \) satisfies the conditions for a complete matching. At least \( m - d \) of the edges in a complete matching of \( G * \) belong to \( G \) .
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Yes
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Corollary 10 Let \( G = {G}_{2}\left( {m, n}\right) \) be a bipartite graph. Write \( h \) for the maximal number of independent edges, i for the maximal number of independent vertices, and \( j \) for the minimal number of edges and vertices covering all the vertices. Then\n\n\[ i = j = m + n - h. \]
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Proof. Let \( {E}^{\prime } \cup {V}^{\prime } \) be a set of \( j \) edges and vertices covering all vertices, with \( {E}^{\prime } \subset E \) and \( {V}^{\prime } \subset V \) . If \( e, f \in {E}^{\prime } \) share a vertex, then in the cover \( {E}^{\prime } \cup {V}^{\prime } \) we may replace \( f \) by its other endvertex. Hence we may assume that \( {E}^{\prime } \) consists of independent edges. This shows that \( j = m + n - h \) .\n\nAlso, \( m + n - i \geq h \), since if \( I \) is a set of \( i \) independent vertices (in any graph), then every edge is incident with at least one vertex not in \( I \) .\n\nFinally, let \( S \subset {V}_{1} \) be such that \( \left| {\Gamma \left( S\right) }\right| = \left| S\right| - \left( {m - h}\right) \), as guaranteed by Corollary 9. Then, with \( T = {V}_{2} - \Gamma \left( S\right) \), the set \( S \cup T \) is a set of \( \left| S\right| + n - \left| {\Gamma \left( S\right) }\right| = \) \( m + n - h \) independent vertices, proving that \( i \geq m + n - h \) .
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Yes
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Corollary 11 Let \( G \) be a bipartite graph with vertex classes \( {V}_{1} = \left\{ {{x}_{1},\ldots ,{x}_{m}}\right\} \) and \( {V}_{2} = \left\{ {{y}_{1},\ldots ,{y}_{n}}\right\} \) . Then \( G \) contains a subgraph \( H \) such that \( {d}_{H}\left( {x}_{i}\right) = {d}_{i} \) and \( 0 \leq {d}_{H}\left( {y}_{j}\right) \leq 1 \) iff\n\n\[ \left| {\Gamma \left( S\right) }\right| \geq \mathop{\sum }\limits_{{{x}_{i} \in S}}{d}_{i} \]\n\nfor every \( S \subset {V}_{1} \) .
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Proof. Replace each vertex \( {x}_{i} \) by \( {d}_{i} \) vertices joined to every vertex in \( \Gamma \left( {x}_{i}\right) \) . Then \( G \) has such a subgraph \( H \) iff the new graph has a matching from the new first vertex class to \( {V}_{2} \) . The result follows from Theorem 7.
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Yes
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Theorem 12 Let \( G = {G}_{2}\left( {m, n}\right) \) be a bipartite graph with vertex classes \( {V}_{1} = \) \( \left\{ {{x}_{1},\ldots ,{x}_{m}}\right\} \) and \( {V}_{2} = \left\{ {{y}_{1},\ldots ,{y}_{n}}\right\} \) . For \( S \subset {V}_{1} \) and \( 1 \leq j \leq n \) denote by \( {S}_{j} \) the number of edges from \( {y}_{j} \) to \( S \) . Let \( {d}_{1},\ldots ,{d}_{m} \) and \( {e}_{1},\ldots ,{e}_{n} \) be natural numbers and let \( d \geq 0 \) . Then there exists a subgraph \( H \) of \( G \) with\n\n\[ e\left( H\right) \geq \mathop{\sum }\limits_{{i = 1}}^{m}{d}_{i} - d \]\n\n\[ {d}_{H}\left( {x}_{i}\right) \leq {d}_{i},\;1 \leq i \leq m \]\n\nand\n\n\[ {d}_{H}\left( {y}_{j}\right) \leq {e}_{j},\;1 \leq j \leq n, \]\n\niff for every \( S \subset {V}_{1} \) we have\n\n\[ \mathop{\sum }\limits_{{{x}_{i} \in S}}{d}_{i} \leq \mathop{\sum }\limits_{{j = 1}}^{n}\min \left\{ {{S}_{j},{e}_{j}}\right\} + d. \]
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Proof. Turn \( G \) into a directed graph \( \overrightarrow{G} \) by sending each edge from \( {V}_{1} \) to \( {V}_{2} \) . Give each edge capacity 1, a vertex \( {x}_{i} \) capacity \( {d}_{i} \), and a vertex \( {y}_{j} \) capacity \( {e}_{j} \) . Then there is a subgraph \( H \) with the required properties iff in \( \overrightarrow{G} \) there is a flow from \( {V}_{1} \) to \( {V}_{2} \) with value at least \( \mathop{\sum }\limits_{1}^{m}{d}_{i} - d \), and by the max-flow min-cut theorem, this happens iff every cut has capacity at least \( \mathop{\sum }\limits_{1}^{m}{d}_{i} - d \) . Now, minimal cuts are of the form \( T \cup U \cup E\left( {{V}_{1} - T,{V}_{2} - U}\right) \), where \( T \subset {V}_{1} \) and \( U \subset {V}_{2} \) . Given a set \( T \), the capacity of such a cut will be minimal if a vertex \( {y}_{j} \) belongs to \( U \) iff its capacity is smaller than the number of edges from \( S = {V}_{1} - T \) to \( {y}_{j} \) . With this choice of \( U \) the capacity of the cut is exactly\n\n\[ \mathop{\sum }\limits_{{{x}_{i} \in T}}{d}_{i} + \mathop{\sum }\limits_{1}^{n}\min \left\{ {{S}_{j},{e}_{j}}\right\} \]\n\nThe condition that this is at least \( \mathop{\sum }\limits_{1}^{m}{d}_{i} - d \) is clearly the condition in the theorem.
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Yes
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Theorem 13 If every antichain in a (finite) partially ordered set \( P \) has at most \( m \) elements, then \( P \) is the union of \( m \) chains.
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Proof. Let us apply induction on \( \left| P\right| \) . If \( P = \varnothing \), there is nothing to prove, so we suppose that \( \left| P\right| > 0 \) and the theorem holds for sets with fewer elements.\n\nLet \( C \) be a maximal chain in \( P \) . (Thus if \( x \notin C \), then \( C \cup \{ x\} \) is no longer a chain.) If no antichain of \( P - C \) has \( m \) elements, then we are home by induction. Therefore, we may assume that \( P - C \) contains an antichain \( A = \left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{m}}\right\} \) .\n\nDefine the lower shadow of \( A \) as\n\n\[ \n{S}^{ - } = \left\{ {x \in P : x \leq {a}_{i}\text{ for some }i}\right\} \n\] \n\nand define the upper shadow \( {S}^{ + } \) of \( A \) analogously. Then \( P \) is the union of the two shadows, since otherwise \( A \) could be extended to an antichain with \( m + 1 \) elements. Furthermore, neither shadow is the whole of \( P \), since the maximal element of \( C \) does not belong to \( {S}^{ - } \) and the minimal element of \( C \) does not belong to \( {S}^{ + } \) . By the induction hypothesis both shadows can be decomposed into \( m \) chains, say\n\n\[ \n{S}^{ - } = \mathop{\bigcup }\limits_{{i = 1}}^{m}{C}_{i}^{ - }\;\text{ and }\;{S}^{ + } = \mathop{\bigcup }\limits_{{i = 1}}^{m}{C}_{i}^{ + } \n\] \n\nSince different \( {a}_{i} \) belong to different chains, we may assume that \( {a}_{i} \in {C}_{i}^{ - } \) and \( {a}_{i} \in {C}_{i}^{ + } \) .\n\nThe proof will be completed if we show that \( {a}_{i} \) is the maximal element of \( {C}_{i}^{ - } \) and the minimal element of \( {C}_{i}^{ + } \), since in that case the chains \( {C}_{i}^{ - } \) and \( {C}_{i}^{ + } \) can be strung together to give a single chain \( {C}_{i} \), and then \( P = \mathop{\bigcup }\limits_{1}^{m}{C}_{i} \) .\n\nSuppose then that, say, \( {a}_{i} \) is not the maximal element of \( {C}_{i}^{ - } : {a}_{i} < x \) for some \( x \in {C}_{i}^{ - } \) . Since \( x \) is in the lower shadow of \( A \), there is an \( {a}_{j} \in A \) with \( x \leq {a}_{j} \) . However, this implies the contradiction \( {a}_{i} < {a}_{j} \) .
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Yes
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Corollary 15 A graph \( G \) contains a set of independent edges covering all but at most \( d \) of the vertices iff\n\n\[ q\left( {G - S}\right) \leq \left| S\right| + d \] \n\nfor every \( S \subset V\left( G\right) \) .
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Proof. Since the number of vertices not covered by a set of independent edges is congruent to \( \left| G\right| \) modulo 2, we may assume that\n\n\[ d \equiv \left| G\right| \left( {\;\operatorname{mod}\;2}\right) \]\n\nPut \( H = G + {K}_{d} \) ; that is, let \( H \) be obtained from \( G \) by adding to it a set \( W \) of \( d \) vertices, and joining every new vertex to every other vertex, old and new. Then \( G \) contains a set of independent edges covering all but \( d \) of the vertices iff \( H \) has a 1-factor. When does (1) hold for \( H \) ? If \( \varnothing \neq {S}^{\prime } \subset V\left( H\right) \) and \( W - {S}^{\prime } \neq \varnothing \), then \( H - {S}^{\prime } \) is connected, so \( q\left( {H - {S}^{\prime }}\right) \leq 1 \), and then (1) does hold; if \( W \subset {S}^{\prime } \) then, setting \( S = {S}^{\prime } - W \), we have \( q\left( {H - {S}^{\prime }}\right) = q\left( {G - \left\{ {{S}^{\prime } \smallsetminus W}\right\} }\right) = q\left( {G - S}\right) \), so (1) is equivalent to\n\n\[ q\left( {G - S}\right) \leq \left| {S}^{\prime }\right| = \left| S\right| + d. \]
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Yes
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Theorem 16 For every assignment of preferences in a bipartite graph, there is a stable matching.
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Proof. Let us describe a variant of the fundamental algorithm we have just mentioned, in which all boys and all girls act simultaneously, in rounds. In every odd round \( \left( {1\mathrm{{st}},3\mathrm{{rd}},\ldots }\right) \), each boy proposes to his highest preference among those girls whom he knows and who have not yet refused him, and in every even round (2 nd,4 th, \( \ldots \) ), each of the \( m \) girls refuses all but her highest suitor. The process ends when no girl refuses a suitor: then every girl marries her (only) suitor, if she has one.\n\nThe algorithm terminates after at most \( {2nm} \) rounds, since at most \( m\left( {n - 1}\right) \) proposals are refused.\n\nWe claim that this fundamental algorithm produces a stable matching. It clearly produces a (partial) matching \( M \), since at every stage each boy proposes to at most one girl, and each girl rejects all but at most one boy. To see that \( M \) is stable, let \( {aB} \in E\left( G\right) - M \) . Then either a never proposed to \( B \), or \( a \) was refused by \( B \) during the algorithm. In the former case \( a \) marries a girl he prefers to \( B \), as he never goes as low as \( B \), and in the latter case \( B \) refused \( a \) for a boy she prefers to \( a \), so eventually ends up with a husband she prefers to her suitor \( a \) .
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Yes
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Lemma 17 Let \( M \) and \( {M}^{\prime } \) be two stable matchings in a bipartite graph with certain preferences, and let \( C \) be a component of the subgraph \( H \) formed by the edges of \( M \cup {M}^{\prime } \) . If \( C \) has at least three vertices, then it is a preference-oriented cycle. In particular, if \( {aA},{bB} \in M \) and \( {aB} \in {M}^{\prime } \), then a prefers \( A \) to \( B \) iff \( B \) prefers a to \( b \) .
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Proof. In this proof it is best not to distinguish between boys and girls: we shall write \( {x}_{1},{x}_{2},\ldots \) for either of them. We know that \( C \) is either a path of length at least two or a cycle of length at least four.\n\nSuppose that \( C \) contains a path \( {x}_{1}{x}_{2}{x}_{3}{x}_{4} \), with \( {x}_{2} \) preferring \( {x}_{3} \) to \( {x}_{1} \) . Assuming, as we may, that \( {x}_{2}{x}_{3} \notin M \), we see that \( {x}_{3} \) prefers \( {x}_{4} \) to \( {x}_{2} \), since \( M \) is stable.\n\nThis simple observation implies that if \( C \) is a \( {cycle} \) , then it is preference-oriented. Indeed, if \( {x}_{1}{x}_{2}{x}_{3}\cdots {x}_{k} \) is a cycle and \( {x}_{2} \) prefers \( {x}_{3} \) to \( {x}_{1} \), then looking at the path \( {x}_{1}{x}_{2}{x}_{3}{x}_{4} \) we see that \( {x}_{3} \) prefers \( {x}_{4} \) to \( {x}_{2} \) . Next, looking at the path \( {x}_{2}{x}_{3}{x}_{4}{x}_{5} \) we see that \( {x}_{4} \) prefers \( {x}_{5} \) to \( {x}_{3} \) . Continuing in this way, we find that \( {x}_{k} \) prefers \( {x}_{1} \) to \( {x}_{k - 1} \) and \( {x}_{1} \) prefers \( {x}_{2} \) to \( {x}_{k} \) .\n\nAlso, if \( C \) is a path \( {x}_{1}{x}_{2}\cdots {x}_{\ell } \) with \( l \geq 3 \) and \( {x}_{1}{x}_{2} \notin M \), say, then \( {x}_{2} \) prefers \( {x}_{3} \) to \( {x}_{1} \), since \( M \) is stable and \( {x}_{1}{x}_{2} \notin M \) . Similarly, \( {x}_{\ell - 1} \) prefers \( {x}_{\ell - 2} \) to \( {x}_{\ell } \) . However, this is impossible, since, arguing as above, \( {x}_{2} \) prefers \( {x}_{3} \) to \( {x}_{1},{x}_{3} \) prefers \( {x}_{4} \) to \( {x}_{2},{x}_{4} \) prefers \( {x}_{5} \) to \( {x}_{3} \), and so on, \( {x}_{\ell - 1} \) prefers \( {x}_{\ell } \) to \( {x}_{\ell - 2} \) .\n\nThe second assertion is immediate from the fact that the component of \( H \) containing the path \( {AaBb} \) is a preference-oriented cycle.
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Yes
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Theorem 18 For every assignment of preferences in a bipartite graph with bipartition \( \left( {{V}_{1},{V}_{2}}\right) \), there are subsets \( {U}_{1} \subset {V}_{1} \) and \( {U}_{2} \subset {V}_{2} \) such that every stable matching is a complete matching from \( {U}_{1} \) to \( {U}_{2} \) . In particular, all stable matchings have the same cardinality.
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Proof. Suppose that the assertion fails. Then we may assume that some edge \( {aA} \) of \( M \) is such that \( a \) is not incident with any edge of \( {M}^{\prime } \) . As \( {M}^{\prime } \) is a maximal matching, \( {bA} \in {M}^{\prime } \) for some \( b \in {V}_{1}, b \neq a \) . But then the component of \( a \) in the subgraph formed by the edges of \( M \cup {M}^{\prime } \) contains \( a, A \), and \( b \), and is not a cycle, contradicting Lemma 17.
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No
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Theorem 20 For every assignment of preferences in an \( n \) by \( n \) complete bipartite graph with bipartition \( \left( {{V}_{1},{V}_{2}}\right) \), there is a \( {V}_{1} \) -optimal stable matching.
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Proof. Let us denote by \( S\left( a\right) \) the set of girls a boy \( a \) could marry in some stable matching: this is the set of possible girls for \( a \) . We claim that in the fundamental algorithm no girl in \( S\left( a\right) \) refuses \( a \), so every boy marries his favourite possible girl, and thus the stable matching is optimal for the boys.\n\nSuppose that this is not the case. Let us stop the algorithm when it happens for the very first time that a boy, say \( a \), is refused by one of his possible girls, say \( A \) . By definition, this happens because \( A \) prefers another of her suitors at the time, say \( b \) . At that time \( b \) prefers \( A \) to all others that have not yet refused him. Hence, a fortiori, \( b \) prefers \( A \) to all others that are possible for him. As \( A \) is possible for \( a \), there is a stable matching \( M \) in which \( a \) marries \( A \) and \( b \) marries a girl \( B \) . But this is impossible, since \( b \) prefers \( A \) to \( B \), and \( A \) prefers \( b \) to \( a \) . This contradiction completes the proof.
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Yes
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Theorem 21 An incomplete system \( \left( {{V}_{1},{V}_{2}, L}\right) \) with \( \left| {V}_{1}\right| = \left| {V}_{2}\right| \) has a stable complete matching iff the associated complete system \( \left( {{V}_{1}^{\prime },{V}_{2}^{\prime },{L}^{\prime }}\right) \) has a stable matching in which the widow marries the widower.
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Proof. Let \( M \) be a complete matching from \( {V}_{1} \) to \( {V}_{2} \), and let \( {M}^{\prime } \) be the complete matching from \( {V}_{1}^{\prime } \) to \( {V}_{2}^{\prime } \) obtained from \( M \) by adding to it the edge \( {wW} \) . To prove the theorem, we shall show that \( M \) is a stable matching in the incomplete system \( \left( {{V}_{1},{V}_{2}, L}\right) \) iff \( {M}^{\prime } \) is a stable matching in the associated complete system \( \left( {{V}_{1}^{\prime },{V}_{2}^{\prime },{L}^{\prime }}\right) \) .\n\nSuppose \( M \) is a stable matching in \( \left( {{V}_{1},{V}_{2}, L}\right) \) . Then \( {M}^{\prime } \) is stable, since if \( {aA} \in M \) then \( A \) prefers \( a \) to \( w \) and \( a \) prefers \( A \) to \( W \) .\n\nAlso, if \( {M}^{\prime } \) is a stable matching in \( \left( {{V}_{1}^{\prime },{V}_{2}^{\prime },{L}^{\prime }}\right) \), then \( M \) is a stable matching in \( \left( {{V}_{1},{V}_{2}, L}\right) \) . Indeed, if \( {aA} \in M \), then \( A \in L\left( a\right) \), since otherwise \( a \) prefers \( W \) to \( A \) , and \( W \) prefers \( a \) to \( w \) . Similarly, \( a \in L\left( A\right) \), since otherwise \( A \) prefers \( w \) to \( a \) and \( w \) prefers \( A \) to \( W \) . Hence every edge \( {aA} \) of \( M \) satisfies \( A \in L\left( a\right) \) and \( a \in L\left( A\right) \) , so \( M \) is a stable matching in \( \left( {{V}_{1},{V}_{2}, L}\right) \) .\n\nSince the \( {V}_{1}^{\prime } \) -optimal stable matching in \( \left( {{V}_{1}^{\prime },{V}_{2}^{\prime },{L}^{\prime }}\right) \) is precisely the \( {V}_{2}^{\prime } \) -pessimal stable matching, if some stable matching in \( \left( {{V}_{1}^{\prime },{V}_{2}^{\prime },{L}^{\prime }}\right) \) contains \( {wW} \), then every stable matching contains \( {wW} \) . This gives us the following slightly stronger form of Theorem 21.\n\nTheorem \( {\mathbf{{21}}}^{\prime } \) . Let \( \left( {{V}_{1},{V}_{2}, L}\right) \) be an incomplete system, with associated complete system \( \left( {{V}_{1}^{\prime },{V}_{2}^{\prime },{L}^{\prime }}\right) \), and let \( {M}^{\prime } \) be a stable matching for \( \left( {{V}_{1}^{\prime },{V}_{2}^{\prime },{L}^{\prime }}\right) \) . Then there is a stable complete matching for \( \left( {{V}_{1},{V}_{2}, L}\right) \) iff \( {M}^{\prime } \) contains \( {wW} \) .
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Yes
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Theorem 22 No matter what the orders of preferences are, there is always an optimal stable admissions scheme.
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Proof. For the sake of argument, call the students boys, and replace each college \( {A}_{i} \) by \( {n}_{i} \) girls, say \( {A}_{i}^{\left( 1\right) },{A}_{i}^{\left( 2\right) },\ldots ,{A}_{i}^{\left( {n}_{i}\right) } \), with each \( {A}_{i}^{\left( j\right) } \) having the same order of preferences among the boys \( {a}_{1},\ldots ,{a}_{n} \) as \( {A}_{i} \) . Also, each boy orders the girls by taking the girls corresponding to the highest-rated college first (in any order) followed by the girls corresponding to the second college (in any order), and so on. In the bipartite graph we have just defined, take a stable matching that is optimal for \( \left\{ {{a}_{1},\ldots ,{a}_{n}}\right\} \) : the admissions scheme this induces is clearly optimal (for the applicants).
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Yes
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Theorem 1 For \( g \geq 3 \) and \( \delta \geq 3 \) put\n\n\[ \n{n}_{0}\left( {g,\delta }\right) = \left\{ \begin{array}{ll} 1 + \frac{\delta }{\delta - 2}\left\{ {{\left( \delta - 1\right) }^{\left( {g - 1}\right) /2} - 1}\right\} & \text{ if }g\text{ is odd,} \\ \frac{2}{\delta - 2}\left\{ {{\left( \delta - 1\right) }^{g/2} - 1}\right\} & \text{ if }g\text{ is even. } \end{array}\right.\n\]\n\nThen a graph \( G \) with minimal degree \( \delta \) and girth \( g \) has at least \( {n}_{0}\left( {g,\delta }\right) \) vertices.
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Proof. Suppose first that \( g \) is odd, say \( g = {2d} + 1, d \geq 1 \) . Pick a vertex \( x \) . There is no vertex \( z \) for which \( g \) contains two distinct \( z - x \) paths of length at most \( d \), since otherwise \( G \) has a cycle of length at most \( {2d} \) . Consequently, there are at least \( \delta \) vertices at distance 1 from \( x \), at least \( \delta \left( {\delta - 1}\right) \) vertices at distance 2, and so on, and at least \( \delta {\left( \delta - 1\right) }^{d - 1} \) vertices at distance \( d \) from \( x \) (Fig. IV.1).\n\n \n\nFIGURE IV. 1. The cases \( \delta = g = 5 \) and \( \delta = 4, g = 6 \) .\n\nThus\n\n\[ \nn \geq 1 + \delta + \delta \left( {\delta - 1}\right) + \cdots + \delta {\left( \delta - 1}\right) }^{d - 1},\n\]\n\nas claimed.\n\nSuppose now that \( g \) is even, say \( g = {2d} \) . Pick two adjacent vertices, say \( x \) and \( y \) . Then there are \( 2\left( {\delta - 1}\right) \) vertices at distance 1 from \( \{ x, y\} ,2{\left( \delta - 1}\right) }^{2} \) vertices at distance 2, and so on, and \( 2{\left( \delta - 1}\right) }^{d - 1} \) vertices at distance \( d - 1 \) from \( \{ x, y\} \) , implying the required inequality.
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Yes
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Theorem 2 Let \( G \) be a connected graph of order \( n \geq 3 \) such that for any two non-adjacent vertices \( x \) and \( y \) we have\n\n\[ d\left( x\right) + d\left( y\right) \geq k \]\n\nIf \( k = n \) then \( G \) is Hamiltonian, and if \( k < n \) then \( G \) contains a path of length \( k \) and a cycle of length at least \( \left( {k + 2}\right) /2 \) .
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Proof. Assume that \( G \) is not Hamiltonian and let \( P = {x}_{1}{x}_{2}\cdots {x}_{\ell } \) be a longest path in \( G \) . The maximality of \( P \) implies that the neighbours of \( {x}_{1} \) and \( {x}_{\ell } \) are vertices of \( P \) . As \( G \) does not contain a cycle of length \( \ell ,{x}_{1} \) is not adjacent to \( {x}_{\ell } \) . Even more, the path \( P \) cannot contain vertices \( {x}_{i} \) and \( {x}_{i + 1} \) such that \( {x}_{1} \) is adjacent to \( {x}_{i + 1} \) and \( {x}_{\ell } \) is adjacent to \( {x}_{i} \), since otherwise \( {x}_{1}{x}_{2}\cdots {x}_{i}{x}_{\ell }{x}_{\ell - 1}\cdots {x}_{i + 1} \) is a cycle of length \( \ell \) (Fig. IV.2).\n\nConsequently, the sets\n\n\[ \Gamma \left( {x}_{1}\right) = \left\{ {{x}_{j} : {x}_{1}{x}_{j} \in E\left( G\right) }\right\} \text{ and }{\Gamma }^{ + }\left( {x}_{\ell }\right) = \left\{ {{x}_{i + 1} : {x}_{i}{x}_{\ell } \in E\left( G\right) }\right\} \]\n\nare disjoint subsets of \( \left\{ {{x}_{2},{x}_{3},\ldots ,{x}_{\ell }}\right\} \), and so\n\n\[ k \leq d\left( {x}_{1}\right) + d\left( {x}_{\ell }\right) = \left| {\Gamma \left( {x}_{1}\right) }\right| + \left| {{\Gamma }^{ + }\left( {{x}_{\ell } \mid \leq \ell - 1 \leq n - 1.}\right. }\right| \]\n\nNow, if \( k = n \) then this is a contradiction, so \( G \) is Hamiltonian. Also, if \( k < n \) , then this relation implies that \( G \) has a path of length \( \ell - 1 \geq k \) . This proves the first two assertions of the theorem.\n\nFinally, the assertion about cycles is even simpler. Assume that \( d\left( {x}_{1}\right) \geq d\left( {x}_{\ell }\right) \) , so \( d\left( {x}_{1}\right) \geq \lceil k/2\rceil \) . Put \( t = \max \left\{ {i : {x}_{1}{x}_{i} \in E\left( G\right) }\right\} \) . Then \( t \geq d\left( {x}_{1}\right) + 1 \geq \) \( \lceil k/2\rceil + 1 \), and \( G \) contains the cycle \( {x}_{1}{x}_{2}\cdots {x}_{t} \) of length \( t \) .
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Yes
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Theorem 3 Let \( G \) be a graph of order \( n \) without a path of length \( k\left( { \geq 1}\right) \) . Then\n\n\[ e\left( G\right) \leq \frac{k - 1}{2}n \]
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Proof. We fix \( k \) and apply induction on \( n \) . The assertion is clearly true if \( n \leq k \) . Assume now that \( n > k \) and the assertion holds for smaller values of \( n \) .\n\nIf \( G \) is disconnected, then the induction hypothesis implies the result. Now, if \( G \) is connected, then it contains no \( {K}_{k} \) and, by Theorem 2, it has a vertex \( x \) of degree at most \( \left( {k - 1}\right) /2 \) . Since \( G - x \) is not an extremal graph,\n\n\[ e\left( G\right) \leq d\left( x\right) + e\left( {G - x}\right) < \frac{k - 1}{2} + \frac{k - 1}{2}\left( {n - 1}\right) = \frac{k - 1}{2}n. \]
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Yes
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Theorem 4 Let \( k \geq 2 \) and let \( G \) be a graph of order \( n \) in which every cycle has length at most \( k \) . Then\n\n\[ e\left( G\right) \leq \frac{k}{2}\left( {n - 1}\right) \]\n\nA graph is extremal iff it is connected and all its blocks are complete graphs of order \( k \) .
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The proof of this result is somewhat more involved than that of Theorem 3. Since a convenient way of presenting it uses \
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No
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Theorem 5 Let \( G \) be a graph with \( n \) vertices and at least \( {t}_{r}\left( n\right) \) edges, and let \( x \) be a vertex of maximal degree, say, \( d\left( x\right) = n - k = \Delta \left( G\right) \) . Set \( W = \Gamma \left( x\right) \) , \( U = V\left( G\right) \smallsetminus W \) and \( H = G\left\lbrack W\right\rbrack \) . Then \( e\left( H\right) \geq {t}_{r - 1}\left( {n - k}\right) \), and the inequality is strict unless \( k = \lfloor n/r\rfloor \) and \( U \) is an independent set of vertices, each of degree \( n - k \) .
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Proof. As we noted above, \( k \leq \lfloor n/r\rfloor \) . Assume that \( e\left( H\right) \leq {t}_{r - 1}\left( {n - k}\right) \) . Then\n\n\[ \n{t}_{r}\left( n\right) \leq e\left( G\right) = e\left( H\right) + \frac{1}{2}\mathop{\sum }\limits_{{u \in U}}d\left( u\right) + \frac{1}{2}e\left( {U, W}\right) \n\]\n\n\[ \n\leq e\left( H\right) + k\left( {n - k}\right) \leq {t}_{r - 1}\left( {n - k}\right) + k\left( {n - k}\right) .\n\]\n\nConsequently, \( k = \lfloor n/s\rfloor, e\left( {U, W}\right) = k\left( {n - k}\right) \), and so \( G = H + \overline{{K}_{k}} \), as claimed.
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Yes
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Theorem 6 Let \( G \) be a graph with \( n \) vertices and at least \( {t}_{r}\left( n\right) \) edges. Consider the following simple algorithm for finding a complete subgraph of order \( r + 1 \) . Pick a vertex \( {x}_{1} \) of maximal degree in \( {G}_{1} = G \), then a vertex \( {x}_{2} \) of maximal degree in the subgraph \( {G}_{2} \) of \( {G}_{1} \) spanned by the neighbours of \( {x}_{1} \), then a vertex \( {x}_{3} \) of maximal degree in the subgraph \( {G}_{3} \) of \( {G}_{2} \) spanned by the neighbours of \( {x}_{2} \) (in \( {G}_{2} \) ), and so on, stopping with \( {x}_{\ell } \) if it has no neighbours in \( {G}_{\ell } \) . Then either \( G \) is a Turán graph \( {T}_{r}\left( n\right) \), or else the procedure above constructs at least \( r + 1 \) vertices, \( {x}_{1},{x}_{2},\ldots ,{x}_{r + 1} \), which then span a complete subgraph.
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Proof. We apply induction on \( r \), noting that for \( r = 1 \) there is nothing to prove. Set \( n - k = d\left( {x}_{1}\right) = \Delta \left( G\right) \) . If \( e\left( {G}_{2}\right) > {t}_{r - 1}\left( {n - k}\right) \), then we are done by the induction hypothesis, since \( {G}_{2} \) cannot be isomorphic to \( {T}_{r - 1}\left( {n - k}\right) \), and \( {x}_{1} \), followed by the vertices \( {x}_{2},{x}_{3},\ldots ,{x}_{r + 1} \) we find in \( {G}_{2} \), gives the sequence as claimed. Otherwise, by Theorem \( 5, k = \lfloor n/r\rfloor, e\left( {G}_{2}\right) = {t}_{r - 1}\left( {n - k}\right) \), and \( G = {G}_{2} + \overline{{K}_{k}} \) . Hence, by another application of the induction hypothesis to \( {G}_{2} \), we see that either our procedure constructs \( {x}_{2},\ldots ,{x}_{r + 1} \), or else \( {G}_{2} \cong {T}_{r - 2}\left( {n - k}\right) \) and \( {G}_{1} \cong {T}_{r - 1}\left( n\right) \) , as claimed.
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Yes
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Theorem 7 Let \( G \) be a graph with vertex set \( V \) that does not contain \( {K}_{r + 1} \), a complete graph of order \( r \) . Then there is an \( r \) -partite graph \( H \) with vertex set \( V \) such that for every vertex \( z \in V \) we have\n\n\[ \n{d}_{G}\left( z\right) \leq {d}_{H}\left( z\right) \n\]\n\nIf \( G \) is not a complete r-partite graph, then there is at least one vertex \( z \) for which the inequality above is strict.
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Proof. We shall apply induction on \( r \) . For \( r = 1 \) there is nothing to prove, since \( G \) is the empty graph \( \overline{{K}_{n}} \), which is 1-partite. Assume now that \( r \geq 2 \) and the assertion holds for smaller values of \( r \) .\n\nPick a vertex \( x \in V \) for which \( {d}_{G}\left( x\right) \) is maximal and denote by \( W \) the set of vertices of \( G \) that are joined to \( x \) . Then \( {G}_{0} = G\left\lbrack W\right\rbrack \) does not contain a \( {K}_{r} \) for otherwise with \( x \) it would form a \( {K}_{r + 1} \) . By the induction hypothesis we can replace \( {G}_{0} \) by an \( \left( {r - 1}\right) \) -partite graph \( {H}_{0} \) with vertex set \( W \) in such a way that \( {d}_{{G}_{0}}\left( y\right) \leq {d}_{{H}_{0}}\left( y\right) \) for every \( y \in W \) and strict inequality holds for at least one \( y \) unless \( {G}_{0} \) is a complete \( \left( {r - 1}\right) \) -partite graph. Add to \( {H}_{0} \) the vertices in \( V - W \) and join each vertex in \( V - W \) to each vertex in \( W \) . To complete the proof let us check that the \( r \) -partite graph \( H \) obtained in this way has the required properties.\n\nIf \( z \in U = V - W \), then \( {d}_{H}\left( z\right) = {d}_{H}\left( x\right) = {d}_{G}\left( x\right) \geq {d}_{G}\left( z\right) \), and if \( z \in W \), then \( {d}_{H}\left( z\right) = {d}_{{H}_{0}}\left( z\right) + n - \left| W\right| \geq {d}_{{G}_{0}}\left( z\right) + n - \left| W\right| \geq {d}_{G}\left( z\right) \) . Thus \( {d}_{G}\left( z\right) \leq {d}_{H}\left( z\right) \) holds for every \( z \in V \) .
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Yes
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Theorem 8 For \( r, n \geq 2 \) we have \( \operatorname{ex}\left( {n;{K}_{r + 1}}\right) = {t}_{r}\left( n\right) \) and \( \operatorname{EX}\left( {n;{K}_{r + 1}}\right) = \left\{ {{T}_{r}\left( n\right) }\right\} \) . In words, every graph of order \( n \) with more than \( {t}_{r}\left( n\right) \) edges contains a \( {K}_{r + 1} \) . Also, \( {T}_{r}\left( n\right) \) is the unique graph of order \( n \) and size \( {t}_{r}\left( n\right) \) that does not contain a \( {K}_{r + 1} \) .
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Proof. The theorem is contained in Theorem 6, and it is also an immediate consequence of Theorem 7, since \( {T}_{r}\left( n\right) \) is the unique \( r \) -partite graph of order \( n \) and maximal size.\n\nNevertheless, let us give two more proofs of the theorem itself, based again on the properties of \( {T}_{r}\left( n\right) \) .\n\n\( {3rd} \) Proof. For \( r = 1 \) there is nothing to prove, so fix \( r \geq 2 \) and apply induction on \( n \) . For \( n \leq r + 1 \) the assertion is trivial, so suppose that \( n > r + 1 \) and the theorem holds for smaller values of \( n \) .\n\nSuppose \( G \) has \( n \) vertices, \( {t}_{r}\left( n\right) \) edges, and it contains no \( {K}_{r + 1} \) . As \( {T}_{r}\left( n\right) \) is a maximal graph without a \( {K}_{r + 1} \) (that is, no edge can be added to it without creating a \( {K}_{r + 1} \) ), the induction step will follow if we show that \( G \) is exactly a \( {T}_{r}\left( n\right) \) . Since the degrees in \( {T}_{r}\left( n\right) \) differ by at most 1, we have\n\n\[ \delta \left( G\right) \leq \delta \left( {{T}_{r}\left( n\right) }\right) \leq \Delta \left( {{T}_{r}\left( n\right) }\right) \leq \Delta \left( G\right) . \]\n\nLet \( x \) be a vertex of \( G \) with degree \( d\left( x\right) = \delta \left( G\right) \leq \delta \left( {{T}_{r}\left( n\right) }\right) \) . Then\n\n\[ e\left( {G - x}\right) = e\left( G\right) - d\left( x\right) \geq e\left( {{T}_{r}\left( {n - 1}\right) }\right) ,\]\n\nso by the induction hypothesis \( {G}_{x} = G - x \) is exactly a \( {T}_{r}\left( {n - 1}\right) \) .\n\nA smallest vertex class of \( {G}_{x} \) contains \( \lfloor \left( {n - 1}\right) /r\rfloor \) vertices, and the vertex \( x \) is joined to all but\n\n\[ n - 1 - \left( {n - \left\lceil \frac{n}{r}\right\rceil }\right) = \left\lfloor \frac{n - 1}{r}\right\rfloor \]\n\nvertices of \( {G}_{x} \) . Since \( x \) cannot be joined to a vertex in each class of \( {G}_{x} \), it has to be joined to all vertices of \( {G}_{x} \) save the vertices in a smallest vertex class. This shows that \( G = {T}_{r}\left( n\right) \), as required.
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Yes
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Lemma 9 Let \( m, n, s, t, k, r \) be non-negative integers, \( 2 \leq s \leq m,2 \leq t \leq n \) , \( 0 \leq r < m \), and let \( G = {G}_{2}\left( {m, n}\right) \) be an \( m \) by \( n \) bipartite graph of size \( z = {my} = {km} + r \) without a \( {K}_{s, t} \) subgraph having \( s \) vertices in the first class and t in the second. Then\n\n\[ m\left( \begin{array}{l} y \\ t \end{array}\right) \leq \left( {m - r}\right) \left( \begin{array}{l} k \\ t \end{array}\right) + r\left( \begin{matrix} k + 1 \\ t \end{matrix}\right) \leq \left( {s - 1}\right) \left( \begin{array}{l} n \\ t \end{array}\right) . \]
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Proof. Denote by \( {V}_{1} \) and \( {V}_{2} \) the vertex classes of \( G \) . We shall say that a \( t \) -set (i.e., a set with \( t \) elements) \( T \) of \( {V}_{2} \) is covered by a vertex \( x \in {V}_{1} \) if \( x \) is joined to every vertex in \( T \) . The number of \( t \) -sets covered by a vertex \( x \in {V}_{1} \) is \( \left( \begin{matrix} d\left( x\right) \\ t \end{matrix}\right) \) . Since the assumption on \( G \) is exactly that each \( t \) -set in \( {V}_{2} \) is covered by at most \( s - 1 \) vertices of \( {V}_{1} \), we find that\n\n\[ \mathop{\sum }\limits_{{x \in {V}_{1}}}\left( \begin{matrix} d\left( x\right) \\ t \end{matrix}\right) \leq \left( {s - 1}\right) \left( \begin{array}{l} n \\ t \end{array}\right) \]\n\n(3)\n\nAs \( \mathop{\sum }\limits_{{x \in {V}_{1}}}d\left( x\right) = z = {my} = {km} + r,0 \leq r < m \), and \( f\left( u\right) = \left( \begin{array}{l} u \\ t \end{array}\right) \) is a convex function of \( u \) for \( u \geq t \), inequality (3) implies (2).\n\nThe proof of Lemma 9 is the simple but powerful double counting argument; as this is perhaps the most basic combinatorial argument, let us spell it out again, this time in terms of the edges of a bipartite graph \( H \) . One of the vertex classes of \( H \) is just \( {V}_{1} \), but the other is \( {V}_{2}^{\left( t\right) } \), the set of all \( t \) -subsets of \( {V}_{2} \) . In our new graph \( H \), join \( x \in {V}_{1} \) to \( A \in {V}_{2}^{\left( t\right) } \) if in \( G \) the vertex \( x \) is joined to all \( t \) vertices of \( A \) . Now, counting from \( {V}_{1} \), we see that\n\n\[ e\left( H\right) = \mathop{\sum }\limits_{{x \in {V}_{1}}}\left( \begin{matrix} d\left( x\right) \\ t \end{matrix}\right) \]\n\nOn the other hand, as \( G \) contains no \( {K}_{s, t} \), in \( H \) every vertex \( A \in {V}_{2}^{\left( t\right) } \) has at most \( s - 1 \) neighbours. Thus\n\n\[ e\left( H\right) \leq \left( {s - 1}\right) \left( \begin{array}{l} n \\ t \end{array}\right) \]\n\nand the rest is simple algebra.
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Yes
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Theorem 10 For all natural numbers \( m, n, s \) and \( t \) we have\n\n\[ z\left( {m, n;s, t}\right) \leq {\left( s - 1\right) }^{1/t}\left( {n - t + 1}\right) {m}^{1 - 1/t} + \left( {t - 1}\right) m. \]
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Proof. Let \( G = {G}_{2}\left( {m, n}\right) \) be an extremal graph for the function \( z\left( {m, n;s, t}\right) = \) \( {my} \) without a \( K\left( {s, t}\right) \) subgraph. As \( y \geq n \), inequality (2) implies\n\n\[ {\left( y - \left( t - 1\right) \right) }^{t} \leq \left( {s - 1}\right) {\left( n - \left( t - 1\right) \right) }^{t}{m}^{-1}. \]
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No
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Theorem 11 Let \( n, s, t, k \) and \( r \) be non-negative integers, and let \( G \) be a graph of order \( z = {ny}/2 = \frac{1}{2}\left( {{kn} + r}\right) \), containing no \( {K}_{s, t} \) . Then\n\n\[ \n n\left( \begin{array}{l} y \\ t \end{array}\right) \leq \left( {n - r}\right) \left( \begin{array}{l} k \\ t \end{array}\right) + r\left( \begin{matrix} k + 1 \\ t \end{matrix}\right) \leq \left( {s - 1}\right) \left( \begin{array}{l} n \\ t \end{array}\right) .\n\]\n\nFurthermore,\n\n\[ \n\operatorname{ex}\left( {n,{K}_{s, t}}\right) \leq \frac{1}{2}{\left( s - 1\right) }^{1/t}{n}^{2 - 1/t} + \frac{1}{2}\left( {t - 1}\right) n.\n\]
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Proof. As in Lemma 9, let us say that a \( t \) -set of the vertices is covered by a vertex \( x \) if \( x \) is joined to every vertex of the \( t \) -set. Since \( G \) does not contain a \( {K}_{s, t} \), every \( t \) -set is covered by at most \( s - 1 \) vertices. Therefore, if \( G \) has degree sequence \( {\left( {d}_{i}\right) }_{1}^{n} \) then\n\n\[ \n\mathop{\sum }\limits_{1}^{n}\left( \begin{matrix} {d}_{i} \\ t \end{matrix}\right) \leq \left( {s - 1}\right) \left( \begin{array}{l} n \\ t \end{array}\right)\n\]\n\nand the rest is as in Lemma 9 and Theorem 10.
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Yes
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Theorem 12 For \( n \geq 1 \), we have\n\n\[ z\left( {n, n;2,2}\right) \leq \frac{1}{2}n\left\{ {1 + {\left( 4n - 3\right) }^{1/2}}\right\} \]\n\nand equality holds for infinitely many values of \( n \) . Furthermore,\n\n\[ \operatorname{ex}\left( {n,{C}_{4}}\right) \leq \frac{n}{4}\left( {1 + \sqrt{{4n} - 3}}\right) \]
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Proof. Since \( \operatorname{lex}\left( {n,{K}_{s, t}}\right) \leq z\left( {n, n;s, t}\right) \), the second inequality is immediate from the first. Moreover, the first inequality is just the case \( s = 2 \) of (4). In fact, the proof of Lemma 9 tells us a considerable amount about the graphs \( G \) for which equality is attained. We must have \( {d}_{1} = {d}_{2} = \cdots = {d}_{n} = d \), and any two vertices in the second vertex class \( {V}_{2} \) have degree \( d \) and any two vertices in \( {V}_{1} \) have exactly one common neighbour. Also, precisely the same assertions hold with \( {V}_{1} \) and \( {V}_{2} \) interchanged.\n\nCall the vertices in \( {V}_{2} \) points and the sets \( \Gamma \left( x\right), x \in {V}_{1} \), lines. By the remarks above there are \( n \) points and \( n \) lines, each point is on \( d \) lines; and each line contains \( d \) points, there is exactly one line through any two points and any two lines meet in exactly one point. Thus we have arrived at the projective plane of order \( d - 1 \) . Since the steps are easy to trace back, we see that equality holds for every \( n \) for which there is a projective plane with \( n \) points. In particular, equality holds for every \( n = {q}^{2} + q + 1 \) where \( q \) is a prime power.\n\nIn conclusion, let us see the actual construction of \( G \) for the above values of \( n \) . Let \( q \) be a prime power and let \( {PG}\left( {2, q}\right) \) be the projective plane over the field of order \( q \) . Let \( {V}_{1} \) be the set of points and \( {V}_{2} \) the set of lines. Then\n\n\[ \left| {V}_{1}\right| = \left| {V}_{2}\right| = {q}^{2} + q + 1 = n. \]\n\nLet \( G \) be the bipartite graph \( {G}_{2}\left( {n, n}\right) \) with vertex classes \( {V}_{1} \) and \( {V}_{2} \) in which we join a point \( P \in {V}_{1} \) to a line \( \ell \in {V}_{2} \) by an edge iff the point \( P \) is on the line \( \ell \) . (For \( q = 2 \) this gives us the Heawood graph, shown in Fig. I.7.) Then \( G \) has \( n\left( {q + 1}\right) = \frac{1}{2}n\left\{ {1 + {\left( 4n - 3\right) }^{1/2}}\right\} \) edges, and it does not contain a quadrilateral.
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Yes
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Lemma 14 Let \( G \) be a graph with vertex set \( V\left( G\right) = \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} \), whose \( k \) -closure \( {C}_{k}\left( G\right) \) contains at most \( t \leq n - 2 \) vertices of degree \( n - 1 \) . Then there are indices \( i, j,1 \leq i < j \leq n \), such that \( {x}_{i}{x}_{j} \in E\left( G\right) \) and each of the following four inequalities holds:\n\n\[ j \geq \max \{ {2n} - k - i, n - t\} \]\n\n\[ d\left( {x}_{i}\right) \leq i + k - n,\;d\left( {x}_{j}\right) \leq j + k - n - 1, \]\n\n(8)\n\n\[ d\left( {x}_{i}\right) + d\left( {x}_{j}\right) \leq k - 1 \]
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Proof. The graph \( H = {C}_{k}\left( G\right) \) is not complete so, we can define two indices \( i \) and \( j \) as follows:\n\n\[ j = \max \left\{ {\ell : {d}_{H}\left( {x}_{\ell }\right) \neq n - 1}\right\} \]\n\n\[ i = \max \left\{ {\ell : {x}_{\ell }{x}_{j} \notin E\left( H\right) }\right\} \]\n\nThen \( {x}_{i}{x}_{j} \notin E\left( H\right) \), so\n\n\[ {d}_{H}\left( {x}_{i}\right) + {d}_{H}\left( {x}_{j}\right) \leq k - 1 \]\n\nwhich implies the fourth inequality in (8). Each of the vertices\n\n\[ {x}_{j + 1},{x}_{j + 2},\ldots ,{x}_{n} \]\n\nhas degree \( n - 1 \) in \( H \), so\n\n\[ n - j \leq t \]\n\nand\n\n\[ n - j \leq \delta \left( H\right) \leq {d}_{H}\left( {x}_{i}\right) \]\n\nThe vertex \( {x}_{j} \) is joined to the \( n - j \) vertices following it and to the \( j - i - 1 \) vertices preceding it, so\n\n\[ {d}_{H}\left( {x}_{j}\right) \geq \left( {n - j}\right) + \left( {j - i - 1}\right) = n - i - 1. \]\n\nThese inequalities enable us to show that the indices \( i, j,1 \leq i < j \leq n \), satisfy the remaining three inequalities in (8). Indeed,\n\n\[ {d}_{G}\left( {x}_{i}\right) \leq {d}_{H}\left( {x}_{i}\right) \leq k - 1 - {d}_{H}\left( {x}_{j}\right) \leq k - 1 - \left( {n - i - 1}\right) = i + k - n, \]\n\n\[ {d}_{G}\left( {x}_{j}\right) \leq {d}_{H}\left( {x}_{j}\right) \leq k - 1 - {d}_{H}\left( {x}_{i}\right) \leq k - 1 - \left( {n - j}\right) = j + k - n - 1, \]\n\nand\n\n\[ i + j \geq \left( {n - {d}_{H}\left( {x}_{j}\right) - 1}\right) + \left( {n - {d}_{H}\left( {x}_{i}\right) }\right) \geq {2n} - 1 - \left( {k - 1}\right) = {2n} - k, \]\ncompleting the proof.
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Yes
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Theorem 18 Let \( W \) be the set of vertices of even degree in a graph \( G \) and let \( {x}_{0} \) be a vertex of \( G \) . Then there is an even number of longest \( {x}_{0} \) -paths ending in \( W \) .
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Proof. Let \( H \) be the graph whose vertex set is the set \( \sum \) of longest \( {x}_{0} \) -paths in \( G \) , in which \( {P}_{1} \in \sum \) is joined to \( {P}_{2} \in \sum \) if \( {P}_{2} \) is a simple transform of \( {P}_{1} \) . Since the degree of \( P = {x}_{0}{x}_{1}\cdots {x}_{k} \in \sum \) in \( H \) is \( d\left( {x}_{k}\right) - 1 \), the set of longest paths ending in \( W \) is exactly the set of vertices of odd degree in \( H \) . The number of vertices of odd degree is even in any graph, so the proof is complete.
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Yes
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Theorem 19 Let \( G \) be a graph in which every vertex has odd degree. Then every edge of \( G \) is contained in an even number of Hamilton cycles.
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Proof. Let \( {x}_{0}y \in E\left( G\right) \) . Then in \( {G}^{\prime } = G - {x}_{0}y \) only \( {x}_{0} \) and \( y \) have even degree, so in \( {G}^{\prime } \) there is an even number of longest \( {x}_{0} \) -paths that end in \( y \) . Thus either \( G \) has no Hamilton cycle that contains \( {x}_{0}y \) or it has a positive even number of them.
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Yes
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Lemma 21 Let \( c,\varepsilon > 0 \) . If \( n \) is sufficiently large, say \( n > 3/\varepsilon \), then every graph of order \( n \) and size at least \( \left( {c + \varepsilon }\right) \left( \begin{array}{l} n \\ 2 \end{array}\right) \) contains a subgraph \( H \) with \( \delta \left( H\right) \geq c\left| H\right| \) and \( \left| H\right| \geq {\varepsilon }^{1/2}n \) .
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Proof. Let \( G \) be a graph of order \( n > 3/\varepsilon \) and size \( e\left( G\right) \geq \left( {c + \varepsilon }\right) \left( \begin{array}{l} n \\ 2 \end{array}\right) \) . Note that in this case \( 0 < \varepsilon < \varepsilon + c \leq 1 \) . If the assertion fails then there is a sequence of graphs \( {G}_{n} = G \supset {G}_{n - 1} \supset \cdots \supset {G}_{\ell },\ell = \left\lfloor {{\varepsilon }^{1/2}n}\right\rfloor \), such that \( \left| {G}_{j}\right| = j \) and for \( n \geq j > \ell \) the only vertex of \( {G}_{j} \) not in \( {G}_{j - 1} \) has degree less than \( {cj} \) in \( {G}_{j} \) . Then\n\n\[ e\left( {G}_{\ell }\right) > \left( {c + \varepsilon }\right) \left( \begin{array}{l} n \\ 2 \end{array}\right) - \mathop{\sum }\limits_{{j = \ell + 1}}^{n}{cj} = \left( {c + \varepsilon }\right) \left( \begin{array}{l} n \\ 2 \end{array}\right) - c\left\{ {\left( \begin{matrix} n + 1 \\ 2 \end{matrix}\right) - \left( \begin{matrix} \ell + 1 \\ 2 \end{matrix}\right) }\right\} \]\n\n\[ > \varepsilon \left( \begin{array}{l} n \\ 2 \end{array}\right) + c\left( \begin{matrix} \ell + 1 \\ 2 \end{matrix}\right) - {cn} > \varepsilon \left( \begin{array}{l} n \\ 2 \end{array}\right) > \left( \begin{array}{l} \ell \\ 2 \end{array}\right) ,\]\n\nsince \( 0 < \varepsilon < 1 \) and \( n \geq 3/n \) . This contradiction completes the proof.
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Yes
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Theorem 22 Let \( r \geq 1 \) be an integer and let \( \varepsilon > 0 \) . Then there is an integer \( {n}_{0} = {n}_{0}\left( {r,\varepsilon }\right) \) such that if \( \left| G\right| = n \geq {n}_{0} \) and\n\n\[ e\left( G\right) \geq \left( {1 - \frac{1}{r} + \varepsilon }\right) \left( \begin{array}{l} n \\ 2 \end{array}\right) \]\n\nthen \( G \supset {K}_{r + 1}\left( t\right) \) for some \( t \geq \varepsilon \log n/\left( {{2}^{r + 1}\left( {r - 1}\right) !}\right) \) .
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Proof. If \( n > 3/\varepsilon \) then, by Lemma 21, \( G \) has a subgraph \( H \) with \( \left| H\right| = h \geq {\varepsilon }^{1/2}n \) and \( \delta \left( H\right) \geq \left( {1 - \frac{1}{r} + \varepsilon /2}\right) h \) . Hence if \( n \) is sufficiently large then \( H \) contains a \( {K}_{r + 1}\left( t\right) \) with \( t \geq \frac{\varepsilon }{2}\log h/\left( {{2}^{r - 1}\left( {r - 1}\right) !}\right) \geq \varepsilon \log n/\left( {{2}^{r + 1}\left( {r - 1}\right) !}\right) \), as claimed.
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Yes
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Corollary 24 Let \( {F}_{1},{F}_{2},\ldots ,{F}_{\ell } \) be non-empty graphs. Denote by \( r + 1 \) the minimum of the chromatic numbers of the \( {F}_{i} \), that is, let \( r + 1 \) be the minimal number for which at least one of the \( {F}_{i} \) is contained in an \( F = {K}_{r + 1}\left( t\right) \) for some t. Then the maximal size of a graph of order \( n \) not containing any of the \( {F}_{i} \) is\n\n\[ \operatorname{ex}\left( {n;{F}_{1},{F}_{2},\ldots ,{F}_{\ell }}\right) = \left( {1 - \frac{1}{r}}\right) \left( \begin{array}{l} n \\ 2 \end{array}\right) + o\left( {n}^{2}\right) . \]
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Proof. The Turán graph \( {T}_{r}\left( n\right) \) does not contain any of the \( {F}_{i} \) so, by (1),\n\n\[ \operatorname{ex}\left( {n;{F}_{1},{F}_{2},\ldots ,{F}_{\ell }}\right) \geq e\left( {{T}_{r - 1}\left( n\right) }\right) = {t}_{r - 1}\left( n\right) \leq \left( {1 - \frac{1}{r}}\right) \left( \begin{array}{l} n \\ 2 \end{array}\right) . \]\n\nConversely, since, say \( {F}_{j} \subset F = {K}_{r + 1}\left( t\right) \) for some \( j \) and \( t \),\n\n\[ \operatorname{ex}\left( {n;{F}_{1},{F}_{2},\ldots ,{F}_{\ell }}\right) \leq \operatorname{ex}\left( {n;{F}_{j}}\right) \leq \operatorname{ex}\left( {n;F}\right) = \left( {1 - \frac{1}{r}}\right) \left( \begin{array}{l} n \\ 2 \end{array}\right) + o\left( {n}^{2}\right) . \]
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Yes
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Corollary 25 The upper density of an infinite graph \( G \) is \( 1,\frac{1}{2},\frac{2}{3},\frac{3}{4},\ldots \), or 0 . Each of these values is the upper density of some infinite graph.
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Proof. Let \( {G}_{r} \) be the complete \( r \) -partite graph with infinitely many vertices in each class. Since the density of \( {K}_{r}\left( t\right) \) tends to \( 1 - \frac{1}{r} \) as \( t \) tends to \( \infty \), the upper density of \( {G}_{r} \) is \( 1 - \frac{1}{r} \), proving the second assertion.\n\nNow, let \( \alpha \) be the upper density of \( G \) and suppose that\n\n\[ \alpha > 1 - \frac{1}{r - 1} \]\n\nwhere \( r \geq 2 \) . Then there is an \( \varepsilon > 0 \) such that \( G \) contains graphs \( {H}_{k} \) of order \( {n}_{k} \) with \( {n}_{k} \rightarrow \infty \) satisfying\n\n\[ e\left( {H}_{k}\right) \geq \frac{1}{2}\left( {1 - \frac{1}{r - 1} + \varepsilon }\right) {n}_{k}^{2}. \]\n\nBy Theorem 20 each \( {H}_{k} \) contains a subgraph \( {K}_{r}\left( {t}_{k}\right) \) with \( {t}_{k} \rightarrow \infty \) ; the subgraphs \( {K}_{r}\left( {t}_{k}\right) \) show that \( \alpha \geq \frac{1}{r} \) .
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Yes
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Lemma 26 Suppose that \( X \) and \( Y \) are disjoint sets of vertices of a graph \( G \) , and \( {X}^{ * } \subset X \) and \( {Y}^{ * } \subset Y \) are such that \( \left| {X}^{ * }\right| \geq \left( {1 - \gamma }\right) \left| X\right| > 0 \) and \( \left| {Y}^{ * }\right| \geq \left( {1 - \delta }\right) \left| Y\right| > 0 \) . Then\n\n\[ \left| {d\left( {{X}^{ * },{Y}^{ * }}\right) - d\left( {X, Y}\right) }\right| \leq \gamma + \delta \]\n\n(9)\n\n\n\nand\n\n\[ \left| {{d}^{2}\left( {{X}^{ * },{Y}^{ * }}\right) - {d}^{2}\left( {X, Y}\right) }\right| < 2\left( {\gamma + \delta }\right) .\n\n(10) \]
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Proof. Note that, rather crudely,\n\n\[ 0 \leq e\left( {X, Y}\right) - e\left( {{X}^{ * },{Y}^{ * }}\right) \leq \left( {\gamma + \delta - {\gamma \delta }}\right) \left| X\right| \left| Y\right| < \left( {\gamma + \delta }\right) \left| X\right| \left| Y\right| ,\]\n\nso\n\n\[ d\left( {X, Y}\right) - d\left( {{X}^{ * },{Y}^{ * }}\right) \leq \frac{e\left( {X, Y}\right) - e\left( {{X}^{ * },{Y}^{ * }}\right) }{\left| X\right| \left| Y\right| } < \gamma + \delta .\n\nIf \( G \) is replaced by its complement \( \bar{G} \), then each density \( d \) changes to \( 1 - d \), so\n\n\[ \left. {{d}_{G}\left( {{X}^{ * },{Y}^{ * }}\right) - {d}_{G}\left( {X, Y}\right) = {d}_{\bar{G}}\left( {X, Y}\right) - {d}_{\bar{G}}\left( {{X}^{ * },{Y}^{ * }}\right) }\right) < \gamma + \delta ,\]\n\ncompleting the proof of (9).\n\nInequality (10) is an immediate consequence of (9):\n\n\[ \left| {{d}^{2}\left( {{X}^{ * },{Y}^{ * }}\right) - {d}^{2}\left( {X, Y}\right) }\right|\]\n\n\[ = \left| {d\left( {{X}^{ * },{Y}^{ * }}\right) + d\left( {X, Y}\right) }\right| \left| {d\left( {{X}^{ * },{Y}^{ * }}\right) - d\left( {X, Y}\right) }\right| < 2\left( {\gamma + \delta }\right) . \]
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Yes
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Lemma 27 Let \( {\left( {d}_{i}\right) }_{i = 1}^{s} \subset \mathbb{R},1 \leq t < s, D = \frac{1}{s}\mathop{\sum }\limits_{{i = 1}}^{s}{d}_{i} \), and \( d = \frac{1}{t}\mathop{\sum }\limits_{{i = 1}}^{t}{d}_{i} \). Then \[ \frac{1}{s}\mathop{\sum }\limits_{{i = 1}}^{s}{d}_{i}^{2} \geq {D}^{2} + \frac{t}{s - t}{\left( D - d\right) }^{2} \geq {D}^{2} + \frac{t}{s}{\left( D - d\right) }^{2}. \]
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Proof. With \[ e = \frac{1}{s - t}\mathop{\sum }\limits_{{i = t + 1}}^{s}{d}_{i} = \frac{{sD} - {td}}{s - t} \] the convexity of the function \( {x}^{2} \) implies that \[ \mathop{\sum }\limits_{{i = 1}}^{s}{d}_{i}^{2} = \mathop{\sum }\limits_{{i = 1}}^{t}{d}_{i}^{2} + \mathop{\sum }\limits_{{i = t + 1}}^{s}{d}_{i}^{2} \geq t{d}^{2} + \left( {s - t}\right) {e}^{2} \] \[ = t{d}^{2} + \frac{{s}^{2}{D}^{2} - {2stdD} + {t}^{2}{d}^{2}}{s - t} \] \[ = s{D}^{2} + \frac{st}{s - t}{\left( D - d\right) }^{2}. \]
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Yes
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Lemma 28 Let \( \;G \) be a graph of order \( \;n\; \) with an equitable partition \( \;V = \mathop{\bigcup }\limits_{{i = 0}}^{k}{C}_{i} \) of the vertex set with exceptional class \( {C}_{0} \) and\n\n\[ \n\left| {C}_{1}\right| = \left| {C}_{2}\right| = \cdots = \left| {C}_{k}\right| = c \geq {2}^{{3k} + 1}.\n\]\n\nSuppose that the partition \( \mathcal{P} = {\left( {C}_{i}\right) }_{i = 0}^{k} \) is not \( \varepsilon \) -uniform, where \( 0 < \varepsilon < \frac{1}{2} \) and \( {2}^{-k} \leq {\varepsilon }^{5}/8 \) . Then there is an equitable partition \( {\mathcal{P}}^{\prime } = {\left( {C}_{i}^{\prime }\right) }_{i = 0}^{\ell } \) with \( \ell = \) \( k\left( {{4}^{k} - {2}^{k - 1}}\right) \) and exceptional class \( {C}_{0}^{\prime } \supset {C}_{0} \) such that\n\n\[ \n\left| {C}_{0}^{\prime }\right| \leq \left| {C}_{0}\right| + \frac{n}{{2}^{k}}\n\]\n\nand\n\n\[ \nq\left( {\mathcal{P}}^{\prime }\right) \geq q\left( \mathcal{P}\right) + \frac{{\varepsilon }^{5}}{2}\n\]
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Proof. For a pair \( \left( {{C}_{i},{C}_{j}}\right) \) that is not \( \varepsilon \) -uniform, let \( {C}_{ij} \subset {C}_{i} \) and \( {C}_{ji} \subset {C}_{j} \) be sets showing that \( \left( {{C}_{i},{C}_{j}}\right) \) is not \( \varepsilon \) -uniform: \( \left| {C}_{ij}\right| \geq \varepsilon \left| {C}_{i}\right| = {\varepsilon c},\left| {C}_{ji}\right| \geq \varepsilon \left| {C}_{j}\right| = {\varepsilon c} \) , and\n\n\[ \n\left| {d\left( {{C}_{ij},{C}_{ji}}\right) - d\left( {{C}_{i},{C}_{j}}\right) }\right| \geq \varepsilon .\n\]\n\n(11)\n\nFurthermore, for an \( \varepsilon \) -uniform pair \( \left( {{C}_{i},{C}_{j}}\right) \), set \( {C}_{ij} = {C}_{ji} = \varnothing \) .\n\nIdeally, we would like to partition each \( {C}_{i} \) into a few (according to the statement of the lemma, into \( {4}^{k} - {2}^{k - 1} \) ) sets \( {C}_{h}^{\prime } \) of size \( d \), say, such that each \( {C}_{ij} \) is the exact union of some of these sets \( {C}_{h}^{\prime } \) . In this way, a large difference \( \mid d\left( {{C}_{ij},{C}_{ji}}\right) - \) \( d\left( {{C}_{i},{C}_{j}}\right) \mid \) would guarantee, by Lemma 27, that the part of \( q\left( {\mathcal{P}}^{\prime }\right) \) arising from \( {d}^{2}\left( {{C}_{i},{C}_{j}}\right) \), namely\n\n\[ \n\frac{{d}^{2}}{{c}^{2}}\sum \left\{ {{d}^{2}\left( {{C}_{g}^{\prime },{C}_{h}^{\prime }}\right) : {C}_{g}^{\prime } \subset {C}_{ij},{C}_{h}^{\prime } \subset {C}_{ji}}\right\} ,\n\]\nis appreciably larger than \( {d}^{2}\left( {{C}_{i},{C}_{j}}\right) \) . In turn, this would imply that \( q\left( {\mathcal{P}}^{\prime }\right) \) is considerably larger than \( q\left( \mathcal{P}\right) \) .\n\nAlthough we cannot construct sets \( {C}_{h}^{\prime } \) such that each \( {C}_{ij} \) is the exact union of some of these sets, we can come fairly close to it: we can achieve that each \( {C}_{ij} \) is almost the union of the sets \( {C}_{h}^{\prime } \) it contains. We do this by considering, for each \( {C}_{i} \) , all the sets \( {C}_{ij} \) at once, and choosing the future \( {C}_{R}^{\prime } \) sets (to be denoted by \( {D}_{ih} \) ) such that they do not cut across any \( {C}_{ij} \) . The price we have to pay is simply that we cannot quite partition \( {C}_{i} \) into the sets \( {C}_{R}^{\prime } \), so we have to add the remainder to the \
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Yes
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Theorem 30 Let \( f \geq 2, r \geq 2,0 < \delta < 1/r \) and let \( {V}_{1},{V}_{2},\ldots ,{V}_{r} \) be disjoint subsets of vertices of a graph \( G \) . Suppose \( \left| {V}_{i}\right| \geq {\delta }^{-f} \) for every \( i \), and if \( 1 \leq i < j \leq r \) and \( {W}_{i} \subset {V}_{i},{W}_{j} \subset {V}_{j} \) satisfy \( \left| {W}_{i}\right| \geq {\delta }^{f}\left| {V}_{i}\right| \) and \( \left| {W}_{j}\right| \geq \) \( {\delta }^{f}\left| {V}_{j}\right| \), then \( d\left( {{W}_{i},{W}_{j}}\right) \geq \delta \) . Then for all non-negative integers \( {f}_{1},\ldots ,{f}_{r} \) with \( \mathop{\sum }\limits_{{i = 1}}^{r}{f}_{i} = f \) there are sets \( {U}_{1} \subset {V}_{1},\ldots ,{U}_{r} \subset {V}_{r} \) with \( \left| {U}_{i}\right| = {f}_{i} \) for \( 1 \leq i \leq r \) , such that for \( 1 \leq i < j \leq r \) every vertex of \( {U}_{i} \) is joined to every vertex of \( {U}_{j} \) . In particular, \( G \) contains every \( r \) -partite graph on \( f \) vertices.
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Proof. Let us apply induction on \( f \) . For \( f = 2 \) the assertion is trivial, so suppose that \( f \geq 3 \) and the assertion holds for smaller values of \( f \) . We may assume that \( {f}_{1} \geq 1 \) .\n\nFor \( 2 \leq i \leq r \), let \( {R}_{i} \) be the set of vertices in \( {V}_{1} \) joined to fewer than \( \delta \left| {V}_{i}\right| \) vertices of \( {V}_{i} \) . Then \( \left| {R}_{i}\right| < {\delta }^{f}\left| {V}_{i}\right| \), so \( \left| {\mathop{\bigcup }\limits_{{i = 2}}^{r}{R}_{i}}\right| < \left( {r - 1}\right) {\delta }^{f}\left| {V}_{1}\right| < \left| {V}_{1}\right| \) . Hence there is a vertex \( x \in {V}_{1} \smallsetminus \mathop{\bigcup }\limits_{{i = 2}}^{r}{R}_{i} \) ; set \( {V}_{1}^{\prime } = {V}_{1} \smallsetminus \{ x\} \) and \( {V}_{i}^{\prime } = {V}_{i} \cap \Gamma \left( x\right) \) for \( i = 2,\ldots, r \) . Then \( \left| {V}_{1}^{\prime }\right| \geq {\delta }^{-f} - 1 \geq {\delta }^{-f + 1} \) and \( \left| {V}_{1}^{\prime }\right| \geq \left( {1 - {\delta }^{f}}\right) \left| {V}_{1}\right| \geq \delta \left| {V}_{1}\right| \) ; furthermore, \( \left| {V}_{i}^{\prime }\right| \geq \delta \left| {V}_{i}\right| \geq {\delta }^{-f + 1} \) for \( 2 \leq i \leq r \) . Also, if \( {W}_{i} \subset {V}_{i}^{\prime } \) and \( {W}_{j} \subset \) \( {V}_{j}^{\prime },1 \leq i < j \leq r \), are such that \( \left| {W}_{i}\right| \geq {\delta }^{f - 1}\left| {V}_{i}^{\prime }\right| \) and \( \left| {W}_{j}\right| \geq {\delta }^{f - 1}\left| {V}_{j}^{\prime }\right| \) , then \( \left| {W}_{i}\right| \geq {\delta }^{f}\left| {V}_{i}\right| \) and \( \left| {W}_{j}\right| \geq {\delta }^{f}\left| {V}_{j}\right| \), so \( {V}_{1}^{\prime },\ldots ,{V}_{r}^{\prime } \) satisfy the conditions for \( 0 < \delta < 1/r \) and \( f - 1 \) . Hence, by the induction hypothesis, there are sets \( {U}_{1}^{\prime } \subset {V}_{1}^{\prime },\ldots ,{U}_{r}^{\prime } \subset {V}_{r}^{\prime } \) with \( \left| {U}_{1}^{\prime }\right| = {f}_{1} - 1 \) and \( \left| {U}_{i}^{\prime }\right| = {f}_{i} \) for \( 2 \leq i \leq r \) such that for \( 1 \leq i < j \leq r \), every vertex of \( {U}_{i}^{\prime } \) is joined to every vertex of \( {U}_{j}^{\prime } \) . Clearly, the sets \( {U}_{1} = {U}_{1}^{\prime } \cup \{ x\} ,{U}_{2} = {U}_{2}^{\prime },\ldots ,{U}_{r} = {U}_{r}^{\prime } \) have the required properties.
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Yes
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Theorem 32 Let \( 0 < \varepsilon < 1 \) and \( 0 < \delta < 1 \) be real numbers, let \( m \geq 2 \) be an integer, and let \( M = {M}^{\prime \prime }\left( {\varepsilon, m}\right) \) be as in Theorem \( {29}^{\prime \prime } \) of the previous section. Let \( G \) be a graph of order \( n \geq M \), and let \( H = G\left\lbrack {k;\varepsilon ;d > \delta }\right\rbrack \) be an \( \left( {m;\varepsilon ;d > \delta }\right) \) -piece of \( G \) . Then\n\n\[ e\left( G\right) - e\left( H\right) < \left( {\varepsilon + \delta + \frac{1}{m} + \frac{2M}{n}}\right) {n}^{2}/2. \]
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Proof. Let \( V\left( G\right) = \mathop{\bigcup }\limits_{{i = 0}}^{k}{C}_{i} \) be the partition guaranteed by Theorem \( {29}^{\prime \prime } \) so that \( \left| {C}_{0}\right| \leq k - 1,\left| {C}_{1}\right| = \cdots = \left| {C}_{k}\right|, m \leq k \leq M \), and \( H \) is the appropriate \( k \) -partite graph with classes \( {C}_{1},\ldots ,{C}_{k} \) . Clearly, \( E\left( G\right) - E\left( H\right) \) consists of four types of edges: the edges incident with \( {C}_{0} \), the edges joining vertices in the same class \( {C}_{i} \) , \( 1 \leq i \leq k \), the \( {C}_{i} - {C}_{j} \) edges with \( \left( {{C}_{i},{C}_{j}}\right) \) not \( \varepsilon \) -uniform, \( 1 \leq i < j \leq k \) and the \( {C}_{i} - {C}_{j} \) edges with \( \left( {{C}_{i},{C}_{j}}\right) \varepsilon \) -uniform with density at most \( \delta ,1 \leq i < j \leq k \) . Hence\n\n\[ \left| {E\left( G\right) - E\left( H\right) }\right| < {kn} + k\left( \begin{matrix} n/k \\ 2 \end{matrix}\right) + \varepsilon \left( \begin{array}{l} k \\ 2 \end{array}\right) {\left( \frac{n}{k}\right) }^{2} + \delta \left( \begin{array}{l} n \\ 2 \end{array}\right) \]\n\n\[ < {kn} + \frac{{n}^{2}}{2k} + \frac{\varepsilon {n}^{2}}{2} + \frac{\delta {n}^{2}}{2} \]\n\n\[ \leq {Mn} + \frac{{n}^{2}}{2m} + \frac{\varepsilon {n}^{2}}{2} + \frac{\delta {n}^{2}}{2} \]\n\n\[ \leq \left( {\varepsilon + \delta + \frac{2M}{n} + \frac{1}{m}}\right) \frac{{n}^{2}}{2} \]\n\n as claimed.
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Yes
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Theorem 33 For every \( \varepsilon > 0 \) and graph \( F \), there is a constant \( {n}_{0} = {n}_{0}\left( {\varepsilon, F}\right) \) with the following property. Let \( G \) be a graph of order \( n \geq {n}_{0} \) that does not contain \( F \) as a subgraph. Then \( G \) contains a set \( {E}^{\prime } \) of less than \( \varepsilon {n}^{2} \) edges such that the subgraph \( H = G - {E}^{\prime } \) has no \( {K}_{r} \), where \( r = \chi \left( F\right) \) .
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Proof. We may assume that \( r \geq 2,0 < \varepsilon < 1/r \), and \( f = \left| F\right| \geq 3 \) . Let \( \delta = \varepsilon /2 \) and \( m \geq 8/\varepsilon = 4/\delta \) . Let \( M = {M}^{\prime \prime }\left( {{\delta }^{f}, m}\right) \) be given by Szemerédi’s lemma, as in Theorem \( {29}^{\prime \prime } \) . We claim that \( {n}_{0} = \left\lceil {{8M}{\delta }^{-f}}\right\rceil \) will do. Indeed, let \( H = G\left\lbrack {k;{\delta }^{f};d > \delta + {\delta }^{f}}\right\rbrack \) be an \( \left( {m;{\delta }^{f};d > \delta + {\delta }^{f}}\right) \) -piece of \( G \), with skeleton \( S = S\left\lbrack {k;{\delta }^{f};d > \delta + {\delta }^{f}}\right\rbrack \) . Then, by Theorem 32, \[ e\left( G\right) - e\left( H\right) < \left( {\delta + {\delta }^{f}}\right) {n}^{2} < \varepsilon {n}^{2}. \] Furthermore, by Theorem \( {30}, S \) contains no \( {K}_{r} \) ; therefore, neither does \( H \) .
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Yes
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Theorem 35 Let \( r \geq 3 \) and \( s \geq 2 \) be fixed integers, and \( c \) and \( \gamma \) positive constants. Then if \( n \) is sufficiently large and \( G \) is a graph of order \( n \) that contains neither \( {K}_{r} \) nor \( {K}_{s, t} \), where \( t = \lceil {cn}\rceil \), then\n\n\[ e\left( G\right) \leq {c}^{1/s}{\left( \frac{r - 2}{r - 1}\right) }^{1 - 1/s}\frac{{n}^{2}}{2} + \gamma {n}^{2}. \]\n
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Proof. We may assume that \( 0 < \gamma < 1/2 \) and \( c < \left( {r - 2}\right) /\left( {r - 1}\right) \), since we do know that \( e\left( G\right) \leq {t}_{r - 1}\left( n\right) \leq \frac{r - 2}{2\left( {r - 1}\right) }{n}^{2} \) .\n\nLet \( \delta = \gamma /2, m \geq 4/\delta \), and suppose that \( n \geq {4Ms}/\delta \geq {8M}/\delta \), where \( M = {M}^{\prime \prime }\left( {{\delta }^{r};m}\right) \) . Let \( G \) be a graph of order \( n \) containing neither \( {K}_{r} \) nor \( {K}_{s, t} \) . Let \( H = G\left\lbrack {k;{\delta }^{r};d > \delta + {\delta }^{r}}\right\rbrack \) be an \( \left( {m;{\delta }^{r};d > \delta + {\delta }^{r}}\right) \) -piece of \( G \) with skeleton \( S \) . Then, by Theorem 32, \( e\left( G\right) - e\left( H\right) < \left( {\delta + {\delta }^{r}}\right) {n}^{2} < {2\gamma }{n}^{2}/3 \), and by Theorem 31, \( S \) does not contain a \( {K}_{r} \) . Hence, by Turán’s theorem, \( q = e\left( S\right) \leq \left( {r - 2}\right) {k}^{2}/2\left( {r - 1}\right) \).\n\nAs \( H \) contains no \( {K}_{s, t} \), by Theorem 34 we have, with \( \ell = \lfloor n/k\rfloor \),\n\n\[ {2e}\left( H\right) \leq {\left( \frac{r - 2}{r - 1}\right) }^{1 - 1/s}{\left( k\ell \right) }^{2 - 1/s}{\left( t - 1}\right) }^{1/s} + {k}^{2}\ell s \]\n\n\[ \leq {c}^{1/s}{\left( \frac{r - 2}{r - 1}\right) }^{1 - 1/s}{n}^{2} + {Msn} \]\n\n\[ \leq {c}^{1/s}{\left( \frac{r - 2}{r - 1}\right) }^{1 - 1/s}{n}^{2} + \frac{\gamma {n}^{2}}{8}. \]\n\nTherefore,\n\n\[ e\left( G\right) < e\left( H\right) + \gamma {n}^{2}/2 + < e\left( H\right) + \gamma {n}^{2}, \]\n\nas claimed.
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Yes
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Theorem 1 Let \( k = \mathop{\max }\limits_{H}\delta \left( H\right) \), where the maximum is taken over all induced subgraphs of \( G \) . Then \( \chi \left( G\right) \leq k + 1 \) .
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Proof. The graph \( G \) itself has a vertex of degree at most \( k \) ; let \( {x}_{n} \) be such a vertex, and put \( {H}_{n - 1} = G - \left\{ {x}_{n}\right\} \) . By assumption, \( {H}_{n - 1} \) has a vertex of degree at most \( k \) . Let \( {x}_{n - 1} \) be one of them and put \( {H}_{n - 2} = {H}_{n - 1} - \left\{ {x}_{n - 1}\right\} = G - \left\{ {{x}_{n},{x}_{n - 1}}\right\} \) . Continuing in this way we enumerate all the vertices.\n\nNow, the sequence \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) is such that each \( {x}_{j} \) is joined to at most \( k \) vertices preceding it. Hence the greedy algorithm will never need colour \( k + 2 \) to colour a vertex.
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Yes
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Theorem 3 Let \( G \) be a connected graph with maximal degree \( \Delta \) . Suppose \( G \) is neither a complete graph nor an odd cycle. Then \( \chi \left( G\right) \leq \Delta \) .
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Proof. We know already that we may assume without loss of generality that \( G \) is 2-connected and \( \Delta \) -regular. Furthermore, we may assume that \( \Delta \geq 3 \), since a connected 2-regular 3-chromatic graph is an odd cycle.\n\n\n\nFIGURE V.2. The vertex set of the thick triangle disconnects \( G \), and we find that \( \chi \left( G\right) = \) \( \max \left\{ {\chi \left( {G}_{1}\right) ,\chi \left( {G}_{2}\right) ,\chi \left( {G}_{3}\right) }\right\} \)\n\nIf \( G \) is 3-connected, let \( {x}_{n} \) be any vertex of \( G \) and let \( {x}_{1},{x}_{2} \) be two nonadjacent vertices in \( \Gamma \left( {x}_{n}\right) \) . Such vertices exist since \( G \) is regular and not complete. If \( G \) is not 3-connected, let \( {x}_{n} \) be a vertex for which \( G - {x}_{n} \) is separable, and thus has at least two blocks. Since \( G \) is 2-connected, each endblock of \( G - {x}_{n} \) has a vertex adjacent to \( {x}_{n} \) . Let \( {x}_{1} \) and \( {x}_{2} \) be such vertices belonging to different endblocks.\n\nIn either case, we have found vertices \( {x}_{1},{x}_{2} \) and \( {x}_{n} \) such that \( G - \left\{ {{x}_{1},{x}_{2}}\right\} \) is connected, \( {x}_{1}{x}_{2} \notin E\left( G\right) \), but \( {x}_{1}{x}_{n} \in E\left( G\right) \) and \( {x}_{2}{x}_{n} \in E\left( G\right) \) . Let \( {x}_{n - 1} \in \) \( V - \left\{ {{x}_{1},{x}_{2},{x}_{n}}\right\} \) be a neighbour of \( {x}_{n} \), let \( {x}_{n - 2} \) be a neighbour of \( {x}_{n} \) or \( {x}_{n - 1} \), etc. Then the order \( {x}_{1},{x}_{2},{x}_{3},\ldots ,{x}_{n} \) is such that each vertex other than \( {x}_{n} \) is adjacent to at least one vertex following it. Thus the greedy algorithm will use at most \( \Delta \) colours, since \( {x}_{1} \) and \( {x}_{2} \) get the same colour and \( {x}_{n} \), the only vertex with \( \Delta \) neighbours preceding it, is adjacent to both.
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Yes
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Theorem 4 Let \( H \) be a graph with \( n \geq 1 \) vertices, \( m \) edges and \( k \) components. Then\n\n\[ \n{p}_{H}\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{{n - k}}{\left( -1\right) }^{i}{a}_{i}{x}^{n - i}, \]\n\nwhere \( {a}_{0} = 1,{a}_{1} = m \) and \( {a}_{i} \) is a positive integer for every \( i,0 \leq i \leq n - k \) .
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Proof. We apply induction on \( n + m \) . For \( n + m = 1 \) the assertions are trivial so we pass to the induction step. If \( m = 0 \), we are again done, since in this case \( k = n \) and, as every map \( f : V\left( H\right) \rightarrow \{ 1,2,\ldots, x\} \) is a colouring of \( H \), we have \( {p}_{H}\left( x\right) = {x}^{n} \) . If \( m > 0 \) we pick two adjacent vertices of \( H \), say \( a \) and \( b \) . Putting \( G = H - {ab} \) we find that \( {G}^{\prime } = H \) . Since \( e\left( G\right) = m - 1 \) and \( \left| {G}^{\prime \prime }\right| + e\left( {G}^{\prime \prime }\right) \leq n - 1 + m \), by the induction hypothesis the assertions of the theorem hold for \( {p}_{G}\left( x\right) \) and \( {p}_{{G}^{\prime \prime }}\left( x\right) \) . Note now that \( {G}^{\prime \prime } \) has \( k \) components and \( G \) has at least \( k \) components. Therefore,\n\n\[ \n{p}_{G}\left( x\right) = {x}^{n} - \left( {m - 1}\right) {x}^{n - 1} + \mathop{\sum }\limits_{{i = 2}}^{{n - k}}{\left( -1\right) }^{i}{b}_{i}{x}^{n - i}, \]\n\nwhere \( {b}_{i} \) is a nonnegative integer for each \( i \), and\n\n\[ \n{p}_{{G}^{\prime \prime }}\left( x\right) = {x}^{n - 1} - \mathop{\sum }\limits_{{i = 2}}^{{n - k}}{\left( -1\right) }^{i}{c}_{i}{x}^{n - i}, \]\n\nwhere \( {c}_{i} \) is a positive integer for each \( i \) . Hence, by (3),\n\n\[ \n{p}_{H}\left( x\right) = {p}_{{G}^{\prime }}\left( x\right) = {p}_{G}\left( x\right) - {p}_{{G}^{\prime \prime }}\left( x\right) \]\n\n\[ \n= {x}^{n} - m{x}^{n - 1} + \mathop{\sum }\limits_{{i = 2}}^{{n - k}}{\left( -1\right) }^{i}\left( {{b}_{i} + {c}_{i}}\right) {x}^{n - i} \]\n\n\[ \n= {x}^{n} - m{x}^{n - 1} + \mathop{\sum }\limits_{{i = 2}}^{{n - k}}{\left( -1\right) }^{i}{a}_{i}{x}^{n - i}, \]\n\nwhere \( {a}_{i} \) is a positive integer for each \( i \) .
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Yes
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Theorem 5 Let \( H \) be a graph with \( n \) vertices and edge set \( E\left( H\right) = \left\{ {{e}_{1},{e}_{2},\ldots ,{e}_{m}}\right\} \). Call a subset of \( E\left( H\right) \) a broken cycle if it is obtained from the edge set of a cycle by deleting the edge of highest index. Then the chromatic polynomial of \( H \) is \[ {p}_{H}\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}{\left( -1\right) }^{i}{a}_{i}{x}^{n - i} \] where \( {a}_{i} \) is the number of \( i \) -subsets of \( E\left( H\right) \) containing no broken cycle.
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Proof. Let us apply induction on \( m \). For \( m = 0 \) the assertion is trivial, so suppose that \( m \geq 1 \) and the assertion holds for smaller values of \( m \). Let \( {e}_{1} = {ab} \) and, as before, set \( G = H - {ab} \), so that \( {G}^{\prime } = G + {ab} = H \) and \( {G}^{\prime \prime } = G/{ab} \) satisfy (1). With a slight abuse of notation, we identify not only \( E\left( G\right) = \left\{ {{e}_{2},{e}_{3},\ldots ,{e}_{m}}\right\} \), but also \( E\left( {G}^{\prime \prime }\right) \), with a subset of \( E\left( H\right) \). If an edge of \( E\left( {G}^{\prime \prime }\right) \) comes from only one edge of \( E\left( G\right) \), we keep its notation, and if an edge \( \left( {ab}\right) x \) comes from two edges of \( G \), say \( {e}_{i} = {ax} \) and \( {e}_{h} = {bx} \), then we denote \( \left( {ab}\right) x \) by \( {e}_{k} \), where \( k = \max \{ i, h\} \). As (1) holds, to complete the induction step, all we have to check is that the number of \( i \) -subsets of \( E\left( {G}^{\prime }\right) \) containing no broken cycle of \( {G}^{\prime } \) is precisely the sum of the number of \( i \) -subsets of \( E\left( G\right) \) containing no broken cycle and the number of \( \left( {i - 1}\right) \) -subsets of \( E\left( {G}^{\prime \prime }\right) \) containing no broken cycle. But this is a consequence of the following two simple assertions. (1) Suppose \( {e}_{1} \notin F \subset E\left( {G}^{\prime }\right) \). Then \( F \) contains no broken cycle of \( {G}^{\prime } \) iff \( F \) contains no broken cycle of \( G \). (2) Suppose \( {e}_{1} \in F \subset E\left( {G}^{\prime }\right) \). Then \( F \) contains no broken cycle of \( {G}^{\prime } \) if \( F - \left\{ {e}_{1}\right\} \subset E\left( {G}^{\prime \prime }\right) \) and \( F - \left\{ {e}_{1}\right\} \) contains no broken cycle of \( {G}^{\prime \prime } \).
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Yes
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Theorem 9 The chromatic number of a graph \( G \) drawn on a closed surface of Euler characteristic \( \chi \leq 1 \) is at most \[ h\left( \chi \right) = \lfloor \left( {7 + \sqrt{{49} - {24\chi }}}\right) /2\rfloor . \]
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Proof. Let \( k \) be the chromatic number of \( G \) . We may and shall assume that \( G \) is a minimal graph of chromatic number \( k \) ; otherwise, we may replace it by a subgraph. But then \( \delta \left( G\right) \geq k - 1 \), so all we need is that if, for \( h = h\left( \chi \right), G \) has \( n \geq h + 1 \) vertices then its minimal degree is at most \( h - 1 \) . Now, if \( n \geq h + 1 \) then \( e\left( G\right) \leq {3n} - {3\chi } \) implies that \[ \delta \left( G\right) \leq 6 - {6\chi }/\left( {h + 1}\right) \] Hence if we had \( \delta \left( G\right) \geq h \) then we would have \[ h \leq 6 - {6\chi }/\left( {h + 1}\right) \] that is, \[ {h}^{2} - {5h} + 6\left( {\chi - 1}\right) \leq 0 \] But this would imply the contradiction \[ h \leq \frac{1}{2}\left( {5 + \sqrt{{49} - {24\chi }}}\right) \]
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Yes
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Theorem 10 Let \( \chi \leq 0, h = h\left( \chi \right) = \lfloor \left( {7 + \sqrt{{49} - {24\chi }}}\right) /2\rfloor \), and let \( G \) be a minimal h-chromatic graph drawn on a surface of Euler characteristic \( \chi \) . If \( \chi \neq - 1, - 2 \) or -7 then \( G = {K}_{h} \) .
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Proof. All we shall use is inequality (8): a graph of order \( n \) drawn on a surface of Euler characteristic \( \chi \) has at most \( 3\left( {n - \chi }\right) \) edges.\n\nSuppose \( G \neq {K}_{h} \) . Then \( n \geq h + 2 \) . Furthermore, if \( n = h + 2 \) then, as claimed by Exercise 38,\n\n\[ e\left( G\right) = \left( \begin{matrix} h + 2 \\ 2 \end{matrix}\right) - 5 \]\n\nwhich is easily checked to be greater than \( 3\left( {h + 2 - \chi }\right) \) . Hence \( n \geq h + 3 \) . Our graph \( G \) is a minimal \( h \) -chromatic graph, so \( \delta \left( G\right) \geq h - 1 \geq 6 \) and, by Brooks’ theorem, \( G \) is not \( \left( {h - 1}\right) \) -regular. Therefore\n\n\[ e\left( G\right) > \frac{n\left( {h - 1}\right) }{2} \]\n\nand so\n\n\[ n\left( {h - 1}\right) + 1 \leq 6\left( {n - \chi }\right) . \]\n\n(9)\n\nSince \( h \geq 1 \), inequality (9) has to hold for \( n = h + 3 \), that is,\n\n\[ {h}^{2} - {4h} - {20} + {6\chi } \leq 0. \]\n\nThis implies that\n\n\[ h \leq 2 + \sqrt{{24} - {6\chi }} \]\n\n(10)\n\nSimple calculations show that (10) fails for \( \chi \leq - {20} \), and it is easily checked that for \( - {19} \leq \chi \leq 0 \) inequality (10) fails unless \( \chi = - 1, - 2 \) or -7 .
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Yes
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Theorem 11 The torus, the projective plane and the Klein bottle have chromatic numbers \( s\left( {S}_{1}\right) = 7, s\left( {N}_{1}\right) = 6 \) and \( s\left( {N}_{2}\right) = 6 \) .
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Proof. The Euler characteristics of these surfaces are \( \chi \left( {N}_{1}\right) = 1 \) and \( \chi \left( {S}_{1}\right) = \) \( \chi \left( {N}_{2}\right) = 0 \), therefore Theorem 9 implies that \( s\left( {N}_{1}\right) \leq 6 \) and \( s\left( {S}_{1}\right), s\left( {N}_{2}\right) \leq 7 \) . Fig. V. 6 shows that \( {K}_{6} \) triangulates \( {N}_{1} \) and \( {K}_{7} \) triangulates \( {S}_{1} \), so \( s\left( {N}_{1}\right) = 6 \) , \( s\left( {S}_{1}\right) = 7 \) and \( 6 \leq s\left( {N}_{2}\right) \leq 7. \) Our problem is then to decide whether the chromatic number of the Klein bottle is 6 or 7 . We know from Theorem 10 that \( s\left( {N}_{2}\right) = 7 \) iff \( {K}_{7} \) can be drawn on \( {N}_{2} \), and so \( {K}_{7} \) triangulates \( {N}_{2} \) . To complete the proof, we shall show that \( {K}_{7} \) triangulates a unique closed surface, the torus, so that \( s\left( {N}_{2}\right) = 6 \) . Suppose then that we have a triangulation by \( {K}_{7} \) of a closed surface (of Euler characteristic 0 ). Then every vertex of \( {K}_{7} \) is on the boundary of six triangular faces, and the third sides of these triangles form a 6-cycle. Writing \( 0,1,\ldots ,6 \) for the vertices, we may assume that the 6-cycle 'surrounding' 0 is 123456 . Then vertex 1 is surrounded by \( {602x} \cdot y \), vertex 2 by \( {301x}\cdots \), and so on (see Fig. V.7). But then \( x \) has to be 4 or 5 : by symmetry, we may assume that it is 4 . Having made this choice, everything else is determined: looking at the neighbourhoods of 1 and 6, namely the cycles \( {y6024} \) . and \( {501y}\cdots \), we see that \( y = 3 \), then we get \( z = 2, u = 1 \), and so on, as shown in Fig. V.7. What we have proved is that if \( {K}_{7} \) triangulates a surface then this triangulation is unique (up to reflection) and is as in Fig. V.7. But this labelling is easily seen to be consistent and to give a triangulation of the torus. (As it happens, we already know that \( {K}_{7} \) triangulates the torus, but in this proof we were forced to find that triangulation.) In particular, \( {K}_{7} \) cannot be drawn on the Klein bottle, so \( s\left( {N}_{2}\right) = 6 \), and we are done.
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Yes
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Theorem 12 Let \( G \) be a near-triangulation with outer cycle \( C = {x}_{1}{x}_{2}\cdots {x}_{k} \) , and for each \( x \in V\left( G\right) \) let \( L\left( x\right) \) be a list of colours assigned to \( x \), such that \( L\left( {x}_{1}\right) = \{ 1\}, L\left( {x}_{2}\right) = \{ 2\} ,\left| {L\left( x\right) }\right| \geq 3 \) for \( 3 \leq i \leq k \), and \( \left| {L\left( x\right) }\right| \geq 5 \) for \( x \in V\left( {G - C}\right) \) . Then \( G \) has an \( L \) -colouring.
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Proof. Let us apply induction on the order of \( G \) . For \( \left| G\right| = 3 \) the assertion is trivial, so suppose that \( \left| G\right| > 3 \) and the assertion holds for graphs of order less than \( \left| G\right| \) . We shall distinguish two cases, according to whether \( C \) contains a ’diagonal’ from \( {x}_{k} \) or not.\n\n(i) First suppose that \( G \) contains a ’diagonal’ \( {x}_{k}{x}_{j},2 \leq j \leq k - 2 \), of \( C \) . Then we can apply the induction hypothesis to the graph formed by the cycle \( {x}_{k}{x}_{1}{x}_{2}\cdots {x}_{j} \) and its interior and then, having fixed the colours of \( {x}_{k} \) and \( {x}_{j} \), to the cycle \( {x}_{k}{x}_{j}{x}_{j + 1}\cdots {x}_{k - 1} \) and its interior, to find an \( L \) -colouring of \( G \) .\n\n(ii) Now suppose that \( G \) contains none of the edges \( {x}_{k}{x}_{j},2 \leq j \leq k - 2 \) . Let the neighbours of \( {x}_{k} \) be \( {x}_{k - 1},{y}_{1},{y}_{2},\ldots ,{y}_{\ell } \) and \( {x}_{1} \), in this order, so that \( {x}_{k}{x}_{k - 1}{y}_{1} \) , \( {x}_{k}{y}_{1}{y}_{2},\ldots ,{x}_{k}{y}_{\ell }{x}_{1} \) are internal faces of our plane graph (see Fig. V.10).\n\nLet \( a \) and \( b \) be colours in \( L\left( {x}_{k}\right) \), distinct from 1 . Our aim is to use one of \( a \) and \( b \) to colour \( {x}_{k} \), having coloured the rest of the graph. To this end, let \( {L}^{\prime }\left( x\right) = L\left( x\right) \) if \( x \notin \left\{ {{y}_{1},\ldots ,{y}_{\ell }}\right\} \) and \( {L}^{\prime }\left( {y}_{i}\right) = L\left( {y}_{i}\right) - \{ a, b\} \) for \( 1 \leq i \leq l \) . Then, by the induction hypothesis, the graph \( {G}^{\prime } = G - {x}_{k} \), with outer cycle \( {x}_{1}{x}_{2}\cdots {x}_{k - 1}{y}_{1}{y}_{2}\cdots {y}_{\ell } \), has an \( {L}^{\prime } \) -colouring. Extend this \( {L}^{\prime } \) -colouring of \( {G}^{\prime } \) to an \( L \) -colouring of \( G \) by assigning \( a \) or \( b \) to \( {x}_{k} \) such that \( {x}_{k} \) and \( {x}_{k - 1} \) get distinct colours.
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Yes
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Theorem 13 Let \( G \) be a bipartite graph with total function \( {t}_{G} \) given by a certain assignment of preferences. Then \( G \) is \( \left( {{t}_{G} + 1}\right) \) -choosable.
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Proof. We apply induction on the size of \( G \) . If \( E\left( G\right) = \varnothing \), there is nothing to prove, so suppose \( E\left( G\right) \neq \varnothing \) and the assertion holds for graphs of smaller size.\n\nLet us fix an assignment of preferences for \( G \) . For each edge \( e \in E\left( G\right) \), let \( L\left( e\right) \) be a set of \( {t}_{G}\left( e\right) + 1 \) natural numbers. We have to show that the edges of \( G \) have an \( L \) -colouring.\n\nLet \( I \neq \varnothing \) be the set of edges whose lists contain a certain colour \( i \), and let \( H = \left( {V\left( G\right), I}\right) \) be the subgraph of \( G \) with edge-set \( I \) . By Theorem III. 15, the graph \( H \) contains a stable matching \( M \) . Let \( {G}^{\prime } = G - M \), and for \( e \in E\left( {G}^{\prime }\right) \) set \( {L}^{\prime }\left( e\right) = L\left( e\right) - \{ i\} \) . We claim that\n\n\[ \left| {{L}^{\prime }\left( e\right) }\right| \geq {t}_{{G}^{\prime }}\left( e\right) + 1 \]\n\n(13)\n\nfor every \( e \in E\left( {G}^{\prime }\right) \) . Indeed, if \( e \notin I \) then \( {L}^{\prime }\left( e\right) = L\left( e\right) \) so this is clearly the case. Also, if \( e \in I - M = E\left( H\right) - M \) then, by relations (11) and (12),\n\n\[ {t}_{G}\left( e\right) - {t}_{{G}^{\prime }}\left( e\right) = {t}_{H}\left( e\right) - {t}_{{H}^{\prime }}\left( e\right) \geq 1, \]\n\nso\n\n\[ \left| {{L}^{\prime }\left( e\right) }\right| = \left| {L\left( e\right) }\right| - 1 \geq {t}_{G}\left( e\right) \geq {t}_{{G}^{\prime }}\left( e\right) + 1, \]\n\nproving (13).\n\nBy the induction hypothesis, \( {G}^{\prime } \) has an \( {L}^{\prime } \) -colouring; colouring the edges of \( M \) by \( i \), we get an \( L \) -colouring of the edges of \( G \) .
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Yes
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Theorem 14 The list-chromatic index of a bipartite graph equals its chromatic index.
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Proof. Let \( G \) be a bipartite graph with bipartition \( \left( {{V}_{1},{V}_{2}}\right) \), and let \( \lambda : E\left( G\right) \rightarrow \) \( \left\lbrack k\right\rbrack \) be an edge-colouring of \( G \), where \( k \) is the chromatic index of \( G \) . Define preferences on \( G \) as follows: let \( a \in {V}_{1} \) prefer a neighbour \( A \) to a neighbour \( B \) iff \( \lambda \left( {aA}\right) > \lambda \left( {aB}\right) \), and let \( A \in {V}_{2} \) prefer a neighbour \( a \) to a neighbour \( b \) iff \( \lambda \left( {aA}\right) < \lambda \left( {bA}\right) \) . Note that the total function defined by this assignment of preferences is at most \( k - 1 \) on every edge, since if \( \lambda \left( {aA}\right) = j \) then \( a \) prefers at most \( k - j \) of its neighbours to \( A \), and \( A \) prefers at most \( j - 1 \) of its neighbours to \( a \) . Hence, by Theorem \( {11}, G \) is \( k \) -choosable.
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Yes
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Theorem 16 Let \( G \) be a bipartite graph with line graph \( H = L\left( G\right) \) . Then \( H \) and \( \overline{H} \) are perfect.
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Proof. Once again, all we have to prove is that \( \chi \left( H\right) = \omega \left( H\right) \) and \( \chi \left( \bar{H}\right) = \omega \left( \bar{H}\right) \) . Clearly, \( \omega \left( H\right) = \Delta \left( G\right) \) and \( \chi \left( H\right) = {\chi }^{\prime }\left( G\right) \) . But as \( G \) is bipartite, \( {\chi }^{\prime }\left( G\right) = \) \( \Delta \left( G\right) \) (see the beginning of Section 2), so \( \chi \left( H\right) = \Delta \left( G\right) = \omega \left( H\right) \) .\n\nAnd what is \( \chi \left( \bar{H}\right) \) ? The minimal number of vertices of \( G \) covering all the edges. Finally, what is \( \omega \left( \bar{H}\right) \) ? The maximal number of independent edges of \( G \) . By Corollary III. 10, these two quantities are equal.
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Yes
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Theorem 17 Comparability graphs and their complements are perfect.
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Proof. Once again, it suffices to show that if \( P \) is a partially ordered set then for \( H = C\left( P\right) \) we have \( \chi \left( H\right) = \omega \left( H\right) \) and \( \chi \left( \bar{H}\right) = \omega \left( \bar{H}\right) \) .\n\nTo see the first equality, for \( x \in P \) let \( r\left( x\right) \), the rank of \( x \), be the maximal integer \( r \) for which \( P \) contains a chain of \( r \) elements, with maximal element \( x \) . Then for \( k = \mathop{\max }\limits_{r}r\left( x\right) \) the map \( r : P \rightarrow \left\lbrack k\right\rbrack \) gives a \( k \) -colouring of \( H \), and a chain of size \( k \) gives a \( k \) -clique.\n\nThe second equality is deeper. Indeed, \( \chi \left( \bar{H}\right) \) is the minimal number of chains into which \( P \) can be partitioned, and \( \omega \left( \bar{H}\right) \) is precisely the maximal number of elements in an antichain. Therefore the equality \( \chi \left( \bar{H}\right) = \omega \left( \bar{H}\right) \) is none other than Dilworth's theorem, Theorem III.12.
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Yes
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A necessary and sufficient condition for a graph \( G \) to be perfect is that for every induced subgraph \( H \subset G \) there is an independent set of vertices, \( I \) , such that\n\n\[ \omega \left( {H - I}\right) < \omega \left( H\right) \]\n\nThat is, a graph is perfect iff every induced subgraph \( H \) has an independent set meeting every clique of \( H \) of maximal order \( \omega \left( H\right) \) .
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Proof. The necessity holds with plenty to spare. Indeed, let \( H \) be a graph with \( k = \chi \left( H\right) = \omega \left( H\right) \), and let \( I \) be a colour class of a \( k \) -colouring of \( H \) . Then \( \omega \left( {H - I}\right) \leq \chi \left( {H - I}\right) = \chi \left( H\right) - 1 < \omega \left( H\right) . \)\n\nThe sufficiency of the condition will be proved by induction on \( \omega \left( G\right) \) . For \( \omega \left( G\right) = 1 \) there is nothing to prove, so suppose that \( \omega \left( G\right) > 1 \) and the assertion holds for smaller values of the clique number. Let \( H \) be an induced subgraph of \( G \) and \( I \) an independent set with \( \omega \left( {H - I}\right) < \omega \left( H\right) \) . By the induction hypothesis, we can colour \( H - I \) with \( \omega \left( {H - I}\right) \) colours; colouring the vertices of \( I \) with a new colour, we obtain a colouring of \( H \) with \( \omega \left( {H - I}\right) + 1 \leq \omega \left( H\right) \) colours. Thus \( \chi \left( H\right) \leq \omega \left( H\right) \), and we are done.
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Yes
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Theorem 19 A graph obtained from a perfect graph by replacing its vertices by perfect graphs is perfect.
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Proof. As we may replace the vertices one by one, it suffices to prove that if a vertex \( x \) of a perfect graph \( G \) is replaced by a perfect graph \( {G}_{x} \) then the resulting graph \( {G}^{ * } \) is perfect. Furthermore, since every induced subgraph of \( {G}^{ * } \) is of precisely the same form (obtained from a perfect graph by replacing one of its vertices by a perfect graph), by Lemma 18 it suffices to show that \( {G}^{ * } \) itself contains an independent set of vertices meeting every clique of \( {G}^{ * } \) with \( \omega \left( {G}^{ * }\right) \) vertices.\n\nHaving identified our task, let us get on with the job. By Lemma 18, the graph \( {G}_{x} \) has an independent set \( I \) such that \( \omega \left( {{G}_{x} - I}\right) < \omega \left( {G}_{x}\right) \) . Colour \( G \) with \( \omega \left( G\right) \) colours, and let \( {W}_{x} \) be the colour class containing \( x \) . Then \( J = I \cup \left( {{W}_{x} - x}\right) \) is an independent set in \( {G}^{ * } \) . We claim this set \( J \) will do for the independent set. Let \( K \) be a clique of \( {G}^{ * } \) with \( \omega \left( {G}^{ * }\right) \) vertices, and let us show that \( J \) meets \( K \) .\n\nNote that either \( K \) is a clique in \( G - x \), or it is the union of a clique of \( {G}_{x} \) of order \( \omega \left( {G}_{x}\right) \) and a clique of \( G\left\lbrack {\Gamma \left( x\right) }\right\rbrack \) . Now, if \( K \) is a clique in \( G - x \) then, as it has \( \omega \left( {G * }\right) \geq \omega \left( G\right) \) vertices, it meets every colour class of \( G \) in our \( \omega \left( G\right) \) -colouring, including \( {W}_{x} \), so \( K \cap J = K \cap {W}_{x} \neq \varnothing \) . On the other hand, if \( K \) meets \( {G}_{x} \) then \( K \) meets \( I \), as the part of \( K \) in \( {G}_{x} \) is an \( \omega \left( {G}_{x}\right) \) -clique of \( {G}_{x} \) . Hence \( J \) does meet \( K \) as claimed.
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Yes
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Theorem 1 The function \( R\left( {s, t}\right) \) is finite for all \( s, t \geq 2 \) . If \( s > 2 \) and \( t > 2 \) then\n\n\[ R\left( {s, t}\right) \leq R\left( {s - 1, t}\right) + R\left( {s, t - 1}\right) \]\n\n(1)\n\nand\n\n\[ R\left( {s, t}\right) \leq \left( \begin{matrix} s + t - 2 \\ s - 1 \end{matrix}\right) \]\n\n(2)
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Proof. As we shall prove (1) and (2), it will follow that \( R\left( {s, t}\right) \) is finite.\n\n(i) When proving (1) we may assume that \( R\left( {s - 1, t}\right) \) and \( R\left( {s, t - 1}\right) \) are finite. Let \( n = R\left( {s - 1, t}\right) + R\left( {s, t - 1}\right) \) and consider a colouring of the edges of \( {K}_{n} \) with red and blue. We have to show that in this colouring there is either a red \( {K}_{s} \) or a blue \( {K}_{t} \) . To this end, let \( x \) be a vertex of \( {K}_{n} \) . Since \( d\left( x\right) = n - 1 = \) \( R\left( {s - 1, t}\right) + R\left( {s, t - 1}\right) - 1 \), either there are at least \( {n}_{1} = R\left( {s - 1, t}\right) \) red edges incident with \( x \) or there are at least \( {n}_{2} = R\left( {s, t - 1}\right) \) blue edges incident with \( x \) . By symmetry we may assume that the first case holds. Consider a subgraph \( {K}_{{n}_{1}} \) of \( {K}_{n} \) spanned by \( {n}_{1} \) vertices joined to \( x \) by red edges. If \( {K}_{{n}_{1}} \) has a blue \( {K}_{t} \), we are done. Otherwise, by the definition of \( R\left( {s - 1, t}\right) \), the graph \( {K}_{{n}_{1}} \) contains a red \( {K}_{s - 1} \) which forms a red \( {K}_{s} \) with \( x \) .\n\n(ii) Inequality (2) holds if \( s = 2 \) or \( t = 2 \) (in fact, we have equality since \( R\left( {s,2}\right) = R\left( {2, s}\right) = s) \) . Assume now that \( s > 2, t > 2 \) and (2) holds for every pair \( \left( {{s}^{\prime },{t}^{\prime }}\right) \) with \( 2 \leq {s}^{\prime } + {t}^{\prime } < s + t \) . Then by (1) we have\n\n\[ R\left( {s, t}\right) \leq R\left( {s - 1, t}\right) + R\left( {s, t - 1}\right) \]\n\n\[ \leq \left( \begin{matrix} s + t - 3 \\ s - 2 \end{matrix}\right) + \left( \begin{matrix} s + t - 3 \\ s - 1 \end{matrix}\right) = \left( \begin{matrix} s + t - 2 \\ s - 1 \end{matrix}\right) . \]
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Yes
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Theorem 2 Let \( 1 < r < \min \{ s, t\} \) . Then \( {R}^{\left( r\right) }\left( {s, t}\right) \) is finite and\n\n\[ \n{R}^{\left( r\right) }\left( {s, t}\right) \leq {R}^{\left( r - 1\right) }\left( {{R}^{\left( r\right) }\left( {s - 1, t}\right) ,{R}^{\left( r\right) }\left( {s, t - 1}\right) }\right) + 1.\n\]
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Proof. Both assertions follow immediately if we prove the inequality under the assumption that \( {R}^{\left( r - 1\right) }\left( {u, v}\right) \) is finite for all \( u, v \), and both \( {R}^{\left( r\right) }\left( {s - 1, t}\right) \) and \( {R}^{\left( r\right) }\left( {s, t - 1}\right) \) are also finite.\n\nLet \( X \) be a set with \( {R}^{\left( r - 1\right) }\left( {{R}^{\left( r\right) }\left( {s - 1, t}\right) ,{R}^{\left( r\right) }\left( {s, t - 1}\right) }\right) + 1 \) elements. Given any red-blue colouring \( c \) of \( {X}^{\left( r\right) } \), pick an \( x \in X \) and define a red-blue colouring \( {c}^{\prime } \) of the \( \left( {r - 1}\right) \) -sets of \( Y = X - \{ x\} \) by colouring \( \sigma \in {Y}^{\left( r - 1\right) } \) the colour of \( \sigma \cup \{ x\} \in {X}^{\left( r\right) } \) . By the definition of the function \( {R}^{\left( r - 1\right) }\left( {u, v}\right) \) we may assume that \( Y \) has a red subset \( Z \) (for \( {c}^{\prime } \) ) with \( {R}^{\left( r\right) }\left( {s - 1, t}\right) \) elements.\n\nNow let us look at the restriction of \( c \) to \( {Z}^{\left( r\right) } \) . If it has a blue \( t \) -set, we are done, since \( {Z}^{\left( r\right) } \subset {X}^{\left( r\right) } \), so a blue \( t \) -set of \( Z \) is certainly also a blue \( t \) -set of \( X \) . On the other hand, if there is no blue \( t \) -set of \( Z \) then there is a red \( \left( {s - 1}\right) \) -set. The union of this red \( \left( {s - 1}\right) \) -set with \( \{ x\} \) is then a red \( s \) -set of \( X \), because \( \{ x\} \cup \sigma \) is red for every \( \sigma \in {Z}^{\left( r - 1\right) } \) .
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Yes
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Theorem 3 For \( k,\ell \geq 2 \), every non-degenerate set of \( \left( \begin{matrix} k + \ell - 4 \\ k - 2 \end{matrix}\right) + 1 \) points contains a \( k \) -cup or an \( \ell \) -cap.
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Proof. Let us write \( \phi \left( {k,\ell }\right) \) for the binomial coefficient \( \left( \begin{matrix} k + \ell - 4 \\ k - 2 \end{matrix}\right) \). (i) We shall prove by induction on \( k + \ell \) that every non-degenerate set of \( \phi \left( {k,\ell }\right) + 1 \) points contains a \( k \) -cup or an \( \ell \) -cap. Since a non-degenerate set of 2 points is both a 2-cup and a 2-cap, this is clear if \( \min \{ k,\ell \} = 2 \), since \( \phi \left( {k,2}\right) = \phi \left( {2,\ell }\right) = 1 \) for all \( k,\ell \geq 2 \). Suppose then that \( k,\ell \geq 3 \) and the assertion holds for smaller values of \( k + \ell \). Let \( S \) be a non-degenerate set of \( \phi \left( {k,\ell }\right) + 1 \) points and suppose that, contrary to the assertion, \( S \) contains neither a \( k \) -cup nor an \( \ell \) -cap. Let \( L \subset S \) be the set of last points of \( \left( {k - 1}\right) \) -cups. Then \( S \smallsetminus L \) has neither a \( \left( {k - 1}\right) \) - cup nor an \( \ell \) -cap so, by the induction hypothesis, \( \left| {S \smallsetminus L}\right| \leq \phi \left( {k - 1,\ell }\right) \). Therefore \( \left| L\right| \geq \phi \left( {k,\ell }\right) + 1 - \phi \left( {k - 1,\ell }\right) = \phi \left( {k,\ell - 1}\right) + 1 \) so, again by the induction hypothesis, \( L \) contains an \( \left( {\ell - 1}\right) \) -cap, say \( \left\{ {{q}_{1},\ldots ,{q}_{\ell - 1}}\right\} \), with first point our set \( S \) contains \( {q}_{1} \). Since \( {q}_{1} \in L,\mathrm{a}\left( {k - 1}\right) \) -cup \( \left\{ {{p}_{1},\ldots ,{p}_{k - 1}}\right\} \), whose last point, \( {p}_{k - 1} \) , is precisely \( {q}_{1} \). Now, if \( s\left( {{p}_{k - 2},{p}_{k - 1}}\right) \leq s\left( {{p}_{k - 1},{q}_{2}}\right) \) then \( \left\{ {{p}_{1},\ldots ,{p}_{k - 1},{q}_{2}}\right\} \) is a \( k \) -cup. Otherwise, \( s\left( {{p}_{k - 2},{q}_{1}}\right) > s\left( {{q}_{1},{q}_{2}}\right) \), so \( \left\{ {{p}_{k - 2},{q}_{1},\ldots ,{q}_{\ell - 1}}\right\} \) is an \( \ell \) -cup. This contradiction completes the proof of the induction step, and we are done.
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Yes
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Theorem 4 Let \( 1 \leq r < \infty \) and let \( c : {A}^{\left( r\right) } \rightarrow \left\lbrack k\right\rbrack = \{ 1,2,\ldots, k\} \) be a \( k \) - colouring of the r-tuples of an infinite set \( A \) . Then \( A \) contains a monochromatic infinite set.
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Proof. We apply induction on \( r \) . Note that the result is trivial for \( r = 1 \), so we may assume that \( r > 1 \) and the theorem holds for smaller values of \( r \) .\n\nPut \( {A}_{0} = A \) and pick an element \( {x}_{1} \in {A}_{0} \) . As in the proof of Theorem 2, define a a colouring \( {c}_{1} : {B}_{1}^{\left( r - 1\right) } \rightarrow \left\lbrack k\right\rbrack \) of the \( \left( {r - 1}\right) \) -tuples of \( {B}_{1} = {A}_{0} - \left\{ {x}_{1}\right\} \) by putting \( {c}_{1}\left( \tau \right) = c\left( {\tau \cup \left\{ {x}_{1}\right\} }\right) ,\tau \in {B}_{1}^{\left( r - 1\right) } \) . By the induction hypothesis \( {B}_{1} \) contains an infinite set \( {A}_{1} \) all of whose \( \left( {r - 1}\right) \) -tuples have the same colour, say \( {d}_{1} \), where \( {d}_{1} \in \{ 1,\ldots, k\} \) . Let now \( {x}_{2} \in {A}_{1},{B}_{2} = {A}_{1} - \left\{ {x}_{2}\right\} \) and define a \( k \) -colouring \( {c}_{2} : {B}_{2}^{\left( r - 1\right) } \rightarrow \left\lbrack k\right\rbrack \) by putting \( {c}_{2}\left( \tau \right) = c\left( {\tau \cup \left\{ {x}_{2}\right\} }\right) ,\tau \in {B}_{2}^{\left( r - 1\right) } \) . Then \( {B}_{2} \) has an infinite set \( {A}_{2} \) all of whose \( \left( {r - 1}\right) \) -tuples have the same colour, say \( {d}_{2} \) . Continuing in this way we obtain an infinite sequence of elements: \( {x}_{1},{x}_{2},\ldots \) , an infinite sequence of colours: \( {d}_{1},{d}_{2},\ldots \), and an infinite nested sequence of sets: \( {A}_{0} \supset {A}_{1} \supset {A}_{2} \supset \cdots \), such that \( {x}_{i} \in {A}_{i - 1} \), and for \( i = 0,1,\ldots \), all \( r \) - tuples whose only element outside \( {A}_{i} \) is \( {x}_{i} \) have the same colour \( {d}_{i} \) . The infinite sequence \( {\left( {d}_{n}\right) }_{1}^{\infty } \) must take at least one of the \( k \) values \( 1,2,\ldots, k \) infinitely often, say \( d = {d}_{{n}_{1}} = {d}_{{n}_{2}} = \ldots \) . Then, by the construction, each \( r \) -tuple of the infinite set \( \left\{ {{x}_{{n}_{1}},{x}_{{n}_{2}},\ldots }\right\} \) has colour \( d \) .
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Yes
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Theorem 5 For each \( r \in \mathbb{N} \), colour the set \( {\mathbb{N}}^{\left( r\right) } \) of \( r \) -tuples of \( \mathbb{N} \) with \( {k}_{r} \) colours, where \( {k}_{r} \in \mathbb{N} \) . Then there is an infinite set \( M \subset \mathbb{N} \) such that for every \( r \) any two \( r \) -tuples of \( M \) have the same colour, provided their minimal elements are not less than the \( {r}^{\text{th }} \) element of \( M \) .
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Proof. Put \( {M}_{0} = \mathbb{N} \) . Having chosen infinite sets \( {M}_{0} \supset \cdots \supset {M}_{r - 1} \), let \( {M}_{r} \) be an infinite subset of \( {M}_{r - 1} \) such that all the \( r \) -tuples of \( {M}_{r} \) have the same colour. This way we obtain an infinite nested sequence of infinite sets: \( {M}_{0} \supset {M}_{1} \supset \cdots \) . Pick \( {a}_{1} \in {M}_{1},{a}_{2} \in {M}_{2} - \left\{ {1,\ldots ,{a}_{1}}\right\} ,{a}_{3} \in {M}_{3} - \left\{ {1,\ldots ,{a}_{2}}\right\} \), etc. Clearly, \( M = \left\{ {{a}_{1},{a}_{2},\ldots }\right\} \) has the required properties.
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Yes
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Theorem 6 Let \( r \) and \( k \) be natural numbers, and for every \( n \geq 1 \), let \( {\mathcal{C}}_{n} \) be a non-empty set of \( k \) -colourings of \( {\left\lbrack n\right\rbrack }^{\left( r\right) } \) such that if \( n < m \) and \( {c}_{m} \in {\mathcal{C}}_{m} \) then the restriction \( {c}_{m}^{\left( n\right) } \) of \( {c}_{m} \) to \( {\left\lbrack n\right\rbrack }^{\left( r\right) } \) belongs to \( {\mathcal{C}}_{n} \) . Then there is a colouring \( c : {\mathbb{N}}^{\left( r\right) } \rightarrow \left\lbrack k\right\rbrack \) such that, for every \( n \), the restriction \( {c}^{\left( n\right) } \) of \( c \) to \( {\left\lbrack n\right\rbrack }^{\left( r\right) } \) belongs to \( {\mathcal{C}}_{n} \) .
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Proof. For \( m > n \), write \( {\mathcal{C}}_{n, m} \) for the set of colourings \( {\left\lbrack n\right\rbrack }^{\left( r\right) } \rightarrow \left\lbrack k\right\rbrack \) that are restrictions of colourings in \( {\mathcal{C}}_{m} \) . Then \( {\mathcal{C}}_{n, m + 1} \subset {\mathcal{C}}_{n, m} \subset {\mathcal{C}}_{n} \) and so \( {\widetilde{\mathcal{C}}}_{n} = \mathop{\bigcap }\limits_{{m = n + 1}}^{\infty }{\mathcal{C}}_{n, m} \neq \varnothing \) for every \( n \), since each \( {\mathcal{C}}_{n, m} \) is finite. Let \( {c}_{r} \in {\widetilde{\mathcal{C}}}_{r} \) , and pick \( {c}_{r + 1} \in {\widetilde{\mathcal{C}}}_{r + 1},{c}_{r + 2} \in {\widetilde{\mathcal{C}}}_{r + 2} \), and so on, such that each is in the preimage of the previous one: \( {c}_{n} = {c}_{n + 1}^{\left( n\right) } \) . Finally, define \( c : {\mathbb{N}}^{\left( r\right) } \rightarrow \left\lbrack k\right\rbrack \) by setting, for \( \rho \in {\mathbb{N}}^{\left( r\right) }, \n\n\[ \nc\left( \rho \right) = {c}_{n}\left( \rho \right) = {c}_{n + 1}\left( \rho \right) = \cdots , \n\n\] \n\nwhere \( n = \max \rho \) . This colouring \( c \) is as required.
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Yes
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Theorem 7 Let \( r, k \) and \( s \geq 2 \) . If \( n \) is sufficiently large then for every \( k \) -colouring of \( {\left\lbrack n\right\rbrack }^{\left( r\right) } \) there is a monochromatic set \( S \subset \left\lbrack n\right\rbrack \) such that\n\n\[ \left| S\right| \geq \max \{ s,\min S\} \text{.} \]
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Proof. Suppose that there is no such \( n \), that is, for every \( n \) there is a colouring \( {\left\lbrack n\right\rbrack }^{\left( r\right) } \rightarrow \left\lbrack k\right\rbrack \) without an appropriate monochromatic set. Let \( {\mathcal{C}}_{n} \) be the set of all such colourings. Then \( {\mathcal{C}}_{n} \neq \varnothing \) and, in the earlier notation, \( {\mathcal{C}}_{n, m} \subset {\mathcal{C}}_{n} \) for all \( n < m \) . But then there is a colouring \( c : {\mathbb{N}}^{\left( r\right) } \rightarrow \left\lbrack k\right\rbrack \) such that its restriction \( {c}^{\left( n\right) } \) to \( {\left\lbrack n\right\rbrack }^{\left( r\right) } \) belongs to \( {\mathcal{C}}_{n} \) . Now, by Theorem 4, there is an infinite monochromatic set \( M \subset \mathbb{N} \) . Set \( m = \min M, t = \max \{ m, s\} \), and let \( S \) consist of the first \( t \) elements of \( M \) . Then, with \( n = \max S \), the colouring \( {c}^{\left( n\right) } \) does have an appropriate monochromatic set, namely \( S \), contradicting \( {c}^{\left( n\right) } \in {\mathcal{C}}_{n} \) .
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Yes
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Theorem 10 For \( \ell \geq 1 \) and \( p \geq 2 \) we have\n\n\[ r\left( {\ell {K}_{2},{K}_{p}}\right) = 2\ell + p - 2. \]
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Proof. The graph \( {K}_{2\ell - 1} \cup {E}_{p - 2} \) does not contain \( \ell \) independent edges, and its complement, \( {E}_{2\ell - 1} + {K}_{p - 2} \), does not contain a complete graph of order \( p \) . Hence \( r\left( {\ell {K}_{2},{K}_{p}}\right) \geq 2\ell + p - 2 \) .\n\nOn the other hand, let \( G \) be a graph of order \( n = 2\ell + p - 2 \), containing a maximal set of \( s \leq \ell - 1 \) independent edges. Then the set of \( n - {2s} \geq 2\ell + p - 2 - 2\left( {\ell - 1}\right) = \) \( p \) vertices not on these edges spans a complete graph of order at least \( p \) . Therefore \( r\left( {\ell {K}_{2},{K}_{p}}\right) \leq 2\ell + p - 2. \)
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Yes
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Theorem 11 For all nonempty graphs \( {H}_{1} \) and \( {H}_{2} \) we have\n\n\[ r\left( {{H}_{1},{H}_{2}}\right) \geq \left( {\chi \left( {H}_{1}\right) - 1}\right) \left( {c\left( {H}_{2}\right) - 1}\right) + u\left( {H}_{1}\right) . \]\n\nIn particular, if \( {\mathrm{H}}_{2} \) is connected then\n\n\[ r\left( {{H}_{1},{H}_{2}}\right) \geq \left( {\chi \left( {H}_{1}\right) - 1}\right) \left( {\left| {H}_{2}\right| - 1}\right) + 1. \]
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Proof. Set \( k = \chi \left( {H}_{1}\right), u = u\left( {H}_{1}\right) \) and \( c = c\left( {H}_{2}\right) \) . Trivially, \( r\left( {{H}_{1},{H}_{2}}\right) \geq \) \( \left. {r\left( {{H}_{1},{K}_{2}}\right) = \left| {H}_{1}\right| \geq \chi \left( {H}_{1}\right) u\left( {H}_{1}\right) = {ku}\text{. Hence, if }c \leq u\text{ then }r\left( {{H}_{1},{H}_{2}}\right) \geq {ku} \geq }\right) \) \( \left( {k - 1}\right) c + u \) . On the other hand, if \( c > u \) then the graph \( G = \left( {k - 1}\right) {K}_{c - 1} \cup \) \( {K}_{u - 1} \) does not contain \( {H}_{2} \), and its complement does not contain \( {H}_{1} \) . Therefore, \( r\left( {{H}_{1},{H}_{2}}\right) \geq \left| G\right| + 1 = \left( {k - 1}\right) \left( {c - 1}\right) + u. \)
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Yes
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Theorem 12 Let \( s, t \geq 2 \) . then for every tree \( T \) of order \( t \) we have \( r\left( {{K}_{s}, T}\right) = \) \( \left( {s - 1}\right) \left( {t - 1}\right) + 1 \) .
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Proof. From Theorem 10 we know that \( r\left( {{K}_{s}, T}\right) \geq \left( {s - 1}\right) \left( {t - 1}\right) + 1 \) . To prove the reverse inequality, let \( G \) be a graph of order \( n = \left( {s - 1}\right) \left( {t - 1}\right) + 1 \) whose complement does not contain \( {K}_{s} \) . Then \( \chi \left( G\right) \geq \lceil n/\left( {s - 1}\right) \rceil = t \) so it contains a critical subgraph \( H \) of minimal degree at least \( t - 1 \) (see Theorem V.1). It is easily seen that \( H \) contains (a copy of) \( T \) . Indeed, we may assume that \( {T}_{1} \subset H \) , where \( {T}_{1} = T - x \) and \( x \) is an endvertex of \( T \), adjacent to a vertex \( y \) of \( {T}_{1} \) (and of \( H \) ). Since \( y \) has at least \( t - 1 \) neighbours in \( H \), at least one of its neighbours, say \( z \), does not belong to \( {T}_{1} \) . Then the subgraph of \( H \) spanned by \( {T}_{1} \) and \( z \) clearly contains (a copy of) \( T \) .
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Yes
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Theorem 13 For \( \ell \geq 2 \) we have \( r\left( {{F}_{1},{F}_{\ell }}\right) = r\left( {{K}_{3},{F}_{\ell }}\right) = 4\ell + 1 \) .
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Proof. We know from Theorem 11 that \( r\left( {{K}_{3},{F}_{\ell }}\right) \geq 2\left( {\left| {F}_{\ell }\right| - 1}\right) + 1 = 4\ell + 1 \) . To prove the reverse inequality, suppose that the inequality is false; that is, there is a triangle-free graph \( G \) of order \( n = 4\ell + 1 \) whose complement does not contain \( {F}_{\ell } \) . For \( x \in G \), let \( U = {\Gamma }_{G}\left( x\right) \) . Then \( U \) is a set of independent vertices and since \( \bar{G} \) does not contains \( {F}_{\ell } \), we see that \( {d}_{G}\left( x\right) = \left| U\right| \leq 2\ell \) . On the other hand, how large can the degree of \( x \) be in \( \bar{G} \) ? Set \( W = {\Gamma }_{\bar{G}}\left( x\right) = \) \( V\left( G\right) - \left( {U\cup \{ x\} }\right) \) . Then \( \bar{G}\left\lbrack W\right\rbrack \) does not contain \( \ell \) independent edges, and its complement, \( G\left\lbrack W\right\rbrack \), has no triangle. Hence, by Theorem \( {10},{d}_{\bar{G}}\left( x\right) = \left| W\right| \leq 2\ell \) . This shows that \( {d}_{G}\left( x\right) = {d}_{\bar{G}}\left( x\right) = 2\ell \), that is, \( G \) is a triangle-free \( 2\ell \) -regular graph of order \( 4\ell + 1 \) . But from the result in Exercise IV. 48 we know that this is impossible.
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Yes
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Lemma 14 For all graphs \( G,{H}_{1} \) and \( {H}_{2} \) we have \( r\left( {G,{H}_{1} \cup {H}_{2}}\right) \leq \) \( \max \left\{ {r\left( {G,{H}_{1}}\right) + \left| {H}_{2}\right|, r\left( {G,{H}_{2}}\right) }\right\} \) . In particular, \( r\left( {s{H}_{1},{H}_{2}}\right) \leq r\left( {{H}_{1},{H}_{2}}\right) + \) \( \left( {s - 1}\right) \left| {H}_{1}\right| \) .
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Proof. Let \( n = \max \left\{ {r\left( {G,{H}_{1}}\right) + \left| {H}_{2}\right|, r\left( {G,{H}_{2}}\right) }\right\} \), and suppose that we are given a red-blue colouring of \( {K}_{n} \) without a red \( G \) . Then \( n \geq r\left( {G,{H}_{2}}\right) \) implies that there is a blue \( {H}_{2} \) . Remove it. Since \( n - \left| {H}_{2}\right| \geq r\left( {G,{H}_{1}}\right) \), the remainder contains a blue \( {H}_{1} \) . Hence \( {K}_{n} \) contains a blue \( {H}_{1} \cup {H}_{2} \) .
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Yes
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Theorem 15 If \( s \geq t \geq 1 \) then\n\n\[ r\left( {s{K}_{2}, t{K}_{2}}\right) = {2s} + t - 1. \]
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Proof. The graph \( G = {K}_{{2s} - 1} \cup {E}_{t - 1} \) does not contain \( s \) independent edges and \( \bar{G} = {E}_{{2s} - 1} + {K}_{t - 1} \) does not contain \( t \) independent edges. Hence \( r\left( {s{K}_{2}, t{K}_{2}}\right) \geq \) \( {2s} + t - 1 \) .\n\nTrivially (or, by Theorem 10), \( r\left( {s{K}_{2},{K}_{2}}\right) = {2s} \), so to complete the proof it suffices to show that\n\n\[ r\left( {\left( {s + 1}\right) {K}_{2},\left( {t + 1}\right) {K}_{2}}\right) \leq r\left( {s{K}_{2}, t{K}_{2}}\right) + 3. \]\n\nTo see this, let \( G \) be a graph of order \( n = r\left( {s{K}_{2}, t{K}_{2}}\right) + 3 \geq {2s} + t + 2 \) . If \( G = {K}_{n} \) then \( G \supset \left( {s + 1}\right) {K}_{2} \), and if \( G = {E}_{n} \) then \( \bar{G} \supset \left( {t + 1}\right) {K}_{2} \) . Otherwise, there are three vertices, say \( x, y \) and \( z \), such that \( {xy} \in G,{xz} \notin G \) . Now, either \( G - \{ x, y, z\} \) contains \( s \) independent edges of \( G \) and then \( {xy} \) can be added to them to form \( s + 1 \) independent edges of \( G \), or else \( \bar{G} - \{ x, y, z\} \) contains \( t \) independent edges and then \( {xz} \) can be added to them to form \( t + 1 \) independent edges of \( \bar{G} \) .
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Yes
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Theorem 16 If \( s \geq t \geq 1 \) and \( s \geq 2 \) then \( r\left( {s{K}_{3}, t{K}_{3}}\right) = {3s} + {2t} \) .
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Proof. Let \( G = {K}_{{3s} - 1} \cup \left( {{K}_{1} + {E}_{{2t} - 1}}\right) \) . Then \( G \) does not contain \( s \) independent triangles and \( \bar{G} = {E}_{{3s} - 1} + \left( {{K}_{1} \cup {K}_{{2t} - 1}}\right) \) does not contain \( t \) independent triangles. Hence \( r\left( {s{K}_{3}, t{K}_{3}}\right) \) is at least as large as claimed.\n\nIt is not difficult to show that \( r\left( {2{K}_{3},{K}_{3}}\right) = 8 \) and \( r\left( {2{K}_{3},2{K}_{3}}\right) = {10} \) (Exercise 15). Hence repeated applications of Lemma 14 give\n\n\[ r\left( {s{K}_{3},{K}_{3}}\right) \leq {3s} + 2, \]\n\nand to complete the proof it suffices to show that for \( s \geq 1, t \geq 1 \) we have\n\n\[ r\left( {\left( {s + 1}\right) {K}_{3},\left( {t + 1}\right) {K}_{3}}\right) \leq r\left( {s{K}_{3}, t{K}_{3}}\right) + 5. \]\n\nTo see this, let \( n = r\left( {s{K}_{3}, t{K}_{3}}\right) + 5 \) and consider a red-blue colouring of \( {K}_{n} \) . Select a monochromatic (say red) triangle \( {R}_{3} \) in \( {K}_{n} \) . If \( {K}_{n} - {R}_{3} \) contains a red \( s{K}_{3} \) then we are home. Otherwise, \( {K}_{n} - {R}_{3} \) contains a blue triangle \( {B}_{3} \) (it even contains a blue \( t{K}_{3} \) ). We may assume that at least five of the nine \( {R}_{3} - {B}_{3} \) edges are red. At least two of these edges are incident with a vertex of \( {B}_{3} \), and together with an edge of \( {R}_{3} \) they form a red triangle \( {R}_{3}^{ * } \) meeting \( {B}_{3} \) . Since \( {K}_{n} - {R}_{3}^{ * } - {B}_{3} \) has \( r\left( {s{K}_{3}, t{K}_{3}}\right) \) vertices, it contains either a red \( s{K}_{3} \) or a blue \( t{K}_{3} \) . These are disjoint from both \( {R}_{3}^{ * } \) and \( {B}_{3} \), so \( {K}_{n} \) contains either a red \( \left( {s + 1}\right) {K}_{3} \) or a blue \( \left( {t + 1}\right) {K}_{3} \) .
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No
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Theorem 17 If \( s \geq t \geq 1 \) then\n\n\[ \n{ps} + \left( {q - 1}\right) t - 1 \leq r\left( {s{K}_{p}, t{K}_{q}}\right) \leq {ps} + \left( {q - 1}\right) t + C.\n\]
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Proof. The graph \( {K}_{{ps} - 1} \cup {E}_{\left( {q - 1}\right) t - 1} \) shows the first inequality. As in the proofs of the previous theorems, we fix \( s - t \) and apply induction on \( t \) . By Lemma 14 we have\n\n\[ \nr\left( {s{K}_{p}, t{K}_{q}}\right) \leq \left( {s - t}\right) p + r\left( {t{K}_{p}, t{K}_{q}}\right) \leq {ps} + C,\n\]\n\nprovided that \( t \leq {t}_{0} \) . Assume now that \( t \geq {t}_{0} \) and the second inequality of the theorem holds for \( s, t \) .\n\nLet \( G \) be a graph of order \( n = p\left( {s + 1}\right) + \left( {q - 1}\right) \left( {t + 1}\right) + C \) such that \( G ⊅ \left( {s + 1}\right) {K}_{p} \) and \( \bar{G} ⊅ \left( {t + 1}\right) {K}_{q} \) . We claim that some \( {K}_{p} \) of \( G \) and \( {K}_{q} \) of \( \bar{G} \) share a vertex. Indeed, suppose that this is not the case. By altering \( G \), if necessary, we may assume that \( G \supset s{K}_{p} \) and \( \bar{G} \supset t{K}_{q} \) . Denote by \( {V}_{p} \) the set of vertices of \( G \) that are in \( {K}_{p} \) subgraphs and put \( {V}_{q} = V \smallsetminus {V}_{p},{n}_{p} = \left| {V}_{p}\right| \), and \( {n}_{q} = \left| {V}_{q}\right| \).\n\nBy our assumption, \( {n}_{p} \geq {sp} \) and \( {n}_{q} \geq {tq} \) . In the graph \( G \), a vertex \( x \in {V}_{q} \) is joined to at most \( r\left( {{K}_{p - 1},{K}_{q}}\right) - 1 \) vertices of \( {V}_{p} \), since otherwise there is a \( {K}_{p} \) of \( G \) containing \( x \) or else a \( {K}_{q} \) of \( \bar{G} \) consisting of vertices of \( {V}_{p} \) . Similarly, in the graph \( \bar{G} \) every vertex \( y \in {V}_{q} \) is joined to at most \( r\left( {{K}_{p},{K}_{q - 1}}\right) - 1 \) vertices of \( {V}_{p} \) . Hence, counting the \( {V}_{p} - {V}_{q} \) edges in \( G \) and \( \bar{G} \), we find that\n\n\[ \n{n}_{q}r\left( {{K}_{p - 1},{K}_{q}}\right) + {n}_{p}r\left( {{K}_{p},{K}_{q - 1}}\right) > {n}_{p}{n}_{q}.\n\]\n\nHowever, this is impossible, since \( {n}_{p} \geq {sp} \geq {t}_{0}p \) and \( {n}_{q} \geq {tq} \geq {t}_{0}q \), so \( {n}_{p} \geq {2r}\left( {{K}_{p - 1},{K}_{q}}\right) \) and \( {n}_{q} \geq {2r}\left( {{K}_{p},{K}_{q - 1}}\right) \) . Therefore, we can find a \( {K}_{p} \) of \( G \) and a \( {K}_{q} \) of \( \bar{G} \) with a vertex in common.\n\nWhen we omit the \( p + q - 1 \) vertices of these two subgraphs, we find that the remainder \( H \) is such that \( H ⊅ s{K}_{p} \) and \( \bar{H} ⊅ t{K}_{q} \) . However, \( \left| H\right| = {ps} + \) \( \left( {q - 1}\right) t + C \), so this is impossible.\n\nIn all the results above, we have \( r\left( {{H}_{1},{H}_{2}}\right) \leq C\left( {\left| {H}_{1}\right| + \left| {H}_{2}\right| }\right) \), where \( C \) depends only on the maximal degrees of \( {H}_{1} \) and \( {H}_{2} \) .
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Yes
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Theorem 20 If \( \mathbb{N} \) is coloured with finitely many colours then, for every \( \ell \geq 1 \) , one of the colour classes contains infinitely many translates of the same \( \ell \) -cube.
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Proof. It clearly suffices to prove the following finite version of this result.\n\nThere is a function \( H : \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N} \) such that if \( N \geq H\left( {k, l}\right) \) then every \( k \) -colouring of \( \left\lbrack N\right\rbrack \) contains a monochromatic \( \ell \) -cube.\n\nSince a 1-cube in \( \mathbb{N} \) is just a pair of integers, \( H\left( {k,1}\right) = k + 1 \) will do in this assertion. Therefore, it suffices to show that if we can have \( H\left( {k,\ell }\right) \leq n \) then \( H\left( {k,\ell + 1}\right) = N = k{n}^{\ell + 1} \) will do.\n\nTo see this, let \( c : \left\lbrack N\right\rbrack \rightarrow \left\lbrack k\right\rbrack \) be a \( k \) -colouring, and partition \( \left\lbrack N\right\rbrack \) into \( N/n = k{n}^{\ell } \) intervals, each of length \( n \) :\n\n\[ \left\lbrack N\right\rbrack = \mathop{\bigcup }\limits_{{j = 1}}^{{N/n}}{I}_{j} \]\n\nwhere \( {I}_{j} = \left\lbrack {\left( {j - 1}\right) n + 1,{jn}}\right\rbrack, j = 1,\ldots, N/n \) . Then each \( {I}_{j} \) contains a monochromatic \( \ell \) -cube. But, up to translation, there are at most \( {\left( n - 1\right) }^{\ell } < {n}^{\ell } \) cubes in these intervals, and each monochromatic cube can get one of \( k \) colours. Since there are \( k{n}^{\ell } \) intervals, some two of these intervals, say \( {I}_{j} \) and \( {I}_{h} \), contain translations of the same \( \ell \) -cube \( {C}_{\ell } \) in the same colour. The union of these two translations is a monochromatic \( \left( {\ell + 1}\right) \) -cube.
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Yes
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