hcju/proofwikiqa huggingface dataset:
The size of a $r$-regular graph is its number of edges. The order of a $r$-regular graph is its number of vertices. The degree of each vertex of an $r$-regular graph is $r$. Hence the total of all the degrees of an $r$-regular graph of order $n$ is $nr$. The result follows directly from the Handshake Lemma. {{qed}} Category:Regular Graphs 3c712ia7q15uhflusq9w8y87en80nkh
=== Sufficient Condition === Let $G = \struct {V, E}$ be bipartite. So, let $V = A \cup B$ such that $A \cap B = \O$ and that all edges $e \in E$ are such that $e$ is of the form $\set {a, b}$ where $a \in A$ and $b \in B$. (This is the definition of a bipartite graph.) Suppose $G$ has (at least) one odd cycle $C$. Let the length of $C$ be $n$. Let $C = \tuple {v_1, v_2, \ldots, v_n, v_1}$. {{WLOG}}, let $v_1 \in A$. It follows that $v_2 \in B$ and hence $v_3 \in A$, and so on. Hence we see that $\forall k \in \set {1, 2, \ldots, n}$, we have: :$v_k \in \begin{cases} A : & k \text{ odd} \\ B : & k \text{ even} \end{cases}$ But as $n$ is odd, $v_n \in A$. But $v_1 \in A$, and $v_n v_1 \in C_n$. So $v_n v_1 \in E$ which contradicts the assumption that $G$ is bipartite. Hence if $G$ is bipartite, it has no odd cycles. {{qed|lemma}} === Necessary Condition === It is enough to consider $G$ as being connected, as otherwise we could consider each component separately. Suppose $G$ has no odd cycles. Choose any vertex $v \in G$. Divide $G$ into two sets of vertices like this: :Let $A$ be the set of vertices such that the shortest path from each element of $A$ to $v$ is of odd length :Let $B$ be the set of vertices such that the shortest path from each element of $B$ to $v$ is of even length. Then $v \in B$ and $A \cap B = \O$. Suppose $a_1, a_2 \in A$ are adjacent. Then there would be a closed walk of odd length $\tuple {v, \ldots, a_1, a_2, \ldots, v}$. But from Graph containing Closed Walk of Odd Length also contains Odd Cycle, it follows that $G$ would then contain an odd cycle. This contradicts our initial supposition that $G$ contains no odd cycles. So no two vertices in $A$ can be adjacent. By the same argument, neither can any two vertices in $B$ be adjacent. Thus $A$ and $B$ satisfy the conditions for $G = \struct {A \cup B, E}$ to be bipartite. {{qed}} Category:Bipartite Graphs bfz90ijp1enww187l7qamiya14j3jb3
From Ore Graph is Connected it is not necessary to demonstrate that $G$ is connected. {{AimForCont}} it were possible to construct a graph that fulfils condition $(1)$ which is not Hamiltonian. For a given $n \ge 3$, let $G$ be the graph with the most possible edges such that $G$ is non-Hamiltonian which satisfies $(1)$. Although it does not contain a Hamilton cycle, $G$ has to contain a Hamiltonian path $\left({v_1, v_2, \ldots, v_n}\right)$. Otherwise it would be possible to add further edges to $G$ without making $G$ Hamiltonian. Since $G$ is not Hamiltonian, $v_1$ is not adjacent to $v_n$, otherwise $\left({v_1, v_2, \ldots, v_n, v_1}\right)$ would be a Hamilton cycle. By $(1)$, we have that: :$\deg v_1 + \deg v_n \ge n$ By the Pigeonhole Principle, for some $i$ such that $2 \le i \le n - 1$, $v_i$ is adjacent to $v_1$, and $v_{i-1}$ is adjacent to $v_n$. But the cycle $\tuple {v_1, v_2, \ldots, v_{i - 1}, v_n, v_{n - 1}, \ldots, v_i, v_1}$ is then a Hamilton cycle. So $G$ is Hamiltonian after all. {{Qed}}
=== Necessary Condition === {{:Path in Tree is Unique/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Path in Tree is Unique/Sufficient Condition}} Category:Tree Theory s04brdleiv4hajtybih4400qcihtzy8
By definition: :the order of a tree is how many nodes it has and: :its size is how many edges it has. === Necessary Condition === {{:Size of Tree is One Less than Order/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Size of Tree is One Less than Order/Sufficient Condition}}
Follows directly from the fact that by definition, $T$ has no circuits. As $T$ has no circuits, then nor can $S$ have. Hence $S$ is a connected simple graph with no circuits. Thus by definition, $S$ is a tree. {{qed}} Category:Tree Theory 7wvxys28vwsd2tiykfn225rvz7bg703
Let the proposition we are proving in the case of a tree of order $k$ be denoted $\map P k$. === Basis for the Induction === That $\map P 2$ is true follows as the Unique Tree of Order 2 contains exactly two leaves (viz. its two nodes). This is our basis for the induction. === Induction Hypothesis === Suppose $\map P i$ is true for all $i$ such that $2 \le i \le k$. Then we need to demonstrated that $\map P i$ is true for all $i$ such that $2 \le i \le k + 1$. This is our induction hypothesis. === Induction Step === This is our induction step: Let $T_{k + 1}$ be a tree of order $k + 1$. Pick an edge $e = \set {v_1, v_2}$ of $T_{k + 1}$. Consider the subgraph $T_0$ obtained by removing $e$. By Edge of Tree is Bridge, $e$ is a bridge and removing it results in a subgraph with two components. By Connected Subgraph of Tree is Tree each such component is a tree. Let $T_i$ be the component containing $v_i$. Denote its order by $k_i$, where $i \in \set{1, 2}$ Since no nodes have been removed from $T_{k + 1}$, $k_1 + k_2 = k + 1$. There are two cases: ==== Case 1 ==== '''Neither $T_1$ nor $T_2$ is of order $1$.''' Then $2 \le k_i \le k - 1, i = 1, 2$. By the induction hypothesis, both $T_1$ and $T_2$ have at least two leaves. Either $v_1$ and $v_2$ are both leaves of the respective trees, or one of them is not. Suppose that $v_1$ and $v_2$ are both leaves of the respective trees. Then by adding $e$ back to the graph and thus recreating $T_{k + 1}$, $v_1$ and $v_2$ cease to be leaves. However, at least one other leaf from each subgraph $T_i, i = 1, 2$ survives the addition, becoming a leaf of $T_{k + 1}$. Therefore $T_{k + 1}$ has at least two leaves. Suppose that either $v_1$ or $v_2$ is not a leaf of its respective tree. Then $T_{k + 1}$ has at least three leaves. {{qed|lemma}} ==== Case 2 ==== '''$T_1$ or $T_2$ is of order $1$.''' {{WLOG}}, assume that $T_1$ is of order $1$, that is $k_1 = 1$. This implies that $k_2 = k$. Then, by the induction hypothesis, $T_2$ has at least two leaf nodes. Also $k_1 = 1$ implies that $v_1$ is a leaf of $T_{k + 1}$. Suppose that $v_2$ is one of the leaves of $T_2$. Then by adding $e$ back and thus recreating $T_{k + 1}$, $v_2$ is no longer a leaf, though at least one other leaf of $T_2$ survives the addition, becoming a leaf of $T_{k + 1}$. Therefore $T_{k + 1}$ has at least two leaves. Suppose that $v_2$ is not one of the leaves of $T_2$. Then $T_{k + 1}$ has at least three leaves. {{qed|lemma}} In both cases, the thesis follows by the Second Principle of Mathematical Induction. {{qed}}
Follows immediately from: :The fact that a cycle graph is $2$-regular :The result Number of Edges of Regular Graph which says that an $r$-regular graph with $n$ vertices has $\dfrac {n r} 2$ edges. {{qed}} Category:Graph Theory m0j3yzk2aodufun4ep3pecxq4uuu87p
Let $G_n$ be the graph obtained by removing any edge from the cycle graph $C_n$. As each of those edges lies on a cycle, by Condition for Edge to be Bridge, none of them is a bridge. So removing any edge from $C_n$ leaves the resulting subgraph of $C_n$ connected. From Size of Cycle Graph equals Order, the cycle graph $C_n$ has $n$ edges. So $G_n$ has $n$ nodes and $n - 1$ edges, and is connected. From Size of Tree is One Less than Order, $G_n$ is a tree. But $G_n$ has two nodes of degree $1$, and all the others are of degree $2$. It follows that $G_n$ is traversable, and hence by definition is a path graph. {{qed}} Category:Graph Theory ipjr16lvo1xtfnjspivynxq5zpddpcl
Let $G$ be a balanced signed graph. Since $G$ is balanced, we can colour the vertices black and white, say, so that every edge has a black and a white end. Consider any cycle $C$ in $G$. As we proceed around $C$, there is a change of colour when we traverse a negative edge. Since this is a cycle, the last colour is the same as the first one. So there must be an even number of colour changes. So there must be an even number of negative edges in $C$. {{qed}} Category:Graph Theory s3bdewvf9bmv1ocpj8b4n403mst4zje
Let $G = \struct {V, E}$ be a graph whose vertices are all even. If there is more than one vertex in $G$, then each vertex must have degree greater than $0$. Begin at any vertex $u$. Since the graph is connected (if the graph is not connected then the argument will be applied to separate components), there must be an edge $\set {u, u_1}$ for some vertex $u_1 \ne u$. Since $u_1$ has even degree greater than $0$, there is an edge $\set {u_1, u_2}$. These two edges make a trail from $u$ to $u_2$. Continue this trail, leaving each vertex on an edge that was not previously used, until we reach a vertex $v$ that we have met before. (Note: $v$ may or may not be the same vertex as $u$. It does not matter either way.) The edges of the trail between the two occurrences of $v$ must form a cycle. Call the cycle formed by this process $C_1$. If $C_1$ covers all the edges of $G$, the proof is complete. Otherwise, remove the edges forming $C_1$ from the graph, leaving the graph $G_1$. All the vertices in $G_1$ are still even. So pick some vertex $u'$ in $G_1$. Repeat the same process as before, starting with an edge $\set {u', u'_1}$. By the same argument, we can generate a new cycle $C_2$, which has no edges in common with $C_1$. If $C_2$ covers all the rest of the edges of $G$, then we are done. Otherwise, remove the edges forming $C_2$ from the graph, leaving the graph $G_2$, which again contains only even vertices. We continue in this way until we have used up all the edges of $G$. By this time we have a number of cycles, $C_1, C_2, \ldots, C_k$ which between them contain all the edges of $G$ but no two of them have an edge in common. {{qed}} The converse is trivial: All the vertices of a cycle by definition are  even. So if we were to take all these cycles and add them to the graph in turn, each vertex would increase its order by an even number. {{qed}}
Let $G = \struct {A \mid B, E}$ be a bipartite graph. To be Hamiltonian, a graph $G$ needs to have a Hamilton cycle: that is, one which goes through all the vertices of $G$. As each edge in $G$ connects a vertex in $A$ with a vertex in $B$, any cycle alternately passes through a vertex in $A$ then a vertex in $B$. {{WLOG}}, suppose  that $\card A > \card B$, that is, that there are more vertices in $A$ than in $B$. Let $\card A = m, \card B = n$. Suppose $G$ has a Hamilton cycle $C$. Let that cycle start at $u \in B$. After $2 n$ edges have been traversed, we will have arrived back at $u$ again, and all the vertices of $B$ will have been visited. But there will still be $m - n$ vertices in $A$ which have not been visited. Hence $C$ can not be a Hamilton cycle. {{qed}}
Clearly $K_{1, 1}$ is not Hamiltonian, although it does have a Hamiltonian path. From Condition for Bipartite Graph to be Hamiltonian, if $m \ne n$ then $K_{m, n}$ is not Hamiltonian. So let $m = n$. We note that the degree of any vertex in $K_{n, n}$ is $n$. Then we can use Ore's Theorem, and the result follows directly. {{qed}} Category:Hamiltonian Graphs Category:Complete Graphs Category:Bipartite Graphs p8rn3zisrw3bxrb7mefpdg9bnab4jdu
Let $K_{m, n}$ be a complete bipartite graph. If $m = n = 1$ then $K_{m, n} = K_{1, 1}$ is trivially semi-Hamiltonian: :100px {{qed|lemma}} If $m = n + 1$, we can construct a Hamiltonian path as follows. Let $K_{m, n} = \struct {A \mid B, E}$ such that $\card A = m, A = \set {u_1, u_2, \ldots, u_m}, \card B = n, B = \set {v_1, v_2, \ldots, v_m}$. We start at vertex $u_1 \in A$. As $K_{m, n}$ is complete, there is an edge connecting $u_1$ to $v_1$. From $v_1$, there is an edge connecting $v_1$ to $u_2$. And so on. There is therefore a Hamiltonian path $\tuple {u_1, v_1, u_2, v_2, \ldots, u_{n - 1}, v_{n - 1}, u_n, v_n, u_m}$ (where of course $u_m = u_{n + 1}$). So when $m = n + 1$, $K_{m, n}$ has a Hamiltonian path. However, from Condition for Bipartite Graph to be Hamiltonian, $K_{m, n}$ is only fully Hamiltonian if $m = n$. Hence when $m = n + 1$, $K_{m, n}$ is semi-Hamiltonian. Similarly, if $n = m + 1$ the same argument applies, but this time the Hamiltonian path is: $\tuple {v_1, u_1, v_2, y_2, \ldots, v_{n - 1}, y_{n - 1}, v_m, u_m, v_n}$ where of course $v_n = u_{m + 1}$. {{qed|lemma}} Now, suppose $K_{m, n}$ is such that neither $m = n + 1$ nor $n = m + 1$, and it is not the case that $m = n = 1$. If $m = n$, then from Condition for Bipartite Graph to be Hamiltonian $K_{m, n}$ is Hamiltonian, and therefore not semi-Hamiltonian. If $m > n + 1$, then we can once more trace a path $P$ through $K_{m, n} = \struct {A \mid B, E}$ as before. However, at the end of $P$ there is a spare vertex in $A$ which has not been traversed which you can't get to without going back to one of the vertices in $B$ that you have already visited. Similarly if $n > m + 1$. {{qed|lemma}} Hence the result. {{qed}} Category:Hamiltonian Graphs Category:Bipartite Graphs aixaygvuab6peng9xh9baw0v9glpvfg
Statement $1$ is the usual definition of a tree. === 1 implies 2 === * The fact that $T$ has no circuits is part of statement $1$. * The fact that $T$ has $n-1$ edges is proved in Size of Tree is One Less than Order. {{qed|lemma}} === 2 implies 3 === * The fact that $T$ has $n-1$ edges is part of statement $2$. * The fact that $T$ is connected is proved in Size of Tree is One Less than Order. {{qed|lemma}} === 3 implies 1 === * The fact that $T$ is connected is part of statement $3$. * Given that $T$ has $n-1$ edges, the fact that it has no circuits is proved during the course of the proof of Size of Tree is One Less than Order. {{qed|lemma}} === 1 implies 4 === As $T$ has no circuits, then from Condition for Edge to be Bridge, every edge is a bridge. Thus removing any edge of $T$ will disconnect $T$. {{qed|lemma}} === 4 implies 1 === If by removing any one edge between two vertices of $T$ renders it disconnected, then that means each edge must be a bridge. So by Condition for Edge to be Bridge, $T$ has no circuits. {{qed|lemma}} === 1 implies and is implied by 5 === This is proved by Path in Tree is Unique. {{qed|lemma}} === 1 implies 6 === Suppose $T = \struct {V, E}$ is connected and has no circuits. Let $u, v \in V$ be any two vertices of $T$. Let $P = \tuple {u, u_1, u_2, \ldots, u_{n - 1}, v}$ be a path from $u$ to $v$. Let a new edges $\set {u, v}$ be added. Then $\tuple {u, u_1, u_2, \ldots, u_{n - 1}, v, u}$ is now a cycle, which is by definition also a circuit, in $T$. Note that this applies even when $P = \tuple {u, v}$: $\tuple {u, v, u}$ is still a cycle in $T$, but now $T$ is a multigraph. {{qed|lemma}} === 6 implies 1 === Suppose $T$ has no circuits, but adding one edge creates a cycle, which is by definition also a circuit. If $T$ were disconnected, then it would be possible to add an edge $e$ to connect two components of $T$. By definition, $e$ would be a bridge. From Condition for Edge to be Bridge, it follows that $e$ does not lie on a circuit. So, if the only way to add an edge to $T$ forms a cycle, it follows that $T$ must be connected. So $T$ is connected and has no circuits. {{qed|lemma}} Thus, all the above can be used as a definition for a finite tree. {{Qed}} Category:Tree Theory aj75mg6cfqijwdw5ry6jb3wx18amjyx
It is clear, by definition of a bridge, that removing $e$ increases the number of components. So after $e$ is removed from $G$, the number of components in the remaining graph is at least $m + 1$. Suppose that removing $e$ disconnects $G$ into more than $m + 1$ components. Since $e$ joins only two vertices of $G$, it can link at most two of these components. So there is at least one extra component when $e$ is put back into $G$. So $G$ has more than $m$ components. This contradicts the fact that $G$ has $m$ components. Hence the result. {{qed}} Category:Graph Theory 58lmhnm8i0e8g657bd601uxnuqk9616
From Equivalent Definitions for Finite Tree, adding an edge creates at least one cycle. Suppose that adding an edge $\tuple {u, v}$ to a tree $T$ creates two or more cycles. Let two such cycles be $\tuple {u, v, \ldots, u_1, u_2, \ldots, u}$ and $\tuple {u, v, \ldots, v_1, v_2, \ldots, u}$. By removing the edge $\tuple {u, v}$ from this cycle, we have two paths from $v$ to $u$: :$\tuple {v, \ldots, u_1, u_2, \ldots, u}$ :$\tuple {v, \ldots, v_1, v_2, \ldots, u}$. But that means $T$ has more than one path between two nodes. From Path in Tree is Unique, that means $T$ can not be a tree. Hence the result. {{qed}} Category:Tree Theory j5dlnos6q84vfpjzfbk0ro6e0nwl0op
By definition, a forest is a disconnected graph whose components are all trees. Let the number of nodes in each component of $F$ be $n_1, n_2, \ldots, n_m$ where of course $\displaystyle \sum_{i \mathop = 1}^m n_i = n$. From Size of Tree is One Less than Order, the number of edges in tree $i$ is $n_i - 1$. So the total number of edges in $F$ is: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^m \paren {n_i - 1}       | r = \sum_{i \mathop = 1}^m n_i - \sum_{i \mathop = 1}^m 1        | c = Summation is Linear }} {{eqn | r = n - m       | c = Sum of Identical Terms }} {{end-eqn}} {{qed}}
Follows directly from Bijection between Prüfer Sequences and Labeled Trees. This shows that there is a bijection between the set of labeled trees with $n$ nodes and the set of all Prüfer sequences of the form: :$\tuple {\mathbf a_1, \mathbf a_2, \ldots, \mathbf a_{n - 2} }$ where each of the $\mathbf a_i$'s is one of the integers $1, 2, \ldots, n$, allowing for repetition. Since there are exactly $n$ possible values for each integer $\mathbf a_i$, the total number of such sequences is $\displaystyle \prod_{i \mathop = 1}^{n - 2} n$. The result follows from Equivalence of Mappings between Sets of Same Cardinality. {{qed}} {{Namedfor|Arthur Cayley|cat = Cayley}}
Let $T$ be the set of all labeled trees of order $n$. Let $P$ be the set of all Prüfer sequence of length $n-2$. Let $\phi: T \to P$ be the mapping that maps each tree to its Prüfer sequence. * From Prüfer Sequence from Labeled Tree, $\phi$ is clearly well-defined, as every element of $T$ maps uniquely to an element of $P$. * However, from Labeled Tree from Prüfer Sequence, $\phi^{-1}: P \to T$ is also clearly well-defined, as every element of $P$ maps to a unique element of $T$. Hence the result. {{questionable|How is it immediate that the two constructions are mutually inverse?}} {{qed}} Category:Tree Theory Category:Combinatorics 1nztp06dejnykcd7hdywz33czkukpjk
A tree whose order is $1$ or $2$ is already trivially central or bicentral. Let $T$ be a tree of order at least $3$. First we establish that the construction of a center or bicenter actually works. From Finite Tree has Leaf Nodes, there are always at least two nodes of degree $1$ to be removed.0 By the Handshake Lemma, it is clear that $T$ must also have at least one node whose degree is greater than $1$:  :$\displaystyle \sum_{i \mathop = 1}^n \map {\deg_G} {v_i} = 2 q$ where $q$ is the number of edges in $T$. But $q = n-1$ from Size of Tree is One Less than Order. So if each node has degree $1$, then $n = 2 \paren {n - 1}$ and so $n = 2$. Therefore if $n > 2$ there must be at least one node in $T$ of degree is greater than $1$. Next, from Connected Subgraph of Tree is Tree, after having removed those nodes, what is left is still a tree. Therefore the construction is valid.  We need to show the following: :$(1): \quad T$ has only one center or bicenter :$(2): \quad $T$ has either a center or a bicenter. Suppose $T$ has more than one center or bicenter. It would follow that at least one of the iterations constructing the center or bicenter disconnects $T$ into more than one component. That could only happen if we were to remove an edge between two nodes of degree greater than $1$. Hence $T$ has at most one center or bicenter. Now to show that $T$ has at least one center or bicenter. The proof works by the Principle of Complete Induction. We know that a tree whose order is $1$ or $2$ is already trivially central or bicentral. This is our base case. Suppose that all tree whose order is $n$ have at most one center or bicenter. This is our induction hypothesis. Take a tree $T$ whose order is $n+1$ where $n > 2$. Let $T$ have $k$ nodes of degree $1$. We remove all these $k$ nodes. This leaves us with a tree with $n + 1 - k$ nodes. As we have seen that $T$ has at least one node whose degree is greater than $1$, $n + 1 - k \ge 1$. As there are always at least two nodes of degree $1$, $n+1-k \le n-1$. So after the first iteration, we are left with a tree whose order is between $1$ and $n-1$ inclusive. By the induction hypothesis, this tree has either a center or bicenter. The result follows by the Principle of Complete Induction. {{qed}} Category:Tree Theory ejzfpwkjjgxesz281ibxku024phkk1l
{{begin-eqn}} {{eqn | o =        | r = \paren {x^2 + 2 y^2 + 2 x y} \paren {x^2 + 2 y^2 - 2 x y}       | c =  }} {{eqn | r = \paren {x^2 + 2 y^2}^2 - \paren {2 x y}^2       | c = Difference of Two Squares }} {{eqn | r = x^4 + 2 \cdot x^2 \cdot 2 y^2 + 4 y^2 - 2 x y \cdot 2 x y       | c =  }} {{eqn | r = x^4 + 4 y^4       | c = gathering up terms and cancelling }} {{end-eqn}} {{qed}}
Suppose that Prim's Algorithm produces a tree $T$. Let there exist another spanning tree $S$ with a smaller total weight. Let $e$ be an edge of smallest weight which lies in $T$ but not $S$. If we add $e$ to $S$, we obtain a cycle, from Equivalent Definitions for Finite Tree. This cycle contains an edge $e'$ which is in $S$ but not $T$, otherwise $T$ would not be a tree. So, we replace $e'$ in $S$ with $e$ from $T$, and obtain a new spanning tree $S'$. {{handwaving|This is just hiding the core of the argument in a sneaky sentence. E.g. in Prim's Algo, it may be that neither of the endpoints of $e'$ are in $T$ yet, so that it may not be added yet}} From the method of construction of $T$, it follows that the weight of $e$ can not exceed that of $e'$. So the total weight of $S'$ does not exceed the total weight of $T$. Also, $S'$ has one more edge in common with $T$ than $S$ has. We repeat the above procedure, and repeatedly change edges of $S$ for edges of $T$, and each time the weight of the intermediate graph does not exceed that of $T$. Thus the weight of $T$ does not exceed that of $S$, contradicting the definition of $S$. Hence $T$ must be a minimum spanning tree. {{qed}} Category:Tree Theory e1nmhoqsk1e5c4omceeld4z268nmwu1
In a complete graph, every vertex is adjacent to every other vertex. Therefore, if we were to take all the vertices in a complete graph in any order, there will be a path through those vertices in that order. Joining either end of that path gives us a Hamilton cycle. From Cardinality of Set of Bijections, there are $n!$ different ways of picking the vertices of $G$ in some order. Hence there are $n!$ ways of building such a Hamilton cycle. Not all these are different, though. On any such cycle, there are: :$n$ different places you can start :$2$ different directions you can travel. So any one of these $n!$ cycles is in a set of $2 n$ cycles which all contain the same set of edges. So there are $\dfrac {n!} {2 n} = \dfrac {\paren {n - 1}!} 2$ distinct Hamilton cycles. {{qed}} Category:Hamiltonian Graphs Category:Combinatorics cassjystun4cwnqt6p3p2u5n9c06y1i
Let $\mathcal R \subseteq S \times S$ be a coreflexive relation. Let $\left({x, y}\right) \in \mathcal R$. By definition of coreflexive, it follows that $x = y$, and hence $\left({x, y}\right) = \left({x, x}\right)$. So by definition of the diagonal relation: : $\left({x, y}\right) \in \Delta_S$ Hence the result. {{qed}} Category:Relation Theory t48ufxcamio4dkjrc35argpo28c078u