Plane graphs and planar graphs. Part 1kostochk/math412-1/412Lec5.pdfA face of a plane graph (G;’)...
Transcript of Plane graphs and planar graphs. Part 1kostochk/math412-1/412Lec5.pdfA face of a plane graph (G;’)...
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Plane graphs and planar graphs. Part 1
Lecture 5
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A polygonal curve is a curve composed of finitely many linesegments.
A drawing of a graph G is a function ϕ : V (G) ∪ E(G)→ R2 s.t.(a) ϕ(v) ∈ R2 for every v ∈ V (G);(b) ϕ(v) 6= ϕ(v ′) if v , v ′ ∈ V (G) and v 6= v ′;(c) ϕ(xy) is a polygonal curve connecting ϕ(x) with ϕ(y).
A crossing in a drawing of a graph is a common vertex in theimages of two edges that is not the image of their common end.
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A graph G is planar if it has a drawing ϕ without crossings.
A plane graph is a pair (G, ϕ) where ϕ is a drawing of G withoutcrossings.
Remind me on Gas-Water-Electricity Problem.
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A graph G is planar if it has a drawing ϕ without crossings.
A plane graph is a pair (G, ϕ) where ϕ is a drawing of G withoutcrossings.
Remind me on Gas-Water-Electricity Problem.
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A graph G is planar if it has a drawing ϕ without crossings.
A plane graph is a pair (G, ϕ) where ϕ is a drawing of G withoutcrossings.
Remind me on Gas-Water-Electricity Problem.
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A face of a plane graph (G, ϕ) is a connected component ofR2 − ϕ(V (G) ∪ E(G)).
The length, `(Fi), of a face Fi in a plane graph (G, ϕ) is the totallength of the closed walk(s) bounding Fi .
Restricted Jordan Curve Theorem: A simple closed polygonalcurve C in the plane partitions the plane into exactly two faceseach having C as boundary.
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A face of a plane graph (G, ϕ) is a connected component ofR2 − ϕ(V (G) ∪ E(G)).
The length, `(Fi), of a face Fi in a plane graph (G, ϕ) is the totallength of the closed walk(s) bounding Fi .
Restricted Jordan Curve Theorem: A simple closed polygonalcurve C in the plane partitions the plane into exactly two faceseach having C as boundary.
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A face of a plane graph (G, ϕ) is a connected component ofR2 − ϕ(V (G) ∪ E(G)).
The length, `(Fi), of a face Fi in a plane graph (G, ϕ) is the totallength of the closed walk(s) bounding Fi .
Restricted Jordan Curve Theorem: A simple closed polygonalcurve C in the plane partitions the plane into exactly two faceseach having C as boundary.
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A face of a plane graph (G, ϕ) is a connected component ofR2 − ϕ(V (G) ∪ E(G)).
The length, `(Fi), of a face Fi in a plane graph (G, ϕ) is the totallength of the closed walk(s) bounding Fi .
Restricted Jordan Curve Theorem: A simple closed polygonalcurve C in the plane partitions the plane into exactly two faceseach having C as boundary.
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By F (G, ϕ) we denote the set of faces of the plane graph (G, ϕ).
Proposition 6.1: For each plane graph (G, ϕ),∑Fi∈F (G,ϕ)
`(Fi) = 2|E(G)|. (1)
Proof. By the definition of `(Fi), each edge either contributes 1to the length of two distinct faces or contributes 2 to the lengthof one face.
Theorem 6.2 (Euler’s Formula): For every connected planegraph (G, ϕ),
|V (G)| − |E(G)|+ |F (G, ϕ)| = 2.
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By F (G, ϕ) we denote the set of faces of the plane graph (G, ϕ).
Proposition 6.1: For each plane graph (G, ϕ),∑Fi∈F (G,ϕ)
`(Fi) = 2|E(G)|. (1)
Proof. By the definition of `(Fi), each edge either contributes 1to the length of two distinct faces or contributes 2 to the lengthof one face.
Theorem 6.2 (Euler’s Formula): For every connected planegraph (G, ϕ),
|V (G)| − |E(G)|+ |F (G, ϕ)| = 2.
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By F (G, ϕ) we denote the set of faces of the plane graph (G, ϕ).
Proposition 6.1: For each plane graph (G, ϕ),∑Fi∈F (G,ϕ)
`(Fi) = 2|E(G)|. (1)
Proof. By the definition of `(Fi), each edge either contributes 1to the length of two distinct faces or contributes 2 to the lengthof one face.
Theorem 6.2 (Euler’s Formula): For every connected planegraph (G, ϕ),
|V (G)| − |E(G)|+ |F (G, ϕ)| = 2.
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Proof of Euler’s FormulaFor given n, we use induction of m = |E(G)|.Base of induction: m = n − 1. Let (G, ϕ) be a plane drawingof an n-vertex connected graph G with n − 1 edges. By theCharacterization Theorem for trees, G is a tree.
Since G has no cycles, (G, ϕ) has only one face. Hence,
|V (G)| − |E(G)|+ |F (G, ϕ)| = n − (n − 1) + 1 = 2,
as claimed.
Induction step: Suppose the formula holds for all planardrawings of all n-vertex connected graphs with m − 1 edges.Let (G, ϕ) be a plane drawing of an n-vertex connected graphG with m edges.
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Proof of Euler’s FormulaFor given n, we use induction of m = |E(G)|.Base of induction: m = n − 1. Let (G, ϕ) be a plane drawingof an n-vertex connected graph G with n − 1 edges. By theCharacterization Theorem for trees, G is a tree.
Since G has no cycles, (G, ϕ) has only one face. Hence,
|V (G)| − |E(G)|+ |F (G, ϕ)| = n − (n − 1) + 1 = 2,
as claimed.
Induction step: Suppose the formula holds for all planardrawings of all n-vertex connected graphs with m − 1 edges.Let (G, ϕ) be a plane drawing of an n-vertex connected graphG with m edges.
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Proof of Euler’s FormulaFor given n, we use induction of m = |E(G)|.Base of induction: m = n − 1. Let (G, ϕ) be a plane drawingof an n-vertex connected graph G with n − 1 edges. By theCharacterization Theorem for trees, G is a tree.
Since G has no cycles, (G, ϕ) has only one face. Hence,
|V (G)| − |E(G)|+ |F (G, ϕ)| = n − (n − 1) + 1 = 2,
as claimed.
Induction step: Suppose the formula holds for all planardrawings of all n-vertex connected graphs with m − 1 edges.Let (G, ϕ) be a plane drawing of an n-vertex connected graphG with m edges.
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Since m ≥ n, G has a cycle, say C.In drawing ϕ, C forms a closed simple polygonal curve. ByRestricted Jordan Curve Theorem, C divides R2 into twocomponents. Hence each face of (G, ϕ) is either outside of Cor inside of C.
Let e be an edge of C. Let (G′, ϕ′) be obtained from (G, ϕ) bydeleting e. Then the two faces of (G, ϕ) containing e on theboundary (one inside C and one outside of C) merge into oneface of (G′, ϕ′).
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Since m ≥ n, G has a cycle, say C.In drawing ϕ, C forms a closed simple polygonal curve. ByRestricted Jordan Curve Theorem, C divides R2 into twocomponents. Hence each face of (G, ϕ) is either outside of Cor inside of C.
Let e be an edge of C. Let (G′, ϕ′) be obtained from (G, ϕ) bydeleting e. Then the two faces of (G, ϕ) containing e on theboundary (one inside C and one outside of C) merge into oneface of (G′, ϕ′).
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Since e was not a cut edge, G′ is connected. By the inductionassumption,
|V (G′)| − |E(G′)|+ |F (G′, ϕ′)| = 2. (2)
We know that |V (G”)| = |V (G)|, |E(G′)| = |E(G)| − 1 and|F (G′, ϕ′)| = |F (G, ϕ)| − 1. Plugging all this into (2), we get
|V (G)| − (|E(G)| − 1) + (|F (G, ϕ)| − 1) = 2,
which yields the theorem.
Corollary 6.3: For n ≥ 3, every simple planar n-vertex graph Ghas at most 3n − 6 edges. Moreover, if G is triangle-free, thenG has at most 2n − 4 edges.
Corollary 6.4: Graphs K5 and K3,3 are not planar.
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Since e was not a cut edge, G′ is connected. By the inductionassumption,
|V (G′)| − |E(G′)|+ |F (G′, ϕ′)| = 2. (2)
We know that |V (G”)| = |V (G)|, |E(G′)| = |E(G)| − 1 and|F (G′, ϕ′)| = |F (G, ϕ)| − 1. Plugging all this into (2), we get
|V (G)| − (|E(G)| − 1) + (|F (G, ϕ)| − 1) = 2,
which yields the theorem.
Corollary 6.3: For n ≥ 3, every simple planar n-vertex graph Ghas at most 3n − 6 edges. Moreover, if G is triangle-free, thenG has at most 2n − 4 edges.
Corollary 6.4: Graphs K5 and K3,3 are not planar.
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Since e was not a cut edge, G′ is connected. By the inductionassumption,
|V (G′)| − |E(G′)|+ |F (G′, ϕ′)| = 2. (2)
We know that |V (G”)| = |V (G)|, |E(G′)| = |E(G)| − 1 and|F (G′, ϕ′)| = |F (G, ϕ)| − 1. Plugging all this into (2), we get
|V (G)| − (|E(G)| − 1) + (|F (G, ϕ)| − 1) = 2,
which yields the theorem.
Corollary 6.3: For n ≥ 3, every simple planar n-vertex graph Ghas at most 3n − 6 edges. Moreover, if G is triangle-free, thenG has at most 2n − 4 edges.
Corollary 6.4: Graphs K5 and K3,3 are not planar.
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Since e was not a cut edge, G′ is connected. By the inductionassumption,
|V (G′)| − |E(G′)|+ |F (G′, ϕ′)| = 2. (2)
We know that |V (G”)| = |V (G)|, |E(G′)| = |E(G)| − 1 and|F (G′, ϕ′)| = |F (G, ϕ)| − 1. Plugging all this into (2), we get
|V (G)| − (|E(G)| − 1) + (|F (G, ϕ)| − 1) = 2,
which yields the theorem.
Corollary 6.3: For n ≥ 3, every simple planar n-vertex graph Ghas at most 3n − 6 edges. Moreover, if G is triangle-free, thenG has at most 2n − 4 edges.
Corollary 6.4: Graphs K5 and K3,3 are not planar.
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Proof of Corollary 6.3It is enough to prove the corollary for connected planar simplegraphs. For the main part, even 2-connected.
Claim 1: For each planar drawing ϕ of G,
3|F (G, ϕ)| ≤ 2|E(G)|. (3)
Proof: Since `(Fi) ≥ 3 for each Fi , by Proposition 6.1,
2|E(G)| =∑
Fi∈F (G,ϕ)
`(Fi) ≥ 3|F (G, ϕ)|.
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Proof of Corollary 6.3It is enough to prove the corollary for connected planar simplegraphs. For the main part, even 2-connected.
Claim 1: For each planar drawing ϕ of G,
3|F (G, ϕ)| ≤ 2|E(G)|. (3)
Proof: Since `(Fi) ≥ 3 for each Fi , by Proposition 6.1,
2|E(G)| =∑
Fi∈F (G,ϕ)
`(Fi) ≥ 3|F (G, ϕ)|.
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Proof of Corollary 6.3It is enough to prove the corollary for connected planar simplegraphs. For the main part, even 2-connected.
Claim 1: For each planar drawing ϕ of G,
3|F (G, ϕ)| ≤ 2|E(G)|. (3)
Proof: Since `(Fi) ≥ 3 for each Fi , by Proposition 6.1,
2|E(G)| =∑
Fi∈F (G,ϕ)
`(Fi) ≥ 3|F (G, ϕ)|.
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Let ϕ be a planar drawing of G. By Euler’s Formula andClaim 1,
2 = |V (G)− |E(G)|+ |F (G, ϕ)| ≤ |V (G)| − |E(G)|+ 23|E(G)|.
Hence13|E(G)| ≤ |V (G)| − 2, as claimed.
For the ”Moreover” part, we modify the claim:Claim 2: For each planar drawing ϕ of triangle-free G,
|F (G, ϕ)| ≤ |E(G)|2
. (4)
Proof: Since `(Fi) ≥ 4 for each Fi , by Proposition 6.1,
2|E(G)| =∑
Fi∈F (G,ϕ)
`(Fi) ≥ 4|F (G, ϕ)|.
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Let ϕ be a planar drawing of G. By Euler’s Formula andClaim 1,
2 = |V (G)− |E(G)|+ |F (G, ϕ)| ≤ |V (G)| − |E(G)|+ 23|E(G)|.
Hence13|E(G)| ≤ |V (G)| − 2, as claimed.
For the ”Moreover” part, we modify the claim:Claim 2: For each planar drawing ϕ of triangle-free G,
|F (G, ϕ)| ≤ |E(G)|2
. (4)
Proof: Since `(Fi) ≥ 4 for each Fi , by Proposition 6.1,
2|E(G)| =∑
Fi∈F (G,ϕ)
`(Fi) ≥ 4|F (G, ϕ)|.
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Let ϕ be a planar drawing of G. By Euler’s Formula andClaim 1,
2 = |V (G)− |E(G)|+ |F (G, ϕ)| ≤ |V (G)| − |E(G)|+ 23|E(G)|.
Hence13|E(G)| ≤ |V (G)| − 2, as claimed.
For the ”Moreover” part, we modify the claim:Claim 2: For each planar drawing ϕ of triangle-free G,
|F (G, ϕ)| ≤ |E(G)|2
. (4)
Proof: Since `(Fi) ≥ 4 for each Fi , by Proposition 6.1,
2|E(G)| =∑
Fi∈F (G,ϕ)
`(Fi) ≥ 4|F (G, ϕ)|.
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Let ϕ be a planar drawing of G. By Euler’s Formula andClaim 1,
2 = |V (G)− |E(G)|+ |F (G, ϕ)| ≤ |V (G)| − |E(G)|+ 23|E(G)|.
Hence13|E(G)| ≤ |V (G)| − 2, as claimed.
For the ”Moreover” part, we modify the claim:Claim 2: For each planar drawing ϕ of triangle-free G,
|F (G, ϕ)| ≤ |E(G)|2
. (4)
Proof: Since `(Fi) ≥ 4 for each Fi , by Proposition 6.1,
2|E(G)| =∑
Fi∈F (G,ϕ)
`(Fi) ≥ 4|F (G, ϕ)|.
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Finishing proof of Corollary 6.3Again, let ϕ be a planar drawing of our triangle-free G. ByEuler’s Formula and Claim 1,
2 = |V (G)− |E(G)|+ |F (G, ϕ)| ≤ |V (G)| − |E(G)|+ 12|E(G)|.
Hence12|E(G)| ≤ |V (G)| − 2, as claimed.
Questions: 1. What happens for |V (G)| ≤ 2?2. In which part(s) of the proof have I used |V (G)| ≥ 3?
Proof of Corollary 6.4. |V (K5)| = 5 and|E(K5)| = 10 = 3|V (K5)| − 5. So, by Cor. 6.3, K5 is not planar.
Similarly, K3,3 is triangle-free, |V (K3,3)| = 6 and|E(K3,3)| = 9 = 2|V (K3,3)| − 3.Again, by Cor. 6.3, K3,3 is not planar.
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Finishing proof of Corollary 6.3Again, let ϕ be a planar drawing of our triangle-free G. ByEuler’s Formula and Claim 1,
2 = |V (G)− |E(G)|+ |F (G, ϕ)| ≤ |V (G)| − |E(G)|+ 12|E(G)|.
Hence12|E(G)| ≤ |V (G)| − 2, as claimed.
Questions: 1. What happens for |V (G)| ≤ 2?2. In which part(s) of the proof have I used |V (G)| ≥ 3?
Proof of Corollary 6.4. |V (K5)| = 5 and|E(K5)| = 10 = 3|V (K5)| − 5. So, by Cor. 6.3, K5 is not planar.
Similarly, K3,3 is triangle-free, |V (K3,3)| = 6 and|E(K3,3)| = 9 = 2|V (K3,3)| − 3.Again, by Cor. 6.3, K3,3 is not planar.
![Page 31: Plane graphs and planar graphs. Part 1kostochk/math412-1/412Lec5.pdfA face of a plane graph (G;’) is aconnected componentof R2 ’(V(G)[E(G)). Thelength, ‘(Fi), of aface Fi in](https://reader033.fdocument.pub/reader033/viewer/2022042915/5f52688321ea2856d945d236/html5/thumbnails/31.jpg)
Finishing proof of Corollary 6.3Again, let ϕ be a planar drawing of our triangle-free G. ByEuler’s Formula and Claim 1,
2 = |V (G)− |E(G)|+ |F (G, ϕ)| ≤ |V (G)| − |E(G)|+ 12|E(G)|.
Hence12|E(G)| ≤ |V (G)| − 2, as claimed.
Questions: 1. What happens for |V (G)| ≤ 2?2. In which part(s) of the proof have I used |V (G)| ≥ 3?
Proof of Corollary 6.4. |V (K5)| = 5 and|E(K5)| = 10 = 3|V (K5)| − 5. So, by Cor. 6.3, K5 is not planar.
Similarly, K3,3 is triangle-free, |V (K3,3)| = 6 and|E(K3,3)| = 9 = 2|V (K3,3)| − 3.Again, by Cor. 6.3, K3,3 is not planar.
![Page 32: Plane graphs and planar graphs. Part 1kostochk/math412-1/412Lec5.pdfA face of a plane graph (G;’) is aconnected componentof R2 ’(V(G)[E(G)). Thelength, ‘(Fi), of aface Fi in](https://reader033.fdocument.pub/reader033/viewer/2022042915/5f52688321ea2856d945d236/html5/thumbnails/32.jpg)
Finishing proof of Corollary 6.3Again, let ϕ be a planar drawing of our triangle-free G. ByEuler’s Formula and Claim 1,
2 = |V (G)− |E(G)|+ |F (G, ϕ)| ≤ |V (G)| − |E(G)|+ 12|E(G)|.
Hence12|E(G)| ≤ |V (G)| − 2, as claimed.
Questions: 1. What happens for |V (G)| ≤ 2?2. In which part(s) of the proof have I used |V (G)| ≥ 3?
Proof of Corollary 6.4. |V (K5)| = 5 and|E(K5)| = 10 = 3|V (K5)| − 5. So, by Cor. 6.3, K5 is not planar.
Similarly, K3,3 is triangle-free, |V (K3,3)| = 6 and|E(K3,3)| = 9 = 2|V (K3,3)| − 3.Again, by Cor. 6.3, K3,3 is not planar.
![Page 33: Plane graphs and planar graphs. Part 1kostochk/math412-1/412Lec5.pdfA face of a plane graph (G;’) is aconnected componentof R2 ’(V(G)[E(G)). Thelength, ‘(Fi), of aface Fi in](https://reader033.fdocument.pub/reader033/viewer/2022042915/5f52688321ea2856d945d236/html5/thumbnails/33.jpg)
Finishing proof of Corollary 6.3Again, let ϕ be a planar drawing of our triangle-free G. ByEuler’s Formula and Claim 1,
2 = |V (G)− |E(G)|+ |F (G, ϕ)| ≤ |V (G)| − |E(G)|+ 12|E(G)|.
Hence12|E(G)| ≤ |V (G)| − 2, as claimed.
Questions: 1. What happens for |V (G)| ≤ 2?2. In which part(s) of the proof have I used |V (G)| ≥ 3?
Proof of Corollary 6.4. |V (K5)| = 5 and|E(K5)| = 10 = 3|V (K5)| − 5. So, by Cor. 6.3, K5 is not planar.
Similarly, K3,3 is triangle-free, |V (K3,3)| = 6 and|E(K3,3)| = 9 = 2|V (K3,3)| − 3.Again, by Cor. 6.3, K3,3 is not planar.