Graphs - UMass Amherstphaas/INFO150/slides/slecture15-anno… · circuit C 2. Observe that G0 is...

GraphsReading: EC 7.1–7.2

Peter J. Haas

INFO 150Fall Semester 2019

Lecture 15 1/ 21

GraphsIntroductionThe Konigsberg Bridge ProblemDefinitions and TerminologyGraph Notation for Konigsberg ProblemEulerian GraphsGraphs with Eulerian TrailsProofs About Graphs

Lecture 15 2/ 21

Introduction

You have already seen lots of graphs—we will now study their abstract properties

Lecture 15 3/ 21

Directed graph

HTTTHH

TreeHS Friends

College Friends

Girlfriend's Friends

University FriendsOther AcademicFriends

griffsgraphs.wordpress.com/2012/07/02/a-facebook-network/

Facebook networkWatson drug discovery

Origins and Euler

Konigsberg bridge problem

I Consider a walk around town: A, 6, B, 5, D, 4, A

I Eulerian trail: Crosses each bridge once

I Eulerian circuit: ... and comes back to starting point

I Does there exist an Eulerian circuit?

Euler’s non-existence proof

I Suppose A is an intermediate vertex(not starting or ending)

I Bridges to A must come in pairs

I So need an even number of edges

I But every vertex has an odd number of edges

I Hence none of the four regions can be an interior vertex

I So Eulerian trail cannot exist

Lecture 15 4/ 21

7 bridges of Konigsberg

Abstract 7 bridges of Konigsberg

7Konigsberg graph

Origins and Euler

Konigsberg bridge problem

I Consider a walk around town: A, 6, B, 5, D, 4, A

I Eulerian trail: Crosses each bridge once

I Eulerian circuit: ... and comes back to starting point

I Does there exist an Eulerian circuit?

Euler’s non-existence proof

I Suppose A is an intermediate vertex(not starting or ending)

I Bridges to A must come in pairs

I So need an even number of edges

I But every vertex has an odd number of edges

I Hence none of the four regions can be an interior vertex

I So Eulerian trail cannot exist

Lecture 15 4/ 21

7 bridges of Konigsberg

Abstract 7 bridges of Konigsberg

7Konigsberg graph

Eulerian Trails

Proposition 1

In any graph, if there are an odd number of edges connected to a vertex x , then x

cannot be an interior vertex of an Eulerian trail.

Example: Explain why the following graph does not have an Eulerian trail

a b c d e

Question: Can we give conditions under which there exists an Eulerian trail?

Lecture 15 5/ 21

vertices gf, h , j haveodd #

Definitions and Terminology

Definitions

1. A graph G consists of a set E of edges and a set V of vertices (also called nodes).I An edge is associated with one or two vertices, called endpoints.I Two nodes joined by an edge are called adjacent nodes.I An edge with one vertex is called a loop.I Two edges having the same endpoints are called multiple edges or parallel edges.

2. A walk is a sequence v1e1v2e2 · · · vnenvn+1 of alternating vertices and edges.I Each edge in the list lies between its endpoints.I If beginning and end vertices are the same, the walk is closed.I The length of a walk is the number of edges (n in the above example).I A walk of length 0 is called a trivial walk.

3. A trail is a walk with no repeated edges; a path is a walk with no repeated vertices.I A circuit is a closed trailI A circuit having one vertex and no edges is called a trivial circuit.I A trail or circuit is Eulerian if it uses every edge in the graph.I A cycle is a nontrivial circuit in which the only repeated node is the first/last one.

Lecture 15 6/ 21

a.t.DE !

Definitions: Example

1. The graph has 3 nodes and 5 edges

2. Edge 1 is a loop; edges 3 and 4 are parallel

3. Some walks in the graph:

(a) B, 5, C, 5, B, 2, A, 2, B is a closed walk that repeats edge 5 (so not a trail)(b) A, 2, B, 5, C is a path (no repeated vertices)(c) B is a trivial walk and a trivial circuit(d) A, 1, A, 2, B, 5, C, 3, A is a circuit starting and ending at A(e) A, 2, B, 5, C, 3, A, 1, A, 4, C is an Eulerian trail(f) The first part of (e), A, 2, B, 5, C, 3, A, is a cycle—unlike (d), node A appears

only twice

Challenge: Add one edge to the graph to create an Eulerian circuit

Lecture 15 7/ 21

Graphs in Applications

The graph that you use depends upon the application

Madrid

Rome London

Paris Madrid

Rome London

1173421

G1: Simple Graph G2: Directed Graph with Parallel Edges

Flight planning

I How do I get from Madrid to London? (G1 or G2)I Note: For G1, it su�ces just to list the sequence of vertices

I What flights will take me from Madrid to London? (G2)

Lecture 15 8/ 21

More Graph Notation

Definitions

1. A simple graph is a graph with no loops and no multiple (i.e., parallel) edges.

2. Notation [a, b]: an undirected edge with endpoints a and b

3. Notation (a, b): an directed edge going from a to b

Madrid

Rome London

Paris Madrid

Rome London

1173421

G1: Simple Graph G2: Directed Graph with Parallel Edges

Examples

I G1 edges are: [Madrid, Paris], [London, Paris], [Madrid, Rome], [Paris, Rome]

I G2 directed edges include: (Rome, Paris), (Paris, Rome), (Rome, Madrid)

I Note: There is never any ambiguity in a simple graph

Lecture 15 9/ 21

Graph Notation for Konigsberg ProblemDefinitions

1. An edge e is incident with a node v if and only if v is an endpoint of e

2. The degree of node v , denoted deg(v), is the number of times v appears as theendpoint of an edge. (It equals the number of edges that are incident with v ,except that loops are counted twice.)

3. A graph G is connected if there is a walk between any two nodes.

4. A graph H is is a subgraph of a graph G if all nodes and edges in H are alsonodes and edges in G .

5. A connected component of a graph G is a connected subgraph H of G suchthat no other connected subgraph of G containing H exists.

Example

I The degree of node 3 is 4: itappears in [1, 3], [3, 2], [5, 3], and[3, 4]. The degree of node 6 is 3

I There is no walk from node 4 tonode 6, so the graph is notconnected

I The graph has two connectedcomponents

Lecture 15 10/ 21

e- • v If

Graph Notation for Konigsberg ProblemDefinitions

Example

I The degree of node 3 is 4: itappears in [1, 3], [3, 2], [5, 3], and[3, 4]. The degree of node 6 is 3

I There is no walk from node 4 tonode 6, so the graph is notconnected

I The graph has two connectedcomponents

Lecture 15 10/ 21

Konigsberg Graph Notation: Example

Definitions

Example

I What is the degree of node 1?

I Find an Eulerian trail in thecomponent on the right

I Find an Eulerian circuit in thecomponent on the left

I Is the red subgraph a connectedcomponent?

Lecture 15 11/ 21

Eulerian Graphs

Definition

A graph G is Eulerian if it contains an Eulerian circuit.

Theorem 2

Let G be a connected graph. The graph G is Eulerian if and only if every node in G

has even degree.

The proof of this theorem uses induction. The basic ideas are illustrated in the nextexample. We reduce the problem of finding an Eulerian circuit in a big graph tofinding Eulerian circuits in several smaller graphs.

Lecture 15 12/ 21

How to Find an Eulerian Circuit

Given a connected graph G with all nodes of even degree:

1. Find any circuit C and form a graph G0 by removing from G all edges in the

circuit C

2. Observe that G 0 is not connected; call its components H1 and H2

3. H1 and H2 are smaller connected graphs, with all nodes having even degree(a) H1 has an Eulerian circuit C1 = 3, 4, 5, 6, 7, 11, 12, 13, 6, 3, 11, 13, 14, 3(b) H2 has an Eulerian circuit C2 = 8, 9, 10, 8

4. Piece together C , C1, and C2 to get the Eulerian circuit:I In C , replace 3 with C1 and 8 with C2:

1, 2, 3, 4, 5, 6, 7, 11, 12, 13, 6, 3, 11, 13, 14, 3| {z }

, 7, 8, 9, 10, 8| {z }

, 11, 14, 1

Lecture 15 13/ 21

circuit C2. Observe that G 0 is not connected; call its components H1 and H2

1, 2, 3, 4, 5, 6, 7, 11, 12, 13, 6, 3, 11, 13, 14, 3| {z }

, 7, 8, 9, 10, 8| {z }

, 11, 14, 1

Lecture 15 13/ 21

1, 2, 3, 4, 5, 6, 7, 11, 12, 13, 6, 3, 11, 13, 14, 3| {z }

, 7, 8, 9, 10, 8| {z }

, 11, 14, 1

Lecture 15 13/ 21

1, 2, 3, 4, 5, 6, 7, 11, 12, 13, 6, 3, 11, 13, 14, 3| {z }

, 7, 8, 9, 10, 8| {z }

, 11, 14, 1

Lecture 15 13/ 21

1, 2, 3, 4, 5, 6, 7, 11, 12, 13, 6, 3, 11, 13, 14, 3| {z }

, 7, 8, 9, 10, 8| {z }

, 11, 14, 1

Lecture 15 13/ 21

We can apply this idearecursively!

Graphs with Eulerian Trails

Can you draw the figure without lifting pencil from paper or retracing a line?

I Same as asking: Can you find an Eulerian trail?

I We are not asking if graph is Eulerian (we’re not trying to find a circuit)

Lecture 15 14/ 21

Two Simple Facts About GraphsTheorem 3

In any graph, the sum of the degrees of the vertices is equal to twice the number ofedges. I.e.,

i=1 deg(vi ) = 2m, where v1, v2, . . . , vn are the vertices and m is thenumber of edges.

Proof:I Each edge has two end points (not necessarily distinct)I Degree of a node counts # of times it appears as endpoint of an edgeI So when summing the degrees of nodes, we count each edge exactly twice

Corollary 4

In any graph, the number of nodes with odd degree is even.

Proof:

I The sum of two evens is even, sum of two odds is even, sum of even and odd is odd

I When calculating sum of all node degrees, the even numbers sum to an even number

I By Theorem 3, all the numbers sum to an even number

I Therefore, the odd numbers sum to an even number

I Therefore, there must be an even number of these

Lecture 15 15/ 21

Corollary 4

Proof:

Lecture 15 15/ 21

Corollary 4

Proof:

Lecture 15 15/ 21

Corollary 4

Proof:

Lecture 15 15/ 21

General Solution for Eulerian Trails

Theorem 5

A connected, non-Eulerian graph G has an Eulerian trail if and only if G has exactlytwo nodes of odd degree. The train must begin and end at these nodes.

Proof:

I Suppose that G is a connected graph that is not Eulerian but has an Eulerian trail from

v to w

I By Proposition 1, every node (except maybe v and w) has even degreeI Since graph is non-Eulerian, at least one node has odd degree (Theorem 2)I Therefore, at least two nodes have odd degree (Corollary 4)I These nodes must be v and w

I Now suppose that G is a connected graph with exactly two nodes, x and y , of odd

degree

I Form G0 by adding the edge [x , y ] to G

I All nodes in G0 have even degree, so G

0 has an Eulerian circuit (Theorem 2)I Deleting edge [x , y ] from the circuit gives an Eulerian trail in G from x to y

Lecture 15 16/ 21

C = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4, 3T = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4

Theorem 5

Proof:

v to w

degree

Lecture 15 16/ 21

C = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4, 3T = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4

Theorem 5

Proof:

v to w

I By Proposition 1, every node (except maybe v and w) has even degree

I Since graph is non-Eulerian, at least one node has odd degree (Theorem 2)I Therefore, at least two nodes have odd degree (Corollary 4)I These nodes must be v and w

degree

Lecture 15 16/ 21

C = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4, 3T = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4

Theorem 5

Proof:

v to w

I By Proposition 1, every node (except maybe v and w) has even degreeI Since graph is non-Eulerian, at least one node has odd degree (Theorem 2)

I Therefore, at least two nodes have odd degree (Corollary 4)I These nodes must be v and w

degree

Lecture 15 16/ 21

C = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4, 3T = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4

Theorem 5

Proof:

v to w

I By Proposition 1, every node (except maybe v and w) has even degreeI Since graph is non-Eulerian, at least one node has odd degree (Theorem 2)I Therefore, at least two nodes have odd degree (Corollary 4)

I These nodes must be v and w

degree

Lecture 15 16/ 21

C = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4, 3T = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4

Theorem 5

Proof:

v to w

degree

Lecture 15 16/ 21

C = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4, 3T = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4

Theorem 5

Proof:

v to w

degree

Lecture 15 16/ 21

C = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4, 3T = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4

Theorem 5

Proof:

v to w

degreeI Form G

0 by adding the edge [x , y ] to G

Lecture 15 16/ 21

C = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4, 3T = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4

Theorem 5

Proof:

v to w

degreeI Form G

0 has an Eulerian circuit (Theorem 2)

I Deleting edge [x , y ] from the circuit gives an Eulerian trail in G from x to y

Lecture 15 16/ 21

C = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4, 3T = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4

Theorem 5

Proof:

v to w

degreeI Form G

Lecture 15 16/ 21

C = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4, 3T = 3, 2, 1, 5, 4, 6, 2, 5, 6, 3, 4

Proofs about Graphs

We will look at two example proofs

I A direct proof

I An inductive proof of Theorem 2

We will use the techniques that we have learned, but in a new, slightly morecomplicated context

Lecture 15 17/ 21

A Result on Cycles and Vertices

Proposition 4

For every connected graph G with at least one edge, if G has no cycles, then G has atleast one vertex of degree 1.

Lecture 15 18/ 21

The Contrapositive ResultProposition 40 (Contrapositive)

For every connected graph G with at least one edge, if every vertex is of degree atleast 2, then G has a cycle.

ProofI Let G be a connected graph with at least one edge and every vertex of degree

� 2. Let n be the number of vertices in G , and choose v0 to be any vertex in G .

I Choose vertices to build a walk:

I Since v0 has degree � 2, we can pick v1 to be the other endpoint of anedge incident with v0.

I Since v1 has degree � 2, we can pick v2 to be the other endpoint of anedge incident with v1 other than [v0, v1].

I ...and so on, until vertex vn has been chosen.

I This walk uses n+ 1 vertices, but G only has n vertices, so there must be a firstvalue j such that vi = vj for some i < j . (The “pigeonhole principle”.)

I The cycle is then C = vi , vi+1, vi+2, . . . , vj .

Lecture 15 19/ 21

I Choose vertices to build a walk:

I Since v0 has degree � 2, we can pick v1 to be the other endpoint of anedge incident with v0.

Lecture 15 19/ 21

I Choose vertices to build a walk:I Since v0 has degree � 2, we can pick v1 to be the other endpoint of an

edge incident with v0.

Lecture 15 19/ 21

Inductive Proof of Theorem 2

Theorem 2

Let G be a connected graph with at least one edge. The graph G is Eulerian if andonly if every node in G has even degree.

1. If G is Eulerian, then every node has even degree

I Suppose G has an Eulerian circuit, and let x be any node in the graphI Write the Eulerian circuit as a walk v1e1v2e2 · · · vnenv1I Since G is connected, x must be one of the vertices in the walkI Each time (if any) that x appears in the middle of the walk, it is the endpoint for

two di↵erent edgesI If x = v1, then it is the endpoint for e1 and en

I So the degree of x is even

2. P(n): Every connected graph G with n edges and all even-degree nodes is Eulerian

I P(1) is trivially true (only such graph exists)I Assume P(1), . . .P(m � 1) all hold, and consider a graph G with m edgesI By Proposition 40, there is a cycle C in G

I Form graph G0 by removing the edges in C

I Each connected component H1, . . . ,Hk has nodes of even degree and—byinductive assumption—has an Eulerian circuit C1, . . . ,Ck

I Each of these must meet C , say, at v1, v2, . . . , vkI “Insert” each Ci into C to obtain desired circuit

(Arrange C to start and end at vi and write Ci in place of vi in C)

Lecture 15 20/ 21

Theorem 2

1. If G is Eulerian, then every node has even degree

I Suppose G has an Eulerian circuit, and let x be any node in the graphI Write the Eulerian circuit as a walk v1e1v2e2 · · · vnenv1I Since G is connected, x must be one of the vertices in the walkI Each time (if any) that x appears in the middle of the walk, it is the endpoint for

Lecture 15 20/ 21

Theorem 2

1. If G is Eulerian, then every node has even degreeI Suppose G has an Eulerian circuit, and let x be any node in the graph

I Write the Eulerian circuit as a walk v1e1v2e2 · · · vnenv1I Since G is connected, x must be one of the vertices in the walkI Each time (if any) that x appears in the middle of the walk, it is the endpoint for

Lecture 15 20/ 21

Theorem 2

1. If G is Eulerian, then every node has even degreeI Suppose G has an Eulerian circuit, and let x be any node in the graphI Write the Eulerian circuit as a walk v1e1v2e2 · · · vnenv1

I Since G is connected, x must be one of the vertices in the walkI Each time (if any) that x appears in the middle of the walk, it is the endpoint for

Lecture 15 20/ 21

Theorem 2

1. If G is Eulerian, then every node has even degreeI Suppose G has an Eulerian circuit, and let x be any node in the graphI Write the Eulerian circuit as a walk v1e1v2e2 · · · vnenv1I Since G is connected, x must be one of the vertices in the walk

I Each time (if any) that x appears in the middle of the walk, it is the endpoint fortwo di↵erent edges

I If x = v1, then it is the endpoint for e1 and en

Lecture 15 20/ 21

Theorem 2

1. If G is Eulerian, then every node has even degreeI Suppose G has an Eulerian circuit, and let x be any node in the graphI Write the Eulerian circuit as a walk v1e1v2e2 · · · vnenv1I Since G is connected, x must be one of the vertices in the walkI Each time (if any) that x appears in the middle of the walk, it is the endpoint for

two di↵erent edges

I If x = v1, then it is the endpoint for e1 and en

Lecture 15 20/ 21

Theorem 2

Lecture 15 20/ 21

Theorem 2

Lecture 15 20/ 21

Theorem 2

Lecture 15 20/ 21

Theorem 2

2. P(n): Every connected graph G with n edges and all even-degree nodes is EulerianI P(1) is trivially true (only such graph exists)

I Assume P(1), . . .P(m � 1) all hold, and consider a graph G with m edgesI By Proposition 40, there is a cycle C in G

Lecture 15 20/ 21

~ a loop•Q

Theorem 2

2. P(n): Every connected graph G with n edges and all even-degree nodes is EulerianI P(1) is trivially true (only such graph exists)I Assume P(1), . . .P(m � 1) all hold, and consider a graph G with m edges

I By Proposition 40, there is a cycle C in G

Lecture 15 20/ 21

Theorem 2

2. P(n): Every connected graph G with n edges and all even-degree nodes is EulerianI P(1) is trivially true (only such graph exists)I Assume P(1), . . .P(m � 1) all hold, and consider a graph G with m edgesI By Proposition 40, there is a cycle C in G

Lecture 15 20/ 21

Theorem 2

Lecture 15 20/ 21

Theorem 2

Lecture 15 20/ 21

Theorem 2

I Each of these must meet C , say, at v1, v2, . . . , vk

I “Insert” each Ci into C to obtain desired circuit(Arrange C to start and end at vi and write Ci in place of vi in C)

Lecture 15 20/ 21

Theorem 2

Lecture 15 20/ 21

Reminder of Earlier Example

Lecture 15 21/ 21

Graphs - UMass Amherstphaas/INFO150/slides/slecture15-anno… · circuit C 2. Observe that G0 is...

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