CS250: Discrete Math for Computer Science. L20: Complete Induction and Proof of Euler s Characterization of Eulerian-Walks

Transcription

1 CS250: Discrete Math for Computer Science L20: Complete Induction and Proof of Euler s Characterization of Eulerian-Walks

2 Last time: Eulerian Graphs Def. An Eulerian walk in a graph G is a walk from o t that traverses every edge exactly once and every vertex at least once. EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd Thm. [Euler] G has an Eulerian walk from o t iff EC(G).

3 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd

4 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd Claim G has an Eulerian walk EC(G).

5 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd Claim G has an Eulerian walk EC(G). Proof: For all vertices, v, besides s and t, the walk must leave v the same number of timehat it enters v. Thus, deg(v) is even.

6 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd Claim G has an Eulerian walk EC(G). Proof: For all vertices, v, besides s and t, the walk must leave v the same number of timehat it enters v. Thus, deg(v) is even. If s = t, then deg(s) = deg(t) is even for the same reason.

7 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd Claim G has an Eulerian walk EC(G). Proof: For all vertices, v, besides s and t, the walk must leave v the same number of timehat it enters v. Thus, deg(v) is even. If s = t, then deg(s) = deg(t) is even for the same reason. If s t, then s is left once more than it is reached and t is reached once more than it is left, so their degrees are both odd.

8 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd Claim G has an Eulerian walk EC(G). Proof: For all vertices, v, besides s and t, the walk must leave v the same number of timehat it enters v. Thus, deg(v) is even. If s = t, then deg(s) = deg(t) is even for the same reason. If s t, then s is left once more than it is reached and t is reached once more than it is left, so their degrees are both odd. Since an Eulerian walk visits every vertex, G is connected.

9 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd

10 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd Claim EC(G) G has an Eulerian walk.

11 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd Claim EC(G) G has an Eulerian walk. We will prove the Claim by induction: N = x α(x), where α(x) def = G ( E G x EC(G) G has an Eulerian walk )

12 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd Claim EC(G) G has an Eulerian walk. We will prove the Claim by induction: N = x α(x), where α(x) def = G ( E G x EC(G) G has an Eulerian walk ) base case: α(0): Let G be an arbitrary graph with 0 edges and EC(G). Since G is connected and has no edges it must consist of a single vertex, s = t. Thus, the empty walk is an Eulerian walk from o t.

13 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd α(x) def = G ( E G x EC(G) G has an Eulerian walk ) Claim N = x α(x). We are proving this by induction.

14 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd α(x) def = G ( E G x EC(G) G has an Eulerian walk ) Claim N = x α(x). We are proving this by induction. inductive case: Assume α(x 0 ) and try to prove α(x 0 + 1).

15 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd α(x) def = G ( E G x EC(G) G has an Eulerian walk ) Claim N = x α(x). We are proving this by induction. inductive case: Assume α(x 0 ) and try to prove α(x 0 + 1). Let G be an arbitrary graph with x edges and EC(G)

16 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd α(x) def = G ( E G x EC(G) G has an Eulerian walk ) Claim N = x α(x). We are proving this by induction. inductive case: Assume α(x 0 ) and try to prove α(x 0 + 1). Let G be an arbitrary graph with x edges and EC(G). Using α(x 0 ), construct an Eulerian walk through G

17 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd α(x) def = G ( E G x EC(G) G has an Eulerian walk ) Claim N = x α(x). We are proving this by induction. inductive case: Assume α(x 0 ) and try to prove α(x 0 + 1). Let G be an arbitrary graph with x edges and EC(G). Using α(x 0 ), construct an Eulerian walk through G An inductive proof is a recursive algorithm: using callo α(x 0 ), constructing Eulerian walks on graphs with at most x 0 edges, we must construct an Eulerian walk through G.

18 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd α(x 0 ) G ( E G x 0 EC(G) G has an Eulerian walk ) G is an arbitrary graph with x edges and EC(G).

19 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd α(x 0 ) G ( E G x 0 EC(G) G has an Eulerian walk ) G is an arbitrary graph with x edges and EC(G)

20 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd α(x 0 ) G ( E G x 0 EC(G) G has an Eulerian walk ) G is an arbitrary graph with x edges and EC(G) Let w be an exhaustive bb walk from s.

21 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd α(x 0 ) G ( E G x 0 EC(G) G has an Eulerian walk ) G is an arbitrary graph with x edges and EC(G) Let w be an exhaustive bb walk from s. w must end at t.

22 EC(G) def = G is connected v {s, t} deg(v) is even ( s = t deg(s) = deg(t) is even ) s t deg(s), deg(t) are odd α(x 0 ) G ( E G x 0 EC(G) G has an Eulerian walk ) G is an arbitrary graph with x edges and EC(G) Let w be an exhaustive bb walk from s. w must end at t

23 Assume EC(G).

24 Assume EC(G). From s, take an exhaustive bb walk on G.

25 Assume EC(G). From s, take an exhaustive bb walk on G t s

26 Assume EC(G). From s, take an exhaustive bb walk on G t s t s

27 Assume EC(G). From s, take an exhaustive bb walk on G t s t s t s

28 Assume EC(G). From s, take an exhaustive bb walk on G. You must end at t t s t s t s

29 Let w be an exhaustive bb walk from s. w must end at t

30 Let w be an exhaustive bb walk from s. w must end at t Let H def = G w

31 Let w be an exhaustive bb walk from s. w must end at t Let H def = G w E H x 0, and EC(C) for each connected component of H.

32 Let w be an exhaustive bb walk from s. w must end at t Let H def = G w E H x 0, and EC(C) for each connected component of H. By indhyp each C has an Eulerian walk.

33 Let w be an exhaustive bb walk from s. w must end at t Let H def = G w E H x 0, and EC(C) for each connected component of H. By indhyp each C has an Eulerian walk. (recursive call)

34 Let w be an exhaustive bb walk from s. w must end at t Let H def = G w E H x 0, and EC(C) for each connected component of H. By indhyp each C has an Eulerian walk. (recursive call) The union of these Eulerian walks is an Eulerian walk for G.

35 Let w be an exhaustive bb walk from s. w must end at t Let H def = G w E H x 0, and EC(C) for each connected component of H. By indhyp each C has an Eulerian walk. (recursive call) The union of these Eulerian walks is an Eulerian walk for G

36 iclicker 20.1 Did we need the assumption EC(G) to know that in G, a bb walk from s always ends at t? A: Yes B: No

37 iclicker 20.2 In the proof of Euler s characterization of Eulerian Graphs, why did we need α to say E G x 0 as opposed to E G = x 0? A: We didn t, E G = x 0 would have been enough. B: We wanted to apply the inductive hypothesio graphs with fewer than x vertices. We didn t have control of exactly how many edges were removed in going from G to the smaller graphs.

38 In complete induction, we strengthen the inductive hypothesis from α(x 0 ) to z x 0 (α(z)).

39 In complete induction, we strengthen the inductive hypothesis from α(x 0 ) to z x 0 (α(z)). Why ihis allowed?

40 In complete induction, we strengthen the inductive hypothesis from α(x 0 ) to z x 0 (α(z)). Why ihis allowed? In order to prove x α(x) by complete induction, we just prove x α c (x) by normal induction, where, α c (x) def = z x (α(z)).

41 In complete induction, we strengthen the inductive hypothesis from α(x 0 ) to z x 0 (α(z)). Why ihis allowed? In order to prove x α(x) by complete induction, we just prove x α c (x) by normal induction, where, α c (x) def = z x (α(z)). This little trick allows uo use the stronger complete inductive hypothesis whenever we want to!

42 In complete induction, we strengthen the inductive hypothesis from α(x 0 ) to z x 0 (α(z)). Why ihis allowed? In order to prove x α(x) by complete induction, we just prove x α c (x) by normal induction, where, α c (x) def = z x (α(z)). This little trick allows uo use the stronger complete inductive hypothesis whenever we want to! In the proof of Euler s characterization of Eulerian graphs, we used complete induction: α(x 0 ) G ( E G x 0 EC(G) G has an Eulerian walk )

43 iclicker 20.3 What ihe difference between complete induction and ordinary induction? A: Complete induction just consists of certain instances of ordinary induction, in which α(x) is of the form z x (ψ(z)). B: Complete induction is stronger than ordinary induction in the sense that there are worldhat are total orderings satisfying all of ordinary induction, but not complete induction.

44 v = V e = E f = # of faces connected plane graphs G 1 e f c a b d G v e f v e + f T 1 ɛ T 0 ɛ 0 1

45 v = V e = E f = # of faces connected plane graphs c G 1 e a d G v e f v e + f G f b T 1 ɛ T 0 ɛ 0 1

46 v = V e = E f = # of faces connected plane graphs c G 1 e f a d G v e f v e + f G T b T 1 ɛ T 0 ɛ 0 1

47 v = V e = E f = # of faces connected plane graphs c G 1 e f a d G v e f v e + f G T T b T 1 ɛ T 0 ɛ 0 1

48 T 2 ɛ

49 T 3 ɛ G

50 Thm: Euler s Formula [1750] Let G be an undirected, connected graph, drawn in the plane. Then v e + f = 2. We will prove this by induction next time.