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PROFESSOR: So today
we're going to continue

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00:00:24,680 --> 00:00:26,960
our course on the graph theory.

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00:00:26,960 --> 00:00:30,110
It's going to be a mixture
of all kinds of topics.

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We first start off
with Euler tour,

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and then we get
into directed graphs

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and cover the definitions.

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And we'll talk about
a special type, which

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are called tournament graphs.

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00:00:42,030 --> 00:00:45,040
And we'll do a whole
bunch of proofs,

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00:00:45,040 --> 00:00:50,220
and hopefully you will all
contribute to make this work

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00:00:50,220 --> 00:00:52,750
and think about how to do this.

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00:00:52,750 --> 00:00:56,110
So next week we will
continue with graph theory,

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and we will discuss
a very special type.

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We will use directed graphs
in communication networks.

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And on Thursday, we'll actually
use these special types

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00:01:09,070 --> 00:01:11,330
of graphs that we'll talk
about in a moment, DAGs.

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00:01:13,960 --> 00:01:17,250
So let's talk about Euler tour.

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00:01:17,250 --> 00:01:21,280
Euler, a famous mathematician,
he asked the question

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00:01:21,280 --> 00:01:24,865
like-- he lived in Konigsberg.

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00:01:24,865 --> 00:01:26,180
And there were seven bridges.

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He was wondering, can you
traverse all the bridges

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00:01:29,940 --> 00:01:30,810
exactly once.

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So he would start walking and
try to cross all those bridges.

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And apparently this little
islands in sort of the river

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00:01:37,310 --> 00:01:38,620
as well, and so on.

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00:01:38,620 --> 00:01:40,980
So how can you do that?

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00:01:40,980 --> 00:01:43,390
This is actually the
birth of graph theory.

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00:01:43,390 --> 00:01:47,030
And this particular
problem is named after him.

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00:01:47,030 --> 00:01:48,570
So what is an Euler tour?

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00:01:52,830 --> 00:01:59,920
It's actually defined
as a special walk.

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00:02:06,380 --> 00:02:14,270
It's a walk that traverses
every edge exactly once.

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00:02:23,110 --> 00:02:25,640
So it turns out the you
can actually characterize

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00:02:25,640 --> 00:02:28,210
these types of graphs.

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00:02:28,210 --> 00:02:31,330
So we are talking here
about undirected graph,

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00:02:31,330 --> 00:02:34,990
so continuation of last time.

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00:02:34,990 --> 00:02:37,690
So the edges that we
consider have no direction.

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00:02:37,690 --> 00:02:39,840
We'll come back to
that in a moment.

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00:02:39,840 --> 00:02:42,210
We will start
defining those later.

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00:02:42,210 --> 00:02:45,220
So Euler tour is a walk that
traverses every edge exactly

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00:02:45,220 --> 00:02:46,240
once.

48
00:02:46,240 --> 00:02:50,840
And at the same time,
it's also a tour.

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00:02:50,840 --> 00:02:58,700
So that means that it
actually starts and finishes

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00:02:58,700 --> 00:02:59,980
at the same vertex.

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00:03:10,050 --> 00:03:12,485
Now it turns out that
undirected graphs,

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00:03:12,485 --> 00:03:16,530
the graphs that we've
been talking about,

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00:03:16,530 --> 00:03:19,680
those that Euler tours can
be easily characterized.

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00:03:19,680 --> 00:03:23,460
And that's the theorem that
we're going to prove next.

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00:03:23,460 --> 00:03:28,390
The theorem states
that actually,

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00:03:28,390 --> 00:03:42,590
if a connected graph
has an Euler tour,

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00:03:42,590 --> 00:03:53,140
if and only if, every vertex
of the graph has even degree.

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00:03:56,120 --> 00:03:59,730
So that's a very nice and
simple and straightforward

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00:03:59,730 --> 00:04:00,560
characterization.

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00:04:00,560 --> 00:04:02,202
So how can we prove this?

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00:04:04,980 --> 00:04:06,070
So let's have a look.

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00:04:08,820 --> 00:04:14,140
So first of all, we have an
if and only if statement.

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00:04:14,140 --> 00:04:16,930
So we need to prove
two directions.

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We need to proof that if I
have a connected graph that

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00:04:20,200 --> 00:04:23,550
has an Euler tour, I need to
show that every vertex has

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00:04:23,550 --> 00:04:24,470
even degree.

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00:04:24,470 --> 00:04:28,000
And also the reverse-- if I have
a graph for which every vertex

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00:04:28,000 --> 00:04:31,080
has even degree, I need to
show that it has an Euler tour.

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So the proof consists
of two parts.

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00:04:34,590 --> 00:04:38,430
So let's first do
this implication,

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where we assure that
we have connected

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00:04:40,280 --> 00:04:42,660
graph that has an Euler tour.

73
00:04:42,660 --> 00:04:47,030
So assume we have a graph.

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00:04:47,030 --> 00:04:52,370
So we have G with the
vertex at V, edge is at E,

75
00:04:52,370 --> 00:04:55,810
and it has an Euler tour.

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00:04:55,810 --> 00:04:57,730
So what does it mean?

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00:04:57,730 --> 00:05:01,810
Well, it's really
means that we can walk

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00:05:01,810 --> 00:05:05,960
from some start vertex, V0.

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We can go all the way around to
V1, and some more, and so on,

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00:05:11,440 --> 00:05:15,170
to V, say, k minus 1.

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00:05:15,170 --> 00:05:19,510
And then end vertex is the
same as the start vertex.

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00:05:19,510 --> 00:05:24,060
So we have a walk that goes
all around the whole graph.

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00:05:24,060 --> 00:05:30,770
So every edge in this
walk are exactly the edges

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00:05:30,770 --> 00:05:31,740
that are in the graph.

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00:05:31,740 --> 00:05:36,700
And each edge in the graph
is offered exactly once.

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00:05:36,700 --> 00:05:38,200
So what does it mean?

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00:05:38,200 --> 00:05:40,240
So let's write
this down actually.

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00:05:40,240 --> 00:05:57,720
So since every edge is
traversed once-- every edge in E

89
00:05:57,720 --> 00:06:01,020
is traversed once-- what
can we conclude from that?

90
00:06:01,020 --> 00:06:11,980
Can we say something about,
given such as walk of length k,

91
00:06:11,980 --> 00:06:14,560
what can we say about the
number of edges, for example?

92
00:06:14,560 --> 00:06:19,777
What can we say about the
degree of the vertices?

93
00:06:19,777 --> 00:06:21,610
Because that's what you
want to show, right?

94
00:06:21,610 --> 00:06:25,270
You want to show that every
vertex has even degree.

95
00:06:25,270 --> 00:06:27,690
Does anyone know how
to go ahead here?

96
00:06:27,690 --> 00:06:31,670
So we know that every
single edge in E

97
00:06:31,670 --> 00:06:34,097
is referred ones
within this whole walk.

98
00:06:34,097 --> 00:06:35,805
So what kind of
properties can we derive?

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00:06:41,998 --> 00:06:44,430
AUDIENCE: [INAUDIBLE]

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00:06:44,430 --> 00:06:46,291
PROFESSOR: That's true.

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00:06:46,291 --> 00:06:47,166
AUDIENCE: [INAUDIBLE]

102
00:06:51,657 --> 00:06:54,152
PROFESSOR: Maybe someone
else can pick up here.

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00:06:54,152 --> 00:06:59,467
So every vertex that I
have here, I will enter it.

104
00:06:59,467 --> 00:07:00,300
But I also leave it.

105
00:07:00,300 --> 00:07:04,760
So if I see a vertex
in here, I can

106
00:07:04,760 --> 00:07:07,120
see at least two
contributing edges

107
00:07:07,120 --> 00:07:10,235
that are coming from this.

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00:07:10,235 --> 00:07:12,360
AUDIENCE: Then you know
that that note has at least

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00:07:12,360 --> 00:07:14,440
degree 2, but it
can't be more than 2

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00:07:14,440 --> 00:07:17,147
because otherwise that
would be the endpoint.

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00:07:17,147 --> 00:07:17,730
PROFESSOR: Ah.

112
00:07:17,730 --> 00:07:23,010
But is it true though
that-- so for example,

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00:07:23,010 --> 00:07:26,120
this particular note here may
repeat itself somewhere else.

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00:07:26,120 --> 00:07:29,830
I only have a condition
on the Euler tour

115
00:07:29,830 --> 00:07:34,340
that every edge
occurs exactly ones.

116
00:07:34,340 --> 00:07:37,062
So how could I formally
sort of continue to prove?

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00:07:37,062 --> 00:07:39,020
AUDIENCE: Well, it will
intersect itself again,

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00:07:39,020 --> 00:07:41,360
and you'll leave again
and it will still

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00:07:41,360 --> 00:07:42,394
have the even number.

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00:07:42,394 --> 00:07:43,060
PROFESSOR: Yeah.

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00:07:43,060 --> 00:07:43,770
That's true.

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00:07:43,770 --> 00:07:45,410
That's right.

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00:07:45,410 --> 00:07:46,895
Any other additions?

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00:07:46,895 --> 00:07:48,875
AUDIENCE: You can define
the number of edges

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00:07:48,875 --> 00:07:50,855
that it will test by
how many you enter it,

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00:07:50,855 --> 00:07:52,835
so the number of
edges that it has

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00:07:52,835 --> 00:07:56,129
is twice the number
of [INAUDIBLE].

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00:07:56,129 --> 00:07:56,795
PROFESSOR: Yeah.

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00:07:56,795 --> 00:08:00,260
That's correct. [INAUDIBLE]

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00:08:00,260 --> 00:08:06,200
AUDIENCE: [INAUDIBLE] traverse
once, then every time you enter

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00:08:06,200 --> 00:08:07,685
a node you have to leave it.

132
00:08:07,685 --> 00:08:09,912
So you can say that
you're never going

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00:08:09,912 --> 00:08:18,410
to leave a node for a node that
it already left for from there.

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00:08:18,410 --> 00:08:21,410
And you're never going to enter
from a node you already entered

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00:08:21,410 --> 00:08:22,743
from or left to from there.

136
00:08:22,743 --> 00:08:24,742
So that's how you can say
that you're only going

137
00:08:24,742 --> 00:08:28,412
to increment your degree by 2.

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00:08:28,412 --> 00:08:29,370
PROFESSOR: That's true.

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00:08:29,370 --> 00:08:30,790
So what he's
essentially saying is

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00:08:30,790 --> 00:08:32,419
that every edge as
I start to count

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00:08:32,419 --> 00:08:35,299
here, in this particular tour,
well they're all different.

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00:08:35,299 --> 00:08:39,179
So they all counts towards
a degree of the node.

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00:08:39,179 --> 00:08:43,590
So everything that you have
been saying is right on.

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00:08:43,590 --> 00:08:48,560
What we can conclude here is
that if you look at a vertex,

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00:08:48,560 --> 00:09:00,230
say vertex U-- for example,
somewhere over here-- well,

146
00:09:00,230 --> 00:09:02,240
it may repeat itself over here.

147
00:09:02,240 --> 00:09:04,990
And it has an incoming and
outgoing edge, incoming

148
00:09:04,990 --> 00:09:07,140
and outgoing edge.

149
00:09:07,140 --> 00:09:08,780
And I may have it
somewhere else,

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00:09:08,780 --> 00:09:11,040
but say it's just at those two.

151
00:09:11,040 --> 00:09:13,570
Well, I know that all
the edges are different.

152
00:09:13,570 --> 00:09:16,220
So I can clearly count this now.

153
00:09:16,220 --> 00:09:20,290
I can say that the
degree of U must

154
00:09:20,290 --> 00:09:27,020
be equal to the number
of times that U actually

155
00:09:27,020 --> 00:09:30,900
appears in this tour.

156
00:09:37,400 --> 00:09:41,540
So in the tour from V0
all the way to Vk minus 1.

157
00:09:41,540 --> 00:09:45,447
And then I have to
multiply times 2.

158
00:09:45,447 --> 00:09:47,780
And what I said, it's exactly
what you have been saying,

159
00:09:47,780 --> 00:09:51,640
all the edges occur exactly
once of the whole graph.

160
00:09:51,640 --> 00:09:54,470
So I just count the number
of times I see U in here.

161
00:09:54,470 --> 00:09:57,620
I come the number of edges that
go into it and leave from it.

162
00:09:57,620 --> 00:10:01,860
Well, that's twice times
the number of times

163
00:10:01,860 --> 00:10:04,060
that U actually
appears in this tour.

164
00:10:04,060 --> 00:10:07,560
So this implication is
pretty straightforward

165
00:10:07,560 --> 00:10:12,630
because now we actually note
that the degree of U is even.

166
00:10:12,630 --> 00:10:14,130
So that's great.

167
00:10:14,130 --> 00:10:17,390
So let's talk about
the other implication

168
00:10:17,390 --> 00:10:21,490
and see whether we can find
a trick to make that happen.

169
00:10:21,490 --> 00:10:25,150
So what do we need
to start off with?

170
00:10:25,150 --> 00:10:29,020
Well, now we have to assume that
every vertex has even degree.

171
00:10:29,020 --> 00:10:31,010
So let's write this down.

172
00:10:31,010 --> 00:10:32,210
So say we have the graph.

173
00:10:35,030 --> 00:10:40,010
And for this graph
we actually issue

174
00:10:40,010 --> 00:10:45,805
that the degree of
every vertex V is even.

175
00:10:52,670 --> 00:10:53,780
Well, what can we do?

176
00:10:58,179 --> 00:10:59,720
Well, this is sort
of creative trick,

177
00:10:59,720 --> 00:11:01,970
so let me just continue here.

178
00:11:01,970 --> 00:11:07,870
So what we have is if you
start to consider a walk, W,

179
00:11:07,870 --> 00:11:15,830
that touches, say, a lot
of vertices in such a way

180
00:11:15,830 --> 00:11:20,500
that W is actually
the longest walk.

181
00:11:20,500 --> 00:11:23,130
So that's often what
you do in graph theory.

182
00:11:23,130 --> 00:11:25,866
You think about, say,
the shortest path

183
00:11:25,866 --> 00:11:28,240
or whatever with a certain
property, or the longest walk,

184
00:11:28,240 --> 00:11:29,960
or whatever with a
certain property.

185
00:11:29,960 --> 00:11:31,760
So that's what
you're doing here.

186
00:11:31,760 --> 00:11:33,670
So let W be the longest walk.

187
00:11:38,020 --> 00:11:42,420
And well, we want to prove
something about an Euler tour.

188
00:11:42,420 --> 00:11:44,480
An Euler tour has a
very specific property

189
00:11:44,480 --> 00:11:46,420
that all the edges
occur only once.

190
00:11:46,420 --> 00:11:51,220
So it makes sense
to look at walks

191
00:11:51,220 --> 00:11:56,280
in which all the edges of
which it walks are unique,

192
00:11:56,280 --> 00:11:57,980
are occurring only once.

193
00:11:57,980 --> 00:12:07,950
So we're interested in the
longest walk that traverses

194
00:12:07,950 --> 00:12:15,930
no edge more than once.

195
00:12:15,930 --> 00:12:18,430
So that's important.

196
00:12:22,060 --> 00:12:25,510
Now, let's think about
this a little bit.

197
00:12:25,510 --> 00:12:29,230
So the first property that you
may want to consider it-- you

198
00:12:29,230 --> 00:12:32,230
really want to show that W
is going to be an Euler tour.

199
00:12:32,230 --> 00:12:34,880
That's what we're going
to try to prove here.

200
00:12:34,880 --> 00:12:36,560
So the first thing
we may want to show

201
00:12:36,560 --> 00:12:39,270
is that actually
it goes all around.

202
00:12:39,270 --> 00:12:42,090
So V0 is equal to Vk.

203
00:12:42,090 --> 00:12:44,540
That will be great
if we can do this.

204
00:12:44,540 --> 00:12:48,720
So we want to show
that this is true.

205
00:12:48,720 --> 00:12:52,645
So what would happen if
this would not be the case?

206
00:12:55,580 --> 00:13:00,380
So suppose these two would
not be equal to one another.

207
00:13:05,860 --> 00:13:11,010
Well, actually, I'm skipping
a little bit ahead here

208
00:13:11,010 --> 00:13:14,570
in the proof, I notice.

209
00:13:14,570 --> 00:13:17,380
So this will be the conclusion
of another observation.

210
00:13:17,380 --> 00:13:22,940
So let me start with
the first observation.

211
00:13:22,940 --> 00:13:27,820
So first of all, let us
consider this end node over

212
00:13:27,820 --> 00:13:30,300
here, this end vertex, Vk.

213
00:13:30,300 --> 00:13:37,800
Now if there is an edge that
is not covered in this walk--

214
00:13:37,800 --> 00:13:39,860
and I can lengthen
the walk, right?

215
00:13:39,860 --> 00:13:44,080
So if I have, say,
another edge, which

216
00:13:44,080 --> 00:13:51,730
looks like Vk, an edge
from Vk to some U, where

217
00:13:51,730 --> 00:13:56,970
this edge is not in the walk.

218
00:13:59,600 --> 00:14:02,820
Or the other one
is not in the walk.

219
00:14:02,820 --> 00:14:06,450
Well, it doesn't matter in
an indirect graph, of course.

220
00:14:06,450 --> 00:14:08,630
So if this is not in the
walk, then what can I do?

221
00:14:08,630 --> 00:14:10,140
I can just lengthen it, right?

222
00:14:10,140 --> 00:14:15,190
I can just say V0 goes
to V1 goes to Vk goes

223
00:14:15,190 --> 00:14:19,590
to U. Now I have a longer walk.

224
00:14:19,590 --> 00:14:22,340
And that's a contradiction.

225
00:14:22,340 --> 00:14:31,660
So I know that all the edges
in which Vk is participating

226
00:14:31,660 --> 00:14:33,910
are actually
covered by the walk.

227
00:14:33,910 --> 00:14:35,360
So that's a great property.

228
00:14:35,360 --> 00:14:36,930
So let's write this down.

229
00:14:36,930 --> 00:14:52,150
So all edges that
are incident to Vk

230
00:14:52,150 --> 00:15:00,370
are actually used--
traversed-- in the walk, W.

231
00:15:00,370 --> 00:15:02,950
So now let's have
a look whether we

232
00:15:02,950 --> 00:15:08,710
can show that it's a walk
that goes all the way around.

233
00:15:08,710 --> 00:15:12,460
So let's try to
prove this statement.

234
00:15:12,460 --> 00:15:17,060
So let me repeat this over here.

235
00:15:17,060 --> 00:15:22,200
So what we want to show
is that Vk equals V0.

236
00:15:22,200 --> 00:15:25,260
So given this statement,
do you have any idea

237
00:15:25,260 --> 00:15:27,210
how we could
possibly prove this,

238
00:15:27,210 --> 00:15:31,170
also given what we have
been doing over here?

239
00:15:31,170 --> 00:15:33,230
Suppose this would not
be the case, right?

240
00:15:35,910 --> 00:15:42,240
So suppose the start and
the end node in this walk

241
00:15:42,240 --> 00:15:45,970
are not equal to one another.

242
00:15:45,970 --> 00:15:50,180
So what can we do here?

243
00:15:50,180 --> 00:15:53,500
So maybe there's
some suggestions?

244
00:15:53,500 --> 00:15:56,560
I already had you once.

245
00:15:56,560 --> 00:15:57,430
Maybe someone else?

246
00:16:01,624 --> 00:16:05,060
AUDIENCE: It has to
attach to another vertex.

247
00:16:05,060 --> 00:16:07,460
And you know that all
the degrees are even.

248
00:16:07,460 --> 00:16:13,322
So it means that Vk
must attach to V0.

249
00:16:13,322 --> 00:16:15,280
PROFESSOR: So what you're
saying, first of all,

250
00:16:15,280 --> 00:16:19,660
is that if this would
not be the case, then

251
00:16:19,660 --> 00:16:26,006
you know that the
degrees must be even.

252
00:16:26,006 --> 00:16:27,630
But suppose this
would not be the case.

253
00:16:27,630 --> 00:16:32,091
That would mean actually that
the degree would be odd, right?

254
00:16:32,091 --> 00:16:32,590
OK.

255
00:16:32,590 --> 00:16:34,600
So let's think about
that a little bit.

256
00:16:34,600 --> 00:16:36,380
So let's write this down.

257
00:16:36,380 --> 00:16:42,010
So otherwise the
Vk has odd degree.

258
00:16:42,010 --> 00:16:46,330
But the only node has odd
degree in this particular walk,

259
00:16:46,330 --> 00:16:50,140
W. So that's what we do now.

260
00:16:50,140 --> 00:16:56,410
Because we really
consider this walk.

261
00:17:01,490 --> 00:17:07,260
So we see that if I have
V0 unequal to Vk and Vk

262
00:17:07,260 --> 00:17:10,240
maybe repeated in this
walk a number of times,

263
00:17:10,240 --> 00:17:12,400
each time it is repeated
it has an incoming

264
00:17:12,400 --> 00:17:14,390
and an outgoing edge.

265
00:17:14,390 --> 00:17:18,510
And all these edges are
different because they do not

266
00:17:18,510 --> 00:17:19,790
occur more than once.

267
00:17:19,790 --> 00:17:24,680
So whenever Vk enters in
this middle part over here--

268
00:17:24,680 --> 00:17:28,520
it's partaking over here--
it gives an even contribution

269
00:17:28,520 --> 00:17:30,830
to the number of
edges in this walk,

270
00:17:30,830 --> 00:17:34,050
and it has one extra
incoming edge over here.

271
00:17:34,050 --> 00:17:38,750
So if V0 is not
equal to Vk, we will

272
00:17:38,750 --> 00:17:45,500
have an odd degree
of edges in W,

273
00:17:45,500 --> 00:17:49,010
in which Vk is participating.

274
00:17:49,010 --> 00:17:52,500
So now we can go ahead and
we can have a look over here

275
00:17:52,500 --> 00:17:57,100
because we just showed here that
all the edges incident to Vk

276
00:17:57,100 --> 00:18:01,790
are actually used in W. So
we can now conclude that this

277
00:18:01,790 --> 00:18:04,920
means that Vk also
has odd degree in G

278
00:18:04,920 --> 00:18:09,540
because all the edges in G,
in which Vk is participating,

279
00:18:09,540 --> 00:18:11,810
are actually used in the walk.

280
00:18:11,810 --> 00:18:12,920
So I hope that is clear.

281
00:18:15,670 --> 00:18:18,840
So let's write this down.

282
00:18:18,840 --> 00:18:24,070
So we'll use the first
statement over here, so by 1.

283
00:18:24,070 --> 00:18:32,410
We have that Vk has
odd degree in G.

284
00:18:32,410 --> 00:18:34,160
And now we come to
what you were saying.

285
00:18:34,160 --> 00:18:38,490
We have assumed over here
that the degree is even.

286
00:18:38,490 --> 00:18:42,000
So this cannot be the case,
so that's a contradiction.

287
00:18:42,000 --> 00:18:45,740
So we know that the otherwise
situation cannot occur.

288
00:18:45,740 --> 00:18:48,960
So we know that Vk
must be equal to V0.

289
00:18:48,960 --> 00:18:51,690
So that's proving
that we actually have

290
00:18:51,690 --> 00:18:54,380
a walk that goes all around.

291
00:18:54,380 --> 00:18:56,710
Now, we are not yet done.

292
00:18:56,710 --> 00:19:00,790
Because why would
this be an Euler tour?

293
00:19:00,790 --> 00:19:02,465
For an Euler tour,
we really want

294
00:19:02,465 --> 00:19:08,390
that every single
edge that G has

295
00:19:08,390 --> 00:19:11,450
is actually used in the tour.

296
00:19:11,450 --> 00:19:14,850
So we need much more.

297
00:19:14,850 --> 00:19:19,608
So let's use this board.

298
00:19:35,420 --> 00:19:42,460
So suppose that W is
not an Euler tour.

299
00:19:42,460 --> 00:19:45,720
We're going to
use contradiction.

300
00:19:45,720 --> 00:19:48,630
But we already have shown
this particular property,

301
00:19:48,630 --> 00:19:50,060
so we can use this.

302
00:19:50,060 --> 00:19:52,720
So suppose W is
not an Euler tour.

303
00:19:52,720 --> 00:19:56,710
Well, we know that
G is connected.

304
00:19:56,710 --> 00:20:00,020
So that's a good
thing because that

305
00:20:00,020 --> 00:20:03,740
means that if it is
not an Euler tour,

306
00:20:03,740 --> 00:20:12,050
there is an edge that is
not used in W, in the walk,

307
00:20:12,050 --> 00:20:28,730
but it is incident to some
vertex that is used in W.

308
00:20:28,730 --> 00:20:30,510
So this is where the
connectivity here

309
00:20:30,510 --> 00:20:31,800
comes in play.

310
00:20:31,800 --> 00:20:37,580
Now let us call
this edge-- so let

311
00:20:37,580 --> 00:20:40,570
U-Vi be this particular edge.

312
00:20:43,710 --> 00:20:48,170
So U is not used in
the walk-- well, it's

313
00:20:48,170 --> 00:20:52,130
maybe used in the walk, but the
edge is not used in the walk.

314
00:20:52,130 --> 00:20:54,350
And Vi is part of the walk.

315
00:20:54,350 --> 00:20:57,490
So what can we do here?

316
00:20:57,490 --> 00:21:00,030
So does anybody see
what we could do here?

317
00:21:00,030 --> 00:21:03,140
Can we somehow get
a contradiction?

318
00:21:03,140 --> 00:21:06,760
So when we started
out here, we actually

319
00:21:06,760 --> 00:21:12,480
assumed that W is the longest
walk in which the edges do not

320
00:21:12,480 --> 00:21:16,270
occur more than once.

321
00:21:16,270 --> 00:21:19,660
Can we create a longer
walk by simply using this?

322
00:21:19,660 --> 00:21:26,606
Now, notice that we did
prove that V0 goes all around

323
00:21:26,606 --> 00:21:30,410
to Vk, and so on.

324
00:21:30,410 --> 00:21:34,670
So can we find a longer walk
that uses this extra edge?

325
00:21:38,020 --> 00:21:40,447
Any ideas?

326
00:21:40,447 --> 00:21:45,630
What would happen if we, for
example, started a walk with U?

327
00:21:45,630 --> 00:21:47,610
Let's see how that
would work out.

328
00:21:47,610 --> 00:21:51,570
We can have U. We walk to Vi.

329
00:21:51,570 --> 00:21:55,720
Well, let's just simply
start walking all around.

330
00:21:55,720 --> 00:22:00,865
So essentially we have Vi
over here and U over here.

331
00:22:03,630 --> 00:22:08,620
So we could do is we can
start walking like this

332
00:22:08,620 --> 00:22:10,290
and then end over here.

333
00:22:10,290 --> 00:22:13,310
And notice we have
used one extra edge.

334
00:22:13,310 --> 00:22:15,070
So we get a longer walk.

335
00:22:15,070 --> 00:22:18,940
So we go to Vi plus 1,
and we go all the way

336
00:22:18,940 --> 00:22:23,230
up to Vk, which is equal to V0.

337
00:22:23,230 --> 00:22:27,870
And then we go up to V1,
all the way up to Vi.

338
00:22:27,870 --> 00:22:29,960
This is a longer walk.

339
00:22:29,960 --> 00:22:34,240
And therefore we
have a contradiction.

340
00:22:34,240 --> 00:22:37,600
And now the proof is
finished because that means

341
00:22:37,600 --> 00:22:39,740
that this assumption was wrong.

342
00:22:39,740 --> 00:22:41,560
So W actually is an Euler tour.

343
00:22:41,560 --> 00:22:44,390
So now we have shown the
existence of an Euler tour,

344
00:22:44,390 --> 00:22:46,390
and that was the other
direction of the theorem.

345
00:22:49,800 --> 00:22:56,130
So let's now start with a
new topic on directed graphs.

346
00:22:56,130 --> 00:23:01,140
And later on, we will talk
about how Hamiltonian paths.

347
00:23:01,140 --> 00:23:03,390
And these are
different from an Euler

348
00:23:03,390 --> 00:23:07,990
tour in that in an Euler tour,
every edge is used exactly

349
00:23:07,990 --> 00:23:08,490
once.

350
00:23:08,490 --> 00:23:11,270
In a Hamiltonian path, we
will have that every vertex

351
00:23:11,270 --> 00:23:13,160
is used exactly once.

352
00:23:13,160 --> 00:23:17,100
So we'll do that a bit later.

353
00:23:17,100 --> 00:23:18,660
So what are directed graphs?

354
00:23:18,660 --> 00:23:22,860
Directed graphs are
graphs where we have edges

355
00:23:22,860 --> 00:23:24,410
that have a specific direction.

356
00:23:24,410 --> 00:23:29,030
So we can walk from one
vertex to the other.

357
00:23:32,760 --> 00:23:34,425
Let me just depict an example.

358
00:23:38,050 --> 00:23:42,420
For example, we could have
something that looks like this.

359
00:23:42,420 --> 00:23:43,960
We have V1.

360
00:23:43,960 --> 00:23:47,950
We have V2, V3.

361
00:23:47,950 --> 00:23:51,360
We may have an edge that
goes like this with an arrow.

362
00:23:51,360 --> 00:23:53,720
We use arrows in this case.

363
00:23:53,720 --> 00:23:55,710
I can go to V2 and V3.

364
00:23:55,710 --> 00:24:00,920
From V2 I may be able to go
to V3, and from V3 to V2.

365
00:24:00,920 --> 00:24:04,100
So this will be a directed
graph, as an example.

366
00:24:04,100 --> 00:24:05,640
We also call these digraphs.

367
00:24:09,680 --> 00:24:14,820
And as you can see, we may have
an edge that goes from V2 to V3

368
00:24:14,820 --> 00:24:17,330
and also one that
goes from V3 to V2.

369
00:24:17,330 --> 00:24:19,690
So these are two separate edges.

370
00:24:19,690 --> 00:24:21,960
So every edge has a direction.

371
00:24:21,960 --> 00:24:27,140
And we usually say that
if we have an edge that

372
00:24:27,140 --> 00:24:35,210
points from V2 to, say, V3,
then we call this the tail

373
00:24:35,210 --> 00:24:37,530
and over here we have the head.

374
00:24:37,530 --> 00:24:39,600
So this is some notation
that you may use.

375
00:24:43,840 --> 00:24:45,520
We also have now
a different notion

376
00:24:45,520 --> 00:24:49,420
for the degree of a vertex
because we have different types

377
00:24:49,420 --> 00:24:51,260
of edges, essentially.

378
00:24:51,260 --> 00:24:53,330
Take, for example, V2.

379
00:24:53,330 --> 00:24:56,235
You have incoming edges
and outgoing edges.

380
00:24:56,235 --> 00:24:57,610
So that's why
we're going to talk

381
00:24:57,610 --> 00:25:03,120
about an indegree
and an outdegree.

382
00:25:03,120 --> 00:25:07,210
So, for example, the indegree
of V2 is equal to, well, I

383
00:25:07,210 --> 00:25:10,380
have one, two incoming edges.

384
00:25:10,380 --> 00:25:17,060
And the outdegree of V2 is
actually something different.

385
00:25:17,060 --> 00:25:18,790
It's just one outgoing edge.

386
00:25:18,790 --> 00:25:22,010
So this is equal to 1.

387
00:25:22,010 --> 00:25:24,560
So this is some notation.

388
00:25:24,560 --> 00:25:28,930
So let's talk about walks.

389
00:25:28,930 --> 00:25:30,850
Let's figure out
where we can enumerate

390
00:25:30,850 --> 00:25:35,940
walks in a directed
graph and compute those.

391
00:25:35,940 --> 00:25:38,830
So how many walks
do I have, say,

392
00:25:38,830 --> 00:25:43,180
of length k that go exactly
from, say, V1 to V3.

393
00:25:43,180 --> 00:25:45,290
How can I compute this?

394
00:25:45,290 --> 00:25:48,840
So I'd like to introduce
adjacency matrices.

395
00:25:48,840 --> 00:25:50,540
You've probably
seen them before.

396
00:25:50,540 --> 00:25:53,510
But let's go over this once
more because induction is also

397
00:25:53,510 --> 00:25:55,470
important.

398
00:25:55,470 --> 00:25:57,420
So let's do the theorem.

399
00:25:57,420 --> 00:26:00,110
So the theorem is this.

400
00:26:00,110 --> 00:26:01,260
Suppose we have a graph.

401
00:26:05,560 --> 00:26:07,495
And suppose it has end nodes.

402
00:26:16,170 --> 00:26:22,250
And we have the
vertices V1 up to Vn.

403
00:26:22,250 --> 00:26:33,650
And now we let the matrix that
contains the entries aij denote

404
00:26:33,650 --> 00:26:46,860
the adjacency matrix for G.

405
00:26:46,860 --> 00:26:49,910
And what does it mean?

406
00:26:49,910 --> 00:26:52,550
It's actually means
that in this case,

407
00:26:52,550 --> 00:27:01,360
we say that aij is equal
to 1 if we actually

408
00:27:01,360 --> 00:27:06,520
have an edge that
goes from Vi to Vj--

409
00:27:06,520 --> 00:27:15,000
so this is an edge-- and
0 if this is not the case.

410
00:27:15,000 --> 00:27:17,630
So this is the adjacency matrix.

411
00:27:17,630 --> 00:27:22,460
And now we can state something
about the number of directed

412
00:27:22,460 --> 00:27:28,150
walks in a directed graph.

413
00:27:28,150 --> 00:27:31,300
So it turns out
that this is easily

414
00:27:31,300 --> 00:27:36,455
computed by taking powers
of the adjacency matrix.

415
00:27:47,890 --> 00:27:53,800
Let aij with the
superscript k be

416
00:27:53,800 --> 00:28:06,130
equal to the number of
directed walks of length k

417
00:28:06,130 --> 00:28:13,530
that go from Vi to Vj.

418
00:28:13,530 --> 00:28:18,110
So you want to compute
this and it turns out

419
00:28:18,110 --> 00:28:22,000
that-- wait a minute.

420
00:28:22,000 --> 00:28:25,140
So what did I do here?

421
00:28:25,140 --> 00:28:29,190
So actually let p
kij be the number

422
00:28:29,190 --> 00:28:34,440
of directed walks of length
k from phi i to phi j.

423
00:28:34,440 --> 00:28:38,600
Then what you can show is
if you look at matrix A

424
00:28:38,600 --> 00:28:44,110
to the power k, this
actually is the matrix that

425
00:28:44,110 --> 00:28:50,190
contains all these numbers.

426
00:28:50,190 --> 00:28:51,710
So let me give a few examples.

427
00:28:54,710 --> 00:28:57,670
So let's take the
matrix over here.

428
00:28:57,670 --> 00:29:00,240
Look at this ajc matrix,
take a few powers

429
00:29:00,240 --> 00:29:02,030
and see what happens.

430
00:29:02,030 --> 00:29:04,210
So the matrix looks like this.

431
00:29:04,210 --> 00:29:09,590
If you just label the rows
by, say, V1, V2, and V3.

432
00:29:09,590 --> 00:29:13,340
The columns by V1, V2, and V3.

433
00:29:13,340 --> 00:29:16,020
Well, V1 has an edge to V1.

434
00:29:16,020 --> 00:29:19,360
V1 has an edge to V2
and also one to V3.

435
00:29:19,360 --> 00:29:24,560
V2 has only one edge to
V3, so we have zeros here.

436
00:29:24,560 --> 00:29:26,710
And we have this.

437
00:29:26,710 --> 00:29:30,210
So this is the matrix A.

438
00:29:30,210 --> 00:29:33,180
For example, if you
compute a squared.

439
00:29:33,180 --> 00:29:34,510
So how do we do this?

440
00:29:34,510 --> 00:29:37,210
How many of you know about
matrix multiplication?

441
00:29:37,210 --> 00:29:38,160
Everybody knows?

442
00:29:41,310 --> 00:29:44,005
Well anyway, so let's assume
that you do know actually.

443
00:29:47,390 --> 00:29:48,560
So that's pretty simple.

444
00:29:48,560 --> 00:29:49,580
So you take the column.

445
00:29:49,580 --> 00:29:52,110
You take the inners product
with the first row, and so on.

446
00:29:52,110 --> 00:29:54,840
And you can easily
compute this otherwise

447
00:29:54,840 --> 00:29:57,810
you may want to do
it yourself later on.

448
00:29:57,810 --> 00:30:02,270
And you get this
particular matrix.

449
00:30:02,270 --> 00:30:05,410
And, for example,
a to the power of 3

450
00:30:05,410 --> 00:30:12,900
is something very similar--
1, 3, 3, 0, 0, 1, 0, 1, 0.

451
00:30:12,900 --> 00:30:15,700
Well, it turns out that
if you look at this graph

452
00:30:15,700 --> 00:30:17,970
and we, for example, want
to know the number of walks

453
00:30:17,970 --> 00:30:22,380
of length 3 that go
from V1 to, say, V2,

454
00:30:22,380 --> 00:30:29,370
let's see whether we
can see those over here.

455
00:30:29,370 --> 00:30:33,830
Well, if I travel from
V1 to V2 in three steps,

456
00:30:33,830 --> 00:30:36,600
I can go like this.

457
00:30:36,600 --> 00:30:39,350
I can go one, two, three.

458
00:30:39,350 --> 00:30:41,010
So that's one option.

459
00:30:41,010 --> 00:30:44,240
I can go one, and
another one time,

460
00:30:44,240 --> 00:30:46,470
so it's a second
step, and three.

461
00:30:46,470 --> 00:30:49,920
I can also go directly over here
and go back here, and back over

462
00:30:49,920 --> 00:30:50,960
here.

463
00:30:50,960 --> 00:30:53,620
So it turns out that's
I can compute this.

464
00:30:53,620 --> 00:30:55,580
So how do we prove this?

465
00:30:55,580 --> 00:30:58,370
We'll use induction.

466
00:30:58,370 --> 00:31:06,130
And the steps are
pretty straightforward.

467
00:31:06,130 --> 00:31:09,020
So let's first
define, because we

468
00:31:09,020 --> 00:31:12,730
want to prove that kth power
of a is equal to that matrix

469
00:31:12,730 --> 00:31:19,350
up there, where all the entries
represent the number of walks.

470
00:31:19,350 --> 00:31:32,690
So let's say that aijk,
that this denote the i jth

471
00:31:32,690 --> 00:31:38,680
entry in a to the power k.

472
00:31:42,610 --> 00:31:45,510
And you want to show that
this number is actually

473
00:31:45,510 --> 00:31:51,930
equal to pijk for
all the entries.

474
00:31:51,930 --> 00:31:55,620
So if you're going
to use induction,

475
00:31:55,620 --> 00:32:03,440
it makes sense to assume that
the theorem is true for k.

476
00:32:03,440 --> 00:32:10,000
So the induction hypothesis
could be something like this.

477
00:32:10,000 --> 00:32:16,700
The theorem is true for k.

478
00:32:16,700 --> 00:32:20,900
And this is really
the same as stating

479
00:32:20,900 --> 00:32:32,370
that all these entries
for all i and j-- aij

480
00:32:32,370 --> 00:32:34,208
is equal to the pijk.

481
00:32:37,310 --> 00:32:39,710
So this is what
you want to prove.

482
00:32:39,710 --> 00:32:41,780
And this is pretty
straightforward

483
00:32:41,780 --> 00:32:44,220
because now what
we can do is we can

484
00:32:44,220 --> 00:32:52,020
start to look at how we compute
walks of length k, or k plus 1.

485
00:32:52,020 --> 00:32:54,110
So let's first do the base case.

486
00:32:54,110 --> 00:32:56,680
That's how we always start.

487
00:32:56,680 --> 00:32:59,210
The base case is k equals 1.

488
00:32:59,210 --> 00:33:03,850
And we have essentially
two options.

489
00:33:03,850 --> 00:33:06,560
So we want to prove this,
so take an i and a j,

490
00:33:06,560 --> 00:33:08,340
whatever you want.

491
00:33:08,340 --> 00:33:14,030
So suppose we has an
edge, Vi, that goes to Vj.

492
00:33:14,030 --> 00:33:16,820
Well in that case, we
know that the number

493
00:33:16,820 --> 00:33:23,700
of walks of length
1, from i to j,

494
00:33:23,700 --> 00:33:26,170
is exactly 1 because
there's this edge.

495
00:33:26,170 --> 00:33:27,510
So this is equal to 1.

496
00:33:27,510 --> 00:33:31,270
But this is also the definition
of my adjacency matrix.

497
00:33:31,270 --> 00:33:35,480
So I can just write
out here aij 1.

498
00:33:35,480 --> 00:33:36,380
So that's great.

499
00:33:36,380 --> 00:33:39,780
This case definitely works.

500
00:33:39,780 --> 00:33:45,530
Now, if there is no
edge of this type,

501
00:33:45,530 --> 00:33:50,210
then you know that in one step,
we can never achieve Vj from Vi

502
00:33:50,210 --> 00:33:51,630
because there's no edge.

503
00:33:51,630 --> 00:33:54,420
So we know this is equal to 0.

504
00:33:54,420 --> 00:33:56,050
There's no such walk.

505
00:33:56,050 --> 00:33:58,490
And this is, by definition
of the adjacency matrix,

506
00:33:58,490 --> 00:34:01,045
also equal to the aij.

507
00:34:01,045 --> 00:34:01,860
So this works.

508
00:34:01,860 --> 00:34:04,000
So the base case it's easy.

509
00:34:04,000 --> 00:34:11,658
And induction step always
starts by issuing pk.

510
00:34:16,429 --> 00:34:18,443
As you can see,
these types of proofs

511
00:34:18,443 --> 00:34:19,734
always have the same structure.

512
00:34:26,340 --> 00:34:33,822
In this case, let
me again assume pk.

513
00:34:39,300 --> 00:34:42,300
We want to prove pk plus 1.

514
00:34:42,300 --> 00:34:43,770
So what you want
to know is what's

515
00:34:43,770 --> 00:34:46,050
the number of walks
of length k plus 1.

516
00:34:46,050 --> 00:34:49,110
So how can we express those?

517
00:34:49,110 --> 00:34:53,360
pij k plus 1.

518
00:34:53,360 --> 00:34:58,260
How can we use this
assumption over here?

519
00:34:58,260 --> 00:35:00,550
Do you have an idea so we can--

520
00:35:04,790 --> 00:35:08,450
AUDIENCE: Any walk
of length k plus 1

521
00:35:08,450 --> 00:35:15,505
can be got by taking one
of-- let's say from Vs

522
00:35:15,505 --> 00:35:25,360
to Vx. [INAUDIBLE] You go
from Vs to V in k steps

523
00:35:25,360 --> 00:35:27,312
and then V to Vf in one step.

524
00:35:27,312 --> 00:35:28,270
PROFESSOR: That's true.

525
00:35:28,270 --> 00:35:30,570
We could do that.

526
00:35:30,570 --> 00:35:31,960
So let's write it down.

527
00:35:31,960 --> 00:35:33,710
So what you're
essentially saying

528
00:35:33,710 --> 00:35:37,150
is that you can
enumerate all the walks

529
00:35:37,150 --> 00:35:43,390
by going of length k plus 1
by first going from i in, say,

530
00:35:43,390 --> 00:35:49,620
k steps, to whatever
V, and then in one

531
00:35:49,620 --> 00:35:55,791
step to, say, Vi to V in k steps
and then in one step to Vj.

532
00:35:58,320 --> 00:36:01,310
So let's write this down
because V can be anything.

533
00:36:01,310 --> 00:36:05,320
So what to do is we have
a sum over all the Vs

534
00:36:05,320 --> 00:36:08,600
such that-- well, we
will use indices here

535
00:36:08,600 --> 00:36:11,030
let me do that a
little bit differently.

536
00:36:11,030 --> 00:36:12,800
So let's say I have
an h over here.

537
00:36:12,800 --> 00:36:20,110
So all the indices
h such that Vh to Vj

538
00:36:20,110 --> 00:36:26,110
is actually an edge in G.

539
00:36:26,110 --> 00:36:32,020
So then we can write out here
that we go in k steps to Vh.

540
00:36:32,020 --> 00:36:35,110
So how many walks are there?

541
00:36:35,110 --> 00:36:37,610
Well, we can use the induction
hypothesis now, right?

542
00:36:37,610 --> 00:36:43,130
So we can say you go
from i to h in k steps.

543
00:36:43,130 --> 00:36:49,360
And then, well, we can
use this particular edge

544
00:36:49,360 --> 00:36:55,960
to complete it to a k
plus 1th walk from i to j.

545
00:36:55,960 --> 00:36:58,110
So now what is this equal to?

546
00:36:58,110 --> 00:37:00,550
Can we simplify this sum?

547
00:37:00,550 --> 00:37:04,830
We can, right, because
we know that there's

548
00:37:04,830 --> 00:37:09,310
an edge if and only if the
adjacency matrix has a 1

549
00:37:09,310 --> 00:37:11,300
in a particular position.

550
00:37:11,300 --> 00:37:19,910
So we could rewrite it and
sum over all h from 1 to n.

551
00:37:19,910 --> 00:37:32,510
And write pij k times, and then
aj-- oh, this should be an h.

552
00:37:32,510 --> 00:37:34,100
So we have the same over here.

553
00:37:34,100 --> 00:37:39,610
And then we have one
edge from h to 2j.

554
00:37:39,610 --> 00:37:44,360
So I only count this number
over here if this is equal to 1,

555
00:37:44,360 --> 00:37:46,500
and that happens exactly
if there's an edge.

556
00:37:46,500 --> 00:37:49,230
I do not count this if
there's a 0 over here,

557
00:37:49,230 --> 00:37:51,930
that is as if there's no edge.

558
00:37:51,930 --> 00:37:56,650
But now we can use induction
hypothesis because I know that

559
00:37:56,650 --> 00:37:59,620
these numbers are
equal to the a's.

560
00:37:59,620 --> 00:38:03,040
So we rewrite this.

561
00:38:03,040 --> 00:38:07,760
And we see that we get aih k.

562
00:38:07,760 --> 00:38:11,920
So this is where we
use the induction step.

563
00:38:11,920 --> 00:38:18,110
So it's like we
assume pk over here.

564
00:38:18,110 --> 00:38:20,530
And I need to
finish this formula.

565
00:38:20,530 --> 00:38:21,970
So we have this.

566
00:38:21,970 --> 00:38:29,400
Now, by the definition
of matrix multiplication,

567
00:38:29,400 --> 00:38:41,700
we actually see that this
is equal to a k plus 1 ij

568
00:38:41,700 --> 00:38:45,390
to the ij-th entry in
the k plus oneth power

569
00:38:45,390 --> 00:38:47,050
of the adjacency matrix.

570
00:38:47,050 --> 00:38:50,630
Here we have this represents
the matrix of the kth power.

571
00:38:50,630 --> 00:38:53,440
This represents the matrix of a.

572
00:38:53,440 --> 00:38:56,140
So we multiply essentially
a to the power k times a

573
00:38:56,140 --> 00:38:58,190
and get a to the power k plus 1.

574
00:38:58,190 --> 00:39:00,260
And that's what we see here.

575
00:39:00,260 --> 00:39:04,000
So this is the induction step,
and so this proves the theorem

576
00:39:04,000 --> 00:39:04,500
up here.

577
00:39:07,480 --> 00:39:11,670
So let's talk about a few
more definitions concerning

578
00:39:11,670 --> 00:39:13,150
directed graphs.

579
00:39:13,150 --> 00:39:18,730
And then you go a step
into a very special type

580
00:39:18,730 --> 00:39:23,480
of graph, which are the
tournament graphs up here.

581
00:39:31,840 --> 00:39:34,240
One of the things that
we have in undirected

582
00:39:34,240 --> 00:39:37,750
graphs, so where we
have no directed edges,

583
00:39:37,750 --> 00:39:42,720
we talked last time a lot about
cyclicity and stuff like that.

584
00:39:42,720 --> 00:39:47,530
And we talked about
acyclic connected graphs.

585
00:39:47,530 --> 00:39:51,350
And as we categorized those,
so we defined them as trees,

586
00:39:51,350 --> 00:39:54,130
and they have a very
special structure.

587
00:39:54,130 --> 00:40:04,950
So what would happen here, if
we look at a directed graph,

588
00:40:04,950 --> 00:40:08,260
and we wonder, what does
it mean to be connected.

589
00:40:08,260 --> 00:40:10,007
Can we really talk about that?

590
00:40:10,007 --> 00:40:10,840
What does that mean?

591
00:40:14,350 --> 00:40:17,720
For example, if I look at
this particular example

592
00:40:17,720 --> 00:40:20,220
graph over here,
I can say, well,

593
00:40:20,220 --> 00:40:24,200
V1 has an edge that
points towards V1 itself,

594
00:40:24,200 --> 00:40:28,220
and towards V2, and towards V3.

595
00:40:28,220 --> 00:40:31,230
So maybe I would call
this graph connected.

596
00:40:31,230 --> 00:40:35,370
But if I look at V2,
I only see an edge

597
00:40:35,370 --> 00:40:38,510
that goes from V2 to
V3, and not to V2.

598
00:40:38,510 --> 00:40:41,270
So maybe I do not
call it connected.

599
00:40:41,270 --> 00:40:48,610
So that's why we define a
stronger notion for a digraph.

600
00:40:48,610 --> 00:41:09,570
So a digraph, G, VE is
called strongly connected

601
00:41:09,570 --> 00:41:18,180
if we know that for every
pair-- so if for all vertices

602
00:41:18,180 --> 00:41:25,190
U and V in the vertex
set-- there exists

603
00:41:25,190 --> 00:41:39,550
a directed path that starts
in U and ends up in V in G.

604
00:41:39,550 --> 00:41:41,940
So this is what we would
call strongly connected.

605
00:41:46,180 --> 00:41:49,900
So now in undirected
graphs, we had connectivity

606
00:41:49,900 --> 00:41:53,540
and then we said, well, if a
connected, undirected graph

607
00:41:53,540 --> 00:41:54,740
has no cycles.

608
00:41:54,740 --> 00:41:56,910
We have trees and they
have special properties

609
00:41:56,910 --> 00:41:57,790
and all that.

610
00:41:57,790 --> 00:42:00,370
So what about this over here?

611
00:42:00,370 --> 00:42:06,820
Suppose you have an acyclic
strongly connected digraph.

612
00:42:06,820 --> 00:42:09,064
What does that look like?

613
00:42:09,064 --> 00:42:10,105
So let's give an example.

614
00:42:12,800 --> 00:42:18,070
It's not completely clear
what kind of structure it has.

615
00:42:18,070 --> 00:42:25,940
So for example, I
may have a graph that

616
00:42:25,940 --> 00:42:27,645
looks like this, for example.

617
00:42:34,690 --> 00:42:39,760
Say this is an acyclic graph.

618
00:42:39,760 --> 00:42:50,400
but It does not have at
all a tree structure.

619
00:42:50,400 --> 00:42:54,500
So actually, the type of
graph the we have here

620
00:42:54,500 --> 00:42:57,872
is called a directed
acyclic graph.

621
00:42:57,872 --> 00:42:59,330
As you can see,
there are no cycles

622
00:42:59,330 --> 00:43:01,970
because I only go
forward, essentially.

623
00:43:01,970 --> 00:43:05,180
I can never go backward
in this particular way

624
00:43:05,180 --> 00:43:07,370
that I depicted the graph.

625
00:43:07,370 --> 00:43:12,930
So this is an example for
directed acyclic graph.

626
00:43:12,930 --> 00:43:16,540
But it doesn't look
like a tree at all.

627
00:43:16,540 --> 00:43:18,870
So it's worth to
define it separately,

628
00:43:18,870 --> 00:43:23,600
and we will use this
on Thursday when we'll

629
00:43:23,600 --> 00:43:25,700
talk about partial orderings.

630
00:43:25,700 --> 00:43:29,780
And it turns out that, as
you can see here-- well,

631
00:43:29,780 --> 00:43:31,980
at that case, a
directed acyclic graph

632
00:43:31,980 --> 00:43:33,790
has really nice
properties, and one of them

633
00:43:33,790 --> 00:43:37,505
is that you can order these
vertices in such a way

634
00:43:37,505 --> 00:43:43,050
that you go from, say, left to
right in a directed fashion.

635
00:43:43,050 --> 00:43:44,930
And that will lead
to partial ordering.

636
00:43:44,930 --> 00:43:49,880
So that's something that
we will talk about as well.

637
00:43:49,880 --> 00:43:52,200
So what's the definition?

638
00:43:52,200 --> 00:44:09,890
A directed graph is called
a directed acyclic graph,

639
00:44:09,890 --> 00:44:14,250
and we appreciate this by DAG.

640
00:44:14,250 --> 00:44:15,690
We call them DAGs.

641
00:44:15,690 --> 00:44:18,250
They're used
everywhere, actually.

642
00:44:26,660 --> 00:44:33,390
If it does not contain
any directed cycles.

643
00:44:38,570 --> 00:44:39,945
So these kinds of
graphs are used

644
00:44:39,945 --> 00:44:43,465
in scheduling and
optimization, and we

645
00:44:43,465 --> 00:44:47,310
will use them next week in the
lecture on partial orderings.

646
00:44:50,680 --> 00:44:52,510
Now we have done a
lot of definitions

647
00:44:52,510 --> 00:44:57,330
concerning directed graphs.

648
00:44:57,330 --> 00:45:00,080
So now let's talk about these
very special ones, tournament

649
00:45:00,080 --> 00:45:03,230
graph, and see whether we
can prove a few really nice

650
00:45:03,230 --> 00:45:04,280
theorems about them.

651
00:45:08,780 --> 00:45:12,040
Let's see whether we can
figure that out together.

652
00:45:22,150 --> 00:45:23,590
So what is a tournament graph?

653
00:45:36,790 --> 00:45:39,760
In a tournament graph, I
have a bunch of vertices.

654
00:45:39,760 --> 00:45:43,930
And essentially we want to
represent like a tournament.

655
00:45:43,930 --> 00:45:47,770
So every vertex, say,
represents a team.

656
00:45:47,770 --> 00:45:51,070
And a team can play
against another team,

657
00:45:51,070 --> 00:45:54,660
and beat them or lose
against the other team.

658
00:45:54,660 --> 00:45:58,670
So we want to use the
directed edges to indicate

659
00:45:58,670 --> 00:46:01,050
who is winning from home.

660
00:46:01,050 --> 00:46:05,270
And such types of graphs
have very special property.

661
00:46:05,270 --> 00:46:08,250
So let me first depict one.

662
00:46:08,250 --> 00:46:19,060
So for example, we have E
goes to A, goes through B,

663
00:46:19,060 --> 00:46:26,700
incoming edge from
C, one coming from B.

664
00:46:26,700 --> 00:46:29,240
And over here we have
this directed edge.

665
00:46:33,080 --> 00:46:36,160
We have this one.

666
00:46:36,160 --> 00:46:37,790
We have this one.

667
00:46:37,790 --> 00:46:44,020
Let me see [INAUDIBLE] don't
make any mistakes here.

668
00:46:44,020 --> 00:46:45,790
And this one.

669
00:46:45,790 --> 00:46:47,025
And over here is another one.

670
00:46:49,560 --> 00:46:51,470
So what do we see in this graph?

671
00:46:51,470 --> 00:47:02,840
We see that either say,
team U, beats team V.

672
00:47:02,840 --> 00:47:06,600
And that means that we have a
directed edge from U pointing

673
00:47:06,600 --> 00:47:12,200
at V. Or it's other way around.

674
00:47:12,200 --> 00:47:19,970
V actually beats U, and we have
a directed edge from V to U.

675
00:47:19,970 --> 00:47:25,140
Let's have a look at this
graph and see how this works.

676
00:47:25,140 --> 00:47:30,360
So for example, we have
that B is beating E,

677
00:47:30,360 --> 00:47:33,150
and E is beating A, and so on.

678
00:47:33,150 --> 00:47:34,030
So let's have a look.

679
00:47:34,030 --> 00:47:39,540
Maybe we can figure out who's
the best player in here.

680
00:47:39,540 --> 00:47:41,414
So this is sort of
a general question

681
00:47:41,414 --> 00:47:42,580
if who would like to answer.

682
00:47:42,580 --> 00:47:44,690
Maybe you cannot answer this.

683
00:47:44,690 --> 00:47:46,910
So let's have an example.

684
00:47:46,910 --> 00:47:55,980
Has anybody seen an example
where we start with A,

685
00:47:55,980 --> 00:48:00,350
then we may beat another vertex,
and maybe another vertex,

686
00:48:00,350 --> 00:48:04,540
and so on, until we have covered
all the different vertices.

687
00:48:04,540 --> 00:48:07,530
Do you see a path
that works like that?

688
00:48:07,530 --> 00:48:10,760
And that could gives
us an ordering on who

689
00:48:10,760 --> 00:48:14,840
is the best player,
like the one at the top,

690
00:48:14,840 --> 00:48:20,030
like A, is able to,
for example, beat B.

691
00:48:20,030 --> 00:48:27,170
And B is, for example, able
to beat D and this one, E,

692
00:48:27,170 --> 00:48:30,660
and this one, C. So you
would say, OK, that's great.

693
00:48:30,660 --> 00:48:33,850
Now I know that this one
is the strongest player.

694
00:48:33,850 --> 00:48:36,810
But there's a little
problem here, right?

695
00:48:36,810 --> 00:48:39,600
Because I can produce
many such paths.

696
00:48:39,600 --> 00:48:45,060
And actually, if I look at
C, then C beats A as well.

697
00:48:45,060 --> 00:48:47,260
So that's kind of weird.

698
00:48:47,260 --> 00:48:48,565
So wait a minute.

699
00:48:51,820 --> 00:48:56,950
We have that there's a
directed edge from C to A.

700
00:48:56,950 --> 00:48:59,120
It's like teams
beats one another.

701
00:48:59,120 --> 00:49:05,800
And it's not very clear how we
can talk about a best player.

702
00:49:05,800 --> 00:49:07,180
Well, we would
have a best player

703
00:49:07,180 --> 00:49:15,090
if one player sort of
wins from everybody else.

704
00:49:15,090 --> 00:49:16,820
But there's many examples here.

705
00:49:16,820 --> 00:49:19,840
So let's look at another walk.

706
00:49:19,840 --> 00:49:25,550
For example, C can go to B,
to D, and then to E, and then

707
00:49:25,550 --> 00:49:28,280
to A. So there are many
possibilities here.

708
00:49:28,280 --> 00:49:30,630
So this leads us to a concept.

709
00:49:33,640 --> 00:49:40,140
And we call that's a
directed Hamiltonian path.

710
00:49:40,140 --> 00:49:43,260
And we're going to show
that, in a tournament graph,

711
00:49:43,260 --> 00:49:46,120
you can always find such a
directed Hamiltonian path.

712
00:49:46,120 --> 00:49:48,060
So what's a Hamiltonian path?

713
00:49:48,060 --> 00:49:50,360
This is actually
an example of it.

714
00:49:50,360 --> 00:49:53,710
There's a walk that
goes around the graph

715
00:49:53,710 --> 00:49:57,240
and visits every
vertex exactly once.

716
00:49:57,240 --> 00:50:00,870
So we're going to prove
that a tournament graph has

717
00:50:00,870 --> 00:50:02,213
this beautiful property.

718
00:50:18,070 --> 00:50:21,190
So let's first write out
a definition of this.

719
00:50:26,620 --> 00:50:48,140
A directed Hamiltonian
path is a directed walk

720
00:50:48,140 --> 00:51:05,570
that visits every
vertex exactly once.

721
00:51:08,090 --> 00:51:12,630
So as I said already, here
we have such an example.

722
00:51:12,630 --> 00:51:16,140
We can go from A to B,
to D to E, to C. Maybe

723
00:51:16,140 --> 00:51:18,440
there are even other examples.

724
00:51:18,440 --> 00:51:20,160
I did not actually see them.

725
00:51:20,160 --> 00:51:22,540
Maybe you can have a
look at them as well.

726
00:51:22,540 --> 00:51:25,800
So maybe there's something that
starts with B going to E maybe.

727
00:51:25,800 --> 00:51:30,660
That's a very different
direction, like this.

728
00:51:30,660 --> 00:51:35,960
There will be one as
well, and so forth.

729
00:51:35,960 --> 00:51:38,960
So what's the theorem
that you want to prove?

730
00:51:38,960 --> 00:51:49,160
The theorem in that you want
to show that every tournament

731
00:51:49,160 --> 00:52:00,260
graph actually contains such
a directed Hamiltonian path.

732
00:52:16,710 --> 00:52:19,190
So let's have a look at
how we can prove this.

733
00:52:38,849 --> 00:52:40,890
What kind of proof technique
are we going to use?

734
00:52:45,220 --> 00:52:46,245
Do you have an idea?

735
00:52:48,595 --> 00:52:50,470
Well, usually we use
induction so that's what

736
00:52:50,470 --> 00:52:52,310
we're going to do here as well.

737
00:52:52,310 --> 00:52:55,770
But what kind of induction
hypothesis can we do?

738
00:52:55,770 --> 00:52:57,990
So what would we
induct on, you think?

739
00:53:00,834 --> 00:53:01,782
Someone else?

740
00:53:01,782 --> 00:53:04,626
Maybe someone up there?

741
00:53:04,626 --> 00:53:06,060
AUDIENCE: The number of nodes.

742
00:53:06,060 --> 00:53:08,670
PROFESSOR: The number of nodes.

743
00:53:08,670 --> 00:53:16,050
So we use induction on
the number of nodes.

744
00:53:16,050 --> 00:53:17,930
And why would that
be of interest?

745
00:53:17,930 --> 00:53:21,130
So let's have a look at
how we can think of that.

746
00:53:21,130 --> 00:53:24,500
So we start thinking about
such a problem, this really

747
00:53:24,500 --> 00:53:29,070
sort of-- one parameter here
that's the number of nodes

748
00:53:29,070 --> 00:53:31,960
in a tournament graph.

749
00:53:31,960 --> 00:53:33,580
I also have edges.

750
00:53:33,580 --> 00:53:41,290
But if I think about
edges, then, the edges

751
00:53:41,290 --> 00:53:45,340
is always directly
related to the number

752
00:53:45,340 --> 00:53:47,430
of nodes in a tournament graph.

753
00:53:47,430 --> 00:53:50,640
So I really have just
that one parameter.

754
00:53:50,640 --> 00:53:55,040
So it makes sense to use
induction on number of nodes.

755
00:53:55,040 --> 00:54:03,947
So induct on n, where Pn
is going to be that every--

756
00:54:03,947 --> 00:54:06,280
and essentially, the theorem
holds true for a tournament

757
00:54:06,280 --> 00:54:19,800
graph on n nodes-- so every
tournament graph on n nodes

758
00:54:19,800 --> 00:54:30,270
actually contains a
directed Hamiltonian path.

759
00:54:32,820 --> 00:54:36,440
So this is semi
induction hypothesis.

760
00:54:36,440 --> 00:54:38,954
So when you think
about that, well, I

761
00:54:38,954 --> 00:54:40,370
feel pretty confident
because if I

762
00:54:40,370 --> 00:54:45,430
look at the base case-- that's
how we always start-- well,

763
00:54:45,430 --> 00:54:46,730
n equals 1.

764
00:54:46,730 --> 00:54:49,440
If n equals 1, I have
just a single vertex.

765
00:54:49,440 --> 00:54:50,420
There's no edges.

766
00:54:50,420 --> 00:54:53,130
Everything is fine
because the single edge

767
00:54:53,130 --> 00:54:55,660
is a directed Hamiltonian path.

768
00:54:55,660 --> 00:54:57,280
So this is great.

769
00:54:57,280 --> 00:54:59,250
So this works.

770
00:54:59,250 --> 00:55:00,570
So what about inductive step?

771
00:55:03,460 --> 00:55:05,700
Now with inductive
step, we always

772
00:55:05,700 --> 00:55:07,800
perceived in the same way.

773
00:55:07,800 --> 00:55:13,540
We start to assume
that Pn is true.

774
00:55:13,540 --> 00:55:16,000
Essentially, the theorem
holds for a tournament graph

775
00:55:16,000 --> 00:55:17,524
on n vertices.

776
00:55:20,510 --> 00:55:23,050
Actually, let me
keep this over here.

777
00:55:29,570 --> 00:55:33,150
So now let's just think about
how we can prove this induction

778
00:55:33,150 --> 00:55:35,090
step.

779
00:55:35,090 --> 00:55:38,570
I need to prove P of n plus 1.

780
00:55:38,570 --> 00:55:40,050
So how do I start?

781
00:55:40,050 --> 00:55:42,640
I have to start with
a tournament graph

782
00:55:42,640 --> 00:55:44,950
on n plus 1 vertices.

783
00:55:44,950 --> 00:55:50,360
And somehow I got to be
able to use this property

784
00:55:50,360 --> 00:55:53,060
because that's what I assume.

785
00:55:53,060 --> 00:55:55,410
And the property only holds
for a tournament graph

786
00:55:55,410 --> 00:55:56,330
on end points.

787
00:55:56,330 --> 00:55:57,930
So what will be a
really good strategy

788
00:55:57,930 --> 00:56:00,900
to do here to sort
of proceed our proof?

789
00:56:00,900 --> 00:56:08,380
So let me first write out
what we want to do here.

790
00:56:08,380 --> 00:56:15,040
Maybe you can think about
how to advance here.

791
00:56:15,040 --> 00:56:18,460
So we have shown Pn.

792
00:56:18,460 --> 00:56:22,840
Now we want to actually
prove something

793
00:56:22,840 --> 00:56:25,765
about tournament graphs
on n plus 1 nodes,

794
00:56:25,765 --> 00:56:28,250
so let's consider one.

795
00:56:28,250 --> 00:56:41,310
Consider a tournament
graph on n plus 1 nodes.

796
00:56:41,310 --> 00:56:47,970
Now how can I use my
induction hypothesis?

797
00:56:47,970 --> 00:56:52,390
So I start with this,
and I want to use

798
00:56:52,390 --> 00:56:55,780
something that talks about the
tournament graph on n nodes.

799
00:56:55,780 --> 00:56:57,270
So how could I proceed here?

800
00:57:02,616 --> 00:57:04,560
Any suggestions?

801
00:57:04,560 --> 00:57:07,130
So what do you
usually do, right?

802
00:57:07,130 --> 00:57:09,850
If I have like an
n plus 1 nodes,

803
00:57:09,850 --> 00:57:13,000
I have to somehow look
at least maybe there

804
00:57:13,000 --> 00:57:15,240
exists a subgraph in
this bigger graph.

805
00:57:15,240 --> 00:57:17,330
So this is really
how you always think

806
00:57:17,330 --> 00:57:21,140
about these types of proofs
or also other problems.

807
00:57:21,140 --> 00:57:23,340
So there must be
some kind of subgraph

808
00:57:23,340 --> 00:57:26,400
that already has this property.

809
00:57:26,400 --> 00:57:31,440
Well, let's take
out one node and see

810
00:57:31,440 --> 00:57:35,470
what happens because then
we have one node less

811
00:57:35,470 --> 00:57:41,380
and maybe we will be able
to apply our induction step.

812
00:57:41,380 --> 00:57:51,550
So let's take out one node V.

813
00:57:51,550 --> 00:57:54,740
And what can we say
about remaining graph

814
00:57:54,740 --> 00:57:56,040
if we take out one node?

815
00:57:56,040 --> 00:57:59,670
For example, if I take
out the node E over here

816
00:57:59,670 --> 00:58:01,400
and I just look
at all the rest, I

817
00:58:01,400 --> 00:58:05,900
can still see that for all
the other nodes either,

818
00:58:05,900 --> 00:58:09,100
say, U beats V or V
beats U. So I still

819
00:58:09,100 --> 00:58:15,130
have an edge in one direction
between each two nodes.

820
00:58:15,130 --> 00:58:17,350
So actually I still
have a tournament graph,

821
00:58:17,350 --> 00:58:18,140
so that's great.

822
00:58:23,340 --> 00:58:25,995
So this gives a
tournament graph.

823
00:58:29,490 --> 00:58:31,810
So essentially,
so far, we really

824
00:58:31,810 --> 00:58:35,660
haven't done anything
creative or anything

825
00:58:35,660 --> 00:58:40,420
that we had to make a big leap
in order to prove this theorem.

826
00:58:40,420 --> 00:58:42,890
We started out with, if
you want to prove something

827
00:58:42,890 --> 00:58:45,770
like this you have to really
look at the number of vertices.

828
00:58:45,770 --> 00:58:49,140
And then we start to write
down this stuff over here

829
00:58:49,140 --> 00:58:50,290
that makes total sense.

830
00:58:50,290 --> 00:58:53,880
And then we are going
to figure out where

831
00:58:53,880 --> 00:58:55,720
we can use this induction step.

832
00:58:55,720 --> 00:58:57,150
And so we just
take out one node.

833
00:58:57,150 --> 00:59:04,180
And yes, it is a tournament
graph on n nodes.

834
00:59:04,180 --> 00:59:05,630
So this is very systematic.

835
00:59:05,630 --> 00:59:07,850
That's what I try
to get at here.

836
00:59:07,850 --> 00:59:14,060
So by the induction step because
by the induction hypothesis,

837
00:59:14,060 --> 00:59:20,120
we know now that we actually
have a directed Hamiltonian

838
00:59:20,120 --> 00:59:21,070
path.

839
00:59:21,070 --> 00:59:27,835
So let V1 to V2 to
Vn be such a path.

840
00:59:31,170 --> 00:59:35,230
So now that we can use this,
we apply it and we get a path.

841
00:59:35,230 --> 00:59:37,030
Now what do we want to do?

842
00:59:37,030 --> 00:59:40,670
We want to show that we can
create a new path, which

843
00:59:40,670 --> 00:59:44,140
is also a directed Hamiltonian
path, but now one that also

844
00:59:44,140 --> 00:59:46,460
includes V in the bigger graph.

845
00:59:46,460 --> 00:59:48,020
If we can do that, we are done.

846
00:59:48,020 --> 00:59:50,220
So now we have to
start really looking

847
00:59:50,220 --> 00:59:52,580
at how we can make that happen.

848
00:59:59,590 --> 01:00:06,550
In order to do this, we have
to see how we can somehow

849
01:00:06,550 --> 01:00:10,120
plug V, the vertex
that we have removed,

850
01:00:10,120 --> 01:00:11,640
in this path over here.

851
01:00:11,640 --> 01:00:15,460
If you can do this, we
are in really good shape.

852
01:00:18,080 --> 01:00:21,200
So far, we haven't used at
all-- so there's also something

853
01:00:21,200 --> 01:00:22,580
that you can look
at if you start

854
01:00:22,580 --> 01:00:27,530
solving these types of
problems-- we haven't used

855
01:00:27,530 --> 01:00:30,430
at all the property that
the tournament graph has,

856
01:00:30,430 --> 01:00:33,900
which I just wiped out.

857
01:00:33,900 --> 01:00:36,030
So let's figure out
what we can do here.

858
01:00:38,870 --> 01:00:41,070
We should be able
to use something.

859
01:00:41,070 --> 01:00:43,740
So of course, we have
a few simple cases.

860
01:00:43,740 --> 01:00:51,170
For example, if V has a
directed edge into V1,

861
01:00:51,170 --> 01:00:55,970
and I have a Hamiltonian path
that goes from V to V1 to V2

862
01:00:55,970 --> 01:00:59,500
all the way to Vn, and then I
cover all the vertices exactly

863
01:00:59,500 --> 01:01:00,610
once.

864
01:01:00,610 --> 01:01:03,540
And I will have a
direct Hamiltonian path.

865
01:01:03,540 --> 01:01:04,480
So this is great.

866
01:01:04,480 --> 01:01:07,420
So this is definitely easy.

867
01:01:07,420 --> 01:01:10,615
So this is case one.

868
01:01:10,615 --> 01:01:19,300
In case two, suppose that
V1 has a directed edge to V.

869
01:01:19,300 --> 01:01:24,790
And So now we have a little
problem because somehow there

870
01:01:24,790 --> 01:01:30,470
is an edge like this to V,
but they cannot go like this.

871
01:01:30,470 --> 01:01:31,870
This is not a Hamiltonian path.

872
01:01:31,870 --> 01:01:35,530
I need to have a
directed walk that covers

873
01:01:35,530 --> 01:01:36,980
all the vertices exactly once.

874
01:01:36,980 --> 01:01:38,690
So now we have a little problem.

875
01:01:38,690 --> 01:01:40,400
So now we have to
start really thinking

876
01:01:40,400 --> 01:01:43,550
about how we can solve this.

877
01:01:43,550 --> 01:01:47,566
So are there any suggestions
to make this happen?

878
01:01:47,566 --> 01:01:49,190
So let's think a
little bit about this.

879
01:01:52,760 --> 01:01:55,160
So somehow if I start
thinking about this,

880
01:01:55,160 --> 01:01:59,450
I would like to plug V
somewhere in this sequence.

881
01:01:59,450 --> 01:02:02,955
That will be like a pretty
obvious way to-- go ahead.

882
01:02:02,955 --> 01:02:03,830
AUDIENCE: [INAUDIBLE]

883
01:02:16,270 --> 01:02:17,015
PROFESSOR: Yeah.

884
01:02:17,015 --> 01:02:18,440
That's true.

885
01:02:18,440 --> 01:02:23,160
So for example, I may
have that for example,

886
01:02:23,160 --> 01:02:27,110
V2 beats V as well.

887
01:02:27,110 --> 01:02:32,820
But suppose that I have V3 over
here, and this one beats V3.

888
01:02:32,820 --> 01:02:35,790
Then, as you say, I could
sort of plug V in here

889
01:02:35,790 --> 01:02:38,860
in this sequence and I may
have a longer sequence.

890
01:02:38,860 --> 01:02:46,250
Or if this is not the case,
then maybe the next one, V4,

891
01:02:46,250 --> 01:02:49,450
may have the property
and I can plug V in here.

892
01:02:49,450 --> 01:02:54,620
So essentially what I want
to show is that I can plug V

893
01:02:54,620 --> 01:03:00,200
somewhere in the middle
of two of these Vi's.

894
01:03:00,200 --> 01:03:02,490
And the property
that you were using

895
01:03:02,490 --> 01:03:07,040
is actually that you said,
well, I know that V2 either

896
01:03:07,040 --> 01:03:09,730
beats V, or V beats V2.

897
01:03:09,730 --> 01:03:12,320
So that's the property
of the tournament graph

898
01:03:12,320 --> 01:03:15,600
that we're going to use
here in order to prove this.

899
01:03:15,600 --> 01:03:19,430
So yes, that's a
great observation.

900
01:03:19,430 --> 01:03:23,220
So the way you formulated it,
we can maybe use induction,

901
01:03:23,220 --> 01:03:24,770
for example, to prove this.

902
01:03:24,770 --> 01:03:30,900
We can sort of go
recursively through this

903
01:03:30,900 --> 01:03:33,850
until you find the right spot.

904
01:03:33,850 --> 01:03:38,390
Maybe they can immediately
precisely indicate such a spot.

905
01:03:38,390 --> 01:03:39,750
So how do we usually do that?

906
01:03:39,750 --> 01:03:43,690
If we are thinking
about other theorems

907
01:03:43,690 --> 01:03:44,978
that we have tried to prove.

908
01:03:51,850 --> 01:03:53,540
AUDIENCE: What's the
smallest value of i

909
01:03:53,540 --> 01:03:56,652
such that V beats Vi?

910
01:03:56,652 --> 01:03:57,600
PROFESSOR: OK.

911
01:03:57,600 --> 01:04:02,350
So what's the smallest
value of i where V beats Vi.

912
01:04:14,930 --> 01:04:21,780
So usually we have words in our
mind like largest or smallest,

913
01:04:21,780 --> 01:04:25,357
et cetera, and sort of
an extreme precision.

914
01:04:25,357 --> 01:04:27,690
And then we like to find out
that we can make it happen.

915
01:04:27,690 --> 01:04:32,880
And then we say, well,
if something goes wrong,

916
01:04:32,880 --> 01:04:36,090
we violate that
smallest condition.

917
01:04:36,090 --> 01:04:37,300
So let's do that here.

918
01:04:37,300 --> 01:04:41,380
So let's indicate
a specific spot.

919
01:04:41,380 --> 01:04:54,883
Let's considered the smallest
i such that V beats Vi.

920
01:04:58,860 --> 01:05:00,710
Well, let's have a look
how this would work.

921
01:05:04,030 --> 01:05:07,760
So this is a little bit
of a different proof

922
01:05:07,760 --> 01:05:11,129
than what I had but this
should work fine as well.

923
01:05:11,129 --> 01:05:12,420
So let's just see how it works.

924
01:05:16,210 --> 01:05:17,210
So we have V1.

925
01:05:22,640 --> 01:05:26,650
First of all, we notice that,
of course, if i equals 1,

926
01:05:26,650 --> 01:05:30,970
which that cannot be the case,
so we know that i is larger

927
01:05:30,970 --> 01:05:31,850
than 1.

928
01:05:31,850 --> 01:05:38,310
So there's really
somewhere a Vi minus 1--

929
01:05:38,310 --> 01:05:39,560
we have to check that, right?

930
01:05:39,560 --> 01:05:41,780
If that exists,
this index is not

931
01:05:41,780 --> 01:05:46,050
equal to zero-- that goes to
Vi and then goes all the way up

932
01:05:46,050 --> 01:05:49,320
to Vn.

933
01:05:54,580 --> 01:06:03,680
So now we say we can use this,
that V actually beats Vi.

934
01:06:03,680 --> 01:06:09,870
Now what's about-- maybe
someone else can help me here--

935
01:06:09,870 --> 01:06:13,285
I would like to
have that Vi minus 1

936
01:06:13,285 --> 01:06:19,310
beats V. That would be fantastic
because then I have a path that

937
01:06:19,310 --> 01:06:21,300
goes from V1 all
the way up here,

938
01:06:21,300 --> 01:06:24,470
goes here, goes there, and
all the way up to here.

939
01:06:24,470 --> 01:06:26,535
So we have a directed
Hamiltonian path

940
01:06:26,535 --> 01:06:29,950
that covers all the
vertices exactly once,

941
01:06:29,950 --> 01:06:32,210
and that's what
you want to prove.

942
01:06:32,210 --> 01:06:36,933
But why would there be an
edge that goes this way?

943
01:06:36,933 --> 01:06:38,224
So how do we reason about this?

944
01:06:44,390 --> 01:06:46,300
Someone else?

945
01:06:46,300 --> 01:06:52,270
AUDIENCE: Because Vi is the
smallest number that V beats,

946
01:06:52,270 --> 01:06:56,729
then anything smaller
than i must have beat V.

947
01:06:56,729 --> 01:06:58,780
PROFESSOR: Exactly.

948
01:06:58,780 --> 01:07:04,110
So if V would beat
Vi minus 1, that

949
01:07:04,110 --> 01:07:07,130
would contradict the
smallest property over here.

950
01:07:07,130 --> 01:07:09,010
So that's a contradiction.

951
01:07:09,010 --> 01:07:12,070
Now we use a property
of the tournament graph.

952
01:07:12,070 --> 01:07:16,150
So now I know that Vi
minus 1 must beat V.

953
01:07:16,150 --> 01:07:19,920
So I really have
an edge over here.

954
01:07:19,920 --> 01:07:21,450
So that works.

955
01:07:21,450 --> 01:07:24,040
So that's the end of the proof.

956
01:07:24,040 --> 01:07:27,480
Now another version
could have been

957
01:07:27,480 --> 01:07:31,280
in which we would do
something similar like this.

958
01:07:31,280 --> 01:07:32,800
But we could also
have used, say,

959
01:07:32,800 --> 01:07:38,480
the largest i-- just something
that you may want to look at.

960
01:07:38,480 --> 01:07:41,580
You can also use
the largest i such

961
01:07:41,580 --> 01:07:49,660
that Vi beats V. We have a
completely symmetrical argument

962
01:07:49,660 --> 01:07:52,630
here, but you could use
this solution as well.

963
01:07:52,630 --> 01:07:54,570
So I'm just trying
to sketch here

964
01:07:54,570 --> 01:07:56,920
the way of thinking that
you may want to consider

965
01:07:56,920 --> 01:07:58,550
in these types of problems.

966
01:07:58,550 --> 01:08:00,220
So why would this
work, by the way?

967
01:08:00,220 --> 01:08:02,780
Well, we have the same
kind of argument like this.

968
01:08:02,780 --> 01:08:06,610
We plug V right after Vi.

969
01:08:06,610 --> 01:08:09,560
We know there's an edge
from V to Vi plus 1.

970
01:08:09,560 --> 01:08:10,340
Why?

971
01:08:10,340 --> 01:08:14,400
If it's not the case, there
will be a large index, i,

972
01:08:14,400 --> 01:08:16,490
that contradicts our
assumption that we

973
01:08:16,490 --> 01:08:18,470
have the largest i already.

974
01:08:18,470 --> 01:08:20,960
It's a tournament graph,
so we know that there's

975
01:08:20,960 --> 01:08:24,890
an edge from V to Vi plus 1.

976
01:08:24,890 --> 01:08:27,534
And we get a directed
Hamiltonian path as well.

977
01:08:27,534 --> 01:08:29,200
So you may want to
look at that as well.

978
01:08:34,729 --> 01:08:36,370
So this is about
tournament graphs.

979
01:08:36,370 --> 01:08:41,439
So let's talk about an
interesting tournament graph

980
01:08:41,439 --> 01:08:42,450
with a funny game.

981
01:08:46,040 --> 01:08:50,569
And this is actually
a chicken tournament,

982
01:08:50,569 --> 01:08:54,870
like the chickens here
represent the vertices

983
01:08:54,870 --> 01:08:59,630
and they are pecking one
another, but in a certain rule

984
01:08:59,630 --> 01:09:01,643
that defines a chicken
to be the king chicken.

985
01:09:01,643 --> 01:09:02,809
So let's see how that works.

986
01:09:07,399 --> 01:09:11,159
So that's a great
application of graph theory.

987
01:09:15,189 --> 01:09:17,210
So what do we have?

988
01:09:17,210 --> 01:09:30,130
We have that either a chicken,
U, pecks a chicken, V.

989
01:09:30,130 --> 01:09:33,540
And we said that U has
a direct edge to V,

990
01:09:33,540 --> 01:09:36,250
so we're actually defining
a tournament graph here.

991
01:09:36,250 --> 01:09:42,500
Or we have a chicken, V,
that pecks a chicken, U,

992
01:09:42,500 --> 01:09:44,950
and we get V has a
direct edge to U.

993
01:09:44,950 --> 01:09:47,000
So we have a tournament graph.

994
01:09:47,000 --> 01:09:50,680
But now we define something new.

995
01:09:50,680 --> 01:10:02,660
We say that U virtually pecks
V if one of the two conditions

996
01:10:02,660 --> 01:10:06,710
holds-- one of these two
conditions-- either U,

997
01:10:06,710 --> 01:10:09,950
of course, pecks
V. That's great.

998
01:10:09,950 --> 01:10:11,060
He's in good shape.

999
01:10:14,520 --> 01:10:18,660
Or there exists another
chicken, W, such

1000
01:10:18,660 --> 01:10:27,760
that U actually pecks
W and W, in turn,

1001
01:10:27,760 --> 01:10:35,740
pecks V. So this is very special
kind of first relationship.

1002
01:10:35,740 --> 01:10:45,350
So we are wondering now can
we now define something--

1003
01:10:45,350 --> 01:10:46,250
is there a question?

1004
01:10:46,250 --> 01:10:49,250
AUDIENCE: [INAUDIBLE]
in between?

1005
01:10:49,250 --> 01:10:52,540
To be virtually
pecked, is one chicken

1006
01:10:52,540 --> 01:10:56,282
in between U and the other one?

1007
01:10:56,282 --> 01:10:58,240
PROFESSOR: Well, there
can be multiple chickens

1008
01:10:58,240 --> 01:11:01,649
in between here.

1009
01:11:01,649 --> 01:11:04,728
I have several friends who help
me out pecking someone else.

1010
01:11:10,410 --> 01:11:15,080
So when we were looking at
these tournament graphs,

1011
01:11:15,080 --> 01:11:21,520
we were wondering, can we really
indicate a winning player?

1012
01:11:21,520 --> 01:11:24,440
Well in this case, if you start
to talk about virtual pecking,

1013
01:11:24,440 --> 01:11:26,580
we look at the pecking order.

1014
01:11:26,580 --> 01:11:33,350
Then we can actually define
something like a chicken king.

1015
01:11:33,350 --> 01:11:36,737
So let me write that down.

1016
01:11:42,986 --> 01:11:47,040
Let me first explain
what I mean here.

1017
01:11:47,040 --> 01:11:48,610
And I give an
example of a graph.

1018
01:11:59,260 --> 01:12:07,180
So a chicken that is able
to virtually peck everyone

1019
01:12:07,180 --> 01:12:10,870
else, we will call a king.

1020
01:12:14,100 --> 01:12:28,670
So chicken that virtual
pecks every other chicken

1021
01:12:28,670 --> 01:12:36,490
is called a king chicken.

1022
01:12:41,250 --> 01:12:45,340
So let's give an
example of a graph.

1023
01:12:45,340 --> 01:12:51,200
So, for example, suppose
I have four chickens that

1024
01:12:51,200 --> 01:12:54,655
know how to pick one
another in this order.

1025
01:13:02,200 --> 01:13:06,668
So who in the pecking party
here is going to be king?

1026
01:13:06,668 --> 01:13:11,350
Do you see some solutions here?

1027
01:13:14,740 --> 01:13:16,760
So take, for example, this one.

1028
01:13:16,760 --> 01:13:19,220
So this one pecks this one.

1029
01:13:22,830 --> 01:13:24,890
Because we talk about
virtually pecking,

1030
01:13:24,890 --> 01:13:27,390
it can also peck
this one over here.

1031
01:13:27,390 --> 01:13:28,780
It does, right?

1032
01:13:28,780 --> 01:13:32,000
It pecks this one and this
one helps out, and can peck

1033
01:13:32,000 --> 01:13:33,590
both this one and this one.

1034
01:13:33,590 --> 01:13:34,610
That's cool.

1035
01:13:34,610 --> 01:13:37,480
So this one is king.

1036
01:13:37,480 --> 01:13:41,650
And this one, actually--
let's have a look.

1037
01:13:41,650 --> 01:13:44,490
It pecks this one this
one, and it virtually

1038
01:13:44,490 --> 01:13:45,450
also pecked this one.

1039
01:13:45,450 --> 01:13:45,950
Yay.

1040
01:13:45,950 --> 01:13:49,640
He has a friend over here
that is doing that for him.

1041
01:13:49,640 --> 01:13:51,730
So this one is
actually also a king.

1042
01:13:51,730 --> 01:13:55,740
So you can have multiple
king chickens in here.

1043
01:13:55,740 --> 01:13:56,820
What about this one?

1044
01:13:56,820 --> 01:14:05,550
The same story-- pecks
this one, pecks this one,

1045
01:14:05,550 --> 01:14:07,710
and virtually-- wait a minute.

1046
01:14:12,780 --> 01:14:16,400
The one on the left--
oh yeah, over here.

1047
01:14:16,400 --> 01:14:18,170
This, this.

1048
01:14:18,170 --> 01:14:20,480
So this one is king as well.

1049
01:14:20,480 --> 01:14:22,510
Now what about this one?

1050
01:14:22,510 --> 01:14:25,570
Well, it can peck this one.

1051
01:14:25,570 --> 01:14:28,050
And then in one
more step from here,

1052
01:14:28,050 --> 01:14:30,280
because there's only
one outgoing edge,

1053
01:14:30,280 --> 01:14:33,360
it can virtually peck this
one, but not this one.

1054
01:14:33,360 --> 01:14:36,740
So this one is definitely
the loser of the four.

1055
01:14:36,740 --> 01:14:41,610
So he's not the king.

1056
01:14:41,610 --> 01:14:46,200
So now what we want to
prove this a theorem

1057
01:14:46,200 --> 01:14:49,221
sort of trans-identify
one of the chickens

1058
01:14:49,221 --> 01:14:50,970
that we know for sure
is going to be king.

1059
01:14:50,970 --> 01:14:52,950
So you can have multiple kings.

1060
01:14:52,950 --> 01:14:56,060
But maybe there's one chicken
that, from our intuition,

1061
01:14:56,060 --> 01:15:00,890
we may feel is definitely
going to be king.

1062
01:15:00,890 --> 01:15:03,420
So what will be
a good intuition?

1063
01:15:03,420 --> 01:15:06,130
So we're talking here
about virtual pecking.

1064
01:15:06,130 --> 01:15:07,670
We have this definition.

1065
01:15:07,670 --> 01:15:13,560
So what kind of node
in a tournament graph

1066
01:15:13,560 --> 01:15:21,727
essentially would be-- can we
know for sure that it's a king?

1067
01:15:21,727 --> 01:15:23,310
Do you have an
intuition for a theorem

1068
01:15:23,310 --> 01:15:25,830
that you may want to prove?

1069
01:15:25,830 --> 01:15:28,460
So that's often what
we do in mathematics.

1070
01:15:28,460 --> 01:15:31,080
We have some kind of
funny new structure,

1071
01:15:31,080 --> 01:15:32,741
and then we want to
find out whether we

1072
01:15:32,741 --> 01:15:34,490
can prove interesting
properties about it.

1073
01:15:34,490 --> 01:15:39,180
So we start to search for actual
nice properties and theorems.

1074
01:15:39,180 --> 01:15:44,350
So in this case, it makes
sense that the vertex

1075
01:15:44,350 --> 01:15:49,490
that has the most outgoing
edges may be always king.

1076
01:15:49,490 --> 01:15:51,820
Can we prove this?

1077
01:15:51,820 --> 01:15:53,920
So that's what
we're going to do.

1078
01:15:53,920 --> 01:15:55,950
So that's the theorem.

1079
01:15:55,950 --> 01:16:03,050
And let's see whether we can
do this in an elegant way.

1080
01:16:03,050 --> 01:16:06,030
So the theorem is
that even though there

1081
01:16:06,030 --> 01:16:07,730
are multiple
kings-- as indicated

1082
01:16:07,730 --> 01:16:09,970
in that particular
example, that may happen--

1083
01:16:09,970 --> 01:16:19,592
but I certainly know that the
chicken that has the highest

1084
01:16:19,592 --> 01:16:24,506
outdegree is definitely a king.

1085
01:16:30,590 --> 01:16:35,820
And the way we're going to
prove this is by contradiction.

1086
01:16:35,820 --> 01:16:38,080
Let's assume that's
not the case.

1087
01:16:38,080 --> 01:16:39,860
That must be really,
really weird.

1088
01:16:39,860 --> 01:16:43,140
If you are the one who
has the largest outdegree,

1089
01:16:43,140 --> 01:16:46,130
it means that you can
directly, just by yourself,

1090
01:16:46,130 --> 01:16:47,760
peck the most others.

1091
01:16:47,760 --> 01:16:50,180
So suppose you're not king.

1092
01:16:53,272 --> 01:17:03,440
So by contradiction,
first of all,

1093
01:17:03,440 --> 01:17:06,650
let U have the
highest outdegree.

1094
01:17:10,910 --> 01:17:16,010
And we want to show
that this U is king.

1095
01:17:16,010 --> 01:17:19,250
So let's assume the contrary.

1096
01:17:19,250 --> 01:17:24,475
So let's suppose
that U is not king.

1097
01:17:29,810 --> 01:17:31,960
So what does that mean?

1098
01:17:31,960 --> 01:17:35,930
So let's look at a
definition over there

1099
01:17:35,930 --> 01:17:41,790
and see what it means
that U is not king.

1100
01:17:44,730 --> 01:17:50,470
So that means that both
those conditions are violated

1101
01:17:50,470 --> 01:17:55,390
because if one of
those two holds,

1102
01:17:55,390 --> 01:17:57,880
I know there must
be one vertex, V,

1103
01:17:57,880 --> 01:18:03,050
such that U does not virtually
peck V. So I know that.

1104
01:18:03,050 --> 01:18:07,860
So let's see what that implies.

1105
01:18:11,020 --> 01:18:13,665
So I know that there
must be a V such

1106
01:18:13,665 --> 01:18:22,060
that U does not virtually peck
V. So maybe can help me out.

1107
01:18:22,060 --> 01:18:23,900
What does that mean?

1108
01:18:23,900 --> 01:18:26,775
It means that both these
conditions are not true.

1109
01:18:26,775 --> 01:18:27,900
So let's look at the first.

1110
01:18:27,900 --> 01:18:31,320
So U pecks V. If
that's not true,

1111
01:18:31,320 --> 01:18:33,350
and we're in the
tournament graph,

1112
01:18:33,350 --> 01:18:40,350
we know that V must
peck U. So we have this.

1113
01:18:40,350 --> 01:18:43,560
and we also know that the
other condition, the second one

1114
01:18:43,560 --> 01:18:45,580
over here, does not hold.

1115
01:18:45,580 --> 01:18:51,120
So what's the negation
of this second condition?

1116
01:18:51,120 --> 01:18:53,500
Maybe you can help me out.

1117
01:18:53,500 --> 01:18:57,720
So we have here, there
exists a W such that U pecks

1118
01:18:57,720 --> 01:19:02,470
W and W pecks V. So how do we
negate that logical expression?

1119
01:19:06,059 --> 01:19:08,930
AUDIENCE: [INAUDIBLE]

1120
01:19:08,930 --> 01:19:14,325
PROFESSOR: For all W, what's
over there is not through true.

1121
01:19:14,325 --> 01:19:16,200
So can we formulate that
a little bit better?

1122
01:19:16,200 --> 01:19:23,820
So it's not true that U pecks
W and W pecks V. So that

1123
01:19:23,820 --> 01:19:32,520
means that either U
pecks W is not true,

1124
01:19:32,520 --> 01:19:35,579
or W pecks V is not true.

1125
01:19:35,579 --> 01:19:36,620
So let's write that down.

1126
01:19:40,550 --> 01:19:41,780
Let's just write it all out.

1127
01:19:41,780 --> 01:19:54,460
So not U pecks W or
not W pecks V. Well,

1128
01:19:54,460 --> 01:19:56,520
how can we rewrite this?

1129
01:19:56,520 --> 01:19:58,110
Well, it's a
tournament graph, so we

1130
01:19:58,110 --> 01:20:09,110
know that W pecks U. Or this
particular condition, which

1131
01:20:09,110 --> 01:20:14,640
is V pecks W.

1132
01:20:14,640 --> 01:20:15,649
So what do I have here?

1133
01:20:15,649 --> 01:20:17,190
Is this going in
the right direction?

1134
01:20:20,540 --> 01:20:28,780
Well, I want to prove something
about-- if I use contradiction

1135
01:20:28,780 --> 01:20:32,110
and I suppose that
U is not the king,

1136
01:20:32,110 --> 01:20:34,600
I've assumed that U has
the highest outdegree.

1137
01:20:34,600 --> 01:20:36,690
So I want to show that
I somehow violated.

1138
01:20:36,690 --> 01:20:41,680
So somehow I'm able to
construct some vertex, V.

1139
01:20:41,680 --> 01:20:44,030
And by negating
that U is not a king

1140
01:20:44,030 --> 01:20:47,880
it seems that this vertex, V,
makes a really good candidate

1141
01:20:47,880 --> 01:20:50,830
to show that there's a higher
degree outdegree than U.

1142
01:20:50,830 --> 01:20:52,430
So let's see whether
we can do this.

1143
01:20:55,040 --> 01:20:58,070
We can rewrite this logical
expression once more.

1144
01:20:58,070 --> 01:21:05,990
We can also say that, well,
if U pecks W-- so this

1145
01:21:05,990 --> 01:21:10,090
is not true-- then
it must be true

1146
01:21:10,090 --> 01:21:15,700
that this one holds because
if that's not the case,

1147
01:21:15,700 --> 01:21:17,720
then this condition is not true.

1148
01:21:17,720 --> 01:21:21,280
So if U pecks W, then
this is not true.

1149
01:21:21,280 --> 01:21:23,200
So then it must be the
case that that is true.

1150
01:21:23,200 --> 01:21:28,960
So V pecks W. So now
let's have a look at V.

1151
01:21:28,960 --> 01:21:35,130
We noticed that for all
outgoing edges from U,

1152
01:21:35,130 --> 01:21:39,890
there exists a similar
outgoing edge for V.

1153
01:21:39,890 --> 01:21:42,570
But V has one more
outgoing edge.

1154
01:21:42,570 --> 01:21:45,030
V, actually, is an
outgoing edge to U.

1155
01:21:45,030 --> 01:21:46,870
So what do we see here?

1156
01:21:46,870 --> 01:21:57,180
That the outdegree of V is
actually at least the outdegree

1157
01:21:57,180 --> 01:22:02,040
of U-- which is this
particular condition over here

1158
01:22:02,040 --> 01:22:04,100
that we show here,
for all the W we

1159
01:22:04,100 --> 01:22:08,690
have that this is true-- plus
and we have an extra one.

1160
01:22:08,690 --> 01:22:09,990
It's this one.

1161
01:22:09,990 --> 01:22:11,500
Oh, but now we have
a contradiction

1162
01:22:11,500 --> 01:22:14,830
because we said that U
has the highest degree.

1163
01:22:14,830 --> 01:22:16,750
But it turns out that
you have constructed one

1164
01:22:16,750 --> 01:22:18,750
that has a higher outdegree.

1165
01:22:18,750 --> 01:22:20,810
So that's a contradiction.

1166
01:22:20,810 --> 01:22:25,260
That means that our original
assumption is actually wrong.

1167
01:22:25,260 --> 01:22:29,010
So suppose that U is not the
king was a wrong assumption,

1168
01:22:29,010 --> 01:22:31,387
and U must be king.

1169
01:22:31,387 --> 01:22:32,720
So that's the end of this proof.

1170
01:22:36,290 --> 01:22:38,780
This is the end of this lecture.

1171
01:22:38,780 --> 01:22:40,710
So see you tomorrow
at recitation

1172
01:22:40,710 --> 01:22:43,540
and next week we will continue
with communication graphs

1173
01:22:43,540 --> 01:22:45,970
and partial orderings.