1
00:00:01,310 --> 00:00:03,740
PROFESSOR: Now we can use the
unique path characterization

2
00:00:03,740 --> 00:00:05,520
of trees to very
quickly figure out

3
00:00:05,520 --> 00:00:07,280
that every tree is 2-colorable.

4
00:00:07,280 --> 00:00:10,900
So we know that a tree is
a graph with unique paths

5
00:00:10,900 --> 00:00:13,020
between every pair of vertices.

6
00:00:13,020 --> 00:00:15,620
And as a consequence,
the chromatic number

7
00:00:15,620 --> 00:00:21,620
of a tree with two or
more vertices is 2.

8
00:00:21,620 --> 00:00:24,120
The proof is just to
show you how to color it.

9
00:00:24,120 --> 00:00:26,380
You clearly can't
get by with one color

10
00:00:26,380 --> 00:00:29,060
if you've got any two
adjacent vertices.

11
00:00:29,060 --> 00:00:34,220
The 2-colorable way is that you
just choose an arbitrary vertex

12
00:00:34,220 --> 00:00:37,850
and-- call it the root-- but
you make the arbitrary choice

13
00:00:37,850 --> 00:00:39,020
on what the root is.

14
00:00:39,020 --> 00:00:46,080
And there's a unique path
from the root to every vertex,

15
00:00:46,080 --> 00:00:48,340
using this unique
path characterization.

16
00:00:48,340 --> 00:00:50,210
And so we're just
going to color vertices

17
00:00:50,210 --> 00:00:54,370
by whether the path from the
root is of odd or even length.

18
00:00:54,370 --> 00:00:56,330
If it's of even
length, color it red.

19
00:00:56,330 --> 00:00:59,410
And if it's odd
length, color it green.

20
00:00:59,410 --> 00:01:02,420
And so we wind up
alternating red and green.

21
00:01:02,420 --> 00:01:05,330
And the fact is
that adjacent nodes

22
00:01:05,330 --> 00:01:07,800
are going to be at a distance
where one is an odd distance

23
00:01:07,800 --> 00:01:09,216
and one is an even
distance, which

24
00:01:09,216 --> 00:01:11,600
is why this method of
coloring is going to work.

25
00:01:11,600 --> 00:01:13,470
A general property
of 2-coloring is

26
00:01:13,470 --> 00:01:15,870
that to figure out whether
or not a graph is 2-colorable

27
00:01:15,870 --> 00:01:18,670
and how to do it,
is you just start.

28
00:01:18,670 --> 00:01:21,300
Pick an arbitrary
vertex, color it red.

29
00:01:21,300 --> 00:01:24,040
And then color all the
vertices adjacent to it green.

30
00:01:24,040 --> 00:01:28,620
And keep going in that
way, coloring a vertex

31
00:01:28,620 --> 00:01:32,440
with a color different from an
adjacent vertex that's colored,

32
00:01:32,440 --> 00:01:33,340
until you get stuck.

33
00:01:33,340 --> 00:01:35,020
If you don't get stuck
it's 2-colorable.

34
00:01:35,020 --> 00:01:37,960
And if it's not 2-colorable,
you're guaranteed to get stuck.

35
00:01:37,960 --> 00:01:40,160
So it's a very easy
way to figure out

36
00:01:40,160 --> 00:01:41,950
if a graph is 2-colorable.

37
00:01:41,950 --> 00:01:46,630
Another characterization of
2-colorability in general,

38
00:01:46,630 --> 00:01:51,110
is that a graph is
2-colorable providing

39
00:01:51,110 --> 00:01:54,820
that all the cycles that it has,
if any, are of even lengths.

40
00:01:54,820 --> 00:01:56,780
Of course, a tree has
no cycles, so that

41
00:01:56,780 --> 00:02:00,110
makes it 2-colorable for sure.