1
00:00:00,080 --> 00:00:02,430
The following content is
provided under a Creative

2
00:00:02,430 --> 00:00:03,820
Commons license.

3
00:00:03,820 --> 00:00:06,050
Your support will help
MIT OpenCourseWare

4
00:00:06,050 --> 00:00:10,150
continue to offer high quality
educational resources for free.

5
00:00:10,150 --> 00:00:12,690
To make a donation or to
view additional materials

6
00:00:12,690 --> 00:00:16,600
from hundreds of MIT courses,
visit MIT OpenCourseWare

7
00:00:16,600 --> 00:00:17,305
at ocw.mit.edu.

8
00:00:26,920 --> 00:00:28,900
PROFESSOR: Today, we
begin a new topic, which

9
00:00:28,900 --> 00:00:30,185
is parameterized complexity.

10
00:00:32,900 --> 00:00:38,730
This is one of the
most recent areas

11
00:00:38,730 --> 00:00:41,430
of hardness that
has been invented

12
00:00:41,430 --> 00:00:44,915
that we will cover so far.

13
00:00:44,915 --> 00:00:48,940
It started around 1999,
and then since the 2000s,

14
00:00:48,940 --> 00:00:50,780
it's been getting
really big, originally

15
00:00:50,780 --> 00:00:52,960
by Downey and Fellows.

16
00:00:52,960 --> 00:00:54,510
And the general
idea is we're going

17
00:00:54,510 --> 00:00:56,570
to take problems,
which normally we

18
00:00:56,570 --> 00:00:59,990
think of as being parameterized
by the problem size, n,

19
00:00:59,990 --> 00:01:02,995
and thinking about polynomial
versus exponential growth of n.

20
00:01:02,995 --> 00:01:05,330
And we're going to
add another parameter.

21
00:01:05,330 --> 00:01:07,350
So you could say this
is a two parameter

22
00:01:07,350 --> 00:01:08,910
analysis in some sense.

23
00:01:08,910 --> 00:01:12,140
Usually, we call problem
size n and then we

24
00:01:12,140 --> 00:01:14,980
call the parameter k.

25
00:01:14,980 --> 00:01:17,530
In general, a parameter
is just some way

26
00:01:17,530 --> 00:01:19,690
of measuring instances.

27
00:01:19,690 --> 00:01:26,520
So parameter, k, is a
function from instances

28
00:01:26,520 --> 00:01:31,950
to non-negative
integers, 0, 1, 2.

29
00:01:31,950 --> 00:01:38,030
And usually, this number k
is just part of the instance,

30
00:01:38,030 --> 00:01:49,060
for example-- well, let me
go over here to examples.

31
00:01:49,060 --> 00:01:57,260
Let's say problem that's
called k-vertex cover--

32
00:01:57,260 --> 00:02:00,450
and usually, I'll just call it
vertex cover for simplicity.

33
00:02:00,450 --> 00:02:03,230
You know the problem
vertices to cover edges.

34
00:02:03,230 --> 00:02:06,900
And usually, the decision
version of that is

35
00:02:06,900 --> 00:02:10,759
are there k vertices
that cover all edges?

36
00:02:18,240 --> 00:02:20,200
So k is an input
to this problem,

37
00:02:20,200 --> 00:02:22,580
and it appears in the
statement of the problem.

38
00:02:22,580 --> 00:02:24,360
But in particular,
it's an input.

39
00:02:24,360 --> 00:02:27,480
And so this function, k,
of a vertex cover instance

40
00:02:27,480 --> 00:02:31,580
just pulls out that one number
and throws away the graph.

41
00:02:31,580 --> 00:02:34,560
That's the typical parameter.

42
00:02:34,560 --> 00:02:37,350
But in general, a parameter
could be some hard to compute

43
00:02:37,350 --> 00:02:38,560
thing.

44
00:02:38,560 --> 00:02:41,610
Maybe k-- you think
of the parameter--

45
00:02:41,610 --> 00:02:43,570
you take vertex cover
as your problem,

46
00:02:43,570 --> 00:02:45,030
but you define your
parameter to be

47
00:02:45,030 --> 00:02:48,770
the smallest independent set--
or largest independent set.

48
00:02:48,770 --> 00:02:53,810
Smallest is pretty small, 0.

49
00:02:53,810 --> 00:02:55,750
Or, pick your favorite.

50
00:02:55,750 --> 00:03:01,700
You could, for example, think
about vertex cover with respect

51
00:03:01,700 --> 00:03:07,780
to-- this is how we'll write
the parameter-- crossing number.

52
00:03:14,220 --> 00:03:19,590
So the idea here is-- sorry
I forgot to covet-- we always

53
00:03:19,590 --> 00:03:21,090
measure our instances
with respect

54
00:03:21,090 --> 00:03:23,340
to the problem size, which
is the size of the graph

55
00:03:23,340 --> 00:03:25,860
for vertex cover instance.

56
00:03:25,860 --> 00:03:27,840
But we're also going
to parameterize

57
00:03:27,840 --> 00:03:29,351
by some other quantity here.

58
00:03:29,351 --> 00:03:31,600
Maybe it's the minimum
crossing number for your graph,

59
00:03:31,600 --> 00:03:34,000
so 0 if it's planar and so on.

60
00:03:34,000 --> 00:03:37,390
We prove that that's
NP-complete to compute.

61
00:03:37,390 --> 00:03:41,610
So this parameter may not be
easy to compute in general.

62
00:03:41,610 --> 00:03:44,690
But a lot of the times it's
just part of the problem.

63
00:03:44,690 --> 00:03:49,386
And in general, this is called
the natural parameterization.

64
00:03:54,840 --> 00:03:56,450
If you have an
optimization problem,

65
00:03:56,450 --> 00:03:59,320
like vertex cover is--
minimum vertex cover--

66
00:03:59,320 --> 00:04:01,930
and you convert the optimization
problem into a decision

67
00:04:01,930 --> 00:04:04,580
problem, you're asking is
opt less than or equal to k.

68
00:04:04,580 --> 00:04:07,790
That k is the natural parameter
for an optimization problem.

69
00:04:07,790 --> 00:04:10,360
So most the time, we'll
be thinking about that.

70
00:04:10,360 --> 00:04:13,820
But things like this also arise.

71
00:04:13,820 --> 00:04:21,300
So just to be a
little more precise

72
00:04:21,300 --> 00:04:27,095
a parameterized problem
is a decision problem

73
00:04:27,095 --> 00:04:27,920
plus the parameter.

74
00:04:36,270 --> 00:04:39,370
So when I said vertex cover
with respect to crossing number,

75
00:04:39,370 --> 00:04:42,330
I meant my decision
problem is vertex cover,

76
00:04:42,330 --> 00:04:45,070
and I'm going to choose to
use this particular parameter

77
00:04:45,070 --> 00:04:48,210
function to parameterize
those instances.

78
00:04:48,210 --> 00:04:52,850
And in general, we'll call them
parameter k and problem size n.

79
00:04:52,850 --> 00:04:57,280
And our goal is to get a
very good dependence on n

80
00:04:57,280 --> 00:05:00,420
at the cost of a
bad dependence on k.

81
00:05:00,420 --> 00:05:07,930
So let me start with a
so-so dependents, XP, would

82
00:05:07,930 --> 00:05:19,570
be the set of all prioritized
problems solvable in n

83
00:05:19,570 --> 00:05:23,450
to the f of k time for some
function k-- for any function

84
00:05:23,450 --> 00:05:26,150
k, any function f.

85
00:05:26,150 --> 00:05:31,370
And FPT is the set of
parameterized problem solvable

86
00:05:31,370 --> 00:05:37,250
in f of k times
n to the order 1.

87
00:05:37,250 --> 00:05:39,000
So this is considered
a good running time.

88
00:05:39,000 --> 00:05:42,820
This is considered
a bad running time.

89
00:05:42,820 --> 00:05:44,620
F is presumably exponential.

90
00:05:44,620 --> 00:05:46,010
Assuming your
problem is NP-hard,

91
00:05:46,010 --> 00:05:48,040
you have to have
exponential in something.

92
00:05:48,040 --> 00:05:50,890
And the goal is to get the
exponential away from n

93
00:05:50,890 --> 00:05:52,410
because n is hard to control.

94
00:05:52,410 --> 00:05:55,837
We like to solve big problems,
but maybe we can somehow

95
00:05:55,837 --> 00:05:57,670
characterize that the
problems we care about

96
00:05:57,670 --> 00:06:02,280
have some small measure, k--
for some interesting measure k.

97
00:06:02,280 --> 00:06:04,626
If such an algorithm is
possible and, in practice,

98
00:06:04,626 --> 00:06:06,250
your k's are small,
then you're golden.

99
00:06:06,250 --> 00:06:07,624
These give really
good algorithms

100
00:06:07,624 --> 00:06:09,750
for solving lots of problems.

101
00:06:09,750 --> 00:06:13,660
Any problem in FPT, if k
is somewhat reasonable,

102
00:06:13,660 --> 00:06:15,810
even when n is huge, we
can solve the problem.

103
00:06:18,840 --> 00:06:23,130
This running time is also
polynomial for fixed k,

104
00:06:23,130 --> 00:06:26,140
but the polynomial
changes depending on k.

105
00:06:26,140 --> 00:06:28,420
So here, we typically get
linear time algorithms

106
00:06:28,420 --> 00:06:29,890
for any fixed k.

107
00:06:29,890 --> 00:06:31,940
Here, as k increases,
the polynomial

108
00:06:31,940 --> 00:06:32,990
gets bigger and bigger.

109
00:06:32,990 --> 00:06:34,720
And in practice,
you probably can't

110
00:06:34,720 --> 00:06:37,960
handle more than an n squared
algorithm for a large n.

111
00:06:37,960 --> 00:06:41,340
So this is considered not
useful even for small k.

112
00:06:41,340 --> 00:06:43,860
This is considered very
useful for small k,

113
00:06:43,860 --> 00:06:47,740
depending on your notion
of small, of course.

114
00:06:47,740 --> 00:06:50,690
And parameterized complexity,
the main name of the game

115
00:06:50,690 --> 00:06:53,320
is distinguishing between FPT.

116
00:06:53,320 --> 00:06:57,310
And XP is actually rather large,
but distinguishing between when

117
00:06:57,310 --> 00:06:59,445
this is possible and
when it is impossible.

118
00:07:03,280 --> 00:07:03,780
Cool.

119
00:07:03,780 --> 00:07:07,551
So for example,
vertex cover is FPT.

120
00:07:07,551 --> 00:07:09,800
Vertex cover with respect
to crossing number, I think,

121
00:07:09,800 --> 00:07:10,630
is FPT.

122
00:07:10,630 --> 00:07:13,080
I have to think about
that a little bit more.

123
00:07:13,080 --> 00:07:18,960
AUDIENCE: Can you say what
XP and FPT roughly stand for?

124
00:07:18,960 --> 00:07:21,450
PROFESSOR: Oh, right.

125
00:07:21,450 --> 00:07:25,790
FPT is fixed
parameter tractable.

126
00:07:25,790 --> 00:07:27,150
That's the good word.

127
00:07:30,750 --> 00:07:33,790
intuitive--
etymological that means

128
00:07:33,790 --> 00:07:37,360
when you fix the parameter k,
you get a tractable problem

129
00:07:37,360 --> 00:07:40,300
although it's a strong
sense of tractable.

130
00:07:40,300 --> 00:07:43,410
You can say it is n
squared for any fixed k.

131
00:07:43,410 --> 00:07:45,570
Or, it's n to the
fifth for any fixed k.

132
00:07:45,570 --> 00:07:48,320
This constant does
not depend on k.

133
00:07:51,720 --> 00:07:52,980
But the lead constant does.

134
00:07:56,720 --> 00:07:58,740
So that's FPT.

135
00:07:58,740 --> 00:08:03,980
XP, experience points.

136
00:08:10,047 --> 00:08:10,630
Good question.

137
00:08:13,230 --> 00:08:14,760
One other good thing to know.

138
00:08:17,440 --> 00:08:21,840
You might wonder why f
of k times polynomial n.

139
00:08:21,840 --> 00:08:26,350
Maybe I could hope for better
like f of k plus polynomial n.

140
00:08:26,350 --> 00:08:29,540
That actually is the same class.

141
00:08:29,540 --> 00:08:35,440
So FPT also equals the set
of all parameterized problems

142
00:08:35,440 --> 00:08:44,310
solvable in f of k
plus n to order 1.

143
00:08:44,310 --> 00:08:45,560
That's a nice general theorem.

144
00:08:45,560 --> 00:08:47,020
I won't bother proving it here.

145
00:08:47,020 --> 00:08:50,180
I think the f ends up becoming
square of the original f.

146
00:08:50,180 --> 00:08:52,410
It's a very easy case
analysis based on whether

147
00:08:52,410 --> 00:08:55,376
which of these two
things is bigger.

148
00:08:55,376 --> 00:08:56,750
I will leave that
as an exercise.

149
00:08:56,750 --> 00:08:58,375
It would be a good
problem set problem.

150
00:09:01,460 --> 00:09:07,390
Cool, so that answers
one plausible question.

151
00:09:07,390 --> 00:09:09,520
Let me give you an example
of an FPT algorithm,

152
00:09:09,520 --> 00:09:12,760
let's say, for vertex cover.

153
00:09:12,760 --> 00:09:17,240
So back over here
to vertex cover.

154
00:09:17,240 --> 00:09:20,240
How about an XP algorithm
for vertex cover.

155
00:09:20,240 --> 00:09:23,570
So that's really easy,
not very interesting.

156
00:09:23,570 --> 00:09:27,610
You could just guess a vertex
cover, try all possible vertex

157
00:09:27,610 --> 00:09:28,690
covers of size k.

158
00:09:28,690 --> 00:09:31,023
Remember, we want to know
whether there's a vertex cover

159
00:09:31,023 --> 00:09:34,370
size k. k is given
to us, so we could

160
00:09:34,370 --> 00:09:38,980
try all vertex covers--
or, I shouldn't

161
00:09:38,980 --> 00:09:42,310
say that-- all vertex
sets of size k.

162
00:09:46,660 --> 00:09:49,120
There's only n to the k of them.

163
00:09:49,120 --> 00:09:52,470
And for each of them,
we'll spend linear time

164
00:09:52,470 --> 00:09:54,270
to check whether
it's a vertex cover.

165
00:09:54,270 --> 00:09:57,240
So this proves that
the problem is in XP.

166
00:09:57,240 --> 00:09:58,140
That's trivial.

167
00:09:58,140 --> 00:09:59,980
Most problems you'll
find are in XP

168
00:09:59,980 --> 00:10:02,210
with reasonable
parameterization.

169
00:10:02,210 --> 00:10:04,650
On some, it's not obvious.

170
00:10:04,650 --> 00:10:07,714
And now, let's prove
that it's in FPT.

171
00:10:07,714 --> 00:10:09,130
This is a little
more interesting.

172
00:10:09,130 --> 00:10:11,230
There are lots of FPT
algorithms for vertex cover.

173
00:10:11,230 --> 00:10:12,600
It's an active area.

174
00:10:12,600 --> 00:10:15,630
Every year there's a better
algorithm pretty much.

175
00:10:15,630 --> 00:10:16,488
Question?

176
00:10:16,488 --> 00:10:19,764
AUDIENCE: So in the inputs of
the problem, the k is an input?

177
00:10:22,399 --> 00:10:24,190
PROFESSOR: For vertex
cover, k is an input.

178
00:10:24,190 --> 00:10:26,242
AUDIENCE: Oh, because you
can also think of vertex

179
00:10:26,242 --> 00:10:27,382
cover where k is not
an input, but you

180
00:10:27,382 --> 00:10:28,294
can analyze the run time.

181
00:10:28,294 --> 00:10:28,750
PROFESSOR: Yes.

182
00:10:28,750 --> 00:10:29,249
Right.

183
00:10:29,249 --> 00:10:36,990
So you could also think about
vertex cover with respect

184
00:10:36,990 --> 00:10:40,180
to vertex cover.

185
00:10:40,180 --> 00:10:44,850
So here, let's say-- this
is a little bit funny,

186
00:10:44,850 --> 00:10:50,510
but here the idea
is k, the parameter,

187
00:10:50,510 --> 00:10:52,440
is the minimum vertex cover.

188
00:10:52,440 --> 00:10:54,750
Let me write minimum
vertex cover.

189
00:10:54,750 --> 00:10:56,920
And here, you're given
some other value.

190
00:10:56,920 --> 00:10:58,320
Let's call it j.

191
00:10:58,320 --> 00:11:01,960
And you want to decide whether
vertex cover is at most j,

192
00:11:01,960 --> 00:11:03,510
but you're not told what k is.

193
00:11:03,510 --> 00:11:06,010
You're not told what
the optimal value is.

194
00:11:06,010 --> 00:11:07,440
So this is a little bit subtle.

195
00:11:07,440 --> 00:11:08,990
It's essentially
the same problem.

196
00:11:08,990 --> 00:11:14,810
You're just not told
the critical value.

197
00:11:14,810 --> 00:11:19,680
But if you can solve k
vertex cover when k is given,

198
00:11:19,680 --> 00:11:23,410
you can just run that
algorithm for k equals 1, 2, 4,

199
00:11:23,410 --> 00:11:25,290
and you'll end up
solving this problem.

200
00:11:25,290 --> 00:11:27,830
So even though--
so in this problem,

201
00:11:27,830 --> 00:11:30,560
you can say, well, if I discover
there's no vertex cover of size

202
00:11:30,560 --> 00:11:33,030
k, I can just return no.

203
00:11:33,030 --> 00:11:35,440
Here, it's a little
more annoying, but just

204
00:11:35,440 --> 00:11:38,266
by iterating and stopping when
you get the first vertex cover,

205
00:11:38,266 --> 00:11:39,640
your algorithm
will end up having

206
00:11:39,640 --> 00:11:41,520
a good running time
with respect to k

207
00:11:41,520 --> 00:11:43,440
even though you didn't know it.

208
00:11:43,440 --> 00:11:44,620
So it's a subtlety.

209
00:11:44,620 --> 00:11:46,580
In general, this is
messy to work with,

210
00:11:46,580 --> 00:11:49,150
so we usually think
about k being an input.

211
00:11:49,150 --> 00:11:52,320
But that reduction shows
they're more or less the same.

212
00:11:52,320 --> 00:11:54,660
But the complexity theory
for this is annoying.

213
00:11:54,660 --> 00:11:58,800
In the same way that for NP, we
restricted decision problems,

214
00:11:58,800 --> 00:12:00,460
here we're going
to usually restrict

215
00:12:00,460 --> 00:12:01,580
to the natural parameter.

216
00:12:04,900 --> 00:12:05,700
So good question.

217
00:12:08,270 --> 00:12:12,240
Let me prove vertex cover is in
FPT, so a much better running

218
00:12:12,240 --> 00:12:13,160
time than n to the k.

219
00:12:13,160 --> 00:12:17,820
I think I will get something
like 2 to the k times n.

220
00:12:17,820 --> 00:12:20,260
Look at an edge of
the graph, any edge.

221
00:12:20,260 --> 00:12:22,000
Pick an edge, any edge.

222
00:12:22,000 --> 00:12:24,330
And we know for this
to be a vertex cover,

223
00:12:24,330 --> 00:12:26,480
one of the two endpoints
has to be in there.

224
00:12:26,480 --> 00:12:27,120
Which one?

225
00:12:27,120 --> 00:12:27,680
I don't know.

226
00:12:27,680 --> 00:12:28,920
Guess.

227
00:12:28,920 --> 00:12:33,680
So I'm going to guess either
the left guy or the right vertex

228
00:12:33,680 --> 00:12:36,490
is in the vertex
cover, and then repeat.

229
00:12:36,490 --> 00:12:38,600
So when I put somebody
in the vertex cover,

230
00:12:38,600 --> 00:12:41,575
I delete all of
the incident edges.

231
00:12:45,360 --> 00:12:46,800
That takes linear time.

232
00:12:46,800 --> 00:12:48,820
Over here, I would
delete whatever

233
00:12:48,820 --> 00:12:51,600
edges are incident to this
guy and the vertex itself,

234
00:12:51,600 --> 00:12:54,300
of course.

235
00:12:54,300 --> 00:12:58,010
And so I get a smaller graph.

236
00:12:58,010 --> 00:13:04,430
I do this k times, so in
general, my execution tree

237
00:13:04,430 --> 00:13:09,000
will have two branches, every
time I pick an edge and say,

238
00:13:09,000 --> 00:13:11,550
do I put the left guy
or the right guy in.

239
00:13:11,550 --> 00:13:16,246
But I only need to worry
about a tree of height k

240
00:13:16,246 --> 00:13:17,620
because after I
make k decisions,

241
00:13:17,620 --> 00:13:19,900
I'm supposed to get a
vertex cover of size k.

242
00:13:19,900 --> 00:13:21,400
After I make k
decisions, if there's

243
00:13:21,400 --> 00:13:22,941
anything left to
the graph that means

244
00:13:22,941 --> 00:13:25,720
that wasn't a vertex cover of
size k and I should backtrack.

245
00:13:25,720 --> 00:13:27,670
So I just explore
this tree, look

246
00:13:27,670 --> 00:13:29,030
at all the different leaves.

247
00:13:29,030 --> 00:13:32,890
If there's any vertex cover
of size k, I will find it.

248
00:13:32,890 --> 00:13:37,340
And so I only spend 2 to
the k to explore this tree.

249
00:13:37,340 --> 00:13:39,270
2 to the k is the size
of the tree times n

250
00:13:39,270 --> 00:13:43,380
because I maybe delete
n things at every node.

251
00:13:43,380 --> 00:13:45,964
And that's one of
the simplest FPT

252
00:13:45,964 --> 00:13:47,130
algorithms for vertex cover.

253
00:13:47,130 --> 00:13:48,759
There are much
cooler, fancier ones.

254
00:13:48,759 --> 00:13:50,550
But this is now going
to be about hardness,

255
00:13:50,550 --> 00:13:51,841
so I will stop with algorithms.

256
00:13:55,470 --> 00:14:01,770
Also related is a
stronger notion of PTAS.

257
00:14:04,910 --> 00:14:09,470
So I mentioned
before, with PTAS, you

258
00:14:09,470 --> 00:14:14,501
could have a running time
something like n to the 2

259
00:14:14,501 --> 00:14:16,390
to the 1 over epsilon--
which is actually

260
00:14:16,390 --> 00:14:20,120
fairly common in a lot
of PTASes, not so great.

261
00:14:20,120 --> 00:14:23,100
If epsilon is anything
good, then this

262
00:14:23,100 --> 00:14:26,400
is going to be a
really huge polynomial.

263
00:14:26,400 --> 00:14:30,090
Well, the corresponding
notion in FPT land

264
00:14:30,090 --> 00:14:37,770
is a running time
of f of epsilon--

265
00:14:37,770 --> 00:14:39,020
or, I'll write 1 over epsilon.

266
00:14:39,020 --> 00:14:43,630
It doesn't matter--
times n to the order 1.

267
00:14:43,630 --> 00:14:50,000
So this is really saying that
the approximation scheme is FPT

268
00:14:50,000 --> 00:14:52,750
with respect to 1 over epsilon.

269
00:14:52,750 --> 00:14:56,630
This is to get a 1 plus
epsilon approximation.

270
00:14:56,630 --> 00:14:58,470
So this is relating
the two worlds.

271
00:14:58,470 --> 00:15:01,262
If you want a FPT algorithm
for approximating a problem

272
00:15:01,262 --> 00:15:03,470
and getting within a factor
of 1 over epsilon, if you

273
00:15:03,470 --> 00:15:06,190
get this kind of
running time verses

274
00:15:06,190 --> 00:15:10,940
this kind of running time, I
consider this good, this bad.

275
00:15:10,940 --> 00:15:16,690
And in general, these are called
Efficient PTASes, or EPTAS.

276
00:15:16,690 --> 00:15:20,420
And in general, if
you have an EPTAS,

277
00:15:20,420 --> 00:15:25,510
you also have an FPT
algorithm with respect

278
00:15:25,510 --> 00:15:36,480
to the natural parameter,
which is to basically compute

279
00:15:36,480 --> 00:15:42,180
what the optimal value is,
because you can let epsilon be

280
00:15:42,180 --> 00:15:50,060
something like 1 over 2k--
that ought to do it-- something

281
00:15:50,060 --> 00:15:52,800
less than 1 over k.

282
00:15:52,800 --> 00:15:55,070
You're trying to
decide whether you

283
00:15:55,070 --> 00:15:57,640
have a solution of less
than or equal to k or not,

284
00:15:57,640 --> 00:15:59,880
whether optimal is
strictly greater than k.

285
00:15:59,880 --> 00:16:03,940
And I said k is an
integer, so this

286
00:16:03,940 --> 00:16:06,940
is an integer valued problem
you're trying to optimize.

287
00:16:06,940 --> 00:16:09,080
And so the difference
became k and k plus 1,

288
00:16:09,080 --> 00:16:13,762
multiplicatively, is about
1 plus 1 over k factor.

289
00:16:13,762 --> 00:16:15,770
It's exactly 1 plus 1 over k.

290
00:16:15,770 --> 00:16:18,470
So if we let epsilon be
smaller than 1 over k,

291
00:16:18,470 --> 00:16:21,130
then to get within a
multiplicative factor of 1

292
00:16:21,130 --> 00:16:24,200
plus 1/2 1 over k,
you actually have

293
00:16:24,200 --> 00:16:28,010
to be within an additive plus
1 of the right answer at k.

294
00:16:28,010 --> 00:16:36,957
So this will decide whether
opt is less than or equal to k

295
00:16:36,957 --> 00:16:39,040
based on whether that
approximation algorithm will

296
00:16:39,040 --> 00:16:42,720
give you something of value
at most k or strictly bigger.

297
00:16:42,720 --> 00:16:46,480
Strictly bigger
would be k plus 1.

298
00:16:46,480 --> 00:16:49,200
So this is particularly
interesting

299
00:16:49,200 --> 00:16:54,630
because it means if a
problem is not in FPT,

300
00:16:54,630 --> 00:16:59,510
then it does not have
an Efficient PTAS.

301
00:16:59,510 --> 00:17:03,340
So we can use-- before,
in previous lectures,

302
00:17:03,340 --> 00:17:05,690
PTAS was considered gold.

303
00:17:05,690 --> 00:17:08,740
Now, we can distinguish
between EPTAS and PTAS.

304
00:17:08,740 --> 00:17:10,660
And there are problems
that have PTASs

305
00:17:10,660 --> 00:17:14,460
that do not have FPT
algorithms assuming

306
00:17:14,460 --> 00:17:17,810
FPT does not equal w1-- an
assumption we will get to

307
00:17:17,810 --> 00:17:19,220
in a moment.

308
00:17:19,220 --> 00:17:21,690
So if you can, in general,
establish something is probably

309
00:17:21,690 --> 00:17:24,190
not in FPT, then we get
it probably does not

310
00:17:24,190 --> 00:17:25,480
have an Efficient PTAS.

311
00:17:25,480 --> 00:17:28,020
So this is useful for lower
bounds about approximation.

312
00:17:28,020 --> 00:17:30,710
Even if you don't care about
fixed parameter tractability,

313
00:17:30,710 --> 00:17:33,584
this will give us cool
things about approximability.

314
00:17:33,584 --> 00:17:38,312
AUDIENCE: Can you give
a few examples of EPTAS?

315
00:17:38,312 --> 00:17:39,770
PROFESSOR: A few
examples of EPTAS.

316
00:17:42,870 --> 00:17:45,880
For example, most of the planar
results I've talked about,

317
00:17:45,880 --> 00:17:48,840
planar independence set, planar
vertex cover, those things all

318
00:17:48,840 --> 00:17:50,230
have Efficient PTASs.

319
00:17:52,724 --> 00:17:54,890
But that gets us into the
world of bounded treewidth

320
00:17:54,890 --> 00:17:56,510
algorithms.

321
00:17:56,510 --> 00:17:59,240
But in general, if you have like
a-- a lot of bounded treewidth

322
00:17:59,240 --> 00:18:00,990
algorithms are FPT.

323
00:18:00,990 --> 00:18:04,640
There will be some exponential
in treewidth times polynomial n

324
00:18:04,640 --> 00:18:07,490
So a lot of times,
you can use those

325
00:18:07,490 --> 00:18:11,300
to get EPTASs for planar
and h-minor free problems.

326
00:18:11,300 --> 00:18:13,170
So that's one set of examples.

327
00:18:13,170 --> 00:18:15,260
There's no short
example I can give you,

328
00:18:15,260 --> 00:18:18,370
but that's a big class.

329
00:18:18,370 --> 00:18:21,870
So for either reason,
either you want

330
00:18:21,870 --> 00:18:24,180
to solve problems
for small k or you

331
00:18:24,180 --> 00:18:26,820
want to figure out
whether there's an EPTAS,

332
00:18:26,820 --> 00:18:29,900
how can we prove that problems
are hard in this world using

333
00:18:29,900 --> 00:18:33,200
reductions like usual?

334
00:18:33,200 --> 00:18:36,120
So in this context,
luckily, there's

335
00:18:36,120 --> 00:18:37,770
only one type of
reduction we need

336
00:18:37,770 --> 00:18:41,110
to learn unlike approximation
where there were a lot.

337
00:18:45,860 --> 00:18:47,560
And I'll call it
parameterized reduction

338
00:18:47,560 --> 00:18:49,690
although throughout this
lecture and the next one

339
00:18:49,690 --> 00:18:52,860
I'll just say reduction usually
because we're always talking

340
00:18:52,860 --> 00:18:54,960
about parameterized problems.

341
00:18:54,960 --> 00:18:58,920
So in general, we have some
problem A-- decision problem A

342
00:18:58,920 --> 00:19:00,330
and parameter k.

343
00:19:00,330 --> 00:19:03,160
And we want to convert into
some decision problem B

344
00:19:03,160 --> 00:19:06,810
with parameter k prime.

345
00:19:06,810 --> 00:19:11,800
And of course, as usual,
the set up is we're

346
00:19:11,800 --> 00:19:13,720
given an instance x.

347
00:19:13,720 --> 00:19:17,570
This is going to look almost
identical to NP Karp-style

348
00:19:17,570 --> 00:19:21,590
reductions, but then we're going
to have one extra condition.

349
00:19:21,590 --> 00:19:24,460
So instance x of
A gets mapped to

350
00:19:24,460 --> 00:19:31,540
by a function f to an
instance x prime of B. X prime

351
00:19:31,540 --> 00:19:36,140
is f o x as usual.

352
00:19:36,140 --> 00:19:42,150
This needs to be a
polynomial time function

353
00:19:42,150 --> 00:19:46,460
just like for NP reductions,
which means, in particular,

354
00:19:46,460 --> 00:19:49,840
x prime has polynomials
size reduced back to x.

355
00:19:49,840 --> 00:19:51,200
It should be answer preserving.

356
00:19:57,180 --> 00:20:03,340
So x is yes instance
for A if and only

357
00:20:03,340 --> 00:20:07,440
x prime is a yes instance for B.

358
00:20:07,440 --> 00:20:10,480
So, so far, exactly
NP reductions.

359
00:20:10,480 --> 00:20:13,350
And then when we
need one extra thing

360
00:20:13,350 --> 00:20:15,970
which is parameter preserving.

361
00:20:21,740 --> 00:20:43,880
This is there's some function
g, which I'll call the parameter

362
00:20:43,880 --> 00:20:46,490
blow up-- or, I guess you
call it parameter growth

363
00:20:46,490 --> 00:20:49,990
for g-- such that
the new parameter

364
00:20:49,990 --> 00:20:52,450
value for the
converted instance is,

365
00:20:52,450 --> 00:20:54,510
at most, that function
of the original parameter

366
00:20:54,510 --> 00:20:56,440
value of the original instance.

367
00:20:56,440 --> 00:20:56,940
Question?

368
00:20:56,940 --> 00:20:59,106
AUDIENCE: Are there any
limits on the amount of time

369
00:20:59,106 --> 00:21:01,930
that g can take to compute?

370
00:21:01,930 --> 00:21:05,240
PROFESSOR: g should be
a computable function.

371
00:21:05,240 --> 00:21:08,290
I think that's all we need.

372
00:21:08,290 --> 00:21:10,970
Probably polynomial
time is also a fine,

373
00:21:10,970 --> 00:21:16,660
but usually this is going to
be like linear or polynomial

374
00:21:16,660 --> 00:21:17,940
or exponential.

375
00:21:17,940 --> 00:21:20,560
It's rarely some
insanely large thing,

376
00:21:20,560 --> 00:21:22,500
but computable would be nice.

377
00:21:28,730 --> 00:21:29,490
Cool.

378
00:21:29,490 --> 00:21:33,520
So that is our notion of
parameterized reduction.

379
00:21:33,520 --> 00:21:37,350
And the consequence,
if this exists

380
00:21:37,350 --> 00:21:40,160
and B is fixed
parameter tractable,

381
00:21:40,160 --> 00:21:47,140
then A is because we can
take an instance of A

382
00:21:47,140 --> 00:21:50,900
converting it to B. If
the original parameter was

383
00:21:50,900 --> 00:21:53,720
founded by some k,
this new parameter

384
00:21:53,720 --> 00:21:56,446
will be bounded by g of k.

385
00:21:56,446 --> 00:21:58,070
New instance will be
bounded of g of k.

386
00:21:58,070 --> 00:22:01,120
So we run the FPT
algorithm for B,

387
00:22:01,120 --> 00:22:06,890
and that gives us the answer
to the original instance of A.

388
00:22:06,890 --> 00:22:15,020
So if we don't care about
what this function is,

389
00:22:15,020 --> 00:22:16,950
we are basically
composing functions.

390
00:22:16,950 --> 00:22:21,310
So there's some f dependence
on k in this algorithm,

391
00:22:21,310 --> 00:22:25,470
and we're taking that function
of g of k is our new function.

392
00:22:25,470 --> 00:22:30,610
And we get a new dependence on
k and the running time over FPT.

393
00:22:30,610 --> 00:22:32,700
So what that means
is if we believe

394
00:22:32,700 --> 00:22:36,820
it A does not have an FPT, then
B does not have an FPT if we

395
00:22:36,820 --> 00:22:37,820
can do these reductions.

396
00:22:37,820 --> 00:22:39,360
So same style, we're
going to reduce

397
00:22:39,360 --> 00:22:42,240
from a problem we know is
hard, A, into a problem

398
00:22:42,240 --> 00:22:46,280
that we don't know about, and
that proves B is hard as well.

399
00:22:46,280 --> 00:22:46,780
Yeah?

400
00:22:46,780 --> 00:22:48,196
AUDIENCE: What is
the relationship

401
00:22:48,196 --> 00:22:49,200
between FPT and XP?

402
00:22:51,912 --> 00:22:53,037
Is there like a difference?

403
00:22:53,037 --> 00:22:53,661
PROFESSOR: Yeah

404
00:22:53,661 --> 00:22:54,910
AUDIENCE: If there's overlap.

405
00:22:54,910 --> 00:22:58,600
PROFESSOR: Well, FPT
is contained in XP.

406
00:22:58,600 --> 00:23:02,530
And they are different if you
believe the exponential time

407
00:23:02,530 --> 00:23:03,760
hypothesis.

408
00:23:03,760 --> 00:23:06,440
If you believe SAT
does not have two

409
00:23:06,440 --> 00:23:13,290
to the little of n algorithms--
if that's not possible for SAT,

410
00:23:13,290 --> 00:23:17,457
then XP and FPT are different,
and much, much more things

411
00:23:17,457 --> 00:23:18,040
are different.

412
00:23:18,040 --> 00:23:21,320
But we'll talk about that
more next class-- relating

413
00:23:21,320 --> 00:23:23,540
to exponential time hypothesis.

414
00:23:23,540 --> 00:23:27,330
So at this moment, we don't
really care what g is.

415
00:23:27,330 --> 00:23:30,020
But if you assume
exponential time hypothesis,

416
00:23:30,020 --> 00:23:34,800
then g matters, and you
can get very explicit lower

417
00:23:34,800 --> 00:23:39,260
bounds about how good an
algorithm you could hope for.

418
00:23:39,260 --> 00:23:42,160
So we won't just prove you're
probably not an-- we won't just

419
00:23:42,160 --> 00:23:45,025
prove you're not an FPT, but we
will give an actual lower bound

420
00:23:45,025 --> 00:23:46,680
and how much running
time you need,

421
00:23:46,680 --> 00:23:48,600
something like n
to the little k.

422
00:23:48,600 --> 00:23:51,180
But the running time we get
will depend on this blow

423
00:23:51,180 --> 00:23:52,340
up function.

424
00:23:52,340 --> 00:23:55,320
So next class, we'll
care about what g is.

425
00:23:55,320 --> 00:23:59,040
So I do try to minimize it,
but so far we don't care.

426
00:23:59,040 --> 00:24:00,016
Any g is fine.

427
00:24:05,620 --> 00:24:07,570
XP will not turn
out to be the class

428
00:24:07,570 --> 00:24:08,890
that we think about very often.

429
00:24:08,890 --> 00:24:11,770
There are things in between that
are a lot easier to work with,

430
00:24:11,770 --> 00:24:13,050
which we will get to.

431
00:24:13,050 --> 00:24:17,540
In particular, W1
is the most common.

432
00:24:17,540 --> 00:24:20,830
So let me do two examples.

433
00:24:20,830 --> 00:24:23,870
One of which is a correct
example and the other is not.

434
00:24:23,870 --> 00:24:26,580
And you can help me
figure out which is which.

435
00:24:26,580 --> 00:24:30,150
Independent set to
vertex cover, these

436
00:24:30,150 --> 00:24:31,840
are reductions
we've seen before.

437
00:24:31,840 --> 00:24:33,790
We have tons of
reductions, but usually we

438
00:24:33,790 --> 00:24:35,830
weren't thinking
about parameters.

439
00:24:35,830 --> 00:24:41,540
And independent set
to clique, these

440
00:24:41,540 --> 00:24:45,090
are basically identical
problems left and right.

441
00:24:45,090 --> 00:24:47,540
And I want the
natural parameter.

442
00:24:47,540 --> 00:24:51,589
Which of these is a parameter
preserving reduction?

443
00:24:51,589 --> 00:24:52,755
I need some quiz show music.

444
00:24:58,242 --> 00:24:59,190
AUDIENCE: The second.

445
00:24:59,190 --> 00:25:00,940
PROFESSOR: The second, yeah.

446
00:25:00,940 --> 00:25:03,580
Because if we take a
graph and a parameter k,

447
00:25:03,580 --> 00:25:06,620
what we convert it to
is the complement graph

448
00:25:06,620 --> 00:25:09,200
with the same parameter.

449
00:25:09,200 --> 00:25:11,340
So that's obviously
parameter preserving.

450
00:25:11,340 --> 00:25:13,590
Whereas independence
set, these are

451
00:25:13,590 --> 00:25:15,230
complementary in
a different sense

452
00:25:15,230 --> 00:25:17,646
that if you take everything
that's not in the independence

453
00:25:17,646 --> 00:25:18,750
set is a vertex cover.

454
00:25:18,750 --> 00:25:21,940
So it's actually the same
graph but with n minus k

455
00:25:21,940 --> 00:25:23,540
as the new parameter.

456
00:25:23,540 --> 00:25:27,950
So this is not a
parameterized reduction.

457
00:25:27,950 --> 00:25:31,070
This is.

458
00:25:31,070 --> 00:25:35,910
And in fact, vertex cover,
we just showed is FPT.

459
00:25:35,910 --> 00:25:41,140
Independence set is not, if you
believe w1 does not equal FPT

460
00:25:41,140 --> 00:25:44,490
or if you believe
exponential time hypothesis.

461
00:25:44,490 --> 00:25:47,340
And that would contradict
this statement.

462
00:25:47,340 --> 00:25:50,506
So this is definitely
not a valid reduction.

463
00:25:50,506 --> 00:25:52,380
But those are some pretty
trivial reductions.

464
00:25:52,380 --> 00:25:58,460
Let's do some more
interesting things.

465
00:25:58,460 --> 00:26:00,810
And I want to start to
introduce the notion of w1.

466
00:26:25,900 --> 00:26:28,540
I won't define w1 yet.

467
00:26:28,540 --> 00:26:31,480
I'd like to wait a little bit
because the definition is not

468
00:26:31,480 --> 00:26:35,900
super intuitive, but here is a
fairly intuitive hard problem

469
00:26:35,900 --> 00:26:41,130
that you should be fairly
convinced is not FPT.

470
00:26:41,130 --> 00:26:49,141
So I'll call this k-step
non-deterministic Turing

471
00:26:49,141 --> 00:26:49,640
machine.

472
00:26:55,270 --> 00:26:58,450
The one downside is I've never
mentioned Turing machines

473
00:26:58,450 --> 00:26:59,140
in this class.

474
00:26:59,140 --> 00:27:01,170
I thought I could
get away without it,

475
00:27:01,170 --> 00:27:02,790
but I'm going to
have mention them.

476
00:27:02,790 --> 00:27:05,075
How many people
know what they are?

477
00:27:05,075 --> 00:27:07,710
Anyone does not?

478
00:27:07,710 --> 00:27:09,460
OK, a couple.

479
00:27:09,460 --> 00:27:11,780
Here's a Turing machine.

480
00:27:11,780 --> 00:27:15,590
I will give you, very briefly,
a very non-standard definition

481
00:27:15,590 --> 00:27:16,480
of a Turing machine.

482
00:27:16,480 --> 00:27:19,820
Let's think of-- so you
have this infinite tape,

483
00:27:19,820 --> 00:27:22,880
infinite memory, and
for our purposes,

484
00:27:22,880 --> 00:27:24,770
these are not just
binary symbols,

485
00:27:24,770 --> 00:27:28,770
you can write up to n different
symbols on each of the squares.

486
00:27:28,770 --> 00:27:30,600
So this is basically
your memory,

487
00:27:30,600 --> 00:27:33,340
but it's in the style of
old-fashioned tape drives

488
00:27:33,340 --> 00:27:37,160
in that you-- In order to get
to position k away from you,

489
00:27:37,160 --> 00:27:38,280
you have to spin k time.

490
00:27:38,280 --> 00:27:40,720
So you can only step
one unit at a time.

491
00:27:40,720 --> 00:27:45,630
And this is basically
computers that we are used to.

492
00:27:45,630 --> 00:27:48,740
So let's say it has a
sequence of instructions,

493
00:27:48,740 --> 00:27:50,570
and it has an
instruction pointer.

494
00:27:50,570 --> 00:27:55,270
And in general, I
want to the machine

495
00:27:55,270 --> 00:28:04,850
to have order n
stateS-- yeah, let's say

496
00:28:04,850 --> 00:28:21,647
order n lines of code and order
n options per cell of the tape.

497
00:28:21,647 --> 00:28:23,230
This will look
actually a little weird

498
00:28:23,230 --> 00:28:25,750
if you're used to
Turing machines,

499
00:28:25,750 --> 00:28:27,370
but I want to use
this definition.

500
00:28:27,370 --> 00:28:29,150
So it's basically a
regular algorithm,

501
00:28:29,150 --> 00:28:30,940
but you're very limited
the number-- in the amount

502
00:28:30,940 --> 00:28:31,690
of internal state.

503
00:28:31,690 --> 00:28:33,900
These are basically
registers, but you

504
00:28:33,900 --> 00:28:36,070
can have only a constant
number of registers

505
00:28:36,070 --> 00:28:37,670
that vary from one to n.

506
00:28:37,670 --> 00:28:40,460
One of them will be what
line of code are you on.

507
00:28:40,460 --> 00:28:44,390
And then you have those lines of
code are instructions like jump

508
00:28:44,390 --> 00:28:48,750
here, compare these things,
write a particular symbol

509
00:28:48,750 --> 00:28:51,640
to the current square of the
tape, move the tape left,

510
00:28:51,640 --> 00:28:52,510
move the tape right.

511
00:28:52,510 --> 00:28:55,850
So regular types of
instructions, let's

512
00:28:55,850 --> 00:28:59,040
say, except you have
this weird tape thing.

513
00:28:59,040 --> 00:29:01,225
Now, I mentioned
these parameters.

514
00:29:01,225 --> 00:29:02,850
And usually we think
of Turing machines

515
00:29:02,850 --> 00:29:05,650
of having constant size,
but here I need an n,

516
00:29:05,650 --> 00:29:07,090
and I need a k.

517
00:29:07,090 --> 00:29:07,840
Question?

518
00:29:07,840 --> 00:29:09,100
AUDIENCE: I was going
to ask what n was.

519
00:29:09,100 --> 00:29:09,683
PROFESSOR: OK.

520
00:29:09,683 --> 00:29:11,062
So n is the input.

521
00:29:11,062 --> 00:29:12,520
I mean basically
the Turing machine

522
00:29:12,520 --> 00:29:14,490
is specified by these things.

523
00:29:14,490 --> 00:29:18,140
You have order n
instructions, and we're

524
00:29:18,140 --> 00:29:21,650
guaranteed there's only order
n possible states, let's say.

525
00:29:21,650 --> 00:29:24,880
And now what we-- and this is
a non-deterministic machine.

526
00:29:24,880 --> 00:29:27,060
Now non-determinism, we
have talked about this

527
00:29:27,060 --> 00:29:28,600
in the context of NP.

528
00:29:28,600 --> 00:29:32,130
A non-deterministic
Turing machine

529
00:29:32,130 --> 00:29:34,640
has a funny
instruction which says

530
00:29:34,640 --> 00:29:38,340
non-deterministically branch
to one of n different locations

531
00:29:38,340 --> 00:29:40,060
in my memory.

532
00:29:40,060 --> 00:29:43,200
Or, let's say choose
a symbol from my-- I

533
00:29:43,200 --> 00:29:47,540
have this alphabet that
I'm using in my cell--

534
00:29:47,540 --> 00:29:51,030
choose a symbol
non-deterministically.

535
00:29:51,030 --> 00:29:58,780
So I have order n choices
made non-deterministically.

536
00:29:58,780 --> 00:30:01,330
In the usual sense of
NP, that if there's

537
00:30:01,330 --> 00:30:03,130
any way for the Turing
machine to output,

538
00:30:03,130 --> 00:30:06,910
yes-- one of the instructions
is output, yes-- then,

539
00:30:06,910 --> 00:30:08,849
I will find it.

540
00:30:08,849 --> 00:30:10,390
These are guesses,
and they're always

541
00:30:10,390 --> 00:30:13,140
lucky guesses, so I always
end up finding the return,

542
00:30:13,140 --> 00:30:14,960
yes, if there is such a path.

543
00:30:14,960 --> 00:30:17,890
Otherwise, all paths return, no,
and then the machine returns,

544
00:30:17,890 --> 00:30:19,451
no.

545
00:30:19,451 --> 00:30:19,950
Yeah?

546
00:30:19,950 --> 00:30:22,116
AUDIENCE: What does it mean
for the number of states

547
00:30:22,116 --> 00:30:23,680
to change as the input changes?

548
00:30:23,680 --> 00:30:28,320
PROFESSOR: I mean the
states are also given to you

549
00:30:28,320 --> 00:30:30,230
as part of the Turing machine.

550
00:30:30,230 --> 00:30:32,243
There's not one
turning machine-- well,

551
00:30:32,243 --> 00:30:34,493
there is actually one Turing
machine to rule them all,

552
00:30:34,493 --> 00:30:36,980
but that's not the point here.

553
00:30:36,980 --> 00:30:41,400
I give you a machine that has--
think of this as firmware built

554
00:30:41,400 --> 00:30:44,310
into the machine, and the
number of states in the machine

555
00:30:44,310 --> 00:30:48,040
includes which instruction
you're currently executing.

556
00:30:48,040 --> 00:30:49,540
So I mean this is
just saying you're

557
00:30:49,540 --> 00:30:53,210
given an arbitrary machine
of size-- of complexity n.

558
00:30:53,210 --> 00:30:54,900
And usual Turing
machine land, there

559
00:30:54,900 --> 00:30:58,640
are n states because I give you
a state diagram with size n.

560
00:30:58,640 --> 00:31:01,230
So you're given everything
about the Turing machine.

561
00:31:01,230 --> 00:31:03,830
So that is the input
to this problem.

562
00:31:03,830 --> 00:31:09,090
And the question is, can
I find a return, yes,

563
00:31:09,090 --> 00:31:12,190
solution that only
is k steps long.

564
00:31:12,190 --> 00:31:14,670
So I'm given this huge
machine, and yet I

565
00:31:14,670 --> 00:31:17,210
want a very short
execution of the machine.

566
00:31:17,210 --> 00:31:20,900
I want the running time to only
be k. k, again, is parameter.

567
00:31:20,900 --> 00:31:23,040
n is big. k is small.

568
00:31:23,040 --> 00:31:31,140
So I want to know is there a
yes path from the initial state

569
00:31:31,140 --> 00:31:33,295
of length k.

570
00:31:38,520 --> 00:31:41,470
So this is basically--
Yeah, question?

571
00:31:41,470 --> 00:31:44,740
AUDIENCE: Are you given the
input to the Turing machine?

572
00:31:44,740 --> 00:31:47,356
PROFESSOR: Yeah, let's
say it has no input.

573
00:31:47,356 --> 00:31:50,150
Input is all
embedded in the code.

574
00:31:50,150 --> 00:31:54,210
So the tape is initially blank.

575
00:31:54,210 --> 00:31:55,717
Good.

576
00:31:55,717 --> 00:31:57,800
There's a reason I need
to do Turing machines here

577
00:31:57,800 --> 00:32:01,500
instead of usual algorithms,
but if you define-- usually

578
00:32:01,500 --> 00:32:03,810
NP is defined in
terms of these things.

579
00:32:03,810 --> 00:32:08,970
So by analogy to NP, we expect
there are no lucky algorithms.

580
00:32:08,970 --> 00:32:11,502
And so we expect
that when you have

581
00:32:11,502 --> 00:32:13,710
non-deterministic branches,
the best thing you can do

582
00:32:13,710 --> 00:32:15,480
is to try all the branches.

583
00:32:15,480 --> 00:32:19,180
And so if I have an execution
time of k, and at each step,

584
00:32:19,180 --> 00:32:21,280
I can potentially make a
non-deterministic choice

585
00:32:21,280 --> 00:32:24,450
among n different
options, then you

586
00:32:24,450 --> 00:32:28,280
would expect the best
algorithm is n to the k.

587
00:32:28,280 --> 00:32:31,920
Try all the branches, that's
just like our vertex cover

588
00:32:31,920 --> 00:32:35,930
algorithm, but in the bad case
where I have branching factor n

589
00:32:35,930 --> 00:32:38,260
instead of branching factor 2.

590
00:32:38,260 --> 00:32:41,980
So presumably, there's no way to
replace a guess among n options

591
00:32:41,980 --> 00:32:45,670
with a guess among two options.

592
00:32:45,670 --> 00:32:47,680
That's an assumption.

593
00:32:47,680 --> 00:32:55,810
And this problem is w1 complete.

594
00:32:55,810 --> 00:32:58,670
I will to define
w1 a little later,

595
00:32:58,670 --> 00:33:00,710
but for now, just
take this as given.

596
00:33:00,710 --> 00:33:03,680
You could define w1
complete to mean problems

597
00:33:03,680 --> 00:33:06,020
that are reducible via
parameterized reductions

598
00:33:06,020 --> 00:33:07,720
to this problem.

599
00:33:07,720 --> 00:33:10,270
And then we'll get lots
of examples of problems

600
00:33:10,270 --> 00:33:11,920
that are as hard as this.

601
00:33:14,610 --> 00:33:18,220
So a modulo the
annoyance of having

602
00:33:18,220 --> 00:33:20,330
to define Turing
machines, I think

603
00:33:20,330 --> 00:33:23,780
this is a pretty
natural assumption.

604
00:33:23,780 --> 00:33:25,620
It's stronger than
p does not equal NP,

605
00:33:25,620 --> 00:33:29,200
so we would imply that.

606
00:33:29,200 --> 00:33:33,426
So let's do some
simple reductions--

607
00:33:33,426 --> 00:33:35,190
some reductions,
maybe not simple.

608
00:33:39,740 --> 00:33:43,240
So I mentioned this
problem independent set.

609
00:33:43,240 --> 00:33:46,980
That's also w1 complete.

610
00:33:46,980 --> 00:33:52,190
And to prove, first of
all, that it's w1 hard,

611
00:33:52,190 --> 00:33:56,650
I'm going to reduce this
problem to independent set.

612
00:33:56,650 --> 00:34:00,090
So I'm given a Turing
machine and a number k.

613
00:34:00,090 --> 00:34:02,530
I want to convert
it into a graph,

614
00:34:02,530 --> 00:34:04,210
and a number k such
that if there's

615
00:34:04,210 --> 00:34:06,330
an independent set of
size k in that graph,

616
00:34:06,330 --> 00:34:07,880
at most k in that
graph if and only

617
00:34:07,880 --> 00:34:11,670
if there's an accept
path of length at most

618
00:34:11,670 --> 00:34:14,820
k in the Turing machine.

619
00:34:14,820 --> 00:34:16,460
So I'm going to skip
a lot of details

620
00:34:16,460 --> 00:34:19,210
because I've been a little vague
about how Turing machines work,

621
00:34:19,210 --> 00:34:22,139
but the idea is nice.

622
00:34:35,739 --> 00:34:40,219
So my graph is actually
going to consist mostly

623
00:34:40,219 --> 00:34:44,480
of k squared cliques.

624
00:34:44,480 --> 00:34:47,170
The cliques are actually quite
large because this graph has

625
00:34:47,170 --> 00:34:51,070
size n, total size, but
there's going to be k squared

626
00:34:51,070 --> 00:34:53,701
clusters, which are cliques.

627
00:34:53,701 --> 00:34:58,850
Plus cliques, not very many
different cliques, only k

628
00:34:58,850 --> 00:34:59,780
squared of them.

629
00:34:59,780 --> 00:35:03,380
And my target independent
set is size k squared,

630
00:35:03,380 --> 00:35:05,830
which means if there's going
to be an independent set,

631
00:35:05,830 --> 00:35:09,140
I must choose exactly one
vertex from each clique.

632
00:35:11,870 --> 00:35:18,420
So one vertex per clique.

633
00:35:18,420 --> 00:35:20,537
Independence that has to
have no edges among them.

634
00:35:20,537 --> 00:35:22,120
So if I chose two
from a clique, there

635
00:35:22,120 --> 00:35:26,420
would be an edge among
them, that's disallowed.

636
00:35:26,420 --> 00:35:29,570
And so the cliques I'm
going to parameterize

637
00:35:29,570 --> 00:35:33,560
by two parameters i and
j between one and k.

638
00:35:33,560 --> 00:35:49,150
And the idea is that ij
represents memory cell i

639
00:35:49,150 --> 00:35:54,880
at time j plus the
state of the machine.

640
00:36:03,580 --> 00:36:09,150
So what I mean is there are
order n states in the machine,

641
00:36:09,150 --> 00:36:11,040
in general, that
completely characterize

642
00:36:11,040 --> 00:36:14,210
what the machine is about to do
and what it's thinking about,

643
00:36:14,210 --> 00:36:15,517
all of its internal state.

644
00:36:15,517 --> 00:36:17,350
And so I'm just going
to take those n states

645
00:36:17,350 --> 00:36:19,050
and plop them in.

646
00:36:19,050 --> 00:36:22,565
And each node in this clique
represents one of those states.

647
00:36:25,180 --> 00:36:31,220
But also I want to keep track
of what symbol is written

648
00:36:31,220 --> 00:36:32,940
on that square of the tape.

649
00:36:32,940 --> 00:36:35,684
And so, in general,
my running time

650
00:36:35,684 --> 00:36:37,850
is bounded by k-- also the
number of squares I write

651
00:36:37,850 --> 00:36:39,016
to the tape is bounded by k.

652
00:36:39,016 --> 00:36:41,190
Space is at most
time-- so I only

653
00:36:41,190 --> 00:36:46,930
have to worry about k different
cells for times 1 up to k.

654
00:36:49,580 --> 00:36:54,240
So there are only order n
states for this, only order n

655
00:36:54,240 --> 00:36:55,770
states for this.

656
00:36:55,770 --> 00:36:59,710
So the size of each clique is
order n squared polynomial.

657
00:36:59,710 --> 00:37:00,900
That's cool.

658
00:37:00,900 --> 00:37:02,600
It's a messy clique.

659
00:37:02,600 --> 00:37:04,711
I won't try to draw what
it really looks like.

660
00:37:04,711 --> 00:37:06,210
And then the general
approach is I'm

661
00:37:06,210 --> 00:37:09,770
going to add some edges
between pairs of vertices

662
00:37:09,770 --> 00:37:13,600
to say, well look, if this cell
was this thing at this time,

663
00:37:13,600 --> 00:37:15,920
and the state of the
machine was move right,

664
00:37:15,920 --> 00:37:19,210
then you better not change what
was on the cell at that time

665
00:37:19,210 --> 00:37:20,460
because nothing changed.

666
00:37:20,460 --> 00:37:22,310
You weren't here.

667
00:37:22,310 --> 00:37:25,720
So we can just forbid
that by saying,

668
00:37:25,720 --> 00:37:27,900
well, here is one state.

669
00:37:27,900 --> 00:37:30,030
These two states should
be mutually exclusive.

670
00:37:30,030 --> 00:37:32,560
So if I choose this vertex
in an independent set,

671
00:37:32,560 --> 00:37:34,160
I can't choose this vertex.

672
00:37:34,160 --> 00:37:36,010
So you just draw all
these connections

673
00:37:36,010 --> 00:37:36,950
of forbidden things.

674
00:37:36,950 --> 00:37:40,650
If something is true
time i, times j,

675
00:37:40,650 --> 00:37:43,680
then something else should
not be true at time j plus 1.

676
00:37:43,680 --> 00:37:47,110
You draw in all those
connections, and it works.

677
00:37:47,110 --> 00:37:50,510
I'll leave it at that because
the details are messy,

678
00:37:50,510 --> 00:37:53,260
but it gives you some idea.

679
00:37:53,260 --> 00:37:58,340
At this point, I can plug
a book on this topic that

680
00:37:58,340 --> 00:38:03,222
is coming out next year, a
little bit far in the future

681
00:38:03,222 --> 00:38:12,560
but it is-- I didn't write
it down here of course.

682
00:38:12,560 --> 00:38:13,160
Sorry.

683
00:38:13,160 --> 00:38:22,230
It is by these guys again,
Fomin, Kowalik, Lockshtanov,

684
00:38:22,230 --> 00:38:26,860
Marx, Pilipczuck
brothers, and Saurabh.

685
00:38:26,860 --> 00:38:31,064
And it does go through the
details, the gory details.

686
00:38:31,064 --> 00:38:33,230
If you're interested in
that, I could share with you

687
00:38:33,230 --> 00:38:35,800
that section.

688
00:38:35,800 --> 00:38:38,490
But unfortunately, the
book is not released yet.

689
00:38:38,490 --> 00:38:39,300
Cool.

690
00:38:39,300 --> 00:38:45,200
So that was a reduction
to independent set, which

691
00:38:45,200 --> 00:38:47,250
shows the independence
set is at least as

692
00:38:47,250 --> 00:38:51,290
hard as this problem, which
we think is hard, w1 hard.

693
00:38:51,290 --> 00:38:52,590
So that proves w1 hardness.

694
00:38:52,590 --> 00:38:56,630
But there's another question
of is independent set in w1.

695
00:38:56,630 --> 00:38:59,690
As you might guess from the name
w1, there's more than just w1.

696
00:38:59,690 --> 00:39:04,500
There's w2 and w3
and w4 and even more.

697
00:39:04,500 --> 00:39:07,260
And there's no one
notion of hardness

698
00:39:07,260 --> 00:39:09,830
like we had with NP here, so
we get a bit of complexity

699
00:39:09,830 --> 00:39:11,590
in that hierarchy.

700
00:39:11,590 --> 00:39:17,360
But in fact, independent set is
w1 hard-- sorry-- w1 complete.

701
00:39:17,360 --> 00:39:21,235
So we can do a reduction
from independence

702
00:39:21,235 --> 00:39:26,970
set to Turing machine.

703
00:39:26,970 --> 00:39:31,640
Let's say k-step
non-deterministic Turing

704
00:39:31,640 --> 00:39:32,140
machine.

705
00:39:39,150 --> 00:39:41,170
And the idea is pretty simple.

706
00:40:02,330 --> 00:40:05,080
We have k-steps, so let's
guess all the vertices.

707
00:40:08,126 --> 00:40:09,500
You can see here
this is where we

708
00:40:09,500 --> 00:40:12,070
need the ability to guess
among n different options

709
00:40:12,070 --> 00:40:13,301
in each guess step.

710
00:40:13,301 --> 00:40:15,300
So we're given-- we want
to know whether there's

711
00:40:15,300 --> 00:40:17,410
an independent set
of size k, so guess

712
00:40:17,410 --> 00:40:20,350
all those vertices of
size k and then check it.

713
00:40:20,350 --> 00:40:22,200
So this was our XP
algorithm, but now

714
00:40:22,200 --> 00:40:27,360
phrased as a non-deterministic
Turing machine algorithm.

715
00:40:27,360 --> 00:40:29,810
And it's a little bit subtle
what it means to check.

716
00:40:41,800 --> 00:40:43,550
Of course, what we
want to do is make sure

717
00:40:43,550 --> 00:40:48,580
there's no edge between any
pair of the chosen vertices.

718
00:40:48,580 --> 00:40:51,620
So what we're going to do is
when we guess these k vertices,

719
00:40:51,620 --> 00:40:54,420
we're going to write them one
at a time on the tape squares.

720
00:40:54,420 --> 00:40:56,770
So we'll use k tape squares
to store the k vertices

721
00:40:56,770 --> 00:40:57,930
that we've chosen.

722
00:40:57,930 --> 00:41:01,380
Then, we want to check
each pair of them.

723
00:41:01,380 --> 00:41:03,050
So this is like a
doubly nested loop

724
00:41:03,050 --> 00:41:06,664
for each vertex among the k,
for each vertex among the k.

725
00:41:06,664 --> 00:41:08,080
So what that means
is you're going

726
00:41:08,080 --> 00:41:11,340
to have to-- sorry,
that's a little bit

727
00:41:11,340 --> 00:41:18,810
annoying-- loop through
the vertices on one tape.

728
00:41:18,810 --> 00:41:22,050
And then the fun
part is the graph

729
00:41:22,050 --> 00:41:24,270
is encoded inside the machine.

730
00:41:24,270 --> 00:41:25,820
You could think of
there being a data

731
00:41:25,820 --> 00:41:29,480
section within the code that
says where all the edges are.

732
00:41:29,480 --> 00:41:32,290
The graph is size n, and so
you can put the entire graph

733
00:41:32,290 --> 00:41:34,110
into the machine.

734
00:41:34,110 --> 00:41:36,190
And then given two
vertices, you could

735
00:41:36,190 --> 00:41:39,140
check that there is indeed
no edge between them

736
00:41:39,140 --> 00:41:41,620
basically by using the
code as a look-up table.

737
00:41:41,620 --> 00:41:46,850
Say hey, is there a
vertex from i to j.

738
00:41:46,850 --> 00:41:50,440
If yes, then we're in
trouble, abort this option,

739
00:41:50,440 --> 00:41:52,270
try the other guesses.

740
00:41:52,270 --> 00:41:56,630
If you look up in the table
and it says there's no edge,

741
00:41:56,630 --> 00:41:57,470
then you keep going.

742
00:41:57,470 --> 00:41:59,440
So you test all the
pairs, make sure there's

743
00:41:59,440 --> 00:42:03,790
no edge by-- the code, as it
is, is a constant size algorithm

744
00:42:03,790 --> 00:42:07,110
which looks like this
plus an order n size

745
00:42:07,110 --> 00:42:11,560
thing, which is the
look-up table for what

746
00:42:11,560 --> 00:42:13,090
edges are in the graph.

747
00:42:13,090 --> 00:42:14,808
So that part's a
little bit weird.

748
00:42:14,808 --> 00:42:15,308
Yeah?

749
00:42:15,308 --> 00:42:19,220
AUDIENCE: So in case of the k
in k vertices aren't the same k?

750
00:42:19,220 --> 00:42:20,260
PROFESSOR: That's right.

751
00:42:20,260 --> 00:42:20,759
I'm sorry.

752
00:42:20,759 --> 00:42:22,200
This is k prime.

753
00:42:25,870 --> 00:42:26,820
Yes.

754
00:42:26,820 --> 00:42:30,406
So probably k prime here
is exactly order k squared.

755
00:42:30,406 --> 00:42:31,664
Thank you.

756
00:42:31,664 --> 00:42:33,210
It makes me a little happier.

757
00:42:33,210 --> 00:42:34,960
Because I'm doing this
doubly nested loop,

758
00:42:34,960 --> 00:42:37,310
I'm going to have to loop
over this thing several times.

759
00:42:37,310 --> 00:42:39,210
I would look at
the first vertex,

760
00:42:39,210 --> 00:42:41,720
memorize it by reloading
it into a register,

761
00:42:41,720 --> 00:42:44,272
and then loop over all the
other guys for each of them

762
00:42:44,272 --> 00:42:45,230
check that they're bad.

763
00:42:45,230 --> 00:42:48,715
So I'm going to have to go
over the tape k squared steps.

764
00:42:53,750 --> 00:42:56,700
And also n prime is whatever.

765
00:42:56,700 --> 00:42:59,750
We get a machine
that's size-- basically

766
00:42:59,750 --> 00:43:06,360
the size of the graph
V plus E. So the point

767
00:43:06,360 --> 00:43:08,670
of doing that was to
show the independence set

768
00:43:08,670 --> 00:43:10,990
is also w1 complete.

769
00:43:10,990 --> 00:43:13,580
So if you were not happy
with Turing machines,

770
00:43:13,580 --> 00:43:16,340
now you can completely
forget about them.

771
00:43:16,340 --> 00:43:18,660
They are mainly used
here as motivation

772
00:43:18,660 --> 00:43:21,130
for why you should expect
this problem to be hard.

773
00:43:21,130 --> 00:43:25,750
But it's equally difficult from
a parameterization perspective.

774
00:43:25,750 --> 00:43:28,080
If you ignore how
big the blow up is,

775
00:43:28,080 --> 00:43:30,210
it's just as hard as
an independence set.

776
00:43:30,210 --> 00:43:32,170
So independence set
is a problem we think

777
00:43:32,170 --> 00:43:34,060
there's no good algorithm.

778
00:43:34,060 --> 00:43:39,750
And by this reduction, and
also the reverse reduction,

779
00:43:39,750 --> 00:43:42,380
clique is also w1 complete.

780
00:43:46,210 --> 00:43:50,090
So a bunch more reductions.

781
00:44:01,800 --> 00:44:06,581
In general, most w1 hardness
results start from clique.

782
00:44:06,581 --> 00:44:09,080
So that's a good problem to
know about, or independent sets,

783
00:44:09,080 --> 00:44:10,680
I think.

784
00:44:10,680 --> 00:44:13,250
And there are some
simpler versions of clique

785
00:44:13,250 --> 00:44:14,800
that are also hard.

786
00:44:14,800 --> 00:44:20,315
So first one is clique
in regular graphs.

787
00:44:24,910 --> 00:44:28,870
So it's helpful to assume
all the vertices have

788
00:44:28,870 --> 00:44:30,680
exactly the same degree.

789
00:44:30,680 --> 00:44:32,750
Is it hard in three
regular graphs?

790
00:44:32,750 --> 00:44:33,450
No.

791
00:44:33,450 --> 00:44:35,830
Because three regular is
only a clique of size three.

792
00:44:35,830 --> 00:44:38,510
So that's pretty easy.

793
00:44:38,510 --> 00:44:42,620
But in s regular graphs, the
degree is going to be huge,

794
00:44:42,620 --> 00:44:45,410
going to be some function of n.

795
00:44:45,410 --> 00:44:50,190
Clique is hard, and
in general, let's say

796
00:44:50,190 --> 00:44:55,160
capital delta is the maximum
degree in your given graph,

797
00:44:55,160 --> 00:44:59,140
and we want to convert it
into a delta regular graph.

798
00:44:59,140 --> 00:45:02,410
So we're going to increase all
the lower degrees up to delta.

799
00:45:02,410 --> 00:45:08,550
What we're going to do is take
delta copies of the graph.

800
00:45:11,890 --> 00:45:16,390
So for every old vertex,
we get delta copies of it.

801
00:45:20,140 --> 00:45:23,250
And then we do this reduction.

802
00:45:23,250 --> 00:45:25,940
So the blue is the original--
is the graph that's

803
00:45:25,940 --> 00:45:27,570
been duplicated delta times.

804
00:45:27,570 --> 00:45:29,807
Here, delta is 5.

805
00:45:29,807 --> 00:45:31,390
But suppose, in the
graph, this vertex

806
00:45:31,390 --> 00:45:34,320
only had degree 3, some
number less than 5.

807
00:45:34,320 --> 00:45:36,440
Then, what we're
going to do is create

808
00:45:36,440 --> 00:45:41,010
delta minus d new vertices
and then add a biclique here,

809
00:45:41,010 --> 00:45:44,930
bipartite clique, connecting all
of those delta minus d things

810
00:45:44,930 --> 00:45:47,150
to all of these delta things.

811
00:45:47,150 --> 00:45:51,250
They're delta copies
of the original vertex.

812
00:45:51,250 --> 00:45:55,004
This is v over here,
v1 through v delta.

813
00:45:55,004 --> 00:45:56,670
And so what that means
is these vertices

814
00:45:56,670 --> 00:46:00,310
will have entered degree delta
because there were delta things

815
00:46:00,310 --> 00:46:01,900
to connect to over there.

816
00:46:01,900 --> 00:46:03,710
And now these things
will have degree delta

817
00:46:03,710 --> 00:46:05,180
because they used
to have degree d,

818
00:46:05,180 --> 00:46:06,800
and now they have
an additional degree

819
00:46:06,800 --> 00:46:09,830
delta minus d on the left.

820
00:46:09,830 --> 00:46:12,290
So cool, everything's
now delta regular.

821
00:46:12,290 --> 00:46:14,740
And I claim that this
reduction preserve cliques

822
00:46:14,740 --> 00:46:19,870
because if you look at
these added vertices,

823
00:46:19,870 --> 00:46:23,830
they do not belong
to any triangles.

824
00:46:23,830 --> 00:46:27,330
This thing here is an
induced bipartite graph,

825
00:46:27,330 --> 00:46:31,170
and so there are no
triangles, no things of size--

826
00:46:31,170 --> 00:46:34,600
no cycles of size 3 because
all cycles are even here,

827
00:46:34,600 --> 00:46:36,262
which means if
you're in a clique--

828
00:46:36,262 --> 00:46:37,720
if you have three
guys in a clique,

829
00:46:37,720 --> 00:46:42,215
then they better have
triangles everywhere.

830
00:46:42,215 --> 00:46:43,840
So if you put one of
these vertices in,

831
00:46:43,840 --> 00:46:46,390
you'll have a very small
clique, namely size 2.

832
00:46:46,390 --> 00:46:49,080
And there are always cliques
of size 2 in a graph.

833
00:46:49,080 --> 00:46:50,240
There's at least one edge.

834
00:46:50,240 --> 00:46:52,190
So if there was at
least one edge before,

835
00:46:52,190 --> 00:46:55,750
afterwards, we do not increase
any of the clique sizes.

836
00:46:55,750 --> 00:46:57,600
So clique size is preserved.

837
00:46:57,600 --> 00:47:00,090
Everything's cool.

838
00:47:00,090 --> 00:47:01,980
So in this situation
k prime equals

839
00:47:01,980 --> 00:47:04,991
k when you blow up anything,
except the graph size,

840
00:47:04,991 --> 00:47:05,490
of course.

841
00:47:05,490 --> 00:47:08,192
That got a little bigger.

842
00:47:08,192 --> 00:47:09,650
And now everything's
delta regular.

843
00:47:09,650 --> 00:47:13,320
Le me give you an example
of why this is useful.

844
00:47:17,857 --> 00:47:20,190
Of course, every time I say
something is true of clique,

845
00:47:20,190 --> 00:47:21,648
it's also true of
independence set.

846
00:47:21,648 --> 00:47:23,940
You can just flip things.

847
00:47:23,940 --> 00:47:25,590
Here, we have to be
a little careful,

848
00:47:25,590 --> 00:47:28,130
but when you complement
a graph, if you

849
00:47:28,130 --> 00:47:30,030
started with a regular
graph, your new graph

850
00:47:30,030 --> 00:47:31,780
will also be regular.

851
00:47:31,780 --> 00:47:34,835
Every vertex would
have degree n minus 1

852
00:47:34,835 --> 00:47:36,918
minus delta or n minus
delta, something like that.

853
00:47:39,650 --> 00:47:48,094
So here's another problem,
eerily similar to vertex cover,

854
00:47:48,094 --> 00:47:49,990
called partial vertex cover.

855
00:47:49,990 --> 00:47:53,770
This is also w1 complete.

856
00:47:53,770 --> 00:48:02,745
We want to know, can I choose
k vertices that cover l edges?

857
00:48:07,010 --> 00:48:09,950
So usually vertex cover, you
need to cover all the edges,

858
00:48:09,950 --> 00:48:12,185
and that let us do some
crazy things because whenever

859
00:48:12,185 --> 00:48:14,310
we looked at an edge, we
knew one of the two things

860
00:48:14,310 --> 00:48:15,670
was in the vertex cover.

861
00:48:15,670 --> 00:48:18,100
Now, we don't know that anymore.

862
00:48:18,100 --> 00:48:20,050
Now, it's a matter
of which vertices get

863
00:48:20,050 --> 00:48:22,210
the most bang for your buck.

864
00:48:22,210 --> 00:48:40,450
And this is easy to reduce from
delta regular independence set

865
00:48:40,450 --> 00:48:44,780
because-- k independence set
if we want to be explicit.

866
00:48:44,780 --> 00:48:47,350
We want to know is there an
independent set of size k.

867
00:48:47,350 --> 00:48:50,210
I just give that to
partial vertex cover

868
00:48:50,210 --> 00:48:55,270
with k prime equal
to delta times k.

869
00:48:55,270 --> 00:49:00,970
Sorry, that's l prime
is delta times k.

870
00:49:00,970 --> 00:49:04,080
So the idea is here I want
to choose k vertices that

871
00:49:04,080 --> 00:49:05,260
are independent.

872
00:49:05,260 --> 00:49:08,340
And if I can do that, I will be
able to choose k vertices that

873
00:49:08,340 --> 00:49:09,839
cover exactly
delta times k edges

874
00:49:09,839 --> 00:49:11,880
because if they're
independent, none of the edges

875
00:49:11,880 --> 00:49:13,710
will be shared among
my independence set.

876
00:49:13,710 --> 00:49:15,180
And this is if and only if.

877
00:49:15,180 --> 00:49:18,630
If I try to choose k vertices
that cover exactly delta k

878
00:49:18,630 --> 00:49:22,190
edges, then they
can't be adjacent.

879
00:49:22,190 --> 00:49:23,840
The vertices can be adjacent.

880
00:49:23,840 --> 00:49:26,490
So these problems become
the same under this mapping,

881
00:49:26,490 --> 00:49:28,000
so I didn't even blow up.

882
00:49:28,000 --> 00:49:29,190
My parameter here is k.

883
00:49:29,190 --> 00:49:30,990
I forgot to mention.

884
00:49:30,990 --> 00:49:34,290
I mean you could guess from the
letter, but parameterized by k,

885
00:49:34,290 --> 00:49:36,210
this problem is w1 complete.

886
00:49:36,210 --> 00:49:39,000
Parameterized by l,
this problem is FPT.

887
00:49:39,000 --> 00:49:41,850
So be a little careful.

888
00:49:41,850 --> 00:49:45,370
But here's the reduction for k.

889
00:49:45,370 --> 00:49:47,690
There is no reduction
for-- there's only

890
00:49:47,690 --> 00:49:51,530
good reductions for l.

891
00:49:51,530 --> 00:49:54,560
So there we are clearly using
that the graph was regular.

892
00:49:54,560 --> 00:49:57,264
Otherwise, it would be
at most k-- at most delta

893
00:49:57,264 --> 00:49:58,930
for everybody, and
then it's hard to get

894
00:49:58,930 --> 00:49:59,929
the actual independence.

895
00:49:59,929 --> 00:50:00,760
Yeah.

896
00:50:00,760 --> 00:50:02,468
AUDIENCE: Sorry, I'm
still a bit confused

897
00:50:02,468 --> 00:50:05,750
about the relationship
between w1 and XP and FPT.

898
00:50:05,750 --> 00:50:07,730
I don't know what w1 is.

899
00:50:07,730 --> 00:50:10,340
PROFESSOR: Yeah,
well we'll get there.

900
00:50:10,340 --> 00:50:12,760
For now, you can think
w1 is all problems they

901
00:50:12,760 --> 00:50:16,030
can be parameterized, reduced to
k-step non-deterministic Turing

902
00:50:16,030 --> 00:50:17,169
machine.

903
00:50:17,169 --> 00:50:18,210
That's a fine definition.

904
00:50:18,210 --> 00:50:20,150
Some people use that.

905
00:50:20,150 --> 00:50:23,485
And in general, FPT
is contained in w1.

906
00:50:23,485 --> 00:50:26,170
It's contained in other
things, which we will get to.

907
00:50:26,170 --> 00:50:30,490
w2 and so on is all
contained in XP.

908
00:50:30,490 --> 00:50:32,880
And these are strict if you
believe exponential time

909
00:50:32,880 --> 00:50:34,010
hypothesis.

910
00:50:34,010 --> 00:50:36,730
So if you believe there are
no sub exponential algorithms

911
00:50:36,730 --> 00:50:41,320
for SAT, then this problem has
no FPT algorithm with respect

912
00:50:41,320 --> 00:50:44,419
to k and all the
w1 complete things.

913
00:50:44,419 --> 00:50:46,210
I mean from a complexity
theory standpoint,

914
00:50:46,210 --> 00:50:48,800
it will be fun to look
at these larger classes.

915
00:50:48,800 --> 00:50:51,030
From is there an FPT
algorithm standpoint,

916
00:50:51,030 --> 00:50:53,540
all you care about
is it's not here.

917
00:50:53,540 --> 00:50:57,090
And any hardness,
w1 or worse, will

918
00:50:57,090 --> 00:51:00,260
imply there's no FPT
algorithm if you assume ETH.

919
00:51:00,260 --> 00:51:02,770
AUDIENCE: Do you know if any
of these inclusions are strict?

920
00:51:02,770 --> 00:51:04,520
PROFESSOR: Like I said,
they're all strict

921
00:51:04,520 --> 00:51:06,270
if you assume exponential
time hypothesis.

922
00:51:08,640 --> 00:51:12,420
If you prove XP is
different from FPT,

923
00:51:12,420 --> 00:51:17,190
then you prove P does not
equal NP, I think, pretty sure.

924
00:51:17,190 --> 00:51:20,640
So we're not going to
non-categorically say

925
00:51:20,640 --> 00:51:24,120
these things are strict, because
these are all stronger versions

926
00:51:24,120 --> 00:51:25,510
of P does not equal NP.

927
00:51:25,510 --> 00:51:28,240
But if you believe
exponential time hypothesis,

928
00:51:28,240 --> 00:51:30,110
then they're all strict.

929
00:51:30,110 --> 00:51:33,225
So that's one
standard assumption

930
00:51:33,225 --> 00:51:36,200
that gives us everything.

931
00:51:36,200 --> 00:51:39,440
Cool, let me give you another
version of clique that's hard.

932
00:51:57,020 --> 00:52:08,840
Mufti-colored clique if you
remember way back to three

933
00:52:08,840 --> 00:52:11,600
partition, we had a
variation on three partition

934
00:52:11,600 --> 00:52:14,520
called numerical
three-dimensional matching

935
00:52:14,520 --> 00:52:17,700
where you had to choose
your triples from three

936
00:52:17,700 --> 00:52:19,390
different sets.

937
00:52:19,390 --> 00:52:21,392
So this is the analog
of that, or the analog

938
00:52:21,392 --> 00:52:22,850
of three-dimensional
matching where

939
00:52:22,850 --> 00:52:25,230
you had three different
sets of vertices.

940
00:52:25,230 --> 00:52:29,410
For clique here, set
the vertices partition

941
00:52:29,410 --> 00:52:35,330
into k clusters.

942
00:52:35,330 --> 00:52:44,910
And the question is, is there
k clique with one vertex

943
00:52:44,910 --> 00:52:47,700
per cluster?

944
00:52:50,410 --> 00:52:53,100
And in fact, so you think of
these things as being colored,

945
00:52:53,100 --> 00:52:56,360
color 1, and these
are color k and so on.

946
00:52:56,360 --> 00:52:59,740
And in fact, we can assume
this is a proper coloring

947
00:52:59,740 --> 00:53:02,960
because if you're not allowed
to choose two vertices from one

948
00:53:02,960 --> 00:53:06,270
class, then there's no reason
to have edges between vertices

949
00:53:06,270 --> 00:53:07,510
of the same color.

950
00:53:07,510 --> 00:53:13,020
So this is in fact a
k-coloring of the graph.

951
00:53:13,020 --> 00:53:15,370
So we are given a k-color graph.

952
00:53:15,370 --> 00:53:18,740
And we want to know,
does it have a k clique?

953
00:53:18,740 --> 00:53:20,790
And if it has a d clique
and it's k-colored, then

954
00:53:20,790 --> 00:53:23,790
in particular, you will
have exactly one vertex

955
00:53:23,790 --> 00:53:24,780
per color class.

956
00:53:28,020 --> 00:53:33,205
We can prove this is hard
by a reduction from clique.

957
00:53:41,400 --> 00:53:44,040
Namely, if we have
a vertex v, we're

958
00:53:44,040 --> 00:53:46,080
going to make k copies of it.

959
00:53:48,699 --> 00:53:50,115
You'll see this
is a common trick.

960
00:53:55,260 --> 00:54:00,760
And we will color
them 1 to up to k.

961
00:54:00,760 --> 00:54:05,890
So good, now we have
vertices in each color class.

962
00:54:05,890 --> 00:54:12,810
And then if we have an edge
(v,w) convert it into edges

963
00:54:12,810 --> 00:54:19,420
(vi, wj) for all
i not equal to j.

964
00:54:19,420 --> 00:54:21,550
So pretty much the
obvious reduction.

965
00:54:21,550 --> 00:54:24,887
Once you said make k copies,
we'll also duplicate the edges

966
00:54:24,887 --> 00:54:27,470
in all versions except, because
it's supposed to be a k-color,

967
00:54:27,470 --> 00:54:31,190
we're not allowed
to connect (vi,wi).

968
00:54:31,190 --> 00:54:34,130
But otherwise, we'll just
throw in all of those edges.

969
00:54:34,130 --> 00:54:36,230
And the point is this,
again, doesn't really

970
00:54:36,230 --> 00:54:38,120
blow up your cliques
because if you

971
00:54:38,120 --> 00:54:40,260
have some clique
in this structure,

972
00:54:40,260 --> 00:54:42,480
you can just forget
about the i indices.

973
00:54:42,480 --> 00:54:45,400
Because you know
it's a clique, you

974
00:54:45,400 --> 00:54:47,440
will never choose two
verses from the same color

975
00:54:47,440 --> 00:54:49,735
class, two vertices
with the same index,

976
00:54:49,735 --> 00:54:51,860
and so you can take any
clique here and collapse it

977
00:54:51,860 --> 00:54:52,647
to a clique here.

978
00:54:52,647 --> 00:54:54,230
Conversely, if you
have a clique here,

979
00:54:54,230 --> 00:54:57,013
you can just assign those
vertices arbitrary numbers

980
00:54:57,013 --> 00:54:59,263
as long as they're all
distinct any permutation of one

981
00:54:59,263 --> 00:55:01,300
through k, and you'll
get a clique down here.

982
00:55:01,300 --> 00:55:02,360
So it's the same problem.

983
00:55:02,360 --> 00:55:02,860
Question?

984
00:55:02,860 --> 00:55:06,390
AUDIENCE: That v is different
from the v above it, right?

985
00:55:06,390 --> 00:55:09,850
PROFESSOR: Yeah, this is
for all v and for all vw.

986
00:55:13,130 --> 00:55:17,820
If there's an edge in the
graph, then we do that.

987
00:55:17,820 --> 00:55:21,980
And if there's a vertex
in the graph, we do that.

988
00:55:21,980 --> 00:55:24,270
Cool.

989
00:55:24,270 --> 00:55:27,712
So k prime here equals
k, no expansion.

990
00:55:31,090 --> 00:55:34,260
This may seem trivial,
but it's actually

991
00:55:34,260 --> 00:55:37,690
a fairly recent innovation to
think about multicolor clique

992
00:55:37,690 --> 00:55:40,250
and, in general, it
simplifies proofs.

993
00:55:40,250 --> 00:55:42,820
I have heard of proofs
simplifying from tens of pages

994
00:55:42,820 --> 00:55:44,950
to one page.

995
00:55:44,950 --> 00:55:47,390
We'll probably get to
some more-- to such

996
00:55:47,390 --> 00:55:49,420
sophisticated example soon.

997
00:55:49,420 --> 00:55:52,865
But I can give you
one simple example.

998
00:55:52,865 --> 00:55:54,240
Before I get to
a simple example,

999
00:55:54,240 --> 00:55:57,430
I want to show
you a fun example.

1000
00:55:57,430 --> 00:56:02,340
Not simple, but let me tell
you what the problem is.

1001
00:56:02,340 --> 00:56:03,805
I won't cover that proof.

1002
00:56:13,280 --> 00:56:23,276
The problem is shortest
common super sequence.

1003
00:56:27,000 --> 00:56:30,490
This is a problem that comes
up in computational biology.

1004
00:56:30,490 --> 00:56:33,540
Your given k strings.

1005
00:56:37,490 --> 00:56:43,250
Let's say alphabet sigma.

1006
00:56:43,250 --> 00:56:52,480
And yes, you're given a number,
which I'm going to write l.

1007
00:56:56,330 --> 00:56:59,970
You want to find
a string of length

1008
00:56:59,970 --> 00:57:14,250
l that's a super sequence
of all input strings.

1009
00:57:20,120 --> 00:57:23,310
So maybe you're given the
DNA sequence of human and DNA

1010
00:57:23,310 --> 00:57:26,160
sequence of mouse,
and you want to know

1011
00:57:26,160 --> 00:57:27,990
what is the shortest
DNA sequence that

1012
00:57:27,990 --> 00:57:30,750
contains all of the
letters of those strings

1013
00:57:30,750 --> 00:57:32,520
in the correct order.

1014
00:57:32,520 --> 00:57:35,080
So this is often
called an alignment.

1015
00:57:35,080 --> 00:57:39,200
If you have ACGG--
this is probably not

1016
00:57:39,200 --> 00:57:41,930
valid, but whatever--
And we have

1017
00:57:41,930 --> 00:57:44,020
some other guy like CAGGAT.

1018
00:57:50,186 --> 00:57:51,560
So I tried to draw
those aligned.

1019
00:57:55,790 --> 00:57:58,280
So then the common super
string here is ACAGGACT.

1020
00:58:03,820 --> 00:58:06,170
This is a super string,
meaning I can drop letters

1021
00:58:06,170 --> 00:58:07,550
from down here and get this.

1022
00:58:07,550 --> 00:58:09,680
Or, I can drop letters from
down here and get this.

1023
00:58:09,680 --> 00:58:11,650
That was for k equals 2.

1024
00:58:11,650 --> 00:58:14,070
In general, this
problem does have

1025
00:58:14,070 --> 00:58:17,030
an n to the k dynamic program.

1026
00:58:17,030 --> 00:58:19,160
But the question is whether
you could get FPT in k.

1027
00:58:19,160 --> 00:58:21,740
Could you get a small dependence
on the-- better dependence

1028
00:58:21,740 --> 00:58:22,940
on the number of strings?

1029
00:58:22,940 --> 00:58:25,610
And the answer is no
because this is w1 complete.

1030
00:58:30,080 --> 00:58:31,840
And this is a
sketch of the proof.

1031
00:58:34,769 --> 00:58:36,560
it's a reduction for
a multicolored clique.

1032
00:58:36,560 --> 00:58:38,650
I think this the paper that
introduced multicolored clique

1033
00:58:38,650 --> 00:58:39,160
in fact.

1034
00:58:39,160 --> 00:58:43,110
It's the earliest one I
could find, and this is 2003.

1035
00:58:43,110 --> 00:58:46,450
It became more
popular since 2009.

1036
00:58:46,450 --> 00:58:48,660
So you're basically
encoding the edges

1037
00:58:48,660 --> 00:58:52,480
by whether there's a 0 here
in between a huge set of ones.

1038
00:58:52,480 --> 00:58:56,210
And then this is intuitively
rep-- and then for every vertex

1039
00:58:56,210 --> 00:58:59,160
you're looking at, all
of the other vertices

1040
00:58:59,160 --> 00:59:01,290
and encoding whether they
have an edge to them.

1041
00:59:01,290 --> 00:59:02,665
And so if you have
a clique, that

1042
00:59:02,665 --> 00:59:05,430
will be a pattern
that you see repeated

1043
00:59:05,430 --> 00:59:07,172
among multiple vertices.

1044
00:59:07,172 --> 00:59:08,880
And so you'll end up
being able to shrink

1045
00:59:08,880 --> 00:59:13,380
your longest-- your shortest
common super sequence.

1046
00:59:13,380 --> 00:59:14,600
But the details are messy.

1047
00:59:14,600 --> 00:59:16,730
It's a few pages to prove it.

1048
00:59:16,730 --> 00:59:18,255
I will skip that
and tell you why

1049
00:59:18,255 --> 00:59:19,880
I wanted to tell you
about this problem

1050
00:59:19,880 --> 00:59:22,680
other than its computational
biology and useful.

1051
00:59:22,680 --> 00:59:27,340
Because it's used to prove
that flood-it is hard.

1052
00:59:27,340 --> 00:59:29,310
Time for some fun.

1053
00:59:29,310 --> 00:59:33,820
So here is flood-it.

1054
00:59:37,700 --> 00:59:39,870
You have a grid of
colored squares.

1055
00:59:39,870 --> 00:59:43,020
And in the top left,
we have colors.

1056
00:59:43,020 --> 00:59:45,467
And this is the special square.

1057
00:59:45,467 --> 00:59:47,550
What I can do is control
the color of that square.

1058
00:59:47,550 --> 00:59:49,970
So for example, I
could set it to red

1059
00:59:49,970 --> 00:59:52,400
and then anything that was
in that connected group

1060
00:59:52,400 --> 00:59:53,010
becomes red.

1061
00:59:53,010 --> 00:59:54,760
And now I've got a
bigger connected group.

1062
00:59:54,760 --> 00:59:59,510
So now I can make it blue,
and then pink, and then red,

1063
00:59:59,510 --> 01:00:03,285
and then blue, and then red.

1064
01:00:03,285 --> 01:00:05,060
Let's see how well I can do.

1065
01:00:05,060 --> 01:00:07,400
And the trouble is I have
a limited number of moves.

1066
01:00:10,310 --> 01:00:18,710
So this is a model for Ebola
virus spreading or zombie

1067
01:00:18,710 --> 01:00:24,000
infection or pick your favorite.

1068
01:00:24,000 --> 01:00:27,145
I mean you imagine these are
different species-- green.

1069
01:00:27,145 --> 01:00:28,990
Thank you.

1070
01:00:28,990 --> 01:00:31,780
A lot easier to play
when I have an oracle.

1071
01:00:31,780 --> 01:00:33,842
Yellow, that's better.

1072
01:00:33,842 --> 01:00:35,580
I don't know.

1073
01:00:35,580 --> 01:00:37,080
I'm actually doing
pretty well here.

1074
01:00:39,820 --> 01:00:43,240
I think now I just
have to do them all.

1075
01:00:43,240 --> 01:00:44,160
I made it!

1076
01:00:44,160 --> 01:00:45,940
25 moves.

1077
01:00:45,940 --> 01:00:48,000
Wow!

1078
01:00:48,000 --> 01:00:48,870
Felicitations!

1079
01:00:48,870 --> 01:00:50,161
That's the first time I've won.

1080
01:00:50,161 --> 01:00:52,610
It's a good demo.

1081
01:00:52,610 --> 01:00:55,660
Only played a few times,
but-- this apparently

1082
01:00:55,660 --> 01:00:58,770
became quite popular in 2006,
a company called LabPixies,

1083
01:00:58,770 --> 01:01:00,495
since bought by Google.

1084
01:01:00,495 --> 01:01:01,870
But there's tons
of free versions

1085
01:01:01,870 --> 01:01:04,290
out there you can play.

1086
01:01:04,290 --> 01:01:11,400
It's w1 complete with respect
to number of colors and number

1087
01:01:11,400 --> 01:01:14,010
of leaves.

1088
01:01:14,010 --> 01:01:16,280
So I don't know if this
is a generalization

1089
01:01:16,280 --> 01:01:19,310
or specialization, but it's a
variation of the problem that's

1090
01:01:19,310 --> 01:01:20,644
been studied.

1091
01:01:20,644 --> 01:01:22,060
It's the only one
I could can that

1092
01:01:22,060 --> 01:01:26,430
talked about parametrized
complexity of flood-it.

1093
01:01:26,430 --> 01:01:29,892
And so this is
flood-it on trees.

1094
01:01:34,590 --> 01:01:38,930
So same set up, but instead
of being a square grid graph,

1095
01:01:38,930 --> 01:01:40,270
I'm on a tree graph.

1096
01:01:40,270 --> 01:01:43,020
And the root of the tree
is the one I can control,

1097
01:01:43,020 --> 01:01:47,100
and my graph is going to
look like-- my tree is

1098
01:01:47,100 --> 01:01:48,100
going to look like this.

1099
01:01:52,955 --> 01:01:54,330
These are going
to be my strings.

1100
01:01:57,130 --> 01:01:59,990
And if I want to be
able to solve this

1101
01:01:59,990 --> 01:02:04,800
in some number of
moves, l moves,

1102
01:02:04,800 --> 01:02:08,790
then I need to-- the
sequence of colors

1103
01:02:08,790 --> 01:02:11,980
I do is a longest common
subsequence, super sequence,

1104
01:02:11,980 --> 01:02:17,280
of these strings because they
all have to get to the bottom.

1105
01:02:17,280 --> 01:02:19,530
They'll sit there and wait,
and so you can essentially

1106
01:02:19,530 --> 01:02:21,430
drop letters.

1107
01:02:21,430 --> 01:02:22,790
And that's cool.

1108
01:02:22,790 --> 01:02:24,160
Happy?

1109
01:02:24,160 --> 01:02:28,180
This is not literally
true because if I have,

1110
01:02:28,180 --> 01:02:32,340
let's say, a zero and a zero
here, both the same color--

1111
01:02:32,340 --> 01:02:35,870
So I'm representing
letters by colors,

1112
01:02:35,870 --> 01:02:38,550
then when I play-- if I
eventually play here and play

1113
01:02:38,550 --> 01:02:42,480
zero, I actually get advanced
two spaces instead of one.

1114
01:02:42,480 --> 01:02:48,530
So for that, you need to
first map every letter a.

1115
01:02:48,530 --> 01:02:54,180
So let's say in string
i, we map every letter

1116
01:02:54,180 --> 01:03:00,160
to a followed by a special
character for that string.

1117
01:03:00,160 --> 01:03:01,860
And so this blows things up.

1118
01:03:01,860 --> 01:03:03,616
In particular, it blows up l.

1119
01:03:03,616 --> 01:03:06,400
l grows by a factor
of the total length

1120
01:03:06,400 --> 01:03:08,820
of all strings because of
all these things that we add.

1121
01:03:08,820 --> 01:03:11,310
These are not
really compressible,

1122
01:03:11,310 --> 01:03:13,732
but it means that we alternate
between regular characters

1123
01:03:13,732 --> 01:03:14,690
and special characters.

1124
01:03:14,690 --> 01:03:16,481
And therefore, we never
have two characters

1125
01:03:16,481 --> 01:03:17,980
in a row that are the same.

1126
01:03:17,980 --> 01:03:20,170
So first you take
this problem, then

1127
01:03:20,170 --> 01:03:22,300
you reduce it to the
version of the problem that

1128
01:03:22,300 --> 01:03:24,750
has no repeated characters.

1129
01:03:24,750 --> 01:03:27,204
And then you can reduce that
to flood-it because that's

1130
01:03:27,204 --> 01:03:27,995
what we care about.

1131
01:03:27,995 --> 01:03:30,235
AUDIENCE: Sorry, what's the
parameter on the shortest

1132
01:03:30,235 --> 01:03:30,822
common super--

1133
01:03:30,822 --> 01:03:31,530
PROFESSOR: Right.

1134
01:03:31,530 --> 01:03:34,160
Parameter is-- they're
actually two parameters,

1135
01:03:34,160 --> 01:03:38,344
so this is a fun
thing, k and sigma.

1136
01:03:38,344 --> 01:03:40,010
So two parameters,
you can think of that

1137
01:03:40,010 --> 01:03:42,100
as being parameterized
by the sum or just

1138
01:03:42,100 --> 01:03:44,517
by the vector k, sigma.

1139
01:03:44,517 --> 01:03:47,100
Basically, we get to assume both
of them are relatively small.

1140
01:03:47,100 --> 01:03:50,200
We could be exponential
in both k and sigma.

1141
01:03:50,200 --> 01:03:52,410
If you put l in there,
it would not be hard.

1142
01:03:52,410 --> 01:03:56,720
But if you put k and sigma
and it's still w1 complete.

1143
01:03:56,720 --> 01:03:59,520
And so over here it's
the number of leaves

1144
01:03:59,520 --> 01:04:02,042
in the tree is one parameter
and the number of colors

1145
01:04:02,042 --> 01:04:03,000
is the other parameter.

1146
01:04:03,000 --> 01:04:05,000
So we're fixed parameter
tractable with respect

1147
01:04:05,000 --> 01:04:06,732
to that joint parameter.

1148
01:04:06,732 --> 01:04:07,232
Yeah?

1149
01:04:07,232 --> 01:04:09,565
AUDIENCE: Why is it obvious
that you can reduce flood-it

1150
01:04:09,565 --> 01:04:10,862
on trees to not flood-it?

1151
01:04:10,862 --> 01:04:12,570
PROFESSOR: It's not
obvious that the grid

1152
01:04:12,570 --> 01:04:14,700
problem is related to trees.

1153
01:04:14,700 --> 01:04:19,160
There are hardness results
for like 2 by n flood-it.

1154
01:04:19,160 --> 01:04:20,410
But they're just NP hardness.

1155
01:04:20,410 --> 01:04:23,270
I didn't see a w1 hardness
for 2 by n flood-it.

1156
01:04:23,270 --> 01:04:26,270
So as far as I
know, 2 by n colored

1157
01:04:26,270 --> 01:04:28,986
is hard-- is open from a
parameterized complexity

1158
01:04:28,986 --> 01:04:29,485
standpoint.

1159
01:04:33,890 --> 01:04:35,410
Cool.

1160
01:04:35,410 --> 01:04:38,510
I want to do one more
reduction, and then I'll

1161
01:04:38,510 --> 01:04:39,550
finally define w1.

1162
01:05:01,780 --> 01:05:08,010
Dominating Set, this problem
is actually w2 complete,

1163
01:05:08,010 --> 01:05:09,780
so this is even harder.

1164
01:05:09,780 --> 01:05:16,190
But before we worry about w2,
let's prove that it's w1 hard.

1165
01:05:16,190 --> 01:05:21,626
So we're going to reduce
from multicolored clique

1166
01:05:21,626 --> 01:05:24,456
to dominating set.

1167
01:05:24,456 --> 01:05:26,610
I have a nice figure.

1168
01:05:26,610 --> 01:05:32,800
This is a preview of this cool
parameterized algorithms book.

1169
01:05:32,800 --> 01:05:41,484
So I'm going to represent
each-- got it right.

1170
01:05:45,348 --> 01:05:48,270
Yeah, funny.

1171
01:05:48,270 --> 01:05:50,030
I'm going to
represent each vertex

1172
01:05:50,030 --> 01:05:52,370
in the multicolored
independence set problem

1173
01:05:52,370 --> 01:05:55,160
by a vertex in the
dominating set problem,

1174
01:05:55,160 --> 01:05:57,660
so vertices mapped to vertices.

1175
01:05:57,660 --> 01:05:59,670
This is the joy of
multicolored cliques.

1176
01:05:59,670 --> 01:06:02,120
We did all this k
duplication stuff

1177
01:06:02,120 --> 01:06:04,990
once so that we don't have to
duplicate our graphic anymore.

1178
01:06:04,990 --> 01:06:05,990
We just take this graph.

1179
01:06:05,990 --> 01:06:07,340
We poured it over.

1180
01:06:07,340 --> 01:06:14,660
Now, what that tells us is that
the vi's form color classes.

1181
01:06:14,660 --> 01:06:17,730
And what I'm going to do is
connect each color class.

1182
01:06:17,730 --> 01:06:19,720
So this is the set of
all things of color one,

1183
01:06:19,720 --> 01:06:21,710
set of all things of color two.

1184
01:06:21,710 --> 01:06:24,270
First of all, I'm going to add
2 dummy vertices to each color

1185
01:06:24,270 --> 01:06:26,030
class just to hang out there.

1186
01:06:26,030 --> 01:06:29,040
And then I'm going to connect
everything in the color class

1187
01:06:29,040 --> 01:06:32,310
by a clique.

1188
01:06:32,310 --> 01:06:36,070
Before we only had edges
between color classes,

1189
01:06:36,070 --> 01:06:39,100
but now I want to color--

1190
01:06:39,100 --> 01:06:40,940
I'm going to also
change those edges.

1191
01:06:40,940 --> 01:06:42,440
But the vertices
mapped to vertices,

1192
01:06:42,440 --> 01:06:44,070
I add two vertices
of each color,

1193
01:06:44,070 --> 01:06:46,730
and then this circle represents
that there's a clique in here,

1194
01:06:46,730 --> 01:06:48,350
clique in here, clique in here.

1195
01:06:48,350 --> 01:06:51,640
Now, my goal is to
find a dominating set.

1196
01:06:51,640 --> 01:06:53,980
And so the intention
is that you need

1197
01:06:53,980 --> 01:06:57,700
to choose one vertex
from each color class

1198
01:06:57,700 --> 01:06:59,827
in the dominant set.

1199
01:06:59,827 --> 01:07:02,160
And that's actually what these
dummy vertices are doing.

1200
01:07:02,160 --> 01:07:04,880
They are only going to be
connect to things in here.

1201
01:07:04,880 --> 01:07:06,910
They have to be
covered by somebody,

1202
01:07:06,910 --> 01:07:09,310
so that means you have to
choose someone in this clique.

1203
01:07:09,310 --> 01:07:11,060
You have to choose
someone in this clique.

1204
01:07:11,060 --> 01:07:12,920
When you do, it covers
everybody in there.

1205
01:07:12,920 --> 01:07:14,753
And there's going to
be no point of choosing

1206
01:07:14,753 --> 01:07:16,910
these dummy vertices because
they're only connected

1207
01:07:16,910 --> 01:07:17,630
to other things in the clique.

1208
01:07:17,630 --> 01:07:19,340
You might as well
choose these things.

1209
01:07:19,340 --> 01:07:25,150
Now, how do I represent an edge
in my independence set graph.

1210
01:07:25,150 --> 01:07:29,300
If I have an edge
between say u and v,

1211
01:07:29,300 --> 01:07:31,300
that's in this color class
and this color class.

1212
01:07:31,300 --> 01:07:33,050
We know they're in
different color classes

1213
01:07:33,050 --> 01:07:36,270
by multicolored
clique property--

1214
01:07:36,270 --> 01:07:38,210
multicolored independent
set property.

1215
01:07:38,210 --> 01:07:40,150
Sorry, same thing.

1216
01:07:40,150 --> 01:07:44,170
So I want to not
choose both u and v.

1217
01:07:44,170 --> 01:07:46,680
So I'm going to represent
that by adding a vertex here,

1218
01:07:46,680 --> 01:07:49,110
which must be dominated,
and connect it

1219
01:07:49,110 --> 01:07:52,480
to everyone except u over
here and everyone except v

1220
01:07:52,480 --> 01:07:54,530
over here, the red patches.

1221
01:07:54,530 --> 01:07:57,990
So there are edges from this
vertex to every red vertex

1222
01:07:57,990 --> 01:07:59,250
here and here.

1223
01:07:59,250 --> 01:08:02,580
So what that means is
if I don't choose u,

1224
01:08:02,580 --> 01:08:04,940
I will cover this added vertex.

1225
01:08:04,940 --> 01:08:09,410
If I don't choose v, I will
cover this added vertex.

1226
01:08:09,410 --> 01:08:13,850
So I will cover it as long as
I don't choose both u and v.

1227
01:08:13,850 --> 01:08:17,330
So it's like I should choose
at most one of these two guys,

1228
01:08:17,330 --> 01:08:18,826
then this will be
covered for free.

1229
01:08:18,826 --> 01:08:20,950
And it has to be covered
for free because we just--

1230
01:08:20,950 --> 01:08:25,040
if I set k prime to be the
number of color classes, which

1231
01:08:25,040 --> 01:08:28,189
is k, then I don't
have any flexibility.

1232
01:08:28,189 --> 01:08:30,720
I can't choose any of
these vertices to dominate.

1233
01:08:30,720 --> 01:08:33,729
And so that simulates
independence using dominating

1234
01:08:33,729 --> 01:08:37,430
set, pretty simple and clean.

1235
01:08:37,430 --> 01:08:40,040
So that gives you a
flavor of nice reductions

1236
01:08:40,040 --> 01:08:45,430
you can do with multicolored
clique or independence set.

1237
01:08:45,430 --> 01:08:49,250
AUDIENCE: What's
two extra vertices?

1238
01:08:49,250 --> 01:08:52,060
PROFESSOR: Why two?

1239
01:08:52,060 --> 01:08:53,667
AUDIENCE: The two
you added, it looks

1240
01:08:53,667 --> 01:08:55,669
like are not adjacent so--

1241
01:08:55,669 --> 01:08:59,069
AUDIENCE: Oh, so
we derived them.

1242
01:08:59,069 --> 01:09:03,300
PROFESSOR: Right We want
them to be there in order

1243
01:09:03,300 --> 01:09:05,779
to force choosing
somebody down here,

1244
01:09:05,779 --> 01:09:07,830
but we don't actually want
to choose one of them.

1245
01:09:07,830 --> 01:09:09,850
And so we emit that edge
so if you choose one,

1246
01:09:09,850 --> 01:09:11,160
you don't cover the other.

1247
01:09:11,160 --> 01:09:11,660
Thanks.

1248
01:09:16,172 --> 01:09:17,630
I think it simplifies
the argument.

1249
01:09:17,630 --> 01:09:19,340
It's probably not
necessary because I

1250
01:09:19,340 --> 01:09:21,359
think you could argue
by replacement argument

1251
01:09:21,359 --> 01:09:26,170
that you won't choose
them, but anyway.

1252
01:09:26,170 --> 01:09:27,920
Time for some definitions.

1253
01:09:27,920 --> 01:09:32,586
I wanted to do some fun problems
before we got to the somewhat

1254
01:09:32,586 --> 01:09:33,335
weird definitions.

1255
01:09:37,850 --> 01:09:38,700
The time has come.

1256
01:09:45,270 --> 01:09:48,560
By the way, dominate set is
w2 complete, so is set cover.

1257
01:09:48,560 --> 01:09:53,500
And most w2 hardness reductions
start from one of these two.

1258
01:09:53,500 --> 01:09:56,910
We already know how to reduce
dominating set to set cover.

1259
01:09:56,910 --> 01:10:00,100
You make a set for every
neighborhood set in the graph.

1260
01:10:00,100 --> 01:10:02,170
So this reduction
is easy, and that

1261
01:10:02,170 --> 01:10:05,580
preserves a parameter so just
from the same reduction we

1262
01:10:05,580 --> 01:10:06,950
did two classes ago, I think.

1263
01:10:10,180 --> 01:10:14,750
Here is a problem which
will look familiar,

1264
01:10:14,750 --> 01:10:17,560
but under a slightly
different name.

1265
01:10:17,560 --> 01:10:21,275
This is what I might
call circuit ones.

1266
01:10:25,660 --> 01:10:29,190
One's meaning you want to
set some number of the inputs

1267
01:10:29,190 --> 01:10:32,120
to be one, just like our
optimization version.

1268
01:10:32,120 --> 01:10:34,406
In this universe, it is
called weighted circuit set.

1269
01:10:34,406 --> 01:10:36,030
I'm not a fan of
weighted because there

1270
01:10:36,030 --> 01:10:38,150
aren't actual weights here.

1271
01:10:38,150 --> 01:10:39,900
The goal is to get
minimum hamming weight,

1272
01:10:39,900 --> 01:10:42,770
meaning the minimum number
of ones in your input.

1273
01:10:42,770 --> 01:10:48,780
So I give you a circuit-- I'll
draw a very simple circuit,

1274
01:10:48,780 --> 01:10:51,010
and it has one output.

1275
01:10:51,010 --> 01:10:53,490
It has some number of inputs.

1276
01:10:53,490 --> 01:11:00,880
My goal is to set k
ones and get a one out.

1277
01:11:00,880 --> 01:11:01,570
Did I do it?

1278
01:11:01,570 --> 01:11:02,465
I did it!

1279
01:11:02,465 --> 01:11:05,020
! wasn't even looking.

1280
01:11:05,020 --> 01:11:07,041
So that's an example of
an input and an output

1281
01:11:07,041 --> 01:11:08,040
to weighted circuit set.

1282
01:11:08,040 --> 01:11:15,514
The parameter here, k, is the
number of ones, one inputs.

1283
01:11:15,514 --> 01:11:16,930
So the question
is can you satisfy

1284
01:11:16,930 --> 01:11:19,900
a circuit using only k ones.

1285
01:11:19,900 --> 01:11:21,550
That's weighted circuit set.

1286
01:11:21,550 --> 01:11:25,400
This problem defines
the class called WP.

1287
01:11:25,400 --> 01:11:27,320
This is the original
definition of WP

1288
01:11:27,320 --> 01:11:31,650
is all problems that reduce
to weighted circuit set

1289
01:11:31,650 --> 01:11:34,020
in a parameterized
reduction sense.

1290
01:11:34,020 --> 01:11:38,400
It's all parameterized
problems that

1291
01:11:38,400 --> 01:11:40,180
reduce to weighted circuit set.

1292
01:11:48,510 --> 01:11:49,870
I lost my hierarchy.

1293
01:11:49,870 --> 01:11:52,600
But this is even bigger
than all the things.

1294
01:11:52,600 --> 01:11:56,740
So we have FPT is
contained in w1.

1295
01:11:56,740 --> 01:12:00,850
It's contained in w2, et cetera.

1296
01:12:00,850 --> 01:12:05,300
And then we have WP,
and then we have XP.

1297
01:12:05,300 --> 01:12:06,845
So it's not bigger
than everything.

1298
01:12:06,845 --> 01:12:10,110
It's not bigger than XP.

1299
01:12:10,110 --> 01:12:14,880
I won't prove this
here, but it is true.

1300
01:12:14,880 --> 01:12:16,910
Now, I want to
specialize this problem.

1301
01:12:16,910 --> 01:12:23,150
I'm going to simplify it
using a notion called weft.

1302
01:12:23,150 --> 01:12:27,300
So first the depth
of the circuit,

1303
01:12:27,300 --> 01:12:28,600
this is the longest path.

1304
01:12:34,400 --> 01:12:35,500
That's a normal notion.

1305
01:12:40,460 --> 01:12:49,130
Then, the weft of a circuit
is the maximum number

1306
01:12:49,130 --> 01:12:54,814
of big gates on an
input to output path.

1307
01:13:03,180 --> 01:13:10,214
Big gates means let's
say more than two inputs.

1308
01:13:10,214 --> 01:13:12,630
In general, it's more than
some constant number of inputs,

1309
01:13:12,630 --> 01:13:16,450
but two is fine because if you
use a bunch of two input gates,

1310
01:13:16,450 --> 01:13:18,990
you can build a ten input gate.

1311
01:13:18,990 --> 01:13:21,420
But there's a distinction
between constant input gates

1312
01:13:21,420 --> 01:13:23,580
and super constant input gates.

1313
01:13:23,580 --> 01:13:25,970
So if you count how many
super constant input gates

1314
01:13:25,970 --> 01:13:30,440
you need in a constant
depth circuit,

1315
01:13:30,440 --> 01:13:38,086
then we get the w
classes, w for weft.

1316
01:13:43,920 --> 01:13:51,500
w(t) is the set of
parameterized problems

1317
01:13:51,500 --> 01:14:05,170
that reduce to constant depth
weft-t weighted circuit set.

1318
01:14:10,070 --> 01:14:12,160
t is not the parameter.

1319
01:14:12,160 --> 01:14:14,740
K is still the parameter.
k is the number of ones.

1320
01:14:14,740 --> 01:14:15,640
t is the thing.

1321
01:14:19,620 --> 01:14:21,170
Let's do some examples.

1322
01:14:21,170 --> 01:14:24,060
AUDIENCE: Depth has to be
larger than weft numbers?

1323
01:14:24,060 --> 01:14:26,230
PROFESSOR: Yes, depth is
always larger than weft.

1324
01:14:26,230 --> 01:14:28,980
So we allow-- so t
is a fixed constant.

1325
01:14:28,980 --> 01:14:31,255
Depth could be 100
times t or something.

1326
01:14:31,255 --> 01:14:34,484
AUDIENCE: Then, w in WP is
a different w than w(t)?

1327
01:14:37,621 --> 01:14:39,370
PROFESSOR: They both
have to do with weft,

1328
01:14:39,370 --> 01:14:42,240
but that's the P
is a polynomial.

1329
01:14:42,240 --> 01:14:45,720
It corresponds to
polynomial weft, which

1330
01:14:45,720 --> 01:14:47,825
is basically unbounded weft.

1331
01:14:47,825 --> 01:14:49,359
w does not stand for weighted.

1332
01:14:49,359 --> 01:14:50,775
In both cases, it
stands for weft.

1333
01:14:54,150 --> 01:14:55,520
What is weft?

1334
01:14:55,520 --> 01:14:57,720
Weft is the opposite of warp.

1335
01:14:57,720 --> 01:15:00,080
When you're weaving,
you've got the warp threads

1336
01:15:00,080 --> 01:15:02,690
and then you've got this weft
thread that goes back and forth

1337
01:15:02,690 --> 01:15:05,260
and ties the whole
circuit together.

1338
01:15:05,260 --> 01:15:09,310
So that's sideways,
so if you imagine

1339
01:15:09,310 --> 01:15:13,980
that constant input gates do
not need a weft to hold them in,

1340
01:15:13,980 --> 01:15:18,880
but super constant ones do, then
it's how many layers of wefting

1341
01:15:18,880 --> 01:15:21,340
do you have to do to
cover all those things?

1342
01:15:21,340 --> 01:15:25,730
That's the larges the longest
path in terms of counting

1343
01:15:25,730 --> 01:15:29,610
the-- that's the term, OK?

1344
01:15:29,610 --> 01:15:32,050
But it's actually fairly
useful to think about.

1345
01:15:32,050 --> 01:15:35,550
So for example, independence
set, inputs are up here.

1346
01:15:35,550 --> 01:15:36,940
Output is down here.

1347
01:15:36,940 --> 01:15:39,050
I need to choose,
for each vertex,

1348
01:15:39,050 --> 01:15:40,550
whether it's in the
independent set.

1349
01:15:40,550 --> 01:15:44,480
And if I negate them, then
I need various constraints.

1350
01:15:44,480 --> 01:15:47,490
This is the graph that's
being represented.

1351
01:15:47,490 --> 01:15:49,650
I want to say either
I don't choose this

1352
01:15:49,650 --> 01:15:52,560
or I don't choose this.

1353
01:15:52,560 --> 01:15:55,410
And then all of those
clauses must be true.

1354
01:15:55,410 --> 01:15:57,170
That's the and of those things.

1355
01:15:57,170 --> 01:15:58,700
This is big.

1356
01:15:58,700 --> 01:15:59,540
These are not big.

1357
01:15:59,540 --> 01:16:00,670
They have two inputs.

1358
01:16:00,670 --> 01:16:02,140
This has one input.

1359
01:16:02,140 --> 01:16:04,410
So these are all free.

1360
01:16:04,410 --> 01:16:07,140
And then we just do one
in terms of the depth

1361
01:16:07,140 --> 01:16:10,380
here where there's one
level of big gates.

1362
01:16:10,380 --> 01:16:12,900
So independent set is in w1.

1363
01:16:12,900 --> 01:16:15,609
That's a proof that it's in w1.

1364
01:16:15,609 --> 01:16:17,900
So that proves all the things
we've been talking about,

1365
01:16:17,900 --> 01:16:22,040
except dominating set, is in w1.

1366
01:16:22,040 --> 01:16:25,970
Now dominating set,
you need two levels.

1367
01:16:25,970 --> 01:16:29,820
So we have the a constraint.

1368
01:16:29,820 --> 01:16:32,080
So this is saying that
a should be covered.

1369
01:16:32,080 --> 01:16:34,680
a is adjacent to a, b, and c.

1370
01:16:34,680 --> 01:16:38,080
I mean if I choose a, b, or
c, then a will be dominated

1371
01:16:38,080 --> 01:16:40,640
by the definition of
dominating set because a

1372
01:16:40,640 --> 01:16:42,430
is adjacent just to b and c.

1373
01:16:42,430 --> 01:16:44,540
But also if I choose
a, then it's dominated.

1374
01:16:44,540 --> 01:16:46,429
So this is an or of
those three things.

1375
01:16:46,429 --> 01:16:48,470
And a should be dominated,
b should be dominated,

1376
01:16:48,470 --> 01:16:49,386
c should be dominated.

1377
01:16:49,386 --> 01:16:50,851
So it's an and of
all those things.

1378
01:16:50,851 --> 01:16:52,350
But if we don't
have bounded degree,

1379
01:16:52,350 --> 01:16:53,933
which we can't because
this problem is

1380
01:16:53,933 --> 01:17:00,140
easy for bounded degree, then
we have two levels of big gates.

1381
01:17:00,140 --> 01:17:02,830
So this problem is in w2.

1382
01:17:02,830 --> 01:17:07,470
Now, funnily enough, w1 and
w2 are the most common things

1383
01:17:07,470 --> 01:17:08,140
you see.

1384
01:17:08,140 --> 01:17:11,710
I've never seen
a paper about w3.

1385
01:17:11,710 --> 01:17:14,290
Some statement, I
think, about problems

1386
01:17:14,290 --> 01:17:18,990
that we tend to care about,
ands of ors are common.

1387
01:17:18,990 --> 01:17:22,510
This, of course,
CNF SAT is in w2.

1388
01:17:22,510 --> 01:17:25,210
I should say weighted
CNF SAT is in w2.

1389
01:17:25,210 --> 01:17:28,800
It's actually w2 complete.

1390
01:17:28,800 --> 01:17:33,390
Weighted three
set is w1 complete

1391
01:17:33,390 --> 01:17:35,850
because things of size 3 are OK.

1392
01:17:35,850 --> 01:17:38,210
We can split those into
two things of size 2.

1393
01:17:38,210 --> 01:17:45,360
So in general, order one
SAT, you have five SAT.

1394
01:17:45,360 --> 01:17:48,640
That's w1 complete.

1395
01:17:48,640 --> 01:17:52,290
And CNF SAT is w2 complete.

1396
01:17:52,290 --> 01:17:54,260
And I think most of the
problems we care about

1397
01:17:54,260 --> 01:17:58,005
can be expressed as--
this should be weighted.

1398
01:17:58,005 --> 01:18:01,820
I'm going tow rite w,
but I mean weighted.

1399
01:18:01,820 --> 01:18:04,460
But this is a
capital W, different.

1400
01:18:04,460 --> 01:18:06,810
That's weft.

1401
01:18:06,810 --> 01:18:11,800
So fine that's-- and most of the
problems we care about can be

1402
01:18:11,800 --> 01:18:13,030
represented as CNF SAT.

1403
01:18:13,030 --> 01:18:15,800
I think that's why we
rarely get outside of w2.

1404
01:18:15,800 --> 01:18:17,190
But there are
other things there.

1405
01:18:17,190 --> 01:18:22,720
You could do an or of ands of
ors or an and of ors of ands.

1406
01:18:22,720 --> 01:18:26,414
Now, we know such formulas
can be converted into CNF form

1407
01:18:26,414 --> 01:18:27,830
and only get a
polynomial blow up.

1408
01:18:27,830 --> 01:18:29,010
What's going on?

1409
01:18:29,010 --> 01:18:32,040
Well, when you do
that inefficient form,

1410
01:18:32,040 --> 01:18:33,370
you add extra variables.

1411
01:18:33,370 --> 01:18:35,420
When you add extra
variables, you are no longer

1412
01:18:35,420 --> 01:18:36,440
preserving the weight.

1413
01:18:36,440 --> 01:18:39,020
It would not be a
parameterized reduction.

1414
01:18:39,020 --> 01:18:41,937
So if you're just caring about
satisfiability, that's fine.

1415
01:18:41,937 --> 01:18:44,020
But if you're considering
weighted satisfiability,

1416
01:18:44,020 --> 01:18:46,340
you want to minimize
the number of ones,

1417
01:18:46,340 --> 01:18:48,540
it totally changes when
you convert to CNF.

1418
01:18:48,540 --> 01:18:49,295
So it matters.

1419
01:18:52,220 --> 01:18:59,100
One more fun result, if you
look at 2-tape non-deterministic

1420
01:18:59,100 --> 01:19:01,930
Turing machines--
the reason I had

1421
01:19:01,930 --> 01:19:03,889
to define 1-tape
non-deterministic Turing

1422
01:19:03,889 --> 01:19:05,430
machines is because
they're different

1423
01:19:05,430 --> 01:19:07,820
from 2-tape non-deterministic
Turing machines.

1424
01:19:07,820 --> 01:19:11,570
These are w2 complete.

1425
01:19:11,570 --> 01:19:15,520
The k-step version of 2-tape.

1426
01:19:15,520 --> 01:19:18,080
2-tapes is you have 2-tapes
that can advance independently.

1427
01:19:18,080 --> 01:19:20,630
Or, you can think of there
are two fingers you can

1428
01:19:20,630 --> 01:19:22,910
independently move on one tape.

1429
01:19:22,910 --> 01:19:25,050
Those are equivalent,
and they give you w2.

1430
01:19:25,050 --> 01:19:28,310
Three tapes, gives w2.

1431
01:19:28,310 --> 01:19:31,337
Any constant number of
tapes, you still get w2.

1432
01:19:31,337 --> 01:19:33,170
And this is another, I
think, natural reason

1433
01:19:33,170 --> 01:19:35,780
why w2 comes up a lot.

1434
01:19:35,780 --> 01:19:40,310
I have seen problems that are
hard for what's called w-star.

1435
01:19:40,310 --> 01:19:43,420
This w-star is wt
for all fixed t.

1436
01:19:43,420 --> 01:19:48,320
You can think of
that as w order a.

1437
01:19:48,320 --> 01:19:49,870
So it's the same thing.

1438
01:19:49,870 --> 01:19:52,310
So I've seen problems that
don't depend on what t is.

1439
01:19:52,310 --> 01:19:53,910
They're hard for all of them.

1440
01:19:53,910 --> 01:19:56,060
But I've never seen a
w3 complete problem.

1441
01:19:56,060 --> 01:19:57,550
Maybe we can think of one.

1442
01:19:57,550 --> 01:20:00,900
I've also seen WP come
up, and there's also

1443
01:20:00,900 --> 01:20:07,240
a problem called W-SAT which
is if you have weighted formula

1444
01:20:07,240 --> 01:20:10,840
sat instead of weighted circuits
sat, you get w-sat instead WP.

1445
01:20:10,840 --> 01:20:15,610
I think it's a little weaker
because in circuit SAT you can

1446
01:20:15,610 --> 01:20:17,664
reuse things and in
formula set, you can't.

1447
01:20:17,664 --> 01:20:19,830
And it turns out to give
slightly different classes.

1448
01:20:19,830 --> 01:20:21,532
So things are messier here.

1449
01:20:21,532 --> 01:20:23,490
But if you just care
about whether a problem is

1450
01:20:23,490 --> 01:20:28,220
fixed parameter tractable,
these are all bad.

1451
01:20:28,220 --> 01:20:29,970
But the point of knowing
about them is you

1452
01:20:29,970 --> 01:20:32,910
want to know if your
problem-- first you

1453
01:20:32,910 --> 01:20:35,230
should check where it
fits in the w hierarchy

1454
01:20:35,230 --> 01:20:37,730
by just thinking about what's
a trivial way to write it down

1455
01:20:37,730 --> 01:20:39,190
in the CNF style thing.

1456
01:20:39,190 --> 01:20:41,190
Figure out whether
it's w1 or w2.

1457
01:20:41,190 --> 01:20:43,780
Then, you'll know which
problems you should start from.

1458
01:20:43,780 --> 01:20:46,497
If it's w1, probably want to
start from a version of clique.

1459
01:20:46,497 --> 01:20:48,080
If it's w2, you
probably want to start

1460
01:20:48,080 --> 01:20:49,455
from a version of
dominating set.

1461
01:20:49,455 --> 01:20:51,030
That's why we talked
about them here.

1462
01:20:51,030 --> 01:20:52,640
And usually it's w1 or w2.

1463
01:20:55,320 --> 01:20:56,210
Cool.

1464
01:20:56,210 --> 01:20:58,820
Questions?

1465
01:20:58,820 --> 01:21:00,360
We'll do more next time.

1466
01:21:00,360 --> 01:21:02,000
This is a fun area.

1467
01:21:02,000 --> 01:21:04,540
And next time we'll talk
about, in particular,

1468
01:21:04,540 --> 01:21:08,400
how the exponential time
hypothesis relates to all this.