1
00:00:00,090 --> 00:00:02,490
The following content is
provided under a Creative

2
00:00:02,490 --> 00:00:04,030
Commons license.

3
00:00:04,030 --> 00:00:06,360
Your support will help
MIT OpenCourseWare

4
00:00:06,360 --> 00:00:10,720
continue to offer high quality
educational resources for free.

5
00:00:10,720 --> 00:00:13,320
To make a donation or
view additional materials

6
00:00:13,320 --> 00:00:17,280
from hundreds of MIT courses,
visit MIT OpenCourseWare

7
00:00:17,280 --> 00:00:18,450
at ocw.mit.edu.

8
00:00:22,100 --> 00:00:24,680
ERIK DEMAINE: All right, welcome
back to Dynamic Optimality.

9
00:00:24,680 --> 00:00:27,020
This is the second
of two lectures.

10
00:00:27,020 --> 00:00:29,770
And today we're going to
focus mainly on lower bounds.

11
00:00:29,770 --> 00:00:32,479
So last time we saw this
geometric connection

12
00:00:32,479 --> 00:00:34,020
to binary search trees.

13
00:00:34,020 --> 00:00:37,760
So again, this is about is there
one best binary search tree.

14
00:00:37,760 --> 00:00:40,070
And we represented binary
search trees, or at least

15
00:00:40,070 --> 00:00:42,080
the execution of
those algorithms,

16
00:00:42,080 --> 00:00:45,920
as point sets in time space.

17
00:00:45,920 --> 00:00:49,340
And of course a point
set corresponded

18
00:00:49,340 --> 00:00:52,410
to a valid execution of a BST
tree where each of these points

19
00:00:52,410 --> 00:00:55,970
represented which nodes got
touched during an access.

20
00:00:55,970 --> 00:00:59,300
If and only if the point
set was arborally satisfied,

21
00:00:59,300 --> 00:01:01,910
meaning you take any two
points in the point set,

22
00:01:01,910 --> 00:01:03,530
if they span a
rectangle that is not

23
00:01:03,530 --> 00:01:05,720
just a horizontal or
vertical line segment,

24
00:01:05,720 --> 00:01:07,640
there must be a
third point somewhere

25
00:01:07,640 --> 00:01:09,410
inside that rectangle,
which in the end

26
00:01:09,410 --> 00:01:14,310
implies that there's a monotone
path between those two points.

27
00:01:14,310 --> 00:01:16,190
And then we saw, on
the upper bound side,

28
00:01:16,190 --> 00:01:19,100
we saw a greedy algorithm, which
was the obvious offline thing

29
00:01:19,100 --> 00:01:22,010
to do, which is as
these points come along,

30
00:01:22,010 --> 00:01:24,170
as you do the accesses,
the white dots,

31
00:01:24,170 --> 00:01:26,540
you add the necessary
red dots in order

32
00:01:26,540 --> 00:01:30,867
to make it arborally
satisfied row by row.

33
00:01:30,867 --> 00:01:33,200
And so that seemed like the
obvious offline thing to do.

34
00:01:33,200 --> 00:01:36,140
Turns out it could be done
online up to constant factors.

35
00:01:36,140 --> 00:01:37,790
I sketched that last time.

36
00:01:37,790 --> 00:01:40,100
And this is conjectured to
be within a constant factor

37
00:01:40,100 --> 00:01:41,060
of optimal.

38
00:01:41,060 --> 00:01:41,810
We can't prove it.

39
00:01:41,810 --> 00:01:43,851
What I'm going to show
today is our best attempts

40
00:01:43,851 --> 00:01:45,166
at proving this is optimal.

41
00:01:45,166 --> 00:01:46,790
In particular, there's
something called

42
00:01:46,790 --> 00:01:49,730
the signed greedy algorithm,
which is almost the same as

43
00:01:49,730 --> 00:01:52,100
greedy, but it's a lower bound.

44
00:01:52,100 --> 00:01:54,007
And greedy is an upper bound.

45
00:01:54,007 --> 00:01:56,090
So all you need to do is
show these two things are

46
00:01:56,090 --> 00:01:59,564
within a constant factor of
each other and we're done.

47
00:01:59,564 --> 00:02:01,230
We're not going to
get there, obviously,

48
00:02:01,230 --> 00:02:02,830
because we haven't
solved that yet.

49
00:02:02,830 --> 00:02:05,900
But along the way,
we're going to see

50
00:02:05,900 --> 00:02:10,639
tango trees, which achieve this
log log n competitive bound.

51
00:02:10,639 --> 00:02:13,120
So we think greedy is
constant competitive.

52
00:02:13,120 --> 00:02:14,840
The best we know right
now is log log n,

53
00:02:14,840 --> 00:02:17,330
this is an improvement
over red black trees,

54
00:02:17,330 --> 00:02:19,340
which achieve log n.

55
00:02:19,340 --> 00:02:21,710
Any balance binary search
tree is within a log n

56
00:02:21,710 --> 00:02:23,030
factor of optimal.

57
00:02:23,030 --> 00:02:27,350
So it's all between constant and
log n that we're trying to do.

58
00:02:27,350 --> 00:02:29,450
Another fun consequence
of lower bounds

59
00:02:29,450 --> 00:02:31,450
is a particular
sense in which log n

60
00:02:31,450 --> 00:02:34,350
is necessary for some
access sequences.

61
00:02:34,350 --> 00:02:37,340
So I argued last time
that if you take, like,

62
00:02:37,340 --> 00:02:39,650
a random access sequence, or--

63
00:02:39,650 --> 00:02:42,487
for example, if you look
at a binary search tree

64
00:02:42,487 --> 00:02:44,570
and you say, oh, I'll just
access the thing that's

65
00:02:44,570 --> 00:02:46,160
deepest in the
tree, there's always

66
00:02:46,160 --> 00:02:48,950
something that's deep in the
tree of depth, at least log n.

67
00:02:48,950 --> 00:02:50,450
And so for any
binary search tree

68
00:02:50,450 --> 00:02:51,710
there is an access sequence.

69
00:02:51,710 --> 00:02:53,543
No matter what that
binary search tree does,

70
00:02:53,543 --> 00:02:56,300
I can choose the next
access to force you

71
00:02:56,300 --> 00:02:58,520
to take log n per operation.

72
00:02:58,520 --> 00:03:01,370
What we're going to see
today in these lower bounds

73
00:03:01,370 --> 00:03:05,600
is one access sequence that
for all binary search trees,

74
00:03:05,600 --> 00:03:08,780
they must spend log n time.

75
00:03:08,780 --> 00:03:11,240
Just changing the
quantifiers around.

76
00:03:11,240 --> 00:03:13,970
So instead of for every
binary search tree

77
00:03:13,970 --> 00:03:15,470
there is an access
sequence, there's

78
00:03:15,470 --> 00:03:17,480
going to be there is
an access sequence

79
00:03:17,480 --> 00:03:19,070
such that for every
binary search tree

80
00:03:19,070 --> 00:03:20,330
you need log n time.

81
00:03:20,330 --> 00:03:22,810
That's something we'll get
easily out of these lower

82
00:03:22,810 --> 00:03:24,470
bounds.

83
00:03:24,470 --> 00:03:26,660
So let's jump into
the lower bounds.

84
00:03:26,660 --> 00:03:29,620
And we're going to cover you
could say three different lower

85
00:03:29,620 --> 00:03:30,120
bounds.

86
00:03:30,120 --> 00:03:32,930
The independent rectangles
is kind of a generic class

87
00:03:32,930 --> 00:03:34,250
of lower bounds.

88
00:03:34,250 --> 00:03:36,260
Then we're going to see
two specific choices

89
00:03:36,260 --> 00:03:38,810
of these independent
rectangles, which are actually

90
00:03:38,810 --> 00:03:41,480
older than this
result. So this is

91
00:03:41,480 --> 00:03:44,450
sort of a modern interpretation
of two older results

92
00:03:44,450 --> 00:03:47,060
and a more general result.

93
00:03:47,060 --> 00:03:50,210
Signed greedy is going to turn
out to be the best lower bound.

94
00:03:50,210 --> 00:03:52,790
Is It's better than all the
ones that we will cover,

95
00:03:52,790 --> 00:03:56,630
but each of them has their
own uses for analysis.

96
00:03:56,630 --> 00:03:59,300
Each of them is going to
let us analyze an algorithm

97
00:03:59,300 --> 00:04:01,340
that we couldn't or
that we don't otherwise

98
00:04:01,340 --> 00:04:04,160
know how to analyze.

99
00:04:04,160 --> 00:04:08,600
So let's do the independent
rectangle lower bound.

100
00:04:08,600 --> 00:04:09,700
The sort of generic one.

101
00:04:25,540 --> 00:04:28,690
So these lower
bounds are all going

102
00:04:28,690 --> 00:04:31,780
to refer to the
original point set,

103
00:04:31,780 --> 00:04:34,720
the white dots, the accesses.

104
00:04:34,720 --> 00:04:37,330
The idea is you're given an
access sequence, a sequence x

105
00:04:37,330 --> 00:04:41,170
i-- x 1 up to x n, and you
want to know some lower bound

106
00:04:41,170 --> 00:04:43,690
that every binary
search tree requires

107
00:04:43,690 --> 00:04:46,930
a certain number of accesses, a
certain number of node touches

108
00:04:46,930 --> 00:04:47,980
for that access sequence.

109
00:04:47,980 --> 00:04:49,021
You know it's at least n.

110
00:04:49,021 --> 00:04:50,910
You want something
bigger than n.

111
00:04:50,910 --> 00:04:54,475
We've got to at least touch the
nodes that are being accessed.

112
00:04:58,983 --> 00:05:00,370
I'm going to drop this.

113
00:05:15,520 --> 00:05:19,090
So I want the notion of
independent rectangles.

114
00:05:19,090 --> 00:05:24,400
And general idea of
dependent rectangles

115
00:05:24,400 --> 00:05:27,340
would be something like this.

116
00:05:30,328 --> 00:05:31,300
Ah, I see.

117
00:05:42,610 --> 00:05:44,170
So these are two rectangles.

118
00:05:44,170 --> 00:05:46,570
I consider them dependent
because one of the corners

119
00:05:46,570 --> 00:05:48,900
is inside the other rectangle.

120
00:05:48,900 --> 00:05:51,320
This is true no matter
where the points are.

121
00:05:51,320 --> 00:05:59,750
So, for example, if I take two
points, they span a rectangle.

122
00:05:59,750 --> 00:06:04,660
If I take these two points, for
example, they span a rectangle.

123
00:06:04,660 --> 00:06:06,180
This corner is inside that one.

124
00:06:06,180 --> 00:06:08,440
So these are considered
dependent rectangles

125
00:06:08,440 --> 00:06:11,127
in either case.

126
00:06:11,127 --> 00:06:13,210
So corner here does not
necessarily mean a point--

127
00:06:13,210 --> 00:06:14,782
any of the four corners.

128
00:06:14,782 --> 00:06:16,240
Rectangle is defined
by two points,

129
00:06:16,240 --> 00:06:19,960
but it has all four corners.

130
00:06:19,960 --> 00:06:23,380
And so, in particular,
independent rectangles--

131
00:06:23,380 --> 00:06:26,080
for example, they might
be completely disjoint.

132
00:06:26,080 --> 00:06:28,700
Those are going
to be independent.

133
00:06:28,700 --> 00:06:31,700
Something like that
is independent.

134
00:06:31,700 --> 00:06:33,140
But there are some other cases.

135
00:06:33,140 --> 00:06:37,180
You can have rectangles
that look like this.

136
00:06:37,180 --> 00:06:38,110
OK?

137
00:06:38,110 --> 00:06:39,901
And it doesn't matter
where the points are.

138
00:06:39,901 --> 00:06:42,000
Maybe here, here,
here, and here.

139
00:06:42,000 --> 00:06:44,210
Or the other way.

140
00:06:44,210 --> 00:06:46,990
These are independent.

141
00:06:46,990 --> 00:06:52,960
And there's one other kind of
special case, which maybe I'll

142
00:06:52,960 --> 00:06:56,680
use color to draw the
other one because they're

143
00:06:56,680 --> 00:06:57,970
right on top of each other.

144
00:07:03,405 --> 00:07:10,000
So I've got a point here, a
point here, and a point here.

145
00:07:10,000 --> 00:07:13,810
These are two rectangles
defined on three points.

146
00:07:13,810 --> 00:07:16,990
So they both use this point.

147
00:07:16,990 --> 00:07:19,930
And if you check, it does
satisfy this condition.

148
00:07:19,930 --> 00:07:23,260
So no corner strictly
inside the other.

149
00:07:23,260 --> 00:07:25,660
But we also need that the
rectangles are unsatisfied.

150
00:07:25,660 --> 00:07:28,660
So this is saying that
there's no other point even

151
00:07:28,660 --> 00:07:30,410
on the boundary
of the rectangle.

152
00:07:30,410 --> 00:07:34,360
So this part says, OK, there's
nothing strictly inside.

153
00:07:34,360 --> 00:07:36,610
But we also need
that on the boundary

154
00:07:36,610 --> 00:07:38,060
there's no other points.

155
00:07:38,060 --> 00:07:40,060
So this is the only
sort of situation other

156
00:07:40,060 --> 00:07:44,200
than reflections where
you get this working

157
00:07:44,200 --> 00:07:45,840
out as independent.

158
00:07:50,632 --> 00:07:52,090
AUDIENCE: Last case
is independent?

159
00:07:52,090 --> 00:07:53,730
ERIK DEMAINE: Last
case is independent.

160
00:07:56,730 --> 00:07:58,800
All right?

161
00:07:58,800 --> 00:08:00,470
So this is a definition.

162
00:08:00,470 --> 00:08:02,470
If I give you a
set of rectangles,

163
00:08:02,470 --> 00:08:05,070
they're independent.

164
00:08:05,070 --> 00:08:06,930
I mean, I was
looking at pairwise.

165
00:08:06,930 --> 00:08:08,640
But if they're are
pairwise independent,

166
00:08:08,640 --> 00:08:10,710
then they will be independent.

167
00:08:10,710 --> 00:08:12,270
No corner of any
rectangle strictly

168
00:08:12,270 --> 00:08:14,590
inside any other rectangle.

169
00:08:14,590 --> 00:08:17,940
And there's no points
of those rectangles

170
00:08:17,940 --> 00:08:21,360
that are inside others.

171
00:08:21,360 --> 00:08:22,166
OK.

172
00:08:22,166 --> 00:08:22,665
Cool.

173
00:08:25,390 --> 00:08:27,790
So what?

174
00:08:32,350 --> 00:08:36,640
Lower bound says the optimal
offline binary search

175
00:08:36,640 --> 00:08:41,799
tree, or the optimal way to add
dots to satisfy your point set,

176
00:08:41,799 --> 00:08:45,490
is going to be at least
the size of the input--

177
00:08:45,490 --> 00:08:48,340
meaning the number of
initial points you have--

178
00:08:48,340 --> 00:08:53,170
plus half the maximum number
of independent rectangles.

179
00:09:02,591 --> 00:09:03,090
OK.

180
00:09:03,090 --> 00:09:05,460
So this is a max
independence set problem.

181
00:09:05,460 --> 00:09:07,350
In general, that's NP-complete.

182
00:09:07,350 --> 00:09:10,260
Turns out we'll be able to at
least approximate the number

183
00:09:10,260 --> 00:09:12,990
of independent rectangles
within a constant factor

184
00:09:12,990 --> 00:09:14,560
by the end of class.

185
00:09:14,560 --> 00:09:16,220
That's going to
be signed greedy.

186
00:09:16,220 --> 00:09:18,180
So signed greedy is
going to be the best

187
00:09:18,180 --> 00:09:21,270
way up to constant factors to
choose independent rectangles.

188
00:09:21,270 --> 00:09:24,780
For now, someone
magically tells you

189
00:09:24,780 --> 00:09:28,770
what's the best way or you
just choose some reasonable--

190
00:09:28,770 --> 00:09:30,534
any choice of
independent rectangles

191
00:09:30,534 --> 00:09:31,450
will be a lower bound.

192
00:09:31,450 --> 00:09:34,600
But you get the best lower
bound by choosing the max.

193
00:09:34,600 --> 00:09:35,100
OK?

194
00:09:35,100 --> 00:09:38,400
So we're going to
prove this theorem,

195
00:09:38,400 --> 00:09:43,432
and then we're going to see
three different ways to choose

196
00:09:43,432 --> 00:09:44,640
those independent rectangles.

197
00:09:44,640 --> 00:09:46,770
And we'll use them
for various things.

198
00:09:46,770 --> 00:09:48,780
Wilber 1, Wilber,
2 and signed greedy

199
00:09:48,780 --> 00:09:51,090
are going to be
the three choices

200
00:09:51,090 --> 00:09:53,530
for independent rectangles.

201
00:09:53,530 --> 00:09:54,030
All right.

202
00:09:54,030 --> 00:09:57,660
To prove this theorem, we're
going to change it a little bit

203
00:09:57,660 --> 00:09:59,850
first.

204
00:09:59,850 --> 00:10:02,700
And this is kind of
the focus of today--

205
00:10:02,700 --> 00:10:04,890
is the idea of
signed rectangles.

206
00:10:09,090 --> 00:10:11,580
If you look at the rectangles
in the world spanned

207
00:10:11,580 --> 00:10:14,800
by two points, there
are two different kinds.

208
00:10:14,800 --> 00:10:20,460
There's the top right,
lower left kind.

209
00:10:20,460 --> 00:10:23,910
And then there's the top
left, lower right kind.

210
00:10:23,910 --> 00:10:26,610
These are positive
slope or negative slope.

211
00:10:29,340 --> 00:10:31,590
Those are the two
kinds of rectangles.

212
00:10:31,590 --> 00:10:36,630
And it's helpful to think about
just the positive rectangles

213
00:10:36,630 --> 00:10:39,690
or the slash rectangles and
just the backslash rectangles

214
00:10:39,690 --> 00:10:41,380
separately.

215
00:10:41,380 --> 00:10:43,920
So we're going to
call a point set--

216
00:10:48,090 --> 00:10:50,400
it's a little hard
to pronounce--

217
00:10:50,400 --> 00:10:53,820
we used to call
this plus satisfied.

218
00:10:53,820 --> 00:10:55,620
So maybe it's easiest
to pronounce it

219
00:10:55,620 --> 00:11:00,990
that way, the symbol formerly
known as plus satisfied,

220
00:11:00,990 --> 00:11:11,370
if all plus rectangles
that are not

221
00:11:11,370 --> 00:11:15,000
on a horizontal or vertical
line contain another point.

222
00:11:23,360 --> 00:11:27,390
So a point set is
arborally satisfied

223
00:11:27,390 --> 00:11:31,310
if and only if it is plus
satisfied and minus satisfied--

224
00:11:31,310 --> 00:11:33,787
just breaking apart that
definition into two parts.

225
00:11:33,787 --> 00:11:36,120
But now, we're going to look
at point sets that are just

226
00:11:36,120 --> 00:11:39,300
plus satisfied or point sets
that are just minus satisfied.

227
00:11:39,300 --> 00:11:42,630
And then we can look
at the optimal solution

228
00:11:42,630 --> 00:11:46,960
if you only care
about plus rectangles.

229
00:11:46,960 --> 00:12:03,840
So this is the smallest
plus satisfied point set

230
00:12:03,840 --> 00:12:08,776
containing all the access
points, all the given points.

231
00:12:08,776 --> 00:12:12,470
So we'll call that the input.

232
00:12:12,470 --> 00:12:15,090
OPT was the smallest
arborally satisfied.

233
00:12:15,090 --> 00:12:17,340
OPT plus is the--

234
00:12:17,340 --> 00:12:19,483
you just look at
plus rectangles.

235
00:12:24,351 --> 00:12:24,850
OK.

236
00:12:24,850 --> 00:12:26,570
Why are we doing this?

237
00:12:26,570 --> 00:12:29,030
Well, for now, we're going to
do it to prove this theorem.

238
00:12:29,030 --> 00:12:35,830
So lemma, which was what
we're actually going to prove,

239
00:12:35,830 --> 00:12:38,650
is if you look at
this OPT plus thing,

240
00:12:38,650 --> 00:12:41,010
it's got to be at least
the size of the input--

241
00:12:41,010 --> 00:12:44,620
everything has to at
least contain the input--

242
00:12:44,620 --> 00:12:54,460
plus maximum number of
independent plus rectangles.

243
00:13:02,260 --> 00:13:04,680
So this is where we're
actually going to prove.

244
00:13:04,680 --> 00:13:06,180
If you want to
get plus satisfied

245
00:13:06,180 --> 00:13:09,240
and you've got k
independent plus rectangles,

246
00:13:09,240 --> 00:13:11,050
you need to add at
least that many points--

247
00:13:11,050 --> 00:13:14,850
so at least one point
per plus rectangle.

248
00:13:14,850 --> 00:13:16,830
If you can prove this,
you prove the theorem

249
00:13:16,830 --> 00:13:18,900
because this holds
for minus just

250
00:13:18,900 --> 00:13:21,120
as well as plus by symmetry.

251
00:13:21,120 --> 00:13:26,400
And so you take your maximum
independent set of rectangles.

252
00:13:26,400 --> 00:13:28,830
At least half of them are
plus or at least half of them

253
00:13:28,830 --> 00:13:30,000
are minus.

254
00:13:30,000 --> 00:13:34,290
You apply this bound, and that's
where you get the 1/2 here.

255
00:13:34,290 --> 00:13:38,182
So this is stronger, I
guess, than the theorem,

256
00:13:38,182 --> 00:13:40,140
and this is what we're
actually going to prove.

257
00:13:40,140 --> 00:13:41,850
And so, in this
world, we just are

258
00:13:41,850 --> 00:13:44,790
thinking about plus rectangles,
which is a little weird.

259
00:13:44,790 --> 00:13:47,340
But it works.

260
00:13:51,120 --> 00:13:54,366
And the proof is going
to be in three steps.

261
00:13:54,366 --> 00:13:56,760
I'm first going to give you
an overview of the steps,

262
00:13:56,760 --> 00:13:58,470
and then we'll actually do them.

263
00:13:58,470 --> 00:14:03,030
So this is like a
two-level proof.

264
00:14:03,030 --> 00:14:07,470
First thing we
do, the top level,

265
00:14:07,470 --> 00:14:13,590
is we're going to
find a rectangle

266
00:14:13,590 --> 00:14:20,370
in the independent
set, and we're

267
00:14:20,370 --> 00:14:27,540
going to find a vertical line
that hits only that rectangle.

268
00:14:33,730 --> 00:14:37,160
So we're going to
have some rectangle

269
00:14:37,160 --> 00:14:43,380
in the independent set, and we
want a vertical line stabbing

270
00:14:43,380 --> 00:14:46,470
it such that no
other rectangle is

271
00:14:46,470 --> 00:14:48,490
stabbed by this vertical line.

272
00:14:48,490 --> 00:14:50,910
So all other rectangles--
that's independence,

273
00:14:50,910 --> 00:14:54,260
so maybe they look
something like this--

274
00:14:54,260 --> 00:14:55,690
but nothing like this.

275
00:14:59,297 --> 00:15:01,380
Not obvious that such a
thing exists, but it does.

276
00:15:01,380 --> 00:15:03,240
Actually, not that hard to find.

277
00:15:03,240 --> 00:15:05,510
We just need some
rectangle with some line.

278
00:15:09,540 --> 00:15:16,230
Then, using that
property, we're going

279
00:15:16,230 --> 00:15:31,200
to be able to find some points
in that rectangle that are also

280
00:15:31,200 --> 00:15:43,530
in the optimal plus
solution in the rectangle

281
00:15:43,530 --> 00:15:44,490
crossing the line.

282
00:15:49,950 --> 00:15:52,770
Let me get another color.

283
00:15:52,770 --> 00:15:56,490
So we're going to find
a point on the left

284
00:15:56,490 --> 00:15:59,964
of the line and a point
on the right of the line.

285
00:15:59,964 --> 00:16:01,380
And they're
horizontally adjacent,

286
00:16:01,380 --> 00:16:03,790
meaning there's no other
point between them.

287
00:16:03,790 --> 00:16:07,164
So we know there's
some point in this box.

288
00:16:07,164 --> 00:16:08,580
Because this is a
plus box, it has

289
00:16:08,580 --> 00:16:09,750
got to be satisfied somehow.

290
00:16:09,750 --> 00:16:11,375
And I claim there's
actually two points

291
00:16:11,375 --> 00:16:12,600
on either side of the line.

292
00:16:12,600 --> 00:16:14,992
One of them could be equal
to this or this, but not both

293
00:16:14,992 --> 00:16:16,950
obviously because they're
horizontally aligned.

294
00:16:21,270 --> 00:16:29,230
And then what we're going to
do is charge the rectangle

295
00:16:29,230 --> 00:16:29,970
to those points.

296
00:16:34,346 --> 00:16:36,720
And then, basically, we're
going to remove that rectangle

297
00:16:36,720 --> 00:16:38,340
and repeat.

298
00:16:38,340 --> 00:16:40,830
And the claim is this charging
sort of only happens once

299
00:16:40,830 --> 00:16:41,910
per point.

300
00:16:41,910 --> 00:16:44,700
And therefore, the number of
points in the optimal solution

301
00:16:44,700 --> 00:16:48,150
has to be at least the
number of rectangles-- number

302
00:16:48,150 --> 00:16:50,940
of plus rectangles in
the independent set.

303
00:16:50,940 --> 00:16:53,740
So, basically, this is a way
of ordering the rectangles.

304
00:16:53,740 --> 00:16:56,640
We're going to take one that
has one of these vertical lines,

305
00:16:56,640 --> 00:17:00,630
find two points that
pay for that rectangle,

306
00:17:00,630 --> 00:17:03,360
and therefore argue that OPT
has to be at least the number

307
00:17:03,360 --> 00:17:04,394
of rectangles.

308
00:17:04,394 --> 00:17:08,250
So we have to argue that at
least one of these points

309
00:17:08,250 --> 00:17:11,470
is not one of the
original points.

310
00:17:11,470 --> 00:17:15,644
And that's where we're
getting the input plus this.

311
00:17:15,644 --> 00:17:17,310
So there's lots of
things to check here.

312
00:17:17,310 --> 00:17:18,476
Let's do them one at a time.

313
00:17:21,560 --> 00:17:23,930
And throughout, I'm
going to assume--

314
00:17:23,930 --> 00:17:27,150
let me write that the bottom--

315
00:17:27,150 --> 00:17:31,700
assume all x- and
y-coordinates are unique.

316
00:17:39,860 --> 00:17:43,490
This is an idea I mentioned
last time as well.

317
00:17:43,490 --> 00:17:46,890
If you have lots of
accesses to the same key,

318
00:17:46,890 --> 00:17:50,870
imagine them being accesses
to slightly different keys.

319
00:17:50,870 --> 00:17:53,330
Just skew them a little
bit, and it doesn't

320
00:17:53,330 --> 00:17:56,420
change any of the bounds much.

321
00:17:56,420 --> 00:18:00,290
I won't to argue that here,
but at the least think of this

322
00:18:00,290 --> 00:18:03,530
as just a simplifying assumption
to make the proofs cleaner.

323
00:18:07,010 --> 00:18:09,230
So how are we going
to do step one?

324
00:18:09,230 --> 00:18:12,710
I need to find some rectangle
and some vertical line that

325
00:18:12,710 --> 00:18:16,580
only stabs that rectangle.

326
00:18:16,580 --> 00:18:18,560
And the way we're going
to do that is just

327
00:18:18,560 --> 00:18:30,190
take the widest rectangle that
just has the maximum x extent.

328
00:18:30,190 --> 00:18:34,810
There might be more than one,
but just take one of them.

329
00:18:34,810 --> 00:18:38,534
So it's very wide.

330
00:18:38,534 --> 00:18:39,950
What this tells
us is that there's

331
00:18:39,950 --> 00:18:42,710
no other rectangle like this.

332
00:18:42,710 --> 00:18:45,110
This would be independent,
but it would be wider.

333
00:18:45,110 --> 00:18:46,040
So that's not allowed.

334
00:18:51,365 --> 00:18:53,490
Now, we have to think about
all sorts of scenarios.

335
00:18:53,490 --> 00:18:55,990
So we've got a point
here and a point here.

336
00:18:55,990 --> 00:18:58,960
It could still be that we
have rectangles like this.

337
00:18:58,960 --> 00:19:00,970
They just can't go
farther to the right.

338
00:19:00,970 --> 00:19:03,670
It could be we have
rectangles that go like this--

339
00:19:03,670 --> 00:19:07,354
just can't go too
far to the left.

340
00:19:07,354 --> 00:19:08,770
These rectangles
that are anchored

341
00:19:08,770 --> 00:19:10,330
in the lower left and
these rectangles that

342
00:19:10,330 --> 00:19:12,610
are anchored in the upper
right can't touch each other

343
00:19:12,610 --> 00:19:16,369
because then one of
them would be satisfied.

344
00:19:16,369 --> 00:19:18,160
This one's going to
have a point down here.

345
00:19:18,160 --> 00:19:20,230
This one is going to
have a point here.

346
00:19:20,230 --> 00:19:23,200
I guess-- yeah, let's
see, how would it

347
00:19:23,200 --> 00:19:24,560
go if they were touching?

348
00:19:31,110 --> 00:19:33,465
We'd have a corner--

349
00:19:33,465 --> 00:19:34,910
hmm, touching is a little weird.

350
00:19:51,000 --> 00:19:52,000
Ah, I see.

351
00:19:52,000 --> 00:19:52,500
Good.

352
00:19:52,500 --> 00:19:53,958
This can't happen
because we assume

353
00:19:53,958 --> 00:19:56,790
the x-coordinates are distinct.

354
00:19:56,790 --> 00:19:59,360
So that's why I did this.

355
00:19:59,360 --> 00:20:01,860
That's the reason.

356
00:20:01,860 --> 00:20:03,580
So this can't happen.

357
00:20:03,580 --> 00:20:07,410
And I also can't have them go
like this because then there's

358
00:20:07,410 --> 00:20:12,120
a corner in the strict interior
of the other rectangle.

359
00:20:12,120 --> 00:20:12,912
Is that clear?

360
00:20:12,912 --> 00:20:14,370
This rectangle
can't come over here

361
00:20:14,370 --> 00:20:18,090
because then that would
be not independent.

362
00:20:18,090 --> 00:20:20,160
Rectangle can't come
right to the same spot

363
00:20:20,160 --> 00:20:21,460
because there is no same spot.

364
00:20:21,460 --> 00:20:24,190
That would be two points
on the same vertical line.

365
00:20:24,190 --> 00:20:26,460
And so what we must
have is a picture more

366
00:20:26,460 --> 00:20:30,810
like this where there's an
empty region in between.

367
00:20:30,810 --> 00:20:34,680
that not hit by-- there can be
many of these rectangles, many

368
00:20:34,680 --> 00:20:35,911
of these rectangles.

369
00:20:35,911 --> 00:20:37,410
They're independent
from each other.

370
00:20:37,410 --> 00:20:40,620
That's like this case here.

371
00:20:40,620 --> 00:20:43,560
There can also be some
rectangles like this.

372
00:20:43,560 --> 00:20:47,100
But by the same argument, these
guys can't touch each other

373
00:20:47,100 --> 00:20:49,380
and they can't
overlap horizontally

374
00:20:49,380 --> 00:20:52,770
because then one of the corners
would be inside the other.

375
00:20:52,770 --> 00:20:54,080
Question?

376
00:20:54,080 --> 00:20:56,455
AUDIENCE: For that picture,
you drew a rectangle

377
00:20:56,455 --> 00:20:59,282
under the other one.

378
00:20:59,282 --> 00:21:00,740
ERIK DEMAINE: This
one or this one?

379
00:21:00,740 --> 00:21:01,531
AUDIENCE: That one.

380
00:21:01,531 --> 00:21:03,480
ERIK DEMAINE: Yeah,
this one cannot happen.

381
00:21:03,480 --> 00:21:04,575
That's what we claim--

382
00:21:04,575 --> 00:21:05,250
haha, right.

383
00:21:05,250 --> 00:21:07,140
So you're right.

384
00:21:07,140 --> 00:21:10,590
So we worry about--

385
00:21:10,590 --> 00:21:11,612
interesting.

386
00:21:11,612 --> 00:21:13,320
Well, we worry about
something like this.

387
00:21:16,194 --> 00:21:19,425
AUDIENCE: Sorry, why
can't that happen?

388
00:21:19,425 --> 00:21:20,800
ERIK DEMAINE:
Yeah, you're right.

389
00:21:20,800 --> 00:21:22,216
I actually drew
the wrong picture.

390
00:21:22,216 --> 00:21:23,215
Sorry.

391
00:21:23,215 --> 00:21:23,715
Kidding.

392
00:21:31,080 --> 00:21:31,580
Yeah.

393
00:21:31,580 --> 00:21:33,250
I really meant
line segment here.

394
00:21:33,250 --> 00:21:34,305
I'm sorry.

395
00:21:39,640 --> 00:21:42,020
Poor choice of wording.

396
00:21:42,020 --> 00:21:43,720
So vertical line is
actually just going

397
00:21:43,720 --> 00:21:47,770
to go the extent
of the rectangle--

398
00:21:47,770 --> 00:21:49,510
something like this.

399
00:21:49,510 --> 00:21:50,230
Sorry.

400
00:21:50,230 --> 00:21:53,910
We can't forbid
rectangles like this.

401
00:21:53,910 --> 00:21:57,260
What we can forbid our
rectangles like this

402
00:21:57,260 --> 00:22:01,632
that also try to
cross that segment.

403
00:22:01,632 --> 00:22:03,340
We'll see why this is
enough in a moment.

404
00:22:03,340 --> 00:22:04,060
Sorry about that.

405
00:22:08,350 --> 00:22:12,070
I really only care about the
interior of this rectangle.

406
00:22:12,070 --> 00:22:15,370
I'm trying to get a vertical
line that only stabs

407
00:22:15,370 --> 00:22:17,530
this rectangle, nothing else--

408
00:22:17,530 --> 00:22:19,600
inside the rectangle.

409
00:22:19,600 --> 00:22:20,740
Sorry, poor wording.

410
00:22:20,740 --> 00:22:22,674
I don't care about
these guys outside

411
00:22:22,674 --> 00:22:24,340
because I can't say
anything about them.

412
00:22:24,340 --> 00:22:29,275
They could be all over the
place in an independent set.

413
00:22:29,275 --> 00:22:31,150
I mean, relative to what
hits this rectangle,

414
00:22:31,150 --> 00:22:32,050
there's stuff on the left.

415
00:22:32,050 --> 00:22:33,050
There's stuff on the right.

416
00:22:33,050 --> 00:22:33,924
There are these guys.

417
00:22:33,924 --> 00:22:37,340
There can also be
things like this.

418
00:22:37,340 --> 00:22:40,600
But still remaining
are these regions

419
00:22:40,600 --> 00:22:42,959
which are not hit
by any rectangles,

420
00:22:42,959 --> 00:22:44,500
and that's because
what I was saying.

421
00:22:44,500 --> 00:22:46,050
These guys can't touch each
other because then there

422
00:22:46,050 --> 00:22:47,680
would be equal x-coordinates.

423
00:22:47,680 --> 00:22:49,660
They can't overlap
because then one of them

424
00:22:49,660 --> 00:22:52,180
would not be independent
from the other.

425
00:22:52,180 --> 00:22:58,540
So I get my vertical lines.

426
00:22:58,540 --> 00:23:01,510
I just need one, but it
could be any of these.

427
00:23:01,510 --> 00:23:06,400
In general, for example, if
you take all of these lower

428
00:23:06,400 --> 00:23:08,470
left anchored
rectangles and take just

429
00:23:08,470 --> 00:23:10,540
to the right of
the rightmost one,

430
00:23:10,540 --> 00:23:13,210
that will be a valid
choice for your line.

431
00:23:13,210 --> 00:23:17,060
Because you can argue none
of these can overlap it.

432
00:23:17,060 --> 00:23:18,550
So that's step one.

433
00:23:18,550 --> 00:23:20,686
We just take a widest rectangle.

434
00:23:20,686 --> 00:23:22,060
The one thing we
needed to forbid

435
00:23:22,060 --> 00:23:25,060
was something going like
this all the way across.

436
00:23:28,120 --> 00:23:28,870
Step two.

437
00:23:31,822 --> 00:23:33,560
Step two is actually
pretty easy.

438
00:23:33,560 --> 00:23:37,490
Once you've identified
this red line--

439
00:23:37,490 --> 00:23:40,370
inside the rectangle, you
know there are some points.

440
00:23:40,370 --> 00:23:43,714
And I'm going to
take the rightmost.

441
00:23:43,714 --> 00:23:45,380
And then among all
the rightmost points,

442
00:23:45,380 --> 00:23:47,560
I'm going to take the
topmost point that

443
00:23:47,560 --> 00:23:50,420
is to the left of the line
and inside the rectangle.

444
00:23:50,420 --> 00:23:59,340
So let p be the topmost,
leftmost point--

445
00:23:59,340 --> 00:24:14,020
sorry, rightmost-- that
is both in the rectangle

446
00:24:14,020 --> 00:24:16,210
and left of the line.

447
00:24:22,240 --> 00:24:28,420
Let me erase this one for
a little bit more room.

448
00:24:28,420 --> 00:24:31,150
So I'm looking at
all of this region

449
00:24:31,150 --> 00:24:33,370
to the left of the
line in the rectangle.

450
00:24:33,370 --> 00:24:39,130
I want to take the
rightmost and then topmost

451
00:24:39,130 --> 00:24:42,100
point-- something like this.

452
00:24:42,100 --> 00:24:43,960
How do I know such
a point exists?

453
00:24:43,960 --> 00:24:46,640
Because this point
is such a point.

454
00:24:46,640 --> 00:24:49,030
And this point is to
the left of the line.

455
00:24:49,030 --> 00:24:52,510
So if there's nothing else in
here, that is a valid choice.

456
00:24:52,510 --> 00:24:54,950
But in general, we go to the
right as much as possible.

457
00:24:54,950 --> 00:24:56,390
Then we go up as
much as possible.

458
00:24:56,390 --> 00:24:59,110
So that's a point, which
we will call them p.

459
00:24:59,110 --> 00:25:03,534
AUDIENCE: Couldn't it be on
the border of the rectangle?

460
00:25:03,534 --> 00:25:05,200
ERIK DEMAINE: It could
be on the border.

461
00:25:05,200 --> 00:25:06,550
It could be interior.

462
00:25:06,550 --> 00:25:08,177
We don't know.

463
00:25:08,177 --> 00:25:11,890
AUDIENCE: When you said
topmost, what is your topmost?

464
00:25:11,890 --> 00:25:15,448
ERIK DEMAINE: Topmost means
of maximum y-coordinate.

465
00:25:15,448 --> 00:25:16,226
AUDIENCE: Oh, OK.

466
00:25:16,226 --> 00:25:16,850
Got it.

467
00:25:16,850 --> 00:25:19,490
ERIK DEMAINE: So it
could be up here.

468
00:25:19,490 --> 00:25:21,380
We don't know.

469
00:25:21,380 --> 00:25:22,760
First, we go rightmost.

470
00:25:22,760 --> 00:25:25,019
Then, among all the
things in that column,

471
00:25:25,019 --> 00:25:26,060
we go to the topmost one.

472
00:25:26,060 --> 00:25:27,140
So it might be on the top.

473
00:25:27,140 --> 00:25:27,806
It might not be.

474
00:25:30,760 --> 00:25:36,265
These are points-- sorry,
this is a point in OPT plus.

475
00:25:40,270 --> 00:25:42,966
And then q is going
to be a similar thing.

476
00:25:42,966 --> 00:25:52,780
It's going to be the
bottom-most, leftmost point

477
00:25:52,780 --> 00:26:03,970
in OPT plus that
is in the rectangle

478
00:26:03,970 --> 00:26:05,230
and right of the line.

479
00:26:08,112 --> 00:26:09,670
Not totally symmetric, though.

480
00:26:09,670 --> 00:26:12,420
We're also going to
say and not below p.

481
00:26:16,570 --> 00:26:25,450
So now we're looking at
this upper region here.

482
00:26:25,450 --> 00:26:28,290
Among all the things
that are not below p--

483
00:26:28,290 --> 00:26:30,280
should have drawn
this more horizontal--

484
00:26:33,430 --> 00:26:35,960
and to the right
of the red line--

485
00:26:35,960 --> 00:26:38,960
so that's up to here, I guess--

486
00:26:38,960 --> 00:26:41,662
I want to take the
leftmost column that

487
00:26:41,662 --> 00:26:43,370
has any points in it
and then among those

488
00:26:43,370 --> 00:26:47,370
take the bottom-most
point in the column.

489
00:26:47,370 --> 00:26:50,539
I claim that's actually
going to be on this line.

490
00:26:50,539 --> 00:26:52,580
First thing to check is
that such a point exists.

491
00:26:52,580 --> 00:26:54,980
Such a point exists
because, in particular, this

492
00:26:54,980 --> 00:26:56,689
is such a point.

493
00:26:56,689 --> 00:26:57,980
It is to the right of the line.

494
00:26:57,980 --> 00:27:00,800
It's above the blue line,
right of the red line,

495
00:27:00,800 --> 00:27:02,660
in the rectangle.

496
00:27:02,660 --> 00:27:05,350
But I claim that if we take the
leftmost, bottom-most one, then

497
00:27:05,350 --> 00:27:06,600
they must actually be aligned.

498
00:27:06,600 --> 00:27:08,010
Why?

499
00:27:08,010 --> 00:27:10,520
So if it was somewhere
else, like up here

500
00:27:10,520 --> 00:27:16,590
or like this point, then I claim
that is an unsatisfied box.

501
00:27:16,590 --> 00:27:24,180
Let me draw that picture,
make a little clearer.

502
00:27:24,180 --> 00:27:27,240
So something like this.

503
00:27:30,760 --> 00:27:39,540
So we've got our red line
and we've got this picture.

504
00:27:43,090 --> 00:27:54,000
This is p, and then actually
we don't know anything

505
00:27:54,000 --> 00:27:55,095
about down here.

506
00:27:59,770 --> 00:28:01,990
This is q.

507
00:28:01,990 --> 00:28:06,230
I claim that these black regions
cannot have any points in them

508
00:28:06,230 --> 00:28:09,010
because, by definition, p
was in the rightmost column.

509
00:28:09,010 --> 00:28:11,677
So there's nothing in this strip
and between p and the red line.

510
00:28:11,677 --> 00:28:14,218
And it was the topmost within
the column, which means there's

511
00:28:14,218 --> 00:28:15,520
nothing above p in the column.

512
00:28:15,520 --> 00:28:17,620
So that's why all the
points are confined

513
00:28:17,620 --> 00:28:19,300
to this region over here.

514
00:28:19,300 --> 00:28:22,360
Similarly, for q, if you look
at the things that are above

515
00:28:22,360 --> 00:28:24,640
or on this horizontal
line, which

516
00:28:24,640 --> 00:28:28,630
was the blue line
over there, then we

517
00:28:28,630 --> 00:28:33,320
know that there's nothing
in this strip in between

518
00:28:33,320 --> 00:28:34,330
because q is leftmost.

519
00:28:34,330 --> 00:28:36,163
And then among leftmost,
it was bottom-most,

520
00:28:36,163 --> 00:28:37,630
so there's nothing down here.

521
00:28:37,630 --> 00:28:40,360
So that means, if these guys
are not horizontally aligned,

522
00:28:40,360 --> 00:28:42,840
there is an
unsatisfied box here.

523
00:28:42,840 --> 00:28:43,940
Contradiction.

524
00:28:43,940 --> 00:28:45,170
It's a plus box.

525
00:28:45,170 --> 00:28:48,227
So in OPT plus, there's got
to be another point, which

526
00:28:48,227 --> 00:28:49,060
was a contradiction.

527
00:28:49,060 --> 00:28:52,360
So in fact, p and q must
be horizontally aligned.

528
00:28:52,360 --> 00:28:54,360
So that was step two.

529
00:28:57,040 --> 00:28:58,300
Finally, we get step three.

530
00:29:03,250 --> 00:29:05,290
So the idea with step three--

531
00:29:05,290 --> 00:29:07,390
now we're going to do
a charging argument.

532
00:29:07,390 --> 00:29:13,060
We want to say, OK, basically,
for every independent

533
00:29:13,060 --> 00:29:15,370
rectangle, we want to
find a point that's

534
00:29:15,370 --> 00:29:19,720
in OPT that was not
in the original input.

535
00:29:19,720 --> 00:29:22,390
And therefore, then
OPT plus has to be

536
00:29:22,390 --> 00:29:25,120
at least the size of the
input plus 1 per each

537
00:29:25,120 --> 00:29:26,350
of these plus rectangles.

538
00:29:29,510 --> 00:29:32,530
So the idea is the following.

539
00:29:32,530 --> 00:29:33,730
Because of all this set-up--

540
00:29:33,730 --> 00:29:39,340
because we made pq
horizontally aligned--

541
00:29:39,340 --> 00:29:40,930
they're inside the rectangle.

542
00:29:40,930 --> 00:29:43,030
And furthermore,
they're adjacent

543
00:29:43,030 --> 00:29:45,040
and they cross
this vertical line.

544
00:29:45,040 --> 00:29:48,880
And that vertical line is not
crossed by any other rectangle.

545
00:29:48,880 --> 00:29:51,780
When I say line, I
mean line segment.

546
00:29:51,780 --> 00:29:54,880
There's no other rectangle
that hits this red thing.

547
00:29:54,880 --> 00:29:56,470
Therefore, these
two points are not

548
00:29:56,470 --> 00:30:00,640
going to get charged
as a pair ever again.

549
00:30:00,640 --> 00:30:05,260
If you remove this rectangle,
repeat this process,

550
00:30:05,260 --> 00:30:09,070
pq is never going to
get charged again.

551
00:30:09,070 --> 00:30:10,990
So we charge to pq.

552
00:30:10,990 --> 00:30:26,230
And the pair never
charged again, never

553
00:30:26,230 --> 00:30:33,860
be charged by another
rectangle because no rectangle

554
00:30:33,860 --> 00:30:39,520
hits the red thing.

555
00:30:46,600 --> 00:30:53,380
So no rectangle
contains the segment pq,

556
00:30:53,380 --> 00:30:54,620
the horizontal segment pq.

557
00:30:57,600 --> 00:30:59,130
So this is almost what we want.

558
00:30:59,130 --> 00:31:01,900
We really want a single point
which is not in the input.

559
00:31:01,900 --> 00:31:03,360
So we have p and q.

560
00:31:03,360 --> 00:31:04,680
They're horizontally aligned.

561
00:31:04,680 --> 00:31:07,080
Now, if they're
horizontally aligned,

562
00:31:07,080 --> 00:31:12,090
we know that not both of them
are in the original input

563
00:31:12,090 --> 00:31:14,564
because all y-coordinates
are distinct.

564
00:31:14,564 --> 00:31:16,230
This is usually true
because you're only

565
00:31:16,230 --> 00:31:19,800
accessing one point
per row, per time step.

566
00:31:19,800 --> 00:31:23,550
So one of these might
be in the input,

567
00:31:23,550 --> 00:31:25,170
but the other one is not.

568
00:31:25,170 --> 00:31:26,970
So that's the one I
want to hold onto.

569
00:31:26,970 --> 00:31:32,040
And say, OK, that's a point
added to OPT plus that pays

570
00:31:32,040 --> 00:31:34,295
for this rectangle.

571
00:31:34,295 --> 00:31:35,670
It's not quite so
simple, though,

572
00:31:35,670 --> 00:31:37,230
because we might
have a whole bunch

573
00:31:37,230 --> 00:31:42,060
of horizontally-aligned things.

574
00:31:42,060 --> 00:31:44,950
And one rectangle
charges to this one.

575
00:31:44,950 --> 00:31:46,980
One rectangle
charges to this one.

576
00:31:46,980 --> 00:31:48,960
One rectangle
charges to this pair.

577
00:31:51,510 --> 00:31:54,660
That's OK, though, because
here we have four points.

578
00:31:54,660 --> 00:31:57,180
Again, one of them
could be in the input.

579
00:31:57,180 --> 00:31:59,160
The other three
have to be added.

580
00:31:59,160 --> 00:32:01,720
And so you've got three
rectangles, three added points,

581
00:32:01,720 --> 00:32:03,190
and we're happy.

582
00:32:03,190 --> 00:32:04,149
Question?

583
00:32:04,149 --> 00:32:05,940
AUDIENCE: Just to make
the argument formal,

584
00:32:05,940 --> 00:32:11,667
wouldn't you want to say that
only when your saying assume

585
00:32:11,667 --> 00:32:13,916
that x and y are always
distinct-- but then,

586
00:32:13,916 --> 00:32:16,050
if you have the
same either x or y--

587
00:32:16,050 --> 00:32:17,490
ERIK DEMAINE: Ah, good point.

588
00:32:17,490 --> 00:32:21,960
So this is distinct in
the input is what I meant.

589
00:32:21,960 --> 00:32:23,832
Obviously, in OPT,
any satisfied set

590
00:32:23,832 --> 00:32:25,290
is not going to
have this property.

591
00:32:25,290 --> 00:32:26,159
Yeah, good.

592
00:32:26,159 --> 00:32:28,200
So I want to assume x-
and y-coordinates are only

593
00:32:28,200 --> 00:32:29,490
distinct in the input.

594
00:32:29,490 --> 00:32:31,020
OPT will not have that property.

595
00:32:31,020 --> 00:32:34,422
And that's why p and q can exist
and have the same y-coordinate.

596
00:32:34,422 --> 00:32:35,130
Another question?

597
00:32:39,397 --> 00:32:40,938
AUDIENCE: Does this
still [INAUDIBLE]

598
00:32:40,938 --> 00:32:43,116
the special case where
your two points are

599
00:32:43,116 --> 00:32:44,932
the points of the rectangle?

600
00:32:44,932 --> 00:32:45,640
ERIK DEMAINE: OK.

601
00:32:45,640 --> 00:32:50,560
So the question is can p and q
be the points of the rectangle?

602
00:32:50,560 --> 00:32:52,180
One of them can be.

603
00:32:52,180 --> 00:32:55,360
Like, p could be here, and then
another point is over here.

604
00:32:55,360 --> 00:32:58,480
So then that will be the
segment that you are using,

605
00:32:58,480 --> 00:33:00,770
between p and q.

606
00:33:00,770 --> 00:33:03,690
Or it could be q is
here, and p is over here.

607
00:33:03,690 --> 00:33:04,875
Then that's the segment.

608
00:33:04,875 --> 00:33:07,000
You can't have them both
equal because p and q have

609
00:33:07,000 --> 00:33:10,320
to be horizontally aligned
and also because there's got

610
00:33:10,320 --> 00:33:13,750
to be another point in there.

611
00:33:13,750 --> 00:33:15,220
Yeah, so that should work.

612
00:33:15,220 --> 00:33:17,710
You have to check that this
boundary case is still OK.

613
00:33:17,710 --> 00:33:23,350
But the claim is no other
rectangle touches this red line

614
00:33:23,350 --> 00:33:24,825
even on the endpoint.

615
00:33:24,825 --> 00:33:26,200
And therefore, no
other rectangle

616
00:33:26,200 --> 00:33:28,760
will wholly contain p and q.

617
00:33:28,760 --> 00:33:32,470
And so that means you're only
charging to this pair once.

618
00:33:32,470 --> 00:33:34,280
And then this pair
charging is OK

619
00:33:34,280 --> 00:33:38,142
because, luckily, there's three
edges here, four vertices.

620
00:33:38,142 --> 00:33:39,850
One of those vertices
we can't charge to,

621
00:33:39,850 --> 00:33:43,840
so there's exactly the right
number of things for the edges,

622
00:33:43,840 --> 00:33:46,210
and we're OK.

623
00:33:46,210 --> 00:33:48,580
Yeah, this can really happen.

624
00:33:48,580 --> 00:33:54,420
In fact our favorite
example of the pinwheel--

625
00:33:54,420 --> 00:33:58,903
if instead of doing the greedy
addition, we do this addition--

626
00:34:03,740 --> 00:34:07,120
these are supposed to
be horizontally aligned.

627
00:34:07,120 --> 00:34:09,770
A little hard without a grid--

628
00:34:09,770 --> 00:34:12,350
a graph blackboard
would a great.

629
00:34:12,350 --> 00:34:14,870
So this is not quite satisfied.

630
00:34:14,870 --> 00:34:19,969
You've got to add some more
points here or something.

631
00:34:19,969 --> 00:34:22,380
But it has the feature that--

632
00:34:22,380 --> 00:34:24,560
here's an independent
set of rectangles.

633
00:34:24,560 --> 00:34:33,199
I can do this one,
this one, and this one.

634
00:34:36,030 --> 00:34:38,330
So this is three
independent rectangles.

635
00:34:38,330 --> 00:34:42,852
As the white points go,
they're independent rectangles.

636
00:34:42,852 --> 00:34:44,810
The corners are not
strictly inside each other,

637
00:34:44,810 --> 00:34:46,393
and none of the white
points satisfies

638
00:34:46,393 --> 00:34:49,590
any of the other rectangles.

639
00:34:49,590 --> 00:34:53,132
And indeed, if you
applied this argument,

640
00:34:53,132 --> 00:34:54,590
first you take the
widest rectangle

641
00:34:54,590 --> 00:34:57,570
and say, OK, here is my
vertical red segment.

642
00:34:57,570 --> 00:35:00,597
I'm going to charge to these
two guys, this segment,

643
00:35:00,597 --> 00:35:02,930
and then eventually this guy
will charge to this segment

644
00:35:02,930 --> 00:35:05,480
and this guy will
charge to this segment.

645
00:35:05,480 --> 00:35:07,460
And luckily, there
are three added points

646
00:35:07,460 --> 00:35:11,140
for exactly the three segments
for the three rectangles.

647
00:35:11,140 --> 00:35:12,390
There had to be another point.

648
00:35:15,080 --> 00:35:16,140
So that's a lower bound.

649
00:35:21,453 --> 00:35:22,930
A lot of work--

650
00:35:22,930 --> 00:35:25,060
but in the end, it
says, look, just

651
00:35:25,060 --> 00:35:28,930
find an independent set of
plus boxes, plus rectangles.

652
00:35:28,930 --> 00:35:30,350
That's a lower bound on OPT.

653
00:35:30,350 --> 00:35:32,620
So now, the question
remains, how

654
00:35:32,620 --> 00:35:36,340
do we find a good independent
set of plus boxes?

655
00:35:36,340 --> 00:35:38,500
And now we'll go through
the three different ways

656
00:35:38,500 --> 00:35:40,230
we know how to do it.

657
00:35:40,230 --> 00:35:42,580
I'll start actually
with Wilber 2.

658
00:35:42,580 --> 00:35:45,220
It's called Wilber 2 because
it was in a paper by Wilber,

659
00:35:45,220 --> 00:35:49,090
and I think he called it lower
bound number 1 and lower bound

660
00:35:49,090 --> 00:35:50,410
number 2.

661
00:35:50,410 --> 00:35:53,620
But for pragmatic reasons, I'm
going to start with number 2.

662
00:36:00,670 --> 00:36:03,460
It's from 1989, so it's
actually an old paper.

663
00:36:03,460 --> 00:36:06,410
And it was sort of
lost for a long time.

664
00:36:06,410 --> 00:36:09,540
I don't think Wilber
wrote any other papers.

665
00:36:09,540 --> 00:36:11,755
It was in SICOMP, a big journal.

666
00:36:14,372 --> 00:36:18,400
so a few years after
splay trees and then

667
00:36:18,400 --> 00:36:21,610
sort of rediscovered
in the early 2000s

668
00:36:21,610 --> 00:36:25,060
and turns out to be really
useful for a lot of theorems.

669
00:36:25,060 --> 00:36:27,970
So here's the lower bound.

670
00:36:27,970 --> 00:36:29,590
Again, we're
looking at the input

671
00:36:29,590 --> 00:36:33,970
point set-- no added points,
just the original points.

672
00:36:33,970 --> 00:36:37,450
Look at every point, and
look at all the points

673
00:36:37,450 --> 00:36:40,970
that you can see from
this point downward.

674
00:36:40,970 --> 00:36:41,830
What does see mean?

675
00:36:41,830 --> 00:36:46,210
I'm interested in
points below p that when

676
00:36:46,210 --> 00:36:49,910
I draw the rectangle
contain no other points.

677
00:36:49,910 --> 00:36:53,740
So this is sort of
like a lower envelope.

678
00:36:53,740 --> 00:36:56,770
It's going to look
something like this--

679
00:36:56,770 --> 00:37:02,180
and maybe some points like this.

680
00:37:02,180 --> 00:37:05,950
So all of these rectangles
have to be empty.

681
00:37:23,540 --> 00:37:37,670
So these are the downward
visible points from p.

682
00:37:41,330 --> 00:37:45,680
And now, among these points, you
can sort them by y-coordinate.

683
00:37:45,680 --> 00:37:48,590
And I want to see
how many times do

684
00:37:48,590 --> 00:37:51,630
they cross this vertical line.

685
00:37:51,630 --> 00:37:54,560
So if I order them
by y-coordinate--

686
00:37:57,590 --> 00:37:59,780
so I start here, and
maybe I go to here.

687
00:37:59,780 --> 00:38:03,470
Then the next one is over
here, so that's across.

688
00:38:03,470 --> 00:38:04,850
Then I go over here.

689
00:38:04,850 --> 00:38:05,520
Then I cross.

690
00:38:05,520 --> 00:38:06,710
Then I go here.

691
00:38:06,710 --> 00:38:08,250
Then I cross.

692
00:38:08,250 --> 00:38:10,670
Go here, here, cross.

693
00:38:10,670 --> 00:38:12,350
So if I visit them
in order, I want

694
00:38:12,350 --> 00:38:15,170
to know how many times do
I cross this vertical line.

695
00:38:19,320 --> 00:38:23,330
So this is the past of p,
all of the accesses before p.

696
00:38:23,330 --> 00:38:25,880
Think of this is how
many times you alternate

697
00:38:25,880 --> 00:38:27,985
between accessing on
the left of the line

698
00:38:27,985 --> 00:38:29,610
and accessing on the
right of the line.

699
00:38:33,594 --> 00:38:44,270
So count number of alternations
left or right of p.

700
00:38:44,270 --> 00:38:46,412
And again, if we assume
that no key is ever

701
00:38:46,412 --> 00:38:47,870
accessed more than
once, then there

702
00:38:47,870 --> 00:38:49,910
will always be left or
right, never exactly on.

703
00:38:54,931 --> 00:38:56,555
And then I want to
sum over all points.

704
00:38:59,820 --> 00:39:03,950
And I claim this
is a lower bound.

705
00:39:03,950 --> 00:39:07,550
Why is it a lower bound?

706
00:39:07,550 --> 00:39:13,580
Essentially, I take each
of these red lines that

707
00:39:13,580 --> 00:39:20,005
cross the p vertical line
and I turn them into a box.

708
00:39:20,005 --> 00:39:28,360
So there's one there, one
there, one there, and one there.

709
00:39:28,360 --> 00:39:33,597
I claim if I do this for
all p, all those boxes

710
00:39:33,597 --> 00:39:34,430
will be independent.

711
00:39:34,430 --> 00:39:36,805
All those rectangles will be
independent from each other.

712
00:39:36,805 --> 00:39:41,810
I won't prove that formally
here, but you can check it.

713
00:39:41,810 --> 00:39:44,240
So it's obvious for one
p because each of these

714
00:39:44,240 --> 00:39:46,250
has a different vertical span.

715
00:39:46,250 --> 00:39:49,270
If you do it for all
p-- all the points p--

716
00:39:49,270 --> 00:39:51,980
these won't conflict.

717
00:39:51,980 --> 00:39:55,560
So by the independent
rectangle lower bound,

718
00:39:55,560 --> 00:40:02,490
this is a lower bound on
OPT up to a factor of 2.

719
00:40:02,490 --> 00:40:04,950
So what?

720
00:40:04,950 --> 00:40:07,080
Wilber 2 is quite interesting.

721
00:40:07,080 --> 00:40:08,520
For a long time,
we've conjectured

722
00:40:08,520 --> 00:40:11,060
that it is the right answer.

723
00:40:11,060 --> 00:40:16,620
So conjecture-- I know
it's a weird lower

724
00:40:16,620 --> 00:40:18,290
bound to even think of.

725
00:40:18,290 --> 00:40:20,850
It's a very hard paper to read.

726
00:40:20,850 --> 00:40:22,950
Without the geometric
view, it's even harder

727
00:40:22,950 --> 00:40:26,800
to imagine the
definition of this bound.

728
00:40:26,800 --> 00:40:28,400
It's sort of an algorithm.

729
00:40:28,400 --> 00:40:29,955
It's a way to assign boxes.

730
00:40:29,955 --> 00:40:31,080
It gives you a lower bound.

731
00:40:31,080 --> 00:40:32,810
It's a little weird.

732
00:40:32,810 --> 00:40:34,560
We conjecture that
it's proportional

733
00:40:34,560 --> 00:40:37,540
to the optimal solution.

734
00:40:37,540 --> 00:40:38,550
We can't prove it.

735
00:40:38,550 --> 00:40:40,440
We've tried many times.

736
00:40:40,440 --> 00:40:47,630
It's a pain to work with,
but it is what it is.

737
00:40:47,630 --> 00:40:49,340
There's one theorem
that uses it,

738
00:40:49,340 --> 00:40:51,150
so I want to tell you
about that theorem.

739
00:40:51,150 --> 00:40:54,200
But I don't want to go
into it in too much detail.

740
00:40:54,200 --> 00:40:56,660
It's a neat theorem.

741
00:40:56,660 --> 00:41:02,900
And it's in a paper
by Iacono, 2002.

742
00:41:02,900 --> 00:41:05,900
And it was the first paper to
revitalize the Wilber stuff.

743
00:41:05,900 --> 00:41:08,390
So it's like, hey, there's
this Wilber 2 bound.

744
00:41:08,390 --> 00:41:10,820
We can use it to solve a
new problem, which is called

745
00:41:10,820 --> 00:41:12,905
key independent optimality.

746
00:41:23,960 --> 00:41:27,100
Briefly, the idea with
key independent optimality

747
00:41:27,100 --> 00:41:29,640
is, suppose you've heard
about dynamic optimality.

748
00:41:29,640 --> 00:41:32,510
You know, it's really cool
because splay trees and whatnot

749
00:41:32,510 --> 00:41:34,940
seem to really adapt to
whatever your inputs are.

750
00:41:34,940 --> 00:41:37,250
But suppose your inputs
really don't have keys.

751
00:41:37,250 --> 00:41:42,140
They're just arbitrary objects
labeled however, just randomly.

752
00:41:42,140 --> 00:41:44,570
In fact, let's assume that
they're labeled randomly.

753
00:41:44,570 --> 00:41:46,340
Suppose the keys are
generated randomly

754
00:41:46,340 --> 00:41:48,797
because they're meaningless
or just arbitrary things.

755
00:41:48,797 --> 00:41:50,630
So you figure, oh, maybe
I'll make it better

756
00:41:50,630 --> 00:41:54,560
and just randomize
them completely.

757
00:41:54,560 --> 00:42:06,380
If keys are random,
then dynamic OPT

758
00:42:06,380 --> 00:42:09,481
is the same thing up to constant
factors as the working set

759
00:42:09,481 --> 00:42:09,980
bound.

760
00:42:15,830 --> 00:42:17,390
That's the theorem.

761
00:42:17,390 --> 00:42:20,120
So this is cool because it
means splay trees are actually

762
00:42:20,120 --> 00:42:22,070
optimal in the setting
where keys are random.

763
00:42:25,580 --> 00:42:30,200
This is in expectation
over the randomized keys.

764
00:42:30,200 --> 00:42:34,070
And the way this theorem
is proved is basically--

765
00:42:34,070 --> 00:42:36,890
so what this is saying is,
if we take a point set--

766
00:42:36,890 --> 00:42:39,290
arbitrary point set--
but then we re-randomize

767
00:42:39,290 --> 00:42:43,350
the x-coordinates-- leave the
y-coordinates as they are--

768
00:42:43,350 --> 00:42:48,106
then you can compute
how Wilber 2 behaves.

769
00:42:48,106 --> 00:42:49,730
Because now you have
a bunch of points,

770
00:42:49,730 --> 00:42:53,940
and you're randomly
shifting their x-coordinate.

771
00:42:53,940 --> 00:42:58,070
So it's like if you're
randomly bouncing around an x

772
00:42:58,070 --> 00:43:00,380
and you're interested
in this envelope

773
00:43:00,380 --> 00:43:02,720
on the left and the right,
you want to know basically

774
00:43:02,720 --> 00:43:03,605
how many times--

775
00:43:08,030 --> 00:43:11,950
I guess since I last
accessed p, which is here.

776
00:43:11,950 --> 00:43:14,330
We didn't do that here, but
in the working set bound

777
00:43:14,330 --> 00:43:16,280
that's part of the deal.

778
00:43:20,960 --> 00:43:23,000
If you look on the
left side, it's

779
00:43:23,000 --> 00:43:26,060
like how many times
does the max change.

780
00:43:26,060 --> 00:43:28,520
And you may know if
you n random numbers

781
00:43:28,520 --> 00:43:33,530
and you want to know how many
times does the max changes

782
00:43:33,530 --> 00:43:35,390
as I go left to right,
as I take larger

783
00:43:35,390 --> 00:43:37,130
and larger prefixes
of those n numbers,

784
00:43:37,130 --> 00:43:40,610
the answer is log
n in expectation.

785
00:43:40,610 --> 00:43:43,040
Because the more points you
have, the less and less likely

786
00:43:43,040 --> 00:43:46,240
it is for the max to change.

787
00:43:46,240 --> 00:43:51,880
So basically, you show
the expected Wilber

788
00:43:51,880 --> 00:43:57,200
2 of a point over
this randomization

789
00:43:57,200 --> 00:44:03,750
is theta log ti, where ti
is the working set bound.

790
00:44:03,750 --> 00:44:08,260
And so, that gives
you the theorem.

791
00:44:08,260 --> 00:44:10,664
This gives you a lower bound
of the working set bound.

792
00:44:10,664 --> 00:44:12,580
We have upper bounds of
the working set bound,

793
00:44:12,580 --> 00:44:15,190
and therefore that's OPT.

794
00:44:15,190 --> 00:44:17,911
So that's just a
very quick sketch.

795
00:44:17,911 --> 00:44:19,660
If you're interested,
check out the paper.

796
00:44:22,230 --> 00:44:25,865
That's unfortunately all we
know what to do with Wilber 2.

797
00:44:25,865 --> 00:44:27,490
But there's this
other bound, Wilber 1,

798
00:44:27,490 --> 00:44:33,030
which seems less good yet we
can do a lot more with it.

799
00:44:33,030 --> 00:44:35,140
So let me go to that.

800
00:44:54,470 --> 00:44:57,220
It's a lot easier to analyze
algorithms with respect

801
00:44:57,220 --> 00:45:00,440
to Wilber 1.

802
00:45:00,440 --> 00:45:01,230
What's Wilber 1?

803
00:45:04,080 --> 00:45:09,870
We're going to fix something
called a lower bound tree.

804
00:45:09,870 --> 00:45:12,660
I'm going to call it
because it's basically

805
00:45:12,660 --> 00:45:16,515
going to be a perfect
binary tree on my keys.

806
00:45:19,200 --> 00:45:20,740
This tree never changes.

807
00:45:20,740 --> 00:45:22,320
That's why I say fix.

808
00:45:22,320 --> 00:45:25,250
It is not the binary search
tree you're looking for.

809
00:45:25,250 --> 00:45:29,040
It is not the binary search
tree that you're interested in.

810
00:45:29,040 --> 00:45:30,540
It's just a thing
to think about.

811
00:45:33,930 --> 00:45:41,400
Now, for each node
of that tree--

812
00:45:41,400 --> 00:45:44,700
let's look at this node,
I'll give the node a name, y.

813
00:45:47,880 --> 00:45:48,720
So here's y.

814
00:45:52,000 --> 00:45:53,960
There's the left subtree
of y, and there's

815
00:45:53,960 --> 00:45:56,360
the right subtree of y.

816
00:45:56,360 --> 00:45:57,766
These are a bunch of keys.

817
00:45:57,766 --> 00:45:59,390
There's keys that
are to the left of y.

818
00:45:59,390 --> 00:46:01,100
There's keys to the right of y.

819
00:46:01,100 --> 00:46:02,840
There's keys
outside the subtree.

820
00:46:02,840 --> 00:46:04,880
We're going to ignore those.

821
00:46:04,880 --> 00:46:08,030
I want to look at the accesses
to these keys and accesses

822
00:46:08,030 --> 00:46:10,220
to these keys and see
how many times do I

823
00:46:10,220 --> 00:46:13,040
switch between left and right.

824
00:46:13,040 --> 00:46:19,730
So count the number
of alternations--

825
00:46:19,730 --> 00:46:22,887
so very similar in
spirit to Wilber 2,

826
00:46:22,887 --> 00:46:24,470
it's just relative
to this weird tree,

827
00:46:24,470 --> 00:46:26,045
which is kind of arbitrary--

828
00:46:29,090 --> 00:46:34,880
in the access sequence--
which is x1 up to xn--

829
00:46:34,880 --> 00:46:48,020
between left and
right subtrees of y

830
00:46:48,020 --> 00:46:50,240
So we're going to ignore
accesses to y itself.

831
00:46:50,240 --> 00:46:53,240
We're going to ignore
accesses to keys outside of y.

832
00:46:53,240 --> 00:46:56,990
Just look at how many times
do I switch between x and y.

833
00:46:56,990 --> 00:46:58,322
That's a lower bound.

834
00:46:58,322 --> 00:46:59,030
That's the claim.

835
00:47:09,550 --> 00:47:11,200
It's a lower bound
for the same reason

836
00:47:11,200 --> 00:47:14,800
we use the independent
rectangle lower bound.

837
00:47:14,800 --> 00:47:17,710
And the claim is, if you
look at these alternations,

838
00:47:17,710 --> 00:47:20,370
draw the corresponding
rectangles--

839
00:47:26,170 --> 00:47:27,820
so over here, we
had a vertical line

840
00:47:27,820 --> 00:47:30,820
which corresponded to the key,
and we see how many times do we

841
00:47:30,820 --> 00:47:32,020
cross the line.

842
00:47:32,020 --> 00:47:38,874
Basically, the same thing
over here except now

843
00:47:38,874 --> 00:47:41,290
there's one big vertical line
that corresponds to the root

844
00:47:41,290 --> 00:47:43,330
node, then there's some
vertical lines that

845
00:47:43,330 --> 00:47:45,670
correspond to this
node and this node,

846
00:47:45,670 --> 00:47:48,350
and you're interested
in the access sequence.

847
00:47:48,350 --> 00:47:51,591
How many times-- let's do some
kind of access sequence like

848
00:47:51,591 --> 00:47:52,090
this--

849
00:47:55,890 --> 00:47:56,920
these are our points--

850
00:48:02,590 --> 00:48:06,220
and you just look at what
lines are you crossing.

851
00:48:08,830 --> 00:48:11,127
So like this crosses
the big line.

852
00:48:11,127 --> 00:48:13,210
So that's going to be one
alternation between left

853
00:48:13,210 --> 00:48:14,157
and right here.

854
00:48:14,157 --> 00:48:16,240
Here's another alternation
between left and right.

855
00:48:16,240 --> 00:48:18,820
Here is another alternation
between left and right.

856
00:48:18,820 --> 00:48:22,750
Here's another alternation
between left and right.

857
00:48:22,750 --> 00:48:24,880
And one more.

858
00:48:24,880 --> 00:48:27,280
So for the big line,
for the root node,

859
00:48:27,280 --> 00:48:29,530
that's how many times you
cross between left and right

860
00:48:29,530 --> 00:48:31,750
relative to the root.

861
00:48:31,750 --> 00:48:34,140
Then, for the left
subtree the root,

862
00:48:34,140 --> 00:48:36,120
there's one crossing here.

863
00:48:36,120 --> 00:48:40,060
There is one crossing
here, one crossing here.

864
00:48:42,940 --> 00:48:45,330
These are touching, but
they're not satisfied.

865
00:48:45,330 --> 00:48:46,200
So it's OK.

866
00:48:46,200 --> 00:48:48,540
The claim is all these
rectangles will be independent.

867
00:48:48,540 --> 00:48:52,010
Again, I won't prove that
formally, but it's true.

868
00:48:55,010 --> 00:48:55,940
OK?

869
00:48:55,940 --> 00:48:58,790
Rough sketch.

870
00:48:58,790 --> 00:49:00,250
So that's Wilber 1.

871
00:49:00,250 --> 00:49:02,600
It's, again, an independent
rectangle lower bound.

872
00:49:02,600 --> 00:49:04,944
It's a little weird because
it depends on this tree.

873
00:49:04,944 --> 00:49:06,860
You could choose it to
be a nice perfect tree.

874
00:49:06,860 --> 00:49:07,970
You could choose it to
be a different tree.

875
00:49:07,970 --> 00:49:10,344
You'll get a different
lower bound each time.

876
00:49:10,344 --> 00:49:12,260
So of course, you take
the max over all trees.

877
00:49:12,260 --> 00:49:16,100
That will give you the
biggest Wilber 1 lower bound.

878
00:49:16,100 --> 00:49:21,140
We don't know much about that
biggest Wilber 1 lower bound.

879
00:49:21,140 --> 00:49:26,960
I guess you could ask the
following open question.

880
00:49:26,960 --> 00:49:30,380
Is it true that for
every access sequence

881
00:49:30,380 --> 00:49:37,700
there exists a tree p such
that Wilber 1 is theta OPT?

882
00:49:37,700 --> 00:49:39,920
Or is theta Wilber
2 or something?

883
00:49:39,920 --> 00:49:41,940
Wilber 2 is a single quantity.

884
00:49:41,940 --> 00:49:43,005
You compute it.

885
00:49:43,005 --> 00:49:44,420
It gives you a bound.

886
00:49:44,420 --> 00:49:46,790
Wilber 1, it depends on this p.

887
00:49:46,790 --> 00:49:48,950
Maybe if you choose the
best p for your sequence

888
00:49:48,950 --> 00:49:50,190
you get the right answer.

889
00:49:50,190 --> 00:49:55,550
But it's definitely the case
that Wilber 1 for a fixed p

890
00:49:55,550 --> 00:49:58,850
is not the right answer.

891
00:49:58,850 --> 00:50:00,740
I recall that's easy to prove.

892
00:50:06,580 --> 00:50:08,690
Well, maybe we'll
come back to that.

893
00:50:08,690 --> 00:50:10,017
Yeah, question?

894
00:50:10,017 --> 00:50:11,686
AUDIENCE: So how
do you construct

895
00:50:11,686 --> 00:50:12,879
this lower bound tree?

896
00:50:12,879 --> 00:50:13,932
Like, is it just--

897
00:50:13,932 --> 00:50:16,140
ERIK DEMAINE: I'll tell you
what we're going to use--

898
00:50:16,140 --> 00:50:17,884
the question is how
do we construct p.

899
00:50:17,884 --> 00:50:19,300
You can make it
whatever you want.

900
00:50:19,300 --> 00:50:21,870
What we're going to use
is the perfect tree,

901
00:50:21,870 --> 00:50:23,730
which is sort of unique.

902
00:50:23,730 --> 00:50:27,076
It's kind of arbitrary,
but it works.

903
00:50:27,076 --> 00:50:28,950
It has the property that
its height is log n.

904
00:50:28,950 --> 00:50:30,300
That's all we need.

905
00:50:30,300 --> 00:50:32,430
We're going to use that
to get tango trees.

906
00:50:32,430 --> 00:50:34,500
Other questions?

907
00:50:34,500 --> 00:50:36,600
All right.

908
00:50:36,600 --> 00:50:39,270
Let me briefly mention
a fun access sequence.

909
00:50:45,315 --> 00:50:48,990
You may recognize this sequence.

910
00:50:48,990 --> 00:50:52,080
This would be in-order
traversal in binary.

911
00:50:52,080 --> 00:50:53,760
But if I take
these bit sequences

912
00:50:53,760 --> 00:51:09,311
and read them backwards, then
I get 0, 4, 2, 6, 1, 5, 3, 7.

913
00:51:09,311 --> 00:51:11,310
This is the number 0
through 7 in a funny order.

914
00:51:11,310 --> 00:51:13,890
It's called the bit
reversal sequence.

915
00:51:13,890 --> 00:51:24,000
If you access 0, 4, 2, 6, 1, 5,
3, 7 in a perfect binary tree,

916
00:51:24,000 --> 00:51:26,410
it maximizes Wilber 1.

917
00:51:26,410 --> 00:51:36,960
So in-order traversal--
0, 1, 2, 3, 4, 5, 6.

918
00:51:36,960 --> 00:51:38,010
Ignore 7.

919
00:51:38,010 --> 00:51:39,647
There's not 7 in this tree.

920
00:51:43,135 --> 00:51:46,050
I do 0, 4--

921
00:51:46,050 --> 00:51:48,640
if you look at the
root, alternate left,

922
00:51:48,640 --> 00:51:53,820
right, left, right, left, right.

923
00:51:53,820 --> 00:51:56,500
Because the high-order bit
is switching every time,

924
00:51:56,500 --> 00:51:58,614
and so whether I go to
the left of the tree here

925
00:51:58,614 --> 00:52:00,780
or the right of the tree,
it's switching every time.

926
00:52:00,780 --> 00:52:02,321
And also, if you
look in any subtree,

927
00:52:02,321 --> 00:52:04,710
like when I'm accessing things
within the subtree of one,

928
00:52:04,710 --> 00:52:06,530
it alternates 0, too.

929
00:52:06,530 --> 00:52:09,030
It's too small a tree to
really see that happening,

930
00:52:09,030 --> 00:52:10,600
but it's true.

931
00:52:10,600 --> 00:52:15,380
And so, if you do
this for k bits,

932
00:52:15,380 --> 00:52:18,060
n equals 2 to the k roughly.

933
00:52:18,060 --> 00:52:21,680
And Wilber 1, the
lower bound, is

934
00:52:21,680 --> 00:52:27,210
log n per [INAUDIBLE] because
the every access alternates.

935
00:52:27,210 --> 00:52:29,622
So if you look at
a subtree, whatever

936
00:52:29,622 --> 00:52:31,080
the size of that
subtree is, that's

937
00:52:31,080 --> 00:52:33,060
how many alternations there are.

938
00:52:33,060 --> 00:52:38,070
And so, number of
alternations is theta n log n

939
00:52:38,070 --> 00:52:41,780
because it's the sum over all
nodes of their subtree sizes.

940
00:52:41,780 --> 00:52:48,300
And so OPT is theta n log n.

941
00:52:48,300 --> 00:52:50,420
We know we can achieve n log n--

942
00:52:50,420 --> 00:52:53,520
this is to do n accesses--

943
00:52:53,520 --> 00:52:55,899
we know we can n log n with
a red-black tree or whatever,

944
00:52:55,899 --> 00:52:57,690
but there's actually
a lower bound of n log

945
00:52:57,690 --> 00:52:59,730
n, meaning all
binary search trees--

946
00:52:59,730 --> 00:53:01,590
if you're given this
access sequence,

947
00:53:01,590 --> 00:53:04,080
doesn't matter what you're
doing-- you have to pay

948
00:53:04,080 --> 00:53:04,770
n log n.

949
00:53:04,770 --> 00:53:06,680
It's kind of cool.

950
00:53:06,680 --> 00:53:07,990
A little side effect--

951
00:53:07,990 --> 00:53:09,840
that's Wilber's paper ended.

952
00:53:09,840 --> 00:53:13,350
It's like, hey, cool, can
find one access sequence that

953
00:53:13,350 --> 00:53:17,340
is bad for everybody.

954
00:53:17,340 --> 00:53:20,480
But now we're going
to use Wilber 1

955
00:53:20,480 --> 00:53:22,890
to get one binary search
tree that's pretty

956
00:53:22,890 --> 00:53:25,770
good for all access sequences.

957
00:53:25,770 --> 00:53:28,590
Pretty good meaning within a
log log n factor of optimal.

958
00:53:51,280 --> 00:53:59,152
And this is tango
trees, which would

959
00:53:59,152 --> 00:54:04,400
be log log n competitive
online binary search trees.

960
00:54:09,380 --> 00:54:11,270
Why are they called tango trees?

961
00:54:11,270 --> 00:54:14,330
People made up all sorts of
reasons, but I can tell you--

962
00:54:14,330 --> 00:54:16,010
because I was there--

963
00:54:16,010 --> 00:54:20,810
they were invented mostly
on a flight from New York

964
00:54:20,810 --> 00:54:24,650
to Buenos Aires, which
is the center of tango.

965
00:54:24,650 --> 00:54:26,900
I bought this T-shirt
I think the day after.

966
00:54:26,900 --> 00:54:29,150
And then that week,
we wrote the paper,

967
00:54:29,150 --> 00:54:30,610
and that was tango trees.

968
00:54:30,610 --> 00:54:35,630
So no particular reason,
but it sounds good.

969
00:54:35,630 --> 00:54:37,130
Always good to have a cool name.

970
00:54:37,130 --> 00:54:39,095
So the secret is revealed.

971
00:54:39,095 --> 00:54:46,100
The true meaning of tango trees
is nothing, but you we'll see.

972
00:54:46,100 --> 00:54:48,630
So how do they work?

973
00:54:48,630 --> 00:54:49,640
It's very simple.

974
00:54:49,640 --> 00:54:55,040
Basically, we take Wilber
1 and we simulate it.

975
00:54:55,040 --> 00:54:59,310
So let me be more precise.

976
00:54:59,310 --> 00:55:05,870
There's one key idea, which
is to look at the preferred

977
00:55:05,870 --> 00:55:10,510
child of a node.

978
00:55:15,870 --> 00:55:19,242
I'm going to say the
preferred child is left.

979
00:55:19,242 --> 00:55:23,818
Let's see, node y in p.

980
00:55:23,818 --> 00:55:38,060
It's left if we accessed some
node in the left subtree of y

981
00:55:38,060 --> 00:55:38,675
most recently.

982
00:55:43,300 --> 00:55:46,760
It's the right
child if we accessed

983
00:55:46,760 --> 00:55:48,960
something in the right
subtree most recently.

984
00:55:48,960 --> 00:55:52,850
So we're just looking at left
and right subtree accesses,

985
00:55:52,850 --> 00:55:54,240
what was most recent?

986
00:55:54,240 --> 00:55:56,390
There is a special case
in the beginning, which

987
00:55:56,390 --> 00:55:58,681
is you don't have a preferred
child because you haven't

988
00:55:58,681 --> 00:56:00,710
accessed either
left or right yet.

989
00:56:00,710 --> 00:56:12,365
So this is if no access
to the left or right yet.

990
00:56:12,365 --> 00:56:14,340
So that just happens
in the beginning.

991
00:56:14,340 --> 00:56:17,180
Once you've touched
everything, everybody

992
00:56:17,180 --> 00:56:19,970
will have a left or
right preferred child.

993
00:56:19,970 --> 00:56:23,780
So this is just what was
your most recent child.

994
00:56:23,780 --> 00:56:26,580
This is like a parent
with a very short memory.

995
00:56:26,580 --> 00:56:31,400
Just whichever child I
most recently talked to,

996
00:56:31,400 --> 00:56:34,292
that is my preferred
child at the moment.

997
00:56:34,292 --> 00:56:35,750
It's kind of like
I don't know when

998
00:56:35,750 --> 00:56:38,575
you're going to job interviews.

999
00:56:38,575 --> 00:56:40,160
You know, the most
recent interview

1000
00:56:40,160 --> 00:56:42,702
is the one you remember
most fondly and so, ah,

1001
00:56:42,702 --> 00:56:44,660
you like that one the
best independent of which

1002
00:56:44,660 --> 00:56:45,570
is the coolest.

1003
00:56:45,570 --> 00:56:48,620
So let me draw a picture.

1004
00:56:48,620 --> 00:56:52,520
And I guess I'm going
to draw a big picture--

1005
00:56:52,520 --> 00:57:00,121
my favorite-- a perfectly
balanced binary search tree

1006
00:57:00,121 --> 00:57:02,810
with eight leaves.

1007
00:57:02,810 --> 00:57:10,840
And so now, suppose that every
node has a preferred child.

1008
00:57:10,840 --> 00:57:12,710
Let's say they all do
just because it makes

1009
00:57:12,710 --> 00:57:13,834
a more interesting picture.

1010
00:57:17,240 --> 00:57:20,780
I'm going to draw that
with a big fat arrow.

1011
00:57:20,780 --> 00:57:23,390
And now, what does that do?

1012
00:57:23,390 --> 00:57:24,950
It decomposes our tree.

1013
00:57:24,950 --> 00:57:27,860
This is the perfect tree. p is
going to be perfectly balanced,

1014
00:57:27,860 --> 00:57:28,700
log n height.

1015
00:57:28,700 --> 00:57:30,290
It could be any
log n height tree,

1016
00:57:30,290 --> 00:57:32,990
but we'll make it perfect.

1017
00:57:32,990 --> 00:57:38,840
And it decomposes
that tree into paths.

1018
00:57:38,840 --> 00:57:40,120
And there's a path here.

1019
00:57:40,120 --> 00:57:43,880
You just keep following parent
pointers, you get a path--

1020
00:57:43,880 --> 00:57:45,680
not parent pointers,
preferred pointers.

1021
00:57:45,680 --> 00:57:47,600
It's also true if you follow
parent pointers you get a path,

1022
00:57:47,600 --> 00:57:49,160
but they'll overlap each other.

1023
00:57:49,160 --> 00:57:50,660
You follow preferred
child pointers,

1024
00:57:50,660 --> 00:57:52,050
you get non-overlapping paths.

1025
00:57:54,515 --> 00:57:55,140
There they are.

1026
00:57:55,140 --> 00:57:58,950
We also get these singleton
paths at the leaves.

1027
00:57:58,950 --> 00:58:01,170
Some of the leaves are
in singleton paths.

1028
00:58:01,170 --> 00:58:02,685
These are called
preferred paths.

1029
00:58:12,820 --> 00:58:15,635
Why do I care?

1030
00:58:15,635 --> 00:58:19,860
So this tells me the most
recently accessed element

1031
00:58:19,860 --> 00:58:21,940
was somebody on this path.

1032
00:58:21,940 --> 00:58:22,911
I don't quite know who.

1033
00:58:22,911 --> 00:58:24,910
It could have been this
one, and that would say,

1034
00:58:24,910 --> 00:58:26,410
OK, this is the most
recent direction

1035
00:58:26,410 --> 00:58:27,110
we went through all of them.

1036
00:58:27,110 --> 00:58:28,270
Let's say it's that one.

1037
00:58:28,270 --> 00:58:30,280
Now suppose I access this node.

1038
00:58:30,280 --> 00:58:32,410
What does that tell me?

1039
00:58:32,410 --> 00:58:35,020
Well, if I most recently
accessed left here

1040
00:58:35,020 --> 00:58:38,650
and now I'm accessing the
right, if you look at this node,

1041
00:58:38,650 --> 00:58:40,840
the Wilber 1 bound goes up by 1.

1042
00:58:40,840 --> 00:58:42,400
Because I just accessed left.

1043
00:58:42,400 --> 00:58:44,170
Now I accessed right.

1044
00:58:44,170 --> 00:58:49,120
Also, if I access this node,
this guy, his Wilber 1 bound

1045
00:58:49,120 --> 00:58:51,430
goes up by 1 because now
he's going to the right,

1046
00:58:51,430 --> 00:58:53,200
whereas last time
he went to the left.

1047
00:58:53,200 --> 00:58:56,170
Also, this node previously
went to the right

1048
00:58:56,170 --> 00:58:57,580
and went to the left.

1049
00:58:57,580 --> 00:59:02,200
So Wilber 1 went up
because of this edge,

1050
00:59:02,200 --> 00:59:03,920
and it went up
because of this edge.

1051
00:59:03,920 --> 00:59:07,240
In general, following
non-preferred edges,

1052
00:59:07,240 --> 00:59:10,240
I can pay for because
Wilber 1 goes up by 1

1053
00:59:10,240 --> 00:59:12,190
every time I use a
non-preferred edge.

1054
00:59:12,190 --> 00:59:15,280
This is another way to
state the Wilber 1 bound.

1055
00:59:15,280 --> 00:59:17,320
This is the cool thing.

1056
00:59:17,320 --> 00:59:20,350
As long as I can go
through a path quickly--

1057
00:59:23,419 --> 00:59:25,210
ideally, if I could do
it in constant time,

1058
00:59:25,210 --> 00:59:27,460
this would be a
dynamically-optimal binary

1059
00:59:27,460 --> 00:59:27,990
search tree.

1060
00:59:27,990 --> 00:59:30,573
If I could instantly transport
to where I need to go on a path

1061
00:59:30,573 --> 00:59:32,680
and then jump off the
path to the next path,

1062
00:59:32,680 --> 00:59:35,900
that I can pay for--

1063
00:59:35,900 --> 00:59:38,590
I can spend constant
time to do that--

1064
00:59:38,590 --> 00:59:40,125
then I'd be OK.

1065
00:59:40,125 --> 00:59:42,250
I'm not going to be able
to do it in constant time,

1066
00:59:42,250 --> 00:59:44,935
but I'm going to be able
to do it log log n time.

1067
00:59:44,935 --> 00:59:47,710
I'm going to be able to jump
through a path in log log n

1068
00:59:47,710 --> 00:59:49,270
time, and then jump--

1069
00:59:49,270 --> 00:59:51,730
figure out where I need
to diverge from the path

1070
00:59:51,730 --> 00:59:53,364
because maybe I'm
accessing this guy.

1071
00:59:53,364 --> 00:59:54,280
Jump to the next path.

1072
00:59:54,280 --> 00:59:56,099
Do that in log log n time.

1073
00:59:56,099 --> 00:59:57,640
I've got to update
the path structure

1074
00:59:57,640 --> 00:59:59,620
because now the preferred
child is to the right.

1075
00:59:59,620 --> 01:00:00,703
It used to be to the left.

1076
01:00:00,703 --> 01:00:05,886
So I've got to do something that
will only cost log log n time.

1077
01:00:05,886 --> 01:00:08,260
If I can do that, the lower
bound is the number of edges.

1078
01:00:08,260 --> 01:00:11,390
The upper bound is the
number of non-preferred edges

1079
01:00:11,390 --> 01:00:13,250
times log log n.

1080
01:00:13,250 --> 01:00:21,190
So we get a lower
bound Wilber 1,

1081
01:00:21,190 --> 01:00:22,930
which is going to be
equal to the number

1082
01:00:22,930 --> 01:00:25,480
of non-preferred edges.

1083
01:00:29,099 --> 01:00:30,640
And we're going to
get an upper bound

1084
01:00:30,640 --> 01:00:35,950
through tango trees,
which is going

1085
01:00:35,950 --> 01:00:39,430
to be order number of
non-preferred edges times

1086
01:00:39,430 --> 01:00:42,080
log log n.

1087
01:00:42,080 --> 01:00:42,580
OK.

1088
01:00:42,580 --> 01:00:44,230
Why is it log log n?

1089
01:00:44,230 --> 01:00:47,710
Because each of these paths
has length only log n.

1090
01:00:47,710 --> 01:00:50,500
So put them in a balanced
binary search tree,

1091
01:00:50,500 --> 01:00:53,350
and it has height log log n.

1092
01:00:53,350 --> 01:00:57,220
So take these paths,
squish them into a tree--

1093
01:00:57,220 --> 01:01:01,430
it's hard, I don't know
which way you're squishing.

1094
01:01:01,430 --> 01:01:02,830
It says log n depth.

1095
01:01:02,830 --> 01:01:03,990
It's a path.

1096
01:01:03,990 --> 01:01:05,680
I'm going to fold
it into a tree.

1097
01:01:05,680 --> 01:01:07,450
So it has height only log log n.

1098
01:01:07,450 --> 01:01:09,886
Then I can jump around
it in log log n time.

1099
01:01:09,886 --> 01:01:11,260
That's the idea
with tango trees.

1100
01:01:11,260 --> 01:01:12,345
You're basically done.

1101
01:01:12,345 --> 01:01:15,550
A few details in how they work.

1102
01:01:15,550 --> 01:01:17,510
I don't want to spend
too much time on them,

1103
01:01:17,510 --> 01:01:18,926
but let's go through
some of them.

1104
01:01:42,410 --> 01:01:53,930
So we're going to store
each preferred path

1105
01:01:53,930 --> 01:01:59,840
as an auxiliary
tree, which is just--

1106
01:01:59,840 --> 01:02:02,650
I don't know-- a
red-black tree, say.

1107
01:02:07,800 --> 01:02:10,200
What is the red-black
tree sorted by?

1108
01:02:10,200 --> 01:02:11,710
I don't have a choice.

1109
01:02:11,710 --> 01:02:13,500
Whatever I do has to
be a binary search

1110
01:02:13,500 --> 01:02:15,610
tree among the original keys.

1111
01:02:15,610 --> 01:02:18,180
So if I take these items
and I just throw them

1112
01:02:18,180 --> 01:02:21,854
into a red-black tree, they
will be sorted by whatever

1113
01:02:21,854 --> 01:02:22,770
their x-coordinate is.

1114
01:02:22,770 --> 01:02:24,910
So this is the max,
this is the min.

1115
01:02:24,910 --> 01:02:26,480
This is somewhere in between.

1116
01:02:26,480 --> 01:02:28,320
This is to the left of that.

1117
01:02:28,320 --> 01:02:29,640
So the order is a little weird.

1118
01:02:29,640 --> 01:02:31,980
I'd really like to store
them sorted by depth,

1119
01:02:31,980 --> 01:02:33,390
but I can't do that.

1120
01:02:33,390 --> 01:02:34,965
They are sorted by
their key values.

1121
01:02:38,430 --> 01:02:42,740
Now, what do I need to do
with these auxiliary trees?

1122
01:02:42,740 --> 01:02:46,110
I mean, the basic thing
I do is a search, right?

1123
01:02:46,110 --> 01:02:47,330
I'm searching for a key.

1124
01:02:47,330 --> 01:02:49,580
It's a binary search tree,
so I can still do a search.

1125
01:02:49,580 --> 01:02:52,740
I can figure out this
tree gets represented

1126
01:02:52,740 --> 01:02:56,040
as something more like this.

1127
01:02:56,040 --> 01:02:59,390
That would be a nicely balanced
version of these four nodes.

1128
01:02:59,390 --> 01:03:05,250
So if I called them, I
don't know, a, b, c, d.

1129
01:03:05,250 --> 01:03:06,900
That's their sorted order.

1130
01:03:06,900 --> 01:03:10,290
It's going to be a, b, c, d.

1131
01:03:10,290 --> 01:03:12,502
That's also their
sorted order over here.

1132
01:03:12,502 --> 01:03:14,460
So if I search for my
key, I'll figure out, oh,

1133
01:03:14,460 --> 01:03:18,090
do I fall off here, here,
here, here, or here?

1134
01:03:18,090 --> 01:03:20,400
Now, each of those
corresponds to another path

1135
01:03:20,400 --> 01:03:21,420
I need to visit.

1136
01:03:21,420 --> 01:03:23,730
So if I fall off
the left side of a,

1137
01:03:23,730 --> 01:03:26,940
then I should have a
pointer to this structure.

1138
01:03:26,940 --> 01:03:29,486
If I fall off the--

1139
01:03:29,486 --> 01:03:32,640
I guess these two are empty.

1140
01:03:32,640 --> 01:03:35,820
Those correspond to
these two places.

1141
01:03:35,820 --> 01:03:40,920
If I fall off here, the right
side of c, which is now here,

1142
01:03:40,920 --> 01:03:45,240
this is going to be a
pointer to my new structure

1143
01:03:45,240 --> 01:03:48,570
which corresponds to this one.

1144
01:03:48,570 --> 01:03:51,652
And then this one is going to
correspond to all this stuff--

1145
01:03:51,652 --> 01:03:52,860
well, in particular this one.

1146
01:03:55,960 --> 01:03:59,970
It's a little hard to draw this
picture, but you get the idea.

1147
01:03:59,970 --> 01:04:02,220
You just rebalance
each of these things.

1148
01:04:02,220 --> 01:04:05,160
Keep that the pointers
between the preferred paths

1149
01:04:05,160 --> 01:04:06,774
just as they were.

1150
01:04:06,774 --> 01:04:08,940
This is uniquely defined
how to do this because it's

1151
01:04:08,940 --> 01:04:11,340
a binary search tree.

1152
01:04:11,340 --> 01:04:20,450
So leaves point to other--

1153
01:04:20,450 --> 01:04:24,775
let's call them child
auxiliary trees.

1154
01:04:24,775 --> 01:04:26,640
It uniquely defines
which ones they

1155
01:04:26,640 --> 01:04:29,160
have to point to in
order to still navigate

1156
01:04:29,160 --> 01:04:30,760
the whole structure.

1157
01:04:30,760 --> 01:04:33,810
So it's a weird way of
rebalancing your tree.

1158
01:04:33,810 --> 01:04:36,360
And the point is each of these
red-black trees has height log

1159
01:04:36,360 --> 01:04:39,750
log n because the number of
nodes in it is only log n.

1160
01:04:39,750 --> 01:04:41,626
And that gives us the bound.

1161
01:04:55,620 --> 01:05:03,330
Now, key thing to think about is
what happens when you change--

1162
01:05:03,330 --> 01:05:05,010
I said I have to
be able to achieve

1163
01:05:05,010 --> 01:05:07,990
number of non-preferred
edges times log log n.

1164
01:05:07,990 --> 01:05:08,490
So fine.

1165
01:05:08,490 --> 01:05:10,920
I do a log log n search in here.

1166
01:05:10,920 --> 01:05:12,850
Maybe I decide I
have to go off here.

1167
01:05:12,850 --> 01:05:14,589
Then I do a log log
n search in here.

1168
01:05:14,589 --> 01:05:16,130
And then maybe I
have to go this way.

1169
01:05:16,130 --> 01:05:18,360
So number of
non-preferred edges was 2.

1170
01:05:18,360 --> 01:05:20,440
I did two, maybe three searches.

1171
01:05:20,440 --> 01:05:21,880
Fine.

1172
01:05:21,880 --> 01:05:23,910
It's going to be number
of non-preferred edges

1173
01:05:23,910 --> 01:05:25,410
plus 1 time log log n.

1174
01:05:25,410 --> 01:05:25,980
No big deal.

1175
01:05:33,600 --> 01:05:35,120
Now I have to update.

1176
01:05:35,120 --> 01:05:37,710
Now this is the preferred
edge from the root,

1177
01:05:37,710 --> 01:05:41,070
and this is the preferred
edge from this node.

1178
01:05:41,070 --> 01:05:43,930
How do I update preferred edges?

1179
01:05:43,930 --> 01:05:45,400
That's something to think about.

1180
01:05:45,400 --> 01:05:49,480
So I've got a path represented
by a red-black tree.

1181
01:05:49,480 --> 01:05:54,360
And now I fall off here, and
there's another path here.

1182
01:05:54,360 --> 01:06:00,960
I need to convert this into
a path that goes like this

1183
01:06:00,960 --> 01:06:02,490
and then does this.

1184
01:06:02,490 --> 01:06:05,040
And separately, a
path that does this.

1185
01:06:05,040 --> 01:06:07,110
That's the new version.

1186
01:06:07,110 --> 01:06:08,120
How do I do that?

1187
01:06:08,120 --> 01:06:11,170
Conceptually, it's
pretty simple.

1188
01:06:11,170 --> 01:06:18,630
I want to cut the path here
and then rejoin along there,

1189
01:06:18,630 --> 01:06:20,490
like that.

1190
01:06:20,490 --> 01:06:23,700
So conceptually, if things
were stored by depth,

1191
01:06:23,700 --> 01:06:26,060
this is what we'd call a
split and a concatenate.

1192
01:06:26,060 --> 01:06:28,680
You should know this from
regular binary search trees.

1193
01:06:28,680 --> 01:06:31,530
This is a standard exercise
for red-black trees.

1194
01:06:31,530 --> 01:06:35,670
Given a query, x, you can
cut this tree into two halves

1195
01:06:35,670 --> 01:06:39,000
and get two red-black trees,
which represent everything

1196
01:06:39,000 --> 01:06:42,960
to the left of x and
everything to the right of x.

1197
01:06:42,960 --> 01:06:44,499
Similarly, given
two trees that are

1198
01:06:44,499 --> 01:06:46,290
sorted like this where
all the elements are

1199
01:06:46,290 --> 01:06:48,112
less than all the
elements over here,

1200
01:06:48,112 --> 01:06:50,070
I can concatenate them
into one red-black tree.

1201
01:06:50,070 --> 01:06:51,809
And all of these
take log n time,

1202
01:06:51,809 --> 01:06:53,100
where n is the number of nodes.

1203
01:06:53,100 --> 01:06:56,790
Here, that would
be log log n time.

1204
01:06:56,790 --> 01:06:58,620
In this world, it's
not quite so simple

1205
01:06:58,620 --> 01:07:00,600
because things are
not sorted by depth.

1206
01:07:00,600 --> 01:07:02,670
They're sorted by key value.

1207
01:07:02,670 --> 01:07:04,860
But it's not so bad.

1208
01:07:04,860 --> 01:07:12,730
Because, if you look at some
path and you want to say,

1209
01:07:12,730 --> 01:07:21,150
OK, I want everything
that's below this key value

1210
01:07:21,150 --> 01:07:24,300
or something, then that's
the same as saying,

1211
01:07:24,300 --> 01:07:27,790
well, take everything that is
within this interval of keys.

1212
01:07:27,790 --> 01:07:29,860
So it's strictly
between here and here.

1213
01:07:32,900 --> 01:07:34,460
Let me redraw this slightly.

1214
01:07:45,020 --> 01:07:54,312
So if you look at the nodes
of depth greater than d,

1215
01:07:54,312 --> 01:07:56,350
I want to cut off
everybody that's

1216
01:07:56,350 --> 01:07:58,120
deeper than a
particular spot in order

1217
01:07:58,120 --> 01:08:01,760
to do this kind of change.

1218
01:08:01,760 --> 01:08:12,970
These are equal to nodes
in subtree of that.

1219
01:08:12,970 --> 01:08:16,850
So let me give it a name.

1220
01:08:16,850 --> 01:08:18,830
Let's say I want to cut here.

1221
01:08:18,830 --> 01:08:21,580
So I'm going to
look at this node y.

1222
01:08:21,580 --> 01:08:24,580
This is nodes in
the subtree of y.

1223
01:08:24,580 --> 01:08:26,290
All of the nodes
that are below y

1224
01:08:26,290 --> 01:08:30,790
are obviously going to have
smaller depth than that path.

1225
01:08:30,790 --> 01:08:31,960
This is nodes in a path.

1226
01:08:35,229 --> 01:08:41,920
And nodes in a subtree
are equal to nodes

1227
01:08:41,920 --> 01:08:50,109
with keys in the
min of that subtree

1228
01:08:50,109 --> 01:08:51,600
to the max of that tree.

1229
01:08:51,600 --> 01:08:53,899
It's an interval.

1230
01:08:53,899 --> 01:08:55,090
So what do I do?

1231
01:08:55,090 --> 01:08:56,590
I split at min of y.

1232
01:08:56,590 --> 01:08:58,660
I split at max of y.

1233
01:08:58,660 --> 01:09:00,010
That gives me the interval.

1234
01:09:00,010 --> 01:09:01,210
So here's the picture.

1235
01:09:01,210 --> 01:09:02,290
I have a tree.

1236
01:09:02,290 --> 01:09:04,229
I want to cut out this
interval of nodes.

1237
01:09:04,229 --> 01:09:07,300
This is like range
queries kind of in 1D.

1238
01:09:07,300 --> 01:09:08,229
So I split here.

1239
01:09:08,229 --> 01:09:09,040
I split here.

1240
01:09:09,040 --> 01:09:10,779
What I will have
are the things I

1241
01:09:10,779 --> 01:09:13,160
care about, the things
to the left of it

1242
01:09:13,160 --> 01:09:15,040
and the things to
the right of it.

1243
01:09:15,040 --> 01:09:17,170
What I wanted was this
and everything else.

1244
01:09:17,170 --> 01:09:18,160
How do I do that?

1245
01:09:18,160 --> 01:09:24,010
I concatenate-- this is y.

1246
01:09:24,010 --> 01:09:32,380
This is in the interval
min of y to max of y.

1247
01:09:32,380 --> 01:09:33,348
So I wanted those guys.

1248
01:09:33,348 --> 01:09:35,139
Those are the nodes
that are deeper than d.

1249
01:09:35,139 --> 01:09:37,240
I also want the nodes
all together that

1250
01:09:37,240 --> 01:09:38,870
are less deep than d.

1251
01:09:38,870 --> 01:09:40,939
That's these nodes
and these nodes.

1252
01:09:40,939 --> 01:09:43,029
So I concatenate
these together, get

1253
01:09:43,029 --> 01:09:45,970
one big tree that represents
things with depth less than d.

1254
01:09:45,970 --> 01:09:49,479
These are the things of
depth greater than d.

1255
01:09:49,479 --> 01:09:49,979
OK?

1256
01:09:49,979 --> 01:09:52,810
So I do two splits,
one concatenate,

1257
01:09:52,810 --> 01:09:56,500
and that simulates this
kind of cut operation.

1258
01:09:56,500 --> 01:09:58,780
Similarly, if I want to
do a joint operation,

1259
01:09:58,780 --> 01:10:01,447
it's a constant number of splits
and concatenates, and I'm done.

1260
01:10:01,447 --> 01:10:03,988
Just dealing with the fact that
things are in the wrong order

1261
01:10:03,988 --> 01:10:05,620
here, but it's not so bad.

1262
01:10:10,510 --> 01:10:15,230
One more thing, which is--

1263
01:10:15,230 --> 01:10:17,100
I basically described
the overall structure

1264
01:10:17,100 --> 01:10:19,350
as a tree of auxiliary trees.

1265
01:10:19,350 --> 01:10:22,680
In reality, we're in the
binary search tree model.

1266
01:10:22,680 --> 01:10:25,830
We can only have one tree.

1267
01:10:25,830 --> 01:10:26,910
Not so hard, though.

1268
01:10:26,910 --> 01:10:30,060
I mean, basically,
you want one tree that

1269
01:10:30,060 --> 01:10:34,030
represents lots of trees that
are kind of pasted together.

1270
01:10:34,030 --> 01:10:36,900
So to do that, you
just mark each node

1271
01:10:36,900 --> 01:10:39,780
that transitions from
one tree to the next.

1272
01:10:39,780 --> 01:10:42,810
So each node will say, I am the
root of a new auxiliary tree

1273
01:10:42,810 --> 01:10:45,720
or just say, no, I'm part
of the same auxiliary tree

1274
01:10:45,720 --> 01:10:46,470
as my parent.

1275
01:10:49,530 --> 01:10:51,790
And then you have to
define these kinds of split

1276
01:10:51,790 --> 01:10:55,080
and concatenate operations in
this setting where you have

1277
01:10:55,080 --> 01:10:56,676
a tree embedded inside a tree.

1278
01:10:56,676 --> 01:10:58,050
But you just ignore
all the nodes

1279
01:10:58,050 --> 01:10:59,883
that are claimed to be
part of another tree.

1280
01:10:59,883 --> 01:11:02,290
Just pretend they weren't
there, and it works.

1281
01:11:02,290 --> 01:11:06,819
So a little hand-wavy there, but
it's kind of a tedious detail.

1282
01:11:06,819 --> 01:11:08,610
You can stick all these
trees into one tree

1283
01:11:08,610 --> 01:11:13,100
just by marking these roots.

1284
01:11:13,100 --> 01:11:14,976
And that's tango trees.

1285
01:11:14,976 --> 01:11:18,970
I already spoiled the climax,
which is this log log n thing,

1286
01:11:18,970 --> 01:11:22,220
but it's pretty obvious
how to get there.

1287
01:11:22,220 --> 01:11:25,150
It's just a lot of
details to actually do it.

1288
01:11:25,150 --> 01:11:27,100
We're just taking
the Wilber 1 bound,

1289
01:11:27,100 --> 01:11:30,070
recasting it in terms of
this preferred path thing

1290
01:11:30,070 --> 01:11:33,370
where it's just the
non-preferred edges.

1291
01:11:33,370 --> 01:11:35,600
Or the non-preferred edges
are what Wilber 1 counts,

1292
01:11:35,600 --> 01:11:37,270
and so we can afford
to spend log log n

1293
01:11:37,270 --> 01:11:38,860
time for each of them.

1294
01:11:38,860 --> 01:11:41,100
And the paths themselves
only have log n nodes,

1295
01:11:41,100 --> 01:11:45,470
so you can search through them
in log log n time pretty easy.

1296
01:11:45,470 --> 01:11:47,110
This also shows you
why Wilber 1 is not

1297
01:11:47,110 --> 01:11:50,890
a good bound with a fixed tree.

1298
01:11:50,890 --> 01:11:53,730
Because here are log n nodes.

1299
01:11:53,730 --> 01:11:58,180
I can just sit there all day
bouncing around all of them

1300
01:11:58,180 --> 01:11:59,322
in random order.

1301
01:11:59,322 --> 01:12:01,780
I'm definitely going to need
log log n time to access them,

1302
01:12:01,780 --> 01:12:04,460
but Wilber 1 is not
changing at all.

1303
01:12:04,460 --> 01:12:08,480
So Wilber 1 stays
constant, like 0.

1304
01:12:08,480 --> 01:12:11,140
I had to warm it up, but
after I test everything,

1305
01:12:11,140 --> 01:12:14,710
I can just sit there and bounce
around these guys randomly.

1306
01:12:14,710 --> 01:12:16,730
I've got to spend log
log n time to do that,

1307
01:12:16,730 --> 01:12:19,090
but Wilber 1 doesn't
justify it for me.

1308
01:12:19,090 --> 01:12:22,720
Wilber 2 will go up, but
Wilber 1 with this tree?

1309
01:12:22,720 --> 01:12:24,389
It's kind of lame.

1310
01:12:24,389 --> 01:12:25,930
So this is the best
tango trees could

1311
01:12:25,930 --> 01:12:28,790
hope to do using Wilber 1.

1312
01:12:31,420 --> 01:12:33,790
I would guess that tango
trees are a log log

1313
01:12:33,790 --> 01:12:37,400
factor away from optimal, though
we don't know that for sure.

1314
01:12:37,400 --> 01:12:40,110
But greedy we're still
pretty sure is good.

1315
01:12:40,110 --> 01:12:43,031
It should be a constant
factor away from optimal.

1316
01:12:43,031 --> 01:12:44,780
So I want to talk a
little bit about that.

1317
01:12:48,170 --> 01:12:50,260
There's one thing
on this outline

1318
01:12:50,260 --> 01:12:51,260
we haven't talked about.

1319
01:12:51,260 --> 01:12:52,450
We did independent rectangles.

1320
01:12:52,450 --> 01:12:53,390
We did Wilber 1 and 2.

1321
01:12:53,390 --> 01:12:55,919
We saw applications of them
in particular tango trees.

1322
01:12:55,919 --> 01:12:57,710
One thing we haven't
done is Signed Greedy.

1323
01:13:01,617 --> 01:13:02,700
So let's do Signed Greedy.

1324
01:13:06,550 --> 01:13:10,260
Still left here is we
have two ways to choose

1325
01:13:10,260 --> 01:13:12,330
rectangles,
independent rectangles.

1326
01:13:12,330 --> 01:13:13,290
They're different.

1327
01:13:13,290 --> 01:13:15,690
It would be kind of nice
to know what the best

1328
01:13:15,690 --> 01:13:17,520
way to choose rectangles is.

1329
01:13:17,520 --> 01:13:19,140
And we actually know that--

1330
01:13:25,340 --> 01:13:26,880
Signed Greedy.

1331
01:13:26,880 --> 01:13:28,500
So there's two kinds
of Signed Greedy.

1332
01:13:28,500 --> 01:13:31,260
There's the plus sign
greedy, and there's

1333
01:13:31,260 --> 01:13:32,420
the minus sign greedy.

1334
01:13:36,640 --> 01:13:38,136
How does plus greedy work?

1335
01:13:38,136 --> 01:13:39,510
It's the same as
greedy, you just

1336
01:13:39,510 --> 01:13:41,970
only look at plus rectangles.

1337
01:13:41,970 --> 01:13:44,140
Remember plus rectangles
and minus rectangles.

1338
01:13:44,140 --> 01:13:49,320
So let's look at our
favorite example here.

1339
01:13:49,320 --> 01:13:52,020
With greedy, I would sweep
up, and every rectangle that

1340
01:13:52,020 --> 01:13:54,390
was unsatisfied, I
would satisfy it.

1341
01:13:54,390 --> 01:13:57,570
Now I'm going to ignore
minus rectangles,

1342
01:13:57,570 --> 01:14:00,220
only look at plus rectangles.

1343
01:14:00,220 --> 01:14:02,160
So I see this rectangle,
and I say, oh, I

1344
01:14:02,160 --> 01:14:05,220
don't care because
that's a minus rectangle.

1345
01:14:05,220 --> 01:14:10,880
Then I see this
one and this one.

1346
01:14:10,880 --> 01:14:13,385
I say, oh, those
are plus rectangles.

1347
01:14:13,385 --> 01:14:14,760
So I'm going to
add a point here.

1348
01:14:14,760 --> 01:14:17,190
I'm going to add a point here.

1349
01:14:17,190 --> 01:14:20,760
Then I go up to here.

1350
01:14:20,760 --> 01:14:23,130
I see this rectangle,
which is a plus rectangle.

1351
01:14:23,130 --> 01:14:24,070
That's bad.

1352
01:14:24,070 --> 01:14:25,290
So I've got to add a point.

1353
01:14:25,290 --> 01:14:29,670
I see this minus rectangle
I don't care about.

1354
01:14:29,670 --> 01:14:31,590
This is plus greedy.

1355
01:14:31,590 --> 01:14:33,150
It does not satisfy the set.

1356
01:14:33,150 --> 01:14:35,790
This rectangle
never got satisfied.

1357
01:14:35,790 --> 01:14:37,930
But it plus satisfies the set.

1358
01:14:37,930 --> 01:14:42,780
If I do plus greedy, it
will be plus satisfied.

1359
01:14:42,780 --> 01:14:45,030
Every rectangle you draw
here, if it's plus rectangle,

1360
01:14:45,030 --> 01:14:47,490
it's got another point in it.

1361
01:14:47,490 --> 01:14:50,430
What's kind of nice,
also, is if you actually

1362
01:14:50,430 --> 01:14:54,600
draw the rectangles
you are satisfying--

1363
01:14:54,600 --> 01:14:55,830
maybe I'm use another color.

1364
01:14:58,530 --> 01:15:00,715
There was one rectangle here.

1365
01:15:00,715 --> 01:15:04,650
There was one rectangle here.

1366
01:15:04,650 --> 01:15:08,379
And there was one
rectangle here.

1367
01:15:08,379 --> 01:15:10,170
That's a little awkward
because they're not

1368
01:15:10,170 --> 01:15:12,690
on the original points.

1369
01:15:12,690 --> 01:15:14,490
So I can change
them a little bit,

1370
01:15:14,490 --> 01:15:21,052
maybe move this one down to here
and move this one over to here.

1371
01:15:21,052 --> 01:15:22,510
You could say that
those rectangles

1372
01:15:22,510 --> 01:15:24,700
came from those points.

1373
01:15:24,700 --> 01:15:27,670
Then this is a set of
independent rectangles

1374
01:15:27,670 --> 01:15:30,210
on the original points.

1375
01:15:30,210 --> 01:15:34,900
Maybe not totally
obvious, but plus greedy

1376
01:15:34,900 --> 01:15:49,220
always gives an independent
set of plus rectangles.

1377
01:15:49,220 --> 01:15:50,629
So it's a lower bound.

1378
01:15:50,629 --> 01:15:53,170
It's not an upper bound because
it's not satisfying the point

1379
01:15:53,170 --> 01:15:55,550
set, but it's a lower bound.

1380
01:15:55,550 --> 01:15:58,418
I claim it's a very
good lower bound.

1381
01:16:07,420 --> 01:16:09,280
It by itself might
not be great, but you

1382
01:16:09,280 --> 01:16:11,686
have to consider both of them.

1383
01:16:11,686 --> 01:16:26,340
So theorem is if I take the
max of plus greedy and minus

1384
01:16:26,340 --> 01:16:26,840
greedy--

1385
01:16:30,290 --> 01:16:32,210
each of them is lower
bound, so the max

1386
01:16:32,210 --> 01:16:35,090
is a lower bound on optimal--

1387
01:16:35,090 --> 01:16:37,760
then this is within
a constant factor

1388
01:16:37,760 --> 01:16:41,650
of the biggest possible
independent rectangle lower

1389
01:16:41,650 --> 01:16:42,150
bound.

1390
01:16:51,860 --> 01:16:53,360
And so this is
the way you should

1391
01:16:53,360 --> 01:16:54,680
choose independent rectangles.

1392
01:16:54,680 --> 01:16:55,430
Run plus greedy.

1393
01:16:55,430 --> 01:16:56,150
Run minus greedy.

1394
01:16:56,150 --> 01:16:57,587
Take the best of the two.

1395
01:16:57,587 --> 01:16:59,420
That will always be
within a constant factor

1396
01:16:59,420 --> 01:17:04,015
of the best independent set
of rectangles, factors like 4

1397
01:17:04,015 --> 01:17:06,540
or something in the worst case.

1398
01:17:06,540 --> 01:17:08,480
So let me prove this to you.

1399
01:17:12,110 --> 01:17:13,610
It's a kind of a weird argument.

1400
01:17:16,307 --> 01:17:17,765
I'm going to define
a new quantity.

1401
01:17:20,670 --> 01:17:23,210
Let's call this OPT x, I guess.

1402
01:17:26,300 --> 01:17:31,940
It's sort of like if you
consider plus rectangles

1403
01:17:31,940 --> 01:17:33,837
separately from minus
rectangles, which

1404
01:17:33,837 --> 01:17:34,670
is what we're doing.

1405
01:17:52,440 --> 01:17:55,220
So I would like a point set--

1406
01:17:55,220 --> 01:17:57,620
first, I'd like a plus
satisfying point set,

1407
01:17:57,620 --> 01:18:01,440
and then I'd also like a
minus satisfying point set.

1408
01:18:01,440 --> 01:18:03,020
And then I take their union.

1409
01:18:03,020 --> 01:18:08,340
And I say the cost of that pair
of plus satisfying and minus

1410
01:18:08,340 --> 01:18:10,350
satisfying is the
size of the union.

1411
01:18:10,350 --> 01:18:13,640
So I get a bonus point if
they happen to overlap.

1412
01:18:13,640 --> 01:18:16,040
Not a big deal,
just a factor of 2.

1413
01:18:16,040 --> 01:18:18,200
So this is not a core
concept, but it turns out

1414
01:18:18,200 --> 01:18:20,510
to be basically what we
were doing over here.

1415
01:18:23,300 --> 01:18:27,200
Let me give you a sequence
of crazy inequalities.

1416
01:18:27,200 --> 01:18:29,900
First one is that this
OPT thing is greater than

1417
01:18:29,900 --> 01:18:33,980
or equal to size of the input.

1418
01:18:33,980 --> 01:18:36,060
Each of these inequalities
is totally obvious,

1419
01:18:36,060 --> 01:18:38,056
but the conclusion
is kind of crazy.

1420
01:18:43,860 --> 01:18:46,520
The independent rectangle
lower bound, which we proved,

1421
01:18:46,520 --> 01:18:49,040
says that if you look at
plus satisfying things that's

1422
01:18:49,040 --> 01:18:51,050
going to be at least
size of the input

1423
01:18:51,050 --> 01:18:52,967
plus the max number of
independent rectangles.

1424
01:18:52,967 --> 01:18:54,758
If you look at the
minus satisfying things,

1425
01:18:54,758 --> 01:18:56,990
that's also going to be
at least size of the input

1426
01:18:56,990 --> 01:19:00,320
plus maximum number of minus
independent rectangles.

1427
01:19:00,320 --> 01:19:01,520
So we already proved this.

1428
01:19:01,520 --> 01:19:03,344
That, if you look
at this union, it's

1429
01:19:03,344 --> 01:19:05,510
going to be at least the
size of the input plus half

1430
01:19:05,510 --> 01:19:08,060
the overall max.

1431
01:19:08,060 --> 01:19:12,320
So that's what we proved
at the beginning a lecture.

1432
01:19:12,320 --> 01:19:17,840
Now, this is the best way to
use independent rectangles.

1433
01:19:17,840 --> 01:19:19,970
This kind of Signed
Greedy, which

1434
01:19:19,970 --> 01:19:22,749
is the max of the
two signs, is a way

1435
01:19:22,749 --> 01:19:24,040
to find independent rectangles.

1436
01:19:24,040 --> 01:19:25,331
So it's only going to be worse.

1437
01:19:25,331 --> 01:19:27,480
It's going to be smaller.

1438
01:19:27,480 --> 01:19:32,000
So we can say is greater
than or equal to half

1439
01:19:32,000 --> 01:19:37,630
the max of plus greedy
and minus greedy.

1440
01:19:42,650 --> 01:19:44,160
This was the max.

1441
01:19:44,160 --> 01:19:46,700
So this is another way to do
it, so it must be smaller.

1442
01:19:49,920 --> 01:19:58,710
Now, greedy computes a
plus satisfying assignment.

1443
01:19:58,710 --> 01:20:02,834
So I could say, well, if you
looked at the optimal plus

1444
01:20:02,834 --> 01:20:05,000
satisfying assignment--
this is something we defined

1445
01:20:05,000 --> 01:20:06,440
at the beginning of lecture--

1446
01:20:06,440 --> 01:20:09,590
and the optimal minus
satisfying assignment, that's

1447
01:20:09,590 --> 01:20:11,870
only going to be smaller
than greedy because greedy

1448
01:20:11,870 --> 01:20:15,290
is an algorithm for
solving OPT plus.

1449
01:20:15,290 --> 01:20:19,302
It can't be better
than the optimum.

1450
01:20:19,302 --> 01:20:21,260
Greedy again has to be
bigger than the optimum.

1451
01:20:24,890 --> 01:20:29,110
Now I just want to turn
this max into a plus

1452
01:20:29,110 --> 01:20:32,550
because the max is always
at least the average.

1453
01:20:32,550 --> 01:20:37,670
So if I take the average, which
turns it into 1/4 OPT plus

1454
01:20:37,670 --> 01:20:39,995
plus OPT minus.

1455
01:20:42,990 --> 01:20:44,090
Then that holds.

1456
01:20:44,090 --> 01:20:46,120
You turn the max into a plus.

1457
01:20:46,120 --> 01:20:48,440
If I look at the
optimal plus satisfying

1458
01:20:48,440 --> 01:20:50,690
plus the optimal
minus satisfying,

1459
01:20:50,690 --> 01:20:54,770
that's only going to be
bigger than this thing

1460
01:20:54,770 --> 01:20:56,900
because this can only
save like a factor of 2

1461
01:20:56,900 --> 01:21:00,436
or whatever over
just adding them up.

1462
01:21:00,436 --> 01:21:02,060
I don't even need to
factor of 2 thing.

1463
01:21:02,060 --> 01:21:05,810
I just need that
if you add them up,

1464
01:21:05,810 --> 01:21:08,270
that's only going to be
worse than just counting them

1465
01:21:08,270 --> 01:21:10,350
as the union.

1466
01:21:10,350 --> 01:21:13,100
So we get what I
call a sandwich.

1467
01:21:13,100 --> 01:21:15,860
On the one side, we have OPT x.

1468
01:21:15,860 --> 01:21:17,489
On the other side,
we have 1/4 OPT x.

1469
01:21:17,489 --> 01:21:19,280
I really don't care
about OPT x personally.

1470
01:21:19,280 --> 01:21:21,690
I mean, it's kind of interesting
to see that it's here.

1471
01:21:21,690 --> 01:21:23,910
But the point is these are
within a constant factor.

1472
01:21:23,910 --> 01:21:25,618
Therefore, all of
these things in between

1473
01:21:25,618 --> 01:21:27,690
are within a constant
factor of each other.

1474
01:21:27,690 --> 01:21:31,970
So in particular, this thing,
max of the two greedys,

1475
01:21:31,970 --> 01:21:34,070
is within a constant
factor of this thing.

1476
01:21:34,070 --> 01:21:37,070
This is the independent
rectangle lower bound,

1477
01:21:37,070 --> 01:21:38,540
the best one.

1478
01:21:38,540 --> 01:21:40,460
It also tells you that
OPT x is basically

1479
01:21:40,460 --> 01:21:43,230
what we're computing here.

1480
01:21:43,230 --> 01:21:44,640
So this is weird.

1481
01:21:44,640 --> 01:21:49,190
I'm going to draw one
more picture, which

1482
01:21:49,190 --> 01:21:56,630
is greedy versus Signed Greedy.

1483
01:21:56,630 --> 01:21:58,760
Remember greedy
from last lecture.

1484
01:21:58,760 --> 01:22:02,930
Greedy says, look, I'm going
to fix plus rectangles,

1485
01:22:02,930 --> 01:22:04,640
and I'm going to fix
minus rectangles.

1486
01:22:04,640 --> 01:22:06,740
It does them both
at the same time.

1487
01:22:06,740 --> 01:22:08,600
Signed Greedy says,
look, I'm going

1488
01:22:08,600 --> 01:22:11,270
to do the plus
rectangles separately,

1489
01:22:11,270 --> 01:22:14,510
and then I'm going to the
minus rectangles separately,

1490
01:22:14,510 --> 01:22:17,605
and then add them up or take
the union or take the max.

1491
01:22:17,605 --> 01:22:18,370
It doesn't matter.

1492
01:22:18,370 --> 01:22:19,760
It's a constant factor.

1493
01:22:19,760 --> 01:22:22,670
Just add them separately.

1494
01:22:22,670 --> 01:22:24,920
This one is an upper bound.

1495
01:22:24,920 --> 01:22:27,680
It is a binary search tree.

1496
01:22:27,680 --> 01:22:29,030
This thing is a lower bound.

1497
01:22:29,030 --> 01:22:32,780
All binary search trees
must take at least this.

1498
01:22:32,780 --> 01:22:35,720
Are they equal up
to constant factors?

1499
01:22:35,720 --> 01:22:36,500
We don't know.

1500
01:22:36,500 --> 01:22:37,970
That's the big question.

1501
01:22:37,970 --> 01:22:40,400
They look almost identical.

1502
01:22:40,400 --> 01:22:43,280
But what greedy has to deal with
is sort of the interrelations.

1503
01:22:43,280 --> 01:22:45,980
When I fix some
plus rectangles, I

1504
01:22:45,980 --> 01:22:49,305
might get new minus rectangles
that I have to fix with greedy.

1505
01:22:49,305 --> 01:22:51,180
Signed Greedy doesn't
have to deal with that.

1506
01:22:51,180 --> 01:22:52,989
It's just the plus rectangles.

1507
01:22:52,989 --> 01:22:54,530
They might make more
plus rectangles,

1508
01:22:54,530 --> 01:22:56,120
but that's all I
have to deal with.

1509
01:22:56,120 --> 01:22:57,620
It doesn't deal
with the interaction

1510
01:22:57,620 --> 01:22:59,420
between plus and
minus rectangles.

1511
01:22:59,420 --> 01:23:03,290
Seems like the interaction
kind of fades away

1512
01:23:03,290 --> 01:23:04,207
as a geometric series.

1513
01:23:04,207 --> 01:23:05,873
And therefore, these
things are the same

1514
01:23:05,873 --> 01:23:07,040
up to constant factors.

1515
01:23:07,040 --> 01:23:08,900
But we have no
way to prove that.

1516
01:23:08,900 --> 01:23:12,380
It could be the interaction
blows you out of the water

1517
01:23:12,380 --> 01:23:14,750
somehow.

1518
01:23:14,750 --> 01:23:19,271
That's the best we know
for dynamic optimality.

1519
01:23:19,271 --> 01:23:21,770
Maybe next time I teach this
class we'll have a final answer

1520
01:23:21,770 --> 01:23:25,730
and it'll be constant, but
that's where we are today.