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PROFESSOR: So you
guys might have heard,

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00:00:22,460 --> 00:00:24,590
we have a quiz tomorrow.

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00:00:24,590 --> 00:00:26,880
So we're going to do
a review for that.

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00:00:26,880 --> 00:00:29,690
Did everyone turn in P set six?

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00:00:29,690 --> 00:00:30,200
Yes?

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00:00:30,200 --> 00:00:30,900
Good.

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00:00:30,900 --> 00:00:34,060
All right, so
what's on this quiz?

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00:00:34,060 --> 00:00:37,470
Numerics and graph stuff.

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00:00:37,470 --> 00:00:39,920
So are there any
specific pain points?

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00:00:39,920 --> 00:00:41,065
I have problems prepared.

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00:00:41,065 --> 00:00:42,750
Now, and then at
8:00 PM is going

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00:00:42,750 --> 00:00:44,700
to be a review
session for concepts.

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00:00:44,700 --> 00:00:47,660
So now problems, 8:00 PM
concepts, tomorrow [? Yakim ?]

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00:00:47,660 --> 00:00:50,580
has office hours at
5:00 PM, conveniently

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00:00:50,580 --> 00:00:51,700
placed before the quiz.

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00:00:51,700 --> 00:00:52,360
Poor guy.

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00:00:52,360 --> 00:00:55,230
So take advantage of that.

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00:00:55,230 --> 00:00:56,085
So yeah.

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00:00:56,085 --> 00:00:59,460
AUDIENCE: Numerics, on
P set we were mostly

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00:00:59,460 --> 00:01:02,140
asked-- you said five,
four, whichever came before.

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00:01:02,140 --> 00:01:05,140
You mostly asked for the running
times and things like that.

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00:01:05,140 --> 00:01:08,570
But the numeric lecture
notes are kind of detailed

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00:01:08,570 --> 00:01:13,320
and go into Newtonian
method and all that.

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00:01:13,320 --> 00:01:15,750
How much detail do
we need to know?

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00:01:15,750 --> 00:01:18,130
PROFESSOR: So on the P set
we make your life easier

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00:01:18,130 --> 00:01:20,800
by giving you the
pseudocode of the code

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00:01:20,800 --> 00:01:23,040
that you needed to implement.

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00:01:23,040 --> 00:01:26,380
And we ask you one
algorithm design question,

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00:01:26,380 --> 00:01:28,620
the problem of
finding the k-th root.

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00:01:28,620 --> 00:01:30,820
And that required
you to understand

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00:01:30,820 --> 00:01:32,760
almost everything
that was going on.

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00:01:32,760 --> 00:01:35,410
We'll want you to understand
things that are going on

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00:01:35,410 --> 00:01:39,951
to come up with something fairly
similar to what you've seen.

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00:01:39,951 --> 00:01:41,450
So yeah, I know
that I told you guys

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00:01:41,450 --> 00:01:42,777
before that ignore numerics.

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00:01:42,777 --> 00:01:44,860
Just close your eyes and
pretend it didn't happen.

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00:01:44,860 --> 00:01:47,950
Actually it turns out there's a
significant amount of numerics

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00:01:47,950 --> 00:01:48,450
on the quiz.

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00:01:48,450 --> 00:01:50,450
So we're going to start
with a numerics problem,

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00:01:50,450 --> 00:01:51,770
and go through it.

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00:01:51,770 --> 00:01:53,500
I'm really sorry about that.

49
00:01:53,500 --> 00:01:55,850
Clearly I didn't have
my way for this, right?

50
00:01:55,850 --> 00:02:00,090
OK so we started two
big things in numerics.

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00:02:00,090 --> 00:02:02,470
Karatsuba, which is
really easy compared

52
00:02:02,470 --> 00:02:07,230
to Newton which requires
a lot of understanding.

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00:02:07,230 --> 00:02:11,200
So we're going to spend
our time on Newton today.

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00:02:11,200 --> 00:02:12,990
OK, so anything aside from that?

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00:02:12,990 --> 00:02:13,813
Yes.

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00:02:13,813 --> 00:02:15,729
AUDIENCE: The thing we
were just talking about

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00:02:15,729 --> 00:02:17,757
before you were here
was what kind of edges

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00:02:17,757 --> 00:02:21,160
are produced by depth
first search versus breadth

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00:02:21,160 --> 00:02:21,701
first search?

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00:02:21,701 --> 00:02:23,334
AUDIENCE: And why
do they matter?

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00:02:23,334 --> 00:02:24,750
PROFESSOR: And why
do they matter.

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00:02:24,750 --> 00:02:24,950
OK.

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00:02:24,950 --> 00:02:26,010
AUDIENCE: Because it's probably
going to be a question on the--

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PROFESSOR: So I'm going to
go on this very quickly now,

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00:02:28,830 --> 00:02:30,290
because I want to
cover problems,

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00:02:30,290 --> 00:02:33,941
but tonight-- this is on
the outline for the review.

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00:02:33,941 --> 00:02:35,440
So if you guys can
make it tonight--

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00:02:35,440 --> 00:02:37,106
AUDIENCE: It's late,
so I can't make it.

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00:02:37,106 --> 00:02:38,212
PROFESSOR: Sorry.

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00:02:38,212 --> 00:02:39,920
Well, they are also
in the lecture notes.

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But I'll go through
them very quickly now.

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00:02:42,110 --> 00:02:43,040
OK, anything else?

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00:02:47,780 --> 00:02:49,450
AUDIENCE: Making trees
with DFS and BFS,

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00:02:49,450 --> 00:02:51,049
this might be tonight also.

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00:02:51,049 --> 00:02:52,090
PROFESSOR: Making trees--

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AUDIENCE: Or your search trees.

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00:02:54,050 --> 00:02:58,700
So instead of-- do you
know what I mean by that?

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00:02:58,700 --> 00:03:01,255
PROFESSOR: Do you mean
graph transformation?

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00:03:01,255 --> 00:03:03,630
AUDIENCE: So you've got a tree
and you search through it.

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00:03:03,630 --> 00:03:05,627
So then when you're
searching through it,

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00:03:05,627 --> 00:03:07,460
the result of searching
it, you can put that

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into almost a binary tree.

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PROFESSOR: OK so
how BFS and DFS work

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00:03:12,980 --> 00:03:15,480
and how they produce
trees, I'm assuming, right?

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00:03:15,480 --> 00:03:17,040
Because if you start
out with a tree,

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00:03:17,040 --> 00:03:19,980
everything's going to be
really nice and simple.

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00:03:19,980 --> 00:03:21,410
AUDIENCE: Right.

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00:03:21,410 --> 00:03:22,360
PROFESSOR: Yes?

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00:03:22,360 --> 00:03:23,797
AUDIENCE: Bidirectional search.

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00:03:23,797 --> 00:03:24,380
PROFESSOR: OK.

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00:03:24,380 --> 00:03:25,960
AUDIENCE: Is that
going to be included?

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00:03:25,960 --> 00:03:27,501
PROFESSOR: At a very
conceptual level

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00:03:27,501 --> 00:03:31,460
because we didn't give
you a P set on it.

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00:03:31,460 --> 00:03:33,140
AUDIENCE: Was that
what rubrics was?

95
00:03:33,140 --> 00:03:34,160
AUDIENCE: Yeah.

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00:03:34,160 --> 00:03:34,810
AUDIENCE: It was bidirectional?

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00:03:34,810 --> 00:03:36,250
AUDIENCE: Well it's not going
to be on it so, we're happy.

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00:03:36,250 --> 00:03:38,041
PROFESSOR: Well I mean
we didn't give you--

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00:03:38,041 --> 00:03:40,910
you kind of had to solve a
problem just by-- we have

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00:03:40,910 --> 00:03:42,950
to tell you this
is bidirectional

101
00:03:42,950 --> 00:03:44,850
BFS, go code it up.

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00:03:44,850 --> 00:03:47,250
This is bi-directional
Dijkstra, go code it up.

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00:03:47,250 --> 00:03:51,110
So we're not going to ask you
anything too fancy on them.

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00:03:51,110 --> 00:03:53,204
So nothing past the P set.

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00:03:53,204 --> 00:03:54,620
AUDIENCE: Will you
have a question

106
00:03:54,620 --> 00:03:57,050
like when you were
studying BFS, saying

107
00:03:57,050 --> 00:04:00,313
how there will be
levels of graphs?

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00:04:00,313 --> 00:04:00,938
PROFESSOR: Yep.

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00:04:00,938 --> 00:04:02,890
AUDIENCE: Could there
be anything about that?

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00:04:02,890 --> 00:04:05,420
PROFESSOR: So levels of graphs
as in the levels in BFS?

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00:04:05,420 --> 00:04:08,227
Or the levels that we're using
when we're building graphs?

112
00:04:08,227 --> 00:04:09,768
AUDIENCE: Copies of
graphs and then--

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00:04:09,768 --> 00:04:10,210
AUDIENCE: Transformations.

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00:04:10,210 --> 00:04:10,570
PROFESSOR: Yep.

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00:04:10,570 --> 00:04:11,980
OK, transformations and layers.

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00:04:11,980 --> 00:04:13,447
Yeah, I think this is important.

117
00:04:13,447 --> 00:04:15,988
AUDIENCE: Any particular tips
on what kind of transformations

118
00:04:15,988 --> 00:04:17,014
we might encounter?

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00:04:17,014 --> 00:04:17,680
PROFESSOR: Sure.

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00:04:17,680 --> 00:04:19,882
Problem one has this
transformation, problem two--

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00:04:19,882 --> 00:04:20,866
[AUDIENCE LAUGHTER]

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00:04:20,866 --> 00:04:24,690
AUDIENCE: Basically,
when do we use these?

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00:04:24,690 --> 00:04:29,440
Is it when we are given more
than one parameter for a cause.

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00:04:29,440 --> 00:04:31,396
PROFESSOR: So I have
some problems on this

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00:04:31,396 --> 00:04:32,520
but we have to have a deal.

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00:04:32,520 --> 00:04:34,560
You guys have to help me
go through the problems

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00:04:34,560 --> 00:04:36,680
fast enough so that we get here.

128
00:04:36,680 --> 00:04:37,180
OK?

129
00:04:37,180 --> 00:04:39,830
So if you know the
answer, say the answer.

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00:04:39,830 --> 00:04:42,910
Don't let me wait.

131
00:04:42,910 --> 00:04:43,665
OK.

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00:04:43,665 --> 00:04:46,660
AUDIENCE: Also, when you run
Dijkstra and Bellman-Ford

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00:04:46,660 --> 00:04:50,340
on a graph, with
positive edges, they

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00:04:50,340 --> 00:04:52,970
produce the same
shortest path weight

135
00:04:52,970 --> 00:04:55,570
but might have different trees?

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00:04:55,570 --> 00:04:57,837
PROFESSOR: Yes,
that's conceivable.

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00:04:57,837 --> 00:04:58,420
Because they--

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00:04:58,420 --> 00:05:00,650
AUDIENCE: It's one of the
questions on spring 2011.

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00:05:00,650 --> 00:05:01,233
PROFESSOR: OK.

140
00:05:01,233 --> 00:05:04,420
So first off, they both
have positive edges, so

141
00:05:04,420 --> 00:05:06,170
they both work, right?

142
00:05:06,170 --> 00:05:09,247
If you had to choose,
which one would you choose?

143
00:05:09,247 --> 00:05:11,080
If you had to choose
which are going to run,

144
00:05:11,080 --> 00:05:11,750
which one would you run?

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00:05:11,750 --> 00:05:12,541
AUDIENCE: Dijkstra.

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00:05:12,541 --> 00:05:13,735
PROFESSOR: OK, why Dijkstra?

147
00:05:13,735 --> 00:05:15,070
AUDIENCE: It's faster.

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00:05:15,070 --> 00:05:15,960
PROFESSOR: Faster.

149
00:05:15,960 --> 00:05:27,464
OK so positive edges,
Dijkstra is faster.

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00:05:27,464 --> 00:05:29,130
Now if you're going
to run both of them,

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00:05:29,130 --> 00:05:31,470
they relax edges in
different orders.

152
00:05:31,470 --> 00:05:31,970
Right?

153
00:05:31,970 --> 00:05:34,310
So let's say we have this.

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00:05:36,950 --> 00:05:40,560
I know you guys remember the
sequence of diamonds, right?

155
00:05:40,560 --> 00:05:43,710
My nice example that shows that
the number of paths in a graph

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00:05:43,710 --> 00:05:46,080
is exponential in the
number of vertices.

157
00:05:46,080 --> 00:05:47,376
So let's have one diamond.

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00:05:47,376 --> 00:05:50,490
1, 1, 1, 1.

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00:05:50,490 --> 00:05:57,490
S, A, B, T. So say we're
looking for shortest path

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00:05:57,490 --> 00:05:58,550
from S. Right?

161
00:05:58,550 --> 00:06:02,660
So the distance
from S to S is zero,

162
00:06:02,660 --> 00:06:05,730
and everything else
starts out as infinity.

163
00:06:05,730 --> 00:06:07,270
SA is infinity.

164
00:06:07,270 --> 00:06:10,150
SB is infinity.

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00:06:10,150 --> 00:06:12,630
ST is infinity.

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00:06:12,630 --> 00:06:14,650
Suppose I relax
edges in this order.

167
00:06:14,650 --> 00:06:18,150
One, two, three, four.

168
00:06:18,150 --> 00:06:19,190
What am I going to get?

169
00:06:19,190 --> 00:06:22,540
If I relax SA, then the
distance from S to A

170
00:06:22,540 --> 00:06:23,730
is going to become 1, right?

171
00:06:23,730 --> 00:06:27,090
Because it's 0 plus 1 1.

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00:06:27,090 --> 00:06:30,610
And then the parent
of A is going to be S.

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00:06:30,610 --> 00:06:34,300
Now say I relax AT.

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00:06:34,300 --> 00:06:38,440
The distance from S
to T is going to be 2.

175
00:06:38,440 --> 00:06:39,640
Sorry.

176
00:06:39,640 --> 00:06:46,020
Parent of T is going
to be A. Right?

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00:06:46,020 --> 00:06:50,080
OK now I'm relaxing this one.

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00:06:50,080 --> 00:06:53,400
This is going to become
one and the parent of B

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00:06:53,400 --> 00:06:56,820
is going to become S. And
as I'm relaxing this one,

180
00:06:56,820 --> 00:07:00,460
nothing happens because I
already had a distance of 2.

181
00:07:00,460 --> 00:07:03,010
Now if I relax them
in this order instead,

182
00:07:03,010 --> 00:07:07,210
so the bottom one first and
the top one is afterwards,

183
00:07:07,210 --> 00:07:09,220
I'm going to get
the different path.

184
00:07:09,220 --> 00:07:09,760
Right?

185
00:07:09,760 --> 00:07:11,850
OK so what path do I
have now from S to T?

186
00:07:15,440 --> 00:07:16,900
AUDIENCE: Right
now you have SAT.

187
00:07:16,900 --> 00:07:19,250
PROFESSOR: OK, why SAT?

188
00:07:19,250 --> 00:07:20,520
How do you compute the path?

189
00:07:20,520 --> 00:07:21,910
Parent pointers, right?

190
00:07:21,910 --> 00:07:25,820
Start at T, look at
the parent pointer.

191
00:07:25,820 --> 00:07:26,980
Look at the parent pointer.

192
00:07:29,940 --> 00:07:32,550
So because I relax
them in this order,

193
00:07:32,550 --> 00:07:35,260
this path was considered
before this path.

194
00:07:35,260 --> 00:07:37,820
So the parent pointer of
T is A but if I relax them

195
00:07:37,820 --> 00:07:40,542
in the other order, the
parent pointer could be B.

196
00:07:40,542 --> 00:07:42,420
So by the way, this
could happen if you're

197
00:07:42,420 --> 00:07:44,102
running Dijkstra
or Bellman-Ford.

198
00:07:44,102 --> 00:07:45,560
Or if you're running
Dijkstra, it's

199
00:07:45,560 --> 00:07:48,610
just about how you're going
to have the ties separated.

200
00:07:48,610 --> 00:07:50,320
So your priority
queue implementation

201
00:07:50,320 --> 00:07:52,346
might give you different paths.

202
00:07:52,346 --> 00:07:53,840
Does it make sense now?

203
00:07:53,840 --> 00:07:55,730
So if the order in
which edges are relaxed.

204
00:07:55,730 --> 00:07:58,453
That's what defines with path
you're going to end up with.

205
00:07:58,453 --> 00:07:59,140
AUDIENCE: OK.

206
00:07:59,140 --> 00:08:00,050
PROFESSOR: OK.

207
00:08:00,050 --> 00:08:02,480
Hopefully useful.

208
00:08:02,480 --> 00:08:04,560
All right.

209
00:08:04,560 --> 00:08:06,485
I saw four hands before.

210
00:08:06,485 --> 00:08:08,720
AUDIENCE: Oh, just
overall, how similar

211
00:08:08,720 --> 00:08:11,370
is this test compared
to previous tests?

212
00:08:11,370 --> 00:08:15,070
PROFESSOR: Less algorithm
design, more concepts.

213
00:08:15,070 --> 00:08:15,695
AUDIENCE: Nice.

214
00:08:15,695 --> 00:08:17,120
AUDIENCE: Yes.

215
00:08:17,120 --> 00:08:19,660
PROFESSOR: Well OK,
not less algorithm--

216
00:08:19,660 --> 00:08:22,090
so it's not going to be as
much out-of-the-box thinking,

217
00:08:22,090 --> 00:08:25,260
it's going to be a lot more
of what we drilled already.

218
00:08:25,260 --> 00:08:27,120
So as a hint, what
kind of problem

219
00:08:27,120 --> 00:08:29,405
did we practice over,
and over, and over?

220
00:08:29,405 --> 00:08:30,590
AUDIENCE: States

221
00:08:30,590 --> 00:08:31,510
PROFESSOR: OK.

222
00:08:31,510 --> 00:08:33,360
Graph transformation, right?

223
00:08:33,360 --> 00:08:35,452
So chances are you'll see that.

224
00:08:35,452 --> 00:08:37,340
AUDIENCE: Wait, what's
graph transformation?

225
00:08:37,340 --> 00:08:42,659
[LAUGHTER]

226
00:08:42,659 --> 00:08:44,700
PROFESSOR: So we start
with a complicated problem

227
00:08:44,700 --> 00:08:48,420
and reduce it to a shortest path
problem or to a shortest length

228
00:08:48,420 --> 00:08:49,090
path problem.

229
00:08:52,200 --> 00:08:52,810
OK.

230
00:08:52,810 --> 00:08:54,660
Cool.

231
00:08:54,660 --> 00:08:57,430
So that being said, let's go.

232
00:08:57,430 --> 00:09:00,480
Numerics, everyone's
favorite topic.

233
00:09:00,480 --> 00:09:04,920
So suppose we're
trying to compute this.

234
00:09:04,920 --> 00:09:08,200
Cube root of 6,006.

235
00:09:08,200 --> 00:09:11,030
And the problem
statement says this,

236
00:09:11,030 --> 00:09:17,080
you start out with an
initial approximation of one,

237
00:09:17,080 --> 00:09:25,810
and it wants two
approximations after that.

238
00:09:25,810 --> 00:09:28,680
The problem also tells you which
function we're going to use.

239
00:09:28,680 --> 00:09:30,540
But why don't we
make this fun and try

240
00:09:30,540 --> 00:09:34,415
to guess what the
function will be?

241
00:09:34,415 --> 00:09:45,330
AUDIENCE: x cubed minus 6,006?

242
00:09:45,330 --> 00:09:47,350
PROFESSOR: OK. x
cubed minus 6,006

243
00:09:47,350 --> 00:09:49,289
why did I choose this function?

244
00:09:49,289 --> 00:09:49,830
Well, not me.

245
00:09:49,830 --> 00:09:51,378
Why did you choose
this function?

246
00:09:51,378 --> 00:09:52,378
AUDIENCE: It's the root.

247
00:09:55,130 --> 00:10:02,940
PROFESSOR: OK so reason
number one-- right?

248
00:10:02,940 --> 00:10:03,897
Important.

249
00:10:03,897 --> 00:10:05,730
Has to be zero at the
answer, because that's

250
00:10:05,730 --> 00:10:07,740
what Newton's method gives us.

251
00:10:07,740 --> 00:10:08,990
And what else?

252
00:10:08,990 --> 00:10:10,460
AUDIENCE:
Multiplication is easy.

253
00:10:14,390 --> 00:10:16,370
PROFESSOR: OK.

254
00:10:16,370 --> 00:10:19,160
So it's easy to compute
whatever Newton's method wants

255
00:10:19,160 --> 00:10:20,620
us to compute.

256
00:10:20,620 --> 00:10:23,570
What does Newton's method
want us to compute?

257
00:10:23,570 --> 00:10:28,230
So given an approximation,
xi, how do we

258
00:10:28,230 --> 00:10:31,320
get to the next
approximation, xi plus 1.

259
00:10:31,320 --> 00:10:32,187
Formula.

260
00:10:32,187 --> 00:10:34,520
By the way, this formula
should be on your cheat sheets.

261
00:10:34,520 --> 00:10:37,564
If it's not, no
sympathy for you.

262
00:10:37,564 --> 00:10:39,516
AUDIENCE: I don't remember.

263
00:10:39,516 --> 00:10:44,884
xi plus f over f prime.

264
00:10:51,250 --> 00:10:52,440
AUDIENCE: Where's the prime?

265
00:10:52,440 --> 00:10:54,240
AUDIENCE: Prime's on the bottom.

266
00:10:54,240 --> 00:10:56,320
PROFESSOR: So yeah,
where's the prime?

267
00:10:56,320 --> 00:10:58,560
And is this right?

268
00:11:01,976 --> 00:11:05,392
AUDIENCE: Isn't
there a constant?

269
00:11:05,392 --> 00:11:06,860
AUDIENCE: Isn't it minus?

270
00:11:06,860 --> 00:11:08,341
PROFESSOR: Isn't it minus?

271
00:11:08,341 --> 00:11:11,287
AUDIENCE: Maybe.

272
00:11:11,287 --> 00:11:12,270
AUDIENCE: There you go.

273
00:11:12,270 --> 00:11:13,390
PROFESSOR: So yeah,
if you guys start out

274
00:11:13,390 --> 00:11:15,640
with the wrong formula
you'd get the wrong answer.

275
00:11:15,640 --> 00:11:17,490
Again, no sympathy.

276
00:11:17,490 --> 00:11:20,500
So get that on your sheets.

277
00:11:20,500 --> 00:11:23,570
OK so this is what we
need to compute, right?

278
00:11:23,570 --> 00:11:26,330
You start with an approximation
and you have to subtract this.

279
00:11:26,330 --> 00:11:29,590
So this has to be
easy to compute.

280
00:11:29,590 --> 00:11:35,450
What if I chose instead
a very nice function.

281
00:11:35,450 --> 00:11:37,450
x-- sorry, x minus.

282
00:11:40,750 --> 00:11:44,432
This is clearly zero where
I want it to be, right?

283
00:11:44,432 --> 00:11:46,056
AUDIENCE: But we
don't know what six--

284
00:11:46,056 --> 00:11:47,680
PROFESSOR: But if
I'm trying to compute

285
00:11:47,680 --> 00:11:51,120
the successive approximations,
very good, I'm getting what?

286
00:11:51,120 --> 00:11:59,670
f of x over f prime of x is
x minus square root 3 6,006

287
00:11:59,670 --> 00:12:02,815
over-- what's the
derivative of this?

288
00:12:02,815 --> 00:12:03,490
AUDIENCE: One.

289
00:12:03,490 --> 00:12:04,880
PROFESSOR: One.

290
00:12:04,880 --> 00:12:07,140
So I haven't done
anything, right?

291
00:12:07,140 --> 00:12:09,370
In order to compute the
successive approximation,

292
00:12:09,370 --> 00:12:11,450
I have to know this number.

293
00:12:11,450 --> 00:12:15,020
So this doesn't get me anywhere,
that's why it's a bad function.

294
00:12:15,020 --> 00:12:18,290
By the way, it's not enough
that this is easy to compute.

295
00:12:18,290 --> 00:12:22,920
What you want is this guy
has to be easy to compute.

296
00:12:22,920 --> 00:12:24,770
In the case of division,
we had a function

297
00:12:24,770 --> 00:12:26,280
that looked pretty wacky.

298
00:12:26,280 --> 00:12:29,360
But then when we divided
that by the derivative,

299
00:12:29,360 --> 00:12:31,690
we got something
reasonably nice.

300
00:12:31,690 --> 00:12:34,240
So this has to be
easy to compute.

301
00:12:34,240 --> 00:12:36,990
AUDIENCE: So we changed it
from that one to this one just

302
00:12:36,990 --> 00:12:38,994
because this one is easier or--

303
00:12:38,994 --> 00:12:40,410
PROFESSOR: So we're
using this one

304
00:12:40,410 --> 00:12:44,476
because this one-- let's see
how easy this one is to compute.

305
00:12:44,476 --> 00:12:46,720
Computing this one requires
that you know the answer

306
00:12:46,720 --> 00:12:48,600
to your original problem,
which is not cool.

307
00:12:48,600 --> 00:12:50,536
Right.

308
00:12:50,536 --> 00:12:53,310
So you're trying to
compute this number.

309
00:12:53,310 --> 00:12:55,420
But if you use this,
Newton's approximation

310
00:12:55,420 --> 00:12:57,860
requires that you compute this.

311
00:12:57,860 --> 00:13:00,262
So you have to know this here.

312
00:13:00,262 --> 00:13:02,470
Which is what you're trying
to do in the first place.

313
00:13:02,470 --> 00:13:03,678
AUDIENCE: Wait, I'm confused.

314
00:13:03,678 --> 00:13:05,650
We went from x
cubed minus 6 over 6

315
00:13:05,650 --> 00:13:07,920
to x minus cube
root of 6 over 6.

316
00:13:07,920 --> 00:13:08,657
So what's your--

317
00:13:08,657 --> 00:13:10,240
PROFESSOR: So this
is a good function.

318
00:13:10,240 --> 00:13:11,100
We can use this.

319
00:13:11,100 --> 00:13:12,890
This is a bad function
and I was asking

320
00:13:12,890 --> 00:13:14,080
why is this a bad function.

321
00:13:14,080 --> 00:13:15,912
AUDIENCE: Oh, OK.

322
00:13:15,912 --> 00:13:16,830
PROFESSOR: OK.

323
00:13:16,830 --> 00:13:18,950
So lets see, if we try
to compute it using this,

324
00:13:18,950 --> 00:13:21,215
what do we get?

325
00:13:21,215 --> 00:13:23,125
xi minus

326
00:13:23,125 --> 00:13:25,544
AUDIENCE: x cubed minus 6,006.

327
00:13:25,544 --> 00:13:27,488
PROFESSOR: X cubed minus 6,006.

328
00:13:27,488 --> 00:13:28,641
Divided by--

329
00:13:28,641 --> 00:13:29,432
AUDIENCE: 3x cubed.

330
00:13:32,360 --> 00:13:34,510
PROFESSOR: OK.

331
00:13:34,510 --> 00:13:38,063
So someone simplify this for me.

332
00:13:38,063 --> 00:13:47,432
AUDIENCE: 6i minus x over 3
plus 6,006 over 3 squared.

333
00:13:47,432 --> 00:13:48,816
Oh, well.

334
00:13:48,816 --> 00:13:49,515
It's 2,002.

335
00:13:49,515 --> 00:13:51,056
PROFESSOR: I'll
choose the easy step.

336
00:13:53,998 --> 00:13:54,498
OK.

337
00:14:01,970 --> 00:14:02,845
So what's x1?

338
00:14:10,030 --> 00:14:12,510
AUDIENCE: You plug in x0.

339
00:14:12,510 --> 00:14:14,316
PROFESSOR: OK.

340
00:14:14,316 --> 00:14:16,785
I heard a 2,002.

341
00:14:16,785 --> 00:14:19,160
I don't think it's--
it's very close.

342
00:14:19,160 --> 00:14:20,585
AUDIENCE: 2,002 and two thirds.

343
00:14:20,585 --> 00:14:22,490
0.67.

344
00:14:22,490 --> 00:14:24,310
PROFESSOR: OK, cool.

345
00:14:24,310 --> 00:14:28,230
So 0.67, let's
get to that later.

346
00:14:28,230 --> 00:14:31,210
I don't like fractional numbers.

347
00:14:31,210 --> 00:14:32,960
I mean this is the
right answer math-wise,

348
00:14:32,960 --> 00:14:35,430
but we'll get to that because
if we want to code this up,

349
00:14:35,430 --> 00:14:37,180
we probably don't want
fractional numbers.

350
00:14:37,180 --> 00:14:38,730
So very good point there.

351
00:14:38,730 --> 00:14:40,940
What if I want to compute x2?

352
00:14:40,940 --> 00:14:43,350
I take this guy and
I plug it into where?

353
00:14:47,020 --> 00:14:49,730
This thing over here, right?

354
00:14:49,730 --> 00:14:51,750
So this way I can
compute approximations

355
00:14:51,750 --> 00:14:56,240
that get closer and closer
to my original value.

356
00:14:56,240 --> 00:14:56,740
OK.

357
00:14:56,740 --> 00:14:57,820
Does this ring a bell?

358
00:15:01,430 --> 00:15:03,051
Yes, everyone is
happy with this?

359
00:15:03,051 --> 00:15:03,550
Cool.

360
00:15:06,099 --> 00:15:07,390
So how about the initial guess?

361
00:15:07,390 --> 00:15:10,530
Here we started with
a known initial guess.

362
00:15:10,530 --> 00:15:11,971
Is this a good guess?

363
00:15:11,971 --> 00:15:14,326
AUDIENCE: It's a
really bad guess.

364
00:15:14,326 --> 00:15:16,450
2,002?

365
00:15:16,450 --> 00:15:17,450
PROFESSOR: No, I mean 1.

366
00:15:17,450 --> 00:15:17,950
AUDIENCE: 1?

367
00:15:17,950 --> 00:15:19,900
That's also a really bad guess.

368
00:15:19,900 --> 00:15:21,370
AUDIENCE: Why
would you even ask?

369
00:15:21,370 --> 00:15:23,330
AUDIENCE: 11 and a half.

370
00:15:23,330 --> 00:15:25,020
PROFESSOR: All right.

371
00:15:25,020 --> 00:15:29,000
So you're saying 11 and a
half would be a nice guess.

372
00:15:29,000 --> 00:15:30,920
So, first off, let's
see, what do we

373
00:15:30,920 --> 00:15:34,640
need to-- what does Newton's
algorithm guarantee?

374
00:15:34,640 --> 00:15:38,120
What-- if the guess
is really, really bad,

375
00:15:38,120 --> 00:15:40,030
would we get an
answer all the time?

376
00:15:40,030 --> 00:15:42,020
Maybe takes forever but
we'll get an answer.

377
00:15:42,020 --> 00:15:43,375
True or false?

378
00:15:43,375 --> 00:15:44,000
AUDIENCE: True.

379
00:15:46,477 --> 00:15:47,310
PROFESSOR: Not true.

380
00:15:47,310 --> 00:15:48,587
Sorry.

381
00:15:48,587 --> 00:15:50,920
AUDIENCE: But it has to be
right for your answer to be--

382
00:15:50,920 --> 00:15:53,456
PROFESSOR: OK, so what does
Newton's method guarantee?

383
00:15:53,456 --> 00:15:54,887
[INTERPOSING VOICES]

384
00:15:54,887 --> 00:15:55,720
PROFESSOR: Yep, yep.

385
00:15:55,720 --> 00:15:57,785
So what is the fancy
math name for that?

386
00:15:57,785 --> 00:15:59,660
AUDIENCE: Quadratic
convergence or something.

387
00:15:59,660 --> 00:16:01,540
PROFESSOR: Quadratic
convergence, all right.

388
00:16:01,540 --> 00:16:03,770
So this is what
Newton guarantees.

389
00:16:06,530 --> 00:16:11,970
Quadratic-- so what
does this mean?

390
00:16:11,970 --> 00:16:15,740
For all your approximations,
they're all going to be xi.

391
00:16:15,740 --> 00:16:21,180
If we write them as true
answer times 1 plus error, i.

392
00:16:21,180 --> 00:16:25,640
So this error is the relative
error of the approximation.

393
00:16:25,640 --> 00:16:32,170
Newtons method guarantees
that the error at each step

394
00:16:32,170 --> 00:16:35,060
is squared.

395
00:16:35,060 --> 00:16:41,120
So if this guy-- if
E0 is greater than 1,

396
00:16:41,120 --> 00:16:43,329
would we get an answer?

397
00:16:43,329 --> 00:16:43,870
AUDIENCE: No.

398
00:16:47,820 --> 00:16:50,380
PROFESSOR: OK so what we need?

399
00:16:50,380 --> 00:16:54,080
What's the minimum that we need?

400
00:16:54,080 --> 00:16:58,140
E0 has to be between
what and what?

401
00:16:58,140 --> 00:17:00,540
AUDIENCE: Negative 1?

402
00:17:00,540 --> 00:17:02,460
AUDIENCE: Is it negative
1, 1 or 0 and 1?

403
00:17:02,460 --> 00:17:04,001
AUDIENCE: It would
be 0 and 1, right?

404
00:17:04,001 --> 00:17:05,820
Because won't it
be less than one?

405
00:17:05,820 --> 00:17:07,280
AUDIENCE: Negative 1.

406
00:17:07,280 --> 00:17:09,200
AUDIENCE: Oh, absolute value.

407
00:17:09,200 --> 00:17:11,254
PROFESSOR: I think if you
score something smaller

408
00:17:11,254 --> 00:17:12,670
than negative 1
you get a positive

409
00:17:12,670 --> 00:17:15,490
and then-- so, let's
say this works.

410
00:17:15,490 --> 00:17:21,000
So then x0 has to be
between what and what

411
00:17:21,000 --> 00:17:22,485
in relation to the real answer?

412
00:17:29,910 --> 00:17:35,355
AUDIENCE: The real answer
minus x0 over [INAUDIBLE].

413
00:17:37,479 --> 00:17:39,187
AUDIENCE: 2 times the
actual answer is 0.

414
00:17:42,770 --> 00:17:45,580
PROFESSOR: OK so this
is a sucky guess.

415
00:17:45,580 --> 00:17:47,320
We can do a lot
better, but at least it

416
00:17:47,320 --> 00:17:48,528
will make the thing converge.

417
00:17:51,320 --> 00:17:51,820
Fair enough?

418
00:17:55,710 --> 00:17:58,870
OK now what if I
want a good guess?

419
00:17:58,870 --> 00:18:03,520
First off, let's backtrack
to a more general problem

420
00:18:03,520 --> 00:18:05,330
because this is easy, right?

421
00:18:05,330 --> 00:18:08,370
To plug it into our calculator
and you have the answer.

422
00:18:08,370 --> 00:18:13,973
So what if I'm trying to
compute cube root of r.

423
00:18:13,973 --> 00:18:16,177
Can everyone see, by the way?

424
00:18:16,177 --> 00:18:16,760
AUDIENCE: Yes.

425
00:18:16,760 --> 00:18:17,015
AUDIENCE: Mhmm.

426
00:18:17,015 --> 00:18:17,598
PROFESSOR: OK.

427
00:18:17,598 --> 00:18:20,345
So I'm trying to
compute cube root of r.

428
00:18:20,345 --> 00:18:21,810
r is a number.

429
00:18:21,810 --> 00:18:26,046
Suppose I'm trying to compute,
for example, cube root of 2.

430
00:18:29,330 --> 00:18:33,130
What's the first
thing I need to do?

431
00:18:33,130 --> 00:18:35,020
I'm going to code
this up, by the way.

432
00:18:35,020 --> 00:18:37,061
Well, not me but suppose
we want to code this up.

433
00:18:37,061 --> 00:18:39,156
What's the first
thing we want to do?

434
00:18:39,156 --> 00:18:42,000
AUDIENCE: I think we want an
initial guess at a function.

435
00:18:42,000 --> 00:18:44,250
PROFESSOR: OK so we're going
to need an initial guess.

436
00:18:44,250 --> 00:18:44,930
That's good.

437
00:18:49,770 --> 00:18:52,259
And what else do we want?

438
00:18:52,259 --> 00:18:54,550
AUDIENCE: We need to decide
how much precision we want.

439
00:18:54,550 --> 00:18:55,175
PROFESSOR: Yep.

440
00:18:55,175 --> 00:18:56,335
And why is that?

441
00:18:56,335 --> 00:18:59,245
AUDIENCE: Because this
cube root of 2 is 1.

442
00:18:59,245 --> 00:19:01,670
If you round down.

443
00:19:01,670 --> 00:19:04,580
That doesn't tell us very much.

444
00:19:04,580 --> 00:19:07,022
I don't know, we don't
like the fractions.

445
00:19:07,022 --> 00:19:08,480
PROFESSOR: We don't
like fractions.

446
00:19:08,480 --> 00:19:10,670
OK, why don't we like fractions?

447
00:19:10,670 --> 00:19:11,835
It's a lot easier to code.

448
00:19:11,835 --> 00:19:12,990
AUDIENCE: Decimal point there.

449
00:19:12,990 --> 00:19:14,640
PROFESSOR: Yep, you have to
deal with the decimal point

450
00:19:14,640 --> 00:19:15,598
and we don't like that.

451
00:19:15,598 --> 00:19:17,912
So we want to take
this problem, and we

452
00:19:17,912 --> 00:19:20,000
want to move it
into integer land.

453
00:19:20,000 --> 00:19:21,750
So what we did
here has fractions.

454
00:19:21,750 --> 00:19:23,980
We don't like that because
that's a pain to code.

455
00:19:23,980 --> 00:19:26,780
We didn't make you code
fractions on the P set, right?

456
00:19:26,780 --> 00:19:29,190
So then we better know
how to avoid fractions

457
00:19:29,190 --> 00:19:32,670
if you want to solve
real life problems.

458
00:19:32,670 --> 00:19:36,310
So we're going to say that
we're happy with some amount

459
00:19:36,310 --> 00:19:38,110
of precision in the answer.

460
00:19:38,110 --> 00:19:39,920
And then we're going
to do something

461
00:19:39,920 --> 00:19:42,676
that lets us do all our
computations in integer land.

462
00:19:49,340 --> 00:19:51,720
Because that's a happy place.

463
00:19:51,720 --> 00:20:00,020
So how do I-- suppose
I'm working in base B

464
00:20:00,020 --> 00:20:03,720
and I want d digits
of precision.

465
00:20:03,720 --> 00:20:06,300
What do I want to compute?

466
00:20:06,300 --> 00:20:08,806
How do I transform the
problem in that way?

467
00:20:08,806 --> 00:20:09,660
Yes?

468
00:20:09,660 --> 00:20:15,262
AUDIENCE: You multiply 2
root 3 times B to the d.

469
00:20:15,262 --> 00:20:16,720
PROFESSOR: OK so
we want the answer

470
00:20:16,720 --> 00:20:22,146
to be 2 times B to the d.

471
00:20:22,146 --> 00:20:24,790
And if we're integer land, what
are we actually going to get?

472
00:20:30,830 --> 00:20:37,210
So suppose B is 10-- let's work
with familiar numbers-- d is 2.

473
00:20:37,210 --> 00:20:40,950
So we're going to
want to compute

474
00:20:40,950 --> 00:20:45,005
square root of 3 2 times 100.

475
00:20:45,005 --> 00:20:46,880
Does anyone have a
calculator and can tell me

476
00:20:46,880 --> 00:20:49,390
what square root of 3 is?

477
00:20:49,390 --> 00:20:50,279
First two digits?

478
00:20:50,279 --> 00:20:51,320
AUDIENCE: Cube root of 2.

479
00:20:51,320 --> 00:20:52,695
PROFESSOR: Sorry,
cube root of 2.

480
00:20:57,720 --> 00:20:59,985
AUDIENCE: 1.2.

481
00:20:59,985 --> 00:21:06,770
PROFESSOR: So let's say that
square root of 3 is 1.2345.

482
00:21:06,770 --> 00:21:08,780
Hopefully someone will
fix this up for us.

483
00:21:08,780 --> 00:21:10,020
But let's say it's this.

484
00:21:10,020 --> 00:21:13,310
If I multiply it by
100 what do I get?

485
00:21:13,310 --> 00:21:14,750
123 times 45.

486
00:21:18,780 --> 00:21:21,860
So multiply by 100 means
you move the decimal point

487
00:21:21,860 --> 00:21:22,640
to the right.

488
00:21:22,640 --> 00:21:23,140
Right?

489
00:21:23,140 --> 00:21:24,820
So it's still not an integer.

490
00:21:24,820 --> 00:21:27,010
If we work in integer
land, our algorithm

491
00:21:27,010 --> 00:21:29,450
will return an integer.

492
00:21:29,450 --> 00:21:30,950
What's the integer
that we're hoping

493
00:21:30,950 --> 00:21:33,464
our algorithm will return?

494
00:21:33,464 --> 00:21:35,260
AUDIENCE: 1.23, right?

495
00:21:35,260 --> 00:21:37,090
PROFESSOR: We're
hoping so, right?

496
00:21:37,090 --> 00:21:37,590
Why?

497
00:21:37,590 --> 00:21:38,730
What can you do with 123?

498
00:21:41,950 --> 00:21:42,520
Shift right.

499
00:21:42,520 --> 00:21:47,060
So if you divide this
by 100, you get 1.23.

500
00:21:47,060 --> 00:21:49,895
And this is our answer with
two digits of precision.

501
00:21:49,895 --> 00:21:56,582
AUDIENCE: Do you want
the actual-- it's 1.2599.

502
00:21:56,582 --> 00:21:57,790
PROFESSOR: You guys are good.

503
00:22:04,800 --> 00:22:07,460
OK.

504
00:22:07,460 --> 00:22:11,320
So our algorithm will
give us this, right?

505
00:22:11,320 --> 00:22:16,530
So it's not going to
give us exactly cube

506
00:22:16,530 --> 00:22:18,690
root of 2 times 100.

507
00:22:18,690 --> 00:22:20,630
It's not going to give
us exactly this value.

508
00:22:20,630 --> 00:22:23,424
What's it going to give us?

509
00:22:23,424 --> 00:22:24,965
AUDIENCE: Two digits--
close to that.

510
00:22:29,720 --> 00:22:32,830
PROFESSOR: So it's going to
give us an integer, right?

511
00:22:32,830 --> 00:22:34,870
So what kind of integer?

512
00:22:34,870 --> 00:22:35,757
Math.

513
00:22:35,757 --> 00:22:36,590
AUDIENCE: Truncated?

514
00:22:36,590 --> 00:22:38,480
PROFESSOR: OK, what's
the math for that?

515
00:22:38,480 --> 00:22:39,050
Truncated.

516
00:22:39,050 --> 00:22:40,530
Floor.

517
00:22:40,530 --> 00:22:42,070
Very good.

518
00:22:42,070 --> 00:22:44,740
So this is what we're
actually going to compute.

519
00:22:44,740 --> 00:22:47,010
Actually, because
we're in integer land,

520
00:22:47,010 --> 00:22:48,830
we're going to have
floors at every step

521
00:22:48,830 --> 00:22:50,810
when we're doing
that approximation.

522
00:22:50,810 --> 00:22:52,310
And there's a fancy
proof for Newton

523
00:22:52,310 --> 00:22:54,268
that says that even though
you're taking floors

524
00:22:54,268 --> 00:22:58,430
all the time, it will still
converge to the right thing.

525
00:22:58,430 --> 00:23:01,739
So in order to compute
this, how are we

526
00:23:01,739 --> 00:23:03,030
going to transform the problem?

527
00:23:06,060 --> 00:23:09,250
AUDIENCE: Slide
the Bd into that.

528
00:23:09,250 --> 00:23:10,066
PROFESSOR: OK.

529
00:23:10,066 --> 00:23:12,190
AUDIENCE: You're going to
have to put a 3 in there.

530
00:23:12,190 --> 00:23:12,734
B to the 3d.

531
00:23:12,734 --> 00:23:14,150
PROFESSOR: So we're
actually going

532
00:23:14,150 --> 00:23:18,466
to compute cube root
of 2 times B to the 3d.

533
00:23:25,580 --> 00:23:26,630
Fair enough?

534
00:23:26,630 --> 00:23:28,379
And if we slide this
right by the digits

535
00:23:28,379 --> 00:23:30,795
afterwards we're going to get
our two digits of precision.

536
00:23:33,590 --> 00:23:35,370
OK.

537
00:23:35,370 --> 00:23:38,910
So now I figured
out my precision.

538
00:23:38,910 --> 00:23:41,740
I want an initial guess--
and actually I lied,

539
00:23:41,740 --> 00:23:44,250
there's one more problem.

540
00:23:44,250 --> 00:23:45,060
Running time.

541
00:23:48,547 --> 00:23:50,505
Do you guys want to start
with the running time

542
00:23:50,505 --> 00:23:53,825
or initial guess?

543
00:23:53,825 --> 00:23:56,570
AUDIENCE: Initial guess.

544
00:23:56,570 --> 00:24:01,245
PROFESSOR: OK so what I
want a good initial guess?

545
00:24:01,245 --> 00:24:06,117
AUDIENCE: Between 0 and 2x.

546
00:24:06,117 --> 00:24:08,450
PROFESSOR: OK so if I don't
have that the algorithm will

547
00:24:08,450 --> 00:24:09,580
crash and burn.

548
00:24:09,580 --> 00:24:11,050
When we did Newton
in recitation,

549
00:24:11,050 --> 00:24:13,600
I said that I would
like two things.

550
00:24:13,600 --> 00:24:17,110
And I backed up, I coded
that into P set five

551
00:24:17,110 --> 00:24:18,596
when I did the guess.

552
00:24:18,596 --> 00:24:19,970
So what are the
two things that I

553
00:24:19,970 --> 00:24:21,645
want from a good initial guess?

554
00:24:21,645 --> 00:24:23,070
AUDIENCE: Order of magnitude?

555
00:24:23,070 --> 00:24:24,770
PROFESSOR: OK.

556
00:24:24,770 --> 00:24:26,280
Perfect.

557
00:24:26,280 --> 00:24:29,000
AUDIENCE: Oh, god.

558
00:24:29,000 --> 00:24:36,000
[LAUGHS]

559
00:24:36,000 --> 00:24:37,700
PROFESSOR: OK, so
order of magnitude

560
00:24:37,700 --> 00:24:43,390
is a fancy CSE way to say,
right number of digits.

561
00:24:43,390 --> 00:24:46,470
Suppose we're computing this.

562
00:24:46,470 --> 00:24:49,190
Or suppose we're
computing the cube

563
00:24:49,190 --> 00:24:57,880
root of some number
in base B. So

564
00:24:57,880 --> 00:25:03,850
supposed r is somewhere
between 0 and B.

565
00:25:03,850 --> 00:25:06,140
It's one digit in base
B. How many digits

566
00:25:06,140 --> 00:25:08,160
is this result is going to have?

567
00:25:11,590 --> 00:25:13,440
AUDIENCE: The amount
of digits divided by 3.

568
00:25:13,440 --> 00:25:16,380
The amount of digits
in 2 times B to the 3d.

569
00:25:16,380 --> 00:25:19,810
Which is whatever
B to the 3d is.

570
00:25:19,810 --> 00:25:21,310
AUDIENCE: D plus 1?

571
00:25:21,310 --> 00:25:25,150
PROFESSOR: OK so roughly
either d or d plus 1.

572
00:25:25,150 --> 00:25:28,590
See, d plus 1.

573
00:25:28,590 --> 00:25:33,500
So a decent approximation would
be a 1 followed by these zeros.

574
00:25:33,500 --> 00:25:35,180
An even better
approximation is what

575
00:25:35,180 --> 00:25:39,050
if we can get the first
digit right or almost right?

576
00:25:39,050 --> 00:25:40,030
How would we do that?

577
00:25:43,730 --> 00:25:48,300
AUDIENCE: Pick up the
first digit of our r

578
00:25:48,300 --> 00:25:50,640
PROFESSOR: OK and I promise
that r is one digit.

579
00:25:50,640 --> 00:25:52,740
Just So let's say we pick up r.

580
00:25:52,740 --> 00:25:54,260
And?

581
00:25:54,260 --> 00:25:56,256
AUDIENCE: We do
Newtons method on that.

582
00:25:56,256 --> 00:25:57,552
And we just guess one?

583
00:25:57,552 --> 00:25:59,912
Is that a thing?

584
00:25:59,912 --> 00:26:01,286
PROFESSOR: Uhh--

585
00:26:01,286 --> 00:26:02,744
AUDIENCE: Take the
cube root of it.

586
00:26:02,744 --> 00:26:04,785
AUDIENCE: It should be
some kind of inefficient--

587
00:26:04,785 --> 00:26:07,800
PROFESSOR: Yeah what's some
kind of inefficient way

588
00:26:07,800 --> 00:26:09,302
that we can think of quickly?

589
00:26:09,302 --> 00:26:10,170
AUDIENCE: Can you--

590
00:26:10,170 --> 00:26:11,040
AUDIENCE: Binary?

591
00:26:11,040 --> 00:26:12,498
PROFESSOR: Did you
say divide and--

592
00:26:12,498 --> 00:26:14,435
AUDIENCE: --divide it by 2r?

593
00:26:14,435 --> 00:26:15,310
PROFESSOR: Oh, sorry.

594
00:26:15,310 --> 00:26:16,260
I thought You were>you
were going to say divide

595
00:26:16,260 --> 00:26:17,580
and conquer.

596
00:26:17,580 --> 00:26:20,450
So I heard binary search,
so binary search wins.

597
00:26:20,450 --> 00:26:21,950
So we're going to
the binary search.

598
00:26:21,950 --> 00:26:26,630
Binary search from 0 to r.

599
00:26:26,630 --> 00:26:27,130
Right?

600
00:26:27,130 --> 00:26:30,610
In order to the binary search,
suppose I have my guess,

601
00:26:30,610 --> 00:26:37,522
g, I want to see how g
compares to cube root of r.

602
00:26:37,522 --> 00:26:38,230
How do I do this?

603
00:26:41,870 --> 00:26:49,440
So this is equivalent to
g cubed compared to r.

604
00:26:49,440 --> 00:26:54,029
g cubed is to multiplication,
so I know how to do that.

605
00:26:54,029 --> 00:26:54,529
OK?

606
00:27:00,120 --> 00:27:01,120
Make sense for everyone?

607
00:27:04,460 --> 00:27:07,060
Sorry, r.

608
00:27:07,060 --> 00:27:07,560
My bad.

609
00:27:11,700 --> 00:27:14,800
So this is going to be one
digit in base, B, so hopefully

610
00:27:14,800 --> 00:27:17,740
not too much work.

611
00:27:17,740 --> 00:27:18,720
OK.

612
00:27:18,720 --> 00:27:21,234
But remember this, you can;t
do the comparison directly

613
00:27:21,234 --> 00:27:22,900
because you can't
compute this directly.

614
00:27:22,900 --> 00:27:26,730
So you need to do it this way.

615
00:27:26,730 --> 00:27:30,197
AUDIENCE: So r is just the
first digit of the number?

616
00:27:30,197 --> 00:27:30,822
PROFESSOR: Yep.

617
00:27:35,260 --> 00:27:37,510
So we're going for a
slightly easier problem.

618
00:27:37,510 --> 00:27:40,570
We're trying to compute the cube
root of a single digit number

619
00:27:40,570 --> 00:27:44,360
to an arbitrary precision.

620
00:27:44,360 --> 00:27:46,810
You can make it harder, but
there's no point right now.

621
00:27:46,810 --> 00:27:48,140
So we'll stick to this.

622
00:27:48,140 --> 00:27:50,390
So how much time will this take?

623
00:27:50,390 --> 00:27:54,070
How many approximations
will I need to do?

624
00:27:54,070 --> 00:27:56,620
By the way, so suppose I
compute my approximations.

625
00:27:56,620 --> 00:27:57,120
Right?

626
00:27:57,120 --> 00:28:02,300
So I compute x0, x1, x2,
x3, so on and so forth.

627
00:28:02,300 --> 00:28:04,970
When do I stop?

628
00:28:04,970 --> 00:28:07,265
AUDIENCE: When both
digits are the same?

629
00:28:07,265 --> 00:28:08,890
PROFESSOR: All the
digits are the same.

630
00:28:08,890 --> 00:28:15,120
So when one approximation is
the same with the next one.

631
00:28:15,120 --> 00:28:15,990
That's what you're?

632
00:28:15,990 --> 00:28:16,490
Right.

633
00:28:19,170 --> 00:28:20,640
So these are all integers.

634
00:28:20,640 --> 00:28:22,802
The moment I get two
identical integers,

635
00:28:22,802 --> 00:28:24,760
I know there's no point
to continue because I'm

636
00:28:24,760 --> 00:28:27,300
going to get the
same value forever.

637
00:28:27,300 --> 00:28:29,360
Also this means
that Newton's method

638
00:28:29,360 --> 00:28:30,770
has converged on this integer.

639
00:28:30,770 --> 00:28:32,530
So hopefully I will
get the answer,

640
00:28:32,530 --> 00:28:37,150
which I can convert
into the right answer.

641
00:28:37,150 --> 00:28:39,050
OK so how fast does this work?

642
00:28:39,050 --> 00:28:40,740
What is n?

643
00:28:40,740 --> 00:28:43,740
When am I going to stop?

644
00:28:43,740 --> 00:28:46,215
AUDIENCE: Log base 2 of d?

645
00:28:46,215 --> 00:28:49,185
PROFESSOR: OK.

646
00:28:49,185 --> 00:28:59,100
So n is-- so the number of
digits in my approximation--

647
00:28:59,100 --> 00:29:02,690
the number of correct
digits-- doubles every time.

648
00:29:02,690 --> 00:29:05,600
So after log these steps,
I have the right answer.

649
00:29:05,600 --> 00:29:09,420
Now what if instead of doing
this I did binary search?

650
00:29:09,420 --> 00:29:14,640
So instead of doing Newton's
method I do binary search on 0

651
00:29:14,640 --> 00:29:15,866
to r?

652
00:29:15,866 --> 00:29:16,740
I can do that, right?

653
00:29:16,740 --> 00:29:19,040
Because I can get
a guess and then

654
00:29:19,040 --> 00:29:24,117
I can use this to see if
my guess is right or wrong.

655
00:29:24,117 --> 00:29:25,950
So in theory I could
do it in binary search.

656
00:29:25,950 --> 00:29:27,100
How fast would that be?

657
00:29:27,100 --> 00:29:28,600
AUDIENCE: And what
is the difference

658
00:29:28,600 --> 00:29:29,724
between big r and little r?

659
00:29:32,350 --> 00:29:34,895
PROFESSOR: Big R is-- oh,
I didn't write the big r.

660
00:29:34,895 --> 00:29:35,896
Big r is this guy, here.

661
00:29:35,896 --> 00:29:36,936
AUDIENCE: Oh, OK, gotcha.

662
00:29:36,936 --> 00:29:37,830
OK.

663
00:29:37,830 --> 00:29:39,220
PROFESSOR: And this is little r.

664
00:29:39,220 --> 00:29:39,880
Sorry.

665
00:29:39,880 --> 00:29:42,550
Good question, I
apologize for that.

666
00:29:42,550 --> 00:29:47,960
So R is little r
times B to the 3d.

667
00:29:47,960 --> 00:29:51,400
AUDIENCE: D times log 2 of B--

668
00:29:51,400 --> 00:29:52,400
AUDIENCE: To the d.

669
00:29:52,400 --> 00:29:53,164
Log r to the d.

670
00:29:53,164 --> 00:29:54,580
AUDIENCE: Oh, it's
the same thing.

671
00:29:54,580 --> 00:29:55,540
Right.

672
00:29:55,540 --> 00:29:58,400
AUDIENCE: Yeah.

673
00:29:58,400 --> 00:29:58,900
On the y.

674
00:30:02,760 --> 00:30:09,110
PROFESSOR: OK so the
number of guesses is

675
00:30:09,110 --> 00:30:10,410
AUDIENCE: Log big r.

676
00:30:10,410 --> 00:30:12,440
PROFESSOR: Log big r.

677
00:30:12,440 --> 00:30:19,170
And someone was one step ahead
and said it's order of B times

678
00:30:19,170 --> 00:30:21,710
log little r.

679
00:30:21,710 --> 00:30:25,300
Or I think it's d plus little r.

680
00:30:25,300 --> 00:30:30,590
So this is order of d.

681
00:30:30,590 --> 00:30:32,928
Does this make
sense for everyone?

682
00:30:32,928 --> 00:30:36,704
AUDIENCE: I didn't quite
follow what happened there.

683
00:30:36,704 --> 00:30:38,870
PROFESSOR: So I'm doing a
binary search on the range

684
00:30:38,870 --> 00:30:40,540
from 0 to r.

685
00:30:40,540 --> 00:30:41,240
How many steps?

686
00:30:41,240 --> 00:30:43,018
How many guesses?

687
00:30:43,018 --> 00:30:43,946
Log r.

688
00:30:43,946 --> 00:30:44,470
Right?

689
00:30:44,470 --> 00:30:47,750
This is binary search.

690
00:30:47,750 --> 00:30:53,620
R is little r times b to the 3d.

691
00:30:53,620 --> 00:31:00,420
So if we log this,
log R is log little r

692
00:31:00,420 --> 00:31:10,760
plus 3d log B. B is order
one, so we get order of log B.

693
00:31:10,760 --> 00:31:14,300
OK so Newton's method
is exponentially faster

694
00:31:14,300 --> 00:31:17,920
in terms of the number of digits
compared to binary search.

695
00:31:17,920 --> 00:31:20,630
So if you can pull it
off with Newton's method

696
00:31:20,630 --> 00:31:22,370
and you have big
numbers, you probably

697
00:31:22,370 --> 00:31:24,299
want to use Newton's method.

698
00:31:24,299 --> 00:31:26,590
Because binary search will
give you a slower algorithm.

699
00:31:30,238 --> 00:31:32,420
OK, how are we doing
with these concepts?

700
00:31:37,480 --> 00:31:39,530
I see everyone's unhappy.

701
00:31:39,530 --> 00:31:41,430
Is it because of Newton
or something I said?

702
00:31:41,430 --> 00:31:43,770
AUDIENCE: Can you
summarize the difference

703
00:31:43,770 --> 00:31:45,520
between binary search
and Newton's, again?

704
00:31:45,520 --> 00:31:47,930
So you said binary
search takes kind

705
00:31:47,930 --> 00:31:50,060
of long versus Newton's method.

706
00:31:50,060 --> 00:31:50,790
PROFESSOR: OK.

707
00:31:50,790 --> 00:31:53,391
So how do you do binary search?

708
00:31:53,391 --> 00:31:55,140
AUDIENCE: Divide and
conquer, which is--

709
00:31:55,140 --> 00:31:56,681
PROFESSOR: Divide
and conquer, right?

710
00:31:56,681 --> 00:31:59,459
You start out with an interval,
you guess somewhere in half,

711
00:31:59,459 --> 00:32:02,000
and then you see which half of
the interval you recurse onto.

712
00:32:02,000 --> 00:32:05,040
And your interval halves
in size every time.

713
00:32:05,040 --> 00:32:08,700
So given the initial size of the
interval, the number of guesses

714
00:32:08,700 --> 00:32:12,350
you have is log of
that interval size.

715
00:32:12,350 --> 00:32:16,410
In this case we're guessing
a number between 0 and r.

716
00:32:16,410 --> 00:32:18,975
So the big goal is to
compute cube root of R.

717
00:32:18,975 --> 00:32:21,100
And we're guessing that
hey, it has to be somewhere

718
00:32:21,100 --> 00:32:25,250
between 0 and R. So we're going
to do binary search on this.

719
00:32:25,250 --> 00:32:28,340
And in order to see
how our guess compares,

720
00:32:28,340 --> 00:32:31,031
we're going to use
this trick, here.

721
00:32:33,740 --> 00:32:37,640
So the number of guesses
we make is log R. Now

722
00:32:37,640 --> 00:32:39,900
R is that number, over there.

723
00:32:39,900 --> 00:32:43,070
It's little r times B to the 3d.

724
00:32:43,070 --> 00:32:44,520
So how many digits does it have?

725
00:32:44,520 --> 00:32:47,530
Roughly d digits.

726
00:32:47,530 --> 00:32:50,390
So log of R is
going to be order d.

727
00:32:53,650 --> 00:32:57,600
OK so if you do binary search
to guess a d digit number,

728
00:32:57,600 --> 00:32:59,780
the running time will
be roughly order d.

729
00:32:59,780 --> 00:33:02,200
Order of the number of digits.

730
00:33:02,200 --> 00:33:05,150
If you use Newton's
method and your function

731
00:33:05,150 --> 00:33:06,600
is good and
everything converges,

732
00:33:06,600 --> 00:33:08,400
then you have
quadratic convergence.

733
00:33:08,400 --> 00:33:11,840
Which guarantees that you'll
have log of d guesses.

734
00:33:11,840 --> 00:33:14,080
Log of d approximations.

735
00:33:14,080 --> 00:33:16,780
So Newton's method is a
lot faster for big numbers,

736
00:33:16,780 --> 00:33:20,450
because you have a log
there instead of d.

737
00:33:20,450 --> 00:33:22,810
So this is the big
difference between them.

738
00:33:22,810 --> 00:33:24,680
But for Newton, you
need a lot of things.

739
00:33:24,680 --> 00:33:24,870
Right?

740
00:33:24,870 --> 00:33:26,953
You need that function,
you need an initial guess,

741
00:33:26,953 --> 00:33:29,900
you need a lot of
things to make it work.

742
00:33:29,900 --> 00:33:30,400
Yes?

743
00:33:30,400 --> 00:33:32,858
AUDIENCE: So is that just like
the number of times it runs?

744
00:33:32,858 --> 00:33:35,985
Because don't you have a
multiplication and a division?

745
00:33:35,985 --> 00:33:36,610
PROFESSOR: Yep.

746
00:33:36,610 --> 00:33:38,830
So this is the
number of guesses.

747
00:33:38,830 --> 00:33:40,720
The total running time
is quite different.

748
00:33:40,720 --> 00:33:43,136
You need to jump through a few
more hoops to analyze that.

749
00:33:43,136 --> 00:33:48,280
But this is the number of
guesses or approximations

750
00:33:48,280 --> 00:33:51,362
that each method makes.

751
00:33:51,362 --> 00:33:52,820
So if your math
would be order one,

752
00:33:52,820 --> 00:33:54,111
this would be the running time.

753
00:33:54,111 --> 00:33:59,150
If the math isn't order one,
then it gets more complicated.

754
00:33:59,150 --> 00:34:02,432
AUDIENCE: So what's the
running time of a division?

755
00:34:02,432 --> 00:34:04,640
PROFESSOR: What is running
time of the division using

756
00:34:04,640 --> 00:34:07,750
Newton's method?

757
00:34:07,750 --> 00:34:10,309
It's the same as the running
time of the multiplication

758
00:34:10,309 --> 00:34:12,350
that you're using to
compute your approximations.

759
00:34:14,949 --> 00:34:17,889
And it's like two pages
of notes to prove that.

760
00:34:17,889 --> 00:34:20,764
So let's not go over that now.

761
00:34:20,764 --> 00:34:22,889
But the idea is that every
time you're multiplying,

762
00:34:22,889 --> 00:34:24,409
you start multiplying
small numbers

763
00:34:24,409 --> 00:34:26,590
and the numbers double in size
when you're multiplying them.

764
00:34:26,590 --> 00:34:28,214
So only the last
multiplication counts.

765
00:34:28,214 --> 00:34:30,420
All the others are
tiny compared to that.

766
00:34:34,679 --> 00:34:35,305
OK.

767
00:34:35,305 --> 00:34:36,429
Anything else about Newton?

768
00:34:43,110 --> 00:34:44,840
Sorry guys, we have
Newton on the exam

769
00:34:44,840 --> 00:34:46,090
so we have to go through this.

770
00:34:46,090 --> 00:34:49,900
Let's go through some
fun graph problems.

771
00:34:49,900 --> 00:34:54,507
OK so we kicked off these two.

772
00:34:54,507 --> 00:34:56,465
Graphs the BFS edges and
graph transformations.

773
00:35:07,532 --> 00:35:09,490
OK I'm going to do a
graph transformation first

774
00:35:09,490 --> 00:35:12,166
because I think that's
more useful than BFS edges,

775
00:35:12,166 --> 00:35:13,790
but hopefully I'm
going to get to both.

776
00:35:25,650 --> 00:35:28,460
So suppose I have a graph.

777
00:35:28,460 --> 00:35:31,220
Each path is either red or blue.

778
00:35:35,110 --> 00:35:38,440
By the way, no political
connection whatsoever.

779
00:35:38,440 --> 00:35:40,410
And you have weights
on each path.

780
00:35:47,280 --> 00:35:52,430
So we want a path from S to T
with the following constraint.

781
00:35:52,430 --> 00:35:55,650
So these weights are
the costs of, maybe,

782
00:35:55,650 --> 00:35:59,250
how much gas you're
spending to go on that road.

783
00:35:59,250 --> 00:36:01,977
So in order to switch from
a red edge to a blue edge,

784
00:36:01,977 --> 00:36:03,310
you also have to pay some money.

785
00:36:03,310 --> 00:36:06,620
You have to pay a cost of 5.

786
00:36:06,620 --> 00:36:08,920
So if you're going
to go-- if you're

787
00:36:08,920 --> 00:36:12,520
going to go from S
to A-- let's call

788
00:36:12,520 --> 00:36:16,790
this guy A-- if you're
going to go this way,

789
00:36:16,790 --> 00:36:23,220
then your total cost is 3, 1, 4.

790
00:36:23,220 --> 00:36:28,600
If you're going to go from S
to B, then if you go this way,

791
00:36:28,600 --> 00:36:31,620
your total cost is
going to be 3 plus 1

792
00:36:31,620 --> 00:36:35,380
plus 5 for switching from
a red to a blue edge.

793
00:36:35,380 --> 00:36:38,640
So the cost for switching
from red to blue is 5.

794
00:36:38,640 --> 00:36:42,602
The cost for switching
from blue to red is 5.

795
00:36:42,602 --> 00:36:44,393
And I'll let you guys
think of it for a bit

796
00:36:44,393 --> 00:36:45,393
while I erase the board.

797
00:37:09,030 --> 00:37:13,090
AUDIENCE: What's your goal in
playing the best path from--

798
00:37:13,090 --> 00:37:16,025
PROFESSOR: Best path from S
to T. Smallest total cost.

799
00:37:16,025 --> 00:37:16,566
AUDIENCE: OK.

800
00:37:26,030 --> 00:37:28,840
PROFESSOR: So any thoughts?

801
00:37:28,840 --> 00:37:31,289
Yes.

802
00:37:31,289 --> 00:37:33,245
AUDIENCE: It's
kind of on the run,

803
00:37:33,245 --> 00:37:38,135
but create a supernode
connect all the red paths.

804
00:37:38,135 --> 00:37:41,069
And create another one
connecting all the blue paths.

805
00:37:41,069 --> 00:37:45,470
And then try to find the
shortest length to T just using

806
00:37:45,470 --> 00:37:49,140
those two graphs, I guess.

807
00:37:49,140 --> 00:37:50,640
PROFESSOR: So you're
going to create

808
00:37:50,640 --> 00:37:52,780
a supernode
connecting the paths.

809
00:37:52,780 --> 00:37:55,006
So how does this work?

810
00:37:55,006 --> 00:38:02,476
AUDIENCE: You create
the path from S

811
00:38:02,476 --> 00:38:04,470
to T just using the reds.

812
00:38:04,470 --> 00:38:07,060
PROFESSOR: OK so I'm going
to copy everything and use

813
00:38:07,060 --> 00:38:08,970
the reds.

814
00:38:08,970 --> 00:38:18,170
So S, C, D, T, and I'm
only going to use the reds.

815
00:38:18,170 --> 00:38:23,806
All right, red, red, red.

816
00:38:27,300 --> 00:38:29,470
OK.

817
00:38:29,470 --> 00:38:33,470
AUDIENCE: And then-- wait,
is that all the right paths?

818
00:38:33,470 --> 00:38:35,858
PROFESSOR: I hope so.

819
00:38:35,858 --> 00:38:37,358
AUDIENCE: All right,
and then you're

820
00:38:37,358 --> 00:38:41,380
going to try doing that
with the blue paths as well.

821
00:38:41,380 --> 00:38:45,326
PROFESSOR: So am I creating
another copy of this?

822
00:38:45,326 --> 00:38:46,080
AUDIENCE: Yeah.

823
00:38:46,080 --> 00:38:46,663
PROFESSOR: OK.

824
00:38:59,571 --> 00:39:01,050
Let's see, what else, what else?

825
00:39:08,940 --> 00:39:13,122
OK

826
00:39:13,122 --> 00:39:14,978
AUDIENCE: From S to T, right?

827
00:39:14,978 --> 00:39:16,230
PROFESSOR: A path from S to T?

828
00:39:16,230 --> 00:39:21,564
AUDIENCE: Yeah, so you
just compare the path of S

829
00:39:21,564 --> 00:39:23,556
to the path of T.
So you completely

830
00:39:23,556 --> 00:39:27,540
take that out, the
additional cost [INAUDIBLE]

831
00:39:27,540 --> 00:39:29,282
just by comparing
those two and seeing

832
00:39:29,282 --> 00:39:30,688
which ones are
the shortest path.

833
00:39:30,688 --> 00:39:32,271
AUDIENCE: It's correct
for this graph,

834
00:39:32,271 --> 00:39:35,010
but it's not correct in general.

835
00:39:35,010 --> 00:39:37,520
[INTERPOSING VOICES]

836
00:39:37,520 --> 00:39:42,510
AUDIENCE: And the last
option would be combining--

837
00:39:42,510 --> 00:39:44,559
PROFESSOR: OK so if
I do this so far.

838
00:39:44,559 --> 00:39:47,100
So I've taken the original graph
and I've created two copies.

839
00:39:47,100 --> 00:39:48,520
One that has all
of the red paths,

840
00:39:48,520 --> 00:39:50,007
one that has all the blue paths.

841
00:39:50,007 --> 00:39:51,254
AUDIENCE: Right.

842
00:39:51,254 --> 00:39:52,920
PROFESSOR: By the
way, let's label this.

843
00:39:52,920 --> 00:39:54,760
The nodes are--

844
00:39:54,760 --> 00:39:56,760
AUDIENCE: So it would
work in this specific case

845
00:39:56,760 --> 00:40:00,700
because you [INAUDIBLE].

846
00:40:00,700 --> 00:40:03,390
PROFESSOR: So if you compute
the two shortest paths here,

847
00:40:03,390 --> 00:40:06,820
you'll get the path using only
reds, path using only blues.

848
00:40:06,820 --> 00:40:09,090
But you're not
expressing the fact

849
00:40:09,090 --> 00:40:12,580
that you're allowed to alternate
between reds and blues.

850
00:40:12,580 --> 00:40:14,190
How do you express that?

851
00:40:14,190 --> 00:40:16,110
AUDIENCE: Make a
lot of corrections.

852
00:40:16,110 --> 00:40:17,920
AUDIENCE: Put two layers in--

853
00:40:17,920 --> 00:40:18,900
PROFESSOR: Two layers.

854
00:40:18,900 --> 00:40:20,110
OK so these are two layers.

855
00:40:20,110 --> 00:40:24,240
This is the red layer, and
this is the blue layer.

856
00:40:27,370 --> 00:40:29,380
And how do I connect them?

857
00:40:29,380 --> 00:40:31,340
AUDIENCE: If there's
an edge connecting them

858
00:40:31,340 --> 00:40:35,750
on the upper graph, we're
just connecting them.

859
00:40:35,750 --> 00:40:38,220
AUDIENCE: Or connect
the respective nodes

860
00:40:38,220 --> 00:40:40,279
and make it a 5.

861
00:40:40,279 --> 00:40:41,695
PROFESSOR: Both
answers work, they

862
00:40:41,695 --> 00:40:43,000
have the same number of edges.

863
00:40:43,000 --> 00:40:44,930
The last answer is
easier to visualize,

864
00:40:44,930 --> 00:40:47,000
so I'm going to go with that.

865
00:40:47,000 --> 00:40:49,420
So I can go from the red
world to the blue world,

866
00:40:49,420 --> 00:40:51,190
and back, if I'm
willing to pay 5.

867
00:40:55,690 --> 00:40:56,615
So this is what--

868
00:41:02,842 --> 00:41:07,261
AUDIENCE: And you just use
those with the new edges.

869
00:41:07,261 --> 00:41:09,426
Or not DFS, but Dijkstra's.

870
00:41:09,426 --> 00:41:12,050
PROFESSOR: All right so all the
edges here are positive, right?

871
00:41:12,050 --> 00:41:13,523
So I'm going to use Dijkstra.

872
00:41:18,360 --> 00:41:20,564
AUDIENCE: Is there any
case in a positive graph

873
00:41:20,564 --> 00:41:24,860
that Dijkstra's slower
than Bellman-Ford?

874
00:41:24,860 --> 00:41:27,681
Because one is like b log b,
and the other one is like b e.

875
00:41:27,681 --> 00:41:28,180
Right?

876
00:41:28,180 --> 00:41:29,860
So if you have a lot of--

877
00:41:29,860 --> 00:41:31,502
PROFESSOR: OK so
what is Dijkstra?

878
00:41:31,502 --> 00:41:32,980
AUDIENCE: B log b-

879
00:41:32,980 --> 00:41:34,132
PROFESSOR: Almost.

880
00:41:34,132 --> 00:41:36,020
AUDIENCE: T log
b, I think, right?

881
00:41:36,020 --> 00:41:39,170
That's the theoretical
best case ever,

882
00:41:39,170 --> 00:41:42,569
PROFESSOR: So actually,
the theoretical is E plus--

883
00:41:42,569 --> 00:41:44,068
AUDIENCE: Yeah,
that's what I meant.

884
00:41:44,068 --> 00:41:45,042
AUDIENCE: Oh.

885
00:41:45,042 --> 00:41:45,770
Oh--

886
00:41:45,770 --> 00:41:47,561
PROFESSOR: This is what
you get with heaps,

887
00:41:47,561 --> 00:41:50,722
so this is what we
got in the P set.

888
00:41:50,722 --> 00:41:52,500
AUDIENCE: What
was the first one?

889
00:41:52,500 --> 00:41:55,375
AUDIENCE: The first one
is like Fibonacci heaps.

890
00:41:55,375 --> 00:41:57,000
It's got a really
high constant factor,

891
00:41:57,000 --> 00:41:59,160
so you never actually use it.

892
00:41:59,160 --> 00:42:01,400
PROFESSOR: So this is a
nice theoretical thing,

893
00:42:01,400 --> 00:42:03,680
and this is what
you get in practice.

894
00:42:03,680 --> 00:42:06,100
If you use the
regular binary heaps.

895
00:42:06,100 --> 00:42:08,290
And so you compare
this to Bellman-Ford

896
00:42:08,290 --> 00:42:16,340
which is-- whoops--
V times E. I'm

897
00:42:16,340 --> 00:42:19,410
going to guess that this
is faster no matter what.

898
00:42:19,410 --> 00:42:21,110
AUDIENCE: Yes, OK.

899
00:42:21,110 --> 00:42:25,660
So the theoretical one could
be slower, then, possible.

900
00:42:25,660 --> 00:42:28,430
The Fibonacci one?

901
00:42:28,430 --> 00:42:32,499
Because if you have a ton of
vertices without many edges.

902
00:42:32,499 --> 00:42:34,290
PROFESSOR: OK If you
have a ton of vertices

903
00:42:34,290 --> 00:42:35,580
but not too many edges.

904
00:42:35,580 --> 00:42:36,590
AUDIENCE: Yeah.

905
00:42:36,590 --> 00:42:38,500
PROFESSOR: Then you're
saying that this

906
00:42:38,500 --> 00:42:42,122
is going to be bigger than--
so what do you do in that case?

907
00:42:42,122 --> 00:42:45,129
AUDIENCE: I mean you could just
ignore the vertices, I guess.

908
00:42:45,129 --> 00:42:46,670
PROFESSOR: OK so if
you have a ton of

909
00:42:46,670 --> 00:42:48,070
vertices and not
that many edges.

910
00:42:48,070 --> 00:42:51,790
So if you have edges
smaller-- sorry-- yeah,

911
00:42:51,790 --> 00:42:53,750
it's just smaller
than vertices, then

912
00:42:53,750 --> 00:42:56,070
some vertices have
to be unconnected.

913
00:42:56,070 --> 00:42:57,091
So what to do?

914
00:42:57,091 --> 00:42:58,382
AUDIENCE: You just ignore them.

915
00:42:58,382 --> 00:42:59,380
AUDIENCE: You're not going
to reach them anyway.

916
00:42:59,380 --> 00:43:00,930
PROFESSOR: How do
you ignore them?

917
00:43:00,930 --> 00:43:01,730
How do you ignore them in--

918
00:43:01,730 --> 00:43:02,750
AUDIENCE: Oh, I see your point.

919
00:43:02,750 --> 00:43:04,624
So you're never going
to get to them anyways.

920
00:43:04,624 --> 00:43:05,125
So--

921
00:43:05,125 --> 00:43:07,207
AUDIENCE: Well no, but
once you've started at one,

922
00:43:07,207 --> 00:43:08,888
you're going to have
to touch them all.

923
00:43:08,888 --> 00:43:11,260
They're all going
to be in the heap.

924
00:43:11,260 --> 00:43:12,240
Yeah.

925
00:43:12,240 --> 00:43:13,906
AUDIENCE: No, don't
put them in the heap

926
00:43:13,906 --> 00:43:16,620
until you look at them.

927
00:43:16,620 --> 00:43:19,210
AUDIENCE: Well, there's
a starting point, right?

928
00:43:19,210 --> 00:43:21,710
If you can't go anywhere in the
starting point, you're done.

929
00:43:21,710 --> 00:43:26,704
PROFESSOR: OK so you take the
starting point and you do--

930
00:43:26,704 --> 00:43:28,090
[INTERPOSING VOICES]

931
00:43:28,090 --> 00:43:31,740
PROFESSOR: Ok so I would take
the starting point and do a BFS

932
00:43:31,740 --> 00:43:33,850
and ignore everything else.

933
00:43:33,850 --> 00:43:37,355
And that is clearly order of
how many vertices and edges that

934
00:43:37,355 --> 00:43:38,590
are reachable from that.

935
00:43:38,590 --> 00:43:41,310
So that will give
you a nice graph.

936
00:43:41,310 --> 00:43:43,570
Either that, or if you
initialization your Dijkstra

937
00:43:43,570 --> 00:43:45,765
carefully enough, this
actually only reflects

938
00:43:45,765 --> 00:43:48,630
the reachable
edges and vertices.

939
00:43:48,630 --> 00:43:50,610
So you never have this.

940
00:43:50,610 --> 00:43:52,730
So either one works.

941
00:43:52,730 --> 00:43:55,235
In theory mode, it might be
better to say I'll do a BFS

942
00:43:55,235 --> 00:43:57,210
and reduce the graph.

943
00:43:57,210 --> 00:43:58,826
Because that's easier to argue.

944
00:43:58,826 --> 00:43:59,326
Yes.

945
00:43:59,326 --> 00:44:03,830
AUDIENCE: So, for
this, when you make all

946
00:44:03,830 --> 00:44:06,714
that-- I guess my
question is are we

947
00:44:06,714 --> 00:44:09,410
going to be asked to
talk about running times

948
00:44:09,410 --> 00:44:10,880
for different transformation?

949
00:44:10,880 --> 00:44:12,772
Or the algorithms?

950
00:44:12,772 --> 00:44:14,730
PROFESSOR: Well so at
the very least, last time

951
00:44:14,730 --> 00:44:17,430
we said, this is the problem.

952
00:44:17,430 --> 00:44:19,560
Give us an algorithm
with this running time.

953
00:44:19,560 --> 00:44:19,840
AUDIENCE: Right.

954
00:44:19,840 --> 00:44:21,548
PROFESSOR: So you have
to be able to know

955
00:44:21,548 --> 00:44:22,470
your transformations.

956
00:44:22,470 --> 00:44:24,636
So not the running for
computing the transformation,

957
00:44:24,636 --> 00:44:27,650
but you have to be able to know
how much bigger the graph gets

958
00:44:27,650 --> 00:44:29,811
and what that implies
for the running time.

959
00:44:29,811 --> 00:44:31,810
Turns out that usually
for these transformations

960
00:44:31,810 --> 00:44:33,310
you can compute
them in linear time.

961
00:44:33,310 --> 00:44:35,330
So the time for computing
a transformation

962
00:44:35,330 --> 00:44:37,190
is much smaller than
the time for running

963
00:44:37,190 --> 00:44:38,450
Dijkstra or Bellman-Ford.

964
00:44:38,450 --> 00:44:43,312
So that's why we don't
really go into that.

965
00:44:43,312 --> 00:44:45,440
This one is definitely
linear time.

966
00:44:45,440 --> 00:44:50,030
So what happens to the
graph in this case?

967
00:44:50,030 --> 00:44:52,503
Number of edges and
number of vertices.

968
00:44:52,503 --> 00:44:54,868
AUDIENCE: Number of
edges increases by V

969
00:44:54,868 --> 00:44:56,760
and number of vertices.

970
00:44:56,760 --> 00:44:59,130
So increase by V.

971
00:44:59,130 --> 00:45:02,692
PROFESSOR: That's
V, and times is 2.

972
00:45:02,692 --> 00:45:04,650
So you plug this into
Bellman-Ford or Dijkstra,

973
00:45:04,650 --> 00:45:06,290
you find out the
new running time.

974
00:45:06,290 --> 00:45:08,990
So every time you see what is
the size of your new graph,

975
00:45:08,990 --> 00:45:10,531
and you plug that
into the algorithm.

976
00:45:14,354 --> 00:45:16,020
OK so what is the
intuition behind this?

977
00:45:16,020 --> 00:45:17,370
Did everyone get this problem?

978
00:45:20,530 --> 00:45:22,530
So it's the old problem
that we're going through

979
00:45:22,530 --> 00:45:26,050
again and again, where you
have a graph that's 2D,

980
00:45:26,050 --> 00:45:28,977
and we want to compute
something that Dijkstra

981
00:45:28,977 --> 00:45:30,060
can't compute on it's own.

982
00:45:30,060 --> 00:45:32,680
Or that Bellman-Ford
can't compute on it's own.

983
00:45:32,680 --> 00:45:34,800
So in order to be able
to compute those things,

984
00:45:34,800 --> 00:45:37,350
we need to add additional
states to the graph.

985
00:45:37,350 --> 00:45:39,720
And the way we do that is
we make copies of the graph

986
00:45:39,720 --> 00:45:40,590
that we call layers.

987
00:45:40,590 --> 00:45:44,550
Because we're thinking that if
you take that 2D map, and you

988
00:45:44,550 --> 00:45:46,330
create copies of
it, you basically

989
00:45:46,330 --> 00:45:49,760
have a 3D graph where there
is the original graph.

990
00:45:52,560 --> 00:45:54,520
That's the science part
of the problem, the art

991
00:45:54,520 --> 00:45:57,150
part of solving the
problem is figuring out

992
00:45:57,150 --> 00:45:59,550
what those layers are
and how you connect them.

993
00:45:59,550 --> 00:46:01,880
Because by doing that you
can solve a ton of problems,

994
00:46:01,880 --> 00:46:05,310
as we have seen in this class.

995
00:46:05,310 --> 00:46:07,760
AUDIENCE: Is there a
possibility that you

996
00:46:07,760 --> 00:46:09,991
have a very large
number of layers needed

997
00:46:09,991 --> 00:46:14,598
to be permutations
of-- different choice

998
00:46:14,598 --> 00:46:17,286
to make instead of
just a red or a blue?

999
00:46:17,286 --> 00:46:18,190
PROFESSOR: Yep.

1000
00:46:18,190 --> 00:46:21,536
So is there a problem where
we had a ton of layers?

1001
00:46:21,536 --> 00:46:22,160
AUDIENCE: Mhmm.

1002
00:46:22,160 --> 00:46:23,430
PROFESSOR: Yep.

1003
00:46:23,430 --> 00:46:25,220
OK so two problems, right?

1004
00:46:25,220 --> 00:46:26,942
AUDIENCE: One for every second.

1005
00:46:26,942 --> 00:46:28,810
[LAUGHS]

1006
00:46:28,810 --> 00:46:31,876
PROFESSOR: And so these
are the layers, presumably.

1007
00:46:31,876 --> 00:46:33,750
We had two problems that
had a ton of layers.

1008
00:46:33,750 --> 00:46:35,208
One of them was
the highway problem

1009
00:46:35,208 --> 00:46:37,070
where you had timetables.

1010
00:46:37,070 --> 00:46:39,172
And there, the number
of layers was the number

1011
00:46:39,172 --> 00:46:40,880
of minutes in the day
you're considering.

1012
00:46:40,880 --> 00:46:44,010
So roughly, 1440.

1013
00:46:44,010 --> 00:46:47,410
And the other one was StarCraft
where the number of layers

1014
00:46:47,410 --> 00:46:50,400
was something ginormous, right?

1015
00:46:50,400 --> 00:46:53,106
So as long as that fits
the theory, that's fine.

1016
00:46:53,106 --> 00:46:55,730
As long as the number of layers
fits whatever the problem wants

1017
00:46:55,730 --> 00:46:57,894
you to compute, its OK.

1018
00:46:57,894 --> 00:47:01,303
AUDIENCE: So transforming
another layer,

1019
00:47:01,303 --> 00:47:07,150
is that like in V time because
copying the number of vertices?

1020
00:47:07,150 --> 00:47:09,946
PROFESSOR: Yeah,
so you should be

1021
00:47:09,946 --> 00:47:12,640
able to implement this in
order of V plus E, which

1022
00:47:12,640 --> 00:47:15,670
is what you need to output
the new graph description.

1023
00:47:15,670 --> 00:47:19,405
I claim that for this you can.

1024
00:47:19,405 --> 00:47:24,240
AUDIENCE: So for E prime, I
guess, how'd you get that?

1025
00:47:24,240 --> 00:47:26,375
PROFESSOR: So for
each vertex you

1026
00:47:26,375 --> 00:47:29,610
are going to create a
red copy and a blue copy.

1027
00:47:29,610 --> 00:47:31,080
2V, this is easy.

1028
00:47:31,080 --> 00:47:34,770
Now, the red edges stay in here,
the blue edges stay in here.

1029
00:47:34,770 --> 00:47:36,540
So nothing changed so far.

1030
00:47:36,540 --> 00:47:39,280
But each red vertex needs to
be connected to the blue vertex

1031
00:47:39,280 --> 00:47:42,320
by an edge of weight 5.

1032
00:47:42,320 --> 00:47:46,280
So that means that we're going
to copy over the original edges

1033
00:47:46,280 --> 00:47:49,656
and we're going to add the
edges that connect the vertices.

1034
00:47:49,656 --> 00:47:51,520
AUDIENCE: [INAUDIBLE].

1035
00:47:51,520 --> 00:47:53,770
PROFESSOR: So we're not going
to ask about the running

1036
00:47:53,770 --> 00:47:55,450
time of the
transformations in general.

1037
00:47:55,450 --> 00:47:57,990
Because we assume that they
can be done in linear time.

1038
00:47:57,990 --> 00:48:01,150
But you need to at least have a
sense of whether your graph is

1039
00:48:01,150 --> 00:48:02,880
going to double,
whether it's going

1040
00:48:02,880 --> 00:48:06,060
to increase exponentially, or
what's going to happen to it.

1041
00:48:06,060 --> 00:48:08,100
So let's make a small
tweak to this problem.

1042
00:48:08,100 --> 00:48:10,150
Suppose that the
instead of having this,

1043
00:48:10,150 --> 00:48:13,460
I can go from red to
blue, but once I've

1044
00:48:13,460 --> 00:48:16,360
gone from red to
blue, I can't go back.

1045
00:48:16,360 --> 00:48:20,850
So I can start out either red
or blue, I can go red to blue,

1046
00:48:20,850 --> 00:48:23,794
but once I'm in blue I
can't go back to red.

1047
00:48:23,794 --> 00:48:25,522
AUDIENCE: Make those
directed edges.

1048
00:48:25,522 --> 00:48:27,390
PROFESSOR: From where to where?

1049
00:48:27,390 --> 00:48:28,785
AUDIENCE: Red to blue.

1050
00:48:32,510 --> 00:48:36,600
PROFESSOR: OK so this is
how I express constraints,

1051
00:48:36,600 --> 00:48:39,570
in constraints among layers.

1052
00:48:39,570 --> 00:48:42,540
Before, we had two
layers, red and blue.

1053
00:48:42,540 --> 00:48:45,216
And we connected them by
an edge of weight 5, which

1054
00:48:45,216 --> 00:48:47,090
says you can go from
red world to blue world,

1055
00:48:47,090 --> 00:48:50,460
and back, all you
have to do is pay 5.

1056
00:48:50,460 --> 00:48:53,510
If we have directed edges,
then this is a constraint.

1057
00:48:53,510 --> 00:48:56,690
If the constraint
said there's no cost,

1058
00:48:56,690 --> 00:48:58,480
but you can only go
from red to blue,

1059
00:48:58,480 --> 00:49:00,440
you still have to do two layers.

1060
00:49:00,440 --> 00:49:02,790
And keep track of
which layer you're in.

1061
00:49:02,790 --> 00:49:05,060
But then your edge
of the weight 0.

1062
00:49:05,060 --> 00:49:07,440
Since we have to pay,
the edge is weight 5.

1063
00:49:10,590 --> 00:49:12,660
So we can use layers to
express additional costs,

1064
00:49:12,660 --> 00:49:13,960
or just to express constraints.

1065
00:49:17,150 --> 00:49:18,395
Does this make sense?

1066
00:49:18,395 --> 00:49:20,770
AUDIENCE: Does this include
back edges, or forward edges,

1067
00:49:20,770 --> 00:49:21,702
as well?

1068
00:49:21,702 --> 00:49:22,910
PROFESSOR: Let's get to that.

1069
00:49:22,910 --> 00:49:25,320
OK, so does this
make sense so far?

1070
00:49:25,320 --> 00:49:28,330
One more question, what shortest
path do I want to compute here?

1071
00:49:31,054 --> 00:49:34,640
To make sure that
you guys got it.

1072
00:49:34,640 --> 00:49:37,500
AUDIENCE: Oh, you have to
do two of them maybe, four--

1073
00:49:37,500 --> 00:49:39,100
PROFESSOR: OK, I
like four of them.

1074
00:49:39,100 --> 00:49:46,290
So I have to do SR, TR,
SR, TB, SB, TR, SB, TB.

1075
00:49:46,290 --> 00:49:48,066
Fortunately, our
algorithms give us

1076
00:49:48,066 --> 00:49:49,440
the shortest path
from one source

1077
00:49:49,440 --> 00:49:50,648
to all of the other vertices.

1078
00:49:50,648 --> 00:49:54,080
So I'd only have to run
Dijkstra or Bellman-Ford twice.

1079
00:49:54,080 --> 00:49:56,250
What if I want to run
the algorithm once?

1080
00:49:56,250 --> 00:49:57,123
What do I do?

1081
00:49:57,123 --> 00:50:00,020
AUDIENCE: You put
vertices on the edges.

1082
00:50:00,020 --> 00:50:03,530
PROFESSOR: OK so supernode
as a source, supernode

1083
00:50:03,530 --> 00:50:05,550
as a destination.

1084
00:50:05,550 --> 00:50:07,480
And I connect them to what?

1085
00:50:07,480 --> 00:50:09,048
AUDIENCE: Zero weight.

1086
00:50:09,048 --> 00:50:09,964
PROFESSOR: To?

1087
00:50:09,964 --> 00:50:10,880
AUDIENCE: The two.

1088
00:50:10,880 --> 00:50:11,800
AUDIENCE: Final V.

1089
00:50:11,800 --> 00:50:13,700
PROFESSOR: OK, very good.

1090
00:50:13,700 --> 00:50:15,785
And?

1091
00:50:15,785 --> 00:50:16,410
AUDIENCE: Same.

1092
00:50:20,115 --> 00:50:21,490
PROFESSOR: So what
happened here?

1093
00:50:21,490 --> 00:50:24,460
This says you can go anywhere.

1094
00:50:24,460 --> 00:50:26,494
And this says you can
come back from anywhere.

1095
00:50:26,494 --> 00:50:28,864
AUDIENCE: Bridge
between the worlds.

1096
00:50:28,864 --> 00:50:29,812
PROFESSOR: Yep.

1097
00:50:29,812 --> 00:50:32,190
[AUDIENCE LAUGHING]

1098
00:50:32,190 --> 00:50:35,550
PROFESSOR: So do we want this
to be directed or undirected,

1099
00:50:35,550 --> 00:50:37,384
by the way, does it matter?

1100
00:50:37,384 --> 00:50:38,175
AUDIENCE: Directed.

1101
00:50:38,175 --> 00:50:39,231
You can only go one way--

1102
00:50:39,231 --> 00:50:41,730
AUDIENCE: Oh, depends on which
problem you're talking about.

1103
00:50:45,592 --> 00:50:47,300
PROFESSOR: I think if
they're undirected,

1104
00:50:47,300 --> 00:50:49,508
it shouldn't matter too
much, because there shouldn't

1105
00:50:49,508 --> 00:50:54,630
be a path where you go back to
the source and then you switch.

1106
00:50:54,630 --> 00:50:56,510
But this would be
if you're on a quiz

1107
00:50:56,510 --> 00:50:57,884
and you don't want
to think about

1108
00:50:57,884 --> 00:51:00,488
that, I'd make them directed
just to be on the safe side.

1109
00:51:00,488 --> 00:51:01,229
AUDIENCE: What were
the four shortest

1110
00:51:01,229 --> 00:51:02,958
paths you said we
needed to calculate

1111
00:51:02,958 --> 00:51:05,430
for the original problem,
before you added the supernodes?

1112
00:51:05,430 --> 00:51:07,760
PROFESSOR: So we don't
know whether we start out

1113
00:51:07,760 --> 00:51:08,714
with a red edge--

1114
00:51:08,714 --> 00:51:09,380
AUDIENCE: Right.

1115
00:51:09,380 --> 00:51:10,629
PROFESSOR: --with a blue edge.

1116
00:51:10,629 --> 00:51:13,911
We don't know whether we end up
with a red edge or a blue edge.

1117
00:51:13,911 --> 00:51:15,910
AUDIENCE: Stay in red
world, stay in blue world,

1118
00:51:15,910 --> 00:51:17,320
red to blue and blue to red.

1119
00:51:17,320 --> 00:51:21,060
PROFESSOR: Yeah OK, cool.

1120
00:51:21,060 --> 00:51:24,332
So let's talk about BFS and
DFS very briefly, I guess.

1121
00:51:27,110 --> 00:51:28,145
So BFS and DFS.

1122
00:51:31,200 --> 00:51:32,450
What does BFS give us?

1123
00:51:32,450 --> 00:51:35,770
Why is it useful?

1124
00:51:35,770 --> 00:51:37,700
AUDIENCE: Shortest paths.

1125
00:51:37,700 --> 00:51:39,250
PROFESSOR: Shortest
paths in terms

1126
00:51:39,250 --> 00:51:40,410
of number of edges, right?

1127
00:51:40,410 --> 00:51:42,710
No weight on the engines.

1128
00:51:42,710 --> 00:51:52,370
So shortest paths using
the number of edges.

1129
00:51:52,370 --> 00:51:55,770
What does DFS give us?

1130
00:51:55,770 --> 00:51:57,802
AUDIENCE: Topological start.

1131
00:51:57,802 --> 00:51:59,760
PROFESSOR: Hey, man, you
had this on the P set,

1132
00:51:59,760 --> 00:52:00,885
I wouldn't call it nothing.

1133
00:52:05,090 --> 00:52:07,800
OK, how does BFS
look at the graph?

1134
00:52:07,800 --> 00:52:10,810
How does it partition the graph?

1135
00:52:10,810 --> 00:52:12,080
Nodes are grouped into--

1136
00:52:12,080 --> 00:52:12,850
AUDIENCE: Levels.

1137
00:52:12,850 --> 00:52:13,522
PROFESSOR: Levels.

1138
00:52:13,522 --> 00:52:15,355
AUDIENCE: The light
distance and the source.

1139
00:52:17,267 --> 00:52:18,850
PROFESSOR: So you
start with a source,

1140
00:52:18,850 --> 00:52:21,310
and then all the nodes
that are one edge

1141
00:52:21,310 --> 00:52:23,910
away are at level one,
all the nodes that

1142
00:52:23,910 --> 00:52:28,960
are two edges away are level
two, so on and so forth.

1143
00:52:28,960 --> 00:52:30,350
What does DFS give us?

1144
00:52:34,470 --> 00:52:34,970
A mess?

1145
00:52:34,970 --> 00:52:36,760
A mess, right?

1146
00:52:36,760 --> 00:52:39,890
So it gives us edge types.

1147
00:52:43,650 --> 00:52:46,775
And it gives us exit times.

1148
00:52:51,020 --> 00:52:53,415
Exit times are useful for
topological sort, right?

1149
00:52:53,415 --> 00:52:53,915
Yes.

1150
00:52:53,915 --> 00:52:56,770
AUDIENCE: It also
gives us a tree.

1151
00:52:56,770 --> 00:53:00,360
PROFESSOR: OK, does
BFS give us a tree?

1152
00:53:00,360 --> 00:53:01,310
AUDIENCE: Yes.

1153
00:53:01,310 --> 00:53:03,020
AUDIENCE: Yes, they
both give us trees.

1154
00:53:03,020 --> 00:53:04,936
PROFESSOR: So both of
them will give us trees.

1155
00:53:10,380 --> 00:53:11,910
So what are those trees?

1156
00:53:11,910 --> 00:53:16,450
For each node, BFS or
DFS discovered that node

1157
00:53:16,450 --> 00:53:19,880
by going from some parent
node across an edge.

1158
00:53:19,880 --> 00:53:23,610
So that edge belongs to
the BFS or to the DFS tree.

1159
00:53:23,610 --> 00:53:25,170
How do we compute those trees?

1160
00:53:27,948 --> 00:53:29,502
AUDIENCE: Just do it?

1161
00:53:29,502 --> 00:53:31,210
PROFESSOR: So how do
you, in BFS and DFS,

1162
00:53:31,210 --> 00:53:32,697
what do you compute to--

1163
00:53:32,697 --> 00:53:33,780
AUDIENCE: Parent pointers?

1164
00:53:33,780 --> 00:53:35,160
PROFESSOR: Parent pointers.

1165
00:53:35,160 --> 00:53:37,260
So that's what you use
to keep track of them.

1166
00:53:37,260 --> 00:53:38,760
Sorry the question
is bad but that's

1167
00:53:38,760 --> 00:53:42,080
what I want you
to get out of it.

1168
00:53:42,080 --> 00:53:45,090
Parent pointers.

1169
00:53:45,090 --> 00:53:47,620
So DFS and BFS will both
compute trees of the graph.

1170
00:53:47,620 --> 00:53:51,852
Are they the same trees
or different trees?

1171
00:53:51,852 --> 00:53:53,060
AUDIENCE: Probably different.

1172
00:53:53,060 --> 00:53:53,950
PROFESSOR: Different
trees, right?

1173
00:53:53,950 --> 00:53:54,616
Let's take this.

1174
00:53:58,160 --> 00:54:02,470
S, A, B, right?

1175
00:54:02,470 --> 00:54:07,072
BFS will give us this,
DFS will give us this.

1176
00:54:11,140 --> 00:54:15,010
OK, what edge types
do we have in DFS?

1177
00:54:15,010 --> 00:54:16,320
AUDIENCE: Forward edges.

1178
00:54:16,320 --> 00:54:17,530
PROFESSOR: OK, forward.

1179
00:54:17,530 --> 00:54:18,030
Cool.

1180
00:54:18,030 --> 00:54:19,290
What are they?

1181
00:54:19,290 --> 00:54:21,675
AUDIENCE: Those are
just from parent

1182
00:54:21,675 --> 00:54:26,210
to children that
are being presented.

1183
00:54:26,210 --> 00:54:27,430
PROFESSOR: OK.

1184
00:54:27,430 --> 00:54:29,530
AUDIENCE: So, I don't
know how else to put it.

1185
00:54:29,530 --> 00:54:32,460
PROFESSOR: OK so you should
have this on your cheat sheet,

1186
00:54:32,460 --> 00:54:34,104
right?

1187
00:54:34,104 --> 00:54:36,020
If nobody knows the
answer, you should have it

1188
00:54:36,020 --> 00:54:37,620
on your cheat sheet.

1189
00:54:37,620 --> 00:54:43,000
So tree edges, the ones that
show up in the DFS tree.

1190
00:54:43,000 --> 00:54:44,572
Backwards edges,

1191
00:54:44,572 --> 00:54:46,300
AUDIENCE: Cross edges.

1192
00:54:46,300 --> 00:54:47,785
PROFESSOR: Cross edges.

1193
00:54:47,785 --> 00:54:50,330
AUDIENCE: Forward edges.

1194
00:54:50,330 --> 00:54:53,300
PROFESSOR: And forward edges.

1195
00:54:53,300 --> 00:54:56,790
If your graph is undirected,
what types of edges

1196
00:54:56,790 --> 00:54:59,690
do you not have?

1197
00:54:59,690 --> 00:55:01,825
AUDIENCE: Forward edges.

1198
00:55:01,825 --> 00:55:04,740
Forward edges and
backward edges.

1199
00:55:04,740 --> 00:55:06,374
Or no, forward edges.

1200
00:55:10,712 --> 00:55:12,170
PROFESSOR: OK.

1201
00:55:12,170 --> 00:55:14,910
So this is a lecture
notes, and we're

1202
00:55:14,910 --> 00:55:17,680
going to go over them tonight.

1203
00:55:17,680 --> 00:55:20,199
But if you can't make it,
they're in lecture notes.

1204
00:55:20,199 --> 00:55:21,990
So you should have this
on your cheat sheet

1205
00:55:21,990 --> 00:55:24,040
if you don't know them.

1206
00:55:24,040 --> 00:55:24,540
Right?

1207
00:55:24,540 --> 00:55:26,748
Nobody answered today, so
if you guys don't have them

1208
00:55:26,748 --> 00:55:29,700
on your cheat sheet
I will be upset.

1209
00:55:29,700 --> 00:55:32,125
OK.

1210
00:55:32,125 --> 00:55:34,550
Cool.

1211
00:55:34,550 --> 00:55:36,591
More questions?

1212
00:55:36,591 --> 00:55:38,591
AUDIENCE: Edges do not
exist in BFS, edge types?

1213
00:55:41,550 --> 00:55:44,160
PROFESSOR: DFS uses-- so
out of BFS we get levels,

1214
00:55:44,160 --> 00:55:48,390
and out DFS we get
those edge types.

1215
00:55:48,390 --> 00:55:54,360
AUDIENCE: Is there--
about the edge types?

1216
00:55:54,360 --> 00:55:57,400
PROFESSOR: Some algorithms
use them in their proofs.

1217
00:55:57,400 --> 00:55:59,610
So the proof for
topological sort

1218
00:55:59,610 --> 00:56:02,030
uses the fact that in
directed acyclic graphs,

1219
00:56:02,030 --> 00:56:04,360
you won't have some
types of edges.

1220
00:56:04,360 --> 00:56:06,120
For the other types
of edges it argues

1221
00:56:06,120 --> 00:56:08,350
that the order that you
get from topological sort

1222
00:56:08,350 --> 00:56:09,810
is the right order.

1223
00:56:09,810 --> 00:56:12,320
So it's mostly theoretical
but since we taught you

1224
00:56:12,320 --> 00:56:15,295
about edge types, we
might ask you about them.

1225
00:56:15,295 --> 00:56:15,794
Yes.

1226
00:56:15,794 --> 00:56:18,280
AUDIENCE: So does
DFS actually-- is

1227
00:56:18,280 --> 00:56:21,050
it able to distinguish
between back-crossing over,

1228
00:56:21,050 --> 00:56:24,590
or is it only-- does it only
see that the node has already

1229
00:56:24,590 --> 00:56:26,930
been visited by something?

1230
00:56:26,930 --> 00:56:32,559
PROFESSOR: So how would you make
it distinguish between them?

1231
00:56:32,559 --> 00:56:33,350
You have to, right?

1232
00:56:33,350 --> 00:56:35,190
Because otherwise, why
are we studying them?

1233
00:56:35,190 --> 00:56:37,780
There has to be a way to
distinguish between them.

1234
00:56:37,780 --> 00:56:40,604
AUDIENCE: Because the
DFS does recursive calls

1235
00:56:40,604 --> 00:56:43,300
and you know where the back is.

1236
00:56:43,300 --> 00:56:45,690
PROFESSOR: So let's
do a DFS tree quickly.

1237
00:56:51,370 --> 00:56:53,630
So suppose we went like this.

1238
00:56:53,630 --> 00:57:02,140
And then like this,
this, and then this.

1239
00:57:02,140 --> 00:57:06,110
Sorry, I'm trying really hard to
make up an example on the spot

1240
00:57:06,110 --> 00:57:09,028
such that I won't
discount myself.

1241
00:57:17,400 --> 00:57:19,122
How do I get that, do you know?

1242
00:57:19,122 --> 00:57:22,008
AUDIENCE: Maybe an edge
from the second one

1243
00:57:22,008 --> 00:57:25,870
to the last one on the right.

1244
00:57:25,870 --> 00:57:26,670
PROFESSOR: Here?

1245
00:57:26,670 --> 00:57:29,818
AUDIENCE: Yeah and this
will be the arrowhead,

1246
00:57:29,818 --> 00:57:33,467
and you can start
from the second one.

1247
00:57:33,467 --> 00:57:34,300
PROFESSOR: This one?

1248
00:57:34,300 --> 00:57:35,734
AUDIENCE: Yeah.

1249
00:57:35,734 --> 00:57:38,819
Wouldn't that be a forward edge?

1250
00:57:38,819 --> 00:57:39,360
PROFESSOR: OK

1251
00:57:39,360 --> 00:57:39,830
AUDIENCE: Because
it's downward--

1252
00:57:39,830 --> 00:57:41,290
PROFESSOR: Yeah, I like that.

1253
00:57:41,290 --> 00:57:45,010
OK, I like that.

1254
00:57:45,010 --> 00:57:51,770
OK so S, A, B, C,
D, E. So assume

1255
00:57:51,770 --> 00:57:54,930
this is the order that they
listed in adjacency list.

1256
00:57:54,930 --> 00:57:56,940
So let's label all of the edges.

1257
00:57:56,940 --> 00:57:59,236
SA and SB are what?

1258
00:57:59,236 --> 00:58:02,230
AUDIENCE: Forward edges.

1259
00:58:02,230 --> 00:58:06,060
PROFESSOR: OK AC?

1260
00:58:06,060 --> 00:58:08,060
AUDIENCE: Forward.

1261
00:58:08,060 --> 00:58:09,060
PROFESSOR: CD?

1262
00:58:09,060 --> 00:58:10,840
AUDIENCE: Forward.

1263
00:58:10,840 --> 00:58:12,147
PROFESSOR: CE?

1264
00:58:12,147 --> 00:58:13,400
AUDIENCE: Forward.

1265
00:58:13,400 --> 00:58:14,325
PROFESSOR: OK, AE?

1266
00:58:17,035 --> 00:58:18,484
AUDIENCE: Aren't
those tree edges?

1267
00:58:18,484 --> 00:58:21,212
AUDIENCE: That's a forward edge.

1268
00:58:21,212 --> 00:58:22,670
PROFESSOR: That
took a while, guys.

1269
00:58:22,670 --> 00:58:24,602
That took a while.

1270
00:58:24,602 --> 00:58:26,051
AUDIENCE: It goes forward.

1271
00:58:26,051 --> 00:58:28,224
AUDIENCE: Well no,
forward means like you're

1272
00:58:28,224 --> 00:58:29,951
skipping a generation.

1273
00:58:29,951 --> 00:58:31,325
PROFESSOR: OK so
the three edges.

1274
00:58:34,476 --> 00:58:35,760
AUDIENCE: [INAUDIBLE]

1275
00:58:35,760 --> 00:58:37,510
PROFESSOR: So the three
edges are the ones

1276
00:58:37,510 --> 00:58:39,320
that DFS uses to go forward.

1277
00:58:39,320 --> 00:58:41,190
So they're the ones
that map the DFS calls.

1278
00:58:44,100 --> 00:58:45,940
So how do we express that?

1279
00:58:45,940 --> 00:58:52,264
So suppose you're at some node
u, and you have an edge, uv.

1280
00:58:52,264 --> 00:58:54,434
AUDIENCE: The parent of v is u.

1281
00:58:54,434 --> 00:58:56,100
PROFESSOR: So you're
at here, right now.

1282
00:58:59,640 --> 00:59:04,570
OK, so v.parent has to be u.

1283
00:59:04,570 --> 00:59:05,940
And?

1284
00:59:05,940 --> 00:59:08,080
v is?

1285
00:59:08,080 --> 00:59:09,702
Is it visited or not?

1286
00:59:09,702 --> 00:59:11,130
AUDIENCE: It's a new vertex.

1287
00:59:11,130 --> 00:59:13,990
AUDIENCE: It's not.

1288
00:59:13,990 --> 00:59:15,450
PROFESSOR: So v is
not visited yet.

1289
00:59:21,160 --> 00:59:23,490
So this is a tree edge.

1290
00:59:23,490 --> 00:59:25,837
Now let's compare this
to a forward edge.

1291
00:59:32,180 --> 00:59:34,059
So what happens
to a forward edge?

1292
00:59:34,059 --> 00:59:36,933
AUDIENCE: Parents of children--

1293
00:59:36,933 --> 00:59:39,345
AUDIENCE: Parent ancestors--

1294
00:59:39,345 --> 00:59:40,970
PROFESSOR: So you're
at here right now.

1295
00:59:44,690 --> 00:59:47,260
And you're looking
at the edge uv.

1296
00:59:47,260 --> 00:59:49,240
Right? so we're
looking at the edge uv.

1297
00:59:49,240 --> 00:59:51,470
uv would point
downward in the tree.

1298
00:59:51,470 --> 00:59:53,801
What's true about--

1299
00:59:53,801 --> 00:59:56,557
AUDIENCE: v.parent is u.

1300
00:59:56,557 --> 00:59:57,140
AUDIENCE: No--

1301
00:59:57,140 --> 00:59:58,223
AUDIENCE: Not necessarily.

1302
00:59:58,223 --> 01:00:00,600
PROFESSOR: OK.

1303
01:00:00,600 --> 01:00:04,560
So v.parent is u.

1304
01:00:04,560 --> 01:00:07,220
So it can be the parent,
or the grandparent,

1305
01:00:07,220 --> 01:00:10,350
or-- so u has to be
somewhere up the tree.

1306
01:00:10,350 --> 01:00:11,020
Right?

1307
01:00:11,020 --> 01:00:14,654
So I can have a ton
of dot parents here.

1308
01:00:14,654 --> 01:00:17,070
AUDIENCE: There has to be more
than one dot parent, right?

1309
01:00:17,070 --> 01:00:19,278
Because otherwise
we get tree edge.

1310
01:00:19,278 --> 01:00:21,760
It can't just be one parent.

1311
01:00:21,760 --> 01:00:22,644
PROFESSOR: OK.

1312
01:00:22,644 --> 01:00:25,097
AUDIENCE: So you'd have to
recurse up more than once.

1313
01:00:25,097 --> 01:00:25,722
PROFESSOR: Yep.

1314
01:00:25,722 --> 01:00:29,560
So for a node you go up until
you find u, and if you found u,

1315
01:00:29,560 --> 01:00:33,770
then it's a forward
edge, otherwise,

1316
01:00:33,770 --> 01:00:35,967
if you find the root of
the tree then you give up.

1317
01:00:35,967 --> 01:00:37,550
AUDIENCE: Well then
it's a cross edge.

1318
01:00:40,826 --> 01:00:44,230
If you go up and you get
to the root but-- oh--

1319
01:00:44,230 --> 01:00:46,626
PROFESSOR: Yeah, so if
you got to the root,

1320
01:00:46,626 --> 01:00:47,750
it is not the forward edge.

1321
01:00:47,750 --> 01:00:49,880
Right now we're looking
at what does it mean.

1322
01:00:49,880 --> 01:00:52,745
So you have to be somewhere
around the ancestor chain of u,

1323
01:00:52,745 --> 01:00:56,070
so if you keep following v's
parents, you have to see u.

1324
01:00:56,070 --> 01:00:59,170
If not, it's not
the forward edge.

1325
01:00:59,170 --> 01:01:00,010
OK.

1326
01:01:00,010 --> 01:01:02,070
And when you see
it, did you visit it

1327
01:01:02,070 --> 01:01:03,807
or did you not visit it?

1328
01:01:03,807 --> 01:01:04,890
AUDIENCE: Already visited.

1329
01:01:10,650 --> 01:01:11,740
PROFESSOR: OK.

1330
01:01:11,740 --> 01:01:15,680
So now we have two more edge
types, we have back edges,

1331
01:01:15,680 --> 01:01:17,040
and we have cross edges.

1332
01:01:17,040 --> 01:01:18,740
Right?

1333
01:01:18,740 --> 01:01:21,608
Show me a back edge here.

1334
01:01:21,608 --> 01:01:22,940
AUDIENCE: DA.

1335
01:01:22,940 --> 01:01:24,720
PROFESSOR: DA is a back edge.

1336
01:01:24,720 --> 01:01:26,570
OK back edges are
also reasonably easy.

1337
01:01:26,570 --> 01:01:27,320
So let's do those.

1338
01:01:30,610 --> 01:01:31,630
What's a back edge.

1339
01:01:31,630 --> 01:01:34,525
So I am at u now.

1340
01:01:34,525 --> 01:01:37,700
I'm looking at the edge uv.

1341
01:01:37,700 --> 01:01:39,028
AUDIENCE: It's a cycle.

1342
01:01:39,028 --> 01:01:40,710
AUDIENCE: Node to ancestor.

1343
01:01:40,710 --> 01:01:41,418
PROFESSOR: Sorry?

1344
01:01:41,418 --> 01:01:43,330
AUDIENCE: It's a
node to an ancestor.

1345
01:01:43,330 --> 01:01:46,295
PROFESSOR: OK so who
is who's ancestor?

1346
01:01:46,295 --> 01:01:47,744
AUDIENCE: A is D's ancestor.

1347
01:01:47,744 --> 01:01:48,910
AUDIENCE: V is u's ancestor.

1348
01:01:48,910 --> 01:01:51,680
PROFESSOR: OK.

1349
01:01:51,680 --> 01:02:00,000
So if I keep going,
u.parent many times over,

1350
01:02:00,000 --> 01:02:05,540
I should eventually
see v. Uhh-- yeah.

1351
01:02:05,540 --> 01:02:09,120
So the difference
between this and this

1352
01:02:09,120 --> 01:02:11,324
is who is who's parent?

1353
01:02:11,324 --> 01:02:16,194
AUDIENCE: Wait, wouldn't
u be a parent of v?

1354
01:02:16,194 --> 01:02:20,000
Because we didn't go from
A to C to D and then at D

1355
01:02:20,000 --> 01:02:23,302
I realize that I
can go to A, but I

1356
01:02:23,302 --> 01:02:27,090
won't go because
it's already visited?

1357
01:02:27,090 --> 01:02:28,970
PROFESSOR: Wait.

1358
01:02:28,970 --> 01:02:30,380
AUDIENCE: But assuming that--

1359
01:02:30,380 --> 01:02:31,790
AUDIENCE: A is an ancestor of D.

1360
01:02:31,790 --> 01:02:33,290
PROFESSOR: So A is
an ancestor of D

1361
01:02:33,290 --> 01:02:35,350
because by the time
I got to D, I've

1362
01:02:35,350 --> 01:02:42,419
already set the parent
pointers for C and D.

1363
01:02:42,419 --> 01:02:43,385
AUDIENCE: Yes.

1364
01:02:43,385 --> 01:02:45,317
But-- wait.

1365
01:02:45,317 --> 01:02:50,114
u.parent is v. Then u is
A here and v is D here--

1366
01:02:50,114 --> 01:02:51,250
AUDIENCE: No u is D here.

1367
01:02:51,250 --> 01:02:53,540
PROFESSOR: I'm at u
right now, so I'm at D.

1368
01:02:53,540 --> 01:02:57,540
And I'm looking at the edge uv.

1369
01:02:57,540 --> 01:02:59,340
AUDIENCE: Do we have
a cross edge in here?

1370
01:02:59,340 --> 01:03:01,220
PROFESSOR: Do we have
a cross edge in here.

1371
01:03:01,220 --> 01:03:01,950
Good question.

1372
01:03:01,950 --> 01:03:03,838
What is the cross edge?

1373
01:03:03,838 --> 01:03:06,680
AUDIENCE: C?

1374
01:03:06,680 --> 01:03:08,430
PROFESSOR: OK so this
guy is a cross edge.

1375
01:03:08,430 --> 01:03:10,346
So what is the difference
between a cross edge

1376
01:03:10,346 --> 01:03:12,790
and a forward edge?

1377
01:03:12,790 --> 01:03:16,015
[CLASS MURMURS]

1378
01:03:16,015 --> 01:03:18,440
AUDIENCE: It's not an ancestor.

1379
01:03:18,440 --> 01:03:22,150
AUDIENCE: Ancester goes up to S.

1380
01:03:22,150 --> 01:03:26,750
PROFESSOR: So a cross edge is
very close to a forward edge

1381
01:03:26,750 --> 01:03:38,670
except u is not an ancestor
of v. v in the DFS tree.

1382
01:03:42,750 --> 01:03:44,434
OK?

1383
01:03:44,434 --> 01:03:47,386
AUDIENCE: [INAUDIBLE]

1384
01:03:47,386 --> 01:03:49,846
PROFESSOR: Yep.

1385
01:03:49,846 --> 01:03:50,840
Yep.

1386
01:03:50,840 --> 01:03:55,998
So the only common ancestor
between B and C is the source.

1387
01:04:04,720 --> 01:04:07,850
OK so please get these
conditions in the notes.

1388
01:04:07,850 --> 01:04:11,145
So as a summary, if it's
an edge that DFS uses,

1389
01:04:11,145 --> 01:04:12,970
it's a tree edge.

1390
01:04:12,970 --> 01:04:16,960
If it's going against the way
of the DFS, it's a back edge.

1391
01:04:16,960 --> 01:04:20,810
If it fast forwards
in DFS, so instead

1392
01:04:20,810 --> 01:04:24,320
of going one level down, it
does multiple levels down,

1393
01:04:24,320 --> 01:04:25,980
then it's a forward edge.

1394
01:04:25,980 --> 01:04:30,520
And otherwise, it's weird
edge so it's a cross edge.

1395
01:04:30,520 --> 01:04:31,524
One way to look at it.

1396
01:04:31,524 --> 01:04:32,024
Yeah.

1397
01:04:32,024 --> 01:04:33,607
AUDIENCE: Is this
only true because we

1398
01:04:33,607 --> 01:04:35,880
were going alphabetically?

1399
01:04:35,880 --> 01:04:37,780
PROFESSOR: Yeah, this
depends on the order

1400
01:04:37,780 --> 01:04:40,430
in which the nodes are
listed in the adjacency list.

1401
01:04:40,430 --> 01:04:43,310
AUDIENCE: So it could've
been pretty different

1402
01:04:43,310 --> 01:04:46,190
if we went from A to E?

1403
01:04:46,190 --> 01:04:47,800
PROFESSOR: Oh Yeah, yeah.

1404
01:04:47,800 --> 01:04:48,724
Yep.

1405
01:04:48,724 --> 01:04:50,682
AUDIENCE: I'm trying to
solve this another way.

1406
01:04:50,682 --> 01:04:54,180
The graph information--

1407
01:04:54,180 --> 01:04:54,973
PROFESSOR: Yeah

1408
01:04:54,973 --> 01:04:57,751
AUDIENCE: --would you say that's
moderate or an easy example?

1409
01:04:57,751 --> 01:05:00,530
The transformation.

1410
01:05:00,530 --> 01:05:03,290
PROFESSOR: This is
on the easy side.

1411
01:05:03,290 --> 01:05:05,950
So look at the problems
that we had so far.

1412
01:05:05,950 --> 01:05:08,160
There are more problems
in this review packet.

1413
01:05:08,160 --> 01:05:11,640
There are like six,
seven, eight problems.

1414
01:05:11,640 --> 01:05:14,310
It's going to be online.

1415
01:05:14,310 --> 01:05:16,060
It's the notes for
this recitation.

1416
01:05:16,060 --> 01:05:17,100
Sorry, bad name.

1417
01:05:17,100 --> 01:05:19,550
So the notes for
this recitation.

1418
01:05:19,550 --> 01:05:21,190
This is on the easy side.

1419
01:05:21,190 --> 01:05:24,010
The StarCraft problem, last time
is on the insanely difficult

1420
01:05:24,010 --> 01:05:25,810
side, so that's not
going to happen.

1421
01:05:25,810 --> 01:05:27,910
AUDIENCE: I so
don't even get that.

1422
01:05:27,910 --> 01:05:30,060
PROFESSOR: That's
on the hard side.