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PROFESSOR: All right, today we
continue our exciting adventure
into dynamic programming.
Are you excited?
I'm excited, super excited.
Dynamic programming,
as you recall way back
before Thanksgiving, is a super
exciting powerful technique
to design algorithms, especially
to solve optimization problems
where you want to maximize
or minimize something.
Last time, we saw how
two algorithms we already
knew-- namely, how to compute
the nth Fibonacci number
and how to compute shortest
paths via Bellman-Ford--
are really dynamic
programs in disguise.
And indeed for, at
least for Bellman-Ford,
that's how they
were invented, was
to apply a general
technique which
we're going to see today
in full generality,
more or less-- most of
this is generality--
in five easy steps.
And we're going to see
that technique applied
to two new problems
which are much more
interesting than the ones
we've already solved-- namely,
how to make your text
look nice in a paragraph,
where to break the lines.
That's text justification.
And how to win and make
loads of money at blackjack.
So lots of practical
stuff here, and we're
going to see one
new technique for
general dynamic programming.
These are some things
I wrote last time.
Actually, one of them I
didn't write last time.
In general, you can think
of dynamic programming
as a carefully executed
brute force search.
So in some sense,
your algorithm is
going to be trying
all the possibilities,
but somehow avoiding
the fact that there
are exponentially many of them.
By thinking of it
in a clever way,
you can reduce the
exponential search space
down to a polynomial one,
even though you're still
not being very intelligent
you're still blindly trying
all possibilities.
So that's the brute force part.
In more detail, the
three main techniques
in dynamic programming are
the idea of guessing, the idea
that, oh, I want to find the
best way to solve a problem.
Let's pick out some
feature of the solution
that I want to know.
I don't know it, so
I'll guess the answer--
meaning I'll try all the
possibilities for that choice
and take the best one.
So guessing is really central
to dynamic programming.
Then we also use a
recursion, some way
to express the solution to our
problem in terms of solutions
to sub-problems.
So it's usually very
easy to get a recursion
for a lot of problems
as long as they
have some kind of substructure.
Like shortest paths, we had that
some paths of shortest paths
were also shortest
paths, so that was handy.
Usually the recursion by
itself is exponential time,
like even with
Fibonacci numbers.
But we add in this
technique of memoization,
which is just once we compute an
answer we've stored in a lookup
table, if we ever
need that answer again
we reuse it instead
of recomputing it.
So we store it.
We write down in our memo
pad anything that we compute.
Those techniques, all
these techniques together
give you, typically,
a polynomial time
dynamic program-- when
they work, of course.
Memoization makes the
recursion polynomial time.
The guessing is what is
doing a brute force search.
And magically, it all
works if you're careful.
Another perspective-- kind
of an orthogonal perspective
or another way of thinking
about it, which I think
should be comfortable
for you because we
spent a lot of time
doing shortest paths
and expressing problems that we
care about in terms of shortest
paths even if they don't look
like it at first glance--
dynamic programming
in some sense
is always computing
shortest paths in a DAG.
So you have some problem
you want to solve,
like you have text you want to
split up into lines so it looks
nice in a paragraph, you
express that problem somehow
as a directed acyclic graph.
And then we know how to
compute shortest path
in directed acyclic
graphs in linear time.
And that's basically what
dynamic programming is doing.
I didn't realize
this until last week,
so this is a new perspective.
It's an experimental
perspective.
But I think it's helpful.
It's actually-- dynamic
programming is not that new.
It's all about how to be
clever in setting up that DAG.
But in the end, the
algorithm is very simple.
And then we had this
other perspective--
back to this
perspective, I guess.
In general, we have-- the
real problem we want to solve,
we generalize it in some
sense by considering
lots of different sub-problems
that we might care about.
Like with Fibonacci, we had
the nth Fibonacci number.
We really just wanted
the nth Fibonacci number.
But along the way, we're going
to compute all f1 up to fn.
So those are our sub-problems.
And if we compute
the amount of time
we need to solve
each sub-problem
and multiply that by the
number of sub-problems we get,
the total time required
by the algorithm.
This is a general true fact.
And the fun part here is we get
to treat any recursive calls
in this recursion as
free, as constant time,
because we really only
pay for it first time.
That's counted out here.
The second time we call
it, it's already memoized,
so we don't have to pay for it.
So this is, in some
sense, an amortization,
if you remember amortization
from table doubling.
We're just changing
around when we
count the cost of
each sub-problem,
and then this is the
total running time.
OK, so that's the
spirit we saw already.
I'm going to give you
the five general steps,
and then we're going to apply
them to two new problems.
So five easy steps to
dynamic programming.
Unfortunately, these are not
necessarily sequential steps.
They're a little
bit interdependent,
and so "easy"
should be in quotes.
This is how you would
express a dynamic program,
and in some sense
how you'd invent one,
but in particular how
you would explain one.
OK, let me get to
the main steps first.
First step is to figure out
what your sub-problems are
going to be.
Second part is to
guess something.
Third step is to relate
sub-problem solutions, usually
with a recurrence.
I guess always
with a recurrence.
Fourth step is to actually
build an algorithm.
And we saw two ways
to do that last time.
One is to use recursion
and memoization, which
is the way I like
to think about it.
But if you prefer, you can
follow the bottom up approach.
And usually that's
called building a table.
And that one's basically to turn
our recursion and memoization,
which is kind of
fancy, into a bunch
of for loops, which
is pretty simple.
And this is going to be more
practical, faster, and so on.
And depending on your
preference, one of them
is more intuitive
than the other.
It doesn't matter.
They have the same running
time, more or less,
in the worst case.
Then the fifth step is to
solve the original problem.
All right, so we've sort
of seen this before.
In fact I have, over
here, a convenient table.
It's called cheating.
The two problems we saw
last time, Fibonacci numbers
and shortest paths.
And I've got steps
one, two, three,
four-- I ran out of room,
so I didn't write five yet.
But we'll get there.
So what are our sub-problems?
Well, for Fibonacci,
they were f1 through fn.
So there were n
different sub-problems.
And in general because
of this formula,
we want to count how many
sub-problems are there.
So number of
sub-problems is-- this
is what we need to
do algorithmically.
And then for analysis,
we want to counter
number of sub-problems
for step one.
And so for Fibonacci
there were n of them.
For shortest paths, we defined
this delta sub k of sv.
This was the
shortest path from s
to v they uses at most k edges.
That was sort of what
Bellman-Ford was doing.
And the number of
different sub-problems
here was v squared, because we
had to do this for every vertex
v and we had to do it for
every value of k between 0
and v minus 1. v minus was
is the number of rounds
we need in Bellman-Ford.
So it's v times v,
different sub-problems,
b squared of them.
OK, second thing was we
wanted to solve our problem.
And we do that by guessing
some feature of the solution.
In Fibonacci, there
was no guessing.
So the number of different
choices for your guess is one.
There's nothing.
There's only one choice,
which is to do nothing.
And for shortest paths,
what we guessed was--
we know we're looking for some
path from s v. B Let's guess
what the last edge is.
There's some last
edge from u to v,
assuming the path has more
than one edge-- or more
than zero edges.
When could the edge possibly be?
Well, it's some
incoming edge to v.
So there's going to be
indegree of v different choices
for that.
And to account for the case that
that's zero, we do a plus 1.
But that's not a big deal.
So that was the number
of different choices.
In general if we're
going to guess something,
we need to write down
the number of choices.
For the guess, how many
different possibilities
are there?
That's our analysis.
OK, the next thing
is the recurrence.
That's step three.
We want to relate all
the sub-problem solutions
to each other.
For Fibonacci,
that's the definition
of Fibonacci numbers.
So it's really easy.
For shortest paths,
we wrote this min.
In general, typically
it's a min or a max,
whatever you're
trying to solve here.
We're doing shortest paths.
You could do longest
paths in the same way.
So you compute them in of
delta sub sk minus 1 of su.
The idea is we want to compute
this part of the path, the s
to u part.
And we know that has one
fewer edge, because we just
guessed what the last edge was.
Except we don't really know
what the last edge was,
so we have to try them all.
We try all the incoming edges
into v-- that's this part--
and for each of
them we compute--
I forgot something here.
This is the cost of the
first part of the path.
Then I also need to do plus
the weight of the uv edge.
That will be the total
cost of that path.
You add those up, you do
it for every incoming edge.
That is, in some sense,
considering all possible paths.
Assuming you find the
shortest path from s to u,
that's going to be the
best way to get there.
And then use some edge from
u to v for some choice of u.
This will try all of them.
So it's really trying
all the possibilities.
So it's pretty clear
this is correct
if there are no
negative weight cycles.
You have to prove some things.
We've already proved them.
It's just slow, but once you
add memoization, it's fast.
Now, how long does it take
to evaluate this recurrence,
constant time, if you
don't count the recursive
calls or count them as constant?
Over here, we're taking a min
over n degree of v things.
So we have to pay
n degree of v time,
again the recursions as free.
But for each one of them,
we have to do an addition.
So it's constant work per guess.
And this is quite common.
Often, the number of
guesses and the running time
per sub-problem are the
same, the constant factors.
Sometimes they're different.
We'll see some examples today.
OK, step four.
Let's see.
So here we evaluate the
time per sub-problem.
Once you have the recurrence,
that becomes clear.
You want to make sure
that's polynomial.
Often these are the same.
And then we add the recursive
memorize or build a DP table.
I'm not going to write those.
We did it for Fibonacci
last time, shortest paths.
Pretty easy.
And in general, what
we need to check here
is that the sub problem
recurrence is acyclic.
In other words, that it
has a topological order so
we can use topological sort.
We don't actually use
topological algorithm usually.
You can just think about it.
In the case of
Fibonacci numbers,
it's clear you want to
start with the smallest one
and end up with the biggest one.
You can't do the reverse,
because then when
you're trying to computer the
nth you don't have the ones
you need, the n minus
1 and n minus 2.
But if you do it in
this order, you always
have the one you need by
the time you get there.
In general, there's
a DAG there--
and for Fibonacci,
it was like this.
Every node depends
on the previous
and the second previous.
But you just choose
a topological order,
which is here left to
right, and you're golden.
And these are
actually the for loops
you get in the bottom of DP.
For shortest paths, you
have to think a little bit.
You have to do the for
loop over k on the outside,
the for loop over
V on the inside.
The reverse does not work.
I won't go through that, but
we drew the DAG last time.
And that's the main thing
you need to do here.
And then, of course,
you use this formula
to compute the overall running
time, which is just multiplying
this quantity with
this quantity.
Total time.
Then there's just
one last step that
usually isn't that big a deal,
but you have think about it.
You need to make sure that
the problem you actually
cared about solving gets solved.
In the case of Fibonacci
and shortest paths,
this is pretty clear.
I didn't write it.
We can do it on here.
Solve the original problem.
Fibonaci, it is Fn.
And this is one of
our sub-problems,
so if we solve all
of them, we're done.
For shortest paths,
it's basically
delta sub v minus 1 of
sv for all v. That's
single source shortest paths.
And by our
Bellman-Ford analysis,
that gives us the
right shortest paths.
There are no negative
weight cycles.
And sometimes this
requires extra time
to combine your solutions
to get the real thing.
Here of course, we just have the
answers, so writing them down
does not take very long.
So that's the dominant running
time-- which I didn't write,
I should have written in
under for here-- this ends up
being n, this ends up being VE.
OK, I don't want to spend
more time on those examples.
Let's go to new things.
So first problem we're
going to look at today
is text justification.
And the informal
statement of this problem
is you're given
some text-- which
means a string, a whole
bunch of characters.
And we want to split
them into good lines.
The rules of the game
here are we're going to,
like in the early lectures
of document distance
where you have some
definition of splitting
a document into words
separated by spaces.
And what we want to do is cut.
We can only cut between
word boundaries.
And we want to write
some text, it's
going to have some spaces in it.
Then there's a new line,
something like that.
And we want to justify our
text on the right here.
And so we'd like to avoid big
gaps like this because they
look ugly, they're hard to read.
Now, if you use
Microsoft Word-- at least
before the latest
versions-- they
follow a greedy strategy,
which is very simple.
You pack as many words as
you can on the first line,
then you go to the next
line, pack as many words
as you can on the second line.
Keep going like that.
And that strategy
is not optimal.
If you use LaTeX--
as some of you
have been doing on problem sets,
and I think also new versions
of Word but I'm
not sure-- then it
uses dynamic programming
to solve this problem.
And that's what we're
going to do here.
So let me specify a little bit
more about what we mean here.
So the text we're going to
think of as a list of words.
And we're going to define
a quantity badness.
And this is an anesthetic
quantity, if you will.
I'm going to tell
you what LaTeX uses.
But this is sort
of how bad it is
to use-- or let's say, yeah,
words i through j as a line.
So this is python notation.
So it starts at i and
ends at J minus 1.
That'll be convenient.
So I have this list of words.
And if I look at words
i through j minus 1
and I think of what happens if
I pack them in a line, well,
they may fit or
they may not fit.
So there are going
to be two cases.
If they don't fit, I'm
going to write infinity.
So that's really bad.
So I have some notion of
how wide my line can be.
And if the sum of the lengths
of those words plus the sum
of the lengths of the
spaces as small as possible
is bigger than the width
of my screen-- or page,
I guess-- then I
say they don't fit,
and then I define badness
to be infinity-- meaning,
I never want to do that.
This is actually
LaTeX sloppy mode,
if you want to be technical.
Otherwise, it's going to be page
width minus total width cubed.
Why cubed?
Who knows.
This is the LaTeX rule.
And squared would
probably also be fine.
So this is the width of the
page minus the total width
of those words,
which you also have
to include the spaces here.
You take the difference.
You cube it.
And so when this
is small-- I mean,
when these are very
close-- then this
is going to be close to zero.
That's good.
That means you use
most of the line.
When the total width is much
smaller than the page width,
then this will be a large value.
You cube it, it
will be even larger.
So this will highly
discourage big gaps like this.
And it will very much
discourage not fitting.
So there's a
tradeoff, of course.
And the idea is you might--
in the greedy algorithm,
you make the first line
as good as you can.
But it might actually
be better to leave out
some of the words
that would fit here
in order to make the
next line better.
In general, it's hard
to tell, where should I
cut the lines in order to get
the best overall strategy?
What I'd like to
minimize is the sum
of the badnesses of the lines.
So it's a sum of cubes,
and that's really hard
to think about.
But that's what dynamic
programming is for.
You don't have to think.
It's great because
it's brute force.
OK, so the first thing we need
to do is define sub-problems.
This is, in some
sense, the hard part.
The rest will follow easily.
So I think actually
it might be easier
to think about, for
this problem, what
would be the brute
force strategy?
How would you try all
possibilities, exponential
time?
Suggestions?
Yeah?
AUDIENCE: Try all partitions
of the words that don't fit?
PROFESSOR: Try all
partitions of the word,
so-- of the string of words.
So I mean, it could be it
all fits in on one line.
It could be it's
split into two lines.
I try all possible splits there.
In general, I'm
guessing for every word,
does this start a line or not?
That would be all ways.
And so there are 2 to the n.
If I have n words, there's
2 to the n different splits.
For every word I say yes or
no, does this is begin a line?
So what I'd like to figure out
is where those lines begin.
That was the point
of that exercise.
So any suggestions?
Maybe it's actually easier
to jump ahead and think,
what would I guess
in my solution
if I have this big
string of words?
What's the natural
first thing to guess?
Yeah?
AUDIENCE: Guess how
long the first line is?
PROFESSOR: Guess how long
the first line is, yeah.
We know that the first
word begins a line.
But where does the
second line begin?
So I'd like to guess where
the second line begins.
That's-- so you know, I have
the beginning of a line here
and then I have a beginning of
a line here at the fourth word.
Where does the
second line begin?
I don't know.
Guess.
So I'm going to try
all the possible words
after the first word.
And say, well, what if I
started my second line here?
At some point I'm
going to be packing
too much into the first
line, and so I abort.
But I'll try them all.
Why not?
OK, that's good.
The issue is that
once I've chosen
where the second line is, of
course the next thing I want
to guess is where the
third line begins.
And then I want I guess
where the fourth line begins,
and so on.
In general, I need to
set up my sub-problems
so that after I
do the first guess
I have the problem
of the original type.
So originally I
have all the words.
But after I guess where
the second line begins,
I have the remaining words.
What's a good word for
the remaining words?
If I give you a list of words
and I want from here on,
it's called-- what?
A sub-problem, yes.
That's what we want to define.
It's called a
suffix of the array.
That's the word I
was looking for.
It's tough when I only
have one word answers.
So my sub-problems are
going to be suffixes.
Which is, in python
notation, i colon.
They call it splices.
And how many sub-problems
are there if I have n words?
Two?
Sorry?
AUDIENCE: 2 to the n.
PROFESSOR: 2 the n?
That would be a problem
if it's 2 to the n.
I hope it's only n.
Originally, we said, OK, for
every word, we're going to say,
is this in our out?
Is this the beginning or not?
That's 2 to the n.
But here, the idea is we're
only thinking about, well,
what are the words that remain?
And it could be you've dealt
with the first 100 words
and then you've got
n minus 100 left,
or it could be you've dealt
with the first thousand words
and you've got n minus 1,000.
There's only n choices for that.
We're only remembering
one line, this is the key.
Even though we may have
already guessed several lines,
we're just going to
remember, well, OK.
This is what we have left to do.
So let's forget about the past.
This is what makes dynamic
programming efficient.
And we're just going to solve
it, solve these sub-problems,
forgetting about the past.
So the sub-problem--
I'm not going
to write it here-- is if
I give you these words,
never mind the other words,
how do I pack them optimally
into a paragraph?
I don't care about the other
words, just these words.
So this is a different
version of the same problem.
Initially, we have
n words to do.
Now I have n minus
i words to do.
But it's again
text justification.
I want to solve this
problem on those words.
That's just how I'm
going to define it.
This will work if I can
specify a recurrence relation.
As we said, what
we guess is where
to break the first
line, where to start
the second line for those words.
OK, so this is-- it could
be the i plus first line.
It could be the i plus
second line-- or sorry, word.
Some word after i is where
we guess the second word.
The number of choices for the
guess is at most n minus i.
I'm just going to think
of that as order n.
It won't matter.
The third part is we need
a recurrence relation.
I claim this is very easy.
I'm going to-- I didn't
give this problem a name,
so I'm just going to
write it as DP of i.
So this is going to be the
solution to that suffix, words
from i onward.
And I'd like to--
what I want to do
is consider all
possible guesses.
So I mean this is going to be
pretty formulaic at this point.
After I've set up these
ideas there's pretty much
only one thing I can
write here, which
is I want to do a for loop.
That would be the
for loop of where
the second line can start.
I can't start at
i, because that's
where the first line starts.
But it could start at i plus 1.
And this special
value of n will mean
that there is no second line.
OK, so DP of i-- now I
want to do this for loop
in order to try all
the possible guesses.
j will be the word where
the next thing starts.
So then what do I write up here?
If I make this guess--
all right, so I have word
i is the first word
of the first line.
And then word j is the first
word of the second line.
And then there's more
stuff down below.
I don't know what that is.
But how can I use
recursion to specify this?
DP of j, exactly.
I guess if I'm
doing recursion, I
should use parentheses
instead of brackets.
But if you're
doing it bottom up,
it would be square brackets.
So that's just DP of j.
That's the cost of the
rest of the problem.
And I can assume that
that's free to compute.
This is the magic of
dynamic programming.
But then I also have
to think about, well,
what about the first line?
How much does that cost?
Well, that's just badness of ij.
And we've already defined that.
We can compute it
in constant time.
Dynamic programming doesn't
really care what this is.
It could be anything.
As long as you're
trying to minimize
the sum of the badnesses,
whatever function is in here,
we just compute it here.
That's the power of
dynamic programming.
It works for all
variations of this problem,
however you define badness.
So you might say, oh,
that's a weird definition.
I want to use
something else instead.
That's fine, as long
as you can compute it
in terms of just i and j
and looking at those words.
OK, now I need to do a
min over the whole thing.
So I want to minimize
the sum of the badnesses.
So I compute for
every guess of j,
I compute the cost of
the rest of the problem
plus the cost of
that first line.
And this, is in some sense,
checking all possible solutions
magically.
OK.
That's the recurrence.
We need to check some things.
I guess right now we just want
to compute how much time does
this cost, time per sub-problem.
To do this for loop, basically
I do constant work-- all of this
is constant work--
for each choice.
So there's order n choices,
so this is order n.
Now we have to
check that there's
a topological order
for this problem
or for these sub-problems.
And this is easy, but
a little different
from what we've done
before because we
have to actually work
from the end backwards,
because we're expressing
DP of i in terms
of DP of larger values of i.
j is always bigger than i.
And so we have to do it
from the right end back
to the beginning.
And n minus 1 down to 0.
I didn't actually
define DP of n.
There's a base case here
which is DP of n equals 0.
Because the meaning of DP
of n is I have zero words,
the nth word onward.
There is no nth word.
It's 0 to n minus
1 in this notation.
So I don't pay anything
for a blank line.
OK, so that's our
top logical order.
This one, of course,
is instantaneous.
And then we work backwards.
And always whenever we
need to compute something,
we already have the value.
The total time we
get is going to be
the number of sub
problems-- which
is n times the running
time per sub-problem.
which is order n, which
is order n squared.
And in the worst case, it
is indeed theta n squared.
Although in practice it's going
to work better, because lines
can't be too long.
So that's the running time.
Then finally we have to check
that the original problem
actually gets solved.
And in this case, the original
problem we need to solve
is DP of 0 because DP of 0 means
I take words from 0 onwards.
That's everybody.
So that's the actual
problem I want to solve.
So we work backwards.
We solve all these sub-problems
that we don't directly
care about, but then the
first one is the one we want.
And we're done.
So in quadratic time,
we can find the best way
to pack words into lines.
Question?
AUDIENCE: [INAUDIBLE]
PROFESSOR: DP of j is returning.
It's like this.
So DP of-- this is a
recursive definition.
Imagine this is a
recursive function.
I wrote equals, which is
Haskell notation, if you will.
But normally, you think of
this as like def DP of i
is return min of this.
This is python.
So it's returning the cost.
What was the best way to pack
those lines from j onwards?
That's what DP of j returns.
So it's a number.
It's going to be a
sum of badness values.
Then we add on one
new badness value.
It's still a sum
of badness values.
We return the best
one that we find.
Now, this does not
actually pack the words.
That's a good-- maybe
your implicit question.
It's not telling you
how to pack the words.
It's telling you how much
it costs to pack the words.
This is a lot like shortest
paths where we didn't-- it was
annoying to actually figure
out what the shortest path was.
Not that annoying,
but that's not
what we were usually
aiming to do.
We were just trying to figure
out the shortest path weight.
And then once we knew
the shortest path weight,
it was pretty easy to
reconstruct the paths.
So maybe I'll take a
little diversion to that
and talk about parent pointers.
The idea with parent
pointers is just
remember which guess was best.
it's a very simple idea, but it
applies to all dynamic programs
and lets you find the
actual solution, not just
the cost of the solution.
We did the same thing
with shortest paths.
We even called them parent.
So when we compute this min,
were trying all choices of j.
One of them-- or maybe more than
one, but at least one of them
actually gave you the min.
That's usually called the
arg min in mathematics.
It's what was the value
of j that gave you
the minimum value of this thing.
So I mean, when you
compute the min,
you're iterating over
every single one.
Just keep track of
which one was the best.
That's it.
Call that the parent pointer.
Do I need to write that?
Here, parent--
parent of i is going
to be arg min of
that same thing.
So it's a j value.
It's the best j value for i.
And so we store that for each i.
It cost no more work,
just a constant factor
more work than
computing the min.
We also write down the arg min.
So we're already storing
the min in the DP table.
DP of i would get
sorted to be that.
We also store parent of i.
And then once we're done, we
start with our original problem
and we follow parent
pointers to figure out
what the best choices were.
So we start at 0 because we
know word zero begins a line.
And then 0 will
be the first line.
Then we go to parent of 0.
That will be where the
second line begins.
Then we go to parent
of parent of 0.
That will be where
the third line begins.
OK, because these
were the best choices
for where the
second line begins,
this is the best place where
the second line begins.
Given that this
is the first line,
this is the best line where
the second line begins
given that this
was the first line.
So that's really the third line
given this was the second line.
Little confusing, but you
just a simple for loop.
You start with 0 because
that's our original problem.
You keep calling parent of
the thing you currently have.
In linear time, you
will reconstruct
where the lines break.
So you can use this
technique in any DP.
It's very simple.
It's totally automatic.
Just like memoization
is a technique
that you can apply
without thinking,
you could even write a program,
given a recursive algorithm,
would turn into a memorized
recursive algorithm.
It's totally automated.
Same thing with the
bottom up DP table.
As long as you know what
the topological order is,
just make those for
loops and then put
exactly the recursive call but
turn it into an array call.
Boom, you've got a
bottom up algorithm.
Totally automatic,
no thinking required.
Parent pointers also,
no thinking required.
As long as you're following
the structure of trial guesses
compute some value--
just remember
what the guess was-- you
reconstruct your solution.
That's the great thing
about dynamic programming
is how much of it is automatic.
The hard part is figuring
out what to guess
and then what your sub-problems
are, or the other order.
Whatever works.
Any other questions about text?
I would like to move
on to blackjack.
OK, now I brought some
cards, because some of you
may not know the
rules to blackjack.
How many people know blackjack?
OK.
How many people do not and
are willing to admit it?
A few, all right.
So this is for you and
for fun, entertainment.
So I'm going to bring Victor
up to help demonstrate
the rules of blackjack.
We're going to play
standard Casino blackjack
as in the movie 21, or whatever.
So I'm going to just do a random
cut here so I can't sheet.
You have a tablet, that's scary.
You're going to
look at strategy.
VICTOR: Nothing special.
PROFESSOR: All right.
Hopefully you do not
have x-ray vision.
So the way it works
is there's a dealer
player and one or more players.
We're just going to do it with
one player to keep it simple.
I'm going to be the dealer.
So my strategy is actually
totally deterministic,
there's nothing interesting.
Victor has the hard
part of winning.
So to start out, I believe
we deal to you first, then
to me, then to you, then to me.
So let's hold up
these cards, Victor,
so that people can see them.
You don't get to
see one of my cards.
That's some peculiarity
of the rule.
And if the sum of our cards
goes over 21, we lose the game.
Victor first.
I cannot have a value more
than 21 in these hands,
because I only have two cards.
You have a value of-- ha, ace.
Great.
An ace can be a 1 or an 11.
That's the fun rule.
So this is either an 8 or an 18.
And so Victor has a choice of
whether to take another card
or not.
What would you like to do?
VICTOR: Standard
strategy says stand.
PROFESSOR: He stands.
So he's going to stick to that.
At this point, my
cards flip over.
I have 17, which
is same you, which
I believe means-- I
forget about tie rules.
VICTOR: I have 18.
PROFESSOR: You have 18.
All right.
VICTOR: See?
The strategy works.
PROFESSOR: So that's good.
I'm going to hit in the hope
that I have a small card that
will push me right above you.
But I do not.
I lose.
I'm sad.
VICTOR: It says
always stand on a 17.
PROFESSOR: Oh,
always stand on 17?
Huh.
All right, never mind.
Thanks.
Yeah, I still lose.
The game is over.
My strategy is always
stand on a value--
VICTOR: Stand on 17.
PROFESSOR: 17 or higher.
And if I have a value less than
17, I always take another card.
So let's do it one more
time to get it right.
So I'm going to deal to you,
deal to me, deal to you,
deal to me.
So hold up your cards.
You have 18 again.
Are you cheating?
VICTOR: I still have to stand.
PROFESSOR: You still
stand, according to tablet.
So I, in this case, have a 20.
And so this I win.
So you get the idea.
Let's say in each
case we're betting $1.
So at this point, we'd be even.
He won $1, I won $1.
But in general, slight--
I think it's balanced.
VICTOR: For these rules, there's
a 1% advantage for the house.
PROFESSOR: 1% advantage
for the house.
Interesting.
All right, well, that's
beyond this class.
What we're going to see is
how to cheat in blackjack.
So this is going to be-- I
encourage you to try this out
at casinos.
Just kidding.
This is a little bit difficult
to actually do in a casino
unless you have an inside man.
So if you have an
inside man, go for it.
It's guaranteed to
win you lots of money
because it's going
to play optimally.
In perfect
information blackjack,
I suppose that I already
know the entire deck.
Suppose somehow either I
get to put the deck there,
or I have some x-ray vision.
I get to see the entire
deck ahead of time.
And then somebody's
going to play
through a game over and over
with me-- or not over and over,
but until the deck
is depleted-- and I
want to know in each
case, should I hit,
or should I stand?
And I claim with
dynamic programming
you can figure that out--
using exactly the same strategy
as text, actually.
It's really for
each word, should I
start a new line or not?
Same problem here.
It's slightly more
complicated to write down.
So let's say the deck
is a sequence of cards.
And I'm going to call it c0,
c1 up to cn minus 1, n cards.
And you are one player.
First is the dealer.
I don't know how to
solve this for two
players, interesting
open problem.
But for one player I can do it.
Let's say $1 bet per hand,
I think they're called.
I'm not sure.
Per play?
Per box?
Whatever.
You're not allowed to double.
You're not allowed to split.
All these fancy rules are
harder to think about,
although you might be able
to solve them as well.
So the idea is I
have some cards.
Should I hit or should I stand?
I don't know.
I'll guess.
So our guessing-- let's jump
ahead to the guessing part--
is whether we want to hit
or stand given a card.
Actually, it would
be easier to think
about an entire
play, an entire hand.
We're going to guess,
how many times should I
hit in the first play?
So initially, four
cards are dealt.
I look at my hands.
Actually, I don't
really look at my hand.
I'm just going to
guess ahead of time.
I think I'll hit
five times this time.
I think I'll hit
zero times this time.
I mean, I'm just
going to try them all.
So I don't really have to
be intelligent here, OK?
It's kind of crazy but it works.
Our sub-problems,
can anyone tell me
what our sub-problems would
be, In one word or less?
Less would be impressive.
Yeah?
AUDIENCE: Where you
start the new hand.
PROFESSOR: Where do
you start the new hand?
Yeah.
So it's going to be
suffixes of the cards.
So at some point we do a play,
and then we get to ith card.
And then the rest of the game
will be from the ith card on.
So it's going to
be suffix ci colon,
I guess would be
the notation here.
It's a bit awkward.
These are the cards that remain.
And so the sub-problem
is, what is the best play?
What's the best
outcome given $1 bets?
How much money can I make--
maximize my winning, say--
given these cards onward?
Who knows what happened to
their earlier cards, but just
these are the cards.
I'm left with.
Number of sub-problems is-- hmm?
n.
How many choices of i
are there? n choices.
This really important.
It's really useful that we're
thinking about suffixes.
It's not that some subset of
the cards have been played.
That would be really hard,
because there's exponentially
many different subsets
that could be left.
It's always a prefix
that gets played,
and therefore suffix is left.
And there's only n
suffixes, remember that.
We're going to use it over and
over in dynamic programming.
So now we need to
solve the sub-problem.
Starting from ci, what's
the best way to play?
Well, the first four
cards are fixed,
and then we guess how
many hits are left.
So it's going to be
something like n minus i
minus four different
possibilities for-- I mean,
that would be the
maximum number of hits
I could take all
the remaining cards.
That would be the most.
And let's see, so the number of
choices-- I'll just say it's,
at most, n.
I don't have to be fancy here.
OK, now we go to the recurrence.
So I'm going to call
this blackjack of i.
It's going to be the solution.
I want to solve this
sub-problem from i onwards.
What's the best play?
And I guess it's going to be a
max if I'm measuring winnings.
And what's the winnings if I
decide to hit this many times?
It's a little bit hard to
write down the exact formula.
I'm going to write a
rough version which
is the outcome of
that first play.
It's going to be either I
lose $1, we tie, or I win $1.
So if we end up
with the same value,
you actually-- in
most versions--
you get your money
back, nothing changes.
The bet is nullified.
So that's a zero outcome.
But if we're only
betting $1, these
are the three possible outcomes.
You can compute this, right?
If I told you how
many times you hit,
then you just execute
through those cards
and you compute the
values of my hand,
of your hand versus
the dealer's hand.
You see, did anyone bust?
If so, they lose.
Otherwise you compare
the values and you
see which is bigger or smaller.
This is easy to
do in linear time.
No biggie.
What's useful here is
that the dealer strategy
is deterministic.
So after you know how many cards
you take, what the dealer does
is force, because he just looks.
Do I have 17 or greater?
If not, take another card
and keep repeating that.
So it's a
deterministic strategy.
In linear time, you can figure
out what the outcome is.
Then you also have to add the
outcome of all the remaining
cards, which is just BG of j.
This is recursion, super easy.
We do this for all choices of j.
It's like a range of i
plus 4 up to n, I think.
Sure, that'll work.
I should probably
put an if here,
which is if it's a valid play.
There are some constraints here.
If I've already busted,
I can't hit again.
So in fact what you have
to do in this for loop
is say, well, maybe
I take another hit.
Maybe I take another hit.
At some point I go
over 21, and then you
have to stop the for loop.
So I'm writing that as an if.
You can also do it with a
break, however you want.
But that's-- you're considering
all possible options,
all valid options of play.
For each of them, you
see what the outcome
was after the dealer
takes some more cards.
This is actually a
little bit funny.
Sorry, this should really be
the number of hits in range
from, let's say, 0 to n.
Maybe you don't hit at all.
And then j is a
little bit tricky,
because this is actually i
plus 4 plus the number of hits
plus the number of dealer hits.
OK, so you have to
run this algorithm
to compute what
happened, which computes
how many times a
dealer took a card.
That's how many
cards got consumed.
And so that's-- if you do i
plus 4 plus that plus that,
that's how many cards are left,
or where the cards resume.
And then you call BG on that.
So we're, in general,
from BG of i--
if you think of the DAG--
there's some position,
maybe i plus 4 happens.
Maybe it doesn't happen.
It depends on what
the dealer does.
We're going to depend on
i plus 6, i plus 5 maybe.
It's going to be all
of these possibilities.
These are all different plays.
And then on each of
these edges, we're
going to have plus
1, 0, or minus 1.
Those are the outcomes,
whether I won or lost or tied.
And then we're just computing
a shortest path in this DAG.
It's actually really easy if
you think about it that way.
This is just how
many cards are left.
From that position, you just see
what are all the possibilities?
What are all the edges
that I could go to?
What states could
I to go to next?
How many cards are remaining?
How much did it
cost me or win me?
And then take longest
paths in that DAG.
That will give you
the exact same answer.
That's what this dynamic
programming is doing.
In the lecture notes, there's
more details where I actually
tried to write out this
function, this recurrence
as an algorithm.
You could do it, assuming
I've got everything right.
It's not that hard.
The order here is just the same
as the order we did before.
The running time is going to
be cubic in the worst case,
because we have-- it's
a little non-obvious,
but we have n sub-problems.
For each of them,
we have n choices.
And for each choice we have
to run the dealer strategy.
And so that conceivably
could take linear time.
Here I'm assuming a
general value of 21.
If 21 is actually
constant, it only
be constant time to
play out a single hand,
and then it's quadratic time.
So it depends on your model
of generalized blackjack.
But that's it.
And get some flavor of the
power of dynamic programming,
we're going to see it's
even more powerful than this
in the next two lectures.
