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  • CHARLES LEISERSON: It is my great pleasure

  • to welcome Jon Bentley, now retired from Bell Labs.

  • Jon was my PhD thesis supervisor at Carnegie Mellon.

  • I actually had two supervisors.

  • The other one was HT Kung, who is now at Harvard.

  • I guess people flee Carnegie Mellon like the plague

  • or something.

  • So Jon is, as you know because you've

  • studied some of his work, is a pioneer in software performance

  • engineering.

  • And he's going to talk today about a particularly neat piece

  • of algorithmic engineering sets that

  • centers around the so-called traveling salesperson problem,

  • which is an NP-hard problem.

  • NP-complete problem in fact.

  • And so, without further ado, Jon, why don't you

  • tell us what you've got to say?

  • JON BENTLEY: As Charles mentioned,

  • I want to talk with you--

  • I want to tell you a story about a cool problem.

  • This is a problem that I first heard when I was a young nerd--

  • not much older than this little pile of nerds in front of me--

  • in high school, the traveling salesperson problem.

  • Who here has heard of the TSP at some point?

  • I heard about this in high school, one of the things

  • you read about it in the math books.

  • And a few years later, I had a chance to work on it.

  • In 1980, I was doing some consulting,

  • and I said, well, what you need to do is solve a TSP.

  • Then I went home and realized that all

  • of the stuff that I learned about it was sort of relevant

  • but it didn't solve the problem, so I

  • started working on it then.

  • Our colleague, Christos Papadimitriou,

  • who's been at Berkeley for a long time

  • after being at a lot of other places,

  • once told me the TSP is not a problem.

  • It is an addiction.

  • So I've been hooked for coming up on 40 years now.

  • And I want to tell you one story about a really cool program I

  • wrote.

  • Because this is one of the--

  • I've been paid to be a computer programmer for coming up

  • on 50 years, since I've been doing it for 48 years now.

  • This is probably the most fun, the coolest program

  • I've written over a couple day period.

  • I want to tell you a story.

  • Start off with what recursive generation is.

  • Then the TSP, what it is.

  • Then I'll start with one program,

  • and we'll make it faster and faster and faster.

  • Again, I spend my whole life squeezing performance.

  • This is the biggest squeeze ever.

  • And then some principles behind that.

  • We'll start, though, with how do you enumerate

  • all the elements in a set?

  • If I want to count--

  • enumerate the guys between 1 and a hundred, I just count.

  • That's no big deal.

  • But how can I, for instance, enumerate

  • all subsets of the set from the integers from 1 to 5?

  • How many subsets are there of integers from 1 to 5?

  • AUDIENCE: 2 to the 5.

  • JON BENTLEY: Pardon me?

  • AUDIENCE: 2 to the 5.

  • JON BENTLEY: 2 to the 5, 32.

  • But how do you say which ones they are?

  • How do you go through and count them?

  • Well, you have to decide how you represent it.

  • You guys know all about set representations.

  • We'll stick with bit vectors for the time being.

  • An iterative solution is you just count--

  • 0, 1, 2, 3, 4, 5, up to 31.

  • That's pretty easy.

  • But what does it mean to count?

  • What does it mean to go from one integer to the next?

  • How do you go from a given integer to the next one?

  • What's the rule for that?

  • It's pretty easy, actually.

  • You just scan over all the 0's, turning the-- you start

  • at the right-hand side, the least significant digit,

  • scan over all the 0's, turn it to 1.

  • Oh, I lied to you.

  • You scan over all the 1's, turning them to 0

  • until you get to the first 0.

  • And then you turn that to a 1.

  • So this one goes to 10.

  • This one goes to 11.

  • This one goes-- that one becomes 0, that one becomes 0.

  • Then it becomes 00100.

  • So a pretty easy algorithm.

  • You could do it that way.

  • Just scan over all the 1's, turn them

  • to 0's, take that first one and flip it around.

  • But that doesn't generalize nicely.

  • We're going to see a method that generalizes very nicely.

  • This is a recursive solution to enumerate all 2

  • to the n subsets of a set of size n.

  • And the answer is this all sets of size m

  • is just put a 0 at this end, enumerate

  • all sets of size m minus 1.

  • How many of these will there be?

  • 2 to the m minus 1.

  • How many of those 2 to the m minus 1?

  • What do they add up to?

  • 2 to the m.

  • But all of these have the 0 at that end,

  • and the one at that end.

  • Everyone see that recursive sketch and how that works?

  • Here's the example.

  • A period with 0's at this end and you fill it out.

  • You have the 1 at that and you fill that out.

  • If you do that, you notice that in fact we're

  • just counting backwards--

  • 000, 001, 010, 3, 4, 5, 6, 7.

  • That's the algorithm.

  • And the cool thing is the code is really simple.

  • I could probably write a program like that in most languages

  • and get it correct.

  • So if m equals 0 in generate all subsets of size m,

  • this doesn't occur at 1.

  • You have a pointer going down the array.

  • Otherwise, set the rightmost bit to 0,

  • generate all subsets recursively, set it to 1,

  • do it again recursively.

  • That's a starting program.

  • If you understand this, everything else

  • is going to be pretty straightforward.

  • If you don't, please speak up.

  • One thing that-- you people have suffered

  • the tragedy of 14 or 15 or 16 years of educational system

  • that has sort of beaten the creativity out of you

  • and you're afraid to speak up.

  • So even if something-- even if I'm up here spouting

  • total bullshit, you'll ignore that fact and just

  • sort of politely stare at me and nod.

  • But this is important.

  • I want you to understand this.

  • If you don't understand this, speak now or forever hold it.

  • Anyone have any questions?

  • Please, please.

  • AUDIENCE: What does mean, [INAUDIBLE]??

  • JON BENTLEY: I'm sorry.

  • Why did we set p to the--

  • AUDIENCE: [INAUDIBLE].

  • JON BENTLEY: So here, first I go out to the extreme rightmost

  • and I set it to 0.

  • Then I recursively fill those in.

  • Then I change it from a 0 to a 1 there, and I fill all those in.

  • So this is a program that will go through,

  • and as it enumerates a subset, it

  • will call the visit procedure a total of 2 to the m times,

  • then it comes down to the bottom of the recursion.

  • Thank you, great question.

  • Any other questions about how this works?

  • OK, we'll come back to this.

  • The traveling salesperson problem.

  • I apologize.

  • I will really try to say the traveling salesperson problem,

  • but I will slip because I was raised with this being

  • the traveling salesman problem.

  • No connotations, no intentionality there,

  • just senility galloping along.

  • It's a cool problem.

  • Abraham Lincoln faced this very problem in the years

  • 1847 to 1853 when he--

  • everyone here has heard of circuit courts?

  • Why do they call them circuit courts?

  • Because the court used to go out and ride

  • a circuit to go to a whole bunch of cities.

  • Now people in the cities come to the court.

  • But back in the day, in 1847 to 1853, Lincoln

  • and all of his homies would hop on their horses--

  • a judge, defense lawyers, prosecutors-- and go around

  • and ride the circuit here.

  • And so this is the actual route that they rode where they

  • wanted to do this effectively.

  • It would be really stupid to start here in Springfield

  • and go over there, then to come back here,

  • then to go over there back and forth.

  • What they did was try to find a circuit

  • that minimized the total amount of distance traveled.

  • That is the traveling salesperson problem.

  • We're given a set of n things.

  • It might be a general graph.

  • These happen to be in the plane.

  • But you really-- helicopter service was really bad

  • in those days, so they didn't fly there from point to point.

  • Whether they stayed on roads, what really matters here

  • is the graph embedded in here.

  • I'm going to speak at this.

  • Everything I say will be true for a graph.

  • It will be true for geometry.

  • I'll be sloppy about that.

  • We'll see interfaces, how you handle both, but just cut

  • me some slack for that.

  • I have actually face this myself when I worked on a system

  • where we had a big mechanical plotter

  • and we wanted to draw these beautiful maps where

  • the maps would fill in dots.

  • They happened to be precincts.

  • Some of them were red, some of them were blue.

  • And you wanted to draw all the red dots first and go around

  • here.

  • And, in fact, the plotter would draw this red dot,

  • then that red dot, then this one, then that one.

  • The plotter took an hour to draw the map.

  • I was consulted on this.

  • Aha, you have a traveling salesperson problem.

  • I went down.

  • I reduced the length to about 1/10 of the original length.

  • If it took an hour before, how long would it take now?

  • Well, it took about half an hour.

  • And the reason is that the plotter took about half

  • of its time moving around about, 30 minutes moving around,

  • and about 30 minutes just drawing the symbols.

  • I didn't reduce the time drawing the symbols at all,

  • but I reduced the time moving things around

  • from about 30 minutes to about 3 minutes.

  • That was still a big difference.

  • So I fixed it there.

  • When I worked at Bell Labs, we had

  • drills that would go around, laser drills,

  • move around on printed circuit boards to drill holes.

  • They wanted to move it that way

  • I talked to people at General Motors at one point.

  • On any day, they might have a thousand cars

  • go through an assembly line.

  • Some of the cars are red, some are white, some are green,

  • some are yellow.

  • You have to change the paint.

  • Some of them have V6, some of them have V8.

  • Some of them are two doors, some of them are four doors.

  • In what order do you want to build those cars?

  • Well, every time you change one part of the car,

  • you have to change the line.

  • And what you want to do is examine,

  • as it were, all n factorial permutations of putting

  • the cars through and choose the one that involves

  • the minimum amount of change.

  • And the change from one car to another

  • is a well-defined function.

  • Everyone see how that weird TSP is in fact a TSP?

  • So all of these are cool problems.

  • Furthermore, as a computer scientist,

  • it's the prototypical problem.

  • It's the E. coli of algorithmic problems.

  • It was literally one of the first problems

  • to be proven to be NP-hard.

  • Held-Karp gave a polynomial time algorithm for it.

  • There are approximation algorithms of this.

  • Kernighan-Lin have given heuristics.

  • It's a really famous problem.

  • It's worth studying.

  • But here is what really happened to me.

  • Here's why I'm standing in front of you

  • today talking about this.

  • My friend Mike Shamos, in his 1978 PhD thesis

  • on computational geometry, talked

  • about a number of problems.

  • One of them was the TSP.

  • And he shows us and he gives an example of this tour.

  • He says, here's a set of 16 points.

  • Here's a tour through them.

  • Here's a traveling salesperson tour through them.

  • And then he says in a footnote, in fact,

  • I'm not sure if it's a really optimal tour.

  • I applied a heuristic several times.

  • I'm not positive it's the shortest tour.

  • If you wrote a thesis, it would be

  • sort of nice to know what's going on there.

  • Can you solve a problem that was--

  • this tiny little 16-element problem,

  • 16 points in the plane.

  • Can you really figure out what the TSP is to that?

  • At the time, my colleague, our colleague, a really smart guy,

  • couldn't do it.

  • It was computationally beyond the bounds for him.

  • Well, in 1997 I came back to this,

  • and I really wondered is it possible now?

  • Computers are a whole lot faster in the 20 years.

  • We were talking about that earlier today.

  • 20 years, computers got faster.

  • A lot of things got better.

  • Have things changed enough so I can

  • write a quick little program to solve this?

  • I don't know.

  • We'll see.

  • I did that.

  • I talked about it.

  • I gave a talk at Lehigh University 20 years ago.

  • They liked it.

  • They incorporated it into an algorithms class.

  • The same professor gave it time and time and time again.

  • Eventually, he retired.

  • They asked me to come over and give this talk to them.

  • I can't give a talk about 20-year-old material.

  • Computer science doesn't work that way.

  • So I coded things.

  • I wanted to see how things changed in two years.

  • So this talk is about a lot of things,

  • but especially it's about how has performance

  • changed in 40 years.

  • So that's one of the reasons we were--

  • one of things we were talking about earlier today.

  • I could give a bunch of titles for this talk.

  • For you, the title I give is a sampler

  • of performance engineering.

  • It could be-- next week I'll give it at Lehigh.

  • This is their final class in--

  • one of their final classes in algorithms and data structures.

  • I'm going to try to tie everything they learn together.

  • It could be all these other things--

  • implementing algorithms, a lot of recursive generation,

  • applying algorithms, really--

  • Charles is a fancy dancy academic.

  • He's a professor at the Massachusetts Institute

  • of Technology.

  • I'm just a poor dumb computer programmer,

  • but boy this is a fun program.

  • What it is not is it's not state-of-the-art TSP algorithm.

  • People have studied the problem for well over a century.

  • They have beautiful algorithms.

  • I am not going to tell you about any

  • of those algorithms for the simple reason

  • that I don't know them.

  • I could look them up in books, but I've never really lived

  • the fancy state-of-the-art algorithms.

  • And I'm also going to just show you getting the answer.

  • I could analyze it.

  • I've analyzed much of these.

  • If I had another hour or three, I could do the analysis.

  • But I can't, so I'm just going to do the--

  • show you some anecdotal speeds without really the analysis.

  • Let's talk about some programs.

  • A simple C program.

  • MAXN is a maximum number, n int is going

  • to be n, the number of cities.

  • I'm going to have a permutation vector, where

  • if I have the tour going from city 1 to 7 to 3 to 4,

  • it says 1734.

  • The distance between cities is going

  • to be given by a distance function.

  • There is this distance d of i, j,

  • the distance from city i to city j.

  • Here's the first algorithm.

  • What I'm going to do is generate all intact real permutations,

  • look at them, and find the best one.

  • It's not rocket science.

  • The way I'm going to do this is a recursive function

  • where I happened--

  • I could have done it from left to right.

  • I am a C programmer.

  • I always count down towards 0.

  • So I'm going to count down that way, where all of these cities

  • are already fixed.

  • I'm going to permute these.

  • Here's the program.

  • To search for m--

  • all of these have already been fixed.

  • What I'm going to do is if m equals 1, then I check it.

  • Otherwise, for i equals 0 up to m, for each value from 0

  • to minus 1, I take the ith element.

  • I swap it.

  • swap 3, 7 takes the third and seventh positions

  • and swaps them.

  • I swap that to the final thing.

  • I call it recursively.

  • I then swap it back to leave it in exactly the same state I

  • found it, and I continue.

  • So here it's going to generate, first, all nine permuta--

  • put all nine digits in the last position.

  • Then for each one of those, I'll put

  • all eight digits in the last position, and go on down.

  • This is really interesting, important, and subtle.

  • If you don't follow this part, it's going to be difficult.

  • Are there any questions at all about this?

  • Have I lied to you yet about this?

  • You're honest enough to tell me if I have.

  • AUDIENCE: You're good.

  • JON BENTLEY: Thank you.

  • Anyone else?

  • AUDIENCE: [INAUDIBLE].

  • JON BENTLEY: I'm sorry.

  • Please.

  • AUDIENCE: Sorry.

  • I'm not really understanding the part that's fixed

  • and what you're permuting, and why is that hard to fix.

  • JON BENTLEY: So, so far, as I recur down with--

  • as m moves down, all these are fixed.

  • So I'm going to fix these things,

  • and then I'm going to take care of all these later.

  • So, originally, I'm going to have this array be

  • 0-- if I have a nine-city TSP, it will be 0, 1 2, 3, 4, 5, 6,

  • 7, 8, 9.

  • And first I put 0 in the end and do the rest.

  • Then I put 1 in the end, [INAUDIBLE] 9 in the end,

  • and recur down.

  • But as the program is progressing,

  • if you stop the program at any time

  • and look at a glance at the program,

  • you can see that, given the value of m, this parameter,

  • the recursive function.

  • So this is a way that I'm essentially

  • building this tree where at the top of the tree

  • the branching factor is 9.

  • At each of those nine nodes, the branching factor

  • is 8, then 7 and 6.

  • It's going to be a big tree.

  • If n is 10, how big is that tree going to be?

  • What's 10 factorial?

  • Pardon me?

  • When I was a nerd, we used to try

  • to impress people of appropriate genders

  • by going off saying things like 3628800.

  • You can probably guess how effective that was.

  • So 3.6 million.

  • It's going to be a big tree.

  • Any questions about that?

  • Let's go.

  • When I check things, I just compute the sum there.

  • I start off with the sum being the distance from 0

  • to p n minus first.

  • Then I go through and add up all the pairwise things

  • and save it.

  • What does it mean to say it?

  • If the sum is less than the minimum sum so far,

  • I just copy those over, change the minsum.

  • And to solve the whole thing, I do a search of size n.

  • This is a simple but powerful recursive program.

  • You should all feel very comfortable with this.

  • Is it correct?

  • Does it work?

  • Is it possible to write a program

  • with about two dozen lines of code that works?

  • Not the first time.

  • But after you get rid of a few syntax errors, you check it.

  • How do you make sure it works?

  • I start with n equals 3, and I put 3.

  • Does it give me a tour?

  • Well, it works.

  • Think about it.

  • For 3, 3 factorial, they're all the same tour.

  • That part wasn't hard.

  • 4, now that's interesting.

  • That one works too.

  • This program, in fact, can work.

  • Is it going to be a fast program?

  • How long will it take if n equals 10?

  • How many seconds?

  • I'm sorry.

  • What class have I stumbled into?

  • Is this in fact Greek Art 303?

  • How long will this take for n equals 10?

  • AUDIENCE: [INAUDIBLE].

  • JON BENTLEY: Pardon me?

  • AUDIENCE: 1 second.

  • JON BENTLEY: About a second.

  • Pretty cool.

  • For equal 20, how long will it take?

  • A lot longer.

  • Technically speaking, it's going to take a boatload longer.

  • So what I'm going to do here is-- notice

  • that there are n factorial permutations.

  • You do n of those at each, total of that,

  • on this fairly fast laptop from a few years ago.

  • But now they're all about the same.

  • At 8 seconds, it took that.

  • At 9 seconds, what should be the ratio-- what would

  • you expect to be the ratio between its time at 8

  • and time at 9?

  • Well, about a factor of 9, you'd hope.

  • Is 0.5 times 9 about 0.34?

  • Yes, close enough.

  • Here, going down, for 10 it's 4 seconds, 46 seconds.

  • Yes, it's going up by a factor--

  • so here I've run all my examples.

  • I ran out to 1 minute of CPU time.

  • After that, I estimate.

  • If this one takes 3/4 of a minute,

  • 12 times that is 12 minutes--

  • 3/4 of that is 9 minutes.

  • For 13, it's 2 hours.

  • How long should 14 take, ballpark?

  • A day, ballpark.

  • How long will 15 take if 14 takes a day?

  • AUDIENCE: Two weeks.

  • JON BENTLEY: Two weeks.

  • How long will 16 take?

  • Eight months.

  • You get the idea.

  • Are you going to go out to 20 for this one?

  • No.

  • Are you going to go out to 16 with this one?

  • Can you just put this into a thesis right now?

  • No.

  • The problem is it's fast for really small values of n.

  • As it becomes bigger--

  • how can you make the program faster?

  • If you wanted to make this program faster,

  • what would you do?

  • What are some ideas?

  • Give me some ideas, please.

  • This is performance engineering.

  • You should know this.

  • Ideas for making it faster.

  • Please.

  • AUDIENCE: You can start with arbitrary nodes.

  • So if you take the tour, you can start anywhere, right?

  • JON BENTLEY: OK.

  • So you're saying just choose one start

  • and ignore that, ignore all the others.

  • You don't need to take each random start.

  • Fantastic.

  • A factor of n.

  • My friend in the gray T-shirt just got a factor of n.

  • How else can you make it faster?

  • What ideas do you have?

  • Please.

  • AUDIENCE: You can start by the distance,

  • and then reject things that were [INAUDIBLE]..

  • JON BENTLEY: Be greedy.

  • Follow the pig principle.

  • If it feels good, do it.

  • Do just local optimization.

  • We'll get to that in a long time,

  • but, boy, would that be a powerful technique.

  • Other ideas, please?

  • AUDIENCE: Parallelize [INAUDIBLE]..

  • JON BENTLEY: Ah.

  • Parallelize.

  • I would write that out, but the first I would have to do

  • is remember how many R's and L's there are in various places.

  • So I'll write that much.

  • But we'll have a comment on that at the end.

  • People tried that.

  • Sir?

  • AUDIENCE: Clock the machine.

  • [STUDENTS LAUGH]

  • JON BENTLEY: Unlike you, Charles and I, at one point,

  • attended a real engineering school

  • at Carnegie Mellon, formerly known as CIT,

  • Carnegie Institute of Technology.

  • Charles, do you remember the cheer?

  • CHARLES LEISERSON: The cheer?

  • JON BENTLEY: The cheer.

  • CHARLES LEISERSON: I don't know how to cheer.

  • JON BENTLEY: 3.14159, tangent, secant, cosine, sine.

  • Square root, cube root, log of e.

  • Water-cooled slipstick, CIT.

  • What's a water-cooled slipstick?

  • AUDIENCE: [INAUDIBLE].

  • JON BENTLEY: Pardon me?

  • AUDIENCE: [INAUDIBLE].

  • JON BENTLEY: It's a slide rule that you run so fast.

  • It has to be water-cooled.

  • So if you can just overclock the machine,

  • just spray it with a garden hose.

  • And as long as it makes over the finish line,

  • you don't care if it dies when it collapses.

  • So, sure, you can get faster machines.

  • We'll talk about that.

  • How else can you make this faster?

  • Other ideas?

  • These are all great ideas.

  • We'll try it.

  • Let's see some ideas.

  • Compiler optimizations.

  • I just said gcc and I ran it.

  • What should I have said instead?

  • Instead of just gcc?

  • AUDIENCE: O3.

  • JON BENTLEY: O3.

  • How much difference will that make?

  • I used to know the answers to all these.

  • [INAUDIBLE] turn on optimization, 10%.

  • Sometimes, whoopee-freaking-do, 15%.

  • Does turning on O3 still make it a 15% difference?

  • We'll see.

  • You could do that.

  • A faster hardware.

  • I did this 20 years ago.

  • I had all that number there.

  • I'll show you some of those numbers.

  • Modify the C code.

  • We'll talk about all those options,

  • but let's start with compiler optimizations.

  • With no options there-- how much faster will

  • it be if I turn on optimization?

  • This is a performance engineering class.

  • You should know that thing.

  • Does it matter at all?

  • Is it going to be 15%?

  • Is it going to not matter at all?

  • How much will it matter to turn on optimization?

  • AUDIENCE: [INAUDIBLE] a lot.

  • JON BENTLEY: How much is a lot?

  • I know this isn't the real engineering school of CIT,

  • but pretend like this is kind of a semi--

  • one of the engineering schools.

  • Give me a number for this.

  • AUDIENCE: More than 15%.

  • JON BENTLEY: More than 15% Do I hear more than 16%?

  • I was surprised.

  • If I enabled O3, it went from 4 seconds to 12--

  • I couldn't even time it here.

  • It wasn't enough to time it here.

  • 45 seconds to 1.6 seconds.

  • I can get real times down there.

  • I observed, ballpark here, about a factor of 25.

  • Holy tamale.

  • On a Raspberry Pi, it was only a factor of 6,

  • and on other machines it was somewhere between the two.

  • Turning on optimization really matters.

  • Enabling that really matters.

  • For now on, I'm only going to show you full optimization.

  • It's cheating not to.

  • But just think about that, a factor of 25.

  • How else can I make if faster?

  • Two machines.

  • Back in the day, I happened to have some data laying around

  • of running it on a Pentium Pro at 20 megahertz.

  • Nowadays, I had this.

  • How much faster will this machine be 20 years later?

  • Again, pretend like you're at a real engineering school.

  • What will it be?

  • Please.

  • AUDIENCE: 20 times faster?

  • JON BENTLEY: 20 times faster?

  • How did you get 20 times faster?

  • AUDIENCE: Well, the clock speed is 10 times faster.

  • JON BENTLEY: The clock speed about 10.

  • AUDIENCE: But I'm guessing that it

  • has much better instructions.

  • JON BENTLEY: Here's what I found.

  • On this machine, it went from a factor--

  • there is about a hundred--

  • these factors, I found, consistently

  • were about, over the 20 years, about a factor of 150.

  • From Moore's law, what would it be

  • if you had 20 years if you doubled every two?

  • That's 10 doublings.

  • What is 2 to the 10th?

  • AUDIENCE: It's a thousand.

  • JON BENTLEY: A thousand.

  • So Moore's law predicts a thousand.

  • It's more than a factor of 20.

  • I got a factor of 150 here, which

  • is close to what Moore's law might predict,

  • but there is some slowing down at the end.

  • I'm not at all traumatized by this.

  • A speed-up of about a factor of 150, where does that come from?

  • My guess is you get about a factor

  • of 12 due to a faster clock speed, and another factor of 12

  • due to things like wider data paths.

  • You don't have to try to its cram everything

  • into 16-bit funnel.

  • You have 64-bit data paths there.

  • Deeper pipelines, more appropriate instruction

  • sets, and compilers that exploit those instruction sets

  • if O3 is enabled.

  • If O3 is not enabled, sucks to be you.

  • Questions about that?

  • Let's go.

  • So we have constant factor improvements,

  • external, modern machines, turn on optimization.

  • But a factor of 150 and a factor 25 is a lot.

  • We were starting off with that.

  • That is a good start.

  • Back in the day, if you change things from doubles to floats,

  • it got way faster.

  • From floats, the answer was faster yet.

  • Does that change make much difference nowadays?

  • No.

  • Exactly the same runtime.

  • One thing that does make a difference

  • is-- this is the definition of the geometric distance.

  • My j is the square root of the sum of the squares

  • of the differences.

  • That's doing an array access, a subtraction, a multiplication,

  • multiplication, two array accesses, subtraction,

  • multiplication, addition, and a square root.

  • That used to take a long time.

  • If I replace that with a table lookup

  • by filling out this sort of table,

  • the distance for algorithm 2 is just

  • the distance arrays of i sub j.

  • That gave me a speedup factor of 2 and 1/2 or 3.

  • Back in the day, that was a speedup factor of 25.

  • For you as performance engineers,

  • you have all this intuition.

  • Every piece of intuition you have, that I had,

  • that was really appropriate 10 years ago is irrelevant now.

  • You have to go back and get models to figure out

  • how much each thing costs.

  • But, still, it's another speedup factor of 3 just

  • by replacing this arithmetic with a table lookup.

  • Algorithm 3.

  • What we're going to do is choose the ones we need to start with.

  • So we'll start at city 1.

  • We'll leave 9, if we have a 9-element problem,

  • in that position, and just search of n minus 1.

  • It doesn't matter where you start.

  • You're going to go back to it, so you can just

  • choose one to start with.

  • Not a lot of code.

  • Permutations are now that, distance at each.

  • So now you've reduced n times n factorial to n factorial.

  • Algorithm 4 is I'm computing the same sum each time.

  • Is there a way to avoid computing the same darn sum

  • each time?

  • We'll carry that sum along with you.

  • Instead of recomputing the same thing over and over and over,

  • start off with the sum being 0.

  • The parameters are now m and the distance so far. s Then you

  • just add in these remaining pieces at each point,

  • and you solve it that way.

  • And there it's sort of a nice piece of mathematics.

  • I wish I had the time to analyze it.

  • I did a spreadsheet where I said, what's the ratio of this?

  • And it started off as 3, 3 and 1/2, 3.6, 3.65, 3.7--

  • 3.718281828.

  • What does that mean if you see a constant 3.718281828?

  • It's 1 plus e.

  • And once I knew what the answer was,

  • even I, in my mathematical frailty,

  • was able to prove that it's 1 plus e times n factorial.

  • I'm not giving you the proof, but it's very cool.

  • You run across these things.

  • So here are the four algorithms so far.

  • On an entirely different semi-fast machine,

  • the runtime--

  • here the real clock times on this machine

  • were 10, 11, 12, 13.

  • Real times in bold are measured times.

  • These other times are approximate estimates.

  • And you can see now that for size 13,

  • you go from taking a fraction of an hour to taking

  • a third of a minute.

  • We've made some programs faster.

  • That's pretty cool.

  • We feel good about this.

  • This is what we do.

  • Any questions at all?

  • We got to go faster.

  • How do we go faster?

  • To say precisely, for all these experiments,

  • I took one data set.

  • And if I say that runtime for size 15,

  • I take the first 15 elements of that data set.

  • For 16, I take the first 16 elements.

  • 17, and so on and so forth.

  • It's not great science.

  • I've done the experiments where I did it

  • on lots of random data.

  • The trends are the same.

  • It smooths out some of the curves, but we'll see this.

  • The times are for initial sequence of one random set.

  • It's pretty robust.

  • But the problem has factorial growth. it started factorial.

  • It's still factorial.

  • What does that mean?

  • Each factor of n allows us to increase the problem size by 1

  • in about the same time.

  • Faster machine and all that, we can now push into the teens.

  • What does that mean?

  • You can take Abraham Lincoln's problem,

  • and they got a tour with this length.

  • The optimal tour looks sort of the same on this side,

  • but it's really different over here.

  • Charles, what figure is that?

  • I've mentioned yesterday that if you work on the traveling

  • salesman, every instance you see turns into a Rorschach test.

  • CHARLES LEISERSON: The first one is a bunny hopping,

  • and the second one is just the head of the bunny.

  • JON BENTLEY: The bunny head.

  • Everyone see that?

  • Those are in fact the correct answers.

  • He is a psychologically sound human being.

  • Does anyone else want to give their Rorschach answers?

  • A free diagnosis.

  • Absolutely no charge.

  • I'll completely diagnose you. but the bunny hopping

  • and the bunny head are in fact the correct answers for here.

  • We'll see more later.

  • So Abraham Lincoln, you've solved his problem now.

  • My friend Mike Shamos could solve his problem.

  • Did he get the optimal tour?

  • Well, over here he got a big part of it.

  • But over here it's really sort of a different character.

  • It's a fairly different character.

  • Is it far off?

  • Yes, about a third of a percent off.

  • So his approach was within a third of a percent.

  • I've always worked-- I spent much

  • of my career working on approximate solutions to TSPs.

  • Those are often good enough.

  • This algorithm, you can prove-- that he applied--

  • is within 50%.

  • In the real world, it got within a third of a percent.

  • Wow.

  • But now we can go out and we can solve the whole problem

  • in 16 hours.

  • If you were writing the thesis and you happened to do this,

  • would it be worthwhile now to sink 16 hours of CPU time

  • into this?

  • You're going to go away for a weekend

  • and leave your machine running.

  • At the time, Charles, when we had

  • one big computer for 60 or 70 people in that department,

  • could we have dreamt about using 16 hours for that?

  • On the very border.

  • If you made it a really mellow background process,

  • it might finish in a week or three.

  • All of these things change.

  • The computers get faster.

  • They get more available.

  • You can devote a machine to dump 16 hours down this.

  • But can we make it faster yet?

  • Can we ever analyze, say, all permutations

  • of a deck of cards?

  • How many permutations are there of a deck of cards

  • if you take out those jokers?

  • What's that number?

  • AUDIENCE: 15 zeros?

  • JON BENTLEY: 1 with 15 zeros after it?

  • It's a big number, 2 to the--

  • 52 factorial.

  • I want to teach you how big 52 factorial is.

  • People say, that problem is growing exponentially.

  • What does that mean?

  • It's quick is what people usually mean by it.

  • In mathematics, it's some constant to the n

  • for some defined time period n.

  • Factorial growth-- is factorial growth exponential growth?

  • Why not?

  • Why isn't a factorial exponential?

  • AUDIENCE: It's more than exponential?

  • JON BENTLEY: It's more than exponential.

  • It's super exponential.

  • We'll talk about the details here.

  • By Sterling's approximation, you have seen in other classes

  • that log of n factorial is n log n minus n plus O of log

  • n for the natural log.

  • The log base 2 of n factorial is about n log n minus 1.386n.

  • Where have you seen this number before?

  • n log n minus 1--

  • In an algorithms class, you did a lower bound on a decision

  • tree model of sorting.

  • There were n factorial leaves to sort.

  • A sort algorithm must take at least as much time.

  • So that gives you that bound.

  • And merge sort is n log n minus n, so you're really narrow.

  • Where else have you seen 1.386n?

  • That's the runtime of quick sort.

  • All these things are coming back together here,

  • because it's the natural log of e--

  • I'm sorry-- the log base 2 of e.

  • So n factorial is not 2 to the n.

  • It's 2 to the n log n.

  • It's about n to the n.

  • It's faster than any exponential function.

  • How big is 52 factorial?

  • You guessed 10 to the 15th?

  • Was that--

  • AUDIENCE: Yes.

  • JON BENTLEY: OK.

  • If we see here, it's going to be something like 2

  • to the n log n.

  • n is 52.

  • Log of 52 is about six.

  • So that's 2 to the 300.

  • But there's a minus n term.

  • Maybe 2 to the 250.

  • It's about 2 to the 225, which is 10 to the 67th.

  • That's a big number.

  • How big is it?

  • Let me put it in everyday terms.

  • Try this after class.

  • Set a timer to count down 52 factorial nanoseconds, 10

  • to the 67th.

  • Stand on the equator--

  • watch out where you are--

  • and take one step forward every million years.

  • Don't rush into this.

  • I don't want you to get all hyper about this.

  • Eventually, when you circle the Earth once,

  • take a drop of water from the Pacific Ocean,

  • and keep on going.

  • Be careful about this.

  • But this is an experiment.

  • You're nerds.

  • It's OK.

  • When the Pacific Ocean is empty, at that point

  • lay a sheet of paper down, refill the ocean, and carry on.

  • Now keep on doing that.

  • When you're stack of paper reaches the Moon,

  • check the timer.

  • You're almost done.

  • This is how big 10 to the 52nd is.

  • The age of the universe so far is about 10

  • to the 26th nanoseconds.

  • 10 to the 52nd is a long time.

  • Can we ever solve a problem if we look at all 10

  • to the 52nd options?

  • What do we have to do instead?

  • AUDIENCE: Quantum computing?

  • JON BENTLEY: Pardon me?

  • AUDIENCE: Quantum computing.

  • JON BENTLEY: Quantum computing.

  • OK.

  • That's great.

  • And I have a really cool bridge across this river out here

  • that I'll sell you after class.

  • Let's talk about that.

  • Is there a nice quantum approach to this problem?

  • Maybe.

  • Maybe you could actually phrase this as an optimization problem

  • where you could maybe get some mileage out of that.

  • But we'll see.

  • So one approach is quantum computing.

  • What's another approach?

  • What are we going to have to do to make our program surmount

  • this obstacle?

  • Please.

  • AUDIENCE: Limit the search space?

  • JON BENTLEY: Pardon me?

  • AUDIENCE: [INAUDIBLE].

  • JON BENTLEY: We're going to have to limit our search space.

  • We're going to have to prune the search space.

  • That's the idea.

  • Let's try it.

  • Here's a cool problem.

  • I was at a ceremony a few weeks ago.

  • A friend of mine said here's this cool problem

  • that his daughter just brought home from high school.

  • How do you solve it?

  • Find all permutations of the 10 integer--

  • the nine integers 1 through n such

  • that each initial substring of length m is divisible by m.

  • So the whole darn thing is divisible by 9.

  • Is any permutation of integers 1 through 9 divisible by 9?

  • Well, they all sum up to numbers divisible by 9.

  • You work that.

  • Is it divisible-- are the first eight

  • characters divisible by 8?

  • But let's start with an easy one.

  • If you were doing it for size 3, 321 works.

  • Is 321 divisible by 3?

  • Is 32 divisible by 2?

  • Is 3 divisible by 1?

  • Thinking, then, it works.

  • Is 132 divisible by 3?

  • Yes.

  • Is 13 divisible by 2?

  • [MAKES BUZZER SOUND] That doesn't work.

  • So we're going to try to solve this problem.

  • My friend Greg Conti, a really great computer security guy,

  • gave me this problem.

  • How do you solve it?

  • How would you solve this problem?

  • If this high school kid says, here's

  • a problem I brought home from school, how do I solve it?

  • What would you do?

  • Ideas?

  • I'm sorry.

  • Please.

  • AUDIENCE: Yes.

  • You could write a program where the state could be [INAUDIBLE]..

  • Or actually just like a subset [INAUDIBLE]..

  • Then you iterate over [INAUDIBLE]..

  • JON BENTLEY: Great.

  • So there are two main approaches.

  • One is write a program.

  • So you can either think or you can compute.

  • Who in this room enjoys writing programs?

  • Who enjoys thinking?

  • Oh, that's an easy call.

  • What's the right approach here?

  • Well, the right answer is you think for a while.

  • If you solve it in the first three minutes,

  • don't write a program.

  • If you spend much more than five minutes on it,

  • let's write a program and see what we learn from the program.

  • We'll go back and forth.

  • Never think when you should compute, never

  • compute when you should think.

  • How do you know which one to do?

  • Try each.

  • See which one gets you further faster.

  • If you write a program for this, What

  • are the basic structures you have to deal with?

  • You have to deal with nine-digit strings that

  • are also nine-digit numbers.

  • What's a good language for dealing with that?

  • What would you-- if you had to write a program to do this,

  • what language would you choose?

  • We'll see.

  • How do you generate all intact real permutations

  • of the string?

  • Well, I hope you can see this.

  • Here's the way that I chose to do it.

  • I chose to have a recursive procedure search.

  • And I'm going to have right be the part that's already fixed,

  • left be the part that you're going to vary.

  • I could've done it the other way,

  • but I'll choose to do it this way.

  • I start with left equals that, right equals that.

  • I end when the left is empty.

  • So I have to recur down, just like we've been doing so far,

  • but I'm going to do that with strings instead.

  • And if I get to the call search of 56-- of 356 with 421978--

  • these are all fixed--

  • I'll take each one of these in turn, 3, 5, and 6,

  • put it into here.

  • So I'll call search of 56 with that, search of 36

  • with that, search of 35 with that.

  • Everyone see how that works?

  • How long will the code be in your favorite language?

  • Here's the code in my favorite language.

  • Has anyone here ever used the AWK programming language,

  • written by Aho, Weinberger, and Kernighan?

  • They observed that naming a language

  • after the initials of the authors

  • shows a certain paucity of imagination.

  • But it works.

  • So a function search of left, right,

  • that, if left equals 0-- is null, I'll check it.

  • Otherwise, what will I do here?

  • The details don't matter.

  • For i equals 1 up to the length of the left-hand side

  • of the string, search the substring

  • at the left starting at 1, going for i minus 1,

  • concatenated with the substring at the left

  • starting at i plus 1.

  • And then take the substring in the middle,

  • put it out in front of the right.

  • Do that for all i values.

  • Any questions about that?

  • The details don't matter.

  • It's not a big program.

  • If I do this, and at the end, for i equal 1 to length,

  • if the substring of the right mod i is nonzero, then return.

  • If it's not that, you print it out.

  • If I run this program, how long, ballpark,

  • will this program take for 9 factorial, ballpark?

  • What was your answer before?

  • AUDIENCE: A second.

  • JON BENTLEY: A second.

  • Great.

  • Well, we'll recycle that.

  • Reduce, reuse, recycle.

  • We'll recycle your answers.

  • If I call it originally with that string,

  • it takes about 3 seconds.

  • And it found that there was-- it searched all 9

  • factorial, 362880, 362,000 strings,

  • and found only one string there that matches that.

  • Whoops.

  • Are these divisible by 9?

  • Well, they sum to a multiple of 9, sure.

  • Is the string that ends in 72 divisible by 8?

  • Yes, that works.

  • 7, I'm not going to bother with.

  • All the way down, is 38 divisible by 2?

  • Is 381 divisible by 3?

  • This one works.

  • That's a pretty cool problem for a high school afternoon.

  • Is 3 seconds fast enough?

  • Yes.

  • The trade-off of thinking and programming.

  • Write the darn program.

  • You're done.

  • It's cool.

  • If you wanted to make it faster, how could you make it faster?

  • That's what this course is all about?

  • Always think about how you could make things faster.

  • Please.

  • AUDIENCE: Well, if you just stop searching once you know

  • one number isn't going to work.

  • JON BENTLEY: How early can you stop searching?

  • That's great.

  • So you could get constant factor speedups.

  • Like don't check for divisibility by 1 at the end.

  • You can change language, all that.

  • But those are never going to matter.

  • The big win is going to come from pruning the search.

  • How can you put in the search?

  • Any winning string must have some properties of this string.

  • What are some properties that that string has

  • that you can check for early?

  • Please.

  • AUDIENCE: The second from the left [INAUDIBLE] 2, 4, 6 or 8.

  • JON BENTLEY: The eighth position has to be a multiple of 2.

  • Furthermore, if you really think about it,

  • you can get more than that.

  • It has to be divisible by 4.

  • So an even number has to be in the eighth position.

  • Anywhere else you're going to need an even number?

  • AUDIENCE: [INAUDIBLE].

  • JON BENTLEY: This position has to be even,

  • that has to be even, that has to be even, that has to be even.

  • In general, what's the general rule?

  • AUDIENCE: All the even positions [INAUDIBLE]..

  • JON BENTLEY: Every even position has to contain an even number.

  • There are four even numbers, there are five odd numbers.

  • What other rule might you come up with?

  • AUDIENCE: The fifth position has to be 5.

  • JON BENTLEY: OK.

  • Every odd position has to be an odd number.

  • And, in particular, the fifth position has to be a 5.

  • So those are a few rules.

  • Even digits in even positions, odd digits in odd positions,

  • digit 5 in position 5.

  • Three simple rules.

  • You can test those easily.

  • The code is pretty straightforward.

  • Will that shrink the size of the search space

  • much at all really?

  • How big was the search space before?

  • 9 for the first one.

  • Now how big is the search space?

  • For the first, if you just had the three rules--

  • evens going evens, odds in odds, and 5 in the middle--

  • how many choices do you have for the first one?

  • AUDIENCE: For the first, we have [INAUDIBLE]..

  • JON BENTLEY: Four choices.

  • For the second one, you have?

  • AUDIENCE: [INAUDIBLE].

  • JON BENTLEY: It can't be a 5.

  • It has to be an odd number, not a 5.

  • You have a 4.

  • So it's 4 by 4 times 3 times 3 times 1 times 2 times 2 times

  • 1 times--

  • everyone see that?

  • We've reduce the size of the search space

  • from a third of a million to half a thousand.

  • Isn't it going to be a lot of hassle to code that?

  • I mean, is it going to take a major software development

  • effort to code that?

  • Well, yes, if you define that as a major software development

  • effort.

  • If the parity of the string length

  • is equal to the parity of the digit, then you can continue.

  • If you don't have these things, you can't continue.

  • Three lines of code allow you to do this.

  • That's the story.

  • Factorial grows quickly.

  • You can never visit the entire search space.

  • The key to speed is pruning the search.

  • We're doing just a baby branch-and-bound, it's called.

  • Some fancy algorithms can be implemented in little code.

  • That's our break.

  • We've learned a couple of things.

  • We're going to go back into the fray.

  • Any questions about this diversion

  • before we go back to the TSP?

  • These are important lessons.

  • We'll try to apply them now.

  • I got great advice yesterday from people

  • about how to do this.

  • And I seem to have skipped--

  • OK, here it is.

  • I've got it.

  • How do we prune our search?

  • Here we had these conditions.

  • How can we prune the search?

  • How can I make the program faster?

  • What's the way I can stop doing the search?

  • Simplest way, don't keep doing what doesn't work.

  • If the sum that you have so far is

  • greater than the minimum sum, by adding more to it,

  • are you going to make it less?

  • What can you do?

  • You can stop the search right there.

  • Is the resulting algorithm going to be faster?

  • Maybe.

  • It's a trade-off.

  • I'm doing more work, which takes some time,

  • but I might be able to prune the search space.

  • The question is, is this benefit worth this cost?

  • What do you think?

  • Well, on the same machine, algorithm 4 at size 12

  • took 0.6 seconds.

  • Now it's a factor of 60 faster, a factor of 40

  • faster, a factor of 100 faster.

  • Just by-- if it doesn't work, if you've already screwed up,

  • just don't keep what doesn't work.

  • That makes the thing a whole lot faster.

  • Everyone see that?

  • That's the first big win.

  • Can we do even better than that?

  • Is there any way of stopping the search with more information

  • other than, whoops, I've already gone too far?

  • Please.

  • AUDIENCE: If the nodes you visited previously--

  • JON BENTLEY: Wait.

  • Command voice.

  • Speak loudly,

  • AUDIENCE: If the nodes you visited previously

  • are the same, like the same subset

  • but a different word than a search you've done before,

  • then the answer [INAUDIBLE].

  • JON BENTLEY: That's a really powerful idea

  • that Held and Karp used to reduce it from n

  • factorial time to n squared 2 to the n time.

  • We'll get to that.

  • That's really powerful, but now we're looking for something

  • not quite that sophisticated.

  • But that's a great idea.

  • Can I somehow prune the search if a sum plus a lower bound

  • on the remaining cities is greater than the minimum sum?

  • What kind of lower bound could I get?

  • Well, I could computed a TSP path through them.

  • That's really powerful.

  • That will give me a really good bound,

  • but it's really expensive to compute.

  • So I could-- if this is a city I've done so far,

  • I could compute a TSP path to the rest, which

  • might in this case looks like this, and hook it up.

  • That's going to be a really powerful heuristic,

  • but it's going to be really expensive to compute.

  • On the other hand, I could take just the distance

  • between two random points.

  • I'm going to choose this point and this point

  • I happened to get the diameter of the set.

  • And that's a lower bound.

  • It's going to be at least that long.

  • And it's really cheap to compute,

  • but is it very effective?

  • Nyah.

  • So the first choice is effective but too expensive.

  • The second point is really inexpensive but not

  • very effective.

  • I could also compute the nearest neighbor of each city.

  • From this city, if I just compute its nearest neighbor

  • among here, so it's that.

  • This one is that.

  • That one has its own nearest neighbor.

  • I could compute these distances.

  • And that's pretty inexpensive to compute,

  • and it's a pretty good lower bound.

  • That would work.

  • Who here knows what a minimum spanning tree is?

  • Good.

  • What I'll do here is I'll take here a minimum spanning tree.

  • In cities, a tree is n minus 1 edges.

  • This tree is n minus 1 edges.

  • This is a spanning tree because it touches--

  • it connects all cities.

  • And, furthermore, it's a minimum spanning tree,

  • because, of all spanning trees, This one

  • has the minimum total distance.

  • Now, the tour is going to be less--

  • or greater in distance than the minimum spanning tree.

  • Why is that?

  • If I get a tour of this, I can just

  • knock off the longest edge.

  • And that now becomes a minimum spanning tree.

  • So the minimum spanning tree is a pretty good bound,

  • a lower bound.

  • It's cheap to compute.

  • Who here has ever seen an algorithm for computing minimum

  • spanning trees?

  • Good, good.

  • Some of you are awake in some of their classes.

  • What are the odds of that?

  • I mean, what an amazing coincidence.

  • So what we'll do is say now that a better lower bound

  • is to add the minimum spanning tree's remaining points.

  • So I change this program to if sum plus the MST distance.

  • And now I'm going to do a trick.

  • I'm going to use word parallelism.

  • I'm going to have the representation

  • of the subset of the cities as a mask, a bit

  • mask in which if the appropriate city is on,

  • the bit is turned on.

  • Otherwise, it's turned off.

  • And I just OR bits into it, and say

  • if I compute the minimum spanning tree of this set,

  • I can cut the search and return.

  • And then I just compute the MST and bring this along with me,

  • turning things off and on in the bit mask as I go down.

  • Pretty straightforward.

  • How much code will it cost to compute

  • a minimum spanning tree?

  • Ballpark?

  • Yes.

  • AUDIENCE: 30 or 20 lines of code.

  • JON BENTLEY: About that many lines of code.

  • This is the Prim-Dijkstra method.

  • It takes quadratic time.

  • For computing an MST of n points,

  • it takes n squared time.

  • It's quite simple.

  • You can do it in e log log b time.

  • But this is a simple code.

  • It's pretty straightforward.

  • Will this make the program run slower or faster?

  • What would the argument be that it might run slower?

  • Holy moly.

  • At every node I'm computing an MST.

  • That takes long time and I will run slower.

  • What's the argument to be that it might run faster?

  • Yes, but I'm getting a much more powerful pruning.

  • Is it worth it?

  • I should point out that I'm only showing the wins here to you.

  • When I redid this myself, I went down a few wrong paths.

  • I wish I would have documented them better now.

  • But I might go back and see if I can find them.

  • That would be a good thing.

  • But here it is.

  • It used to take 17 seconds.

  • Now it takes-- or 4 seconds.

  • Now it takes 0.

  • You like algorithms to take 0 seconds.

  • You'd like to live in the rounding error.

  • 4.40 to 0.2.

  • Down here, this program is not only faster,

  • it's a boatload faster.

  • And so now we can go out in this.

  • And notice here that as you go out,

  • the times usually get bigger, but they are bumpy,

  • from 2.4 seconds to 0.7 seconds, to 1.8 seconds.

  • It's because you're doing that one thing.

  • It's just the matter of the geometry.

  • The times that were originally really smooth now turn bumpy.

  • I've done experiments where I do 10 different data sets,

  • randomly drawing each one, and it's a nice smooth line.

  • But I missed doing it here to be easy.

  • Before we can go out to size 17.

  • Now we can go out to size 30.

  • Wow.

  • How cool is that?

  • That's pretty powerful.

  • Can I make this--

  • please.

  • AUDIENCE: So is it possible that the [INAUDIBLE]

  • is chosen in such a way that this thing doesn't actually

  • prune any bad permutations?

  • JON BENTLEY: That's absolutely true.

  • And I've tried this both on random point sets.

  • I've tried it on distance matrices.

  • I've tried on points where they're randomly

  • distributed around the perimeter of a circle.

  • And so this could be a lot of time.

  • Almost always, it's pretty effective.

  • Again, if I had more time, I'd talk about it.

  • But in fact we're going to go until 3:45, Charles?

  • CHARLES LEISERSON: 3:55.

  • JON BENTLEY: 3:55?

  • When the big hand is on the 11?

  • Oh.

  • Sucks to be you.

  • [STUDENTS LAUGH]

  • I profiled this bad boy, and it shows that most of the time

  • is in building minimum spanning trees.

  • Your fear that it might take a long time,

  • it might make it slower, has a basis.

  • That's where all the time is going.

  • How can I reduce the time spent in building

  • minimum spanning trees?

  • As I search this-- please.

  • AUDIENCE: Maybe don't do it every time?

  • JON BENTLEY: I could do some incremental minimum spanning

  • trees because they change a lot.

  • And so there are several responses.

  • One is whenever you're building something over again, rather

  • than building it from scratch, see if you can do

  • an incremental algorithm, where you just

  • change one bit of the minimum spanning tree.

  • If I just add one edge into the graph,

  • always try an incremental algorithm.

  • That's cool.

  • That's one sophisticated approach.

  • What is one-- what was another pretty idiot simpler approach?

  • Whenever you compute something over and over again,

  • what can you do to reduce the time spent computing it?

  • AUDIENCE: Store it?

  • JON BENTLEY: Store it.

  • Do I ever compute the same MST over and over again?

  • I don't know.

  • I think maybe it's worth a try.

  • So what I'll do is return of caching.

  • Store rather than recompute.

  • Cache MST distances rather than computing them.

  • The code looks like this.

  • The new mask is that.

  • If the MST distance array is less than 0,

  • initialize everything to 0.

  • Here I'm just going to store them all in a table of size 2

  • to the n.

  • I can do direct indexing.

  • If it's less than 0, compute it, fill in the value.

  • If sum plus that, return.

  • Not much code.

  • But do you really want to store--

  • to blast it out and to use a lazy--

  • I'm using lazy evaluation of this table here.

  • Only when I need it do I fill in a value.

  • That's not effective.

  • Rather than storing all 2 to the n tables,

  • what can I do instead?

  • What's our favorite data structure for storing stuff?

  • Hash table.

  • A cache via hash.

  • So the key to happiness.

  • You can write that down too.

  • Store them in a hash table.

  • If sum plus MST distance lookup--

  • oh, but I have to implement a hash table now.

  • How much code is that going to be?

  • Ballpark?

  • What does it cost to build a hash table?

  • Roughly.

  • Come on.

  • Yes.

  • About that many lines.

  • So just go down the hash table.

  • If you find it, return it.

  • Otherwise, make a new node, compute the distance,

  • put it in there, fill in the values, and you're done.

  • Is it going to be faster?

  • Oh, we'll see.

  • Who reads xkcd on a regular basis?

  • The rest of you are bad, bad, bad people,

  • and you should feel very guilty until you

  • go to xkcd.com and start reading this on a regular basis.

  • I mean, like wow.

  • This is two deep psychological insights

  • in one lecture for no additional fee.

  • Sir.

  • CHARLES LEISERSON: Were you resolving collisions

  • by chaining them?

  • JON BENTLEY: Right, by chaining, yes.

  • CHARLES LEISERSON: Why bother?

  • Why not just store the place value

  • and keep a key to make sure that it's the value associated

  • with the one that you want?

  • JON BENTLEY: That is a great question,

  • and the answer is left as an exercise for the listener.

  • We've got about 20 minutes, Charles.

  • CHARLES LEISERSON: Code, less code.

  • JON BENTLEY: It would be, yes.

  • And it's well worth a try.

  • All of these things are empirical questions.

  • One thing that's really important

  • to learn as a performance engineer

  • is that your intuition is almost always wrong.

  • Always try to experiment to see.

  • It's a great question.

  • When I get home, I'll actually-- when I leave here,

  • I'm going to go up to try to climb Mount Monadnock.

  • Who here has ever climbed Mount Monadnock?

  • Yes.

  • I finished climbing all 115 4,000-foot peaks

  • in the Northeastern US last year.

  • I've never climbed Monadnock.

  • I'm really eager to give it a try tomorrow.

  • xkcd.

  • Brute force n factorial.

  • The Held-Karp dynamic programming algorithm

  • uses the grown-up version of dynamic programming for n

  • squared 2 to the n, but even better.

  • Algorithm 6 looks like that if I cache the TSPs.

  • Does it have to be faster?

  • No.

  • Is it faster?

  • Oh, by about a factor of 15 there.

  • By about a factor of 25 there, 26 there.

  • You can go out now much further, 6 and 8.

  • So we've done that.

  • Is there any other way to make this program faster?

  • We've pruned the search like crazy.

  • Any other way to make it faster?

  • Please.

  • AUDIENCE: [INAUDIBLE].

  • JON BENTLEY: I forget what happens at 39.

  • Let's see.

  • At 39, it went over a minute.

  • And, like I said, this thing goes up and down.

  • I guess it just hit some weird bumps in the search space.

  • That's something else.

  • The first algorithm is completely predictable.

  • The other algorithms, you have to get more and more

  • into analysis.

  • And now the times go up and down.

  • There is a trend.

  • And, basically, I'm taking an exponent and I'm lowering--

  • I turned it from super exponential to exponential,

  • and then I'm being down on the exponent right now.

  • Can you make this run faster?

  • What we're going to do is take this idea of a greedy search.

  • I've can have smarter researching.

  • Better than a random order, I'm going

  • to do a better starting tour.

  • And what I'm going to do is always at each point sort

  • the points to look at the nearest one to the current one

  • first.

  • Start with a random one.

  • Then for the next one, always look

  • at the nearest point first, then the second nearest,

  • the third nearest, et cetera.

  • So I'll go in that order.

  • That should make the search smarter,

  • and that should guide me rather quickly

  • to the initial starting tour.

  • Rather than just a random tour, I'll

  • have a good one that will give me a better prune of the search

  • space.

  • Will that make a difference?

  • We'll have to include a sort.

  • I'll get two birds with one modification.

  • By a really dumb insertion sort, which takes up

  • that many lines of code, I'll visit the nearest city

  • first, then others in order.

  • If I do that, here it's a factor of 2,

  • there it's a factor of 8, a factor of 4.

  • But it seems to work.

  • It gives you some--

  • as you go out especially.

  • I can now go out further.

  • I lied.

  • I didn't stop my search at 60 seconds there.

  • But I can now go up further, and it seems to be a lot faster.

  • So in 1997, 20 years ago, I was really happy to get out to 30.

  • The question now is, in 20 more years,

  • how much bigger can I go?

  • If I just depend on Moore's law alone, in 20 years

  • a factor of a thousand.

  • At 30, 30 times 31 times 32, that's

  • a factor I can go up by Moore's law.

  • With a [INAUDIBLE] algorithm, it would give me two more cities

  • at this size in 20 years.

  • Can I get from 30 on to anything interesting by combining

  • Moore's law, and compiler technology, and all

  • the algorithms.

  • How far can I go?

  • Well, I was going to give a talk at Lehigh.

  • So I could go out-- in under a minute,

  • I could go to the 45-city tour.

  • Charles answered this yesterday, so he is completely clear.

  • Rorschach test.

  • Who's willing to go out--

  • what do you see there?

  • AUDIENCE: A puppy.

  • JON BENTLEY: Dancing doggy.

  • That was my answer, dancing doggy.

  • I like that a lot.

  • That's the obvious answer.

  • But Charles-- and this shows a profoundly profound mind.

  • Professor Leiserson, what is this?

  • CHARLES LEISERSON: This is a dog doing his business [INAUDIBLE]..

  • JON BENTLEY: OK.

  • So any Freudians, you feel free to go to town on that one.

  • 45-city tour, it's pretty cool.

  • Dancing doggy.

  • How far can it go?

  • I got out to 45 in under a minute.

  • 46-- I broke my rule of this--

  • I went over the minute boundary.

  • This was my Thanksgiving 2016 cycle test.

  • I was just going hog wild.

  • I was willing to spend the--

  • I had to give a--

  • I was doing this Wednesday night.

  • I had to give a lecture on Monday.

  • A hundred hours of CPU time.

  • How far can I go?

  • 47.

  • Yes.

  • Yikes, factor of 5.

  • When do I think?

  • When do I run?

  • Should I go back and [? work on it. ?]

  • 52-- wouldn't it be sweet to be able to go out to 52 factorial?

  • Wouldn't that be cool?

  • 48-- that's not bad.

  • That's looking pretty good there, actually.

  • Oh, ouch, ouch.

  • That's going to take a--

  • so that about 2 hours right there.

  • But 50, whoo, edge of my seat.

  • The turkey was smelling good, but 51.

  • And can I get to 52?

  • Will it make it?

  • Will I have to go back to my--

  • whew.

  • 3 hours and 7 minutes.

  • By combining all of these things,

  • we're able to go out to something that is just obscene.

  • 52 is obscenely huge.

  • We're able to get out there by a combination of all

  • of these things, of some really simple performance engineering

  • techniques.

  • If you're going to work on a real TSP, read the literature,

  • study that.

  • I hope we can come across some things

  • that I've written about approximation algorithms.

  • But if you really need them, forget the approximation

  • algorithms because they're too short.

  • There's a huge literature.

  • I haven't told you any of that.

  • Everything that I've done here are

  • things that you, as a person who has completed this class,

  • should be able to do.

  • All these things are well within your scope of practice,

  • as we say.

  • You will not be sued for malpractice.

  • How much code is the final thing?

  • About that much.

  • You build an MST.

  • You had a hash table.

  • Charles points out you could nuke

  • three or four of those lines.

  • You have the sort here.

  • Altogether about 160 lines.

  • Where have we been?

  • We started we could get out to 11.

  • Store the distances.

  • Out to 12.

  • Fix the starting city.

  • That was a big one.

  • Accumulate distance along the way.

  • These were all good.

  • But then by pruning the search, we

  • started making the big things.

  • Add the MST, store the distances in a hash table,

  • visit the cities in a greedy algorithm.

  • Each one of these gave us more and more and more

  • power as we went out there, till we're finally

  • able to go out pretty far.

  • There are a lot of things you can do.

  • Parallelism, faster machines, more code tuning,

  • better hashing.

  • That malloc is just begging to be removed.

  • Better pruning, a better starting tour, better bounds.

  • I can take the MST length plus the nearest--

  • that's why I do this MST--

  • plus the nearest neighbor to each of the ends.

  • I can get that.

  • Would that make a big difference?

  • Empirical question.

  • Easy to find out.

  • Can I move by pruning tests earlier?

  • Better sorting.

  • This is really cool.

  • Can I maybe just sort once for each city

  • to get that sorted list, then go through that,

  • precompute and sort, and select the things in order?

  • Is that going to be a win in this context?

  • The main ideas here are caching, precomputing, storing this,

  • avoiding the work.

  • Can I change that n squared algorithm to just a linear time

  • selection?

  • All of these things are really fun to look at.

  • I've tried to tell you about incremental software

  • development.

  • I started off with around 30, 40 lines of code.

  • It grew to 160.

  • But altogether all the versions come

  • to about 600 lines of code.

  • You've now seen more than you need

  • for one life about recursive generation.

  • It's a surprisingly powerful technique

  • if you ever need to use it.

  • No excuses now.

  • You're obligated to build it immediately.

  • Storing precomputed results, partial sums, early cut-offs.

  • Algorithms and data structures.

  • These are things that sounded fancy in your algorithms class,

  • but you just pull them out when you need them.

  • Vectors, strings, arrays and bit vectors,

  • minimum spanning trees, hash tables, insertion sort.

  • It's easy.

  • It's a dozen lines of code here. two dozen lines of code there.

  • I believe that Charles may had mentioned

  • earlier that I wrote a book in 1982 about code tuning.

  • At the time, you did these in the smaller programs.

  • Now compilers do all that for you.

  • But these ideas-- some of these ideas still apply.

  • Store precomputed results.

  • Rather than [INAUDIBLE] elimination in an expression,

  • you now put interpoint distances in a matrix

  • or a table of MST lengths.

  • Lazy evaluation.

  • You compute the n squared distances eagerly

  • but only the MSTs that you need.

  • Don't bother computing them all.

  • That's essentially what Held and Karp does.

  • Short-circuiting monotone functions, reordering tests,

  • word parallelism.

  • These are the things that you as performance engineers

  • can do quite readily.

  • I had a lot of tools behind the scenes.

  • I wish I could come back and give you

  • another hour about how I really did this

  • with the analysis and the tools that I used.

  • I had a driver to make the experiments easy,

  • a whole bunch of profilers.

  • Where is the time really going here?

  • What should I focus on?

  • Cost models that allowed me to estimate those,

  • how much does an MST cost.

  • A spreadsheet was my lab notebook

  • for graphs of performance, all sorts of curve fitting.

  • But these are the main things I wanted to tell you about.

  • The big hand is getting about nine minutes away from the 11.

  • Professor Leiserson, is there anything else

  • that these fine, young semi-humanoids need

  • to know about this material?

  • CHARLES LEISERSON: Does anybody happen

  • to see any analogies with the current project 4.

  • Maybe people could chat a little bit

  • about where they see analogies [INAUDIBLE]..

  • JON BENTLEY: I don't know it, but one

  • of my first exposures to MIT was when

  • I had Donovan as a software systems book,

  • and it was dedicated to 6.51 graduate students.

  • I saw that I thought, that bastard.

  • I'm sure that the six students really worked hard on it,

  • but to say that the seventh student worked only a little

  • much more over halfway and then to be

  • so precise, that's just cruel.

  • What a son of a bitch that guy had to be.

  • So I don't know what project 4 is, but is it Leiserchess?

  • Oh, great.

  • I know what that is.

  • So what things-- have you used any of these techniques?

  • Did you ever prune searches?

  • Did you ever store results?

  • What did you do in project 4?

  • You're delegating this.

  • That's a natural leader right there.

  • AUDIENCE: We talked about search pruning-- we already have--

  • JON BENTLEY: Speak up so all of them can hear, please.

  • AUDIENCE: Commander voice.

  • So we already have --everybody in this room

  • knows-- alpha-beta pruning.

  • [INAUDIBLE] It's got search.

  • I don't know how many teams are already working on search

  • but at least my team is working on changing

  • order representation first.

  • So we haven't gotten into pruning search yet,

  • but that's definitely on the horizon [INAUDIBLE]..

  • JON BENTLEY: Is there anyone here

  • from the state of California?

  • I was born in California.

  • When you hear alpha beta, apart from the search,

  • what do you think of?

  • AUDIENCE: The grocery store.

  • JON BENTLEY: There's a grocery store there called Alpha Beta.

  • And when Knuth wrote a paper on that topic,

  • he went out and bought a box of Alpha Beta

  • prunes that he had in his desk.

  • So he was an expert in two senses on alpha beta pruning.

  • So good.

  • Other techniques?

  • Please.

  • AUDIENCE: The hashing.

  • There's one function [INAUDIBLE] takes a long time,

  • and suggested maybe you could somehow

  • keep track of the laser path with a hash table [INAUDIBLE]..

  • JON BENTLEY: Great.

  • Did you resolve collisions at all?

  • Or did you just have one element there with a key?

  • How did you address the problem that Charles mentioned of--

  • what kind of hashing did you use?

  • AUDIENCE: So we haven't used caching yet.

  • JON BENTLEY: Other techniques?

  • CHARLES LEISERSON: Yes.

  • That's a classic example of the fastest way

  • to compute is not to compute at all.

  • JON BENTLEY: In general, in life no problem is so big

  • that it can't be run away from.

  • These things about avoiding work and being lazy

  • are certainly models for organizing your own life.

  • The lazy evaluation really works in the real world.

  • Other questions?

  • Was that a question or a random obscene hand gesture?

  • AUDIENCE: [INAUDIBLE].

  • JON BENTLEY: Please.

  • AUDIENCE: [INAUDIBLE] state-of the-art [INAUDIBLE]??

  • JON BENTLEY: Oh.

  • That's a great question.

  • I worked on this problem a lot in the early 1990s

  • with my colleague David Johnson, who literally wrote

  • the book on NP-completeness.

  • An MIT PhD guy.

  • We were really happy we're in--

  • at the time, in a couple of hours of CPU time

  • we could solve 100,000 city problems to

  • within a few percent.

  • We were able to solve a million city problems in a day of CPU

  • time to within a few percent.

  • And we were ecstatic.

  • That was really big.

  • So we could go out that big to within a few percent.

  • If we worked really, really hard,

  • we can get 10,000 problems down within a half a percent.

  • But if you want to go all the way

  • to have not only the optimal solution but a proof that it's

  • optimal, for a while people bragged about

  • we finally solved that problem.

  • This will let you see about what was done.

  • We solved the problem of all 48 state capitals.

  • So for a while that was the state of the art.

  • And then that number has crept over time.

  • And now you can get exact solutions

  • to some famous problems into the tens of thousands

  • by using lots and lots of really clever searching

  • the branching down with really clever lower bounds

  • to guide it up.

  • And you at one point get a tour, and you can make that tour.

  • But then you get a proof of a lower bound along

  • with it to do that.

  • CHARLES LEISERSON: Hey, old man, I

  • want to let you know that there are actually

  • now 50 states in the union.

  • JON BENTLEY: No.

  • What time did this happen?

  • You can tell that I am much, much, much older than Charles,

  • and he never lets me hear the end of it.

  • I trust that the rest of you-- this

  • is like the third free deep psychology insight,

  • is be kind to old people ignore the example

  • that the kid over there sets and show some class

  • and respect to me and my fellow geezers.

  • CHARLES LEISERSON: We were both born in 1953.

  • JON BENTLEY: But I was born in the good part of 1953.

  • In particular, I was born before Her Majesty

  • the Queen of England assumed the throne.

  • Can you make the same claim?

  • CHARLES LEISERSON: I cannot make the same claim.

  • JON BENTLEY: I'm sorry.

  • He can, but only because he's a sneaky bastard.

  • Can you make it truthfully is the question

  • that I should have asked.

  • Other questions?

  • This class can be very important.

  • Like I said, I spent the past almost half

  • century as a working computer programmer.

  • The majority of that thing I've done most

  • is performance engineering.

  • It's allowed me to do a number of really interesting things.

  • I've been able to dabble in all sorts of computational systems,

  • ranging from automated gerrymandering.

  • Every time you make a telephone call in this country,

  • if it's, say, a call from inside an institution like a hospital

  • of a university, it uses some code that I wrote,

  • some of the performance things.

  • If you make a long-distance call,

  • it uses code that I wrote.

  • If you've ever used something called Google internet

  • search or maps, or stocks or anything else,

  • that uses some algorithms I've done.

  • It's incredibly satisfying.

  • It's been a very, very fulfilling way for me

  • to spend a big chunk of my life.

  • I am grateful.

  • It's allowed me to make friends, whom

  • I've known for almost half a century,

  • and to our wonderful dear people.

  • And it's been a great way for my life.

  • I hope that performance engineering is as good to you

  • as it has been to me.

  • Anything else, professor?

  • CHARLES LEISERSON: Thank you very much, Jon.

  • JON BENTLEY: Thank you.

  • [STUDENTS APPLAUD]

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B1 中級

21.調整TSP算法 (21. Tuning a TSP Algorithm)

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    林宜悉 發佈於 2021 年 01 月 14 日
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