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  • JULIAN SHUN: All right.

  • So we've talked a little bit about caching before,

  • but today we're going to talk in much more detail about caching

  • and how to design cache-efficient algorithms.

  • So first, let's look at the caching hardware

  • on modern machines today.

  • So here's what the cache hierarchy looks

  • like for a multicore chip.

  • We have a whole bunch of processors.

  • They all have their own private L1 caches

  • for both the data, as well as the instruction.

  • They also have a private L2 cache.

  • And then they share a last level cache, or L3 cache,

  • which is also called LLC.

  • They're all connected to a memory controller

  • that can access DRAM.

  • And then, oftentimes, you'll have multiple chips

  • on the same server, and these chips

  • would be connected through a network.

  • So here we have a bunch of multicore chips

  • that are connected together.

  • So we can see that there are different levels of memory

  • here.

  • And the sizes of each one of these levels of memory

  • is different.

  • So the sizes tend to go up as you move up

  • the memory hierarchy.

  • The L1 caches tend to be about 32 kilobytes.

  • In fact, these are the specifications for the machines

  • that you're using in this class.

  • So 32 kilobytes for both the L1 data cache

  • and the L1 instruction cache.

  • 256 kilobytes for the L2 cache.

  • so the L2 cache tends to be about 8 to 10 times

  • larger than the L1 cache.

  • And then the last level cache, the size is 30 megabytes.

  • So this is typically on the order of tens of megabytes.

  • And then DRAM is on the order of gigabytes.

  • So here we have 128 gigabyte DRAM.

  • And nowadays, you can actually get machines

  • that have terabytes of DRAM.

  • So the associativity tends to go up as you move up

  • the cache hierarchy.

  • And I'll talk more about associativity

  • on the next couple of slides.

  • The time to access the memory also tends to go up.

  • So the latency tends to go up as you move up

  • the memory hierarchy.

  • So the L1 caches are the quickest to access,

  • about two nanoseconds, just rough numbers.

  • The L2 cache is a little bit slower--

  • so say four nanoseconds.

  • Last level cache, maybe six nanoseconds.

  • And then when you have to go to DRAM,

  • it's about an order of magnitude slower-- so 50 nanoseconds

  • in this example.

  • And the reason why the memory is further down in the cache

  • hierarchy are faster is because they're

  • using more expensive materials to manufacture these things.

  • But since they tend to be more expensive, we can't fit as much

  • of that on the machines.

  • So that's why the faster memories are smaller

  • than the slower memories.

  • But if we're able to take advantage of locality

  • in our programs, then we can make use of the fast memory

  • as much as possible.

  • And we'll talk about ways to do that in this lecture today.

  • There's also the latency across the network, which

  • tends to be cheaper than going to main memory

  • but slower than doing a last level cache access.

  • And there's a lot of work in trying

  • to get the cache coherence protocols right, as we

  • mentioned before.

  • So since these processors all have private caches,

  • we need to make sure that they all

  • see a consistent view of memory when

  • they're trying to access the same memory

  • addresses in parallel.

  • So we talked about the MSI cache protocol before.

  • And there are many other protocols out there,

  • and you can read more about these things online.

  • But these are very hard to get right,

  • and there's a lot of verification involved

  • in trying to prove that the cache coherence protocols are

  • correct.

  • So any questions so far?

  • OK.

  • So let's talk about the associativity of a cache.

  • So here I'm showing you a fully associative cache.

  • And in a fully associative cache,

  • a cache block can reside anywhere in the cache.

  • And a basic unit of movement here is a cache block.

  • In this example, the cache block size is 4 bytes,

  • but on the machines that we're using for this class,

  • the cache block size is 64 bytes.

  • But for this example, I'm going to use a four byte cache line.

  • So each row here corresponds to one cache line.

  • And a fully associative cache means that each line here

  • can go anywhere in the cache.

  • And then here we're also assuming

  • a cache size that has 32 bytes.

  • So, in total, it can store eight cache line

  • since the cache line is 4 bytes.

  • So to find a block in a fully associative cache,

  • you have to actually search the entire cache,

  • because a cache line can appear anywhere in the cache.

  • And there's a tag associated with each of these cache lines

  • here that basically specify which

  • of the memory addresses in virtual memory space

  • it corresponds to.

  • So for the fully associative cache,

  • we're actually going to use most of the bits of that address

  • as a tag.

  • We don't actually need the two lower order

  • bits, because the things are being

  • moved at the granularity of cache lines, which

  • are four bytes.

  • So the two lower order bits are always going to be the same,

  • but we're just going to use the rest of the bits

  • to store the tag.

  • So if our address space is 64 bits,

  • then we're going to use 62 bits to store the tag in a fully

  • associative caching scheme.

  • And when a cache becomes full, a block

  • has to be evicted to make room for a new block.

  • And there are various ways that you can

  • decide how to evict a block.

  • So this is known as the replacement policy.

  • One common replacement policy is LRU Least Recently Used.

  • So you basically kick the thing out that

  • has been used the farthest in the past.

  • The other schemes, such as second chance and clock

  • replacement, we're not going to talk

  • too much about the different replacement schemes today.

  • But you can feel free to read about these things online.

  • So what's a disadvantage of this scheme?

  • Yes?

  • AUDIENCE: It's slow.

  • JULIAN SHUN: Yeah.

  • Why is it slow?

  • AUDIENCE: Because you have to go all the way [INAUDIBLE]..

  • JULIAN SHUN: Yeah.

  • So the disadvantage is that searching

  • for a cache line in the cache can be pretty slow, because you

  • have to search entire cache in the worst case,

  • since a cache block can reside anywhere in the cache.

  • So even though the search can go on in parallel and hardware

  • is still expensive in terms of power and performance

  • to have to search most of the cache every time.

  • So let's look at another extreme.

  • This is a direct mapped cache.

  • So in a direct mapped cache, each cache block

  • can only go in one place in the cache.

  • So I've color-coded these cache blocks here.

  • So the red blocks can only go in the first row of this cache,

  • the orange ones can only go in the second row, and so on.

  • And the position which a cache block can go into

  • is known as that cache blocks set.

  • So the set determines the location

  • in the cache for each particular block.

  • So let's look at how the virtual memory address is divided up

  • into and which of the bits we're going

  • to use to figure out where a cache block should

  • go in the cache.

  • So we have the offset, we have the set,

  • and then the tag fields.

  • The offset just tells us which position

  • we want to access within a cache block.

  • So since a cache block has B bytes,

  • we only need log base 2 of B bits as the offset.

  • And the reason why we have to offset

  • is because we're not always accessing something

  • at the beginning of a cache block.

  • We might want to access something in the middle.

  • And that's why we need the offset

  • to specify where in the cache block we want to access.

  • Then there's a set field.

  • And the set field is going to determine which position

  • in the cache that cache block can go into.

  • So there are eight possible positions for each cache block.

  • And therefore, we only need log base 2 of 8 bits--

  • so three bits for the set in this example.

  • And more generally, it's going to be log base 2 of M over B.

  • And here, M over B is 8.

  • And then, finally, we're going to use the remaining

  • bits as a tag.

  • So w minus log base 2 of M bits for the tag.

  • And that gets stored along with the cache block in the cache.

  • And that's going to uniquely identify

  • which of the memory blocks the cache block corresponds to

  • in virtual memory.

  • And you can verify that the sum of all these quantities

  • here sums to w bits.

  • So in total, we have a w bit address space.

  • And the sum of those three things is w.

  • So what's the advantage and disadvantage of this scheme?

  • So first, what's a good thing about this scheme compared

  • to the previous scheme that we saw?

  • Yes?

  • AUDIENCE: Faster.

  • JULIAN SHUN: Yeah.

  • It's fast because you only have to check one place.

  • Because each cache block can only

  • go in one place in a cache, and that's only place

  • you have to check when you try to do a lookup.

  • If the cache block is there, then you find it.

  • If it's not, then you know it's not in the cache.

  • What's the downside to this scheme?

  • Yeah?

  • AUDIENCE: You only end up putting the red ones

  • into the cache and you have mostly every [INAUDIBLE],, which

  • is totally [INAUDIBLE].

  • JULIAN SHUN: Yeah.

  • So good answer.

  • So the downside is that you might, for example, just

  • be accessing the red cache blocks

  • and then not accessing any of the other cache blocks.

  • They'll all get mapped to the same location in the cache,

  • and then they'll keep evicting each other,

  • even though there's a lot of empty space in the cache.

  • And this is known as a conflict miss.

  • And these can be very bad for performance

  • and very hard to debug.

  • So that's one downside of a direct map

  • cache is that you can get these conflict misses where you have

  • to evict things from the cache even though there's

  • empty space in the cache.

  • So as we said, finding a block is very fast.

  • Only a single location in the cache has to be searched.

  • But you might suffer from conflict

  • misses if you keep axing things in the same set

  • repeatedly without accessing the things in the other sets.

  • So any questions?

  • OK.

  • So these are sort of the two extremes for cache design.

  • There's actually a hybrid solution

  • called set associative cache.

  • And in a set associative cache, you still sets,

  • but each of the sets contains more than one line now.

  • So all the red blocks still map to the red set,

  • but there's actually two possible locations

  • for the red blocks now.

  • So in this case, this is known as a two-way associate of cache

  • since there are two possible locations inside each set.

  • And again, a cache block's set determines k possible cache

  • locations for that block.

  • So within a set it's fully associative,

  • but each block can only go in one of the sets.

  • So let's look again at how the bits are

  • divided into in the address.

  • So we still have the tag set and offset fields.

  • The offset field is still a log base 2 of b.

  • The set field is going to take log base 2 of M over kB bits.

  • So the number of sets we have is M over kB.

  • So we need log base 2 of that number

  • to represent the set of a block.

  • And then, finally, we use the remaining bits as a tag,

  • so it's going to be w minus log base 2 of M over k.

  • And now, to find a block in the cache,

  • only k locations of it's set must be searched.

  • So you basically find which set the cache block maps too,

  • and then you check all k locations

  • within that set to see if that cached block is there.

  • And whenever you want to whenever

  • you try to put something in the cache because it's not there,

  • you have to evict something.

  • And you evict something from the same set as the block

  • that you're placing into the cache.

  • So for this example, I showed a two-way associative cache.

  • But in practice, the associated is usually bigger

  • say eight-way, 16-way, or sometimes 20-way.

  • And as you keep increasing the associativity,

  • it's going to look more and more like a fully associative cache.

  • And if you have a one way associative cache,

  • then there's just a direct map cache.

  • So this is a sort of a hybrid in between-- a fully mapped cache

  • and a fully associative cache in a direct map cache.

  • So any questions on set associative caches ?

  • OK.

  • So let's go over a taxonomy of different types of cache

  • misses that you can incur.

  • So the first type of cache miss is called a cold miss.

  • And this is the cache miss that you

  • have to incur the first time you access a cache block.

  • And if you need to access this piece of data,

  • there's no way to get around getting a cold miss for this.

  • Because your cache starts out not having this block,

  • and the first time you access it,

  • you have to bring it into cache.

  • Then there are capacity misses.

  • So capacity misses are cache misses

  • You get because the cache is full

  • and it can't fit all of the cache blocks

  • that you want to access.

  • So you get a capacity miss when the previous cache

  • copy would have been evicted even with a fully

  • associative scheme.

  • So even if all of the possible locations in your cache

  • could be used for a particular cache line,

  • that cache line still has to be evicted because there's not

  • enough space.

  • So that's what's called a capacity miss.

  • And you can deal with capacity misses

  • by introducing more locality into your code, both spatial

  • and temporal locality.

  • And we'll look at ways to reduce the capacity

  • misses of algorithms later on in this lecture.

  • Then there are conflict misses.

  • And conflict misses happen in set associate of caches

  • when you have too many blocks from the same set wanting

  • to go into the cache.

  • And some of these have to be evicted,

  • because the set can't fit all of the blocks.

  • And these blocks wouldn't have been

  • evicted if you had a fully associative scheme, so these

  • are what's called conflict misses.

  • For example, if you have 16 things in a set

  • and you keep accessing 17 things that all belong in the set,

  • something's going to get kicked out

  • every time you want to access something.

  • And these cache evictions might not

  • have happened if you had a fully associative cache.

  • And then, finally, they're sharing misses.

  • So sharing misses only happened in a parallel context.

  • And we talked a little bit about true sharing

  • a false sharing misses in prior lectures.

  • So let's just review this briefly.

  • So a sharing miss can happen if multiple processors are

  • accessing the same cache line and at least one of them

  • is writing to that cache line.

  • If all of the processors are just

  • reading from the cache line, then the cache [INAUDIBLE]

  • protocol knows how to make it work so that you don't get

  • misses.

  • They can all access the same cache line at the same time

  • if nobody's modifying it.

  • But if at least one processor is modifying it,

  • you could get either true sharing misses

  • or false sharing misses.

  • So a true sharing miss is when two processors are

  • accessing the same data on the same cache line.

  • And as you recall from a previous lecture,

  • if one of the two processors is writing to this cache

  • line, whenever it does a write it

  • needs to acquire the cache line in exclusive mode

  • and then invalidate that cache line and all other caches.

  • So then when one another processor

  • tries to access the same memory location,

  • it has to bring it back into its own cache,

  • and then you get a cache miss there.

  • A false sharing this happens if two processes

  • are accessing different data that just happened to reside

  • on the same cache line.

  • Because the basic unit of movement

  • is a cache line in the architecture.

  • So even if you're asking different things,

  • if they are on the same cache line,

  • you're still going to get a sharing miss.

  • And false sharing is pretty hard to deal with,

  • because, in general, you don't know what data

  • gets placed on what cache line.

  • There are certain heuristics you can use.

  • For example, if you're mallocing a big memory region,

  • you know that that memory region is contiguous,

  • so you can space your access is far enough apart

  • by different processors so they don't touch the same cache

  • line.

  • But if you're just declaring local variables on the stack,

  • you don't know where the compiler

  • is going to decide to place these variables

  • in the virtual memory address space.

  • So these are four different types of cache

  • misses that you should know about.

  • And there's many models out there

  • for analyzing the cache performance of algorithms.

  • And some of the models ignore some of these different types

  • of cache misses.

  • So just be aware of this when you're looking at algorithm

  • analysis, because not all of the models

  • will capture all of these different types of cache

  • misses.

  • So let's look at a bad case for conflict misses.

  • So here I want to access a submatrix within a larger

  • matrix.

  • And recall that matrices are stored in row-major order.

  • And let's say our matrix is 4,096 columns by 4,096 rows

  • and it still stores doubles.

  • So therefore, each row here is going

  • to contain 2 to the 15th bytes, because 4,096

  • is t2 to the 12th, and we have doubles,

  • which takes eight bytes.

  • So 2 to the 12 times to the 3rd, which is 2 to the 15th.

  • We're going to assume the word width is 64, which is standard.

  • We're going to assume that we have a cache size of 32k.

  • And the cache block size is 64, which, again, is standard.

  • And let's say we have a four-way associative cache.

  • So let's look at how the bits are divided into.

  • So again we have this offset, which

  • takes log base 2 of B bits.

  • So how many bits do we have for the offset in this example?

  • Right.

  • So we have 6 bits.

  • So it's just log base 2 of 64.

  • What about for the set?

  • How many bits do we have for that?

  • 7.

  • Who said 7?

  • Yeah.

  • So it is 7.

  • So M is 32k, which is 2 to the 15th.

  • And then k is 2 to the 2, b is 2 6.

  • So it's 2 to the 15th divided by 2 the 8th, which is to the 7th.

  • And log base 2 of that is 7.

  • And finally, what about the tag field?

  • AUDIENCE: 51.

  • JULIAN SHUN: 51.

  • Why is that?

  • AUDIENCE: 64 minus 13.

  • JULIAN SHUN: Yeah.

  • So it's just 64 minus 7 minus 6, which is 51.

  • OK.

  • So let's say that we want to access a submatrix

  • within this larger matrix.

  • Let's say we want to acts as a 32 by 32 submatrix.

  • And THIS is pretty common in matrix algorithms, where

  • you want to access submatrices, especially in divide

  • and conquer algorithms.

  • And let's say we want to access a column of this submatrix A.

  • So the addresses of the elements that we're going to access

  • are as follows--

  • so let's say the first element in the column

  • is stored at address x.

  • Then the second element in the column

  • is going to be stored at address x plus 2 to the 15th,

  • because each row has 2 to the 15th bytes,

  • and we're skipping over an entire row

  • here to get to the element in the next row of the sub matrix.

  • So we're going to add 2 to the 15th.

  • And then to get the third element,

  • we're going to add 2 times 2 to the 15th.

  • And so on, until we get to the last element,

  • which is x plus 31 times 2 to the 15th.

  • So which fields of the address are

  • changing as we go through one column of this submatrix?

  • AUDIENCE: You're just adding multiple [INAUDIBLE] tag

  • the [INAUDIBLE].

  • JULIAN SHUN: Yeah.

  • So it's just going to be the tag that's changing.

  • The set and the offset are going to stay the same, because we're

  • just using the lower 13 bits to store the set and a tag.

  • And therefore, when we increment by 2 to the 15th,

  • we're not going to touch the set and the offset.

  • So all of these addresses fall into the same set.

  • And this is a problem, because our cache

  • is only four-way associative.

  • So we can only fit four cache lines in each set.

  • And here, we're accessing 31 of these things.

  • So by the time we get to the next column of A,

  • all the things that we access in the current column of A

  • are going to be evicted from cache already.

  • And this is known as a conflict miss,

  • because if you had a fully associative cache

  • this might not have happened, because you could actually

  • use any location in the cache to store these cache blocks.

  • So does anybody have any questions on why

  • we get conflict misses here?

  • So anybody have any ideas on how to fix this?

  • So what can I do to make it so that I'm not

  • incrementing by exactly 2 to the 15th every time?

  • Yeah.

  • AUDIENCE: So pad the matrix?

  • JULIAN SHUN: Yeah.

  • So one solution is to pad the matrix.

  • You can add some constant amount of space

  • to the end of the matrix.

  • So each row is going to be longer than 2

  • to the 15th bytes.

  • So maybe you add some small constant like 17.

  • So add 17 bytes to the end of each row.

  • And now, when you access a column of this submatrix,

  • you're not just incrementing by 2 to the 15th,

  • you're also adding some small integer.

  • And that's going to cause the set and the offset fields

  • to change as well, and you're not

  • going to get as many conflict misses.

  • So that's one way to solve the problem.

  • It turns out that if you're doing a matrix multiplication

  • algorithm, that's a cubic work algorithm,

  • and you can basically afford to copy the submatrix

  • into a temporary 32 by 32 matrix,

  • do all the operations on the temporary matrix,

  • and then copy it back out to the original matrix.

  • The copying only takes quadratic work

  • to do across the whole algorithm.

  • And since the whole algorithm takes cubic work,

  • the quadratic work is a lower order term.

  • So you can use temporary space to make sure that you

  • don't get conflict misses.

  • Any questions?

  • So this was conflict misses.

  • So conflict misses are important.

  • But usually, we're going to be first concerned

  • about getting good spatial and temporal locality,

  • because those are usually the higher order

  • factors in the performance of a program.

  • And once we get good spatial and temporal locality

  • in our program, we can then start

  • worrying about conflict misses, for example,

  • by using temporary space or padding our data

  • by some small constants so that we don't

  • have as if any conflict misses.

  • So now, I want to talk about a model

  • that we can use to analyze the cache

  • performance of algorithms.

  • And this is called the ideal-cache model.

  • So in this model, we have a two-level cache hierarchy.

  • So we have the cache and then main memory.

  • The cache size is of size M, and the cache line size

  • is of B bytes.

  • And therefore, we can fit M over V cache lines inside our cache.

  • This model assumes that the cache is fully associative,

  • so any cache block can go anywhere in the cache.

  • And it also assumes an optimal omniscient replacement policy.

  • So this means that where we want to evict a cache

  • block from the cache, we're going

  • to pick the thing to evict that gives us

  • the best performance overall.

  • It gives us the lowest number of cache

  • misses throughout our entire algorithm.

  • So we're assuming that we know the sequence of memory requests

  • throughout the entire algorithm.

  • And that's why it's called the omniscient mission replacement

  • policy.

  • And if something is in cache, you can operate on it for free.

  • And if something is in main memory,

  • you have to bring it into cache and then

  • you incur a cache miss.

  • So two performance measures that we care about--

  • first, we care about the ordinary work,

  • which is just the ordinary running time of a program.

  • So this is the same as before when

  • we were analyzing algorithms.

  • It's just a total number of operations

  • that the program does.

  • And the number of cache misses is

  • going to be the number of lines we

  • have to transfer between the main memory and the cache.

  • So the number of cache misses just

  • counts a number of cache transfers,

  • whereas as the work counts all the operations that you

  • have to do in the algorithm.

  • So ideally, we would like to come up

  • with algorithms that have a low number of cache misses

  • without increasing the work from the traditional standard

  • algorithm.

  • Sometimes we can do that, sometimes we can't do that.

  • And then there's a trade-off between the work

  • and the number of cache misses.

  • And it's a trade-off that you have

  • to decide whether it's worthwhile as a performance

  • engineer.

  • Today, we're going to look at an algorithm

  • where you can't actually reduce the number of cache

  • misses without increasing the work.

  • So you basically get the best of both worlds.

  • So any questions on this ideal cache model?

  • So this model is just used for analyzing algorithms.

  • You can't actually buy one of these caches at the store.

  • So this is a very ideal cache, and they don't exist.

  • But it turns out that this optimal omniscient replacement

  • policy has nice theoretical properties.

  • And this is a very important lemma that was proved in 1985.

  • It's called the LRU lemma.

  • It was proved by Slater and Tarjan.

  • And the lemma says, suppose that an algorithm incurs

  • Q cache misses on an ideal cache of size M. Then,

  • on a fully associative cache of size 2M, that uses the LRU,

  • or Least Recently Used replacement policy,

  • it incurs at most 2Q cache misses.

  • So what this says is if I can show the number of cache

  • misses for an algorithm on the ideal cache,

  • then if I take a fully associative cache that's twice

  • the size and use the LRU replacement policy,

  • which is a pretty practical policy,

  • then the algorithm is going to incur,

  • at most, twice the number of cache misses.

  • And the implication of this lemma

  • is that for asymptotic analyses, you

  • can assume either the optimal replacement policy or the LRU

  • replacement policy as convenient.

  • Because the number of cache misses

  • is just going to be within a constant factor of each other.

  • So this is a very important lemma.

  • It says that this basically makes

  • it much easier for us to analyze our cache misses in algorithms.

  • And here's a software engineering principle

  • that I want to point out.

  • So first, when you're trying to get good performance,

  • you should come up with a theoretically good algorithm

  • that has good balance on the work and the cache complexity.

  • And then after you come up with an algorithm that's

  • theoretically good, then you start engineering

  • for detailed performance.

  • You start worrying about the details such as real world

  • caches not being fully associative, and, for example,

  • loads and stores having different costs with respect

  • to bandwidth and latency.

  • But coming up with a theoretically good algorithm

  • is the first order bit to getting good performance.

  • Questions?

  • So let's start analyzing the number of cache

  • misses in a program.

  • So here's a lemma.

  • So the lemma says, suppose that a program reads a set of r data

  • segments, where the i-th segment consists of s sub i bytes.

  • And suppose that the sum of the sizes of all the segments

  • is equal to N. And we're going to assume that N is less than M

  • over 3.

  • So the sum of the size of the segments

  • is less than the cache size divided by 3.

  • We're also going to assume that N over r

  • is greater than or equal to B. So recall

  • that r is the number of data segments we have,

  • and N is the total size of the segment.

  • So what does N over r mean, semantically?

  • Yes.

  • AUDIENCE: Average [INAUDIBLE].

  • JULIAN SHUN: Yeah.

  • So N over r is the just the average size of a segment.

  • And here we're saying that the average size of a segment

  • is at least B-- so at least the size of a cache line.

  • So if these two assumptions hold, then all of the segments

  • are going to fit into cache, and the number of cache

  • misses to read them all is, at most, 3 times N over B.

  • So if you had just a single array of size N,

  • then the number of cache misses you

  • would need to read that array into cache

  • is going to be N over B. And this

  • is saying that, even if our data is divided

  • into a bunch of segments, as long as the average length

  • of the segments is large enough, then the number of cache misses

  • is just a constant factor worse than reading a single array.

  • So let's try to prove this cache miss lemma.

  • So here's a proof so.

  • A single segment, s sub i is going

  • to incur at most s sub i over B plus 2 cache misses.

  • So does anyone want to tell me where the s sub i over B plus 2

  • comes from?

  • So let's say this is a segment that we're analyzing,

  • and this is how it's aligned in virtual memory.

  • Yes?

  • AUDIENCE: How many blocks it could overlap worst case.

  • JULIAN SHUN: Yeah.

  • So s sub i over B plus 2 is the number of blocks that could

  • overlap within the worst case.

  • So you need s sub i over B cache misses just

  • to load those s sub i bytes.

  • But then the beginning and the end of that segment

  • might not be perfectly aligned with a cache line boundary.

  • And therefore, you could waste, at most, one block

  • on each side of the segment.

  • So that's where the plus 2 comes from.

  • So to get the total number of cache

  • misses, we just have to sum this quantity from i equals 1 to r.

  • So if I sum s sub i over B from i equals 1 to r,

  • I just get N over B, by definition.

  • And then I sum 2 from i equals 1 to r.

  • So that just gives me 2r.

  • Now, I'm going to multiply the top and the bottom

  • with the second term by B. So 2r B over B now.

  • And then that's less than or equal to N over B

  • plus 2N over B. So where did I get this inequality here?

  • Why do I know that 2r B is less than or equal to 2N?

  • Yes?

  • AUDIENCE: You know that the N is greater than or equal to B r.

  • JULIAN SHUN: Yeah.

  • So you know that N is greater than or equal to B

  • r by this assumption up here.

  • So therefore, r B is less than or equal to N.

  • And then, N B plus 2 N B just sums up to 3 N B.

  • So in the worst case, we're going to incur 3N over B cache

  • misses.

  • So any questions on this cache miss lemma?

  • So the Important thing to remember here is that if you

  • have a whole bunch of data segments and the average length

  • of your segments is large enough--

  • bigger than a cache block size--

  • then you can access all of these segments just

  • like a single array.

  • It only increases the number of cache

  • misses by a constant factor.

  • And if you're doing an asymptotic analysis,

  • then it doesn't matter.

  • So we're going to be using this cache miss lemma later

  • on when we analyze algorithms.

  • So another assumption that we're going to need

  • is called the tall cache assumption.

  • And the tall cache assumption basically

  • says that the cache is taller than it is wide.

  • So it says that B squared is less than c M

  • for some sufficiently small constant c less than

  • or equal to 1.

  • So in other words, it says that the number of cache lines

  • M over B you have is going to be bigger than B.

  • And this tall cache assumption is usually

  • satisfied in practice.

  • So here are the cache line sizes and the cache

  • sizes on the machines that we're using.

  • So cache line size is 64 bytes, and the L1 cache size

  • is 32 kilobytes.

  • So 64 bytes squared, that's 2 to the 12th.

  • And 32 kilobytes is 2 to the 15th bytes.

  • So 2 to the 12th is less than 2 to the 15th,

  • so it satisfies the tall cache assumption.

  • And as we go up the memory hierarchy,

  • the cache size increases, but the cache line length

  • stays the same.

  • So the cache has become even taller

  • as we move up the memory hierarchy.

  • So let's see why this tall cache assumption is

  • going to be useful.

  • To see that, we're going to look at what's

  • wrong with a short cache.

  • So in a short cache, our lines are going to be very wide,

  • and they're wider than the number of lines

  • that we can have in our cache.

  • And let's say we're working with an m

  • by n submatrix sorted in row-major order.

  • If you have a short cache, then even if n squared

  • is less than c M, meaning that you

  • can fit all the bytes of the submatrix in cache,

  • you might still not be able to fit it into a short cache.

  • And this picture sort of illustrates this.

  • So we have m rows here.

  • But we can only fit M over B of the rows in the cache,

  • because the cache lines are so long,

  • and we're actually wasting a lot of space

  • on each of the cache lines.

  • We're only using a very small fraction of each cache line

  • to store the row of this submatrix.

  • If this were the entire matrix, then

  • it would actually be OK, because consecutive rows

  • are going to be placed together consecutively in memory.

  • But if this is a submatrix, then we

  • can't be guaranteed that the next row is going to be placed

  • right after the current row.

  • And oftentimes, we have to deal with submatrices

  • when we're doing recursive matrix algorithms.

  • So this is what's wrong with short caches.

  • And that's why we want us assume the tall cache assumption.

  • And we can assume that, because it's usually

  • satisfied in practice.

  • The TLB be actually tends to be short.

  • It only has a couple of entries, so it might not satisfy

  • the tall cache assumption.

  • But all of the other caches will satisfy this assumption.

  • Any questions?

  • OK.

  • So here's another lemma that's going to be useful.

  • This is called the submatrix caching llama.

  • So suppose that we have an n by m matrix,

  • and it's read into a tall cache that

  • satisfies B squared less than c M for some constant c less than

  • or equal to 1.

  • And suppose that n squared is less than M over 3,

  • but it's greater than or equal to c M. Then

  • A is going to fit into cache, and the number of cache

  • misses required to read all of A's elements into cache is,

  • at most, 3n squared over B.

  • So let's see why this is true.

  • So we're going to let big N denote

  • the total number of bytes that we need to access.

  • So big N is going to be equal to n squared.

  • And we're going to use the cache miss lemma, which

  • says that if the average length of our segments

  • is large enough, then we can read all of the segments

  • in just like it were a single contiguous array.

  • So the lengths of our segments here are going to be little n.

  • So r is going to be a little n.

  • And also, the number of segments is going to be little n.

  • And the segment length is also going

  • to be little n, since we're working with a square submatrix

  • here.

  • And then we also have the cache block size B is less than

  • or equal to n.

  • And that's equal to big N over r.

  • And where do we get this property that B is less than

  • or equal to n?

  • So I made some assumptions up here,

  • where I can use to infer that B is less than or equal to n.

  • Does anybody see where?

  • Yeah.

  • AUDIENCE: So B squared is less than c M,

  • and c M is [INAUDIBLE]

  • JULIAN SHUN: Yeah.

  • So I know that B squared is less than c

  • M. C M is less than or equal to n squared.

  • So therefore, B squared is less than n squared,

  • and B is less than n.

  • So now, I also have that N is less than M

  • over 3, just by assumption.

  • And therefore, I can use the cache miss lemma.

  • So the cache miss lemma tells me that I only

  • need a total of 3n squared over B cache

  • misses to read this whole thing in.

  • Any questions on the submatrix caching lemma?

  • So now, let's analyze matrix multiplication.

  • How many of you have seen matrix multiplication before?

  • So a couple of you.

  • So here's what the code looks like

  • for the standard cubic work matrix multiplication

  • algorithm.

  • So we have two input matrices, A and B,

  • And we're going to store the result in C.

  • And the height and the width of our matrix is n.

  • We're just going to deal with square matrices here,

  • but what I'm going to talk about also

  • extends to non-square matrices.

  • And then we just have three loops here.

  • We're going to loop through i from 0 to n minus 1, j from 0

  • to n minus 1, and k from 0 to n minus 1.

  • And then we're going to let's C of i n plus j

  • be incremented by a of i n plus k times b of k n plus j.

  • So that's just the standard code for matrix multiply.

  • So what's the work of this algorithm?

  • It should be review for all of you.

  • n cubed.

  • So now, let's analyze the number of cache

  • misses this algorithm is going to incur.

  • And again, we're going to assume that the matrix is

  • in row-major order, and we satisfy the tall cache

  • assumption.

  • We're also going to analyze the number of cache

  • misses in matrix B, because it turns out

  • that the number of cache misses incurred

  • by matrix B is going to dominate the number of cache

  • misses overall.

  • And there are three cases we need to consider.

  • The first case is when n is greater than c M

  • over B for some constant c.

  • And we're going to analyze matrix B, as I said.

  • And we're also going to assume LRU, because we can.

  • If you recall, the LRU lemma says

  • that whatever we analyze using the LRU

  • is just going to be a constant factor within what we analyze

  • using the ideal cache.

  • So to do this matrix multiplication,

  • I'm going to go through one row of A and one column of B

  • and do the dot product there.

  • This is what happens in the innermost loop.

  • And how many cache misses am I going

  • to incur when I go down one column of B here?

  • So here, I have the case where n is greater than M over B.

  • So I can't fit one block from each row into the cache.

  • So how many cache misses do I have the first time

  • I go down a column of B?

  • So how many rows of B do I have?

  • n.

  • Yeah, and how many cache misses do I need for each row?

  • One.

  • So in total, I'm going to need n cache misses

  • for the first column of B.

  • What about the second column of B?

  • So recall that I'm assuming the LRU replacement policy here.

  • So when the cache is full, I'm going

  • to evict the thing that was least recently used--

  • used the furthest in the past.

  • Sorry, could you repeat that?

  • AUDIENCE: [INAUDIBLE].

  • JULIAN SHUN: Yeah.

  • So it's still going to be n.

  • Why is that?

  • AUDIENCE: Because there are [INAUDIBLE] integer.

  • JULIAN SHUN: Yeah.

  • It's still going to be n, because I

  • can't fit one cache block from each row into my cache.

  • And by the time I get back to the top of my matrix B,

  • the top block has already been evicted from the cache,

  • and I have to load it back in.

  • And this is the same for every other block that I access.

  • So I'm, again, going to need n cache misses

  • for the second column of B. And this

  • is going to be the same for all the columns of B.

  • And then I have to do this again for the second row of A.

  • So in total, I'm going to need theta of n

  • cubed number of cache misses.

  • And this is one cache miss per entry that I access in B.

  • And this is not very good, because the total work was also

  • theta of n cubed.

  • So I'm not gaining anything from having any locality

  • in this algorithm here.

  • So any questions on this analysis?

  • So this just case 1.

  • Let's look at case 2.

  • So in this case, n is less than c M over B.

  • So I can fit one block from each row of B into cache.

  • And then n is also greater than another constant, c prime time

  • square root of M, so I can't fit the whole matrix into cache.

  • And again, let's analyze the number of cache

  • misses incurred by accessing B, assuming LRU.

  • So how many cache misses am I going

  • to incur for the first column of B?

  • AUDIENCE: n.

  • JULIAN SHUN: n.

  • So that's the same as before.

  • What about the second column of B?

  • So by the time I get to the beginning of the matrix here,

  • is the top block going to be in cache?

  • So who thinks the block is still going to be in cache when

  • I get back to the beginning?

  • Yeah.

  • So a couple of people.

  • Who think it going to be out of cache?

  • So it turns out it is going to be in cache, because I

  • can fit one block for every row of B into my cache

  • since I have n less than c M over B.

  • So therefore, when I get to the beginning of the second column,

  • that block is still going to be in cache, because I loaded it

  • in when I was accessing the first column.

  • So I'm not going to incur any cache misses

  • for the second column.

  • And, in general, if I can fit B columns or some constant

  • times B columns into cache, then I

  • can reduce the number of cache misses I have by a factor of B.

  • So I only need to incur a cache miss the first time I

  • access of block and not for all the subsequent accesses.

  • And the same is true for the second row of A.

  • And since I have m rows of A, I'm

  • going to have n times theta of n squared over B cache misses.

  • For each row of A, I'm going to incur

  • n squared over B cache misses.

  • So the overall number of cache misses is n cubed over B.

  • And this is because inside matrix B

  • I can exploit spatial locality.

  • Once I load in a block, I can reuse it the next time

  • I traverse down a column that's nearby.

  • Any questions on this analysis?

  • So let's look at the third case.

  • And here, n is less than c prime times square root of M.

  • So this means that the entire matrix fits into cache.

  • So let's analyze the number of cache misses for matrix B

  • again, assuming LRU.

  • So how many cache misses do I have now?

  • So let's count the total number of cache

  • misses I have for every time I go through a row of A. Yes.

  • AUDIENCE: Is it just n for the first column?

  • JULIAN SHUN: Yeah.

  • So for the first column, it's going to be n.

  • What about the second column?

  • AUDIENCE: [INAUDIBLE] the second [INAUDIBLE]..

  • JULIAN SHUN: Right.

  • So basically, for the first row of A,

  • the analysis is going to be the same as before.

  • I need n squared over B cache misses to load the cache in.

  • What about the second row of A?

  • How many cache misses am I going to incur?

  • AUDIENCE: [INAUDIBLE].

  • JULIAN SHUN: Yeah.

  • So for the second row of A, I'm not

  • going to incur any cache misses.

  • Because once I load B into cache,

  • it's going to stay in cache.

  • Because the entire matrix can fit in cache,

  • I assume that n is less than c prime times square root of M.

  • So total number of cache misses I

  • need for matrix B is theta of n squared over B since everything

  • fits in cache.

  • And I just apply the submatrix caching lemma from before.

  • Overall, this is not a very good algorithm.

  • Because as you recall, in case 1 I

  • needed a cubic number of cache misses.

  • What happens if I swap the order of the inner two loops?

  • So recall that this was one of the optimizations in lecture 1,

  • when Charles was talking about matrix multiplication

  • and how to speed it up.

  • So if I swapped the order of the two inner loops,

  • then, for every iteration, what I'm doing

  • is I'm actually going over a row of C and a row of B,

  • and A stays fixed inside the innermost iteration.

  • So now, when I analyze the number of cache

  • misses of matrix B, assuming LRU,

  • I'm going to benefit from spatial locality,

  • since I'm going row by row and the matrix is

  • stored in row-major order.

  • So across all of the rows, I'm just

  • going to require theta of n squared over B cache misses.

  • And I have to do this n times for the outer loop.

  • So in total, I'm going to get theta

  • of n cubed over B cache misses.

  • So if you swap the order of the inner two loops

  • this significantly improves the locality of your algorithm

  • and you can benefit from spatial accounting.

  • That's why we saw a significant performance improvement

  • in the first lecture when we swapped the order of the loops.

  • Any questions?

  • So does anybody think we can do better than n

  • cubed over B cache misses?

  • Or do you think that it's the best you can do?

  • So how many people think you can do better?

  • Yeah.

  • And how many people think this is the best you can do?

  • And how many people don't care?

  • So it turns out you can do better.

  • And we're going to do better by using

  • an optimization called tiling.

  • So how this is going to work is instead

  • of just having three for loops, I'm

  • going to have six for loops.

  • And I'm going to loop over tiles.

  • So I've got a loop over s by s submatrices.

  • And within each submatrix, I'm going

  • to do all of the computation I need for that submatrix

  • before moving on to the next submatrix.

  • So the three innermost loops are going

  • to loop inside a submatrix, and a three outermost loops

  • are going to loop within the larger matrix,

  • one submatrix matrix at a time.

  • So let's analyze the work of this algorithm.

  • So the work that we need to do for a submatrix of size

  • s by s is going to be s cubed, since that's just a bound

  • for matrix multiplication.

  • And then a number of times I have to operate on submatrices

  • is going to be n over s cubed.

  • And you can see this if you just consider each submatrix to be

  • a single element, and then using the same cubic work

  • analysis on the smaller matrix.

  • So the work is n over s cubed times s cubed,

  • which is equal to theta of n cubed.

  • So the work of this tiled matrix multiplies the same

  • as the version that didn't do tiling.

  • And now, let's analyze the number of cache misses.

  • So we're going to tune s so that the submatrices just

  • fit into cache.

  • So we're going to set s to be equal to theta

  • of square root of M. We actually need to make this 1/3

  • square root of M, because we need to fit

  • three submatrices in the cache.

  • But it's going to be some constant times square

  • root of M.

  • The submatrix caching level implies that for each submatrix

  • we're going to need x squared over B misses to load it in.

  • And once we loaded into cache, it fits entirely into cache,

  • so we can do all of our computations within cache

  • and not incur any more cache misses.

  • So therefore, the total number of cache

  • misses we're going to incur, it's

  • going to be the number of subproblems, which

  • is n over s cubed and then a number of cache

  • misses per subproblem, which is s squared over B.

  • And if you multiply this out, you're

  • going to get n cubed over B times square root of M.

  • So here, I plugged in square root of M for s.

  • And this is a pretty cool result,

  • because it says that you can actually do better

  • than the n cubed over B bound.

  • You can improve this bound by a factor of square root of M.

  • And in practice, square root of M

  • is actually not insignificant.

  • So, for example, if you're looking at the last level

  • cache, the size of that is on the order of megabytes.

  • So a square root of M is going to be

  • on the order of thousands.

  • So this significantly improves the performance

  • of the matrix multiplication code

  • if you tune s so that the submatrices just

  • fit in the cache.

  • It turns out that this bound is optimal.

  • So this was shown in 1981.

  • So for cubic work matrix multiplication,

  • this is the best you can do.

  • If you use another matrix multiply algorithm,

  • like Strassen's algorithm, you can do better.

  • So I want you to remember this bound.

  • It's a very important bound to know.

  • It says that for a matrix multiplication

  • you can benefit both from spatial locality as well

  • as temporal locality.

  • So I get spatial locality in the B term in the denominator.

  • And then the square root of M term

  • comes from temporal locality, since I'm

  • doing all of the work inside a submatrix

  • before I evict that submatrix from cache.

  • Any questions on this analysis?

  • So what's one issue with this algorithm here?

  • Yes.

  • AUDIENCE: It's not portable, like different architecture

  • [INAUDIBLE].

  • JULIAN SHUN: Yeah.

  • So the problem here is I have to tune s

  • for my particular machine.

  • And I call this a voodoo parameter.

  • It's sort of like a magic number I put into my program

  • so that it fits in the cache on the particular machine I'm

  • running on.

  • And this makes the code not portable,

  • because if I try to run this code on a another machine,

  • the cache sizes might be different there,

  • and then I won't get the same performance

  • as I did on my machine.

  • And this is also an issue even if you're running it

  • on the same machine, because you might

  • have other programs running at the same time

  • and using up part of the cache.

  • So you don't actually know how much of the cache

  • your program actually gets to use in a multiprogramming

  • environment.

  • And then this was also just for one level of cache.

  • If we want to optimize for two levels of caches,

  • we're going to have two voodoo parameters, s and t.

  • We're going to have submatrices and sub-submatrices.

  • And then we have to tune both of these parameters

  • to get the best performance on our machine.

  • And multi-dimensional tuning optimization

  • can't be done simply with binary search.

  • So if you're just tuning for one level of cache,

  • you can do a binary search on the parameter s,

  • but here you can't do binary search.

  • So it's much more expensive to optimize here.

  • And the code becomes a little bit messier.

  • You have nine for loops instead of six.

  • And how many levels of caches do we have on the machines

  • that we're using today?

  • AUDIENCE: Three.

  • JULIAN SHUN: Three.

  • So for three level cache, you have three voodoo parameters.

  • You have 12 nested for loops.

  • This code becomes very ugly.

  • And you have to tune these parameters

  • for your particular machine.

  • And this makes the code not very portable,

  • as one student pointed out.

  • And in a multiprogramming environment,

  • you don't actually know the effective cache size

  • that your program has access to.

  • Because other jobs are running at the same time, and therefore

  • it's very easy to mistune the parameters.

  • Was their question?

  • No?

  • So any questions?

  • Yeah.

  • Is there a way to programmatically get

  • the size of the cache?

  • [INAUDIBLE]

  • JULIAN SHUN: Yeah.

  • So you can auto-tune your program

  • so that it's optimized for the cache sizes

  • of your particular machine.

  • AUDIENCE: [INAUDIBLE] instruction

  • to get the size of the cache [INAUDIBLE]..

  • JULIAN SHUN: Instruction to get the size of your cache.

  • I'm not actually sure.

  • Do you know?

  • AUDIENCE: [INAUDIBLE] in--

  • AUDIENCE: [INAUDIBLE].

  • AUDIENCE: Yeah, in the proc--

  • JULIAN SHUN: Yeah, proc cpuinfo.

  • AUDIENCE: Yeah. proc cpuinfo or something like that.

  • JULIAN SHUN: Yeah.

  • So you can probably get that as well.

  • AUDIENCE: And I think if you google,

  • I think you'll find it pretty quickly.

  • JULIAN SHUN: Yeah.

  • AUDIENCE: Yeah.

  • But even if you do that, and you're

  • running this program when other jobs are running,

  • you don't actually know how much cache your program has access

  • to.

  • Yes?

  • Is cache of architecture and stuff like that

  • optimized around matrix problems?

  • JULIAN SHUN: No.

  • They're actually general purpose.

  • Today, we're just looking at matrix multiply,

  • but on Thursday's lecture we'll actually

  • be looking at many other problems

  • and how to optimize them for the cache hierarchy.

  • Other questions?

  • So this was a good algorithm in terms of cache performance,

  • but it wasn't very portable.

  • So let's see if we can do better.

  • Let's see if we can come up with a simpler design

  • where we still get pretty good cache performance.

  • So we're going to turn to divide and conquer.

  • We're going to look at the recursive matrix multiplication

  • algorithm that we saw before.

  • Again, we're going to deal with square matrices,

  • but the results generalize to non-square matrices.

  • So how this works is we're going to split

  • our [INAUDIBLE] matrices into four submatrices or four

  • quadrants.

  • And then for each quadrant of the output matrix,

  • it's just going to be the sum of two matrix multiplies on n

  • over 2 by n over 2 matrices.

  • So c 1 1 one is going to be a 1 1 times b 1 1,

  • plus a 1 2 times B 2 1.

  • And then we're going to do this recursively.

  • So every level of recursion we're

  • going to get eight multiplied adds of n over 2

  • by n over 2 matrices.

  • Here's what the recursive code looks like.

  • You can see that we have eight recursive calls here.

  • The base case here is of size 1.

  • In practice, you want to coarsen the base case to overcome

  • function call overheads.

  • Let's also look at what these values here correspond to.

  • So I've color coded these so that they correspond

  • to particular elements in the submatrix

  • that I'm looking at on the right.

  • So these values here correspond to the index

  • of the first element in each of my quadrants.

  • So the first element in my first quadrant

  • is just going to have an offset of 0.

  • And then the first element of my second quadrant,

  • that's going to be on the same row

  • as the first element in my first quadrant.

  • So I just need to add the width of my quadrant, which

  • is n over 2.

  • And then to get the first element in quadrant 2 1,

  • I'm going to jump over and over two rows.

  • And each row has a length row size,

  • so it's just going to be n over 2 times row size.

  • And then to get the first element in quadrant 2 2,

  • it's just the first element in quadrant 2 1 plus n over 2.

  • So that's n over 2 times row size plus 1.

  • So let's analyze the work of this algorithm.

  • So what's the recurrence for this algorithm--

  • for the work of this algorithm?

  • So how many subproblems do we have?

  • AUDIENCE: Eight

  • JULIAN SHUN: Eight.

  • And what's the size of each Subproblem n over 2.

  • And how much work are we doing to set up the recursive calls?

  • A constant amount of work.

  • So the recurrences is W of n is equal to 8 W

  • n over 2 plus theta of 1.

  • And what does that solve to?

  • n cubed.

  • So it's one of the three cases of the master theorem.

  • We're actually going to analyze this in more detail

  • by drawing out the recursion tree.

  • And this is going to give us more intuition about why

  • the master theorem is true.

  • So at the top level of my recursion tree,

  • I'm going to have a problem of size n.

  • And then I'm going to branch into eight subproblems of size

  • n over 2.

  • And then I'm going to do a constant amount of work

  • to set up the recursive calls.

  • Here, I'm just labeling this with one.

  • So I'm ignoring the constants.

  • But it's not going to matter for asymptotic analysis.

  • And then I'm going to branch again

  • into eight subproblems of size n over 4.

  • And eventually, I'm going to get down to the leaves.

  • And how many levels do I have until I get to the leaves?

  • Yes?

  • AUDIENCE: Log n.

  • JULIAN SHUN: Yeah.

  • So log n-- what's the base of the log?

  • Yeah.

  • So it's log base 2 of n, because I'm dividing my problem

  • size by 2 every time.

  • And therefore, the number of leaves

  • I have is going to be 8 to the log base 2 of n,

  • because I'm branching it eight ways every time.

  • 8 to the log base 2 of n is the same as n to the log base

  • 2 of 8, which is n cubed.

  • The amount of work I'm doing at the top level is constant.

  • So I'm just going to say 1 here.

  • At the next level, it's eight times, then 64.

  • And then when I get to the leaves,

  • it's going to be theta of n cubed,

  • since I have m cubed leaves, and they're all

  • doing constant work.

  • And the work is geometrically increasing as I go down

  • the recursion tree.

  • So the overall work is just dominated by the work

  • I need to do at the leaves.

  • So the overall work is just going to be theta of n cubed.

  • And this is the same as the looping

  • versions of matrix multiply--

  • they're all cubic work.

  • Now, let's analyze the number of cache misses of this divide

  • and conquer algorithm.

  • So now, my recurrence is going to be different.

  • My base case now is when the submatrix fits in the cache--

  • so when n squared is less than c M. And when that's true,

  • I just need to load that submatrix into cache,

  • and then I don't incur any more cache misses.

  • So I need theta of n squared over B cache

  • misses when n squared is less than c M for some sufficiently

  • small constant c, less than or equal to 1.

  • And then, otherwise, I recurse into 8 subproblems of size n

  • over 2.

  • And then I add theta of 1, because I'm

  • doing a constant amount of work to set up the recursive calls.

  • And I get this state of n squared over B term

  • from the submatrix caching lemma.

  • It says I can just load the entire matrix into cache

  • with this many cache misses.

  • So the difference between the cache analysis here

  • and the work analysis before is that I have a different base

  • case.

  • And I think in all of the algorithms

  • that you've seen before, the base case was always

  • of a constant size.

  • But here, we're working with a base case

  • that's not of a constant size.

  • So let's try to analyze this using the recursion tree

  • approach.

  • So at the top level, I have a problem of size n

  • that I'm going to branch into eight problems of size n

  • over 2.

  • And then I'm also going to incur a constant number of cache

  • misses.

  • I'm just going to say 1 here.

  • Then I'm going to branch again.

  • And then, eventually, I'm going to get to the base case

  • where n squared is less than c M.

  • And when n squared is less than c M, then the number of cache

  • misses that I'm going to incur is going

  • to be theta of c M over B. So I can just plug-in c M here

  • for n squared.

  • And the number of levels of recursion

  • I have in this recursion tree is no longer just log base 2 of n.

  • I'm going to have log base 2 of n minus log base 2

  • of square root of c M number of levels, which

  • is the same as log base 2 of n minus 1/2 times

  • log base 2 of c M. And then, the number of leaves I get

  • is going to be 8 to this number of levels here.

  • So it's 8 to log base 2 of n minus 1/2 of log base 2 of c M.

  • And this is equal to the theta of n cubed over M to the 3/2.

  • So the n cubed comes from the 8 to the log base 2 of n term.

  • And then if I do 8 to the negative 1/2 of log base 2

  • of c M, that's just going to give me M to the 3/2

  • in the denominator.

  • So any questions on how I computed the number of levels

  • of this recursion tree here?

  • So I'm basically dividing my problem size by 2

  • until I get to a problem size that fits in the cache.

  • So that means n is less than square root of c M.

  • So therefore, I can subtract that many levels

  • for my recursion tree.

  • And then to get the number of leaves,

  • since I'm branching eight ways, I

  • just do 8 to the power of the number of levels I have.

  • And then that gives me the total number of leaves.

  • So now, let's analyze a number of cache

  • misses I need each level of this recursion tree.

  • At the top level, I have a constant number

  • of cache misses--

  • let's just say 1.

  • The next level, I have 8, 64.

  • And then at the leaves, I'm going

  • to have theta of n cubed over B times square root of M cache

  • misses.

  • And I got this quantity just by multiplying

  • the number of leaves by the number

  • of cache misses per leaf.

  • So number of leaves is n cubed over M to the 3/2.

  • The cache misses per leaves is theta of c M over B.

  • So I lose one factor of B in the denominator.

  • I'm left with the square root of M at the bottom.

  • And then I also divide by the block size B.

  • So overall, I get n cubed over B times square root of M cache

  • misses.

  • And again, this is a geometric series.

  • And the number of cache misses at the leaves

  • dominates all of the other levels.

  • So the total number of cache misses

  • I have is going to be theta of n cubed

  • over B times square root of M.

  • And notice that I'm getting the same number of cache

  • misses as I did with the tiling version of the code.

  • But here, I don't actually have the tune my code

  • for the particular cache size.

  • So what cache sizes does this code work for?

  • So is this code going to work on your machine?

  • Is it going to get good cache performance?

  • So this code is going to work for all cache sizes,

  • because I didn't tune it for any particular cache size.

  • And this is what's known as a cache-oblivious algorithm.

  • It doesn't have any voodoo tuning parameters,

  • it has no explicit knowledge of the caches,

  • and it's essentially passively auto-tuning itself

  • for the particular cache size of your machine.

  • It can also work for multi-level caches

  • automatically, because I never specified what level of cache

  • I'm analyzing this for.

  • I can analyze it for any level of cache,

  • and it's still going to give me good cache complexity.

  • And this is also good in multiprogramming environments,

  • where you might have other jobs running

  • and you don't know your effective cache size.

  • This is just going to passively auto-tune for whatever

  • cache size is available.

  • It turns out that the best cache-oblivious codes to date

  • work on arbitrary rectangular matrices.

  • I just talked about square matrices,

  • but the best codes work on rectangular matrices.

  • And they perform binary splitting

  • instead of eight-way splitting.

  • And you're split on the largest of i, j, and k.

  • So this is what the best cache-oblivious matrix

  • multiplication algorithm does.

  • Any questions?

  • So I only talked about the serial setting so far.

  • I was assuming that these algorithms

  • ran on just a single thread.

  • What happens if I go to multiple processors?

  • It turns out that the results do generalize

  • to a parallel context.

  • So this is the recursive parallel matrix multiply

  • code that we saw before.

  • And notice that we're executing four sub calls in parallel,

  • doing a sync, and then doing four more sub

  • calls in parallel.

  • So let's try to analyze the number of cache

  • misses in this parallel code.

  • And to do that, we're going to use this theorem, which

  • says that let Q sub p be the number of cache

  • misses in a deterministic cell computation

  • why run on P processors, each with a private cache of size M.

  • And let S sub p be the number of successful steals

  • during the computation.

  • In the ideal cache model, the number of cache

  • misses we're going to have is Q sub p equal to Q sub 1

  • plus big O of number of steals times M over B.

  • So the number of cache misses in the parallel context is

  • equal to the number of cache misses when you run it serially

  • plus this term here, which is the number of steals

  • times M over B.

  • And the proof for this goes as follows-- so every call

  • in the Cilk runtime system, we can

  • have workers steal tasks from other workers

  • when they don't have work to do.

  • And after a worker steals a task from another worker,

  • it's cache becomes completely cold in the worst case,

  • because it wasn't actually working on that subproblem

  • before.

  • But after M over B cold cache misses,

  • its cache is going to become identical to what it would

  • be in the serial execution.

  • So we just need to pay M over B cache

  • misses to make it so that the cache looks the same as

  • if it were executing serially.

  • And the same is true when a worker

  • resumes a stolen subcomputation after a Cilk sync.

  • And the number of times that these two situations can happen

  • is 2 times as S p--

  • 2 times the number of steals.

  • And each time, we have to pay M over b cache misses.

  • And this is where this additive term comes from-- order

  • S sub p times M over B.

  • We also know that the number of steals in a Cilk program

  • is upper-bounded by P times T infinity,

  • in the expectation where P is the number of processors

  • and T infinity is the span of your computation.

  • So if you can minimize the span of your computation,

  • then this also gives you a good cache bounds.

  • So moral of the story here is that minimizing

  • the number of cache misses in a serial elision

  • essentially minimizes them in the parallel execution

  • for a low span algorithm.

  • So in this recursive matrix multiplication algorithm,

  • the span of this is as follows--

  • so T infinity of n is 2T infinity of of n over 2

  • plus theta of 1.

  • Since we're doing a sync here, we

  • have to pay the critical path length of two sub calls.

  • This solves to theta of n.

  • And applying to previous lemma, this gives us

  • a cache miss bound of theta of n cubed over B square root of M.

  • This is just the same as the serial execution

  • And then this additive term is going to be order P times n.

  • And it's a span times M over B

  • So that was a parallel algorithm for matrix multiply.

  • And we saw that we can also get good cache bounds there.

  • So here's a summary of what we talked about today.

  • We talked about associativity and caches, different ways

  • you can design a cache.

  • We talked about the ideal cache model

  • that's useful for analyzing algorithms.

  • We talked about cache-aware algorithms

  • that have explicit knowledge of the cache.

  • And the example we used was titled matrix multiply.

  • Then we came up with a much simpler algorithm

  • that was cache-oblivious using divide and conquer.

  • And then on Thursday's lecture, we'll

  • actually see much more on cache-oblivious algorithm

  • design.

  • And then you'll also have an opportunity

  • to analyze the cache efficiency of some algorithms

  • in the next homework.

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B2 中高級

14.緩存和緩存效率算法 (14. Caching and Cache-Efficient Algorithms)

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    林宜悉 發佈於 2021 年 01 月 14 日
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