# A Gentle Explanation of the Intel Speculative Execution CPU Bug

There’s a lot of talk today about the recent discovery of a significant bug in the Intel CPUs that so many of us use every day. It’s an interesting problem, and understanding it requires knowing a little bit about how CPUs work, so I thought I’d take a shot at writing an explainer.

Let me start with a huge caveat: this involves a lot of details about how CPUs work, and in order to explain it, I’m going to simplify things to an almost ridiculous degree in order to try to come up with an explanation that’s comprehensible to a lay person. I’m never deliberately lying about how things work, but at times, I’m simplifying enough that it will be infuriating to an expert. I’m doing my best to explain my understanding of this problem in a way that most people will be able to understand, but I’m bound to oversimplify in some places, and get details wrong in others. I apologize in advance.

It’s also early days for this problem. Intel is still trying to keep the exact details of the bug quiet, to make it harder for dishonest people to exploit it. So I’m working from the information I’ve been able to gather about the issue so far. I’ll do my best to correct this post as new information comes out, but I can only do that when I’m not at work!

That said: what we know so far is that the Intel bug involves non-kernel code being able to access cached kernel memory through the use of something called speculative execution.

To an average person, that means about as much as a problem in the flux thruster atom pulsar electrical ventury space-time implosion field generator coil.

Let’s start with a quick overview of a modern CPU.

The CPU, in simple terms, the brain of a computer. It’s the component that actually does computations. It reads a sequence of instructions from memory, and then follows those instructions to perform some computation on some values, which are also stored in memory. This is a massive simplification, but basically, you can think of the CPU as a pile of hardware than runs a fixed program:

def simplified_cpu_main_loop():
IP = 0
while true:
(op, in1, in2, out) = fetch(IP)
val1 = fetch(in1)
val2 = fetch(in2)
result, IP = perform(op, in1, in2)
store(result, out)


There’s a variable called the instruction pointer (abbreviated IP) built-in to the CPU that tells it where to fetch the next instruction from. Each time the clock ticks, the CPU fetches an instruction from the memory address stored in the instruction pointer, fetches the arguments to that instruction from cells in memory, performs the operation on those arguments, and then stores the result into another cell in the computer memory. Each individual operation produces both a result, and a new value for the instruction pointer. Most of the time, you just increment the instruction pointer to look at the next instruction, but for comparisons or branches, you can change it to something else.

What I described above is how older computers really worked. But as CPUs got faster, chipmaker ran into a huge problem: the CPU can perform its operations faster and faster every year. But retrieving a value from memory hasn’t gotten faster at the same rate as executing instructions. The exact numbers don’t matter, but to give you an idea, a modern CPU can execute an instruction in less than one nanosecond, but fetching a single value from memory takes more than 100 nanoseconds. In the scheme we described above, you need to fetch the instruction from memory (one fetch), and then fetch two parameters from memory (another two fetches), execute the instruction (1 nanosecond), and then store the result back into memory (one store). Assuming a store is no slower than a fetch, that means that for one nanosecond of computation time, the CPU needs to do 3 fetches and one store for each instruction. That means that the CPU is waiting, idle, for at least 400ns, during which it could have executed another 400 instructions, if it didn’t need to wait for memory.

That’s no good, obviously. There’s no point in making a fast CPU if all it’s going to do is sit around and wait for slow memory. But designers found ways to work around that, by creating ways to do a lot of computation without needing to pause to wait things to be retrieved from/stored to memory.

Expanding on the idea of putting memory into the CPU, they added ways of reducing the cost of working with memory by creating copies of the active memory regions on the CPU. These are called caches. When you try to retrieve something from memory, if it’s in the cache, then you can access it much faster. When you access something from a memory location that isn’t currently in the cache, the CPU will copy a chunk of memory including that location into the cache.

You might ask why, if you can make the cache fast, why not just make all of memory like the cache? The answer is that the time it takes in hardware to retrieve something from memory increases with amount of memory that you can potentially access. Pointing at a cache with 1K of memory is lightning fast. Pointing at a cache with 1 megabyte of memory is much slower that the 1K cache, but much faster that a 100MB cache; pointing at a cache with 100MB is even slower, and so on.

So what we actually do in practice is have multiple tiers of memory. We have the registers (a very small set – a dozen or so memory cells, which can be accessed instantly); a level-0 cache (on the order of 8k in Intel’s chips), which is pretty fast; a level-1 cache (100s of kilobytes), an L2 cache (megabytes), L3 (tens of megabytes), and now even L4 (100s of megabytes). If something isn’t in L0 cache, then we look for it in L1; if we can’t find it in L1, we look in L2, and so on, until if we can’t find it in any cache, we actually go out to the main memory.

There’s more we can do to make things faster. In the CPU, you can’t actually execute an entire instruction all at once – it’s got multiple steps. For a (vastly simplified) example, in the pseudocode above, you can think of each instruction as four phases: (1) decode the instruction (figuring out what operation it performs, and what its parameters are), (2) fetch the parameters, (3) perform the operations internal computation, and (4) write out the result. By taking advantage of that, you can set up your CPU to actually do a lot of work in parallel. If there are three phases to executing an instruction, then you can execute phase one of instruction one in one cycle; phase one of instruction two and phase two of instruction one in the next cycle; phase one of instruction three, phase two of instruction two, and phase three of instruction one in the third cycle. This process is called pipelining.

To really take advantage of pipelining, you need to keep the pipeline full. If your CPU has a four-stage pipeline, then ideally, you always know what the next four instructions you’re going to execute are. If you’ve got the machine code version of an if-then-else branch, when you start the comparison, you don’t know what’s going to come next until you finish it, because there are two possibilities. That means that when you get to phase 2 of your branch instruction, you can’t start phase one of the next instruction. instruction until the current one is finished – which means that you’ve lost the advantage of your pipeline.

That leads to another neat trick that people play in hardware, called branch prediction. You can make a guess about which way a branch is going to go. An easy way to understand this is to think of some numerical code:

def run_branch_prediction_demo():
for i in 1 to 1000:
for j in 1 to 1000:
q = a[i][j] * sqrt(b[i][j])


After each iteration of the inner loop, you check to see if j == 1000. If it isn’t, you branch back to the beginning of that loop. 999 times, you branch back to the beginning of the loop, and one time, you won’t. So you can predict that you take the backward branch, and you can start executing the early phases of the first instructions of the next iteration. That may, most of the time you’re running the loop, your pipeline is full, and you’re executing your computation quickly!

The catch is that you can’t execute anything that stores a result. You need to be able to say “Oops, everything that I started after that branch was wrong, so throw it away!”. Alongside with branch prediction, most CPUs also provide speculative execution, which is a way of continuing to execute instructions in the pipeline, but being able to discard the results if they’re the result of an incorrect branch prediction.

Ok, we’re close. We’ve got to talk about just another couple of basic ideas before we can get to just what the problem is with these Intel chips.

We’re going to jump up the stack a bit, and instead of talking directly about the CPU hardware, we’re going to talk about the operating system, and how it’s implemented on the CPU.

An operating system is just a program that runs on your computer. The operating system can load and run other programs (your end-user applications), and it manages all of the resources that those other programs can work with. When you use an application that allocates memory, it sent a request called a syscall to the operating system asking it to give it some memory. If your application wants to read data from a disk drive, it makes a syscall to open a file and read data. The operating system is responsible for really controlling all of those resources, and making sure that each program that’s running only accesses the things that it should be allowed to access. Program A can only use memory allocated by program A; if it tries to access memory allocated by program B, it should cause an error.

The operating system is, therefore, a special program. It’s allowed to touch any piece of memory, any resource owned by anything on the computer. How does that work?

There are two pieces. First, there’s something called memory protection. The hardware provides a mechanism that the CPU can use to say something like “This piece of memory is owned by program A”. When the CPU is running program A, the memory protection system will arrange the way that memory looks to the program so that it can access that piece of memory; anything else just doesn’t exist to A. That’s called memory mapping: the system memory of the computer is mapped for A so that it can see certain pieces of memory, and not see others. In addition to memory mapping, the memory protection system can mark certain pieces of memory as only being accessible by privileged processes.

Privileged processes get us to the next point. In the CPU, there’s something called an execution mode: programs can run in a privileged mode (sometimes called kernel space execution), or it can run in a non-privileged mode (sometimes called user-space execution). Only code that’s running in kernel-space can do things like send commands to the memory manager, or change memory protection settings.

When your program makes a syscall, what really happens is that your program puts the syscall parameters into a special place, and then sends a signal called an interrupt. The interrupt switches the CPU into system space, and gives control to the operating system, which reads the interrupt parameters, and does whatever it needs to. Then it puts the result where the user space program expects it, switches back to user-space, and then allows the user space program to continue.

That process of switching from the user space program to the kernel space, doing something, and then switching back is called a context switch. Context switches are very expensive. Implemented naively, you need to redo the memory mapping every time you switch. So the interrupt consists of “stop what you’re doing, switch to privileged mode, switch to the kernel memory map, run the syscall, switch to the user program memory map, switch to user mode”.

Ok. So. We’re finally at the point where we can actually talk about the Intel bug.

Intel chips contain a trick to make syscalls less expensive. Instead of having to switch memory maps on a syscall, they allow the kernel memory to be mapped into the memory map of every process running in the system. But since kernel memory can contain all sorts of secret stuff (passwords, data belonging to other processes, among other things), you can’t let user space programs look at it – so the kernel memory is mapped, but it’s marked as privileged. With things set up that way, a syscall can drop the two “switch memory map” steps in the syscall scenario. Now all a syscall needs to do is switch to kernel mode, run the syscall, and switch back to user mode. It’s dramatically faster!

Here’s the problem, as best I understand from the information that’s currently available:

Code that’s running under speculative execution doesn’t do the check whether or not memory accesses from cache are accessing privileged memory. It starts running the instructions without the privilege check, and when it’s time to commit to whether or not the speculative execution should be continued, the check will occur. But during that window, you’ve got the opportunity to run a batch of instructions against the cache without privilege checks. So you can write code with the right sequence of branch instructions to get branch prediction to work the way you want it to; and then you can use that to read memory that you shouldn’t be able to read.

With that as a starting point, you can build up interesting exploits that can ultimately allow you to do almost anything you want. It’s not a trivial exploit, but with a bit of work, you can use a user space program to make a sequence of syscalls to get information you want into memory, and then write that information wherever you want to – and that means that you can acquire root-level access, and do anything you want.

The only fix for this is to stop doing that trick where you map the kernel memory into every user space memory map, because there’s no way to enforce the privileged memory property in speculative execution. In other words, drop the whole syscall performance optimization. That’ll avoid the security issue, but it’s a pretty expensive fix: requiring a full context switch for every syscall will slow down the execution of user space programs by something between 5 and 30 percent.

# Garbage Collection with Semispaces

The roots of most garbage collection ideas come from the Lisp community. Lisp was really the first major garbage collection language that was used to write complicated things. So it’s natural that the first big innovation in the world of GC that we’re going to look at comes from the Lisp community.

In early Lisp systems with garbage collection, the pause that occured when the GC did a mark/sweep to reclaim memory was very long, so it was important to find ways to make the cycle faster. Lisp code had the properly that it tended to allocate a lot of small, short-lived objects: Lisp, particularly early lisp, tended to represent everything using tiny structures called cons cells, and Lisp programs generate bazillions of them. Lots of short-lived cons cells needed to get released in every GC cycle, and the bulk of the GC pause was caused by the amount of time that the GC spend going through all of the dead cons cells, and releasing them.

Beyond just that speed issue, there’s another problem with naive mark-sweep collection when you’re dealing with large numbers of short lived objects, called heap fragmentation. The GC does a pass marking all of the memory in use, and then goes through each unused block of memory, and releases it. What can happen is that you can end up with lots of memory free, but scattered around in lots of small chunks. For an extreme example, imagine that you’re building two lists made up of 8-byte cells. You allocate a cell for list A, and then you do something using A, and generate a new value which you add as a new cell in list B. So you’re alternating: allocate a cell for A, then a cell for B. When you get done, you discard A, and just keep B. After the GC runs, what does your memory look like? If A and B each have 10,000 cells, then what you have is 8 bytes of free memory that used to be part of A, and then 8 bytes of used memory for a cell of B, then 8 bytes of free, then 8 used, etc. You’ve ended up with 80,000 bytes of free memory, none of which can be used to store anything larger than 8 bytes. Eventually, you can wind up with your entire available heap broken into small enough pieces that you can’t actually use it for anything.

What the lisp folks came up with is a way of getting rid of fragmentation, and dramatically reducing the cost of the sweep phase, by using something called semispaces. With semispaces, you do some cleverness that can be summed up as moving from mark-sweep to copy-swap.

The idea is that instead of keeping all of your available heap in one chunk, you divide it into two equal regions, called semispaces. You call one of the semispaces the primary, and the other the secondary. When you allocate memory, you only allocate from the primary space. When the primary space gets to be almost full, you start a collection cycle.

When you’re doing your mark phase, instead of just marking each live value, you copy it to the secondary space. When all of the live values have been copied to the secondary space, you update all of the pointers within the live values to their new addresses in the secondary space.

Then, instead of releasing each of the unused values, you just swap the primary and secondary space. Everything in the old primary space gets released, all at once. The copy phase also compacts everything as it moves it into the secondary space, consolidating all of the free memory in one contiguous chunk. If you implement it well, you can also have beneficial side effect of moving things close in ways that improve the cache performance of your program.

For Lisp programs, semispaces are a huge win: they reduce the cost of the sweep phase to a constant time, at the expense of a nearly linear increase in mark time, which works out really well. And it eliminates the problem of fragmentation. All in all, it’s a great tradeoff!

Of course, it’s got some major downsides as well, which can make it work very poorly in some cases:

1. The copy phase is significantly more expensive than a traditional mark-phase. The time it takes to copy is linear in the total amount of live data, versus linear in the number of live objects in a conventional mark. Whether semispaces will work well for a given application depend on the properties of the data that you’re working with. If you’ve got lots of large objects, then the increase in time caused by the copy instead of mark can significantly outweigh the savings of the almost free swap, making your GC pauses much longer; but if you’ve got lots of short-lived, small objects, then the increase in time for the copy can be much smaller than time savings from the swap, resulting in dramatically shortened GC pauses.
2. Your application needs to have access to twice as much memory as you actually expect it to use, because you need two full semispaces. There’s really no good way around this: you really need to have a chunk of unused memory large enough to store all of your live objects – and it’s always possible that nearly everything is alive for a while, meaning that you really do need two equally sized semispaces.
3. You don’t individually release values, which means that you can’t have any code that runs when an value gets collected. In a conventional mark-sweep, you can have objects provide functions called finalizers to help them clean up when they’re released – so objects like files can close themselves. In a semispace, you can’t do that.

The basic idea of semispaces is beautiful, and it’s adaptable to some other environments where a pure semispace doesn’t make sense, but some form of copying and bulk release can work out well.

For example, years ago, at a previous job, one of my coworkers was working on a custom Java runtime system for a large highly scalable transaction processing system. The idea of this was that you get a request from a client system to perform some task. You perform some computation using the data from the client request, update some data structures on the server, and then return a result to the client. Then you go on to the next request.

The requests are mostly standalone: they do a bunch of computation using the data that they recieved in the request. Once they’re done with a given request, almost nothing that they used processing it will ever be looked at again.

So what they did was integrate a copying GC into the transaction system. Each time they started a new transaction, they’d give it a new memory space to live it. When the transaction finished, they’d do a quick copy cycle to copy out anything that was referenced in the master server data outside the transaction, and then they’d just take that chunk of memory, and make it available for use by the next transaction.

The result? Garbage collection became close to free. The number of pointers into the transaction space from the master server data was usually zero, one, or two. That meant that the copy phase was super-short. The release phase was constant time, just dropping the pointer to the transaction space back into the available queue.

So they were able to go from an older system which had issues with GC pauses to a new system with no pauses at all. It wouldn’t work outside of that specific environment, but for that kind of application, it screamed.

# A Beginner’s Guide to Garbage Collection

I was just reading an interesting paper about garbage collection (GC), and realized that I’d never written about it here, so I thought maybe I’d write a couple of articles about it. I’m going to start by talking about the two most basic techniques: mark and sweep collection, and reference counting. In future posts, I’ll move on to talk about various neat things in the world of GC.

So let’s start at the beginning. What is garbage collection?

When you’re writing a program, you need to store values in memory. Most of the time, if you want to do something interesting, you need to be able to work with lots of different values. You read data from your user, and you need to be able to create a place to store it. So (simplifying a bit) you ask your operating system to give you some memory to work with. That’s called memory allocation.

The thing about memory allocation is that the amount of memory that a computer has is finite. If you keep on grabbing more of your computer’s memory, you’re eventually going to run out. So you need to be able to both grab new memory when you need it, and then give it back when you’re done.

In many languages (for example, C or C++), that’s all done manually. You need to write code that says when you want to grab a chunk of memory, and you also need to say when you’re done with it. Your program needs to keep track of how long it needs to use a chunk of memory, and give it back when it’s done. Doing it that way is called manual memory management.

For some programs, manual memory management is the right way to go. On the other hand, it can be very difficult to get right. When you’re building a complicated system with a lot of interacting pieces, keeping track of who’s using a given piece of memory can become extremely complicated. It’s hard to get right – and when you don’t get it right, your program allocates memory and never gives it back – which means that over time, it will be using more and more memory, until there’s none left. That’s called a memory leak. It’s very hard to write a complicated system using manual memory management without memory leaks.

You might reasonably ask, what makes it so hard? You’re taking resources from the system, and using them. Why can’t you just give them back when you’re done with them?

It’s easiest to explain using an example. I’m going to walk through a real-life example from one of my past jobs.

I was working on a piece of software that managed the configuration of services for a cluster management platform. In the system, there were many subsystems that needed to be configured, but we wanted to have one configuration. So we had a piece of configuration that was used to figure out what resources were needed to run a service. There was another piece that was used to figure out where log messages would get stored. There was another piece that specified what was an error that was serious enough to page an engineer. There was another piece that told the system how to figure out which engineer to page. And so on.

We’d process the configuration, and then send pieces of it to the different subsystems that needed them. Often, one subsystem would then need to grab information from the piece of configuration that was the primary responsibility of a different subsystem. For example, when there’s an major error, and you need to page an engineer, we wanted to include a link to the appropriate log in the page. So the pager needed to be able to get access to the logging configuration.

The set of components that worked as part of this configuration system wasn’t fixed. People could add new components as new things got added to the system. Each component would register which section of the configuration it was interested in – but then, when it received its configuration fragment, it could also ask for other pieces of the configuration that it needed.

So, here’s the problem. Any given piece of the configuration could be used by 1, or 2, or 3, or 4, or 20 different components. Which piece of the system should be responsible for knowing when all of the other components are done using it? How can it keep track of that?

That’s the basic problem with manual memory management. It’s easy in easy cases, but in complex systems with overlapping realms of responsibility, where multiple systems are sharing data structures in memory, it’s difficult to build a system where there’s one responsible agent that knows when everyone is done with a piece of memory.

It’s not impossible, but it’s difficult. In a system like the one above, the way that we made it work was by doing a lot of copying of data. Instead of having one copy of a chunk of evaluated configuration which was shared among multiple readers, we’d have many copies of the same thing – one for each component. That worked, but it wasn’t free. We ended up needing to use well over ten times as much memory as we could have using shared data structures. When you’ve got a system that could work with a gigabyte of data, multiplying it by ten is a pretty big deal! Even if you’ve got a massive amount of memory available, making copies of gigabytes of data takes a lot of time!

The most important point here is to understand just how hard it is to get this stuff right. I’ve been a software engineer for a long time, and I’ve worked on a lot of systems. Until the advent of the Rust programming language, I’d never seen a single long-running system built with manual memory management that didn’t have a memory leak somewhere. (I’ll talk more about Rust and how it manages to accomplish this in a later post.)

So manual memory management is very hard to get right, and it can potentially be pretty expensive. On the other hand, it’s predictable: you know, when you write your code, what the costs of managing memory will be, because you wrote the code that does it. And, if you get it right, you can control exacly how much memory your program is using at any time.

The alternative to manual memory management is to somehow make the program automatically keep track of memory, and get rid of it when it’s no longer used. But how do you know when something is no longer used?

It’s pretty easy. You call a chunk of memory live if it can be reached by any variable in the program. If it can’t, it’s garbage, and you can get rid of it. Garbage collection is any mechanism in a programming language or execution environment that automatically figures out when something is garbage, and releases it.

There are two basic methods that we can use to figure out which chunks of memory contain live values, and which are garbage. They’re called reference counting and mark-sweep. (There’s a pool of people who are going to get angry at this definition because, they argue, reference counting isn’t garbage collection. They insist that reference counting is something fundmentally different, and that only mark-sweep is really garbage collection. Obviously I disagree. The definition that I’m using is that anything which automatically releases unused memory is garbage collection.)

In reference counting, each block of memory that gets allocated includes a counter called its reference count. Every time you create a new way of referring to something – by assigning it to a variable, or passing it as a parameter, assigning it to a field of another data structure – you add one to the reference count of the block of memory that contains it. Every time you remove a way of referencing something – by changing a variable, or returning from a function call, or garbage collecting a data structure that referenced it, you decrement its reference count by one. When the reference count for a block of memory hits zero, you can release it.

It’s simple, and it’s predictable. You know that the moment you stop using something, it’s going to get released. That’s great! But there are some problems with reference counting. First, you need to make sure that every single time you change anything, you correctly update the reference counts. Miss any updates, and either things will get released before you’re done with them, or things won’t get released and you’ll leak memory. The other, potentially bigger problem, is that there are a bunch of data structures where simple reference counting doesn’t work. For example, think of a doubly-linked list. That’s a list of values, stored so that each value in the list contains pointers to both the element ahead of it in the list, and to the element behind it in the list. Every element in that list always has at least one thing pointing to it. So none of their reference counts will ever be zero, and no element of the list will ever get collected! (There are ways around that, which we’ll talk about in a later post.)

The other main garbage collection technique is called mark-sweep. In mark-sweep, you have a two-phase process: in the mark phase, you walk over all of the data structures figuring out what’s still being used, and in the sweep phase, you free up anything that isn’t getting used.

In the marking phase, you start with a set of pointers called the root set. The root set consists of the things that you know are being used: the values of all of the variables in the parts of your program that are running, and anything that’s being referenced by the execution environment.

You create a marking queue, consisting initially of the root-set. Then you start to process the queue. For each element in the queue, if it hasn’t been marked yet, you mark it as alive, and then you add everything that it contains a reference to to the queue. If it was already marked as live, you just skip over it: it’s done.

Once the mark phase is done, everything that can possibly be referenced by your running program has been marked as live. So now you can sweep: go through the memory space of your program, and release anything that wasn’t marked as live. Boom: you’ve just gotten rid of everything that’s no longer needed.

Naive mark-sweep has one really big problem: your program can’t change anything during the mark phase! That means that any time you want to release some unusued memory, you need to stop the execution of your program while the garbage collection is going through memory, figuring out what’s still alive.

Personally, I really love working in garbage collected languages. In modern GC systems, the pauses are relatively non-intrusive, and the execution time cost of them is often significantly less than the additional copy-costs of manual collection. But it’s far from a panacaea: it doesn’t even completely prevent memory leaks! (One of the things that surprised me quite a bit earlier in my career was discovering a huge memory leak in a Java program.)

Anyway, that’s the intro to the general ideas. In subsequent posts, I’ll talk about a lot of different things in the area of memory management and garbage collection. I’m mostly going to focus on mark-sweep: reference counting is a very simple idea, and most of the applications of it focus on maintaining that simplicity. But in the world of mark-sweep, there’s a ton of interesting stuff: semispaces (which make the sweep phase of GC faster and more effective), generational garbage collection (which makes the GC system faster, and reduces the number of pauses), incremental collection (which allows the mark phase to be done without stopping the whole program), and various techniques used to implement all of this, like read-barriers, write barriers, and colored pointers.

# Time in Distributed Systems: Lamport Timestamps

What I do in my day job is working on infrastructure systems at dropbox. That means that I’m neck-deep in distributed systems programming – that’s my bread and butter. One of the problems that comes up over and over again in distributed systems is time.

In distributed systems, time is a problem. Each computer has a clock built in, but those clocks are independent. The clocks on different machines can vary quite a bit. If a human being is setting them, then they’re probably at best accurate to one second. Even using a protocol like NTP, which synchronizes clocks between different computers, you can only get the clocks accurate to within about a millisecond of each other.

That sounds pretty good. In human timescales, a millisecond is a nearly imperceptible time interval. But to a modern computer, which can execute billions of instructions per second, it’s a long time: long enough to execute a million instructions! To get a sense of how much time that is to a computer, I just measured the time it took to take all of the source code for my main project, compile it from scratch, and execute all of its tests: it took 26 milliseconds.

That’s a lot of work. On the scale of a machine running billions of instructions per second, a millisecond is a long time.

Why does that matter?

For a lot of things that we want to do with a collection of computers, we need to know what event happened first. This comes up in lots of different contexts. The simplest one to explain is a shared resource locked by a mutex.

A mutex is a mutual exclusion lock. It’s basically a control that only allows one process to access some shared resource at a time. For example, you could think of a database that a bunch of processes all talk to. To make an update, a process P needs to send a request asking for access. If no one is using it when the server receives the request, it will give a lock to P, and and then block anyone else from accessing it until P is done. Anyone else who asks for access to the the database will have to wait. When P is done, it releases the lock on the mutex, and then if there’s any processes waiting, the database will choose one, and give it the lock.

Here’s where time comes into things. How do you decide who to give the lock to? You could give it to whoever you received the request from first, using the time on the database host. But that doesn’t always work well. It could easily end up with hosts with a lower-bandwidth connection to the server getting far worse service than a a closer host.

You get better fairness by using “send time” – that is, the time that the request was sent to the server by the client. But that’s where the clock issue comes up. Different machines don’t agree perfectly on the current time. If you use their clocktime to determine gets the lock first, then a machine with a slow clock will always get access before one with a fast clock. What you need is some fair way of determining ordering by some kind of timestamp that’s fair.

There are a couple of algorithms for creating some notion of a clock or timestamp that’s fair and consistent. The simplest one, which we’ll look at in this post, is called Lamport timestamps. It’s impressively simple, but it works really well. I’ve seen it used in real-world implementations of Paxos at places like Google. So it’s simple, but it’s serious.

The idea of Lamport timestamps is to come up with a mechanism that defines a partial order over events in a distributed system. What it defines is a causal ordering: that is, for any two events, A and B, if there’s any way that A could have influenced B, then the timestamp of A will be less than the timestamp of B. It’s also possible to have two events where we can’t say which came first; when that happens, it means that they couldn’t possible have affected each other. If A and B can’t have any affect on each other, then it doesn’t matter which one “comes first”.

The way that you make this work is remarkably simple and elegant. It’s based on the simplest model of distributed system, where a distributed system is a collection of processes. The processes only communicate by explicitly sending messages to each other.

1. Every individual process $p$ in the distributed system maintains an integer timestamp counter, $\tau_p$.
2. Every time a process $p$ performs an action, it increments $\tau_p$. Actions that trigger increments of $\tau_p$ include message sends.
3. Every time a process $p$ sends a message to another process, it includes the current value of $\tau_p$ in the message.
4. When a process $p$ receives a message from a process $q$, that message includes the value of $\tau_q$ when the message was sent. So it updates its $\tau_q$ to the $\max(\tau_p, \tau_q)+1$ (one more than the maximum of its current timestamp and the incoming message timestamp).

For any two events A and B in the system, if $A \rightarrow B$ (that is, if A causally occurred before B – meaning that A could have done something that affected B), then we know that the timestamp of A will be smaller than the timestamp of B.

The order of that statement is important. It’s possible for timestamp(A) to be smaller than timestamp(B), but for B to have occurred before A by some wallclock. Lamport timestamps provide a causal ordering: A cannot have influenced or caused B unless $A \rightarrow B$; but A and B can be independent.

Let’s run through an example of how that happens. I’ll write it out by describing the clock-time sequence of events, and following it by a list of the timestamp counter settings for each host. We start with all timestamps at 0: [A(0), B(0), C(0), D(0).

• [Event 1] A sends to C; sending trigger a timestamp increment. [A(1), B(0), C(0), D(0)].
• [Event 2] C receives a message from A, and sets its counter to 2. [A(1), B(0), C(2), D(0).
• [Event 3] C sends a message to A (C increments to 3, and sends.) [A(1), B(0), C(3), D(0).
• [Event 4] A recieves the message from C, and sets its clock to 4. [A(4), B(0), C(3), D(0)]
• [Event 5] B sends a message to D. [A(4), B(1), C(3), D(0)]
• [Event 6] D receives the message. [A(4), B(1), C(3), D(2)].
• [Event 7] D sends a message to C. [A(4), B(1), C(3), D(3)].
• [Event 8] C receives the message, and sets its clock to 4.

According to the Lamport timestamps, in event 5, B sent its message to D at time 1. But by wallclock time, it sent its message after C’s timestamp was already 3, and A’s timestamp was already 4. We know that in our scenario, event 5 happened before event 3 by wallclock time. But in a causal ordering, it didn’t. In causal order, event 8 happened after event 4, and event 7 happened before event 8. In causal comparison, we can’t say whether 7 happened before or after 3 – but it doesn’t matter, because which order they happened in can’t affect anything.

The Lamport timestamp is a partial ordering. It tells us something about the order that things happened in, but far from everything. In effect, if the timestamp of event A is less than the timestamp of event B, it means that either A happened before B or that there’s no causal relation between A and B.

The Lamport timestamp comparisons only become meaningful when there’s an actual causal link between events. In our example, at the time that event 5 occurs, there’s no causal connection at all between the events on host A, and the events on host B. You can choose any arbitrary ordering between causally unrelated events, and as long as you use it consistently, everything will work correctly. But when event 6 happens, now there’s a causal connection. Event 5 could have changed some state on host D, and that could have changed the message that D sent in event 7. Now there’s a causal relationship, timestamp comparisons between messages after 7 has to reflect that. Lamport timestamps are the simplest possible mechanism that captures that essential fact.

When we talk about network time algorithms, we say that what Lamport timestamps do is provide weak clock consistency: If A causally happened before B, then the timestamp of A will be less than the timestamp of B.

For the mutex problem, we’d really prefer to have strong clock consistency, which says that the timestamp of A is smaller than the timestamp of B if and only if A causally occurred before B. But Lamport timestamps don’t give us enough information to do that. (Which is why there’s a more complex mechanism called vector clocks, which I’ll talk about in another post.

Getting back to the issues that this kind of timestamp is meant to solve, we’ve got a partial order of events. But that isn’t quite enough. Sometimes we really need to have a total order – we need to have a single, correct ordering of events by time, with no ties. That total order doesn’t need to be real – by which I mean that it doesn’t need to be the actual ordering in which events occured according to a wallclock. But it needs to be consistent, and no matter which host you ask, they need to always agree on which order things happened in. Pure lamport timestamps don’t do that: they’ll frequently have causally unrelated events with identical timestamps.

The solution to that is to be arbitrary but consistent. Take some extra piece of information that uniquely identifies each host in the distributed system, and use comparisons of those IDs to break ties.

For example, in real systems, every host has a network interface controller (NIC) which has a universally unique identifier called a MAC address. The MAC address is a 48 bit number. No two NICs in the history of the universe will ever have the same MAC address. (There are 281 trillion possible MAC codes, so we really don’t worry about running out.) You could also use hostnames, IP addresses, or just random arbitrarily assigned identifiers. It doesn’t really matter – as long as it’s consistent.

This doesn’t solve all of the problems of clocks in distributed systems. For example, it doesn’t guarantee fairness in Mutex assignment – which is the problem that I used as an example at the beginning of this post. But it’s a necessary first step: algorithms that do guarantee fairness rely on some kind of consistent event ordering.

It’s also just a beautiful example of what good distributed solutions look like. It’s simple: easy to understand, easy to implement correctly. It’s the simplest solution to the problem that works: there is, provably, no simpler mechanism that provides weak clock consistency.

# Technical Interviews: More than a brain teaser?

A request I’ve gotten from a couple of readers (unrelated to the recent charity thing) is to talk about engineering interviews at tech companies. I’ve been at Google, Foursquare, Twitter, and now Dropbox, so I’ve spent some time inside multiple companies that put a lot of effort into recruiting.

Tech interviews get a lot of bad press. Only some of it is deserved.

What you frequently hear in criticism is stuff about “gotcha” questions, or “brain teasers”. Those do happen, and when they do, they deserve condemnation. For example, I have seriously had an interviewer for a software job ask me “Why are manholes round?” That’s stupid. People who like it claim that it’s a test of lateral thinking; I think it’s garbage. But these days, that kind of rubbish seems pretty rare.

Instead, what’s become very common is for interviewers to present a programming problem, and ask the candidate to solve it by writing some code. A lot of people really hate these kinds of interviews, and describe them as just brain-teasers. I disagree, and I’m going to try to explain why.

The underlying problem for tech job interviews is that hiring the right people is really, really hard.

When someone applies for a job as a software engineer with your company, you start off with just their resume. Resumes are not particularly informative. All they give you is a brief, possibly partial history of the candidates work experience. From a resume, can’t tell how much they really contributed to the projects they worked on. You can’t tell how much their work added (or subtracted) from the team they were part of. You can’t tell if they get work done in a reasonable amount of time, or if they’re slower than a snail. You can’t even tell if they can write a simple program at all.

So you start by screening resumes, doing your best to infer as much as you can from them. Next, you often get recommendations. Recommendations can be useful, but let’s be honest: All recommendation letters are positive. You’re not going to ask for a recommendation from someone who isn’t going to say great things about you. So no matter how terrible a candidate is, the recommendations are going to say nice things about them. At best, you can sometimes infer a problem by reading between the lines – but that’s a very subjective process.

So you end up interviewing someone who’s resume looks good on paper, and who got a couple of people to write letters for them. How do you determine whether or not they’re going to be a valuable addition to your team?

You need to do something to decide whether or not to hire a particular person. What can you do?

That’s what the interview is for. It’s a way to try to get more information. Sure, this person has a college degree. Sure, they’ve got N years of experience. But can they program? Can they communicate well with their coworkers? Do they actually know what they’re doing?

A tech interview is generally an attempt to get information about a candidate by watching them work on a problem. The interview isn’t about knowing the right answer. It’s not even about getting the correct solution to the problem. It’s about watching a candidate work.

When I ask a job candidate a technical question, there’s three main things I’m looking for.

1. What’s their process for solving the problem? On this level, I’m trying to figure out: Do they think about it, or do they jump in and start programming? Do they make sure they understand the problem? Do they clearly state their assumptions?
2. Can they write a simple program? Here I’m trying to see if they’ve got any ability to write
code. No one writes great code in an interview setting. But I want to know if they’re
able to sit down with an unfamiliar problem, and work out a solution in code. I want to see if they start coding immediately, or take time to think through their solution before they start writing.
3. How well can they communicate ideas about programming? Can they grasp the problem from my description? If not, can they figure out what questions they need to ask to understand it? Once they start solving the problem, how well can they explain what they’re doing? Can they describe the algorithm that they’ve chosen? Can they explain why it works?

To try to clarify this, I’m going to walk through a problem that I used to use in interviews. I haven’t used this question in about 3 years, and as far as I know, no one is using the question anymore. The problem involves something called Gray code. Gray code is an alternative representation of numbers in binary form that’s useful for a range of applications involving things like switching systems.

Here’s a quick run through one of the reasons to use gray code. Imagine a system that uses physical switches. You’ve got an array of 8 switches representing a number. It’s currently presenting the number 7 in standard binary – so the first 5 switches are off, and last 3 are on. You want to increment the number. To do that, you need to change the position of four switches at exactly the same time. The odds of your being able to do that without even a transient state that appeared to be a number other than 7 or 8 are vanishingly small.

Gray code solves that by changing the representation. In Gray code, the representation of every number N+1 is only different from the representation of N by exacly one bit. That’s a nice property which makes it useful, even nowadays when we’re not using physical switches for much of anything anymore.

The easiest way that you get the gray code of numbers is by writing a table. You start off by writing 0 and 1, which are the same in both gray code and standard binary:

Decimal Standard Binary Gray
0 0 0
1 1 1

There’s the one-bit gray codes. To get the two bit, make two copies of the rows in that table.
To the first copy, prepend a 0. To the second copy, reverse the order of the rows, prepend a 1:

Decimal Standard Binary Gray
0 00 00
1 01 01
2 10 11
3 11 10

To get to the three bit gray codes, you repeat the process. Copy the rows, prepend 0s to
the first copy; reverse the order of the second, and prepend 1s.

Decimal Standard Binary Gray
0 000 000
1 001 001
2 010 011
3 011 010
4 100 110
5 101 111
6 110 101
7 111 100

So, the gray code of 6 is 101, and the gray code of 7 is 100.

What I would ask an interview candidate to do is: implement a recursive function that given an integer $N$, returns a string with the gray code of $N$.

I can understand how some people look at this question, and say, “Yeah, that’s just a stupid puzzle.” On one level, yeah. It’s obvious an artifical question. In fact, in practice, no one ever uses a recursive algorithm for something like this. Even if you have a problem where gray code is part of a practical solution, there’s a better way of converting numbers to gray code than this silly recursive nonsense.

So I agree that it’s artificial. But interview questions have to be artificial. In a typical interview, you’ve got an hour with a candidate. You’re not going to be able to explain a real problem to them in that amount of time, much less have them solve it!

But it’s artificial in a useful way that allowed me, as an interviewer, to learn about the candidate. I wasn’t trying to see if the candidate was number-puzzle wizard who could instantly see the recursion pattern in a problem like this. Most people have never heard of gray code, and to most people (including me, the first time I saw this problem!), the recursion pattern isn’t obvious. But that’s not the point: there’s a lot more to the interview that just the initial problem statement.

I don’t present the problem, and then sit back and watch silently as they try to solve it. If I did, all I’d be learning is whether or not they’re a number-puzzle wizard. I don’t care about that. So I didn’t just leave them floundering trying to somehow come up with a solution. In the beginning, after describing the problem, I set an initial direction. I usually have them start by extending the table themselves, to make sure they understand the process. Then I take their extended table, and add a new column:

Decimal Standard Binary Gray Rec
0 0000 0000
1 0001 0001
2 0010 0011 “1” + gray(1)
3 0011 0010 “1” + gray(0)
4 0100 0110
5 0101 0111
6 0110 0101
7 0111 0100
8 1000 1100 “1” + gray(7)
9 1001 1101 “1” + gray(6)
10 1010 1111 “1” + gray(5)
11 1011 1110
12 1100 1010
13 1101 1011
14 1110 1001
15 1111 1000 “1” + gray(0)

With that on the board/screen, I’d ask them to try to take what I just gave them, and rewrite it a bit. For example, in row 8, instead of “1” + gray(7), come up with an expression using the numeric value “8” of the row, which will produce 7. They should be able to come up with “15 – 8” – and to see that in every row $n$ , where $n \ge 8$ and $n < 16$, the gray code of $n$ is “1” + gray(15 – n).

For most people, that’s enough of a clue to be able to start writing the function. If they can’t get there, it shows me that they’ve got some trouble wrapping their head around this abstraction. I’ve got a few more hints up my sleeve to help, but if without all of the help I can give, they still can’t come up with that, that’s one mark against them.

But even if they can’t come up with the relation at all, it’s not the end of the interview. I have, sometimes, ended up recommending hiring someone who had trouble this far! If they can’t come up with that basic relation, it’s just one piece of information about the candidate. I’d file that away, and move on, by giving them the recurrence relation, and then I would ask them to code it.

There’s one problem that comes up in coding, which is interesting and important. The most naive code for gray is something like:

def gray(n):
print("gray(%s)" % n)
if n == 0:
return "0"
if n == 1:
return "1"
num_digits = math.floor(math.log(n, 2)) + 1
return "1" + gray(int(2**num_digits - 1 - n))


That’s very close, but not right. If you call gray(10), you get the right answer.
If you call gray(4), you get “110”, which is correct. But if you call gray(8), you’d get “110”, when you should have gotten 1010.

Most candidates make this mistake. And then I ask them to trace it through on 8 as an example. They usually see what the problem is. If they don’t, then that’s another red flag.

If they’re really struggling to put together a recursive function, then I’d ask them to just write a function to convert an integer into standard binary. If they can do that, then I start making suggestions about how to convert that to do gray code.

The big indicator here is whether or not they can write a simple recursive function at all. The systems I was working on at the time made heavy use of recursion – if a candidate couldn’t write recursive code, there was simply no way that they’d be able to do the job. (It was depressing to see just how many qualified-looking candidates came in, but who couldn’t write a simple recursive function.)

Through this whole process, how well they were able to talk about what they were doing was as important as the solution they came up with. If they heard the question, and immediately wrote down perfect, beautiful code, but they couldn’t explain how they got it, or how it worked? They’d get a mediocre rating, which wouldn’t get them a job offer. If they made a lot of mistakes in their code, but they were crystal clear about explaining how they worked out a solution, and how it worked? They’d probably get a better rating than the perfect code candidate.

I hope this shines a bit of light on this kind of interview question. While it’s necessarily artificial, it’s artifical in a way that we hope can help us learn more about the candidate. It’s not a trick question that’s irrelevant to the job, like “Why are manholes round?”: this is an attempt to probe at an area of knowledge that any candidate for a software engineering job should have. It’s not an all or nothing problem: even if you start off with no clue of how to approach it, I’m guiding you through. If you can’t solve it without help, this problem gives me some insight into what it is that you don’t understand, and hopefully whether or not that’s going to be a problem if we hire you.

Is it a great way of doing interviews? Honestly, no. But it’s the best way we know of doing it.

As an interview system, it doesn’t do a good job of identifying the very best people to hire. There’s no correlation between outstanding interview performance and outstanding on-the-job performance. But there’s a strong correlation between poor performance on this kind of interview question and poor performance on the job. Great performers on the job show the same distribution of interview performance as average ones; but poor performers on interviews show a significantly negative-shifted job performance distribution.

We haven’t found a way of interviewing people that does a better job than this. It’s the best we have. Statistically, it works far better at selecting people than “open-ended” interviews that don’t involve any kind of practical programming exercise. So for all of its faults, it’s better that the alternatives.

I’m sure there are pedants out there who are asking “So what’s the correct implementation of gray code?” It’s totally not the point of talking about it, but here’s one sloppy but correct implementation. This isn’t the quality of code I would ever use for anything serious at work, but it’s perfectly adequate for an interview.

import math

def gray(n):
def required_digits(n):
"""Compute the number of digits required to
represent a number in binary
"""
return int(math.floor(math.log(n, 2))) + 1

if len(gray) < num_digits:
return "0"*(num_digits - len(gray)) + gray
return gray

if n == 0:
return "0"
if n == 1:
return "1"
num_digits = int(math.floor(math.log(n, 2)) + 1)
return "1" + pad_digits(gray(int(2**num_digits - 1 - n)), num_digits - 1)


# Simpler Consensus with Raft

A few weeks ago, I wrote about Paxos, which is (at least in my experience), the most widely used algorithm for consensus in distributed systems. I’m a huge fan of Paxos – I think that it’s a remarkably elegant system.

But Paxos does have its problem.

1. Paxos has a lot of roles: client, proposer, learner, acceptor, leader, follower. When you want to implement Paxos, you need to figure out all of those roles, and how you’re going to implement them. In general, you end up merging roles – but there are lots of ways of doing that merge. Each particular way of setting up the roles has its own properties, and thus its own tradeoffs that you need to understand.
2. Paxos, as we normally talk about it, is really a single-consensus protocol – that is, the basic protocol is designed to get a group of agents to come to consensus just once. If you want to be able to repeatedly seek new consensus values, you’re actually going to be using an extension to the basic paxos protocol. There are a ton of Paxos extensions that work to add repeated consensus. Paxos itself is simple and elegant, with well-defined formal properties that we care about – the moment we start modifying it, we can no longer count on those properties unless we can also prove them in our extension!
3. Paxos was originally described in a truly awful paper. Leslie Lamport was trying to write a paper that would be less dull than the typical bone-dry technical snoozer – but the way that he wrote it actually makes it much harder to understand.

In short: Paxos has more complexity than it needs, and despite that, it needs to be tweaked to be really useful, and getting those tweaks right is hard. There are, sadly, a lot of incorrect Paxos implementations – and their incorrectness has all-too-often come as a surprise to the people who rely on them.

To avoid those problems, there are other consensus algorithms out there. In this post, we’re going to look at one of the Paxos competitors – a consensus algorithm/protocol called raft.

Raft does away with the role complexity of Paxos. In Raft, you have a collection of cooperating agents. There are no distinct proposers, acceptors, or learners: there are just servers. Communication between the servers in raft is done entirely with synchronous remote procedure calls.

In Raft, the target of consensus is a log containing a sequence of events. The log is the history of the distributed system. The goal of raft is that the log be maintained in a consistent state throughout the raft network. Just like in Paxos, if we have $2n+1$ servers, up to $n$ can fail without the network losing its consistency.

Raft is designed in terms of remote procedure calls between the elements of the network. In Raft, we never talk about single messages – every communication between servers is a pair of messages: a request from caller to callee; and a response from callee to caller. When a message gets lost, we’ll just talk about it as a failed remote procedure call.

Within a Raft network, at any given time, each server has a state. It can be a follower, a leader, or a candidate. Within the network, there is at most one leader. When there is a leader, all of the other servers are in the follower state. The followers are almost entirely passive. Followers don’t talk to clients at all – they just wait for RPCs from the leader. The leader is the only participant that’s allowed to talk to clients: any client request must go through the current leader. The leader is also the only server that’s allowed to add new entries to the consensus log.

Raft divides time into a sequence of terms. In each term, the servers in the raft network need to select a leader using a process called an election. Raft is a strong leader protocol – no interactions with a client can take place except through a leader. If there’s no leader, then we can’t process client requests without a leader.

So, to understand Raft, there’s three processes that we need to
understand:

2. Transitions between terms
3. Appending an entry to a log.

In those processes, the servers have a collection of variables
that they use for the Raft protocol:

currentTerm
the current term for the server.
votedFor
the serverID that this server voted for in the current term, or “none”.
log
the list of entries in the log.
commitIndex
The index of the highest log entry known to be committed by the server.
lastApplied
The index of the highest log entry that’s been added to the log – but not necessarily committed. (It doesn’t become committed until a majority of servers accept it.)

In each term, the Raft cluster needs to have a leader. The way that a leader is selected called election.

Elections are triggered by a term transition. When a server in the cluster decided that it needs to start a new term, it increments its term number, puts itself into the candidate state, and sends a RequestVote(term, candidateId) RPC to each of the other servers in the cluster. This request asks the other servers in the cluster to select it as the leader. If it receives enough “yes” votes, it will become the leader.

When a server receives a RequestVote RPC, it checks the term. If it’s smaller than the server’s current term, then it replies “No” – meaning that it cannot support the requestor as leader.

If the term in the request is greater that the receiver’s term, then the receiver cannot have voted in the new term. So it updates to the term from the request, and then it replies “Yes”.

If the term in the request equals the receiver’s term, then the receiver has already updated its term. If it’s already voted for someone else as leader, then it can’t support the requestor, so it replies “No”. If it hasn’t voted for a leader in the term, then it votes for the requestor, and replies “Yes”.

If the requester receives “Yes” votes from more than 1/2 of the cluster (counting itself), then it becomes the leader, and starts both processing requests from clients, and sending heartbeats to the other servers in the cluster.

If it doesn’t receive enough votes, then it waits to see if anyone else becomes the leader and starts sending heartbeats. If it doesn’t get a heartbeat in time, then it starts over: it would increment its term again, and try to start a fresh election.

### Term Transitions

For a given server, term transitions happen in three ways:

1. Timeout: the leading server needs to periodically communicate with each of the followers. This process is called heartbeat: even if the leader has no updates for its followers, it sends RPC calls to the followers just to say “I’m still here”. If a client goes too long without receiving a heartbeat, it decides that the leader was lost, and it will increment the term number, and trigger a new election.
2. Leader resignation: the current leader can, at any time, decide to stop being the leader. (This is typically done by an implementation as part of a system that says that there’s a maximum period between leader elections. For example, in the Aurora scheduler, we had leader elections at least once per day. In a raft consensus, the leader would trigger this by deciding it was time for it to stop being a leader, and triggering an election by starting a new term.)
3. External term change: every RPC received by a server includes a term number. If any RPC to a server ever includes a term number greater than the current term for that server, the server will update its term to the new number. As a special case of this, when a leader server decides to resign, it does that by sending an RPC to the other servers with an incremented term number.

### Appending to the log

We just spent a fair bit of time talking about leaders and elections. That’s almost beside the point. What we really want to do is just maintain a consistent log across the cluster of servers. Everything except creating log entries is just the book-keeping that’s necessary to make the consistent log work. The log itself is maintained using the AppendEntries RPC call.

When a client request does something that alters the state of the cluster, the leader describes that change by adding an entry to the log. It builds a proposed log entry, and sends it to the other members of the cluster using an RPC. If it gets enough “Yes” votes from other cluster members, then the log entry becomes committed, and the leader updates its commitIndex to the index of the new log entry to reflect that.

The RPC request takes a bunch of parameters:

1. term: the leader’s term.
2. leaderId: the id of the leader.
3. prevLogIndex: the index number of the last log entry in the consensus log preceeding this new entry.
4. prevLogTerm: the number of the term where the last log entry was committed.
5. entries: a set of new log entries to be appended to the log.
6. leaderCommit: the index of the commitlog on the leader after this set of entries has been committed.

When an AppendEntries call is received by a follower, what it does is:

• If the receiver’s term is greater than the request term, then the receiver rejects the request by replying “No”.
• If the the receivers commit index is larger than the commit index of the request, then it rejects the request by replying “No”.
• If the receiver’s log doesn’t contain an entry at prevLogIndex, or that entry’s term doesn’t match the request term, then it rejects the request by replying “No”.
• If there’s an entry in the log with the same index as the new log entries, and the term in the request matches the receiver’s term, then the receiver removes all entries after prevLogIndex from its log.
• The receiver then appends the new entries from the request to its log.
• If the leaderCommit is greater than the commitIndex on the receiver, then the receiver updates its commitIndex.
• Finally, the receiver replies “Yes”.

When a majority of the cluster members have accepted an AppendEntries call, then the log entry gets committed.

The one part of this that’s confusing is how the logs get managed. The leader creates a new log entry, and sends it to the other servers. The complexity comes from dealing with cases where something doesn’t reach consensus.

For example, the leader sends entries 5, 6, and 7 to server S. S adds the entries to its copy of log – it now contains [1, 2, 3, 4, 5, 6, 7]. Meanwhile, the leader also sends those entries to server T, but the RPC to T fails due to a network fault. Another client request happens, and now the leader sends [5, 6, 7, 8] to S. S sees that it’s got entry 5 already: so it discards everything after 5, and then re-appends.

So the trailing segment of the log can change! How do we handle consensus?

The next time that the leader sends an AppendEntries to a follower, it contains the leader’s commitIndex. The follower updates its commit index to that value. Once it’s done that, any request from a leader that tries to modify anything that comes before that commit index will be rejected.

The consensus commit thus doesn’t really occur until the next heartbeat call after a log update.

## Raft versus Paxos

That’s the basics of Raft.

In comparison to Paxos, there’s a couple of things to notice:

1. There’s a lot less confusion around roles. Paxos had a ton of different roles, and rules for interactions between the different roles. Raft doesn’t have any of that: it’s just servers, with one of the servers designated as the leader.
2. Raft explicitly manages a log, and it adds complexity around log management. In Paxos, you’re just managing a single consensus value; in Raft, you’ve got a sequence of log entries.
3. Paxos is defined in terms of messages; Raft is designed in terms of remote procedure calls.

So is Raft really simpler than Paxos? I think that’s up for discussion. Personally, I prefer Paxos. There’s a lot of complexity hidden under the covers of the RPC system. It looks simple on the surface, but all of the complexity of message passing, lost messages, message duplication – it’s still there. It’s just been swept under the carpet, as if that really makes it easier.

The way that the logs get maintained is confusing. That’s inevitable: getting distributed knowledge is never easy. Raft at least makes that part of things explicit, whereas it’s a common part of Paxos implementations, but it’s not really specified in the protocol.

# Mathematical Data Structures Part 1: Binary Heaps

Being a PhD in computer science, it’s only natural that I love data structures. In particular, I’m fascinated by the way that the math factors in to the way we structure data. Data structures fit into a beautiful intersection between information theory and algorithms: the way that a good data structure is built is a reflection of what information it really needs to maintain. The best data structure encodes exactly the information it needs in order to do it’s job – no more, and no less. The mathematical impacts of that are beautiful, and sometimes surprising. To try to demonstrate that, I’m going to take a couple of posts, and work my way through one of my favorite examples of a surprising outcome in a structure called a fibonacci heap.

A heap is a structure designed to solve a common problem. You’ve got a collection of objects, each of which has an associated numeric value. You want, at any time, to be able to find and remove the largest value in the collection, and to be able to add new elements to it. Those two operations are the core of the heap. Some variations also allow you to increase the value of objects inside the heap, or to remove values other than the maximum.

There are a lot of different ways to implement a heap. One obvious one is to just maintain a sorted sequence of objects. The problem with that is performance: some of the common operations are painfully slow!

Using the sorted sequence approach, removing the largest value is easy: you just remove the last element of the sequence. That’s very fast: it’s constant time. But you also need to be able to add values to the heap, and that’s not so good.

There’s two basic ways of doing a sequence: an array, or a linked list. In both cases, the performance isn’t acceptable. If we used an array,then in order to add a new object to the collection, we’d need to:

1. Find the correct position for it in the array. We can do that by doing a binary search, which takes time $O(lg n)$ where $n$ is the length of the array. This step isn’t bad – in general, we’re pretty happy with $O(lg n)$ operations.
2. Insert the value into the array – which means shifting all of the elements that come after it one place to the right. That’s $O(n)$ time, which is pretty crappy.

In the linked list approach, inserting the value isn’t a problem – it’s a constant time operation. But finding the position where it should be inserted is linear time. So we’re still talking about linear time.

Similarly, we could use a linked list, where inserting the element is constant time, but then finding its position is $O(n)$ – again, unacceptable.

The problem with the sorted sequence approach isn’t really related to the kind of structure we use to maintain the sorted list; the problem is that we’re maintaining more information that we need. At any time, we want to be able to find the largest element of the heap quickly – we don’t care about the relative positions of any pair of values that don’t include the largest element of the collection! But if we keep a sorted list, every time we insert an element, we’re spending a lot of time comparing things whose comparison we don’t really care about!

To be able to make it faster, we need to build a data structure that doesn’t waste time and effort computing and maintaining information that we don’t want.

So our goal is to find ways of building structures that always let us both find the largest element quickly, and add new elements quickly, without maintaining more information that is really necessary. We’ll start off with a simple but good version, and then work our way through to better ones.

The most basic implementation of a heap is called a binary heap. A binary heap is a binary tree with two key properties:

1. Every node in the tree is larger than its children.
2. The tree is left-full: every level of the tree is full except for the last; and the last level is filled in from left to right.

The left-full property might seem a bit strange, but it turns out to be pretty straightforward. A binary heap can be implemented using an array. The root node is stored in the first position of the array; its children are in positions 2 and 3; the children of node 2 are stored in positions 4 and 5; the childen of position 3 are stored in positions 6 and 7. Using one-based indices, for any node N, it’s children are stored in positions 2N and 2N+1. Adding a new leaf to the tree can always be done by just appending one value to the array. The left-full property just means that you always extend the array by adding an element onto the left.

Implementing a heap this way is simple:

1. To get the maximum value, you just look at the first element of the array – O(1).
2. To remove the largest element from the array, you get the value from the first element of the array, and save it. Then you remove the last element from the array, and bubble it down – swapping it with one of its children if they’re bigger than it. We’ll look at this in more detail, but the bubble down process is O(lg n) in the worst case.
3. Inserting a new element is done by adding it to the end of the array, and then bubbling up, by comparing it to its parent, and swapping if it’s bigger than its parent. Again, it’s $O(lg n)$.

I’m going to show code for this. For fun, I wrote the code in a language called xtend. Xtend is a Java extension that cleans up the syntax, gets rid of semicolons, improves the type system, adds lambdas, and does a few other really neat things.

The whole beast is just a wrapper around an array:

class BinHeap> {
val ArrayList _contents

new() {
_contents = new ArrayList()
}

...
}


If you know Java, this is mostly clear. In xtend, you write constructors using the name “new” instead of the name of the class being constructed.

Then we’ll set up some utilities to make other stuff easier to write.

  def leftChildPosition(int pos) {
2 * (pos + 1) - 1
}

def rightChildPosition(int pos) {
2 * (pos + 1)
}

def int parentPosition(int pos) {
if (pos == 0) {
throw new MaxHeapException()
} else {
(pos + 1)/ 2 - 1
}
}

def void swap(int one, int two) {
val T first = _contents.get(one)
_contents.set(one, _contents.get(two))
_contents.set(two, first)
}


Again, these should be straightforward. The only tricky thing is that the JVM uses zero-based arrays – so the left child of the node in position $N$ is $(2*(N+1) - 1)$: we need to add one to the node number to shift to one-based position; and then subtract one from the result to switch back to zero-based position. We do a similar thing for each of the other position computations.

Now we can get to the interesting bits. How do we get values into the heap?

def insert(T v) {
val idx = _contents.size()
bubbleUp(idx)
}


Insert is exactly what I described in prose above: append the new value onto the end of the array, and then bubble it up. Bubbling is the interesting part:

  private def void bubbleUp(int pos) {
if (pos > 0) {
val parentPos = parentPosition(pos)
if (_contents.get(pos) > _contents.get(parentPos)) {
swap(pos, parentPos)
bubbleUp(parentPos)
}
}
}


Bubbling up from a position $P$ compares $P$ to its parent. If it’s bigger than its parent, it swaps positions with the parent, and then tries to continue bubbling up from its new position.

For example, imagine we had a tree like:

9
8
5
4
0
6
3
7
2
1


Now, suppose we wanted to add the value “10” to this. We’d add 10 to the end of the array, which would make it a child of 6. That would give us:

  9
8
5
4
0
6
3
10
7
2
1


So, we’d compare 10 to its parent – it’s bigger, so we’d swap:

  9
8
5
4
0
10
3
6
7
2
1


Then we’d compare 10 to its new parent, 8. It’s bigger, so we swap:

  9
10
5
4
0
8
3
6
7
2
1


And finally, we’d compare 10 to its new parent, 9. It’s bigger so we swap, and then we’re done.

  10
9
5
4
0
8
3
6
7
2
1


Appending to the end of the array is constant time, so the dominant time cost is the bubbling. The maximum possible number of swaps in the doubling process is depth of the tree minus 1 – and the depth of a full binary tree with N members is $\lceil ln N \rceil$. So it’s $O(lg n)$ swaps, and the overall cost of inserts is $O(lg n)$.

Getting the largest value is trivial:

def getMax() {
_contents.get(0)
}


Removing the largest value is a lot like adding a value: we really play with the last element of the array, and then do a bubbling process – only this time we’ll bubble in the opposite direction:

  def removeMax() {
if (_contents.size == 0) {
throw new MaxHeapException()
} else {
val result = getMax()
val last = _contents.remove(_contents.size() - 1)
if (_contents.size() > 0) {
_contents.set(0, last)
bubbleDown(0)
}
result
}
}


Bubbling down is similar to bubbling up, but it’s a bit more complicated, because we need to look at both children.

  private def void bubbleDown(int pos) {
val rightChildPos = rightChildPosition(pos)
val leftChildPos = leftChildPosition(pos)
if (leftChildPos >= _contents.size) {
return
}
// Try to bubble left if there is no right child, or if the lift child is
// bigger than the right.
if (rightChildPos >= _contents.size || _contents.get(leftChildPos) > _contents.get(rightChildPos)) {
if (_contents.get(pos) < _contents.get(leftChildPos)) {
swap(pos, leftChildPos)
bubbleDown(leftChildPos)
}

} else {
// Try to bubble right
if (_contents.get(pos) < _contents.get(rightChildPos)) {
swap(pos, rightChildPos)
bubbleDown(rightChildPos)
}
}
}


The process is almost the same as bubbling up, but moving in the opposite direction. We're starting with a parent node, and comparing it to its children. If it's bigger than either of its children, then we swap it with the largest child, and then continue bubbling down.

For example, let's look at the same heap we looked at for insert:

  9
8
5
4
0
6
3
7
2
1


If we want to remove 9, we set the value 9 aside, and then remove 3 from the end of the array, and put it at the root of the tree:

  3
8
5
4
0
6
7
2
1


Then we'd compare 3 against its two children, 8 and 7. Since 8 is the larger child, we swap 8 for 3:

  8
3
5
4
0
6
7
2
1


Now we compare 3 with its new children, 5 and 6. 6 is bigger, so we swap 6 with 3:

  8
6
5
4
0
3
7
2
1


3 has no children, so we're done: it's bubbled down as far as it can go.

Note: I messed up this example in the original version of the post. Thanks to John Armstrong for pointing it out.

The cost here is the same as insert, for the same reason. The dominant cost is the bubbling, and the bubbling is bounded by the depth of the tree. So removing the maximum is also $O(lg n)$.

It's worth noting that heaps can be used to build a very reasonable sorting algorithm. To sort a collection, just insert all of the elements of the collection, and then remove them one by one. It's $O(n lg n)$, and it's conceptually quite simple. It's not widely used, because the old classic quicksort is faster - not in big(0) notation, but it ends up with a smaller constant. (In big-O notation, something that takes 3(lg n) steps and something that takes 6(lg n) steps are both $O(lg n)$, but the one whose constant is 3 is still twice as fast as the one whose constant is 6.)