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Abstract
We are going to talk about some very basic techniques of securing a DHT (Distributed Hash Table). We begin by discussing the kind of Adversary are we going to defend against. We then move on to some ideas regarding the choice of Identities in a DHT. Finally we present a simple idea about bounding the amount of adversarial nodes that manage to get into our network.Most of the talk is going to be pretty informal, though it serves as a useful down to earth introduction to the subject.
DHT Reminder
Before we move on to talk about security , We review again the basic functions of our DHT. In the lower level of the DHT structure, it means the ability to search for node $\floor{v}$ given the value $v$. In the higher level (Thinking about a Distributed Hash table), it is the ability to store pairs of (key,value), or retrieve a correct value given a key.The Adversary
"It is not cowardly, quite the contrary, to seek to meet the adversary and know his intentions. However, it is cowardly, shameful and treasonable to lay down arms." (King Hassan II.)The first thing we want to do when talking about security is to define the adversary. We will discuss it in a very informal manner, because this is just an introduction to this subject. Our informal definitions here will be good enough for this text.
Generally speaking, in our case the adversary is some entity that wants to disrupt the DHT functions. By entity we don't just mean one node, or one computer, or even one person. The adversary could mean many things. More than something realistic, it is an entity that helps us think about security. It's like a game. We give the adversary certain powers, and see if our model can deal with it. In some sense, our Adversarial model is really our understanding of reality. If all of this sounds too abstract to you, don't worry, we are soon going to see some examples.
Our general model of the adversary in the following text will be as follows. The adversary manifests itself in the network in the form of "corrupt" nodes. Those are nodes that are under the full control of the adversary. We then make a distinction: For every node in the network, it is either a node controlled by the adversary  A "corrupt node", or a "correct" node. By "correct" we mean honest, or "not corrupt". It's a node that follows the rules as we defined them. (From now on we will not add the quotes when referring correct or corrupt nodes).
Another important thing to note about the two types of nodes is that it is not possible to tell if a node is a correct node or a corrupt node. Every node knows if he himself is correct or corrupt, but he can never know for sure if another node is correct or corrupt. You could try to ask a node if he is correct, but then, if it's a corrupt node, he might lie. You could also try to watch a node for a really long time, and then try to conclude that it is a correct node. However that node might be just pretending, waiting for you to look away.
Of those reasons, (And several others that will become clearer in the future), we will never try to directly conclude if a node is correct or corrupt. We just know that some nodes are corrupt and some are correct. This is the way of life.
One more thing to note is that for the sake of simplicity, we are going to forget for a while the Churn and Reliability issues. We will not think about nodes that fail in the article. We assume that all the correct nodes are perfectly reliable, and our main problem to deal with is the Adversary.
Let's talk about what the adversary can do, or at least what we "allow" him to do in this model.
The Sybil Attack
We begin with an adversary that can get into the network as many corrupt nodes as he likes. To make things easy to think about, forget about reliability issues (Nodes that fail). We will assume that the correct nodes never fail.
Try to think for a moment what could be done with this ability of inserting as many corrupt nodes as you want into the DHT.
One very philosophical way of thinking about this situation is as follows: Does the network really belong to us if there are more corrupt nodes than correct nodes? Could we really enforce our structure on the network if the nodes we control are in a minority? And if we could, why couldn't the adversary do exactly the same? Thinking about it, what makes us and the adversary so different?
Going back to our physical world, let's try to see some concrete examples of things the adversary can do:

Disconnecting the DHT: The adversary could insert many nodes into the network, and let them run by the DHT rules. Correct nodes will not be able to distinguish between the corrupt nodes and the correct nodes, and they will form connection to the corrupt nodes. Then in one sudden moment, the adversary will make all the corrupt nodes halt. That means: All the corrupt nodes will stop responding. As the adversary has many nodes (Much more than corrupt nodes), With a pretty high probability, the DHT will become disconnected.

Blocking or Changing specific keys: Assume that the adversary doesn't want some key to be accessible (It will not be possible to get the value $v$ for a key $k$). To do this, The adversary could insert many nodes into the network with random IDs (Remember that every node has some ID). If the adversary has a very big amount of nodes, with some high probability the key $k$ will be under the responsibility of a corrupt node. Then whenever that key is asked for, the adversary could return some other value $v'$ instead of $v$, or just not return anything.
As we mentioned above though (In the philosophical paragraph), if the adversary wanted he could make anything out of this network, because really, he owns this network. He was a majority of "corrupt" nodes.
This kind of attack where an adversary inserts a large amount of "corrupt" nodes into a network is also known as Sybil Attack This kind of attack is not specific to DHTs. It is related to any distributed network where all the nodes have symmetric roles. It is probably the first attack to consider when you hear about a fully distributed technology.
The Sybil attack is one of the fundamental issues we have to deal with when designing anything that is really distributed. We will not solve this at this point, but we are going to talk about it more in the future.
Node bounded Adversary
We have seen that we can't defend ourselves at this point from an adversary that inserts unlimited amount of nodes into the network. So let's put a limit to the amount of nodes the adversary can insert.
Assuming that there are $n$ nodes in the network, a reasonable bound might be $\alpha \cdot n$, where $0 < \alpha < 0.5$. In other words, the maximal amount of corrupt nodes is a constant fraction of the total size of the network. Why do we pick $\alpha$ smaller than half? That is because (As we noted above) if the adversary has more than half the nodes im the network, he pretty much owns the network. Just as a sanity check, note that the amount of correct nodes in the network is $(1  \alpha)n$.
The adversary has a collection of $\alpha n$ corrupt nodes. He can insert each of those nodes into the DHT, and also remove each of them from the DHT. The adversary has much freedom to move the corrupt nodes around, however he owns only $\alpha n$ nodes.
You might be thinking how come we get to choose the abilities of our adversary. For example  How can we allow ourselves to bound the amount of corrupt nodes? After all, the adversary doesn't work by our rules. There are two answers. The first is that we can't, but it helps us to think. The second answer is that we might actually create some mechanism where the adversary will not be able to issue too many corrupt nodes. In that case, it could be easier to first think about the DHT construct, and only later about the other mechanism to bound the amount of corrupt nodes.
Just to make sure I don't leave you with abstract promises, let's consider a very naive mechanism to bound the amount of possible corrupt nodes. For every node who wants to join the network, we will ask the computer owner to show up in person and hand over his id card. (In the real world). We will let one node in for every unique id card that we get. In this example, hopefully the adversary will not be able to get too many id cards, and therefore he will not be able to have too many nodes inside the network.
We are not going to use this mechanism (It is not very distributed or efficient), but it serves as a good example of how to bound the amount of corrupt nodes in a network. Another formulation of that example would be: We used the scarcity of id cards in the real world to make Sybil attacks hard to perform.
Security of Identities' choice
In the following sections we will try to deal with various security issues that might arise in a DHT. We have met the Chord DHT but this might apply to a wider range of DHT structures.
Regarding the Adversarial model: We are going to assume a Node Bounded Slow Changing Adversary. To avoid writing those 5 words every time I want to refer the Adversary, we will call it just Adversary on the rest of this text.
Most of our security considerations presented here will relate to the way we choose the Identities of nodes in the network. I remind you that in Chord, every node has an Identity number which is from a set $B_s := {0,1,2,\dots,2^{s}1 }$ (A number of size $s$ bits). Identities in a DHT are very important, as they determine the range of keys a node has responsibility over.
Given that we know the set of possible identities, we are left with the task of choosing the Identity for new nodes who join the network. We will observe a few methods here.
Random Identities
A simple solution for the Identity choice would be to let every new node in the network pick a random Identity number. If we are talking about Chord, that random Identity number will be a random bit string of size $s$ bits. If all the nodes are correct and $s$ is large enough, we can be pretty sure that no two nodes are going to have the exact same Identity number.
However, if we take into consideration an Adversary as discussed above, we are expected to have problems. One thing the adversary can do, for example, is to block or modify the value for any wanted key. Assume that the Adversary wants to block the value of key $k$. It is enough that the Adversary will insert just one corrupt node $z$ into the network, choosing for him the Identity $k$.
Then for sure the node $z$ will have the responsibility of keeping the key $k$ and its corresponding value. Whenever some other node $x$ will search for the key $k$ and ask for its value, the node $z$ could not respond (Thus blocking access to the key $k$), or give a different value $v'$.
The idea of Random Identities gets us started, but obviously it can not deal with our Adversary.
Public keys as Identities
Following the previous attempt, we want to make sure the Adversary could not choose any arbitrary DHT Identity that he wants. We want to make it hard for the adversary to choose arbitrary DHT identities.
One possibility for Identity choice is Public keys, using any form of asymmetric cryptography. (Also known as Public key Cryptography).
In this method, every node that joins the network first generates a key pair of public key and private key. Then the node uses his own public key as his DHT Identity. Whenever a node $y$ comes in contact with a node $x$, $y$ will ask $x$ to prove his claimed identity. $x$ will then prove his identity using the Public Key Cryptography system. This way the Adversary will have to use a valid Identity for his corrupt nodes, if he wants to be able to communicate with correct nodes.
We somehow hope that it is going to be hard for the Adversary to generate an Identity that is both valid and close to a specified wanted number.
Some words about Public Key Cryptography
All those Cryptography stuff might sound very strange to you if you have never heard about Public Key Cryptography before. I will try to explain shortly what Public Key Cryptography feels like, however if you want to really understand things, please take the time to read more about it from some more serious sources. It will really make a difference in your understanding of things.
Public Key cryptography is a system with a few participants. Every participant initially generates somehow two keys: A public key and a private key. Those keys are usually generated randomly (By the participants), and they are related somehow to each other. Then every participant shares his public key with the rest of the participants, but keeps his private key a secret.
What could be done with those public and private keys? Using the public key one can encrypt data. The corresponding private key could be used to decrypt that data.
If for example Bob generates a key pair of Public key and private key, he will share the public key with all the participants in the system. Then every participant can encrypt messages with Bob's public key, but only Bob could decrypt those messages with his secret private key.
Another feature that some Public Key cryptography systems has is the ability to sign. Bob could sign some message, and then every participant could verify the signature Bob's signature, and be sure that Bob wrote the message. It's pretty much like signatures in the real world (Though a bit more secure :) )
In our case we are really interested in the signing ability that we get from Public Key Cryptography. (Hint: We are going to use it later to prove that we own an Identity in the DHT).
I don't want to get too abstract on you, so let me follow with a real example of a Public Key Cryptography system. A famous example is RSA. If Alice wants to generate a key pair, she has to generate two big random prime numbers. Let's call those $p,q$. Then Alice multiplies them, to get the value $N=pq$. $N$ is then called the public key, and $p,q$ are the private keys. Alice can share the public key ($N$) with anyone, however she will never tell $p,q$. This is the private (Or secret) key. Note that concluding the numbers $p,q$ knowing only $N$ is considered to be a difficult computational problem. (See Integer Factorization)
In RSA the public key could be used to Encrypt messages, and the private key could be used to decrypt messages. In a similar way, we could use the private key to sign messages, and the public key to verify those messages. For the interested, the operation of signing a message in RSA is the same as decrypting a message using the private key. The operation of verifying an RSA signature is the same as encrypting a message using the public key.
Using Public keys as Identities
As we mentioned in the beginning of this section, we planned that every node $x$ will generate a key pair, and use the Public key as a DHT Identity. When a node $x$ joins the network and presents his Public key (Which is his claimed DHT Identity), we will send him to prove his identity.
How could we ask $x$ to prove it his identity? A naive method would be to send him some random value $t$, and ask him ask him to sign it. If $x$ manages to create a valid signature over $t$, we will conclude that his identity is confirmed.
Again let's assume that the Adversary wants to block some key $k$ in the DHT, and see what happens in this case. The Adversary has to position some corrupt node with an Identity very close to $k$ (But not bigger than $k$) inside the network. If we are dealing with RSA, then all the Adversary needs to do is find some two prime number $p,q$ such that $pq$ is very close to $k$. Then the adversary will use $N=pq$ as the corrupt node's Identity (Public Key), and $p,q$ as the private key.
There are many prime numbers, so this should not really be a problem. It's a bit harder than the previous case, where the Adversary could just pick and Identity that he wants for his corrupt nodes, however it is not much harder.
(One could also say that it might be not needed to supply a value $N$ that is really made of a multiplication of two primes, but we don't need to get into this. Even if we follow the rules and create $N$ as a multiplication of two primes, it seems to be pretty easy to get close as we want to $k$).
Using Hashed Public Keys as identities
We could make it a bit harder, though. For a node $x$ with public key $N$ and private key $p,q$, we could declare $x$'s identity to be $f(N)$, where $f$ is some [cryptographic hash function](http://en.wikipedia.org/wiki/Cryptographic_hash_function).Let's again go over all the process of $x$ joining the network to make sure that this makes sense. $x$ first generates a pair $p,q$ of random big primes, and then derives $N=pq$ to be his public key. Next, $x$ calculates $Id=f(N)$, and this is $x$'s DHT Identity.
When $x$ joins the network, he has to confirm his identity. He will claim his identity to be $Id=F(N)$, and he will also supply the value $N$. (The verifier will check that $Id=F(N)$). Next, the verifier will send $x$ some random value $t$, and ask $x$ to sign it. $x$ will create a signature over $t$ and send it back to the verifier. The verifier will then make sure that the signature is correct. If it is, $x$'s identity is confirmed (With respect to this verifier).
Now let's see what happens if the adversary wants to take control over a key $k$. The adversary has to create a corrupt node with Identity close to $k$ but no bigger than $k$. Eventually, the Identity is $f(N)$. $N$ has to be created in a specific way (Probably a multiplication of two primes), however we could be generous and assume that the Adversary could get to any $N$ that it wants. All that is left is finding $N$ such that $f(N)$ is close to $k$. If the adversary could do that, he will be able to take control over the key $k$.
We assume that the adversary is computationally bounded. (He can only compute things in polynomial time). So far we didn't discuss this property of our adversary, however this might be a good place to add this assumption. Basically it means that we assume that adversary doesn't have too much computation power.
Without thinking about any specific properties of the cryptographic hash function $f$, a good idea to find a correct $N$ will to be generate random numbers and check if $d(f(a),k)$ is small enough. (Recall that $d(a,b)$ is the distance between two identity numbers on the ring). As we expect $f$ to be somewhat random, we expect that for some number $a$, we will get that $d(f(a),k)$ is distributed uniformly on $[0,2^s)$. In simpler words, it means that $f(a)$ has the same likelihood of being anywhere on the ring. By creating lots of numbers $N=pq$ for many different primes $p,q$, we can generate many values $f(N)$ that distribute uniformly on the ring.
We have some interval that we want to "land on", when getting a random value $f(a)$. This interval is exactly between $\floor{k}$ and $k$  Right between the node that is currently responsible over the key $k$, and the key $k$ itself.
In the picture: The interval that the Adversary wants to land on to take over the key $k$ is marked with the green color.
Now we want to know how many different values $N=pq$ we have to generate before we get that $f(N)$ is inside the wanted interval on the ring. A simple calculation shows that the expected number of tries is going to be The size of the ring divided by the size of the interval. (Think if it makes sense for an interval of size $\frac{1}{4}$ of the ring, for example).
We still don't know the size of the interval, though. We want to evaluate somehow the value $d(\floor{k},k)$ for some key $k$. We could get a rough estimation of this value by thinking about the easy case, where all the nodes are distributed evenly on the ring. In that case, if there are $n$ nodes, we expect that the distance between two consecutive nodes will be exactly $\frac{2^s}{n}$.
Finally to get the amount of tries, we divide the size of the ring by the our rough estimation of the size of the interval. We get $\frac{2^s}{\frac{2^s}{n}}=n$. Therefore we expect about $n$ tries before we get a number $N$ such that $d(f(N),k) < d(\floor{k},k)$. Recall that $n$ is the amount of nodes in the network. This number is probably not larget than $2^40$ in most cases.
As a short summary, the adversary could gain control over any specific key $k$ using this algorithm:
 Generate two random primes $p,q$. Calculate $N=pq$.
 Calculate $f(N)$.
 If $d(f(N),k) < d(\floor{k},k)$ then return (p,q). Else go back to 1.
By our estimations we expect this algorithm to loop about $n$ times, before a suitable pair of $p,q$ is found.
Also note that the Adversary doesn't have to use more than one corrupt node in the network to gain control over the key $k$. Most of the calculation is done offline, ahead of time.
We can conclude that this method can not help us deal with this adversary, but we did make some progress. Recall that in the previous sections the adversary could gain control over some key $k$ without much effort. Here the adversary has to do some effort to take control over some key.
Hashing IPs
We have already considered the idea of hash functions to make it harder for the adversary to control the Identity of corrupt nodes. Another idea would be to use the hash of a the IP address of a node as its DHT Identity. This solution relies on the hierarchical structure of the Internet and the fact that it is hard to obtain many different IP addresses (At least for simple Adversaries).
Let's be more detailed. For a new node $x$ with IP Address $x_a$ that wants to join the DHT, we define $x$'s Identity to be $f(x_a)$, where $f$ is a known cryptographic hash function. Every other node $y$ on the network that contacts $x$ knows $x$'s IP Address (Or else, how could $y$ contact $x$ from the first place). Therefore $y$ knows automatically the value $f(x_a)$ which is $x$'s DHT Identity.
This time, if an Adversary wants to take control over a key $k$ in the DHT, he will have to create a corrupt node with IP address $a$ such that $f(a)$ is close to the value $k$ (But not bigger than $k$).
In IPv4 There are no more than $2^{32}$ possible addresses. By the calculations we have seen in the idea of Public keys as Identities, we conclude that about $\frac{2^{32}}{n}$ Addresses will be suitable for the adversary if he wants to take control over some specific key $k$.
Obtaining specific IPs could be pretty hard these days (Though not impossible). Therefore this method does make it hard to take control over a specific key in the DHT. Thinking about it, the change to IPv6 (Where there are $2^{128}$ possible addresses) might make this method less effective.
Generally speaking, in a network where every node has the freedom to choose his Address in some fast manner (And there are many possibilities for addresses), this method might not work well. The Adversary could generate many Addresses (We have seen that about $n$ addresses should be enough), until one address has a hash value in the correct range. Assuming that $n$ is not too big, this shouldn't be a hard calculation.
Bounding the Adversary
We talked earlier about our assumption of a Node Bounded Adversary  An Adversary that can insert only so many corrupt nodes into the network. At this point I want to show an example of achieving this property  Making sure that the Adversary doesn't have too many corrupt nodes in the network.
Recall that the very naive solution we initially proposed for this was using some kind of external scarcity, like real world ID cards. Every node that wants to join the network should first be represented by a person in the real world, handing over his (real world) ID card to some central authority.
There is probably something deeper about the scarcity idea. I say probably, because we don't have any rigorous theory yet about it, but it seems like most distributed systems today that are able to deal with Sybil attacks are based on some kind of scarcity. That scarcity makes it hard for any participant to insert many nodes into the network, as every node is linked to some amount of that scarce resource.
In this example we are going to use Computing power as a scarce resource. This kind of scarcity is widely used in Bitcoin based crypto currencies.
Recall that a DHT structure is based on links between nodes. Every node is linked to a few other nodes, allowing strong network connectivity and fast network traversal (Fast searching) at the same time. Also recall that to maintain a link between two nodes, we proposed the idea of heartbeats. Messages are sent in constant time interval between two nodes, to make sure that the remote node is alive.
Every node in the network has to maintain a few links with other nodes. Maintaining a link requires sending periodic heartbeat messages, which is not very computationally expensive. It would be interesting if we could make the task of maintaining a link computationally expensive. In that case, the more nodes the adversary inserts into the network, the more links he will have to maintain. A computationally bounded Adversary will not be able to insert too many nodes, as he won't be able to maintain all the links.
One way to make the task of maintaining links more difficult is using a proof of work puzzle. These are special riddles that are very easy to generate, very hard to solve but it is very easy to verify their solution.
We will extend the idea of heartbeat, by adding a difficult riddle to the heartbeat. If $x$ and $y$ are two linked nodes in the DHT, $x$ will send a riddle to $y$ every 10 seconds, for example. $x$ will then expect $y$ to answer the riddle in a short time. If $y$ doesn't manage to solve the riddle, $x$ will disconnect the link to $y$. The same happens the other way: $y$ will send $x$ periodic riddles, and ask $x$ to solve those riddles. If $x$ doesn't manage to solve the riddles in time, $y$ will disconnect $x$.
In the picture: Illustration of the riddles exchange between two linked nodes. Each node sends periodically a riddle to the other, and waits for a correct solution.
This is somehow like the alarm clock applications that make you solve a hard problem, to prove that you are awake and present. Here we use a riddle to make a remote computer prove that he is present, and not scattered as many different nodes in the network.
You might be wondering about the kind of "riddles" we are going to use. As we are dealing with computers here, it must be a hard riddle. We also have to make sure that the riddle is hard to solve, but easy to verify. We could use cryptographic hash functions to create some pretty hard riddles. Assuming that we have a cryptographic hash function $f$, A famous riddle is to find a value $a$ such that $f(ab)$ begins with $k$ binary zeroes. (By  we mean concatenation of strings). We could call this riddle riddle$_b$.
As $f$ is some general cryptographic hash function, we expect that the best strategy to solve this would be to randomize many values $a$ until we get that $f(ab)$ begins with $k$ binary zeroes. We should generate about $2^k$ random values $a$ until we get a good result for $f(ab)$. For a large enough $k$, this would be a hard question for a computer.
Note that it is pretty easy to verify that a solution $s$ is a correct solution to a given riddle$_b$. We just make sure that $f(sb)$ begins with $k$ binary zeroes.
Now let's go back to the Adversary point of view. We want to check if we managed to bound the amount of corrupt nodes an adversary can insert into the network. Assume that some adversary wants to insert many corrupt nodes into the network. Each of those corrupt nodes has to maintain a few links, to be considered as part of the network. If the adversary wants to insert $m$ corrupt nodes, and maintaining links of some node inside the network costs $t$ basic calculation units per second, then the adversary will need about $mt$ basic calculation units per second to maintain $m$ corrupt nodes inside the network.
Assuming that the Adversary is capable of $Q$ calculation units per second, we get that the maximum amount of corrupt nodes that could stay inside the network is about $\frac{Q}{t}$. Note that if the cost of maintaining a node inside the network, $t$, is increased, the adversary can insert less corrupt nodes into the network, however at the same time it becomes less comfortable for a correct node to stay linked in the network. There is some tradeoff here.
One question that I leave you to think about  How could we avoid the following situation (Riddle redirection): Assume that a corrupt node $z$ is connected to two correct nodes $x$ and $y$. $x$ sends a riddle $R$ to $z$, but $z$ (As a corrupt node) doesn't want to invest the time in solving the riddle $R$. Therefore $z$ forwards the riddle $R$ to $y$, asking him to solve that riddle. $y$ is an innocent correct node, and he solves the riddle $R$, sending back the solution $S$ to $z$. $z$ then returns the solution $S$ back to $x$, and $x$ accepts the solution. This way $z$ doesn't have to solve any riddles that $x$ sends.
In the picture: The riddle redirection attack illustrated. How can you solve it?
As a final note about this example  We managed to find a way to bound the amount of corrupt nodes in the network, assuming that our adversary is computationally bounded. We used scarcity of computing power, and linked it to maintenance of a node in the network.