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- How to hand out bridges.
- Divide bridges into 'strategies' as they come in. Do this uniformly
- at random for now.
- For each strategy, we'll hand out bridges in a different way to
- clients. This document describes two strategies: email-based and
- IP-based.
- 0. Notation:
- HMAC(k,v) : an HMAC of v using the key k.
- A|B: The string A concatenated with the string B.
- 1. Email-based.
- Goal: bootstrap based on one or more popular email service's sybil
- prevention algorithms.
- Parameters:
- HMAC -- an HMAC function
- P -- a time period
- K -- the number of bridges to send in a period.
- Setup: Generate two nonces, N and M.
- As bridges arrive, put them into a ring according to HMAC(N,ID)
- where ID is the bridges's identity digest.
- Divide time into divisions of length P.
- When we get an email:
- If it's not from a supported email service, reject it.
- If we already sent a response to that email address (normalized)
- in this period, send _exactly_ the same response.
- If it is from a supported service, generate X = HMAC(M,PS|E) where E
- is the lowercased normalized email address for the user, and
- where PS is the start of the currrent period. Send
- the first K bridges in the ring after point X.
- [If we want to make sure that repeat queries are given exactly the
- same results, then we can't let the ring change during the
- time period. For a long time period like a month, that's quite a
- hassle. How about instead just keeping a replay cache of addresses
- that have been answered, and sending them a "sorry, you already got
- your addresses for the time period; perhaps you should try these
- other fine distribution strategies while you wait?" response? This
- approach would also resolve the "Make sure you can't construct a
- distinct address to match an existing one" note below. -RD]
- [I think, if we get a replay, we need to send back the same
- answer as we did the first time, not say "try again."
- Otherwise we need to worry that an attacker can keep people
- from getting bridges by preemtively asking for them,
- or that an attacker may force them to prove they haven't
- gotten any bridges by asking. -NM]
- [While we're at it, if we do the replay cache thing and don't need
- repeatable answers, we could just pick K random answers from the
- pool. Is it beneficial that a bridge user who knows about a clump of
- nodes will be sharing them with other users who know about a similar
- (overlapping) clump? One good aspect is against an adversary who
- learns about a clump this way and watches those bridges to learn
- other users and discover *their* bridges: he doesn't learn about
- as many new bridges as he might if they were randomly distributed.
- A drawback is against an adversary who happens to pick two email
- addresses in P that include overlapping answers: he can measure
- the difference in clumps and estimate how quickly the bridge pool
- is growing. -RD]
- [Random is one more darn thing to implement; rings are already
- there. -NM]
- [If we make the period P be mailbox-specific, and make it a random
- value around some mean, then we make it harder for an attacker to
- know when to try using his small army of gmail addresses to gather
- another harvest. But we also make it harder for users to know when
- they can try again. -RD]
- [Letting the users know about when they can try again seems
- worthwhile. Otherwise users and attackers will all probe and
- probe and probe until they get an answer. No additional
- security will be achieved, but bandwidth will be lost. -NM]
- To normalize an email address:
- Start with the RFC822 address. Consider only the mailbox {???}
- portion of the address (username@domain). Put this into lowercase
- ascii.
- Questions:
- What to do with weird character encodings? Look up the RFC.
- Notes:
- Make sure that you can't force a single email address to appear
- in lots of different ways. IOW, if nickm@freehaven.net and
- NICKM@freehaven.net aren't treated the same, then I can get lots
- more bridges than I should.
- Make sure you can't construct a distinct address to match an
- existing one. IOW, if we treat nickm@X and nickm@Y as the same
- user, then anybody can register nickm@Z and use it to tell which
- bridges nickm@X got (or would get).
- Make sure that we actually check headers so we can't be trivially
- used to spam people.
- 2. IP-based.
- Goal: avoid handing out all the bridges to users in a similar IP
- space and time.
- Parameters:
- T_Flush -- how long it should take a user on a single network to
- see a whole cluster of bridges.
- N_C
- K -- the number of bridges we hand out in response to a single
- request.
- Setup: using an AS map or a geoip map or some other flawed input
- source, divide IP space into "areas" such that surveying a large
- collection of "areas" is hard. For v0, use /24 address blocks.
- Group areas into N_C clusters.
- Generate secrets L, M, N.
- Set the period P such that P*(bridges-per-cluster/K) = T_flush.
- Don't set P to greater than a week, or less than three hours.
- When we get a bridge:
- Based on HMAC(L,ID), assign the bridge to a cluster. Within each
- cluster, keep the bridges in a ring based on HMAC(M,ID).
- [Should we re-sort the rings for each new time period, so the ring
- for a given cluster is based on HMAC(M,PS|ID)? -RD]
- When we get a connection:
- If it's http, redirect it to https.
- Let area be the incoming IP network. Let PS be the current
- period. Compute X = HMAC(N, PS|area). Return the next K bridges
- in the ring after X.
- [Don't we want to compute C = HMAC(key, area) to learn what cluster
- to answer from, and then X = HMAC(key, PS|area) to pick a point in
- that ring? -RD]
- Need to clarify that some HMACs are for rings, and some are for
- partitions. How rings scale is clear. How do we grow the number of
- partitions? Looking at successive bits from the HMAC output is one way.
- 3. Open issues
- Denial of service attacks
- A good view of network topology
- at some point we should learn some reliability stats on our bridges. when
- we say above 'give out k bridges', we might give out 2 reliable ones and
- k-2 others. we count around the ring the same way we do now, to find them.
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