nim-eth/eth/p2p/discoveryv5/routing_table.nim

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# nim-eth - Node Discovery Protocol v5
# Copyright (c) 2020-2024 Status Research & Development GmbH
# Licensed and distributed under either of
# * MIT license (license terms in the root directory or at https://opensource.org/licenses/MIT).
# * Apache v2 license (license terms in the root directory or at https://www.apache.org/licenses/LICENSE-2.0).
# at your option. This file may not be copied, modified, or distributed except according to those terms.
{.push raises: [].}
import
std/[algorithm, times, sequtils, bitops, sets],
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bearssl/rand, results,
stint, chronicles, metrics, chronos,
../../net/utils,
"."/[node, random2, enr]
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export results
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declareGauge routing_table_nodes,
"Discovery routing table nodes", labels = ["state"]
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type
DistanceProc* =
proc(a, b: NodeId): NodeId {.raises: [], gcsafe, noSideEffect.}
LogDistanceProc* =
proc(a, b: NodeId): uint16 {.raises: [], gcsafe, noSideEffect.}
IdAtDistanceProc* =
proc (id: NodeId, dist: uint16): NodeId {.raises: [], gcsafe, noSideEffect.}
DistanceCalculator* = object
calculateDistance*: DistanceProc
calculateLogDistance*: LogDistanceProc
calculateIdAtDistance*: IdAtDistanceProc
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RoutingTable* = object
localNode*: Node
buckets*: seq[KBucket]
bitsPerHop: int ## This value indicates how many bits (at minimum) you get
## closer to finding your target per query. Practically, it tells you also
## how often your "not in range" branch will split off. Setting this to 1
## is the basic, non accelerated version, which will never split off the
## not in range branch and which will result in log base2 n hops per lookup.
## Setting it higher will increase the amount of splitting on a not in range
## branch (thus holding more nodes with a better keyspace coverage) and this
## will result in an improvement of log base(2^b) n hops per lookup.
ipLimits: IpLimits ## IP limits for total routing table: all buckets and
## replacement caches.
distanceCalculator: DistanceCalculator
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rng: ref HmacDrbgContext
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KBucket = ref object
istart, iend: NodeId ## Range of NodeIds this KBucket covers. This is not a
## simple logarithmic distance as buckets can be split over a prefix that
## does not cover the `localNode` id.
nodes*: seq[Node] ## Node entries of the KBucket. Sorted according to last
## time seen. First entry (head) is considered the most recently seen node
## and the last entry (tail) is considered the least recently seen node.
## Here "seen" means a successful request-response. This can also not have
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## occurred yet.
replacementCache: seq[Node] ## Nodes that could not be added to the `nodes`
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## seq as it is full and without stale nodes. This is practically a small
## LRU cache.
ipLimits: IpLimits ## IP limits for bucket: node entries and replacement
## cache entries combined.
## The routing table IP limits are applied on both the total table, and on the
## individual buckets. In each case, the active node entries, but also the
## entries waiting in the replacement cache are accounted for. This way, the
## replacement cache can't get filled with nodes that then can't be added due
## to the limits that apply.
##
## As entries are not verified (=contacted) immediately before or on entry, it
## is possible that a malicious node could fill (poison) the routing table or
## a specific bucket with ENRs with IPs it does not control. The effect of
## this would be that a node that actually owns the IP could have a difficult
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## time getting its ENR distributed in the DHT and as a consequence would
## not be reached from the outside as much (or at all). However, that node can
## still search and find nodes to connect to. So it would practically be a
## similar situation as a node that is not reachable behind the NAT because
## port mapping is not set up properly.
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## There is the possibility to set the IP limit on verified (=contacted) nodes
## only, but that would allow for lookups to be done on a higher set of nodes
## owned by the same identity. This is a worse alternative.
## Next, doing lookups only on verified nodes would slow down discovery start
## up.
TableIpLimits* = object
tableIpLimit*: uint
bucketIpLimit*: uint
NodeStatus* = enum
Added
LocalNode
Existing
IpLimitReached
ReplacementAdded
ReplacementExisting
NoAddress
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# xor distance functions
func distance*(a, b: NodeId): UInt256 =
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## Calculate the distance to a NodeId.
a xor b
func logDistance*(a, b: NodeId): uint16 =
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## Calculate the logarithmic distance between two `NodeId`s.
##
## According the specification, this is the log base 2 of the distance. But it
## is rather the log base 2 of the distance + 1, as else the 0 value can not
## be used (e.g. by FindNode call to return peer its own ENR)
## For NodeId of 256 bits, range is 0-256.
let a = a.toBytesBE
let b = b.toBytesBE
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var lz = 0
for i in 0..<a.len:
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let x = a[i] xor b[i]
if x == 0:
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lz += 8
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else:
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lz += bitops.countLeadingZeroBits(x)
break
return uint16(a.len * 8 - lz)
func idAtDistance*(id: NodeId, dist: uint16): NodeId =
## Calculate the "lowest" `NodeId` for given logarithmic distance.
## A logarithmic distance obviously covers a whole range of distances and thus
## potential `NodeId`s.
# xor the NodeId with 2^(d - 1) or one could say, calculate back the leading
# zeroes and xor those` with the id.
id xor (1.stuint(256) shl (dist.int - 1))
const
BUCKET_SIZE* = 16 ## Maximum amount of nodes per bucket
REPLACEMENT_CACHE_SIZE* = 8 ## Maximum amount of nodes per replacement cache
## of a bucket
ID_SIZE = 256
DefaultBitsPerHop* = 5
DefaultBucketIpLimit* = 2'u
DefaultTableIpLimit* = 10'u
DefaultTableIpLimits* = TableIpLimits(tableIpLimit: DefaultTableIpLimit,
bucketIpLimit: DefaultBucketIpLimit)
XorDistanceCalculator* = DistanceCalculator(calculateDistance: distance,
calculateLogDistance: logDistance, calculateIdAtDistance: idAtDistance)
func distance*(r: RoutingTable, a, b: NodeId): UInt256 =
r.distanceCalculator.calculateDistance(a, b)
func logDistance*(r: RoutingTable, a, b: NodeId): uint16 =
r.distanceCalculator.calculateLogDistance(a, b)
func idAtDistance*(r: RoutingTable, id: NodeId, dist: uint16): NodeId =
r.distanceCalculator.calculateIdAtDistance(id, dist)
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func new(T: type KBucket, istart, iend: NodeId, bucketIpLimit: uint): T =
KBucket(
istart: istart,
iend: iend,
nodes: @[],
replacementCache: @[],
ipLimits: IpLimits(limit: bucketIpLimit))
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func midpoint(k: KBucket): NodeId =
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k.istart + (k.iend - k.istart) div 2.u256
func len(k: KBucket): int = k.nodes.len
func tail(k: KBucket): Node = k.nodes[high(k.nodes)]
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func ipLimitInc(r: var RoutingTable, b: KBucket, n: Node): bool =
## Check if the ip limits of the routing table and the bucket are reached for
## the specified `Node` its ip.
## When one of the ip limits is reached return false, else increment them and
## return true.
let ip = n.address.get().ip # Node from table should always have an address
# Check ip limit for bucket
if not b.ipLimits.inc(ip):
return false
# Check ip limit for routing table
if not r.ipLimits.inc(ip):
b.ipLimits.dec(ip)
return false
return true
func ipLimitDec(r: var RoutingTable, b: KBucket, n: Node) =
## Decrement the ip limits of the routing table and the bucket for the
## specified `Node` its ip.
let ip = n.address.get().ip # Node from table should always have an address
b.ipLimits.dec(ip)
r.ipLimits.dec(ip)
proc add(k: KBucket, n: Node) =
k.nodes.add(n)
routing_table_nodes.inc()
proc remove(k: KBucket, n: Node): bool =
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let i = k.nodes.find(n)
if i != -1:
routing_table_nodes.dec()
if k.nodes[i].seen:
routing_table_nodes.dec(labelValues = ["seen"])
k.nodes.delete(i)
true
else:
false
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func split(k: KBucket): tuple[lower, upper: KBucket] =
## Split the kbucket `k` at the median id.
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let splitid = k.midpoint
result.lower = KBucket.new(k.istart, splitid, k.ipLimits.limit)
result.upper = KBucket.new(splitid + 1.u256, k.iend, k.ipLimits.limit)
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for node in k.nodes:
let bucket = if node.id <= splitid: result.lower else: result.upper
bucket.nodes.add(node)
# Ip limits got reset because of the KBucket.new, so there is the need to
# increment again for each added node. It should however never fail as the
# previous bucket had the same limits.
doAssert(bucket.ipLimits.inc(node.address.get().ip),
"IpLimit increment should work as all buckets have the same limits")
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for node in k.replacementCache:
let bucket = if node.id <= splitid: result.lower else: result.upper
bucket.replacementCache.add(node)
doAssert(bucket.ipLimits.inc(node.address.get().ip),
"IpLimit increment should work as all buckets have the same limits")
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func inRange(k: KBucket, n: Node): bool =
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k.istart <= n.id and n.id <= k.iend
func contains(k: KBucket, n: Node): bool = n in k.nodes
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func binaryGetBucketForNode*(buckets: openArray[KBucket],
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id: NodeId): KBucket =
## Given a list of ordered buckets, returns the bucket for a given `NodeId`.
## Returns nil if no bucket in range for given `id` is found.
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let bucketPos = lowerBound(buckets, id) do(a: KBucket, b: NodeId) -> int:
cmp(a.iend, b)
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# Prevent cases where `lowerBound` returns an out of range index e.g. at empty
# openArray, or when the id is out range for all buckets in the openArray.
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if bucketPos < buckets.len:
let bucket = buckets[bucketPos]
if bucket.istart <= id and id <= bucket.iend:
result = bucket
proc computeSharedPrefixBits(nodes: openArray[NodeId]): int =
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## Count the number of prefix bits shared by all nodes.
if nodes.len < 2:
return ID_SIZE
var mask = zero(UInt256)
let one = one(UInt256)
for i in 1 .. ID_SIZE:
mask = mask or (one shl (ID_SIZE - i))
let reference = nodes[0] and mask
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for j in 1 .. nodes.high:
if (nodes[j] and mask) != reference: return i - 1
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for n in nodes:
echo n.toHex()
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# Reaching this would mean that all node ids are equal.
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doAssert(false, "Unable to calculate number of shared prefix bits")
func init*(T: type RoutingTable, localNode: Node, bitsPerHop = DefaultBitsPerHop,
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ipLimits = DefaultTableIpLimits, rng: ref HmacDrbgContext,
distanceCalculator = XorDistanceCalculator): T =
## Initialize the routing table for provided `Node` and bitsPerHop value.
## `bitsPerHop` is default set to 5 as recommended by original Kademlia paper.
RoutingTable(
localNode: localNode,
buckets: @[KBucket.new(0.u256, high(UInt256), ipLimits.bucketIpLimit)],
bitsPerHop: bitsPerHop,
ipLimits: IpLimits(limit: ipLimits.tableIpLimit),
distanceCalculator: distanceCalculator,
rng: rng)
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func splitBucket(r: var RoutingTable, index: int) =
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let bucket = r.buckets[index]
let (a, b) = bucket.split()
r.buckets[index] = a
r.buckets.insert(b, index + 1)
func bucketForNode(r: RoutingTable, id: NodeId): KBucket =
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result = binaryGetBucketForNode(r.buckets, id)
doAssert(not result.isNil(),
"Routing table should always cover the full id space")
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func addReplacement(r: var RoutingTable, k: KBucket, n: Node): NodeStatus =
## Add the node to the tail of the replacement cache of the KBucket.
##
## If the replacement cache is full, the oldest (first entry) node will be
## removed. If the node is already in the replacement cache, it will be moved
## to the tail.
## When the IP of the node has reached the IP limits for the bucket or the
## total routing table, the node will not be added to the replacement cache.
let nodeIdx = k.replacementCache.find(n)
if nodeIdx != -1:
if k.replacementCache[nodeIdx].record.seqNum <= n.record.seqNum:
# In case the record sequence number is higher or the same, the new node
# gets moved to the tail.
if k.replacementCache[nodeIdx].address.get().ip != n.address.get().ip:
if not ipLimitInc(r, k, n):
return IpLimitReached
ipLimitDec(r, k, k.replacementCache[nodeIdx])
k.replacementCache.delete(nodeIdx)
k.replacementCache.add(n)
return ReplacementExisting
elif not ipLimitInc(r, k, n):
return IpLimitReached
else:
doAssert(k.replacementCache.len <= REPLACEMENT_CACHE_SIZE)
if k.replacementCache.len == REPLACEMENT_CACHE_SIZE:
# Remove ip from limits for the to be deleted node.
ipLimitDec(r, k, k.replacementCache[0])
k.replacementCache.delete(0)
k.replacementCache.add(n)
return ReplacementAdded
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proc addNode*(r: var RoutingTable, n: Node): NodeStatus =
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## Try to add the node to the routing table.
##
## First, an attempt will be done to add the node to the bucket in its range.
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## If this fails, the bucket will be split if it is eligible for splitting.
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## If so, a new attempt will be done to add the node. If not, the node will be
## added to the replacement cache.
##
## In case the node was already in the table, it will be updated if it has a
## newer record.
## When the IP of the node has reached the IP limits for the bucket or the
## total routing table, the node will not be added to the bucket, nor its
## replacement cache.
# Don't allow nodes without an address field in the ENR to be added.
# This could also be reworked by having another Node type that always has an
# address.
if n.address.isNone():
return NoAddress
if n == r.localNode:
return LocalNode
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let bucket = r.bucketForNode(n.id)
## Check if the node is already present. If so, check if the record requires
## updating.
let nodeIdx = bucket.nodes.find(n)
if nodeIdx != -1:
if bucket.nodes[nodeIdx].record.seqNum < n.record.seqNum:
# In case of a newer record, it gets replaced.
if bucket.nodes[nodeIdx].address.get().ip != n.address.get().ip:
if not ipLimitInc(r, bucket, n):
return IpLimitReached
ipLimitDec(r, bucket, bucket.nodes[nodeIdx])
# Copy over the seen status, we trust here that after the ENR update the
# node will still be reachable, but it might not be the case.
n.seen = bucket.nodes[nodeIdx].seen
bucket.nodes[nodeIdx] = n
return Existing
# If the bucket has fewer than `BUCKET_SIZE` entries, it is inserted as the
# last entry of the bucket (least recently seen node). If the bucket is
# full, it might get split and adding is retried, else it is added as a
# replacement.
# Reasoning here is that adding nodes will happen for a big part from
# lookups, which do not necessarily return nodes that are (still) reachable.
# So, more trust is put in the own ordering by actually contacting peers and
# newly additions are added as least recently seen (in fact they have not been
# seen yet from our node its perspective).
# However, in discovery v5 a node can also be added after a incoming request
# if a handshake is done and an ENR is provided, and considering that this
# handshake needs to be done, it is more likely that this node is reachable.
# However, it is not certain and depending on different NAT mechanisms and
# timers it might still fail. For this reason we currently do not add a way to
# immediately add nodes to the most recently seen spot.
if bucket.len < BUCKET_SIZE:
if not ipLimitInc(r, bucket, n):
return IpLimitReached
bucket.add(n)
else:
# Bucket must be full, but lets see if it should be split the bucket.
# Calculate the prefix shared by all nodes in the bucket's range, not the
# ones actually in the bucket.
let depth = computeSharedPrefixBits(@[bucket.istart, bucket.iend])
# Split if the bucket has the local node in its range or if the depth is not
# congruent to 0 mod `bitsPerHop`
if bucket.inRange(r.localNode) or
(depth mod r.bitsPerHop != 0 and depth != ID_SIZE):
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r.splitBucket(r.buckets.find(bucket))
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return r.addNode(n) # retry adding
else:
# When bucket doesn't get split the node is added to the replacement cache
return r.addReplacement(bucket, n)
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proc removeNode*(r: var RoutingTable, n: Node) =
## Remove the node `n` from the routing table.
let b = r.bucketForNode(n.id)
if b.remove(n):
ipLimitDec(r, b, n)
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proc replaceNode*(r: var RoutingTable, n: Node) =
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## Replace node `n` with last entry in the replacement cache. If there are
## no entries in the replacement cache, node `n` will simply be removed.
# TODO: Kademlia paper recommends here to not remove nodes if there are no
# replacements. However, that would require a bit more complexity in the
# revalidation as you don't want to try pinging that node all the time.
let b = r.bucketForNode(n.id)
if b.remove(n):
ipLimitDec(r, b, n)
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if b.replacementCache.len > 0:
# Nodes in the replacement cache are already included in the ip limits.
b.add(b.replacementCache[high(b.replacementCache)])
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b.replacementCache.delete(high(b.replacementCache))
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func getNode*(r: RoutingTable, id: NodeId): Opt[Node] =
## Get the `Node` with `id` as `NodeId` from the routing table.
## If no node with provided node id can be found,`none` is returned .
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let b = r.bucketForNode(id)
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for n in b.nodes:
if n.id == id:
return Opt.some(n)
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func contains*(r: RoutingTable, n: Node): bool = n in r.bucketForNode(n.id)
# Check if the routing table contains node `n`.
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func bucketsByDistanceTo(r: RoutingTable, id: NodeId): seq[KBucket] =
sortedByIt(r.buckets, r.distance(it.midpoint, id))
func nodesByDistanceTo(r: RoutingTable, k: KBucket, id: NodeId): seq[Node] =
sortedByIt(k.nodes, r.distance(it.id, id))
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func neighbours*(r: RoutingTable, id: NodeId, k: int = BUCKET_SIZE,
seenOnly = false): seq[Node] =
## Return up to k neighbours of the given node id.
## When seenOnly is set to true, only nodes that have been contacted
## previously successfully will be selected.
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result = newSeqOfCap[Node](k * 2)
block addNodes:
for bucket in r.bucketsByDistanceTo(id):
for n in r.nodesByDistanceTo(bucket, id):
# Only provide actively seen nodes when `seenOnly` set.
if not seenOnly or n.seen:
result.add(n)
if result.len == k * 2:
break addNodes
# TODO: is this sort still needed? Can we get nodes closer from the "next"
# bucket?
result = sortedByIt(result, r.distance(it.id, id))
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if result.len > k:
result.setLen(k)
func neighboursAtDistance*(r: RoutingTable, distance: uint16,
k: int = BUCKET_SIZE, seenOnly = false): seq[Node] =
## Return up to k neighbours at given logarithmic distance.
result = r.neighbours(r.idAtDistance(r.localNode.id, distance), k, seenOnly)
# This is a bit silly, first getting closest nodes then to only keep the ones
# that are exactly the requested distance.
keepIf(result, proc(n: Node): bool = r.logDistance(n.id, r.localNode.id) == distance)
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func neighboursAtDistances*(r: RoutingTable, distances: seq[uint16],
k: int = BUCKET_SIZE, seenOnly = false): seq[Node] =
## Return up to k neighbours at given logarithmic distances.
# TODO: This will currently return nodes with neighbouring distances on the
# first one prioritize. It might end up not including all the node distances
# requested. Need to rework the logic here and not use the neighbours call.
if distances.len > 0:
result = r.neighbours(r.idAtDistance(r.localNode.id, distances[0]), k,
seenOnly)
# This is a bit silly, first getting closest nodes then to only keep the ones
# that are exactly the requested distances.
keepIf(result, proc(n: Node): bool =
distances.contains(r.logDistance(n.id, r.localNode.id)))
func len*(r: RoutingTable): int =
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for b in r.buckets: result += b.len
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func moveRight[T](arr: var openArray[T], a, b: int) =
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## In `arr` move elements in range [a, b] right by 1.
var t: T
when declared(shallowCopy):
shallowCopy(t, arr[b + 1])
for i in countdown(b, a):
shallowCopy(arr[i + 1], arr[i])
shallowCopy(arr[a], t)
else:
t = move arr[b + 1]
for i in countdown(b, a):
arr[i + 1] = move arr[i]
arr[a] = move t
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proc setJustSeen*(r: RoutingTable, n: Node) =
## Move `n` to the head (most recently seen) of its bucket.
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## If `n` is not in the routing table, do nothing.
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let b = r.bucketForNode(n.id)
let idx = b.nodes.find(n)
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if idx >= 0:
if idx != 0:
b.nodes.moveRight(0, idx - 1)
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if not n.seen:
b.nodes[0].seen = true
routing_table_nodes.inc(labelValues = ["seen"])
func nodeToRevalidate*(r: RoutingTable): Node =
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## Return a node to revalidate. The least recently seen node from a random
## bucket is selected.
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var buckets = r.buckets
r.rng[].shuffle(buckets)
# TODO: Should we prioritize less-recently-updated buckets instead? Could
# store a `now` Moment at setJustSeen or at revalidate per bucket.
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for b in buckets:
if b.len > 0:
return b.nodes[^1]
proc randomNodes*(r: RoutingTable, maxAmount: int,
pred: proc(x: Node): bool {.raises: [], gcsafe, noSideEffect.} = nil):
seq[Node] =
## Get a `maxAmount` of random nodes from the routing table with the `pred`
## predicate function applied as filter on the nodes selected.
var maxAmount = maxAmount
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let sz = r.len
if maxAmount > sz:
debug "Less peers in routing table than maximum requested",
requested = maxAmount, present = sz
maxAmount = sz
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result = newSeqOfCap[Node](maxAmount)
var seen = HashSet[Node]()
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# This is a rather inefficient way of randomizing nodes from all buckets, but even if we
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# iterate over all nodes in the routing table, the time it takes would still be
# insignificant compared to the time it takes for the network roundtrips when connecting
# to nodes.
# However, "time it takes" might not be relevant, as there might be no point
# in providing more `randomNodes` as the routing table might not have anything
# new to provide. And there is no way for the calling code to know this. So
# while it will take less total time compared to e.g. an (async)
# randomLookup, the time might be wasted as all nodes are possibly seen
# already.
# We check against the number of nodes to avoid an infinite loop in case of a filter.
while len(result) < maxAmount and len(seen) < sz:
let bucket = r.rng[].sample(r.buckets)
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if bucket.nodes.len != 0:
let node = r.rng[].sample(bucket.nodes)
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if node notin seen:
seen.incl(node)
if pred.isNil() or node.pred:
result.add(node)