1. Routing Chapter 4 Shortest path routing Gossip
Iterative Construction of Routing Tables
Shortest Path Spanning Tree
All-pair
Min-hop routing
Coping with changes
Point of Failure re-routing
Compact Routing
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2. Routing: Decision procedure deciding the path of messages through intermediary nodes.
Router: Dedicated processor which automatically determines a route for messages going through it.
The decision is taken according to a piece of information contained in the message (destination address) and local knowledge possessed by the router (specific routing table or algorithm.)
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3. Each identity has a distinct identifier (global names.)
Each link is labelled (and there is local orientation.) 2 5 4 1 1 4 2 2 1 1 3 1 2 4 3 2 6 5 3 3 2
Local orientation: un nodo puede distinguir entre sus vecinos
Each entity has a routing function f(.). For all destinations y, f(y) returns the link on which to send a message to its destination y. ?
dest = “y” ? x ?
f(y) = link
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4. number of bits necessary to hold the local routing function.
Routing memory:
number of bits necessary to hold the local routing function.
Search time:
time to choose the link to be used.
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5. Routing Tables
When a message arrives at a node, the router consults a local table and determines the message’s exit port.
Node a b c
... u z
Link 2 1
... - 4 4 1 u 3 2 n
entries
n entradas -> logn bits para su representación binaria
Binary search -> divide and conquer
Complexity
Routing memory: O(n log n)
Search time: O(log n) (binary search)
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6. Routing table Full routing table Dest h k c d e f Next link (s,h)
(s,c)
(s,e)
The full routing table
describes the whole graph
at each node
Routing table
Asumimos: bidirectional links, conectivity, total reliability, IDs
Dest h k c d e f
Shortest path
(s,h)
(s,h)(h,k)
(s,c)
(s,c)(c,d)
(s,e)
(s,e)(e,f)
Cost 1 4 10 12 5 8 f 5 3 h k 3 1 3 5 s e 10 8 2 d c
Full routing table
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7. Map_Gossip To construct ALL routing tables: input collection (gossip)
Every entity broadcasts its initial information
Map_Gossip
1. Create an arbitrary spanning tree
2. Each entity gets info about its own neighbours
3. Each entity broadcast the info along the tree
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8. x deg(x) (n-1) = 2m (n-1) Message Complexity O(m+ n log n)
Create an arbitrary spanning tree
2. Each entity gets info about its own neighbours
3. Each entity broadcast the info along the tree
e.g., with Megamerger 2m
just an exchange of info
m links, n nodos
x deg(x) (n-1) = 2m (n-1)
2 m n +O(m) =O(mn)
With HUGE MESSAGES
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9. Note: an entity must have enough space to store
the full map of the network.
Es decir: link state routing algorithms
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10. Iterative construction of Routing Tables
When an entity does not have enough space to store
the map of the network.
Initially an entity knows only its neighbours f
Dest h k c d e f
Shortest path
(s,h) ?
(s,c)
(s,e) 5 3 h k 3 1 3 5 s e 10 8 2 d c
Distance vector
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11. At each iteration every entity sends its current distance
vector to its neighbours
Receiving a distance vector from the neighbours,
an entity updates its own
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12. s h c Dest h k c d e f Shortest path (s,h) 1 ? (s,c) 10 (s,e) 5 Dest s3 h k 3 1 3 5 s e 10 8 2 d c
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13. Initial Distance Vectorsh k c d e f - 1  10 5 3 2 8 f 5 3 h k 3 1 3 5 s e 10 8 2 d c
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14. Distance Vectors after 1st Iterationh k c d e f - 1 4 10 12 5 8 3 11  6 2 f 5 3 h k 3 1 3 5 s e 10 8 2 d c
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15. Distance Vectors after 2nd Iterationh k c d e f - 1 4 10 12 5 8 3 11 13 6 2 f 5 3 h k 3 1 3 5 s e 10 8 2 d c
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16. With at most n-1 iteration each node has the correct information
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17. At each iteration each entity x sends the vector to |N(x)| neighbours
Message Complexity
At each iteration each entity x sends the vector
to |N(x)| neighbours
n items of info
Messages per iteration = n x |N(x)| = 2 n m
|N(x)| es deg(x)
Total = 2 (n-1) n m
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18. Map-gossip: a lot of local storage Iterative-construction:
less local storage but more messages
Is there something in between ?
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19. Observation: the full routing table of a node x describes a
Spanning tree (shortest path spanning tree for x PT(x))
Dest h k c d e f
Shortest path
(s,h)
(s,h)(h,k)
(s,c)
(s,c)(c,d)
(s,e)
(s,e)(e,f) 3 5 2 1 10 s
Full routing table
Cost 4 12 8
Se llama también sink tree de x
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20. To construct the routing table of x we just need
to construct the shortest path spanning tree from x
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21. Shortest Path Spanning Tree in Weighted Graphs
- Dijkstra’s Algorithm (classical, serial)
- Distributed Version (PT-Construction)
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22. Dijkstra’s Algorithm for building the shortest path spanning tree
The distance of a vertex v from a vertex s is the length of a shortest path between s and v
Dijkstra’s algorithm computes the distances of all the vertices from a given start vertex s
Assumptions:
the graph is connected
the edges are undirected
the edge weights are nonnegative
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23. Distributed Version PT-Construction
Every node in the tree knows its cost
(shortest path to the root) A 4 8
The root broadcasts in T the
start of a new iteration
(start message) 2 2 3
Ex. 7 1 B C D 3 9 2 5 E 8 F
2) Each entity in T computes locally the shortest distance
from the root to its neighbours outside T
(mycost+edge-cost)
Ex. C B A E D F 3 2 4 8 7 1 5 9 11
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24. 3) The overall minimum is computed at the 8 A
3) The overall minimum is computed at the
root (minimum finding on T) (min message)
and the corresponding link is chosen 5 8 9 C D 11 8 5
4) The root notifies the corresponding node
of the selection (expansion message)
“Join the tree, your cost is 5” C A D 8 11 E B 7 2 9 F 5 3
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25. 4) The new node notifies its neighbours so that the internal
links can be marked B C D E F
“I am in the tree”
And wait for the acknowledgement before
sending a message to the root to notify
the end of the iteration (notification message) A B C D
“OK”
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“OK” E F

26. Complexity start, minimum, expansion, notification:
ni: nodes in the tree at step i
n-1
 (4(ni -1) +2) = 2 n2 -4n + 2
i=1
O(n2)
New link just added
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27. Complexity Neighbour notification 2 |N(x)| = 2  deg(x) = 4 m = O(m)
Final broadcast: n-1
Total: O(m + n2) = O(n2)
x  V
x  V
COST FOR BUILDING JUST ONE
ROUTING TABLE !!!
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28. All-pairs Shortest Path
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29. Algorithm Cost Restriction Map-gossip O(n m ) (m) local storage
Iterative O(n2 m)
PT-constr O(n3)
Sparser-gossip O(n2 log n)
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30. Min-Hop Routing
Breadth First Spanning Tree
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31. At step 1 the root chooses all nodes at distance 1
At step 2, it chooses all the nodes at
distance 2 …
At step i, it chooses the nodes at distance i
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32. Breadth-First Spanning Tree Construction
The root send “start iteration 1”
A node receiving the message mark
itself as a node at distance “1” , mark
the sender as parent and sends back an ack.
In general at step i, all nodes at distance i
from the root are in the tree.
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33. The Algorithm At iteration i+1: The root send “start iteration i+1”
when the message reaches a node at distance i, it
will send an attack to the neighbours (that are
not at distance i-1) and will wait for a reply
A node X will reply with a positive ack only if:
this is the first attack (so X is not part of the tree yet)
and X is not at distance i, otherwise it will reply
with a negative ack.
An acknowledgement “end-of-iteration i+1”
travels up to the root and the root starts the
new iteration
Attack = explore
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34. + Notification end-of- iteration upok ok no ok no no
+ Notification end-of-iteration up
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35. + Notification end-of-iteration upok ok
+ Notification end-of-iteration up no
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36. Complexity Start and end-of iteration: ni: nodes in the tree at step i
1  i < r(root)
(r(root) = eccentricity of the root) depends on the topology)
[ ex. line, O(n2) ]
Attack&Reply
 |N(x)|
x  V
=  deg(x) = 2 m
= O(m)
Total: 2m + (ni -1)  2m+ 2(n-1)d(G)
O(m + n2)= O(n2)
d(g) es el diámetro, y coincide con el número de iteraciones de BF
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37. Multiple Layers
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38. Compact Routing Basic Idea: Use tricks in node and edge labels
to decrease routing memory.
123
dest = 2000 ? 17 4
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39. Compact Routing - An idea...
dest 1 2 3 4 5 6 7 8 9
link 2 - 3 1 4 1 4 3 2 3 n
entries
link 1 2 3 4
interval
[6, 7)
[1, 3)
[4, 6)
[7,1)
d (degree)
entries
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40. Interval Routing
Destination addresses with the same exit are grouped in intervals.
A router finds the direction on which to transmit the message by determining the interval containing the address of the message’s destination.
Idea: Vertex labelling: in 0, ..., n-1
Edge labelling: (cyclic) intervals (cyclic: [a, b) = [a, n)  [0, b) if a  b))
[1, 8)
(message, dest) “5
[8, 11)
[11, ...)
Choose the arc labelled by [a, b) such that dest  [a, b)
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41. Complexity Memory: d * 2 * log n bits Time: log d (binary search)
[1, 8)
[8, 11)
[11, 1)
Memory: d * 2 * log n bits
Time: log d (binary search)
Memory Time
Routing tables O(n log n) O(log n)
Interval O(d log n) O(log d)
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42. Interval Routing in Trees
Choose an arbitrary root v0 (Tv = subtree with root v nv = dimension of Tv (ex. nv0 = n))
Labelling of vertices: preoder (l(v) = label of v)
Labelling of arcs: cyclic intervals (l(v,u) = label of (v, u))
[2,8) 1
[12,1)
[8,12) 2
[3,7) 12 8
[7,3) 3 7
[4,5) 13 9 10 4 5 6 11
There is only one path between each pair of entities.
l(v,u) = [l(u), l(u + nu))
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43. i
[a,b)
[c,i)
[b,c)
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44. In Arbitrary Networks First phase (vertex labelling)
Begin with an arbitrary vertex v.
Build the BFS spanning tree T of G with root v.
Traverse T depth-first (preorder) labeling vertices accordingly. 1 2 7 4 8 3 6 5
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45. More...
Second (edge labelling) 1  2 7 3 6 4 8 3 6 5
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46. More...
Theorem: x, y  V, the distance between x and y in the spanning tree is  2DG, where DG is the diameter of G.
Proof:
ROOT R Y X
dT(x, y)  dT(x, R) + dT(R, y)
= dT(x, R) + dT(y, R)
= dG(x, R) + dG(y, R)
 DG + DG = 2DG
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47. Interval routing in the ring
[1, 4)
[6, 2) 7 1 2
[5, 1)
[2, 5)
[4, 7)
[3, 6)
[7, 3) 3
[4, 7)
[7, 3)
[3, 6)
[5, 1)
[1, 4) 6 5 4
[6, 2)
[2, 5)
WITH INTERVAL
COMPLETE TABLE 1 2 3 4 5 6 7
left
right -
left
right
[6, 2)
[2, 5) i
[i + n/2 + 1), i)
[i+1, i + n/2)
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48. In the mesh [1, 2] 1 2 [3, 8] [0, 2] [0, 2] [4, 5] [3] [5] 3 4 51 2
[3, 8]
[0, 2]
[0, 2]
[4, 5]
[3]
[5] 3 4 5
[6, 8]
[6, 8] 6 7 8
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49. Classical Routing in the mesh (with a compass)00 01 02 10 11 12 20 21 22
From (a,b) to (c,d): let i = a-c and j =b-d
If i <0 take -i times WEST, otherwise take i times EAST
If j <0 take -i times NORTH, otherwise take j times SOUTH
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50. In the Hypercube [7,0] 7 6 [4,5] [1,3] 4 5 [4, 6] [7,0] 1 [5, 7]1
[5, 7]
[2, 3] 3 2
[0, 1]
[3, 4]
Label the nodes according to a particular hamiltonian cycle ...
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51. Classical routing in the Hypercube
From source to dest: follow the path given by the
dimensions in which the two source and dest differ.
001
101
110
111
000
001
010
011
EX: to 100 dim 2 and dim 3
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52. Boolean Routing Idea: Labelling of vertices: boolean values.
Arc labeling: boolean predicates.
Destination addresses with the same exit port satisfy the same condition (boolean predicates.)
A router finds the direction on which to transmit a message by determining the boolean predicate which is true on the message destination.
a  b
(message, dest) “101”
a  c
b  c
Choose the arc labelled by P such that P(dest) is true.
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53. Boolean Routing - Example...01 10 11 00 *0 1* *1 0*
destinations starting with bit 0 0* 1* 10
destinations starting with bit 1
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54. Prefix Routing
Destination addresses with the same exit port have the longest common prefix.
A router finds the direction on which to transmit a message by determining the arc with the longest prefix that is common with the destination.
Vertex labelling: Strings on an alphabet .
Arc labelling: Strings on an alphabet .
111
(message, dest) “101”
1001
???..! 01
Choose the arc labelled by s   such that s is the prefix of maximal length of dest  .
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55. Prefix Routing - More... a  b   a b aa ab ba    aa ab ba aaa aab
1001
101
10101 1
100
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