In a Chord ring using m = 8, nodes with the following peer ids (or node ids) join the system: 45. 32, 132, 234, 99, 199. If node 45 fails, then what is the comma-separated list of all the nodes whose finger tables need to be updated? 

Answer:

Before we answer we need to understand the following.

-In an m-bit identifier space, there are 2m identifiers. Here m=8 means 2⁸=64 identifies.
-Identifiers are ordered on an identifier circle modulo 2^m.
-The identifier ring is called Chord ring.
-Key k is assigned to the first node whose identifier is equal to or follows (the identifier of) k in the identifier space.
-This node is the successor node of key k, denoted by successor(k).
-Each node n’ maintains a routing table with up to m entries (which is in fact the number of bits in identifiers), called finger table.
Since we have m=8, which means each Node i would be having 8 entries in its finger table denoting the 8 successors for that keys from (2⁰ to 2⁸)
-Key K is assigned to node whose identifier is >= Key.

Ex:Finger table for node 32

Keys    | Successor Node
N32+2⁰=32
N32+2¹=45
N32+2²=45
N32+2³=45
N32+2⁴=45
N32+2⁵=99
N32+2⁶=99
N32+2⁷=99

So likewise we can create the finger tables for the rest of the nodes (45,99,132,199,234)

Now if the node 45 is removed, we see that Node 45 is mentioned only as a finder table entries in only Node 32.

So the answer is 32.