Merkle Tree Traversal Techniques - CDC

Step 6: leaf = 5 .
• ( )
5
Figure 2.6: The tree after round 5.
P n with its authentication path { 0, 1, 2}
{ P( n4) , P( n67) , P( n 03)
} are output.
h
• 2 / leaf + 1 has two solutions so we consider both cases:
23
Auth Auth Auth =
Case 1: h = 0 :
• Auth 0 becomes P( n 7)
, because it is the only value in Stack 0 .
becomes empty. startnode = 6 so we run Stack0 . initialize ( 6,0)
.
Stack 0
• For h = 0 we run 0 . Stack update , which only puts P( n 6)
onto the stack
and stops.
Case 2: h = 1:
Auth becomes ( )
• 1
P n 45 , because it is the only value in 1
becomes empty. startnode = 10 , which is too big, so stop.
So the tree, after the sixth round, looks like the one in figure 2.7:
Stack . Stack 1