Merkle Tree Traversal Techniques - CDC

• For h = 0 we run 0 . Stack update , which only computes P( n 5)
, puts it
onto the stack and stops. For h = 1 we run 1 . Stack update , which removes
the previous two values from the stack, computes P( n 67)
and puts it onto
the stack.
So the tree after the third round looks like the one in figure 2.4:
Step 4: leaf = 3.
• ( )
3
Figure 2.4: The tree after round 3.
P n with its authentication path { 0, 1, 2}
{ P( n2) , P( n01) , P( n 47)
} are output.
h
• 2 / leaf + 1 has three solutions so we consider three cases:
21
Auth Auth Auth =
Case 1: h = 0 :
• Auth 0 becomes P( n 5)
, because it is the only value in Stack 0 .
becomes empty. startnode = 4 so we run Stack0 . initialize ( 4,0)
.
Stack 0
• For h = 0 we run 0 . Stack update , which only puts P( n 4)
onto the stack
and then stops.
Case 2: h = 1:
Auth becomes P( n 67)
, because it is the only value in 1
becomes empty. startnode = 4 so we run Stack1 . initialize ( 4,1)
.
h = we run 1 . Stack update two times. First it computes ( 4)
puts it onto the stack, then it puts P( n 5)
onto the stack and stops.
• 1
• For 1
Case 3: h = 2 :
Stack . Stack 1
P n and