CSE390D Notes for Friday, 10/25/24
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structural induction
The set of extended binary trees can be defined recursively by these steps:
basis step: The empty set is an extended binary tree
recursive step: If T1 and T2 are disjoint extended binary trees,
there is an extended binary tree denoted by T1 * T2 consisting of
a root r together with edges connecting the root to each of the
roots of the left subtree T1 and the right subtree T2 when these
trees are nonempty
functions on trees:
N(T) = # of nodes in a tree
N(empty) = 0
N(T1 * T2) = 1 + N(T1) + N(T2)
Ht(T) = height of a tree
Ht(empty) = 0
Ht(T1 * T2) = 1 + max(Ht(T1), Ht(T2))
The set of fully balanced binary trees (FBBT) can be defined
recursively by these steps:
basis step: The empty set is a FBBT
recursive step: If T1 and T2 are FBBTs with Ht(T1) = Ht(T2), then
T1 * T2 is a FBBT
The set of Almost Balanced Binary Trees (ABBT) can be defined
recursively by these steps:
basis step: The empty set is a FBBT
recursive step: If T1 and T2 are ABBTs with Ht(T1) = Ht(T2), then
ht(T1) - 1 <= ht(T2) <= ht(T1) + 1 then T1 * T2 is a ABBT when T1 in ABBT
sample tree:
have to move node at RLLRL to something like RLRR or RRRRL
structural induction:
1) base case: Show P holds for all basis elements
2) recursive case: Show that if P holds for elements used to
construct a new element of S, then P holds for the new element
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If T is a FBBT, then N(T) = 2^Ht(T) - 1
basis step: Ht(empty) = 0, N(empty) = 0 = 2^0 - 1
recursive step:
given an FBBT T with subtrees T1 and T2
Ht(T1) = Ht(T2) (by definition)
N(T) = 1 + N(T1) + N(T2)
applying the inductive hypothesis
= 1 + 2^Ht(T1) - 1 + 2^Ht(T2) - 1
= 2^Ht(T) - 1
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Prove P(T): N(T) >= phi^Ht(T) - 1 for ABBT
Base case:
Prove P(empty):
N(empty) = 0 >= 1 - 1 = phi^0 - 1
Prove P(T) holds when it holds of the components of T:
Let T = T1 * T2
without loss of generality, assume Ht(T1) >= Ht(T2)
let k = Ht(T1)
then Ht(T) = k + 1
Ht(t2) = k or k-1
by i.h., N(T1) >= phi^k - 1
if Ht(t2) = k, then by i.h., N(T2) >= phi^k - 1
if Ht(t2) = k - 1, then by i.h., N(T2) >= phi^(k-1) - 1
phi^k - 1 >= phi^(k-1) - 1
therefore N(T2) >= phi^(k-1) - 1
Need to prove that N(T) >= phi^(k+1) - 1
N(T) = 1 + N(T1) + N(T2)
using results from i.h. above:
>= 1 + phi^k - 1 + phi^(k-1) - 1
= phi^k + phi^(k-1) - 1
= phi^(k-1)(phi + 1) - 1
= phi^(k-1)(phi^2) - 1 (because phi + 1 = phi^2)
= phi^(k+1) - 1
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Stuart Reges
Last modified: Fri Oct 25 15:39:40 PDT 2024