Topics: 1. Trees - properties 2. The master theorem 3. Decoders מבנה המחשב - אביב 2004 תרגול 4#

Tree: Basic properties An undirected graph is a tree if it is: Connected, i.e., there is a path between every two vertices. Contains no cycles. One of these properties can be replaced with the following: |E| = |V|-1

Directed Tree: Basic properties An directed graph is a directed tree if it is: In-degree = 1 for all v, besides the root. In degree of the root is 0. There is a directed path from the root to every node. Tree  directed tree: Pick one of its vertices to be a root Direct all the edges out from the root Repeat it with the vertices as the roots.

Directed Tree (cont.) A leaf is a node with out-degree zero. We define on the vertices of a directed tree: The parent of a node v. Note there is only one such vertices. A child of a node v (there may be many). Ancestor and Descendent of v.  The root (source): Ancestor for all x, but has no parent.  Internal node: has a parent, may have children.  A leaf (sink): has no children.

A claim on Trees In a directed tree, any non-leaf is called an internal node. For a tree: a leaf is a node whose degree is 1. Claim: If the degree of any vertex in the tree  3, then the number of internal nodes  the number of the leaves –2. Note: the underlying graph of a directed binary tree is such a tree, but since the root does not count as a leaf, we have in directed trees: #internal nodes  #internal leaves -1

A claim on Trees (cont.) Proof (Induction on the number of nodes): Basis: trivial for n=1,2, check for all trees with 3 nodes (there is only one such tree). Assumption: correct for all trees of size  n. Step: prove for any tree of size n+1. Given such a tree, it must contain an internal node v. Pick such a v that is connected to a leaf. Define |T|=n = m+k, (m is #leaf, k is # internal.

A claim on Trees (cont.) If v is connected to one leaf u: remove (u,v), to obtain a tree T’ of size n-1  m (# leafs) is the same for T and T’  k (# internal) increased by 1  k  m-2 (induction hypothesis).

A claim on Trees (cont.) If v is connected to two leaves u,w: remove both edges, to obtain a tree T’ of size n-2  m’ = m – (u,w were removed, but v became a leaf in T’)  k’ = k – 1 (v is an internal node in T)  Applying the induction hypothesis, we obtain: k = k’+1  m’-2+1 = m-2  m-2

The master theorem a,b are two constants  1.  is a constant > 0. f(n) is a function. T(n) = aT(n/b) + f(n) 1.If f(n) = O(n (log b a -  ) ) then T(n) =  (n (log b a) ) 2.If f(n) = O(n log b a ) then T(n)=  (n (log b a) logn) 3.If f(n) = O(n (log b a +  ) ) and af(n/b)  cf(n) for some c < 1, then T(n) =  (f(n))

The master theorem - Examples T(n) = 9T(n/3) +n. Set  =1and we are in the first case. Hence T(n) =  (n (log 3 9) ) =  (n 2 ). T(n) = T(2n/3) + 1. a=1, b=3/2 note that log(1) = 0 so f(n) =  (n 0 ) and we are in case 2. Hence T(n) =  (n 0 logn) =  (logn).

The master theorem – Techniques for the proof Two main techniques used in the proof. First we may focus on exact powers of b. This simplifies the arguments. Next we examine the recurrence tree. Finally we extend it to any n. This step is purely technical. In algorithmic cases, we usually may assume that instead of running on an input of size n, we run on a larger input n’ which is a power of b. This would not change the asymptotic.

The master theorem - Examples T(n) = 9T(n/3) +n. Set  =1and we are in the first case. Hence T(n) =  (n (log 3 9) ) =  (n 2 ). T(n) = T(2n/3) + 1. a=1, b=3/2 note that log(1) = 0 so f(n) =  (n 0 ) and we are in case 2. Hence T(n) =  (n 0 logn) =  (logn).

Input: A string {0,1} n. Output: A string {0,1} 2 n. Functionality: For every string s, returns 1 in the bit which is the binary representation of s. Implementation: Construct a device with n inputs and 1 output, that returns 1 when its input is the number k. The device is implemented using up to n NOT gates, and one AND gate of fan-in n. Brute force decoder(n)

Implementation (cont.): For every output bit k, connect the device that implements k. Correctness: Trivial, it is the implementation of the required boolean function. Size: O(n2 n ). Delay: O(1), if we use and gates with non- restricted fan-in. If we need to implement with and gates with fan-in = 2, requires O(log n). Brute force decoder(n) (cont.)

Decoder’(n)

Decoder(n) design shown and proven in class:

for k = n-1: Decoder(1) Decoder’(n)

Decoder’(n) is a private case of Decoder(n) design. (k = n-1 or k = 1) Decoder’(n) is a correct implementation of a decoder. Cost analysis: Solving the recurrence: asymptotics

Linear Delay!!! Delay analysis: Solving the recurrence:

Decoder(n) for k = n/2

Consider the top two recursion steps: each AND-gate in the “AND-gates array” of the second recursion step, i.e. Decoder(n/2), feeds 2 n/2 AND-gates in the “AND-gates array” of the primary recursion step, i.e. Decoder(n).

Decoder’(n)Decoder(n) k=n/2 Design n Decoder’(n)Decoder(n) k=n/2 Design n Cost Delay

The next slides are for the curious students (home reading)

Input: A string {0,1} 2 n. Output: A string {0,1} n. Functionality: For every string s with weight 1, returns the the binary representation of s. Implementation: Connect all the bits with an OR gate of size 2 n, this gate sets bit 0 of the output. Connect bits 1,3,…2 n –1 with an OR gate of size 2 n-1, this gate sets bit 1 of the output. And so on for each power of 2. Brute force encoder(n)

Correctness: Trivial, bit j of the output is 1 if and only if any of the inputs in the set S j is 1. The set S j is the set of all input bits whose index is such that 2 j is 1 in its binary representation. Size: O(n2 n ). Delay: O(1), if we use OR gates with non- restricted fan-in. If we need to implement with and gates with fan-in = 2, requires O(n). Brute force encoder(n) (cont.)

A balanced tree structure minimizes the delay and cost. The tree is a binary tree due to fan-out limitations. Notice: * The leaves of the tree feed the gate outputs. * The root of the tree is fed by the gate input. * The first branching is free → no need for a “root” gate. Buffers

Cost analysis: 1-buf-trees Fanout limitation effect on design

Delay analysis: 1-buf-trees

Cost analysis: Delay analysis: 1-buf-trees

Cost Decoder(n): n=1: 1 n=2: 10 n=4: 52 n=8: 616 n=16: n=32: e+009 n=64: e+019 n=128: e+038 Cost Decoder‘(n): n=1: 1 n=2: 10 n=4: 60 n=8: 1024 n=16: n=32: e+010 n=64: e+019 n=128: e+039 Cost Decoder‘(n), Fanout<=2: n=1: 1 n=2: 11 n=4: 79 n=8: 1511 n=16: n=32: e+010 n=64: e+020 n=128: e+039 Cost Decoder(n), Fanout<=2: n=1: 1 n=2: 10 n=4: 68 n=8: 1096 n=16: n=32: e+010 n=64: e+019 n=128: e+039

Delay Decoder(n): n=1: 1 n=2: 3 n=4: 5 n=8: 7 n=16: 9 n=32: 11 n=64: 13 n=128: 15 Delay Decoder‘(n): n=1: 1 n=2: 3 n=4: 7 n=8: 15 n=16: 31 n=32: 63 n=64: 127 n=128: 255 Delay Decoder‘(n), Fanout<=2: n=1: 1 n=2: 4 n=4: 8 n=8: 16 n=16: 32 n=32: 64 n=64: 128 n=128: 256 Delay Decoder(n), Fanout<=2: n=1: 1 n=2: 3 n=4: 6 n=8: 11 n=16: 20 n=32: 37 n=64: 70 n=128: 135