Digital Logic Design: a rigorous approach c � Chapter 12: Trees Guy Even Moti Medina School of Electrical Engineering Tel-Aviv Univ. May 3, 2020 Book Homepage: http://www.eng.tau.ac.il/~guy/Even-Medina 1 / 31
Preliminary questions: Which Boolean functions are suited for implementation by 1 tree-like combinational circuits? In what sense are tree-like implementations optimal? 2 2 / 31
Reminder: Binary Boolean Functions Definition A binary Boolean function is a function f : { 0 , 1 } 2 → { 0 , 1 } . A binary function is often denoted by a dyadic operator, say ∗ . So instead of writing f ( a , b ), we write a ∗ b . 3 / 31
Reminder: Associative Boolean functions Definition A binary Boolean function ∗ : { 0 , 1 } 2 → { 0 , 1 } is associative if ( x 1 ∗ x 2 ) ∗ x 3 = x 1 ∗ ( x 2 ∗ x 3 ) , for every x 1 , x 2 , x 3 ∈ { 0 , 1 } . One may omit parenthesis: x 1 ∗ x 2 ∗ x 3 is well defined. Consider the function f n : { 0 , 1 } n → { 0 , 1 } defined by f n ( x 1 , . . . , x n ) � x 1 ∗ · · · ∗ x n 4 / 31
Extension of associative function Definition Let f : { 0 , 1 } 2 → { 0 , 1 } denote a Boolean function. The function f n : { 0 , 1 } n → { 0 , 1 } , for n ≥ 1, is defined recursively as follows. If n = 1, then f 1 ( x ) = x . 1 If n = 2, then f 2 = f . 2 If n > 2, then f n is defined based on f n − 1 as follows: 3 △ f n ( x 1 , x 2 , . . . x n ) = f ( f n − 1 ( x 1 , . . . , x n − 1 ) , x n ) . Claim If f : { 0 , 1 } 2 → { 0 , 1 } is an associative Boolean function, then f n ( x 1 , x 2 , . . . x n ) = f ( f n − k ( x 1 , . . . , x n − k ) , f k ( x n − k +1 , . . . , x n )) , for every n ≥ 2 and k ∈ [1 , n − 1] . 5 / 31
Trees of associative Boolean gates To simplify the presentation, consider the Boolean function or n . Definition A combinational circuit H = ( V , E , π ) that satisfies the following conditions is called an or -tree( n ). The graph DG ( H ) is a rooted tree with n sources. 1 Each vertex v in V that is not a source or a sink is labeled 2 π ( v ) = or . The set of labels of leaves of H is { x 0 , . . . , x n − 1 } . 3 x [0] x [1] x [2] x [3] x [0] x [1] or or or x [2] or or x [3] y or y 6 / 31
Correctness of or -tree( n ) Definition A combinational circuit H = ( V , E , π ) that satisfies the following conditions is called an or -tree( n ). Topology. The graph DG ( H ) is a rooted tree with n sources. 1 Each vertex v in V that is not a source or a sink is labeled 2 π ( v ) = or . The set of labels of leaves of H is { x 0 , . . . , x n − 1 } . 3 Claim Every or -tree ( n ) implements the Boolean function or n . 7 / 31
Relation to Boolean Formulas Definition A Boolean formula ϕ is an or ( n ) formula if it satisfies three conditions: (i) it is over the variables X 0 , . . . , X n − 1 , (ii) every variable X i appears exactly once in ϕ , and (iii) the only connective in ϕ is the or connective. Claim A Boolean circuit C is an or ( n ) -tree if and only if its graph (without the input/output gates) is a parse tree of an or ( n ) -formula. 8 / 31
Cost of or -tree( n ) x [0] x [1] x [2] x [3] x [0] x [1] or or or x [2] or or x [3] y or y Claim The cost of every or -tree ( n ) is ( n − 1) · c ( or ) . Lemma Let G = ( V , E ) denote a rooted tree in which the in-degree of each vertex is at most two. Then |{ v ∈ V | deg in ( v ) = 2 }| = |{ v ∈ V | deg in ( v ) = 0 }| − 1 . 9 / 31
Depth of tree delay of an or tree = number of or -gates along the longest path from an input to an output. Definition (depth - nonstandard definition) The depth of a rooted tree T is the maximum number of vertices with in-degree greater than one in a path in T . We denote the depth of T by depth ( T ). Why is this nonstandard? Usually, depth is simply the length of the longest path. Here we count only vertices with in-degree ≥ 2. Why? Input and output gates have zero delay (no computation) Assume inverters are free and have zero delay (we will show that for or ( n ) cost & delay are not reduced even if inverters are free and without delay) 10 / 31
Binary rooted trees Definition A rooted tree is a binary tree if the maximum in-degree is two. A rooted tree is a minimum depth tree if its depth is minimum among all the rooted trees with the same number of leaves. All binary trees with n leaves have the same cost. But, which have minimum depth? if n that is a power of 2, then there is a unique minimum 1 depth tree, namely, the perfect binary tree with log 2 n levels. if n is not a power of 2, then there is more than one minimum 2 depth tree... (balanced trees) 11 / 31
Example: Delay analysis Are these minimum depth trees? or or or or or or or or or or Figure: Two trees with six inputs. 12 / 31
Depth: lower bound Claim If T n is a rooted binary tree with n leaves, then the depth of T n is at least ⌈ log 2 n ⌉ . Suffice to prove depth ≥ log 2 n . 1 Complete induction on n . 2 13 / 31
Min Depth: the case n = 2 k (perfect binary trees) The distance of a vertex v to the root r in a rooted tree is the length of the path from v to r . Definition A rooted binary tree is perfect if: The in-degree of every non-leaf is 2, and All leaves have the same distance to the root. Note that the depth of a perfect tree equals the distance from the leaves to the root (no vertices with in-degree 1). Claim The number of leaves in a perfect tree is 2 k , where k is the distance of the leaves to the root. Claim Let n denote the number of leaves in a perfect tree. Then, the distance from every leaf to the root is log 2 n. 14 / 31
Minimum depth trees We now show that for every n , we can construct a minimum depth tree T ∗ n of depth ⌈ log 2 n ⌉ . In fact, if n is not a power of 2, then there are many such trees. 15 / 31
Balanced partitions Definition Two positive integers a , b are a balanced partition of n if: a + b = n , and 1 max {⌈ log 2 a ⌉ , ⌈ log 2 b ⌉} ≤ ⌈ log 2 n ⌉ − 1. 2 Claim If n = 2 k − r, where 0 ≤ r < 2 k − 1 , then the set of balanced partitions is = { ( a , b ) | 2 k − 1 − r ≤ a ≤ 2 k − 1 and b = n − a } . △ P 16 / 31
Construction of a balanced tree Algorithm 1 Balanced-Tree( n ) - a recursive algorithm for construct- ing a binary tree T ∗ n with n ≥ 1 leaves. The case that n = 1 is trivial (an isolated root). 1 If n ≥ 2, then let a , b be balanced partition of n . 2 Compute trees T ∗ a and T ∗ b . Connect their roots to a new root 3 to obtain T ∗ n . Definition A rooted binary tree T n is a balanced tree if it is a valid output of Algorithm Balanced-Tree(n). 17 / 31
Def: balanced tree Algorithm 2 Balanced-Tree( n ) - a recursive algorithm for construct- ing a binary tree T ∗ n with n ≥ 1 leaves. The case that n = 1 is trivial (an isolated root). 1 If n ≥ 2, then let a , b be balanced partition of n . 2 Compute trees T ∗ a and T ∗ b . Connect their roots to a new root 3 to obtain T ∗ n . Claim The depth of a binary tree T ∗ n constructed by Algorithm Balanced-Tree ( n ) is ⌈ log 2 n ⌉ . Corollary The propagation delay of a balanced or -tree ( n ) is ⌈ log 2 n ⌉ · t pd ( or ) . 18 / 31
Optimality of trees Goals: prove optimality of a balanced or -tree( n ). Theorem Let C n denote a combinational circuit that implements or n . Then, c ( C n ) ≥ n − 1 . Theorem Let C n denote a combinational circuit that implements or n . Let k denote the maximum fan-in of a gate in C n . Then t pd ( C n ) ≥ ⌈ log k n ⌉ . 19 / 31
Flipping bits Definition Let flip i : { 0 , 1 } n → { 0 , 1 } n be the Boolean function defined by △ flip i ( � x ) = � y , where � x j if j � = i △ y j = not ( x j ) if i = j . 20 / 31
The cone of a function Definition (Cone of a Boolean function) The cone of a Boolean function f : { 0 , 1 } n → { 0 , 1 } is defined by △ cone ( f ) = { i : ∃ � v such that f ( � v ) � = f ( flip i ( � v )) } Example cone ( xor ) = { 1 , 2 } . We say that f depends on x i if i ∈ cone ( f ). 21 / 31
Example Consider the following Boolean function: � if � 0 i x i < 3 f ( � x ) = 1 otherwise. Suppose that one reveals the input bits one by one. As soon as 3 ones are revealed, one can determine the value of f ( � x ). Nevertheless, the function f ( � x ) depends on all its inputs, and hence, cone ( f ) = { 1 , . . . , n } . 22 / 31
Constant Functions Claim cone( f ) = ∅ ⇐ ⇒ f is a constant Boolean function. 23 / 31
Composition of Functions Claim x ) � B ( f 1 ( � If g ( � x ) , f 2 ( � x )) , then cone( g ) ⊆ cone( f 1 ) ∪ cone( f 2 ) . 24 / 31
Graphical Cone Definition Let G = ( V , E ) denote a DAG. The graphical cone of a vertex v ∈ V is defined by △ cone G ( v ) = { u ∈ V : deg in ( u ) = 0 and ∃ path from u to v } . In a combinational circuit, every source is an input gate. This means that the graphical cone of v equals the set of input gates from which there exists a path to v . 25 / 31
Functional Cone ⊆ Graphical Cone Claim Let H = ( V , E , π ) denote a combinational circuit. Let G = DG ( H ) . For every vertex v ∈ V , the following holds: cone( f v ) ⊆ cone G ( v ) . Namely, if f v depends on x i , then the input gate u that feeds the input x i must be in the graphical cone of v . 26 / 31
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