Associative dyadic boolean functions Goals Def: A Boolean function f : { 0 , 1 } 2 → { 0 , 1 } is associative if define associative Boolean functions (and classify them). Chapter 3: Trees f ( f ( σ 1 , σ 2 ) , σ 3 ) = f ( σ 1 , f ( σ 2 , σ 3 )) , trees - combinational circuits that implement associative Computer Structure - Spring 2004 Boolean funcs. for every σ 1 , σ 2 , σ 3 ∈ { 0 , 1 } . analyze delay & cost of trees. � Dr. Guy Even c Q: List all the associative Boolean functions prove optimality. Tel-Aviv Univ. f : { 0 , 1 } 2 → { 0 , 1 } . “A”: There are 16 dyadic Boolean functions, only need to list them and check... – p.1 – p.2 – p.3 f n : repeating f : { 0 , 1 } 2 → { 0 , 1 } f n : the associative case Definition of OR -trees Def: A combinational circuit C = �G , N� that satisfies the following conditions is called an OR -tree ( n ) . Def: Let f : { 0 , 1 } 2 → { 0 , 1 } denote a Boolean function. The If f ( x 1 , x 2 ) is associative, then parenthesis are not function f n : { 0 , 1 } n → { 0 , 1 } , for n ≥ 2 is defined by 1. Input: x [ n − 1 : 0] . important. induction as follows. Claim: If f : { 0 , 1 } 2 → { 0 , 1 } is an associative Boolean 2. Output: y ∈ { 0 , 1 } 1. If n = 2 then f 2 ≡ f . function, then 3. Functionality: y = OR ( x [0] , x [1] , · · · , x [ n − 1]) . 2. If n > 2 , then f n is defined based on f n − 1 as follows: 4. Gates: All the gates in G are OR -gates. f n ( x 1 , x 2 , . . . x n ) = f ( f k ( x 1 , . . . , x k ) , f n − k ( x k +1 , . . . , x n )) , 5. Topology: The underlying graph of DG ( C ) (i.e. △ f n ( x 1 , x 2 , . . . x n ) = f ( f n − 1 ( x 1 , . . . , x n − 1 ) , x n ) . for every k ∈ [2 , n − 2] . undirected graph obtained by ignoring edge directions) is a tree. Q: Show that the set of functions f n ( x 1 , . . . , x n ) that are Example: induced by associative dyadic Boolean functions is Note that in the tree: leaves correspond to the inputs x [ n − 1 : 0] and the NOR 4 ( x 1 , x 2 , x 3 , x 4 ) = NOR ( NOR ( NOR ( x 1 , x 2 ) , x 3 ) , x 4 ) . { constant 0 , constant 1 , x 1 , x n , AND , OR , XOR , NXOR } . output y . interior nodes - OR -gates. Note that NOR is not associative! note: only last 4 functions are “interesting”. We focus on OR . Could root the tree, and then the root is the output. – p.4 – p.5 – p.6
Example: OR -tree (4) Recursive definition of OR -trees Cost of OR -trees Def: an OR -tree ( n ) is defined recursively as follows: Claim: The cost of every OR -tree ( n ) is ( n − 1) · c ( OR ) . x [0] x [1] x [2] x [3] x [0] x [1] Proof: By induction on n . or or or basis: a single OR -gate is an or x [2] OR -tree (2) . Induction basis: n = 2 . In this case, OR -tree (2) contains a single OR -gate. or or step: an OR ( n ) -tree is a circuit x [3] in which y or -tree( n 1 ) or -tree( n 2 ) 1. the output is computed by or an OR -gate. y 2. the inputs of this OR - Cost - both trees have 3 gates. gate are the outputs of or OR -tree ( n 1 ) & OR -tree ( n 2 ) , Delay - 2 gates vs. 3 . where n = n 1 + n 2 . – p.7 – p.8 – p.9 Cost of OR -trees - Induction step Delay of OR -trees Are balanced OR -trees optimal? let C denote an OR -tree ( n ) . let g denote the OR -gate that outputs the output of C . Claim: The delay of a balanced OR -tree ( n ) is What is the best (min. cost & delay) choice of a topology for a combinational circuit that implements the g is fed by two wires e 1 and e 2 . ⌈ log 2 n ⌉ · t pd ( OR ) . Boolean function OR n ? Is a tree indeed the best e 1 is the output of C 1 - an OR -tree ( n 1 ) topology? e 2 is the output of C 2 - an OR -tree ( n 2 ) Could one do better if another implementation is used? Proof: homework. Note that the term “balanced tree” can n 1 + n 2 = n be interpreted in more than one way especially if n is not a Ind. Hyp. ⇒ c ( C 1 ) = ( n 1 − 1) · c ( OR ) & c ( C 2 ) = ( n 2 − 1) · c ( OR ) . power of 2 ... c ( C ) = c ( g ) + c ( C 1 ) + c ( C 2 ) = (1 + n 1 − 1 + n 2 − 1) · c ( OR ) = ( n − 1) · c ( OR ) . QED – p.10 – p.11 – p.12
� � � � � Optimality of balanced OR -trees Optimality of balanced OR -trees Restriction of a Boolean function Def: Let f : { 0 , 1 } n → { 0 , 1 } denote a Boolean function. Let Would like to prove that every combinational circuit C that Would like to prove that every combinational circuit C that σ ∈ { 0 , 1 } . The Boolean function g : { 0 , 1 } n − 1 → { 0 , 1 } implements OR n satisfies: implements OR n satisfies: defined by c ( C ) ≥ n − 1 c ( C ) ≥ n − 1 △ t pd ( C ) ≥ log 2 n. t pd ( C ) ≥ log 2 n. g ( w 0 , . . . , w n − 2 ) = f ( w 0 , . . . , w i − 1 , σ, w i , . . . , w n − 2 ) is called the restriction of f with x i = σ . We denote it by We need to be more accurate about the model: Looking for proof also for the case that DG ( C ) is not a tree! f x i = σ . Q: what is the cost/delay of an n -input OR -gate? Examples: △ assumption 1: the fan-in of every gate ≤ 2 , so we have to x 2 =1 ( x 1 ) = XOR ( x 1 , 1) XOR build big gates from basic gates. assumption 2: the cost of every basic gate is ≥ 1 . � (input/output gates are free) if � n − 1 1 i =1 x i + 1 > n/ 2 △ x n =1 ( x 1 , . . . , x n − 1 ) = MAJORITY otherwise. 0 – p.13 – p.14 – p.15 Input-Output reachability Input-Output reachability - cont. Cone of a Boolean function A boolean function f : { 0 , 1 } n → { 0 , 1 } depends on its i th Find vectors w ′ , w ′′ ∈ { 0 , 1 } n such that Claim: If a combinational circuit C implements a Boolean input if function f , then there must be a path in DG ( C ) from every f ( w ′ ) � = f ( w ′′ ) x i =0 �≡ f input in cone ( f ) to the output of f . f x i =1 . w ′ [ i ] � = w ′′ [ i ] . Proof: by contradiction, Def: The cone of a Boolean function f is defined by assume i ∈ cone ( f ) . Proof of Simulation Theorem △ ⇒ C outputs the same value in y when input w ′ and w ′′ . cone ( f ) = { i : f depends on its i th input } . let g i ∈ G denote the input gate that feeds the i th input. assume that in DG ( C ) there is no path from g i to the ⇒ C errs either with w ′ or with w ′′ . QED Claim: The Boolean function OR n depends on all its inputs, output y . namely show that C does not implement f . | cone ( OR n ) | = n. – p.16 – p.17 – p.18
DAG terminology Linear Cost Lower Bound Theorem Leaves and interior vertices in trees Consider the directed acyclic graph (DAG) DG ( C ) . Let T = ( V, E ) denote a tree with at least two vertices. assumptions: deg in ( v ) : in-degree of a vertex v is the number of edges A leaf is a vertex of degree 1 . that enter the vertex v . fan-in of every gate at most 2 . An interior vertex is a vertex that is not a leaf. deg out ( v ) : out-degree of a vertex v : is the number of leaves ( V ) - set of leaves in V . cost of trivial gates (i.e. input/output gates) is zero. edges that emanate from the vertex v . cost of non-trivial gate is at least 1 . interior ( V ) - set of interior vertices in V . source - a vertex with in-degree zero. Claim: sink - a vertex with out-degree zero. Theorem: If C is a combinational circuit that implements a If the degree of every vertex in T is at most three, then Boolean function f , then interior vertex - a vertex that is neither a source or a | interior ( V ) | ≥ | leaves ( V ) | − 2 . sink. c ( C ) ≥ | cone ( f ) | − 1 . sources Corollary: If C n is a combinational circuit that implements OR n , then c ( C n ) ≥ n − 1 . Easy to prove theorem for trees, but what about arbitrary sinks interior vertices DAGs? – p.19 – p.20 – p.21 Underlying graph of DG ( C ) Construction of T Proof of linear cost lower bound theorem C - a combinational circuit & DG ( C ) = ( V, A ) - a DAG If underlying graph of DG ( C ) is a tree, then previous claim proves the theorem. underlying graph of DG ( C ) - undirected graph v v v v 1 2 3 4 G = ( V, E ) , where ( u, v ) ∈ E ⇔ ( u → v ) ∈ A . If DG ( C ) = ( V, E ) is not a tree, then construct a directed “binary tree” T = ( V ′ , E ′ ) such that If fan-in of every gate in C is at most 2 , then degree of V ′ ⊆ V & E ′ ⊆ E every vertex in G is at most 3 . sources ( T ′ ) = cone ( f ) Leaves in G correspond to input and output gates in C . output gate ∈ V ′ . Interior vertices in G correspond to non-trivial gates in in T ′ we have | interior nodes | ≥ | sources | − 1 . C . Case of a tree: But interior nodes of T are also interior in DG ( C ) , and Claim: Assume C has n inputs and a single output. number of sources in T equals | cone ( f ) | . QED. y Assume fan-in of all gates is at most 2 . If G is a tree, then Left to show how T is constructed... c ( C ) ≥ n − 1 . – p.22 – p.23 – p.24
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