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Threshold Logic for Nanotechnologies 1 Zvi Kohavi and Niraj K. Jha Introductory Concepts Threshold element or gate: x1 w1 x2 w2 T y wn xn Example: y = f(x1,x2,x3) = (1,2,3,6,7) = x1’x3 + x2 x1 -1 x2 1_ 2 2 y 1 x3 2 MOBILEs Monostable-bistable transition logic element (MOBILE): a resonant tunneling diode (RTD) and heterostructure field-effect transistor (HFET) nanotechnology based threshold element • Rising edge-triggered, current-controlled gate • Serially-connected load and driver RTDs • RTD-HFET structures in parallel to the load (driver) RTDs perform positive (negative) weighting of inputs • Area of RTDs: corresponds to weight • Difference in the areas of the driver and load RTDs: threshold Clk RTD Positiveweight inputs Load w1 x1 w2 x2 f HFET -w3 T x3 Driver Negativeweight input 3 Majority Gates Majority gate: a special type of threshold element • A three-input majority gate: produces a 1 if a majority of its inputs are 1 M(x1,x2,x3) = x1x2 + x2x3 + x1x3 • Can be implemented as a threshold element: with w1 = w2 = w3 = 1 and T=2 • Acts like an AND (OR) gate when one of its inputs is tied to 0 (1) Nanotechnology implementations: quantum cellular automata (QCA), single-electron box (SEB) Input x1 0 Input x2 1 1 Input x3 Device cell Output cell Node 1 Inputs x1 x2 x3 C C C Vd CL CL C0 Cj Cj Node 2 Output terminal C f3 C f2 C f 1 Output capacitor Input capacitor 1 QCA SEB 4 Minority Gates Minority gate: produces a 1 if a majority of its inputs are 0 m(x1,x2,x3) = x1’x2’ + x2’x3’ + x1’x3’ • Acts like a NAND (NOR) gate when one of its inputs is tied to 0 (1) Nanotechnology implementation: tunneling phase logic (TPL) Clock 1 Ci J1 J2 Clock 2 Cj J4 Pump J3 Pump TPL 5 Capabilities and Limitations of Threshold Logic Threshold gate: generalization of conventional gates • More powerful than conventional gates because it can realize a larger class of functions • Any conventional gate can be realized with a threshold gate • Thus, threshold gates are functionally complete Example: NAND implementation x1 -1 -11_ 2 y -1 x2 6 Is Every Switching Function Realizable by One Threshold Element? Answer: No Example: Let f(x1,x2,x3,x4) = x1x2 + x3x4 • Output value must be 1: for x1x2x3’x4’, x1’x2’x3x4 • Output value must be 0: for x1’x2x3’x4, x1x2’x3x4’ • Since the requirements in the inequalities are conflicting, no threshold value can satisfy them – Thus, the function is not realizable by a single threshold element 7 Basic Problem of Threshold Logic Given a switching function f(x1,x2, …,xn): determine whether it is realizable by a single threshold element, and if it is, find appropriate weights and threshold • Such a function is called a threshold function Straightforward approach: Solve a set of 2n linear, simultaneous inequalities Example: Let f(x1,x2,x3) = (0,1,3) Combination 0: T must be negative Combinations 2, 4: w2, w1 must be negative Combinations 3, 5: w2 must be greater than w1 Combination 1: w3 is greater than or equal to T Thus, w3 T > w2 > w1 w1 = -2, w2 = -1, w3 = 1, T = -1/2 8 Sensitivity to Variations Limitation: Due to variations in input and supply voltages, the weighted sum may deviate from its prescribed value and cause circuit malfunction • Restrictions imposed on the number of inputs and threshold T • Introduce defect tolerances: non-negative and 9 Elementary Properties Weight-threshold vector: V = {w1,w2, …,wn;T} Let f(x1,x2, …,xn) be realized by V1 = {w1,w2, …,wj, …,wn;T}. If xj is complemented, it can be realized by V2 = {w1,w2, …,-wj, …,wn;T-wj} with inputs x1,x2, …,xj’, …,xn From V1: When V2 replaces V1 and xj’ replaces xj: where g is realized by V2 g and f are identical: since the equations reduce to each other for both xj = 0 and xj = 1 10 Important Conclusions If a function is realizable using a single threshold element, then by an appropriate choice of complemented and uncomplemented input variables: a realization with any sign distribution is possible Corollary: if a function is realizable by a single threshold element, then it is realizable by an element with only positive weights 11 Important Property If f(x1,x2, …,xn) is realizable by a single threshold element with V1 = {w1,w2, …,wn;T}, then its complement is realizable by a single threshold element with V2 = {-w1,-w2, …,-wn;-T} From V1: Multiplying both sides by -1: Thus, f’ is realizable by V2 12 Synthesis of Threshold Networks Unate functions: function f(x1,x2, …,xn) is positive (negative) in variable xi if there exists a disjunctive or conjunctive expression for the function in which xi only appears in uncomplemented (complemented) form If f is either positive or negative in xj: it is said to be unate in xi Example: f = x1x2’ + x2x3’ is positive in x1 and negative in x3, but not unate in x2 If f(x1,x2, …,xn) is unate in each of its variables: then it is called unate Example: f = x1’x2 + x1x2x3’ is unate since it can be simplified to x1’x2 + x2x3’ Example: f = x1x2’ + x1’x2 is not unate in either variable 13 Unate Functions If f(x1,x2, …,xn) is positive in xi: then it can be expressed as and vice versa If f(x1,x2, …,xn) is negative in xi: then it can be expressed as and vice versa 14 Geometric Representation n-cube: contains 2n vertices, each of which represents an assignment of values to n variables and thus corresponds to a minterm • a line is drawn between every pair of vertices which differ in just one variable • Vertices for which the function is 1 (0) called: true (false) vertices Example: Three-cube representation for f = x’y’ + xz (1,1,1) (1,1,0) (0,1,1) (1,0,1) (0,1,0) (1,0,0) (0,0,1) (0,0,0) 15 Partial Ordering Partial-ordering relation between vertices of the n-cube: (a1,a2, …,an) (b1,b2, …,bn) if and only if for all i, ai bi • • • • Partially ordered set of vertices: a lattice (0,0, …,0): least vertex (1,1, …,1): greatest vertex Some pair of variables incomparable: e.g., (0,0, …,0,1) and (1,0, …,0,0) Without loss of generality: concentrate on positive unate functions Example: relabel x1’x2x3’ + x2x3’x4 as x1x2x3 + x2x3x4 • By reconverting the latter: possible to determine the original function 16 Unate Function Theorem Theorem 1: f(x1,x2, …,xn) is unate if and only if it is not a tautology and the above partial ordering exists, such that for every pair of vertices, (a1,a2, …,an) and (b1,b2, …,bn), if (a1,a2, …,an) is a true vertex and (b1,b2, …,bn) (a1,a2, …,an), then (b1,b2, …,bn) is also a true vertex of f Minimal true vertex: A true vertex Si is said to be minimal if no other true vertex Sj < Si Maximal false vertex: A false vertex Si is said to be maximal if no other false vertex Sj > Si Example: For x1x2 + x3x4 • Minimal true vertices: S1 = (1,1,0,0), S2 = (0,0,1,1) • Thus, every vertex greater than S1 or S2 must be a true vertex: e.g., (1,1,1,0), (0,1,1,1) – These vertices correspond to x1x2x3 and x2x3x4, which are covered by f 17 Linear Separability For an n-cube representation for threshold functions: linear equation w1x1 + w2x2 + … + wnxn = T corresponds to an (n-1)-dimensional hyperplane that cuts through the n-cube • Since f = 0 when w1x1 + w2x2 + … + wnxn < T • and f = 1 when w1x1 + w2x2 + … + wnxn T the hyperplane separates the true vertices from the false ones Such a function is called a linearly separable function • Thus, every threshold function is linearly separable, and vice versa 18 Theorems Theorem 2: Every threshold function is unate Theorem 3: Given an expression for a unate switching function, f(x1,x2, …,xn), replace xj by xk’, resulting in f(x1,x2, …,xn). If g is not a threshold function, then neither is f Example: Let f = x1x2 + x3x4 • • • • To determine if f is a threshold function: replace x2 by x3’ This results in g = x1x3’ + x3x4 Since g is not unate in x3, it is not a threshold function Hence, neither is f 19 Identification and Realization of Threshold Functions Procedure: 1. 2. 3. 4. Test the given function f for unateness If it is unate, convert it into another function g that is positive in all its variables Find all minimal true and maximal false vertices of g Derive and solve a system of pq inequalities, corresponding to the p minimal true and q maximal false vertices - For minimal true vertex A = {a1,a2, …,an} and maximal false vertex B = {b1,b2, …,bn}, write w1a1 + w2a2 + … + wnan > w1b1 + w2b2 + … + wnbn 20 Identification Example Example: Given f = x1x2x3’x4 + x2x3’x4’ 1. Reduce to f = x1x2x3’ + x2x3’x4’, which is unate 2. g = x1x2x3 + x2x3x4 3. Minimal true vertices: (1,1,1,0), (0,1,1,1); Maximal false vertices: (1,1,0,1), (1,0,1,1), (0,1,1,0) 4. p = 2 and q = 3 yields 6 inequalities: 5. Necessary constraints that must be satisfied: V = {1,2,2,1; 9/2} for g => V = {1,2,-2,-1; 3/2} for f 21 Map-based Synthesis of Two-level Threshold Networks Decomposition of non-threshold functions: into two or more factors that are threshold functions Admissible pattern: a pattern of 1 cells that can be realized by a single threshold element • An admissible pattern may be in any position on the map • An admissible pattern for functions of three variables is also an admissible pattern for functions of four or more variables • Since the complement of a threshold function is also a threshold function, patterns of 0 cells are also admissible • Select a minimal number of admissible patterns such that each 1 cell is covered by at least one admissible pattern 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 22 1 Synthesis Example Example: For f(x1,x2,x3,x4) = (2,3,6,7,10,12,14,15), find a minimal threshold-logic realization x1 x2 x3 x4 00 01 11 10 1 00 01 11 1 1 1 10 1 1 1 1 g h (a) Map for f exhibiting two admissible patterns x1 x1 x2 x2 -2 1 3 1 x3 5 2 2 1 1 -1 g x3 x4 5 2 h x4 (b) Threshold elements realizing the admissible patterns x1 x1 x2 -2 1 3 1 x3 x4 x2 5 2 2 1 3 1 -1 g x3 x4 (c) Threshold-logic realization of f 5 2 f 23 Another Synthesis Example Example: For f(x1,x2,x3,x4) = (3,5,7,10,12,14,15), find a minimal threshold-logic realization x1 x2 x3 x4 00 01 x1 x3 x4 10 x1 x2 x4 1 00 01 11 11 1 1 1 1 1 10 x2 x3 x4 x1 x2 x4 x1 x3 x4 1 (b) AND-OR realization of f (a) Map showing a minimal set of prime implicants which cover f. x1 x2 x3 x4 00 01 01 10 10 1 00 11 11 1 x2 1 1 x1 x1 1 1 f -1 1 1 2 x3 1 x4 x2 1 2 2 2 1 1 32 1 -1 g x3 2 1 2 f x4 (d) A threshold-logic realization of f 24 (c) Map showing the admissible pattern realized by each threshold element. Synthesis of Multi-level Threshold Networks Example: One-to-one map from the following network to a threshold network requires seven threshold elements (including the inverter) and five logic levels – quite sub-optimal • Reason: some nodes can be collapsed into a single threshold node x1 x2 x3 n4 n5 x1 x4 n3 x5 n1 f x6 x7 n2 (a) Switching network Assuming a fanin restriction of four: • • • • Collapse f = n1 + n2 to n3x5 + x6x7 Since f is not threshold: split it into n1 + x6x7, where n1 = n3x5 Since n1 + x6x7 is threshold: synthesize n1 next x1 Since n1 = n4x5 + n5x5 is threshold: x5 1 n 4 x2 1 3 synthesize n4 = x1x2x3 and 2 1 1 3 x3 1 n5 n5 = x1’x4, which are both threshold x 1 x6 n1 1 2 2 1 f x7 -1 1 1 x4 25 (b) Equivalent threshold network General Synthesis Procedure Procedure: 1. Start with a multi-output algebraically-factored switching network G 2. Process each primary output of G • If the node represents a binate function, split into multiple nodes and process recursively • If the node is unate and is also a threshold function, save it in the threshold network and process its input nodes recursively • Else, split the unate node into two or more nodes that are threshold functions 3. Terminate procedure when all the nodes in G are mapped to threshold nodes 26 Mapping Threshold Networks to MOBILEs MOBILE: a self-latching threshold gate because its output is valid only when the clock is high Four-phase clocking: all signals to any threshold element must arrive in the same clock phase • Ensured by inserting buffers as necessary 27 MOBILE Example Full-adder: Level: 2 1 3 a 1 1 1 b c0 2 ci -2 1 1 1 CLK 1 s wa=1 threshold buffer a a f (a) Network before inserting buffers 1 1 2 1 T=2 Level: 2 1 3 a b 1 1 1 c0 2 ci -2 1 1 1 1 s threshold buffer (b) Network after inserting buffers 28 f Synthesis of Multi-level Majority/Minority Networks Realizable pattern: pattern of 1 cells realizable by a majority gate • For three-input positive functions: 10 realizable patterns • Removing the restriction that function be positive: 38 realizable patterns x1x2 00 x3 01 11 10 0 1 1 1 1 1 x1x2 00 x3 11 0 1 1 0 1 1 1 1 x1 = M(x1, 1, 0) = M(x1, 0, 1) x1x2 00 x3 0 1 1 01 11 1 1 1 1 10 x2 = M(1, x2, 0) = M(0, x2,1) x1x2 00 x3 10 1 0 1 1 1 0 1 1 1 1 1 11 10 x1x2 00 x3 1 0 1 1 1 x1x2= M(x1, x2, 0) 01 11 10 0 1 x1x2 00 x3 10 01 1 x2x3= M(0, x2, x3) 10 1 1 1 01 1 1 11 10 1 1 1 1 11 10 1 1 x1x3= M(x1, 0, x3) x1x2 x3 00 01 0 1 11 x1 + x3 = M(x1, 1, x3) x1 + x2 = M(x1, x2, 1) 01 01 x3 = M(1, 0, x3) = M(0, 1, x3) 11 0 x1x2 00 x3 1 01 x2 + x3 = M(1, x2, x3) x1x2 00 x3 x1x2 00 x3 01 1 11 10 1 1 1 1 x1x2 + x1x3 +x2x3 = M(x1, x2, x3) 29 Synthesis Example Example: Consider f = x1’x2’x3’ + x1’x2x3 + x1x2x3’ + x1x2’x3 • Naïve approach: decompose network into two-input AND and OR gates and replace each such gate by a reduced majority gate • However, if we make full use of the three inputs of a majority gate: only four gates necessary • Minority network: can be obtained from a majority network using De Morgan’s theorem x1 x2 x3 x1 x2 x3 x1 x2 x3 x1 x2 x3 f (a) x1 x2 x3 M x1 x2 x3 M x1 x2 x3 M f1 f2 M f3 (b) f x1 x2 x3 m x1 x2 x3 m x1 x2 x3 m f1 f2 m f3 (c) 30 f General Synthesis Procedure Procedure: 1. Start with a multi-output algebraically-factored switching network G 2. Decompose G into a network in which nodes have at most three inputs • If the node represents a majority function, move on to the next node • If a common literal exists in all the product terms of the node function, factor it out and perform AND/OR mapping on it • If a common literal does not exist, check to see if the node can be implemented with fewer than four AND/OR nodes • Else, map the node onto at most four majority gates using a Karnaugh-map based procedure Example: Consider f = x1x2’+ x2’x3 • • With AND/OR mapping, three majority gates are needed: – f1 = x1x2’, f2 = x2’x3, f = f1 + f2 However, since literal x2’ can be factored out: f = f1x2’ where f1 = x1+x3 – This requires only two majority gates 31 K-map based Procedure Given the map of a node function n with at most three inputs: 1. Find a realizable pattern f1 2. Find a second realizable pattern f2 based on f1 and n 3. Find the third realizable pattern f3 based on f1, f2 and n – Realizable patterns chosen such that n = M(f1,f2,f3) = f1f2+f2f3+f1f3 4. f1 may contain makeup minterms that are not minterms of n – A minterm (maxterm) of n must also be a minterm (maxterm) of at least two of the three functions, f1, f2 and f3 – Enforce rule by defining two sets: 1 and 0 » For finding f2: if a minterm (maxterm) of n is not a minterm (maxterm) of f1, add it to 1 ( 0 ) » For finding f3: if a minterm (maxterm) of n is not a minterm (maxterm) of both f1 and f2: add it to 1 ( 0 ) 5. On failure to find f3, backtrack to find new f2 32 Synthesis Example Example: Consider f = x1’x2’x3’ + x1’x2x3 + x1x2x3’ + x1x2’x3 00 x3 0 01 11 1 1 x3 10 0 1 1 00 0 01 11 01 11 1 1 1 10 x3 00 01 1 Step 2: find f2 00 x3 01 0 0 11 1 Update 1 x1x2 10 00 x3 0 1 1 1 01 Update 0 x1x2 00 x3 0 01 11 10 1 0 (g) 10 1 1 x1 x2 x3 M x1 x2 x3 M x1 x2 x3 M f1 f2 M f3 (f) Step 3: find f3 x1x2 00 01 11 10 x3 0 0 11 1 f2 = x1x2+ x2 x3+ x1x3 = M(x1, x2, x3) (e) (d) 10 (c) 1 1 1 11 0 1 x1x2 10 Compute1 x1x2 f1 = x1 x2+ x2x3 + x1x3= M(x1, x2, x3) (b) Compute 0 x1x2 00 1 1 n = x1x2 x3 + x1 x2x3 + x1 x2x3 + x1 x2x3 (a) x3 Step 1: find f1 x1x2 x1x2 1 1 1 1 1 f3 = x1x2 + x2 x3+ x1x3 = M(x1, x2, x3) (h) 33 f