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**Threshold Logic for Nanotechnologies**

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**Introductory Concepts**

Threshold element or gate: Example: y = f(x1,x2,x3) = (1,2,3,6,7) = x1’x3 + x2

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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

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**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) QCA SEB

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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) TPL

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**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

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**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

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**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

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**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

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**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

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**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

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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

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**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

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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

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**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

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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

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**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

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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

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**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

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**Identification and Realization of Threshold Functions**

Procedure: 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

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**Identification Example**

Example: Given f = x1x2x3’x4 + x2x3’x4’ Reduce to f = x1x2x3’ + x2x3’x4’, which is unate g = x1x2x3 + x2x3x4 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

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**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

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Synthesis Example Example: For f(x1,x2,x3,x4) = (2,3,6,7,10,12,14,15), find a minimal threshold-logic realization

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**Another Synthesis Example**

Example: For f(x1,x2,x3,x4) = (3,5,7,10,12,14,15), find a minimal threshold-logic realization

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**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 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 Since n1 = n4x5 + n5x5 is threshold: synthesize n4 = x1x2x3 and n5 = x1’x4, which are both threshold

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**General Synthesis Procedure**

Start with a multi-output algebraically-factored switching network G 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 Terminate procedure when all the nodes in G are mapped to threshold nodes

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**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

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MOBILE Example Full-adder:

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**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

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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

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**General Synthesis Procedure**

Start with a multi-output algebraically-factored switching network G 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

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K-map based Procedure Given the map of a node function n with at most three inputs: Find a realizable pattern f1 Find a second realizable pattern f2 based on f1 and n 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 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: and For finding f2: if a minterm (maxterm) of n is not a minterm (maxterm) of f1, add it to ( ) For finding f3: if a minterm (maxterm) of n is not a minterm (maxterm) of both f1 and f2: add it to ( ) On failure to find f3, backtrack to find new f2

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Synthesis Example Example: Consider f = x1’x2’x3’ + x1’x2x3 + x1x2x3’ + x1x2’x3

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