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State Assignment. The problem: Assign a unique code to each state to produce a minimal binary logic level implementation. Given: |S| states, at least  log |S|  state bits ( minimum width encoding ), at most |S| state bits ( one-hot encoding )

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State Assignment

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State assignment

State Assignment

  • The problem:

    Assign a unique code to each state to produce a minimal binary logic level implementation.

    Given:

    • |S| states,

    • at least log |S| state bits (minimum width encoding),

    • at most |S| state bits (one-hot encoding)

      estimate impact of encoding on the size and delay of the optimized implementation.

      Most known techniques are directed towards reducing the size. Difficult to estimate delay before optimization.

      There are possible state assignments for 2n codes (n   log |S|  ),

      since there are that many ways to select |S| distinct state codes and |S|! ways to permute them among the states.


State assignment1

State Assignment

  • Techniques are in two categories:

    Heuristic techniques try to capture some aspect of the process of optimizing the logic, e.g. common cube extraction.

    Example:

    • Construct a weighted graph of the states. Weights express the gain in keeping the codes of the pair of states ‘close’.

    • Assign codes that minimize a proximity function (graph embedding step)

      Exact techniques model precisely the process of optimizing the resulting logic as an encoding problem.

      Example:

    • Perform an appropriate multi-valued logic minimization(optimization step)

    • Extract a set of encoding constraints, that put conditions on the codes

    • Assign codes that satisfy the encoding constraints


Sa as an encoding problem

SA as an Encoding Problem

  • State assignment is a difficult problem: transform ‘optimally’

  • a cover of multi-valued (symbolic) logic functions into

  • an equivalent cover of two-valued logic functions.

  • Various applications dictate various definitions of optimality.

  • Encoding problems are classified as

  • input (symbolic variables appear as input)

  • output(symbolic variables appear as output)

  • input-output(symbolic variables appear as both input and output)

  • State assignment is the most difficult case of input-output encoding where the state variable is both an input and output.


Encoding problems status

Encoding Problems: status

  • two-level

  • well understood theory

  • efficient algorithms

  • many applications

  • two-level

  • well understood theory

  • no efficient algorithm

  • few applications

  • two-level

  • well understood theory

  • no efficient algorithm

  • few applications

  • Input Encoding

  • Output Encoding

  • Input- Output Encoding

  • multi-level

  • basic theory developed

  • algorithms not yet mature

  • few applications

  • multi-level

  • no theory

  • heuristic algorithms

  • few applications

  • multi-level

  • no theory

  • heuristic algorithms

  • few applications


Input encoding for two level implementations

Input Encoding for Two-Level Implementations

  • Reference:[De Micheli, Brayton, Sangiovanni 1985]

  • To solve the input encoding problem for minimum product term count:

    • represent the function as a multi-valued function

    • apply multi-valued minimization (Espresso-MV)

    • extract, from the minimized multi-valued cover, input encoding constraints

    • obtain codes (of minimum length) that satisfy all input encoding constraints


Input encoding

Input Encoding

  • Example: single primary output of an FSM

  • 3 inputs: state variable, s, and two binary variables,c1and c2

  • 4 states: s0,s1, s2, s3

  • 1 output: y

  • y is 1 when following conditions hold:

  • (state = s0 )and c2 = 1

  • (state = s0 ) or (state = s2 ) and c1 = 0

  • (state = s1 )and c2 = 0 and c1 = 1

  • (state = s3 ) or (state = s2 ) and c1 = 1

  • Pictorially

  • for output

  • Y=1:

Values along the state axis are not ordered.


Input encoding1

Input Encoding

  • Function representation

    • Symbolic variables represented by multiple valued (MV) variables Xi restricted to Pi = { 0,1,…,ni-1 }.

    • Output is a binary valued function f of variables Xi

      f : P1 P2  …  Pn B

    • S  Pi and literalX Sis as:

      XiS = {1 if XiSi{0 otherwise

  • Example: y = X {0}c2+ X {0,2}c1’+ X {1}c1c2’+ X {2,3}c1 (only 1 MV variable)

  • The cover of the function can also be written in positional form:-1 1000 1

  • 0- 1010 1

  • 10 0100 1

  • 1- 0011 1


Input encoding2

Input Encoding

  • y = X {0}c2+ X {0,2}c1’+ X {1}c1c2’+ X {2,3}c1

  • Minimize using classical two-level MV-minimization (Espresso-MV).

  • Minimized result:

  • y = X {0,2,3} c1c2+ X {0,2}c1’+ X {1,2,3}c1c2’


Input encoding3

Input Encoding

  • Extraction of face constraints:

  • y = X {0,2,3} c1c2+ X {0,2}c1’+ X {1,2,3}c1c2’

  • The face constraints are: (assign codes to keep MV-literals in single cubes)

  • (s0,s2, s3)

  • (s0,s2)

  • (s1, s2, s3)

    • two-level minimization results in the fewest product terms for any possible encoding

    • would like to select an encoding that results in the same number of product terms

    • each MV literal should be embedded in a face (subspace or cube) in the encoded space of the associated MV-variable.

    • unused codes may be used as don’t cares


Input encoding4

Input Encoding

  • Satisfaction of face constraints:

    • lower bound can always be achieved (one-hot encoding will do it),

      • but this results in many bits (large code width)

    • for a fixed code width need to find embedding that will result in the fewest cubes

  • Example: A satisfying solution is:

  • s0 = 001, s1 = 111, s2 = 101, s3 = 100

  • The face constraints are assigned to the following faces:

  • (s0,s2, s3)202

  • (s0,s2)201

  • (s1, s2, s3)122

  • Encoded representation:

  • y = X {0,2,3} c1c2+ X {0,2}c1’+ X {1,2,3}c1c2’

  • -> x2’c1c2+ x2’x3c1’+ x1c1c2’

  • Note: needed 3 bits instead of the minimum 2.

dc

1

0

2

x3

dc

dc

x2

dc

3

x1


Procedure for input encoding

Procedure for Input Encoding

  • Summary:

    A face constraint is satisfied by an encoding if the supercube of the codes of the symbolic values in the literal does not contain any code of a symbolic value not in the literal.

    Satisfying all the face constraints guarantees that the encoded minimized binary-valued cover will have size no greater than the multi-valued cover.

    Once an encoding satisfying all face constraints has been found, a binary-valued encoded cover can be constructed directly from the multi-valued cover, by

    • replacing each multi-valued literal by the supercube corresponding to that literal.


Satisfaction of input constraints

Satisfaction of Input Constraints

  • Any set of face constraints can be satisfied by one-hot encoding the symbols. But code length is too long …

  • Finding a minimum length code that satisfies a given set of constraints is NP-hard

  • [Saldanha, Villa, Brayton, Sangiovanni 1991].

  • Many approaches proposed. The above reference describes the best algorithm currently known, based on the concept of encoding dichotomies.

  • [Tracey 1965, Ciesielski 1989].


Satisfying input constraints

Satisfying Input Constraints

  • Ordered Dichotomy:

    • A ordereddichotomy is a 2-block partition of a subset of symbols. Example:(s0 s1;s2 s3)

      • Left block is s0,s1 - associated with 0 bit

      • Right block is s2 s3 - associated with 1 bit

    • A dichotomy is complete if each symbol of the entire symbol set appears in one block

      • (s0;s1 s2) is not complete but (s0 s1;s2 s3) is complete

    • Two dichotomies d1 and d2 are compatible if

      • the left block of d1 is disjoint from the right block of d2, and

      • the right block of d1 is disjoint from the left block of d2.


Satisfying input constraints1

Satisfying Input Constraints

  • The union of two compatible dichotomies, d1 and d2, is (a;b) where

    • a is the union of the left blocks of d1 and d2and

    • b is the union of the right blocks of d1 and d2.

  • The set of all compatibles of a set of dichotomies is obtained by

    • a initial dichotomy is a compatible

    • the union of two compatibles is a compatible

      Example: initial (0 1;2) (1:4) (3:2) (2;4)

      generated: (0 1 ;2 4) (0 1 3;2) (1 3;2 4) (1 2;4)

      (0 1 3;2 4)


Dichotomies

Dichotomies

  • Dichotomy d1 covers d2 if

    • the left block of d1 contains the left block of d2, and

    • the right block of d1 contains the right block of d2

      (s0;s1 s2) is covered by (s0 s3;s1 s2 s4)

  • A dichotomy is a prime compatible of a given set if it is not covered by any other compatible of the set.

  • A set of complete dichotomies generates an encoding: Each dichotomy generates a bit of the encoding

    • The left block symbols are 0 in that bit

    • The right block symbols are 1 in that bit

      Example:(s0 s1;s2 s3) and (s0 s3;s1 s2) ==>

      s0=00, s1=01, s2=11, s3=10


Input constraint satisfaction

Input constraint satisfaction

  • Initial dichotomies:

    • Each face constraint with k symbols generates 2(n-k) initial dichotomies (symbol set (s0 s1 s2 s3 s4 s5))

      • (s1 s2 s4) ==> (s1 s2 s4;s0) (s1 s2 s4;s3) (s1 s2 s4;s5)

        (s0;s1 s2 s4) (s3;s1 s2 s4) (s5;s1 s2 s4)

    • Also require that each symbol is encoded uniquely

      (s0;s1) (s0;s2) … (s0;s5)

      (s1;s0) (s1;s2) … (s1;s5)

      (s5;s0) (s5;s1) … (s5;s4)

      n2-n such additional initial dichotomies

    • All initial dichotomies, { d }, must be satisfied by encoding - a dichotomy, d, is satisfied by a set of encoding dichotomies if there exists a bit where the symbols in left block of d are 0 and the symbols in right block of d are 1.

      Note: can delete any initial dichotomy d that is covered by another e, since since satisfaction of e implies satisfaction of d.


Input constraint satisfaction1

Input constraint satisfaction

  • Procedure:

  • Given set of face constraints, generate initial dichotomies.

  • Generate set of prime dichotomies.

  • Form covering matrix

    • Rows = initial dichotomies

    • Columns = prime dichotomies

    • A 1 is in (i, j) if prime j covers dichotomy i.

  • Solve unate covering problem


Efficient generation of prime dichotomies

Efficient generation of prime dichotomies

  • Introduce symbol for each initial dichotomy

  • Form a 2-clause for each dichotomy a that is incompatible with dichotomy b i.e. (a+b)

  • Multiply product of all 2-clauses to SOP.

  • For each term in SOP, list dichotomies not in term - these are the set of all primes dichotomies.

  • Example: initial dichotomies (a b c d e)

  • ab,ac,bc,cd,de are incompatibles

  • CNF is (a+b)(a+c)(b+c)(c+d)(d+e) = abd+acd+ace+bcd+bce

  • Primes are cUe, bUe, bUd, aUe, aUd


Efficient generation of prime dichotomies1

Efficient generation of prime dichotomies

  • Procedure for multiplying 2-CNF expression E

  • Procedurecs(E)

  • x = splitting variable

  • C = sum terms with x

  • R = sum terms without x

  • p = product of all variables in C except x

  • q = cs(R)

  • result = single_cube_containment(x q + p q)

  • Note: procedure is linear in number of primes


Applications of input encoding

Applications of Input Encoding

  • References:

    • PLA decomposition with multi-input decoders [Sasao 1984, K.C.Chen and Muroga 1988, Yang and Ciesielski 1989]

    • Boolean decomposition in multi-level logic optimization [Devadas, Wang, Newton, Sangiovanni 1989]

    • Communication based logic partitioning [Beardslee 1992, Murgai 1993]

    • Approximation to state assignment [De Micheli, Brayton, Sangiovanni 1985, Villa and Sangiovanni 1989]


Input encoding pla decomposition

y1 y2

x4 x5 x6

x4 x5 x6 x7

Input Encoding: PLA Decomposition

  • Consider the PLA:

  • y1 = x1 x3 x4’x6+ x2 x5 x6’x7 + x1 x4 x7’+ x1’x3’x4 x7’+

  • x3’x5’x6’x7’

  • y2 = x2’x4’x6+ x3 x4’x6 + x2’x5 x6’x7 + x1’x3’x4 x7’ +x1’x3’x5’x6’x7’ + x2’x4’x5’x7’

  • Decompose it into a driving PLA fed by inputs {X4, X5, X6, X7 } and a driven PLA fed by the outputs of the driving PLA and the remaining inputs {X1, X2, X3 } so as to minimize the area.


Input encoding pla decomposition1

Input Encoding: PLA Decomposition

  • y1 = x1 x3 x4’x6+ x2 x5 x6’x7 + x1 x4 x7’+ x1’x3’x4 x7’+

  • x3’x5’x6’x7’

  • y2 = x2’x4’x6+ x3 x4’x6 + x2’x5 x6’x7 + x1’x3’x4 x7’ +x1’x3’x5’x6’x7’ + x2’x4’x5’x7’

  • Projecting onto the selected inputs there are 5 distinct product terms:

  • x4’x6 , x5 x6’x7 , x4 x7’, x5’x6’x7’ ,x4’x5’x7’

  • Product term pairs (x4 x7’,x5’x6’x7’ ) and ( x5’x6’x7’,x4’x5’x7’)are not disjoint.


Pla decomposition

PLA Decomposition

  • y1 = x1 x3 x4’x6+ x2 x5 x6’x7 + x1 x4 x7’+ x1’x3’x4 x7’+ x3’x5’x6’x7’

  • y2 = x2’x4’x6+ x3 x4’x6 + x2’x5 x6’x7 + x1’x3’x4 x7’ +x1’x3’x5’x6’x7’ + x2’x4’x5’x7’

  • To re-encode make disjoint all product terms involving selected inputs:

  • y1 = x1 x3 x4’x6+ x2 x5 x6’x7 + x1 x4 x7’+ x1’x3’x4 x7’+ x3’x4’x5’x6’x7’ (using reduce)

  • y2 = x2’x4’x6+ x3 x4’x6 + x2’x5 x6’x7 + x1’x3’x4 x7’ +x1’x3’x4’x5’x6’x7’ + x2’x4’x5’x6’x7’ (using reduce)

  • Projecting on the selected inputs there are 4 disjoint product terms:

  • x4’x6 , x5 x6’x7 , x4 x7’, x4’x5’x6’x7’

  • View theseas 4 values 1,2,3,4 of an MV variable S:

  • y1 = x1 x3 S {1} + x2 S {2}+ x1 S {3}+ x1’x3’S {3}+ x3’S {4}

  • y2 = x2’S {1} + x3 S {1}+ x2’S {2}+ x1’x3’S {3} + x1’x3’S {4} + x2’S {4}


Input encoding pla decomposition2

Input Encoding: PLA Decomposition

  • y1 = x1 x3 S{1} + x2 S{2}+ x1 S{3}+ x1’x3’S{3}+ x3’S{4}

  • y2 = x2’S{1} + x3 S{1}+ x2’S{2}+ x1’x3’S{3} + x1’x3’S{4} + x2’S{4}

  • Performing MV two-level minimization:

  • y1 = x1 x3 S {1,3} + x2 S {2}+ x3’S {3,4}

  • y2 = x2’S {1,2,4} + x3 S {1}+ x1’x3’S {3,4}

  • Face constraints are:

  • (1,3 ), (3,4 ), (1,2,4 )

  • Codes of minimum length are:

  • enc(1 ) = 001, enc(2 ) = 011, enc(3) = 100, enc(4) = 111.

2

4

x10

1

x9

x8

3


Input encoding pla decomposition3

y1 y2

x8, x9, x10

x4 x5 x6

x4 x5 x6 x7

Input Encoding: PLA Decomposition

  • The driven PLA becomes:

  • y1 = x1 x3 x9’+ x2 x8’x9 x10 + x3’ x8

  • y2 = x2’x10+ x3 x8’x9’x10 + x1’x3’ x8

  • The driving PLA becomes the function:

  • f: {X4, X5, X6, X7} {X8, X9, X10}

  • f(x4’ x6 ) = enc( 1 ) = 001f(x5 x6’ x7 ) = enc( 2 ) = 011

  • f(x4 x7’ ) = enc( 3 ) = 100f(x4’ x5’x6’x7’ ) = enc( 4 ) = 111

  • Represented in SOP as:

  • x8 = x4 x7’+ x4’ x5’ x6’ x7’

  • x9 = x5 x6’ x7 + x4’ x5’ x6’ x7’

  • x10 = x4’ x6 + x5 x6’ x7 + x4’ x5’ x6’ x7’

  • Result: Area before decomposition: 160.

  • Area after decomposition: 128.


Output encoding for two level implementations

Output Encoding for Two-Level Implementations

  • The problem:

    • find binary codes for symbolic outputs in a logic function so as to minimize a two-level implementation of the function.

  • Terminology:

    • Assume that we have a symbolic cover S with a symbolic output assuming n values. The different values are denoted v0, … , vn-1.

    • The encoding of a symbolic value vi is denoted enc(vi).

    • The onset of vi is denoted ONi. Each ONi is a set of Di minterms {mi1, …, miDi}.

    • Each minterm mij has a tag indicating which symbolic value’s onset it belongs to. A minterm can only belong to one symbolic value’s onset.


Output encoding

Output Encoding

  • Consider a function f with symbolic outputs and two different encoded realizations of f :

  • 0001 out1 0001 0010001 10000

  • 00-0 out2 00-0 01000-0 01000

  • 0011 out2 0011 0100011 01000

  • 0100 out3 0100 0110100 00100

  • 1000 out3 1000 0111000 00100

  • 1011 out4 1011 1001011 00010

  • 1111 out5 1111 1011111 00001

    • Two-level logic minimization exploits the sharing between the different outputs to produce a minimum cover.

  • In the second realization no sharing is possible. The first can be minimized to:

1111 001

1-11 100

0100 011

0001 101

1000 011

00-- 010


Dominance constraints

Dominance Constraints

  • Definition: enc(vi) > enc(vj) iff the code of vi bit-wise dominates the code of vj, i.e. for each bit position where vj has a 1, vi also has a 1. Example: 101 > 001

  • If enc(vi ) > enc(vj ) then ONi can be used as a DC set when minimizing ONj.

  • Example:

  • symbolic coverencoded coverminimized encoded cover

  • 0001 out10001 110 0001 110

  • 00-0 out200-0 010 00-- 010

  • 0011 out20011 010

  • Here enc(out1)= 110 > enc(out2)= 010. The input minterm 0001 of out1 has been included into the single cube 00-- that asserts the code of out2.

  • Algorithms to exploit dominance constraints implemented in Cappucino (De Michelli, 1986) and Nova (Villa and Sangiovanni, 1990).


Disjunctive constraints

Disjunctive Constraints

  • If enc(vi ) = enc(vj) + enc(vk), ONi can be minimized using ONj and ONk as don’t cares.

  • Example:

  • symbolic coverencoded coverminimized encoded cover

  • 101 out1 1101 11 1 10- 01 1

  • 100 out2 1100 01 1 1-1 10 1

  • 111 out3 1111 10 1

  • Here enc(out1)= enc(out2) + enc(out3). The input minterm 101 of out1 has been merged with the input minterm 100 of out2 (resulting in 10-) and with the input minterm 111 of out3 (resulting in 1-1). Input minterm 101 now asserts 11 (i.e. the code of out1), by activating both cube 10- that asserts 01 and cube 1-1 that asserts 10.

  • Algorithm to exploit dominance and disjunctive constraints implemented in esp_sa (Villa, Saldanha, Brayton and Sangiovanni, 1995).


Exact output encoding

Exact Output Encoding

  • Proposed by Devadas and Newton, 1991.

  • The algorithm consists of the following:

    • Generate generalized prime implicants (GPIs) from the original symbolic cover.

    • Solve a constrained covering problem,

      • requires the selection of a minimum number of GPIs that form an encodeable cover.

    • Obtain codes (of minimum length) that satisfy the encoding constraints.

    • Given the codes of the symbolic outputs and the selected GPIs, construct trivially a PLA with product term cardinality equal to the number of GPIs.


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