Download presentation

Presentation is loading. Please wait.

Published byMakena Pickron Modified over 2 years ago

1
FUNCTION OPTIMIZATION Switching Function Representations can be Classified in Terms of Levels Number of Levels, k, is Number of Unique Boolean (binary) Operators EXAMPLES 1-Level 2-Level multi-level

2
k-Level functions 2 2 n Possible Switching Functions of n Variables (actually much fewer types obtained by permuting and/or complementing input variables) 1-Level Forms –Cannot Represent all Possible Functions 2-Level Forms –Can Represent all Possible Functions –With Additional Restrictions – CANONICAL k -Level Forms ( k 2 ) –Many Different Ways to Represent a Given Functions If a multi-input Gate Represents a (binary or greater) Boolean Operator –Expression can Represent a Netlist –k Indicates “depth” of Netlist EXAMPLES 1-Level 2-Level Multi-Level a c b c a c b c b a b e d a f b f f

3
k-Level function properties 1-Level Forms –Due to Restriction of Representable Functions, Not as Useful 2-Level Forms –Can be Canonical –Longest PI to PO Path is Always 2 Related to Delay –Minimization Usually Means 1.Minimum Number of “Terms” 2.Minimum Number of “Literals” in Expression k -Level Forms ( k 2 ) –Many More Possible Representations –Can be Optimized for Area –Delay is More Complicated False Path Problem Spatio-temporal Correlations Typically an Area versus Delay Tradeoff (e.g., Ripple versus CL Adder)

4
Common Terms for 2-Level Minimization Literal – A variable in complemented or uncomplemented form Product – The disjunction (AND) of a set of literals; also represents a cube Support Set – Set of all variables that define the domain of a switching function Minterm – A disjunction (AND) containing an instance of each literal corresponding to a variable in the support set that is in the on-set, f on,of a function Don’t Care – The absence of a supporting variable in a product term Implicant – A product term that covers one or more minterms in the on- set, f on, of a function Prime Implicant – An implicant in the on-set, f on, of a function such that it is not a subproduct of any other possible implicant in the set. Essential Prime Implicant – A prime implicant that covers at least one minterm NOT covered by any other implicant in the on-set, f on.

5
SOP Minimization – Terms Example Consider: Each Element of f on is an Implicant Prime Implicants are: abc is a Prime Implicant but NOT an Essential Prime Implicant bcd is an Implicant but NOT a Prime Implicant bd is an Essential Prime Implicant Product abc : – covers minterms: m 4 = abcd and m 5 = abcd – disjunction of literals a, b, c – variable d is a Don’t Care f Represented by f on has Support Set: {a, b, c, d}

6
Karnaugh Map Simplification Using Karnaugh Map a)Use as “few loops” as possible b)Use as “large loop” as possible Prime Implicant Corresponds to as “large loop” as Possible Karnaugh Map Simplification is Finding f on such that: a) f on contains as few prime implicants as possible b) f on must contain all essential prime implicants c) f on contains NO implicants that are NOT prime

7
Tabulation Method

8
Tabulation Method - Notation Consider Support Set of f : S={x 1, x 2, …, x n } x i ci denotes: x i if c i = ‘1’ x i if c i = ‘0’ 1 if c i = ‘-’ If NO c i = -, then we have a minterm –Can be Represented by Decimal Equivalent of c i EXAMPLE( S={x 1, x 2, x 3, x 4 } ) x 1 1 x 0 2 x 0 3 x 1 4 = m 9, a minterm c 1 c 2 c 3 c 4 = 1001 = 9 x 0 1 x - 2 x - 3 x 0 4 = a 4-cube 0--0

9
Cube Merging Basic Operation in Tabulation Method 2 Cubes that Differ in a SINGLE c i can be Merged into a Single Cube EXAMPLE = 1-01 = 0-01 Merge and into = --01 Merging is also called star operator and is a special case of consensus

10
Tabulation Method Input: f on as a set of minterms Output: f on as a set of 1.All Essential Prime Implicants 2.As Few Prime Implicants as Possible Finding as few Prime Implicants as Possible is an NP -Hard Problem!!!!! Reduces to the “Set Covering” Problem for Unate Functions Unate function – a constant or is represented by a SOP using either uncomplemented or complemented literals for each variable Reduces to the “Minimum Cost Assignment” Problem for Binate Functions (ex. EXOR) This is 2-Level (SOP) Optimization (Minimization)

11
Tabulation Method STEP 1: –Convert Minterm List (specifying f on ) to Prime Implicant List STEP 2: –Choose All Essential Prime Implicants –If all minterms are covered HALT Else GO To STEP 3 STEP 3: –Formulate the Reduced Cover Table Omitting the rows/cols of EPI –If Cover Table can be Reduced using Dominance Properties, Go To Step 2 –Else Must Solve the “Cyclic Cover” Problem 1) Use Exact Method (exponentially complex) 2) Use Heuristic Method (possibly non-optimal result) NOTE: “Quine-McCluskey” Refers to Using a “Branch and Bound” Heuristic NOTE: “Petrick’s Method” is Exact Technique – Generates all Solutions Allowing the Best to be Used

12
STEP 1 Tabulation Method – STEP 1 1.Partition Prime Implicants (or minterms) According to Number of 1’s 2.Check Adjacent Classes for Cube Merging Building a New List 3.If Entry in New List Covers Entry in Current List – Disregard Current List Entry 4.If Current List = New List HALT Else Current List New List New List NULL Go To Step 1

13
STEP 1 - EXAMPLE f on = {m 0, m 1, m 2, m 3, m 5, m 8, m 10, m 11, m 13, m 15 } = (0, 1, 2, 3, 5, 8, 10, 11, 13, 15)

14
STEP 1 - EXAMPLE f on = {m 0, m 1, m 2, m 3, m 5, m 8, m 10, m 11, m 13, m 15 } = (0, 1, 2, 3, 5, 8, 10, 11, 13, 15)

15
STEP 1 - EXAMPLE f on = {m 0, m 1, m 2, m 3, m 5, m 8, m 10, m 11, m 13, m 15 } = (0, 1, 2, 3, 5, 8, 10, 11, 13, 15)

16
STEP 1 - EXAMPLE f on = {m 0, m 1, m 2, m 3, m 5, m 8, m 10, m 11, m 13, m 15 } = (0, 1, 2, 3, 5, 8, 10, 11, 13, 15) f on = {A,B,C,D,E,F,G} = {00--, -01-, -0-0, 0-01, -101, 1-11, 11-1} Question: Can this be done on a CCM? How modified?

17
STEP 2 – Construct Cover Table PIs Along Vertical Axis (in order of # of literals) Minterms Along Horizontal Axis NOTE: Table 4.2 in book is incomplete

18
STEP 2 – Finding the Minimum Cover Extract All Essential Prime Implicants, EPI EPIs are the PI for which a Single x Appears in a Column C is an EPI so: f on ={C,...} Row C and Columns 0, 2, 8, and 10 can be Eliminated Giving Reduced Cover Table Examine Reduced Table for New EPIs

19
STEP 2 – Reduced Table Essential row Distinguished Column The Row of an EPI is an Essential row The Column of the Single x in the Essential Row is a Distinguished Column

20
Row and Column Dominance If Row P has x ’s Everywhere Row Q Does Then Q Dominates P if P has fewer x ’s If Column i has x ’s Everywhere j Does Then j Dominates i if i has fewer x ’s If Row P is equal to Row Q and Row Q does not cost more than Row P, eliminate Row P, or if Row P is dominated by Row Q and Row Q Does not cost more than Row P, eliminate Row P If Column i is equal to Column j, eliminate Column i or if Column i dominates Column j, eliminate Column i

21
STEP 3 – The Reduced Cover Table Initially, Columns 0, 2, 8 and 10 Removed No EPIs are Present No Row Dominance Exists No Column Dominance Exists This is Cyclic Cover Table Must Solve Exactly OR Use a Heuristic

22
The Cyclic Cover Table

23
For now, we Arbitrarily Choose a PI Later we will Study Exact and Heuristic Methods Arbitrarily Choose F so: f on ={C, F,...} This Choice May Lead to a Non-Optimal Result!!!! Form Reduced Cover and Go To Step 2

24
STEP 3 – Dominance Initially, Reduced Table has Columns 11 and 15 Removed G is Dominated by E B is Dominated by A Form Reduced Cover Table and Go To Step 2

25
STEP 2 – The Reduced Cover Initially, Table has Rows G and B Removed Secondary EPIs – A and E All Columns Covered Eliminate D f on ={C, F, A, E}

26
Result Check Initial Minterm List f on = {m 0, m 1, m 2, m 3, m 5, m 8, m 10, m 11, m 13, m 15 } = (0, 1, 2, 3, 5, 8, 10, 11, 13, 15) Final Result f on ={A, C, E, F} 1 1 1 1 00011110 00 01 11 10 ab cd 1 1 1 1 1 1 1 1 1 1 00011110 00 01 11 10 ab cd 11 11 1 1

27
Questions 1.Explain Quine-McCluskey Method. 2.How to program QM Method in CCM? 3.Methods to solve Covering Problem. 4.Possible Applications of Covering. 5.Reduction of Covering to Petrick Function. 6.Merging Operator.

Similar presentations

Presentation is loading. Please wait....

OK

Lecture # 5 University of Tehran

Lecture # 5 University of Tehran

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on power system stability studies Ppt on properties of different quadrilaterals Ppt on forest in india Ppt on communication skills in hindi Ppt on deforestation in india Ppt on west central railway Ppt on electric meter testing Mp ppt online Download ppt on market around us Ppt on central limit theorem video