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Petrick.m
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Petrick.m
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function [ minimalBooleanExpression, minimalMatrixExpression ] = Petrick(binaryTruthTableVector)
%lgcExpr return the simplist expression of a logic function
% N: number of elements (<=10)
% m: list of minterms
% d: list of don't care terms
N = log2(length(binaryTruthTableVector));
jIndex = 1;
m = [];
for iIndex = 1:length(binaryTruthTableVector)
if(binaryTruthTableVector(iIndex) == 1)
m(jIndex) = iIndex - 1;
jIndex = jIndex + 1;
end
end
% Double Checked!
d = []; %comment this
if (max(max(m), max(d)) >= 2^N)
disp('Error: N is too small!');
return;
end
% Double Checked!
m = unique(sort(m));
d = unique(sort(d));
mLen = length(m);
dLen = length(d);
if mLen + dLen == 2^N,
minimalBooleanExpression = '1';
minimalMatrixExpression = ones(1, 2*N); % This line!!!
return;
end
%d Double Checked!
if mLen + dLen == 0
minimalBooleanExpression = '0';
minimalMatrixExpression = zeros(1, 2*N);
return;
end
% all k * N matrix except cmb_flg
binstr = [];
cmb_flg = zeros(mLen + dLen, 1);
nextbinstr = [];
final = [];
% initializing
bisntr1 = dec2bin(m, N);
binstr2 = dec2bin(d, N);
binstr = [bisntr1; binstr2];
% combining
while 1,
countnew = 0;
Len = size(binstr, 1);
for p = 1:(Len - 1),
for q = (p + 1):Len,
notEqual = (binstr(p,:) ~= binstr(q,:));
if sum(notEqual) == 1,
countnew = countnew + 1;
cmb_flg(p) = 1;
cmb_flg(q) = 1;
tmp = binstr(p,:);
tmp(notEqual) = '-';
nextbinstr = [nextbinstr; tmp]; % may get repeated binstr
end
end
end
for k = 1:Len,
if cmb_flg(k) == 0,
final = [final; binstr(k,:)];
end
end
if countnew == 0,
break;
end
cmb_flg = zeros(countnew, 1);
binstr = nextbinstr;
binstr = unique(binstr, 'rows'); % eliminate repeated rows in time
nextbinstr = [];
end
final = unique(final, 'rows');
% Petrick's Method
% Forming the PI table----------------------
rslt = [];
rw = size(final, 1);
cl = mLen;
ptk = zeros(rw, cl);
for p = 1:cl,
for q = 1:rw,
vec1 = dec2bin(m(p), N);
vec2 = final(q, :);
neq = (vec1 ~= vec2);
tmp = unique(vec2(neq));
if ((length(tmp) == 1) && (tmp(1) == '-')) || (isempty(tmp)),
ptk(q, p) = 1;
end
end
end
% Check the output first
%dummy = 'Stop point';
ptk = ptk';
ptkCell = cell(size(ptk, 1)+1, size(ptk, 2)+1);
for i=1:size(ptk, 1)
for j=1:size(ptk, 2)
ptkCell{i+1, j+1} = ptk(i,j);
end
end
ptkCell{1, 1} = 'Minterms - PIs';
% Writing the minterms in all rows of the Prime Implicant Table
for i=2:size(ptkCell, 1)
ptkCell{i,1} = m(i-1);
end
% Writing the PIs in all columns of the Prime Implicant Table
for j=2:size(ptkCell, 2)
ptkCell{1, j} = final(j-1, :);
end
PIs = [];
for i=1:size(final, 1)
PIs = [PIs; final(i, :)];
end
%--------------------------------------------------------------------------
% Now the Prime Implicant Table is ready
% We must perform a 3 Step operation
% 1- First, we remove EPI columns and their covered minterm rows from the table
% 2- Second, we remove the row dominating Rows from the table
% 3- Third, we remove the column dominated Columns from the table
% we perform step 1 to 3, until the table is either empty or there are rows
% and columns suitable for removing
% All the removed PIs are stored in the result[]
% 1st Step: Reducing the table---------------------------------------------
result = [];
ptkOld = [];
tempFinal = final;
while ( ~isequal(ptk, ptkOld))
ptkOld = ptk;
% finding and storing the EPIs into result
listOfEPIIndexes = [];
for i=1:size(ptk, 1)
if (sum(ptk(i, :)) == 1) % if true, then the PI covering this minterm is EPI
EPIIndex = find(ptk(i, :));
result = [result; PIs(EPIIndex, :)];
listOfEPIIndexes = [listOfEPIIndexes, EPIIndex];
end
end
listOfEPIIndexes = sort(unique(listOfEPIIndexes));
% removing the EPIs from ptk if it is not empty
if (~isempty(listOfEPIIndexes))
for j=1:size(listOfEPIIndexes, 2)
coveredMintermsIndexes = find(ptk(:, listOfEPIIndexes(j)));
for i=1:size(coveredMintermsIndexes, 1)
ptk(coveredMintermsIndexes(i), :) = [];
coveredMintermsIndexes = [coveredMintermsIndexes(1:i); coveredMintermsIndexes(i+1:end)-1];
end
ptk(:, listOfEPIIndexes(j)) = [];
% Removing its corresponding entry from PI Table
if ( ~isempty(PIs) )
PIs(listOfEPIIndexes(j), :) = [];
end
listOfEPIIndexes = [listOfEPIIndexes(1:j), listOfEPIIndexes(j+1:end)-1];
end
end
if (isempty(ptk))
break;
end
%--------------------------------------------------------------------------
% If ptk is not empty, proceed
% 2nd Step: Row Dominance--------------------------------------------------
% Finding dominating rows
listOfDominatingRows = [];
for i=1:size(ptk, 1)
for j=i+1:size(ptk, 1)
if (dominatingRowIdentifier(ptk(i, :), ptk(j, :)) == 0 )
continue;
elseif (dominatingRowIdentifier(ptk(i, :), ptk(j, :)) == 1 )
listOfDominatingRows = [listOfDominatingRows, i];
elseif (dominatingRowIdentifier(ptk(i, :), ptk(j, :)) == 2 )
listOfDominatingRows = [listOfDominatingRows, j];
end
end
end
listOfDominatingRows = sort(unique(listOfDominatingRows));
tempListOfDominatingRows = listOfDominatingRows;
% Removing dominating rows if its list is not empty
if (~isempty(listOfDominatingRows))
for i=1:size(listOfDominatingRows, 2)
ptk(tempListOfDominatingRows(i), :) = [];
tempListOfDominatingRows = [tempListOfDominatingRows(1:i), tempListOfDominatingRows(i+1:end)-1];
end
end
%--------------------------------------------------------------------------
% 3rd Step: Column Dominance-----------------------------------------------
% Finding dominated columns
listOfDominatedColumns = [];
for i=1:size(ptk, 2)
for j=i+1:size(ptk, 2)
if ( dominatedColumnIdentifier(ptk(:, i), ptk(:, j)) == 0 )
continue;
elseif ( dominatedColumnIdentifier(ptk(:, i), ptk(:, j)) == 1 )
listOfDominatedColumns = [listOfDominatedColumns, i];
elseif ( dominatedColumnIdentifier(ptk(:, i), ptk(:, j)) == 2 )
listOfDominatedColumns = [listOfDominatedColumns, j];
end
end
end
listOfDominatedColumns = sort(unique(listOfDominatedColumns));
tempListOfDominatedColumns = listOfDominatedColumns;
% Removing dominated columns if its list is not empty
if(~isempty(listOfDominatedColumns))
for j=1:size(listOfDominatedColumns, 2)
ptk(:, tempListOfDominatedColumns(j)) = [];
% Removing its corresponding entry from the PI Table
if ( ~isempty(PIs) )
PIs(tempListOfDominatedColumns(j), :) = [];
end
tempListOfDominatedColumns = [tempListOfDominatedColumns(1:j), tempListOfDominatedColumns(j+1:end)-1];
end
end
end
%--------------------------------------------------------------------------
% 4- Solving the table
% Petrick's Method for minimization
% We need PIs array and ptk matrix to solve the table
% first we mark each PI as Pi
PIsCell = cell(size(PIs,1), 2);
for i=1:size(PIs, 1)
PIsCell{i, 1} = PIs(i, :);
PIsCell{i, 2} = ['P', num2str(i)];
end
if(~isempty(ptk))
minimalExpression = formCoveringPIs( ptk(1, :) );
for i=2:size(ptk, 1)
minimalExpression = minimalSolution(minimalExpression, formCoveringPIs(ptk(i, :)) );
end
end
%--------------------------------------------------------------------------
% Finding the minimal Boolean Expression using result and minimalExpression
%INPUTS
%minimalExpression (Matrix)
%PIs (Vertical Array)
%result (Vertical Array)
%OUTPUT
%minimalBooleanEquation
if(~isempty(ptk))
minimalBooleanEquation = findMinimalBooleanEquation(minimalExpression, PIs);
%--------------------------------------------------------------------------
for j=1:size(minimalBooleanEquation, 2)
if( isequal(minimalBooleanEquation(1, j), 1) )
result = [result; PIs(j, :)];
end
end
%--------------------------------------------------------------------------
end
%--------------------------------------------------------------------------
result = unique(result, 'rows');
% Create algebraic boolean equations from the result
% N is the number of vaiables aka maximum number of required alphabets
Symbols = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
minimalBooleanExpression = [];
% minimal Expression in matrix form. This output is used for estimating
% power and area and other electrical parameters
minimalMatrixExpression = zeros(size(result, 1) , 2*N);
for i=1:size(result, 1)
for j=1:size(result, 2) % Or j=1:N
if( isequal(result(i, j), '1') )
minimalBooleanExpression = [minimalBooleanExpression, Symbols(1, j)];
Index = ((j-1)*2) +1;
minimalMatrixExpression(i, Index) = 1;
elseif ( isequal(result(i, j), '0') )
minimalBooleanExpression = [minimalBooleanExpression, Symbols(1, j), ''''];
Index = ((j-1)*2) +2;
minimalMatrixExpression(i, Index) = 1;
end
end
if ( ~isequal(i, size(result, 1)) )
minimalBooleanExpression = [minimalBooleanExpression, '+'];
end
end
%--------------------------------------------------------------------------
%---------------------------------THE END----------------------------------
end