The covering procedure
The covering procedure
The covering procedure
• Remove rows with essential PI’s and any columns with x’s in those rows
The covering procedure
• Remove rows that are covered by other rows• Remove columns that cover other columns• Why?
The covering procedure
• Remove rows that are covered by other rows• Remove columns that cover other columns
The covering procedure
• Rows PI’s– Covering row takes care of more minterms
– Minterms included in a smaller (covered) row are also included in the bigger (covering) one
– Can discard the small ones and use only the covering row; minterm coverage is preserved
• Columns min/max terms– Whenever a min/max term corresponding to a
covered (smaller) column is included by some PI, the min/max term corresponding to the covering (bigger) column also gets included
– Covering column can be dropped
– Reduces # of PI’s that include this min/max term
Cyclic PI charts
• Cyclic PI charts have no essential PI’s– Cannot be reduced by rules 1 and 2
• Example of cyclic PI chart of 3 variables
BC
A 00 01 11 10
0 1 1 1
1 1 1 1
Cyclic PI chart
chosen PI
• Cyclic PI charts have no essential PI’s– Select the row with max number of x’s (randomly if
more than one); PI1 in this example
Cyclic PI chart
After removing PI1,
apply rules 1 and 2.
Remove covered
PI2 and PI6
PI3 and PI5 cover the
resulting chart.
Minimal cover: PI1, PI3, PI5
Cyclic PI charts
• Example of cyclic PI chart of 4 variables
1 1 1
1 1 1
• Q: if PI’s covering 4 minterms are allowed, can one create a cyclic PI chart where no PIs are essential?
Cyclic PI charts
• A: yes
1 1 1
1 1 1
1 1 1
1 1 1
• Q: what about a 4 variable K-map and groups of 8 ones?– In general n variable functions with a K-map and
PIs covering 2n-1 min/max terms – can there be a cyclic PI chart?
Incompletely specified functions
• When some of the minterms can be either 0 or 1, we can denote them by ‘d’ (don’t care)
• When simplifying, we use ‘d’s to generate PIs, but do not include them in the PI chart
Incompletely specified functions
Incompletely specified functions
Multiple simultaneous outputs
Multiple simultaneous outputs
Multiple simultaneous outputs
• In general, # of lists ≤ n+1 (n = # variables)
‘d’s are not in the charts, but are used for PIs
• List 1, group 1
• Group 2, list 2
• Group 3, list 3
Multiple simultaneous outputs
Why select
PI3 over
PI11?
• PIs from higher-numbered list are likely to cover more PIs (not always true: don’t cares)
Multiple simultaneous outputs
• Static hazard or glitch: unwanted output transition when inputs change and the output should have remained the same
• For simplicity consider only a single input changes at a time
• Different gates have different propagation delays
Hazards and K-maps
Hazards and K-maps
t1
t3
t2
t1 = t2 = t3
Hazards and K-maps
t1
t3
t2
t1 > t2 > t3
Hazards and K-maps
• A hazard exists when a changing input requires corresponding minterms/maxterms to be covered by different product/sum terms
• Remove hazards by bridging the gaps on the K-map:
Hazards and K-maps
• Hazard-free circuit:• Cover each pair
of adjacentminterms by adifferent product term
• Deliberate redundancy like this makes circuits impossible to test completely
• Static 1 hazards: in SOP circuits: 1 0 1• Static 0 hazards: in POS circuits: 0 1 0
Hazards and K-maps
• Static 0 hazards in POS circuits:
• Identify the hazard(s): how many? Where?
Hazards and K-maps
• Hazards identified and fixed? What is missing?
Hazards and K-maps
• Dynamic hazards:– When input change requires output change– Occur when output makes more than one transition
• Always result from static hazards elsewhere– Eliminating the static hazards eliminates the
dynamic ones as well
Prime number detector: F = (1, 2, 3, 5, 7, 11, 13)
N3 N2 00 01 11 10
00
01 x x xN1 N0
11 x x x
10 x
0--1
00-1
01-1
00010011
01010111
Karnaugh maps: 2, 3, and 4 variable
F = X’YZ’ + XZ + Y’Z
Example:
Another example: Prime implicants(maximal clusters)
Prime number detector
Prime number detector
Another example:distinguished cell: covered by only one prime implicantessential prime implicant: contains distinguished cell
Another example:primes, distinguished cells, essentials
Selecting essentials leaves an uncovered cellcover with simpler implicant: W’Z
Eclipsing (in reduced map)P eclipses Q if P covers all of Q’s onesif P is no more expensive (same or fewer literals),
then choose P over Q
Alas, no essential prime implicantsbranching: choose a cell and examine all implicants
that cover that cell
Don’t cares....
Multiple functionscan use separate Karnaugh maps
...or can manage to find common terms...
For more than 6 input variables,Karnaugh maps are difficult to manipulate
Need computer program....Quine-McCluskey algorithm
typedef unsigned short WORD; /* assume 16-bit registers */struct cube {
WORD t; /* marks uncomplemented variables */WORD f; /* marks complemented variables */}
typedef struct cube CUBE;
CUBE P1, x, y, z;
0149101215 XXXXXXX
Equation:
w x’ y z’ + w’ x’ y z’ = x’ y z’
Karnaugh map:
wx 00 01 11 10yz 00 01 11 10 1 1
Example in four variables
Cubes (last four bits):
1010 0010 = 1000 ==> single one in common position ==> combinable0101 1101 = 1000
1010 & 0010 = 0010 ==> w now missing, new cube corresponds to z’ y z’0101 & 1101 = 0101
Start with minterms (0-cubes) Combine when possible to form (1-cubes)....
Example: w’xy’z + wxy’z + w’xyz + wxyz = xz
wx 00 01 11 10yz 00 01 1 1 11 1 1 10 Cubes: 0101 1101 0111 1111 1010 0010 1000 0000
0101 1010
1101 0010
0111 1000
1111 0000
0101 1101 = 10001010 0010 = 1000
0101 & 1101 = 01011010 & 0010 = 0010
Start with minterms (0-cubes) Combine when possible to form (1-cubes)....
Example: w’xy’z + wxy’z + w’xyz + wxyz = xz
wx 00 01 11 10yz 00 01 1 1 11 1 1 10 Cubes: 0101 1101 0111 1111 1010 0010 1000 0000
0101 1010
1101 0101 0010 0010
0111 1000
1111 0000
wx 00 01 11 10yz 00 01 1 1 11 1 1 10
0101 1101 = 10001010 0010 = 1000
0101 & 1101 = 01011010 & 0010 = 0010
Start with minterms (0-cubes) Combine when possible to form (1-cubes)....
Example: w’xy’z + wxy’z + w’xyz + wxyz = xz
wx 00 01 11 10yz 00 01 1 1 11 1 1 10 Cubes: 0101 1101 0111 1111 1010 0010 1000 0000
0101 1010
1101 0101 0010 0010
0111 0101 1000 1000
1111 1101 0111 0000 0000 0000
wx 00 01 11 10yz 00 01 1 1 11 1 1 10
Continue to form 2-cubes
Example: w’xy’z + wxy’z + w’xyz + wxyz = xz
wx 00 01 11 10yz 00 01 1 1 11 1 1 10 Cubes: 0101 0101 1101 0111 0010 1000 0000 0000
0101 0010
0101 1000
1101 0101 0000 0000
0111 0101 0000 0000
Read in all minterms (0-cubes);mark all 0-cubes “uncovered”;for (m = 0; m < Nvar; m++) for (j = 0; j < Ncubes[m]; j++) for (k = j + 1; k < Ncubes[m]; k++) if (combinable(cube[m][j], cube[m][k])) { mark cube[m][j] and cube[m][k] “covered” temp = combined cube; if (temp not already at level m + 1) { add temp to level m + 1; mark temp “uncovered”
} }
Quine-McCluskey Algorithm:
Manual algorithm: F = (2, 5, 7, 9, 13, 15) (variables WXYZ)
0010
01011001
01111101
1111
01-1-1011-01
-11111-1
-1-1
uncovered terms
Manual algorithm: F = (2, 5, 7, 9, 13, 15) (variables WXYZ)
0010
0101 x 01-1 x -1-11001 x -101 x
1-01
0111 x -111 x1101 x 11-1 x
1111 xWX
00 01 11 1000
YZ 01 1 1 1
11 1 1
10 1
XZ
WY’Z
W’X’YZ’
MintermsPrime 2 5 7 9 13 15Implicants
0010 x1-01 x x-1-1 x x x x
distinguished minterms (cells): 2, 5, 7, 9, 15essential prime implicants: 0010, 1-01, -1-1 (all)
F = 0010 + 1-01 + -1-1 = W’X’YZ’ + WY’Z + XZ
Not all prime implicants are necessarily essential
distinguished cellsessential implicants
remainder C eclipses B and D
Minimal form: A + E + C
Not all prime implicants are necessarily essential
distinguished cellsessential implicants
remainder C eclipses B and D
Minimal form: A + E + C
Not all prime implicants are necessarily essential
distinguished cellsessential implicants
remainder C eclipses B and D
Minimal form: A + E + C
Static hazard: X = Y = 1, Z falls from 1 to 0
Z’
XZ’ Z’
XZ’
Consensus term
Reconstruct Karnaugh map:F = XZ’ + YZ = XYZ’ + XY’Z’ + XYZ + X’YZ
Solution: add consensus term
Z’
XZ’ Z’
XZ’
Consensus term
Z’
XZ’
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