As the last CC-list represents Maximum Compatible Classes we conclude:
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As the last CC-list represents Maximum Compatible Classes we conclude: MCC0 = {B 4 ,B 7 ,} - PowerPoint PPT Presentation


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As the last CC-list represents Maximum Compatible Classes we conclude: MCC0 = {B 4 ,B 7 ,} MCC1 = {B 4 ,B 5 ,B 6 }, MCC2 = {B 2 ,B 3 ,B 4 ,B 5 }, MCC3 = {B 0 ,B 1 ,B 3 }.

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As the last CC-list represents Maximum Compatible Classes we conclude:

MCC0 = {B4,B7,}

MCC1 = {B4,B5,B6},

MCC2 = {B2,B3,B4,B5},

MCC3 = {B0,B1,B3}.


The next step in calculating conclude:G is the process of selecting only a subset of MCCs that cover the set of all blocks of PG i.e. {B1,...,Bm}.

The MIN(M) represents partition

G = {1,..., k} in the following way:

 = {Bi1,Bi2,...,Biq} forms a block of G if and only if  MCC.


The final conclude:G is a result of adding the objects forming the set of blocks included in one block of G.

For the MCCs from Example 4.5, one of the minimal covers is:

{B7}, {B6}, {B2,B4,B5}, {B0,B1,B3}.

and the corresponding G is

G = (4,5,7; 8,10; 9,11,12,13,1; 14,15,2,3,6).


Coding the mccs

Coding the MCCs conclude:

Coding of MCCs is necessary to generate the H table.


A simple example

A simple example: conclude:

Let A = {x0} and B = {x1, x2}.

P(B) = {(0,6)(1,7)(3,4)(2,5)}

MCC1: 0,1,3

MCC2: 2

Decompose:

100 1

101 1

111 1

110 0

010 0

011 0

000 0

001 0

G table:

00 0

01 0

10 1

11 0

H table:

00 0

10 1

01 0

11 0


Non disjoint decomposition an another example

xx conclude:

02

00 0

01 1

10 1

11 0

xx

01

00 0

01 0

10 1

11 0

Non-Disjoint Decomposition: An another example

xxx y

012

000 0

010 0

001 1

011 1

100 1

110 1

101 1

111 0

00 0

01 1

10 1

11 1

y



Desirable characteristics: conclude:

1. The generated H table should be as simple as possible (should we try to retain as many “don’t cares” as possible? Should we try to introduce more “don’t cares”?).

2. The generated truth table should facilitate “easier” finding of “good” decompositions further down the line.


Coding procedure

Coding procedure: conclude:

Encoding MCCs (we will call it classes (CCs) for simplicity).


Let us decompose the conclude:

table for B = {x0,x1,x3,x4}


A subset of the Maximum Compatibility Classes that cover all the blocks of P(B) is:

CC(0) = {B9} = (7,10)

CC(1) = {B7} = (7,15)

CC(2) = {B6} = (0,8,11)

CC(3) = {B4} = (0,4,11)

CC(4) = {B3,B5,B8} = (0,1,2,3,8,9,13,14,15,16)

CC(5) = {B1} = (2,6,12,13,15,16)

CC(6) = {B0,B2} = (0,1,2,4,5).


There are seven CCs and therefore seven codes are needed to encode each of them. If a cube is a member of two different classes, it is possible to encode the classes in such a manner so that the two codes differ only in one position. This leads to eventually encoding the cube found in both the classes with a code having one "don't care" value.


Such an encoding procedure retains or some times even introduces "don't care" values in the G function and consequently in the H function also. This results in significantly reducing the space required to store the truth tables, and may reduce the number of blocks required to implement a function. A smaller table also reduces the calculation complexities.


Therefore, first we look for the elements which are members of 2l-1 classes, where l is the number of bits in the code. If there is such an element, 2l-1 codes which differ only in l-1 positions are selected and if such codes are available they are allotted to the 2l-1 classes to which the element belongs. Then we look for the elements which are members of 2l-2 classes and continue the process till 2l-i equals 1, when the classes holding elements that belong to only one class are encoded.


For the example under consideration, where l equals 3, the cube 0 belongs to four (2l-1) classes CC(2), CC(3), CC(4) and CC(6).

CC(2) : 010

CC(3) : 011

CC(4) : 001

CC(6) : 000.


As there is no other cube found in four classes, the cubes that belong to two (2l-2) classes are considered for further encoding. The following seven cubes are found in two classes: 1,4,7,8,11,13,16. First, cube number 1 is considered. The cube is found in CC(4) and CC(6). Since both the classes are already encoded and the codes differ only in one (l-2) position, nothing needs to be done.


The next cube is taken for consideration, i.e. cube number 4, which is found in CC(3) and CC(6). Though both the classes are already encoded, the codes differ in more than one position and hence it is decided that cube four can not be encoded at this point and it is added to a set of cubes Z, which are to be encoded later.


The next element, cube number seven is found in CC(0) and CC(1) and as neither CC is encoded, we look for 2(l-2) available codes which differ only in l-2 positions. Such codes are found and the classes are encoded as follows:

CC(0) : 100

CC(1) : 101


The case for the cube 8 is similar to the case of cube 4 and hence it is added to the set Z. Cube 11 is encoded in a similar way as cube 1. The cubes 13 and 16 have one CC encoded already and there is no second code available, which differs only in one position with respect to the accepted code. Therefore both the cubes 13 and 16 are added to the set Z.


Now, the classes holding the elements which belong to only one class are encoded, if they are already not encoded. Thus:

CC(5) : 111


Encoding cubes

Encoding Cubes one class are encoded, if they are already not encoded. Thus:

The encoding method discussed so far has encoded the Compatible Classes in such a way that the cubes 4,8,13 and 16 (elements of the set Z), which belong to more than one CC have inconsistent codes. Moreover, there may be cubes which belong to more than one CC but not to 2(l-i) number of classes. These cubes also have inconsistent codes for obvious reasons. In the example under consideration there are two cubes (2 and 15) each of which belongs to three classes. These cubes are also added to the set Z of non-encoded cubes.


Step 1

Step 1: one class are encoded, if they are already not encoded. Thus:

Two sub-cubes are obtained by splitting a cube at a position where a "-" appears. The truth table of the G function is searched if both the sub-cubes have been encoded already. If not, a different pair of sub-cubes are obtained by splitting the cube at a different position and the search is continued until codes are found for a pair of sub-cubes. If this step does not yield any result step two is executed.


Step 2

Step 2: one class are encoded, if they are already not encoded. Thus:

Step two is similar to step one with a difference that this time instead of searching the truth table G, each sub-cube is compared with the product of all the elements of a CC and if the sub-cube is compatible to the product it is said that the sub-cube is encodable. If both the sub-cubes are encodable, the codes of the respective compatible classes are accepted as the codes for the sub-cubes.


Step 3

Step 3: one class are encoded, if they are already not encoded. Thus:

If step two does not produce any result, step three accepts even if only one sub-cube is encodable and allots the code of the compatible CC to the sub-cube concerned. The other sub-cube is added to the set Z as a non-encoded cube for further splitting.


Step 4

Step 4: one class are encoded, if they are already not encoded. Thus:

If step 3 has also failed to produce any satisfactory result, then the given cube is split at the most significant position where a "-" is found and both the sub-cubes are added to the set of non-encoded cubes Z to be split further.


Step 5

Step 5: one class are encoded, if they are already not encoded. Thus:

Repeated failed execution of steps one through four may produce cubes with no "-". Under such a situation the codes of the Maximum Compatible Classes of all the cubes in the table G compatible to the cube under consideration are collected. Each code is treated as a potential candidate code and if a code is consistent, then it is accepted. It is worth noting that this kind of situation may arise when there are two intersecting cubes.


Step 6

Step 6: one class are encoded, if they are already not encoded. Thus:

If step 5 fails, then the cube ought to belong to a CC which has not been allotted a code by the method described in 4.3. Therefore the CC to which the given cube belongs is given any available code.


If a cube is allotted a code in step three or four, its consistency with the table G is verified before proceeding further. The consistency of the final table is checked as still there may be inconsistency because of intersecting cubes. In case such a pair of inconsistent cubes are detected, the cube or cubes encoded using the step two, three or four are considered as not encoded and the described six step procedure is revoked once again to encode the cube or cubes.


Reverting back to the example, the set of non-encoded cubes in the truth table G is: Z = {2,4,8,13,15,16}. The cube (--10) numbered 13 is initially split into two sub-cubes (0-10) and (1-10). As these sub-cubes are not encoded already, the next pair of sub-cubes (-010) and (-110) are verified. It is found that the cubes are encoded and their codes are:

-010 : 001 (row 14 in Table)

-110 : 111 (row 12 in Table)

Step one does not yield any results for other elements of set Z.


In step two, cube number four (1-00) is split as (1000) and (1100). The sub-cube (1000) is compatible to the product of all the elements of CC(3) and hence its code (011) is acceptable. The sub-cube (1100) is compatible to the product of all the elements of CC(6) and hence its code (000) is also acceptable. Since both the sub-cubes have acceptable codes, they are encoded as such. In the same way cube 8 is also encoded as:

0000 : 010

0001 : 001


Step three codes the cubes 2 and 15 as follows: (1100). The sub-cube (1000) is compatible to the product of all the elements of CC(3) and hence its code (011) is acceptable. The sub-cube (1100) is compatible to the product of all the elements of CC(6) and hence its code (000) is also acceptable. Since both the sub-cubes have acceptable codes, they are encoded as such. In the same way cube 8 is also encoded as:

cube 2.

-1--  01-- : 000

11-- : not coded

cube 15.

1-1-  101- : -01 (compatible to CC(1) and CC(4))

111- : not coded

Execution of step three has left two more cubes to be encoded.


The cube (11--) is encoded in step two as: (1100). The sub-cube (1000) is compatible to the product of all the elements of CC(3) and hence its code (011) is acceptable. The sub-cube (1100) is compatible to the product of all the elements of CC(6) and hence its code (000) is also acceptable. Since both the sub-cubes have acceptable codes, they are encoded as such. In the same way cube 8 is also encoded as:

110- : 000

111- : not coded

Now, the only cube left not encoded is (111-) and it is encoded in step two as:

1110 : 111

1111 : 001


As all the input cubes of the table G are encoded, the table is verified for any inconsistency. It is found that the cubes 2 and 6 are inconsistent:

cube 2. 01-- : 000

cube 6. -110 : 111


As cube 2 was encoded using step 3, annul its code and start encoding (01--) all over again from step one. In step three the following code is found:

010- : 000

011- : not coded

One more run of the entire procedure for the cube (011- ) fetches the following codes for the cube (011- ) in step 5.

0110 : 111

0111 : 001

No more inconsistency is detected and this ends the process of encoding.


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