Karnaugh maps, truth tables, and Boolean expressions

Maurice Karnaugh, a telecommunications engineer, developed the Karnaugh map at Bell Labs in 1953 while designing digital logic based telephone switching circuits.

Now that we have developed the Karnaugh map with the aid of Venn diagrams, let's put it to use. Karnaugh maps reduce logic functions more quickly and easily compared to Boolean algebra. By reduce we mean simplify, reducing the number of gates and inputs. We like to simplify logic to a lowest cost form to save costs by elimination of components. We define lowest cost as being the lowest number of gates with the lowest number of inputs per gate.

Given a choice, most students do logic simplification with Karnaugh maps rather than Boolean algebra once they learn this tool.

We show five individual items above, which are just different ways of representing the same thing: an arbitrary 2-input digital logic function. First is relay ladder logic, then logic gates, a truth table, a Karnaugh map, and a Boolean equation. The point is that any of these are equivalent. Two inputs A and B can take on values of either 0 or 1, high or low, open or closed, True or False, as the case may be. There are 2² = 4 combinations of inputs producing an output. This is applicable to all five examples.

These four outputs may be observed on a lamp in the relay ladder logic, on a logic probe on the gate diagram. These outputs may be recorded in the truth table, or in the Karnaugh map. Look at the Karnaugh map as being a rearranged truth table. The Output of the Boolean equation may be computed by the laws of Boolean algebra and transfered to the truth table or Karnaugh map. Which of the five equivalent logic descriptions should we use? The one which is most useful for the task to be accomplished.

The outputs of a truth table correspond on a one-to-one basis to Karnaugh map entries. Starting at the top of the truth table, the A=0, B=0 inputs produce an output α. Note that this same output α is found in the Karnaugh map at the A=0, B=0 cell address, upper left corner of K-map where the A=0 row and B=0 column intersect. The other truth table outputs β, χ, δ from inputs AB=01, 10, 11 are found at corresponding K-map locations.

Below, we show the adjacent 2-cell regions in the 2-variable K-map with the aid of previous rectangular Venn diagram like Boolean regions.

Cells α and χ are adjacent in the K-map as ellipses in the left most K-map below. Referring to the previous truth table, this is not the case. There is another truth table entry (β) between them. Which brings us to the whole point of the organizing the K-map into a square array, cells with any Boolean variables in common need to be close to one another so as to present a pattern that jumps out at us. For cells α and χ they have the Boolean variable B' in common. We know this because B=0 (same as B') for the column above cells α and χ. Compare this to the square Venn diagram above the K-map.

A similar line of reasoning shows that β and δ have Boolean B (B=1) in common. Then, α and β have Boolean A' (A=0) in common. Finally, χ and δ have Boolean A (A=1) in common. Compare the last two maps to the middle square Venn diagram.

To summarize, we are looking for commonality of Boolean variables among cells. The Karnaugh map is organized so that we may see that commonality. Let's try some examples.

Example:

Transfer the contents of the truth table to the Karnaugh map above.

Solution:

The truth table contains two 1s. the K- map must have both of them. locate the first 1 in the 2nd row of the truth table above.

• note the truth table AB address
• locate the cell in the K-map having the same address
• place a 1 in that cell

Repeat the process for the 1 in the last line of the truth table.

Example:

For the Karnaugh map in the above problem, write the Boolean expression. Solution is below.

Solution:

Look for adjacent cells, that is, above or to the side of a cell. Diagonal cells are not adjacent. Adjacent cells will have one or more Boolean variables in common.

• Group (circle) the two 1s in the column
• Find the variable(s) top and/or side which are the same for the group, Write this as the Boolean result. It is B in our case.
• Ignore variable(s) which are not the same for a cell group. In our case A varies, is both 1 and 0, ignore Boolean A.
• Ignore any variable not associated with cells containing 1s. B' has no ones under it. Ignore B'
• Result Out = B

This might be easier to see by comparing to the Venn diagrams to the right, specifically the B column.

Example:

Write the Boolean expression for the Karnaugh map below.

Solution: (above)

• Group (circle) the two 1's in the row
• Find the variable(s) which are the same for the group, Out = A'

Example:

For the Truth table below, transfer the outputs to the Karnaugh, then write the Boolean expression for the result.

Solution:

Transfer the 1s from the locations in the Truth table to the corresponding locations in the K-map.

• Group (circle) the two 1's in the column under B=1
• Group (circle) the two 1's in the row right of A=1
• Write product term for first group = B
• Write product term for second group = A
• Write Sum-Of-Products of above two terms Output = A+B

The solution of the K-map in the middle is the simplest or lowest cost solution. A less desirable solution is at far right. After grouping the two 1s, we make the mistake of forming a group of 1-cell. The reason that this is not desirable is that:

• The single cell has a product term of AB'
• The corresponding solution is Output = AB' + B
• This is not the simplest solution

The way to pick up this single 1 is to form a group of two with the 1 to the right of it as shown in the lower line of the middle K-map, even though this 1 has already been included in the column group (B). We are allowed to re-use cells in order to form larger groups. In fact, it is desirable because it leads to a simpler result.

We need to point out that either of the above solutions, Output or Wrong Output, are logically correct. Both circuits yield the same output. It is a matter of the former circuit being the lowest cost solution.

Example:

Fill in the Karnaugh map for the Boolean expression below, then write the Boolean expression for the result.

Solution: (above)

The Boolean expression has three product terms. There will be a 1 entered for each product term. Though, in general, the number of 1s per product term varies with the number of variables in the product term compared to the size of the K-map. The product term is the address of the cell where the 1 is entered. The first product term, A'B, corresponds to the 01 cell in the map. A 1 is entered in this cell. The other two P-terms are entered for a total of three 1s

Next, proceed with grouping and extracting the simplified result as in the previous truth table problem.

Example:

Simplify the logic diagram below.

Solution: (Figure below)

• Write the Boolean expression for the original logic diagram as shown below
• Transfer the product terms to the Karnaugh map
• Form groups of cells as in previous examples
• Write Boolean expression for groups as in previous examples
• Draw simplified logic diagram

Example:

Simplify the logic diagram below.

Solution:

• Write the Boolean expression for the original logic diagram shown above
• Transfer the product terms to the Karnaugh map.
• It is not possible to form groups.
• No simplification is possible; leave it as it is.

No logic simplification is possible for the above diagram. This sometimes happens. Neither the methods of Karnaugh maps nor Boolean algebra can simplify this logic further. We show an Exclusive-OR schematic symbol above; however, this is not a logical simplification. It just makes a schematic diagram look nicer. Since it is not possible to simplify the Exclusive-OR logic and it is widely used, it is provided by manufacturers as a basic integrated circuit (7486).

Logic simplification with Karnaugh maps

The logic simplification examples that we have done so could have been performed with Boolean algebra about as quickly. Real world logic simplification problems call for larger Karnaugh maps so that we may do serious work. We will work some contrived examples in this section, leaving most of the real world applications for the Combinatorial Logic chapter. By contrived, we mean examples which illustrate techniques. This approach will develop the tools we need to transition to the more complex applications in the Combinatorial Logic chapter.

We show our previously developed Karnaugh map. We will use the form on the right.

Note the sequence of numbers across the top of the map. It is not in binary sequence which would be 00, 01, 10, 11. It is 00, 01, 11 10, which is Gray code sequence. Gray code sequence only changes one binary bit as we go from one number to the next in the sequence, unlike binary. That means that adjacent cells will only vary by one bit, or Boolean variable. This is what we need to organize the outputs of a logic function so that we may view commonality. Moreover, the column and row headings must be in Gray code order, or the map will not work as a Karnaugh map. Cells sharing common Boolean variables would no longer be adjacent, nor show visual patterns. Adjacent cells vary by only one bit because a Gray code sequence varies by only one bit.

If we sketch our own Karnaugh maps, we need to generate Gray code for any size map that we may use. This is how we generate Gray code of any size.

Note that the Gray code sequence, above right, only varies by one bit as we go down the list, or bottom to top up the list. This property of Gray code is often useful in digital electronics in general. In particular, it is applicable to Karnaugh maps.

Let us move on to some examples of simplification with 3-variable Karnaugh maps. We show how to map the product terms of the unsimplified logic to the K-map. We illustrate how to identify groups of adjacent cells which leads to a Sum-of-Products simplification of the digital logic.

Above we, place the 1's in the K-map for each of the product terms, identify a group of two, then write a p-term (product term) for the sole group as our simplified result.

Mapping the four product terms above yields a group of four covered by Boolean A'

Mapping the four p-terms yields a group of four, which is covered by one variable C.

After mapping the six p-terms above, identify the upper group of four, pick up the lower two cells as a group of four by sharing the two with two more from the other group. Covering these two with a group of four gives a simpler result. Since there are two groups, there will be two p-terms in the Sum-of-Products result A'+B

The two product terms above form one group of two and simplifies to BC

Mapping the four p-terms yields a single group of four, which is B

Mapping the four p-terms above yields a group of four. Visualize the group of four by rolling up the ends of the map to form a cylinder, then the cells are adjacent. We normally mark the group of four as above left. Out of the variables A, B, C, there is a common variable: C'. C' is a 0 over all four cells. Final result is C'.

The six cells above from the unsimplified equation can be organized into two groups of four. These two groups should give us two p-terms in our simplified result of A' + C'.

Below, we revisit the Toxic Waste Incinerator from the Boolean algebra chapter. See Boolean algebra chapter for details on this example. We will simplify the logic using a Karnaugh map.

The Boolean equation for the output has four product terms. Map four 1's corresponding to the p-terms. Forming groups of cells, we have three groups of two. There will be three p-terms in the simplified result, one for each group. See "Toxic Waste Incinerator", Boolean algebra chapter for a gate diagram of the result, which is reproduced below.

Below we repeat the Boolean algebra simplification of Toxic waste incinerator for comparison.

Below we repeat the Toxic waste incinerator Karnaugh map solution for comparison to the above Boolean algebra simplification. This case illustrates why the Karnaugh map is widely used for logic simplification.

The Karnaugh map method looks easier than the previous page of boolean algebra.

Larger 4-variable Karnaugh maps

Knowing how to generate Gray code should allow us to build larger maps. Actually, all we need to do is look at the left to right sequence across the top of the 3-variable map, and copy it down the left side of the 4-variable map. See below.

The following four variable Karnaugh maps illustrate reduction of Boolean expressions too tedious for Boolean algebra. Reductions could be done with Boolean algebra. However, the Karnaugh map is faster and easier, especially if there are many logic reductions to do.

The above Boolean expression has seven product terms. They are mapped top to bottom and left to right on the K-map above. For example, the first P-term A'B'CD is first row 3rd cell, corresponding to map location A=0, B=0, C=1, D=1. The other product terms are placed in a similar manner. Encircling the largest groups possible, two groups of four are shown above. The dashed horizontal group corresponds the the simplified product term AB. The vertical group corresponds to Boolean CD. Since there are two groups, there will be two product terms in the Sum-Of-Products result of Out=AB+CD.

Fold up the corners of the map below like it is a napkin to make the four cells physically adjacent.

The four cells above are a group of four because they all have the Boolean variables B' and D' in common. In other words, B=0 for the four cells, and D=0 for the four cells. The other variables (A, B) are 0 in some cases, 1 in other cases with respect to the four corner cells. Thus, these variables (A, B) are not involved with this group of four. This single group comes out of the map as one product term for the simplified result: Out=B'C'

For the K-map below, roll the top and bottom edges into a cylinder forming eight adjacent cells.

The above group of eight has one Boolean variable in common: B=0. Therefore, the one group of eight is covered by one p-term: B'. The original eight term Boolean expression simplifies to Out=B'

The Boolean expression below has nine p-terms, three of which have three Booleans instead of four. The difference is that while four Boolean variable product terms cover one cell, the three Boolean p-terms cover a pair of cells each.

The six product terms of four Boolean variables map in the usual manner above as single cells. The three Boolean variable terms (three each) map as cell pairs, which is shown above. Note that we are mapping p-terms into the K-map, not pulling them out at this point.

For the simplification, we form two groups of eight. Cells in the corners are shared with both groups. This is fine. In fact, this leads to a better solution than forming a group of eight and a group of four without sharing any cells. Final Solution is Out=B'+D'

Below we map the unsimplified Boolean expression to the Karnaugh map.

Above, three of the cells form into a groups of two cells. A fourth cell cannot be combined with anything, which often happens in "real world" problems. In this case, the Boolean p-term ABCD is unchanged in the simplification process. Result: Out= B'C'D'+A'B'D'+ABCD

Often times there is more than one minimum cost solution to a simplification problem. Such is the case illustrated below.

Both results above have four product terms of three Boolean variable each. Both are equally valid minimal cost solutions. The difference in the final solution is due to how the cells are grouped as shown above. A minimal cost solution is a valid logic design with the minimum number of gates with the minimum number of inputs.

Below we map the unsimplified Boolean equation as usual and form a group of four as a first simplification step. It may not be obvious how to pick up the remaining cells.

Pick up three more cells in a group of four, center above. There are still two cells remaining. the minimal cost method to pick up those is to group them with neighboring cells as groups of four as at above right.

On a cautionary note, do not attempt to form groups of three. Groupings must be powers of 2, that is, 1, 2, 4, 8 ...

Below we have another example of two possible minimal cost solutions. Start by forming a couple of groups of four after mapping the cells.

The two solutions depend on whether the single remaining cell is grouped with the first or the second group of four as a group of two cells. That cell either comes out as either ABC' or ABD, your choice. Either way, this cell is covered by either Boolean product term. Final results are shown above.

Below we have an example of a simplification using the Karnaugh map at left or Boolean algebra at right. Plot C' on the map as the area of all cells covered by address C=0, the 8-cells on the left of the map. Then, plot the single ABCD cell. That single cell forms a group of 2-cell as shown, which simplifies to P-term ABD, for an end result of Out = C' + ABD.

This (above) is a rare example of a four variable problem that can be reduced with Boolean algebra without a lot of work, assuming that you remember the theorems.

Minterm vs maxterm solution

So far we have been finding Sum-Of-Product (SOP) solutions to logic reduction problems. For each of these SOP solutions, there is also a Product-Of-Sums solution (POS), which could be more useful, depending on the application. Before working a Product-Of-Sums solution, we need to introduce some new terminology. The procedure below for mapping product terms is not new to this chapter. We just want to establish a formal procedure for minterms for comparison to the new procedure for maxterms.

A minterm is a Boolean expression resulting in 1 for the output of a single cell, and 0s for all other cells in a Karnaugh map, or truth table. If a minterm has a single 1 and the remaining cells as 0s, it would appear to cover a minimum area of 1s. The illustration above left shows the minterm ABC, a single product term, as a single 1 in a map that is otherwise 0s. We have not shown the 0s in our Karnaugh maps up to this point, as it is customary to omit them unless specifically needed. Another minterm A'BC' is shown above right. The point to review is that the address of the cell corresponds directly to the minterm being mapped. That is, the cell 111 corresponds to the minterm ABC above left. Above right we see that the minterm A'BC' corresponds directly to the cell 010. A Boolean expression or map may have multiple minterms.

Referring to the above figure, Let's summarize the procedure for placing a minterm in a K-map:

• Identify the minterm (product term) term to be mapped.
• Write the corresponding binary numeric value.
• Use binary value as an address to place a 1 in the K-map
• Repeat steps for other minterms (P-terms within a Sum-Of-Products).

A Boolean expression will more often than not consist of multiple minterms corresponding to multiple cells in a Karnaugh map as shown above. The multiple minterms in this map are the individual minterms which we examined in the previous figure above. The point we review for reference is that the 1s come out of the K-map as a binary cell address which converts directly to one or more product terms. By directly we mean that a 0 corresponds to a complemented variable, and a 1 corresponds to a true variable. Example: 010 converts directly to A'BC'. There was no reduction in this example. Though, we do have a Sum-Of-Products result from the minterms.

Referring to the above figure, Let's summarize the procedure for writing the Sum-Of-Products reduced Boolean equation from a K-map:

• Form largest groups of 1s possible covering all minterms. Groups must be a power of 2.
• Write binary numeric value for groups.
• Convert binary value to a product term.
• Repeat steps for other groups. Each group yields a p-terms within a Sum-Of-Products.

Nothing new so far, a formal procedure has been written down for dealing with minterms. This serves as a pattern for dealing with maxterms.

Next we attack the Boolean function which is 0 for a single cell and 1s for all others.

A maxterm is a Boolean expression resulting in a 0 for the output of a single cell expression, and 1s for all other cells in the Karnaugh map, or truth table. The illustration above left shows the maxterm (A+B+C), a single sum term, as a single 0 in a map that is otherwise 1s. If a maxterm has a single 0 and the remaining cells as 1s, it would appear to cover a maximum area of 1s.

There are some differences now that we are dealing with something new, maxterms. The maxterm is a 0, not a 1 in the Karnaugh map. A maxterm is a sum term, (A+B+C) in our example, not a product term.

It also looks strange that (A+B+C) is mapped into the cell 000. For the equation Out=(A+B+C)=0, all three variables (A, B, C) must individually be equal to 0. Only (0+0+0)=0 will equal 0. Thus we place our sole 0 for minterm (A+B+C) in cell A,B,C=000 in the K-map, where the inputs are all0 . This is the only case which will give us a 0 for our maxterm. All other cells contain 1s because any input values other than ((0,0,0) for (A+B+C) yields 1s upon evaluation.

Referring to the above figure, the procedure for placing a maxterm in the K-map is:

• Identify the Sum term to be mapped.
• Write corresponding binary numeric value.
• Form the complement
• Use the complement as an address to place a 0 in the K-map
• Repeat for other maxterms (Sum terms within Product-of-Sums expression).

Another maxterm A'+B'+C' is shown above. Numeric 000 corresponds to A'+B'+C'. The complement is 111. Place a 0 for maxterm (A'+B'+C') in this cell (1,1,1) of the K-map as shown above.

Why should (A'+B'+C') cause a 0 to be in cell 111? When A'+B'+C' is (1'+1'+1'), all 1s in, which is (0+0+0) after taking complements, we have the only condition that will give us a 0. All the 1s are complemented to all 0s, which is 0 when ORed.

A Boolean Product-Of-Sums expression or map may have multiple maxterms as shown above. Maxterm (A+B+C) yields numeric 111 which complements to 000, placing a 0 in cell (0,0,0). Maxterm (A+B+C') yields numeric 110 which complements to 001, placing a 0 in cell (0,0,1).

Now that we have the k-map setup, what we are really interested in is showing how to write a Product-Of-Sums reduction. Form the 0s into groups. That would be a group of two below. Write the binary value corresponding to the sum-term which is (0,0,X). Both A and B are 0 for the group. But, C is both 0 and 1 so we write an X as a place holder for C. Form the complement (1,1,X). Write the Sum-term (A+B) discarding the C and the X which held its' place. In general, expect to have more sum-terms multiplied together in the Product-Of-Sums result. Though, we have a simple example here.

Let's summarize the procedure for writing the Product-Of-Sums Boolean reduction for a K-map:

• Form largest groups of 0s possible, covering all maxterms. Groups must be a power of 2.
• Write binary numeric value for group.
• Complement binary numeric value for group.
• Convert complement value to a sum-term.
• Repeat steps for other groups. Each group yields a sum-term within a Product-Of-Sums result.

Example:

Simplify the Product-Of-Sums Boolean expression below, providing a result in POS form.

Solution:

Transfer the seven maxterms to the map below as 0s. Be sure to complement the input variables in finding the proper cell location.

We map the 0s as they appear left to right top to bottom on the map above. We locate the last three maxterms with leader lines..

Once the cells are in place above, form groups of cells as shown below. Larger groups will give a sum-term with fewer inputs. Fewer groups will yield fewer sum-terms in the result.

We have three groups, so we expect to have three sum-terms in our POS result above. The group of 4-cells yields a 2-variable sum-term. The two groups of 2-cells give us two 3-variable sum-terms. Details are shown for how we arrived at the Sum-terms above. For a group, write the binary group input address, then complement it, converting that to the Boolean sum-term. The final result is product of the three sums.

Example:

Simplify the Product-Of-Sums Boolean expression below, providing a result in SOP form.

Solution:

This looks like a repeat of the last problem. It is except that we ask for a Sum-Of-Products Solution instead of the Product-Of-Sums which we just finished. Map the maxterm 0s from the Product-Of-Sums given as in the previous problem, below left.

Then fill in the implied 1s in the remaining cells of the map above right.

Form groups of 1s to cover all 1s. Then write the Sum-Of-Products simplified result as in the previous section of this chapter. This is identical to a previous problem.

Above we show both the Product-Of-Sums solution, from the previous example, and the Sum-Of-Products solution from the current problem for comparison. Which is the simpler solution? The POS uses 3-OR gates and 1-AND gate, while the SOP uses 3-AND gates and 1-OR gate. Both use four gates each. Taking a closer look, we count the number of gate inputs. The POS uses 8-inputs; the SOP uses 7-inputs. By the definition of minimal cost solution, the SOP solution is simpler. This is an example of a technically correct answer that is of little use in the real world.

The better solution depends on complexity and the logic family being used. The SOP solution is usually better if using the TTL logic family, as NAND gates are the basic building block, which works well with SOP implementations. On the other hand, A POS solution would be acceptable when using the CMOS logic family since all sizes of NOR gates are available.

The gate diagrams for both cases are shown above, Product-Of-Sums left, and Sum-Of-Products right.

Below, we take a closer look at the Sum-Of-Products version of our example logic, which is repeated at left.

Above all AND gates at left have been replaced by NAND gates at right.. The OR gate at the output is replaced by a NAND gate. To prove that AND-OR logic is equivalent to NAND-NAND logic, move the inverter invert bubbles at the output of the 3-NAND gates to the input of the final NAND as shown in going from above right to below left.

Above right we see that the output NAND gate with inverted inputs is logically equivalent to an OR gate by DeMorgan's theorem and double negation. This information is useful in building digital logic in a laboratory setting where TTL logic family NAND gates are more readily available in a wide variety of configurations than other types.

The Procedure for constructing NAND-NAND logic, in place of AND-OR logic is as follows:

• Produce a reduced Sum-Of-Products logic design.
• When drawing the wiring diagram of the SOP, replace all gates (both AND and OR) with NAND gates.
• Unused inputs should be tied to logic High.
• In case of troubleshooting, internal nodes at the first level of NAND gate outputs do NOT match AND-OR diagram logic levels, but are inverted. Use the NAND-NAND logic diagram. Inputs and final output are identical, though.
• Label any multiple packages U1, U2,.. etc.
• Use data sheet to assign pin numbers to inputs and outputs of all gates.

Example:

Let us revisit a previous problem involving an SOP minimization. Produce a Product-Of-Sums solution. Compare the POS solution to the previous SOP.

Solution:

Above left we have the original problem starting with a 9-minterm Boolean unsimplified expression. Reviewing, we formed four groups of 4-cells to yield a 4-product-term SOP result, lower left.

In the middle figure, above, we fill in the empty spaces with the implied 0s. The 0s form two groups of 4-cells. The solid blue group is (A'+B), the dashed red group is (C'+D). This yields two sum-terms in the Product-Of-Sums result, above right Out = (A'+B)(C'+D)

Comparing the previous SOP simplification, left, to the POS simplification, right, shows that the POS is the least cost solution. The SOP uses 5-gates total, the POS uses only 3-gates. This POS solution even looks attractive when using TTL logic due to simplicity of the result. We can find AND gates and an OR gate with 2-inputs.

The SOP and POS gate diagrams are shown above for our comparison problem.

Given the pin-outs for the TTL logic family integrated circuit gates below, label the maxterm diagram above right with Circuit designators (U1-a, U1-b, U2-a, etc), and pin numbers.

Each integrated circuit package that we use will receive a circuit designator: U1, U2, U3. To distinguish between the individual gates within the package, they are identified as a, b, c, d, etc. The 7404 hex-inverter package is U1. The individual inverters in it are are U1-a, U1-b, U1-c, etc. U2 is assigned to the 7432 quad OR gate. U3 is assigned to the 7408 quad AND gate. With reference to the pin numbers on the package diagram above, we assign pin numbers to all gate inputs and outputs on the schematic diagram below.

We can now build this circuit in a laboratory setting. Or, we could design a printed circuit board for it. A printed circuit board contains copper foil "wiring" backed by a non conductive substrate of phenolic, or epoxy-fiberglass. Printed circuit boards are used to mass produce electronic circuits. Ground the inputs of unused gates.

Label the previous POS solution diagram above left (third figure back) with Circuit designators and pin numbers. This will be similar to what we just did.

We can find 2-input AND gates, 7408 in the previous example. However, we have trouble finding a 4-input OR gate in our TTL catalog. The only kind of gate with 4-inputs is the 7420 NAND gate shown above right.

We can make the 4-input NAND gate into a 4-input OR gate by inverting the inputs to the NAND gate as shown below. So we will use the 7420 4-input NAND gate as an OR gate by inverting the inputs.

We will not use discrete inverters to invert the inputs to the 7420 4-input NAND gate, but will drive it with 2-input NAND gates in place of the AND gates called for in the SOP, minterm, solution. The inversion at the output of the 2-input NAND gates supply the inversion for the 4-input OR gate.

The result is shown above. It is the only practical way to actually build it with TTL gates by using NAND-NAND logic replacing AND-OR logic.

(sum) and (product) notation

For reference, this section introduces the terminology used in some texts to describe the minterms and maxterms assigned to a Karnaugh map. Otherwise, there is no new material here.

Σ (sigma) indicates sum and lower case "m" indicates minterms. Σm indicates sum of minterms. The following example is revisited to illustrate our point. Instead of a Boolean equation description of unsimplified logic, we list the minterms.

f(A,B,C,D) = Σ m(1, 2, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15)

or

f(A,B,C,D) = Σ(m₁,m₂,m₃,m₄,m₅,m₇,m₈,m₉,m₁₁,m₁₂,m₁₃,m₁₅)

The numbers indicate cell location, or address, within a Karnaugh map as shown below right. This is certainly a compact means of describing a list of minterms or cells in a K-map.

The Sum-Of-Products solution is not affected by the new terminology. The minterms, 1s, in the map have been grouped as usual and a Sum-OF-Products solution written.

Below, we show the terminology for describing a list of maxterms. Product is indicated by the Greek Π (pi), and upper case "M" indicates maxterms. ΠM indicates product of maxterms. The same example illustrates our point. The Boolean equation description of unsimplified logic, is replaced by a list of maxterms.

f(A,B,C,D) = Π M(2, 6, 8, 9, 10, 11, 14)

or

f(A,B,C,D) = Π(M₂, M₆, M₈, M₉, M₁₀, M₁₁, M₁₄)

Once again, the numbers indicate K-map cell address locations. For maxterms this is the location of 0s, as shown below. A Product-OF-Sums solution is completed in the usual manner.

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