Karnaugh mapping allaboutcircuits
- 1. 1 KarnaughMapping Notes www.allaboutcircuits.com – All About Circuits: Karnaugh Mapping OBJECTIVES Introduction to Karnaugh Mapping The Karnaugh Map simplifies logic faster and more easily in most cases. Booleansimplificationis actuallyfaster thanthe Karnaughmap for a task involving two or fewer Booleanvariables. It is still quite usable at three variables, but a bit slower. At four input variables, Booleanalgebra becomes tedious. Karnaughmaps are both faster andeasier. Karnaughmaps work well for up to six input variables, are usable for upto eight variables. For more thansix to eight variables, simplification should be byCAD (computer automated design). In theoryanyof the three methods will work. However, as a practical matter, the above guidelines work well. Venn Diagrams and Sets The Venndiagram bridges the Booleanalgebra from a previous chapter to the Karnaugh Map. We will relate what you alreadyknowabout Booleanalgebra to Venndiagrams, thentransitionto Karnaughmaps. Knowledge brush-up: Visualize how bothA’B and A’+B are same. Venn diagrams don’t actuallyprove anything. Booleanalgebra is needed for formal proofs. However, Venn diagrams canbe usedfor verification and visualization. We have verified andvisualizedDeMorgan’s theoremwith a Venndiagram. A 3-variableK-Map has 23 = 8 cells. is equivalent andmore simplyrepresented as The columnheaders onthe left B’C’, B’C, BC, BC’ are equivalent to 00, 01, 11, 10 on the right. The row headers A, A’ are equivalent to 0, 1 on the right map. Karnaughmaps reduce logic functions more quicklyand easilycomparedto Boolean algebra. We define lowest cost as being the lowest number of gates withthe lowest number of inputs per gate. We showfive individual items above, whichare just different ways of representing the same
- 2. 2 KarnaughMapping Notes www.allaboutcircuits.com – thing:anarbitrary2-input digital logic function. First is relayladder logic, thenlogic gates, a truth table, a Karnaugh map, anda Booleanequation. The point is that anyof these are equivalent. The outputs of a relayladder logic maybe recordedinthe truth table, or inthe Karnaughmap. Look at the Karnaughmapas being a rearranged truth table. The outputs of a truth tablecorrespond on a one-to-one basisto Karnaugh map entries. Startingat the top of the truth table,the A=0, B=0 inputs produce an output α. The other truth table outputs β, χ, δ from inputs AB=01, 10, 11 arefound at correspondingK-map locations. Which brings us to the whole point ofthe organizing the K-mapintoa square array, cellswith anyBoolean variables incommonneed to be close to one another soas to present a patternthat jumps out at us. For cells α and χ theyhave the Booleanvariable B’ in common. Lest we forget to note down, the rule for adjacencyof two cells (especiallyaroundcorners and top-bottom) is that the cells must varybythe value ofone variable only. If a truth table contains two 1s, the K- mapmust have both of them. Diagonal cells are not adjacent. Example: Fill inthe Karnaughmap for the Booleanexpressionbelow, thenwrite the Boolean expressionfor the result. Note:There will be a 1 entered for each product term. The product term is the addressof the cell where the 1 is entered.
- 3. 3 KarnaughMapping Notes www.allaboutcircuits.com – Logic Simplification With Karnaugh Maps The logic simplification examples that we have done socould have been performed with Boolean algebra about as quickly. Realworldlogic simplificationproblems call for larger Karnaughmaps so that we may do serious work. We will work some contri ved examplesinthis section, leaving most of the real worldapplications for the CombinatorialLogic chapter. Bycontrived, we mean examples which illustrate techniques. This approach willdevelopthe tools we need to transitionto 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 topof the map. It is not inbinarysequence which would be 00, 01, 10, 11. It is 00, 01, 11 10, which is Gray code sequence. Graycode sequence onlychanges one binarybit as we go fromone number to the next inthe sequence, unlike binary. That meansthat adjacent cells will onlyvarybyone bit, or Booleanvariable. Thisis what we needto organize the outputs of a logic function so tha t we may view commonality. Moreover, the column androw headings must be inGraycode order, or the map will not work as a Karnaugh map. Cellssharing common Booleanvariables wouldno longer be adjacent, nor showvisual patterns. Adjacent cells varybyonlyone bit because a Graycode sequence varies byonlyone 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. Some more examples: Above we, place the 1’s inthe K-mapfor eachof the product terms, identifya groupof two, then write a p-term (product term) for the sole group as our simplified result. Mappingthe four product terms above yields a groupof four coveredbyBooleanA’. Mappingthe four p-terms yields a groupof four, whichis coveredbyone variable C. (Note that, crudelyspeaking, it’s not just horizontal groups or vertical groups of 1s that constitute a group, even2x2 groups, as longas theyhave powers of twoelements, canbe considered . One more example below consolidates it). After mappingthe six p-terms above, identifythe upper groupof four, pick upthe lower two cells as a group of four bysharing the two withtwo more from the other group. Covering these twowith a group offour gives a simpler result. Since there are two groups, there will be two p-terms in the Sum-of-Products result A’+B.
- 4. 4 KarnaughMapping Notes www.allaboutcircuits.com – Before we move to 4-variable K-maps, let’s revisit the Toxic waste incinerator example from previous chapter to juxtapose the two methods of simplifying Boolean expressions, andto show how K-maps expedite things (Myobservation:It’s not just the number of variables inthe equation, but the number of product terms inthe SOPexpressionthat make things difficult). Larger 4-variable Karnaugh Maps Knowinghowto generate Graycode shouldallowus to buildlarger maps. Actually, allwe needto dois lookat the left to right sequence across the topof the 3-variable map, andcopyit down the left side of the 4-variable map.See below. The following four variable Karnaughmaps illustrate reduction of Boolean expressions too tedious for Booleanalgebra. Reductions could be done with Boolean algebra. However, the Karnaugh map is faster and easier, especially if there are many logic reducti ons to do. The above Booleanexpressionhas sevenproduct terms. Theyare mappedtop to bottomandleft to right on the K-map above. For example, the first P-term A’B’CD is first row 3rd cell, correspondingto maplocation A=0, B=0, C=1, D=1. The other product terms are placed in a
- 5. 5 KarnaughMapping Notes www.allaboutcircuits.com – similar manner. Encircling the largest groups possible,two groups of four are shown above. The dashedhorizontal group corresponds to the simplified product term AB. The vertical groupcorresponds to BooleanCD. Since there are twogroups, there willbe two product terms in the Sum-Of-Products result of Out=AB+CD. Now look at three VERY important examples below: Example: Fold upthe corners ofthe map belowlike it is a napkin to make the four cells physicallyadjacent. The four cellsabove are a groupof four because theyall have the Booleanvariables B’ andD’ incommon. 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 caseswith respect to the four corner cells. Thus, these variables (A, B) are not involved with this group of four. This single groupcomes out of the mapas one product termfor the simplified result: Out=B’C’ Example: For the K-mapbelow, roll the topandbottomedges intoa cylinder forming eight adjacent cells. The above groupof eight has one Boolean variable incommon: B=0. Therefore, the one groupof eight is coveredbyone p-term: B’. The original eight term Boolean expression simplifiesto Out=B’ Example: The Boolean expressionbelow has nine p-terms, three of which have three Booleans insteadof four. The difference is that while four Booleanvariable product terms cover one cell, the three Booleanp- terms cover a pair of cells each. The six product terms of four Boolean variables mapinthe 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. Cellsinthe corners are sharedwith bothgroups. This is fine. Infact, 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’ Example: Below we map the unsimplified Boolean expression to the Karnaugh map. Above, three of the cells formintoa groups of twocells. A fourth cell cannot be combined with anything, which often happens in“real world” problems. In this case, the Booleanp-term ABCD is unchangedinthe simplification process. Result: Out= B’C’D’+A’B’D’+ABCD Example: Often times there is more thanone minimumcost solutionto a simplificationproblem. Suchis the case illustrated below. Both results above have four product terms of three Booleanvariables each. Bothare equallyvalid minimal cost solutions. Th e difference in the final solution is due to how the cells are groupedas shownabove. A minimal cost solutionis a validlogic design with the minimum number of gates with the minimum number of inputs.
- 6. 6 KarnaughMapping Notes www.allaboutcircuits.com – Example: Belowwe mapthe unsimplifiedBoolean equationas usual andforma groupof four as a first simplificationstep. It maynot be obvious how to pick up the remaining cells. Pick up three more cells in a group offour, center above. There are still twocells remaining. the minimalcost methodto pick upthose is to group themwith neighboring cells as groups of four as at above right. On a cautionarynote, donot attempt to form groups of three. Groupings must be powers of 2, that is, 1, 2, 4, 8 ... Example: Below we have another example of twopossible minimal cost solutions. Start by forming a couple of groups of four after mapping the cells. The two solutions depend onwhether the single remaining cell is groupedwiththe first or the secondgroup offour as a group oftwo 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. Example: Below we have anexample of a simplification using the Karnaughmap at left or Booleanalgebra at right. Plot C’ on the mapas the area ofall cells covered byaddressC=0, the 8-cellson the left ofthe map. Then, plot the single ABCD cell. That single cell forms a groupof 2-cell as shown, whichsimplifies to P-term ABD, for an end result of Out = C’ + ABD. This (above)is a rare example of a four variable problemthat canbe reduced withBooleanalgebra without a lot of work, assuming that you remember the theorems. Minterm vs Maxterm Solution So far we have beenfindingSum-Of-Product (SOP) solutions to logic reduction problems. For eachof these SOPsolutions, there is also a Product-Of-Sums solution (POS), which could be more useful, depending on the application. A minterm is a Booleanexpressionresultingin1 for the output of a single cell, and0s for all other cells in a Karnaughmap, or truth table. The illustrationabove left shows the minterm ABC, a single product term, as a single 1 in a map that is otherwise 0s. We have not shownthe 0s in our Karnaughmaps upto this point, as it is customaryto omit themunless specificallyneeded. Another minterm A’BC’ is shown above right. The point to review is that the address ofthe cell corresponds directlyto the minterm being mapped (A Boolean expression or map may have multiple minterms). A Boolean expression willmore oftenthan not consist of multiple minterms corresponding to multiple cells in a Karnaugh map as shown above. 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.
- 7. 7 KarnaughMapping Notes www.allaboutcircuits.com – Next we attack the Boolean functionwhichis 0 for a single celland1s for all others. A maxterm is a Booleanexpressionresulting in a 0 for the output of a single cellexpression, and 1s for all other cells inthe Karnaugh map, or truth table. The illustrationabove left shows the maxterm (A+B+C), a single sum term, as a single 0 in a map that is otherwise 1s. 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’) inthiscell (1,1,1) of the K-map as shown above. A Boolean Product-Of-Sums expressionor map mayhave multiple maxterms as shownabove. Maxterm(A+B+C) yields numeric 111 wh ich 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-mapsetup, what we are reallyinterestedinis showing howto write a Product-Of-Sums reduction. Form the 0s into groups. That wouldbe a groupof two below. Write the binaryvalue correspondingto the sum-term which is (0,0,X). BothA andB are 0 for the group. But, Cis both0 and 1 so we write anXas 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 writingthe Product-Of-Sums Booleanreductionfor a K-map: Form largest groups of0s possible, covering all maxterms. Groups must be a power of 2. Write binarynumeric value for group. Complement binarynumeric value for group. Convert complement value to a sum-term. Repeat steps for other groups. Eachgroupyields a sum-term withina Product-Of-Sums result. Example: Simplifythe Product-Of-Sums Booleanexpressionbelow, providing a result in POS form. Solution: Transfer the sevenmaxterms to the mapbelowas 0s. Be sure to complement the input variablesinfinding the proper cell location. We map the 0s as theyappear left to right topto bottom onthe mapabove. We locate the last three maxterms withleader lines. Once the cells are inplace above, form groups of cells as shownbelow. Larger groups will give a sum-term withfewer inputs. Fewer groups will yield fewer sum-terms in the result.
- 8. 8 KarnaughMapping Notes www.allaboutcircuits.com – We have three groups, sowe expect to have three sum-terms inour POS result above. The group of4-cells yields a 2-variable sum-term. The two groups of 2-cells give us two 3-variable sum-terms. Details are shownfor how we arrivedat the Sum-terms above. For a group, write the binarygroupinput 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 insteadof 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 inthe implied1s inthe remainingcellsof the mapabove right. Form groups of 1s to cover all 1s. Then write the Sum-Of-Products simplifiedresult as in the previous section of this chapter. This is identical to the below example. Above we show both the Product-Of-Sums solution, from the previous example, and the Sum-Of-Products solutionfrom the current problemfor comparison. Whichis the simpler solution?The POS uses 3-OR gates and 1-ANDgate, while the SOPuses 3-AND gatesand 1-OR gate. Both use four gates each. Taking a closer look, we count the number of gate inputs. The POS uses8-inputs;the SOPuses7-inputs. Bythe definition ofminimal cost solution, the SOPsolutionis simpler. Thisis anexample ofa technicallycorrect answer that is of little use in the real world. Example: Let us revisit a previous probleminvolving anSOPminimization. Produce a Product-Of-Sums solution. Compare the POS solutionto the previous SOP. Solution: Above left we have the originalproblemstarting witha 9-mintermBooleanunsimplifiedexpression. Reviewing, we formed four groups of 4-cellsto yielda 4-product-term SOPresult, lower left. Inthe middle figure, above, we fill in the emptyspaces with the implied 0s. The 0s form two groups of 4-cells. The solid blue groupis (A’+B), the dashedredgroup is (C’+D). This yields two sum-terms in the Product-Of-Sums result, above right Out = (A’+B)(C’+D). Comparing the previous SOPsimplification, left, to the POS simplification, right, shows that the POS is the least cost solution. The SOPuses 5-gates total, the POS uses only 3-gates. SUM AND PRODUCT NOTATION For reference, this sectionintroduces the terminologyusedinsome texts to describe the minterms andmaxterms assigned to a Karnaugh map.
- 9. 9 KarnaughMapping Notes www.allaboutcircuits.com – Σ (sigma) indicatessum and lower case “m” indicates minterms. Σm indicates sumof minterms. The following example is revisitedto illustrate our point. Instead of a Boolean equation description of unsimpl ified logic, we list the minterms. The numbers indicate cell location, or address, withina Karnaughmap as shown belowright. Thisis certainlya compact means of describing a list of minterms or cells in a K-map. The Sum-Of-Products solution is not affectedbythe newterminology. The minterms, 1s, inthe map have beengroupedas usual and a Sum-OF- Products solution written. Below, we show the terminologyfor describing a list of maxterms. Product is indicated by the Greek Π (pi), and upper case “M” indicates maxterms. ΠMindicates product of maxterms. The same example illustrates our point. The Boolean equationdescriptionof unsim plified logic, is replaced by a list of maxterms. Once again, the numbers indicate K-mapcell address locations. For maxterms this is the location of 0s, as shown below. A Product-OF-Sums solution is completed in the usual manner. DON’T CARE CELLS IN THE KARNAUGH MAP Up to this point we have consideredlogic reductionproblems where the input conditions were completely specified. That is, a 3-variable truth table or Karnaughmaphad2n = 23 or 8-entries, a full table or map. It is not always necessaryto fill inthe complete truth table for some real-world problems. We may have a choice to not fill in the complete table. For example, whendealingwith BCD (BinaryCodedDecimal) numbers encodedas four bits, we may not care about any codes above the BCD range of (0, 1, 2…9). The 4-bit binarycodes for the hexadecimal numbers (Ah, Bh, Ch, Eh, Fh)are not validBCD codes. Thus, we do not have to fill in th ose codes at the endof a truth table, or K-map, if we do not care to. We wouldnot normallycare to fill in those codes because those codes (1010, 1011, 1100, 1101, 1110, 1111) will never exist as long as we are dealing onlywith BCD encoded numbers. These six invalid codes are don’t cares as far as we are concerned. That is, we do not care what output our logic circuit produces for these don’t cares. Don’t cares ina Karnaugh map, or truth table, maybe either 1s or 0s, as long as we don’t care what the output is for aninput condition we never expect to see. We plot these cells with anasterisk, *, among the normal 1s and0s. Whenforming groups ofcells, treat the d on’t care cell as either a 1 or a 0, or ignore the don’t cares. This is helpful if it allows us to forma larger group than would otherwise be possible without the don’t cares. There is no requirement to group all or any of the don’t cares. Only use them in a group if it simplifies the logic. Above is anexample of a logic functionwhere the desired output is 1 for input ABC= 101 over the range from 000 to 101. We do not care what the output is for the other possible inputs (110, 111). Map those two as don’t cares. We show twosolutions. The solution on the right Out = AB’C is the more complex solutionsince we didnot use the don’t care cells. The solutionin the middle, Out=AC, is lesscomplex because we groupeda don’t care cell withthe single 1 to form a groupof two. The third solution, a Product-Of-Sums onthe right, results from grouping a don’t care withthree zeros formi ng a group of four 0s. This is the same, less complex, Out=AC. We have illustrated that the don’t care cells may be used as either 1s or 0s , whichever is useful. KARNAUGH MAPS - RULES OF SIMPLIFICATION (http://www.ee.surrey.ac.uk/Projects/Labview/minimisation/karrules.html) The Karnaughmap uses the following rules for the simplification ofexpressions by grouping together adjacent cellscontainingones
- 10. 10 KarnaughMapping Notes www.allaboutcircuits.com – To be Cleared Insteadof twooverlapping groups of 4 elements each, what if we have one of4 elements (2 x 2) and one of two (horizontal)? Associated Content We would not normallyresort to computer automationto simplifya three input logic block. We couldsooner solve the problem with pencil and paper. However, ifwe had seven ofthese problems to solve, sayfor a BCD (BinaryCodedDecimal) to sevensegment decoder, we might want to automate the process. A BCD to seven segment decoder generates the logic signalsto drive a sevensegment LED (light emitting diode) display. Examples of computer automateddesign languages for simplificationof logic are PALASM, ABEL, CUPL, Verilog, andVHDL. These p rograms accept a hardware descriptor language input file whichis basedonBooleanequations andproduce anoutput file describing a reduced (or simplified) Booleansolution. We will not require such tools in this chapter. Let’s move on to Venn diagrams as an introduction to Karnau gh maps.