Karnaugh Map Exercise

Exercise -

Karnaugh mapping

Question 1: Identify each of these logic gates by name, and complete their respective truth tables:

Question 2: A Karnaugh map is map is nothing more than a special form of truth table, useful for reducing logic functions into minimal Boolean expressions. Here is a truth table for a specific three-input logic circuit:

Complete the folloing !arnaugh map, according to the values found in the above truth table:

Question 3: A Karnaugh map is map is nothing more than a special form of truth table, useful for reducing logic functions into minimal Boolean expressions. Here is a truth table for a specific four-input logic circuit:

Complete the folloing !arnaugh map, according to the values found in the above truth table:

Question 4: Here is a truth table for a four-input logic circuit:

If we translate this truth table into a Karnaugh map, we obtain the following result:

Note how the only 1's in the map are clustered together in a group of four:

If you loo" at the input variables #A, B, C, and $%, you should notice that only to of them actually change ithin this cluster of four &'s. (he other to variables hold the same value for each of these conditions here the output is a )&). Identify hich variables change, and hich stay the same, for this cluster.

Question 6: Here is a truth table for a four-input logic circuit:

If we translate this truth table into a Karnaugh map, we obtain the following result:

Note how the only 1's in the map all exist on the same row:

If you loo" at the input variables #A, B, C, and $%, you should notice that only to of them are constant for each of the )&) conditions on the !arnaugh map. Identify these variables, and remember them.  *o, rite an + #+um-of-roducts% expression for the truth table, and use Boolean algebra to reduce that ra expression to its simplest form. hat do you notice about the simplified + expression, in relation to the common variables noted on the !arnaugh map/ 0eveal Anser  1or this cluster of four &'s, variables A and B are the only to inputs that remain constant for the four )&) conditions shon in the !arnaugh map. (he simplified Boolean expression for the truth table is AB. +ee a pattern here/ Notes: (his 2uestion strongly suggests to students that the !arnaugh map is a graphical method of achieving a reduced-form + expression for a truth table. nce students reali3e !arnaugh mapping holds the "ey to escaping arduous Boolean algebra simplifications, their interest ill be  pi2ued4

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Question 7: ne of the essential characteristics of Karnaugh maps is that the input !ariable se"uences are

always arranged in #ray code se"uence$ %hat is, you ne!er see a Karnaugh map with the input combinations arranged in binary order:

%he reason for this is apparent when we consider the use of Karnaugh maps to detect common !ariables in output sets$ &or instance, here we ha!e a Karnaugh map with a cluster of four 1's at the center:

rranged in this order, it is apparent that two of the input !ariables ha!e the same !alues for each of the four (high( output conditions$ )e-draw this Karnaugh map with the input !ariables se"uenced in binary order, and comment on what happens$ *an you still tell which input !ariables remain the same for all four output conditions+

0eveal Anser 

5oo"ing at this, e can still tell that B 6 & and $ 6 & for all four )high) output conditions, but this is not  apparent by proximity as it as before.

Notes: 7ou could simply tell your students that the input variables must be se2uenced according to 8ray code in order for !arnaugh mapping to or" as a simplification tool, but this ouldn't explain to students why it needs to be such. (his 2uestion shos students the purpose of 8ray code se2uencing in !arnaugh maps, by shoing them the alternative #binary se2uencing%, and alloing them to see ho the tas" of see"ing noncontradictory variables is complicated by it.

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Question 8: xamine this truth table and corresponding Karnaugh map:

%hough it may not be ob!ious from first appearances, the four (high( conditions in the Karnaugh map actually belong to the same group$ %o mae this more apparent, I will draw a new .o!ersi/ed0 Karnaugh map template, with the #ray code se"uences repeated twice along each axis:

1ill in this map ith the 9 and & values from the truth table, and then see if a grouping of four )high) conditions becomes apparent. 0eveal Anser 

1ollo-up 2uestion: hat does this problem tell us about grouping/ In other ords, ho can e identify groups of )high) states ithout having to ma"e oversi3ed !arnaugh maps/ Notes: (he concept of bit groups extending past the boundaries of a !arnaugh map tends to confuse students. In fact, it is about the only thing that tends to confuse students about !arnaugh maps4 +imply telling them to group past the borders of the map doesn't really teach them why the techni2ue is valid. Here, they should see ith little difficulty hy the techni2ue or"s. And, if for some reason they ust can't visuali3e bit groups past the boundaries of a !arnaugh map, they "no they can ust dra an oversi3ed map and it ill become obvious4

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Question 9:

A student is as"ed to use !arnaugh mapping to generate a minimal + expression for the folloing truth table: A

B

C

utput

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&

&ollowing the truth table shown, the student plots this Karnaugh map:

(%his is easy,( says the student to himself$ ll the '1' conditions fall within the same group2( %he student then highlights a triplet of 1's as a single group:

3ooing at this cluster of 1's, the student identifies * as remaining constant .10 for all three conditions in the group$ %herefore, the student concludes, the minimal expression for this truth table must simply be *$ Howe!er, a second student decides to use 4oolean algebra on this problem instead of Karnaugh mapping$ 4eginning with the original truth table and generating a 5um-of-6roducts .560 expression for it, the simplification goes as follows:

A

BC ; A

BC#

A

B

C ; ABC

; A% ; A

B

C

BC ; A

B

C#B ; A

C

B

%

C#B ; A%

AC ; BC

b!iously, the answer gi!en by the second student's 4oolean reduction .* 7 4*0 does not match the answer gi!en by the first student's Karnaugh map analysis .*0$ 6erplexed by the disagreement between these two methods, and failing to see a mistae in the 4oolean algebra used by the second student, the first student decides to chec his Karnaugh mapping again$ 8pon reflection, it becomes apparent that if the answer really were *, the Karnaugh map would loo different$ Instead of ha!ing three cells with 1's in them, there would be four cells with 1's in them .the output of the function being (1( any  time * 9 1:

+omehere, there must have been a mista"e made in the first student's grouping of &'s in the !arnaugh map, because the map shon above is the only one proper for an anser of C, and it is not the same as the real map for the given truth table.
6roper grouping of 1's in the Karnaugh map:

Notes: (he purpose of this 2uestion is to illustrate ho it is incorrect to identify clusters of arbitrary si3e in a !arnaugh map. A cluster of three, as seen in this scenario, leads to an incorrect conclusion. f course, one could easily 2uote a textboo" as to the proper numbers and patterns of &'s to identify in a !arnaugh map, but it is so much more informative #in my opinion% to illustrate by example. osing a dilemma such as this ma"es students think  about hy the anser is rong, rather than as"ing them to remember seemingly arbitrary rules.

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Question 10:

+tate the rules for properly identifying common groups in a !arnaugh map. 0eveal Anser  Any good introductory digital textboo" ill give the rules you need to do !arnaugh mapping. I leave you to research these rules for yourself4 Notes: (he anser spea"s for itself here - let your students research these rules, and as" them exactly here they found them #including the page numbers in their textboo"#s%4%.

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Question 11:  seven segment decoder  is a digital circuit designed to dri!e a !ery common type of digital display de!ice: a set of 3 .or 3*0 segments that render numerals ; through < at the command of a four-bit code:

(he behavior of the display driver IC may be represented by a truth table ith seven outputs: one for each segment of the seven-segment display #a through g%. In the folloing table, a )&) output represents an active display segment, hile a ) 9) output represents an inactive segment: $

C

B

A

a

b

c

d

e

f

g

$isplay

9

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)9)

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)&)

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)=)

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)>)

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)?)

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)@)

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))

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))

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))

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)D)

A real-life example such as this provides an exc ellent shocase for techni2ues such as !arnaugh mapping. 5et's ta"e output a for example, shoing it ithout all the other outputs included in the truth table:

$

C

B

A

a

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6lotting a Karnaugh map for output a, we get this result:

Identify ad=acent groups of 1's in this Karnaugh map, and generate a minimal 56 expression from those groupings$ Note that six of the cells are blan because the truth table does not list all the possible input combinations with four !ariables ., 4, *, and 0$ >ith these large gaps in the Karnaugh map, it is difficult to form large groupings of 1's, and thus the resulting (minimal( 56 expression has se!eral terms$ Howe!er, if we do not care about output a's state in the six non-specified truth table rows, we can fill in the remaining cells of the Karnaugh map with (don't care( symbols .usually the letter ?0 and use those cells as (wildcards( in determining groupings:

ith this ne !arnaugh map, identify adacent groups of &'s, and generate a minimal + expression from those groupings. 0eveal Anser 

Karnaugh map groupings with strict (1( groups:

$

B;

$

CA ; $

C B

;

C B A

!arnaugh map groupings ith )don't care) ildcards:

$ ; B ; CA ;

C A

1ollo-up 2uestion: this 2uestion and anser merely focused on the a output for the BC$-to-segment decoder circuit. Imagine if e ere to approach all seven outputs of the decoder circuit in these to fashions, first developing + expressions using strict groupings of )&) ou tputs, and then using )don't care) ildcards. hich of these to approaches do you suppose ould yield the simplest gate circuitry overall/ hat impact ould the to different solutions have on the decoder circuit's behavior for the six unspecified input combinations &9&9, &9&&, &&99, &&9&, &&&9, and &&&&/ Notes: ne of the points of this 2uestion is for students to reali3e that bigger groups are better, in that they yield simpler + terms. Also, students should reali3e that the ability to u se )don't care) states as )ildcard) placeholders in the !arnaugh map cells increases the chances of creating

 bigger groups. (ruth be "non, I chose a pretty bad example to try to ma"e an + expression from, since there are only to non-3ero output conditions out of ten4 1ormulating a + expression ould have  been easier, but that's a subect for another 2uestion4 Hide Anser 

Question 12:

hen designing a circuit to emulate a truth table such as this here nearly all the input conditions result in )&) output states, it is easier to use roduct-of-+ums #+% expressions rather than +um-of-roducts #+% expressions: A

B

C

utput

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9

Is it possible to use a !arnaugh map to generate the appropriate + expression for this truth table, or are !arnaugh maps limited to + expressions only/
than they tend to free3e hen presented ith a problem li"e this. ithout specific instructions on hat to do, the obvious steps of )try it and see) elude them. It is your charge as their instructor to encourage an experimental mindset among your students. $o not simply tell them ho to go about )discovering) the anser on their on, for if you do you ill rob them of an authentic discovery experience. Hide Anser 

Question 13:

Gse a !arnaugh map to generate a simple Boolean expression for this truth table, and dra a relay logic circuit e2uivalent to that expression: A

B

C

utput

9

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0eveal Anser  +imple expression and relay circuit:

B

C

Notes: ne of the things you may ant to have your students share in front of the class is their !arnaugh maps, and ho they grouped common output states to arrive at Boolean expression terms. I have found that an overhead #acetate% or computer-proected image of a blan" !arnaugh map on a hiteboard serves ell to present !arnaugh maps on. (his ay, cell entries may be easily erased and re-dran ithout having to re-dra the map #grid lines% itself. As" your students to compare using a !arnaugh map versus using standard +EBoolean simplifications to arrive at the simplest expression for this truth table. hich techni2ue ould they prefer to use, and hy/

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Question 14:

Gse a !arnaugh map to generate a simple Boolean expression for this truth table, and dra a gate circuit e2uivalent to that expression: A

B

C

$

utput

9

9

9

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&

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&

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0eveal Anser  +imple expression and gate circuit:

AC ; BC

$

Challenge 2uestion: use Boolean algebra techni2ues to simplify the table's ra + expression into minimal form ithout the use of a !arnaugh map. Notes: ne of the things you may ant to have your students share in front of the class is their !arnaugh maps, and ho they grouped common output states to arrive at Boolean expression terms. I have found that an overhead #acetate% or computer-proected image of a blan" !arnaugh map on a hiteboard serves ell to present !arnaugh maps on. (his ay, cell entries may be easily erased and re-dran ithout having to re-dra the map #grid lines% itself.

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Question 1:

Gse a !arnaugh map to generate a simple Boolean expression for this truth table, and dra a gate circuit e2uivalent to that expression:

A

B

C

$

utput

9

9

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&

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0eveal Anser  +imple expression and gate circuit:

AC ; A

B $

Challenge 2uestion: use Boolean algebra techni2ues to simplify the table's ra + expression into minimal form ithout the use of a !arnaugh map. Notes: ne of the things you may ant to have your students share in front of the class is their !arnaugh maps, and ho they grouped common output states to arrive at Boolean expression terms. I have found that an overhead #acetate% or computer-proected image of a blan" !arnaugh map on a hiteboard serves ell to present !arnaugh maps on. (his ay, cell entries may be easily erased and re-dran ithout having to re-dra the map #grid lines% itself. (his is one of those situations here an important group )raps around) the edge of the !arnaugh map, and thus is li"ely to be overloo"ed by students.

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Question 16:

Gse a !arnaugh map to generate a simple Boolean expression for this truth table, and dra a relay circuit e2uivalent to that expression: A

B

C

$

utput

9

9

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&

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0eveal Anser  +imple expression and relay circuit:

B

$

;

C $

1ollo-up 2uestion: although the relay circuit shon above does satisfy the minimal + Boolean expression, there is a ay to ma"e it simpler yet. Hint: done properly, you may eliminate one of the contacts in the circuit4 Challenge 2uestion: use Boolean algebra techni2ues to simplify the table's ra + expression into minimal form ithout the use of a !arnaugh map. Notes: ne of the things you may ant to have your students share in front of the class is their !arnaugh maps, and ho they grouped common output states to arrive at Boolean expression terms. I have found that an overhead #acetate% or computer-proected image of a blan" !arnaugh

map on a hiteboard serves ell to present !arnaugh maps on. (his ay, cell entries may be easily erased and re-dran ithout having to re-dra the map #grid lines% itself. (his is one of those situations here an important group )raps around) the edge of the !arnaugh map, and thus is li"ely to be overloo"ed by students. Hide Anser 

Question 17:

Gse a !arnaugh map to generate a simple Boolean expression for this truth table, and dra a relay circuit e2uivalent to that expression: A

B

C

$

utput

9

9

9

9

&

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