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Simpler sum of products from boolean algebra than from karnaugh map

Boolean Algebra simplification - odd number termsHow can I get a product-of-sums from this sum-of-products?Boolean Simplification for KmapFinding a simple function for a given Karnaugh diagramWhat if I am not given the labels of a Karnaugh map?Do 'sum-of-products' and 'product-of-sums' represent the same function?Minimizing using a Karnaugh map when given as subscripts F4,2655boolean algebra simplify using kmapCircuit to Karnaugh mapConverting Boolean equations to canonical sum of minterms with different number of variables in each minterm?

0

$begingroup$

I was given a question, simplify the expression represented by the sum of minterms 0,1,3 and 7 for the 3 parameter function f. Writing this out I got $f = A'B'C' + A'B'C + A'BC + ABC = A'B' + BC$. However, using a karnaugh map

BC 00 10 11 01
A
0 1 0 1 1
1 0 0 1 0

I found 3 groups of 2 ones, which gave me the expression $A'B' + A'C + BC$. I know the two expressions are equivalent, but I was under the impression that a karnaugh map would produce the simplest possible sum of products, which it demonstrably did not do here. Did I make a mistake somewhere or are my assumptions wrong?

Note: the second answer was also the answer given.

boolean-algebra

share|cite|improve this question

edited Mar 17 at 22:12

Ben

asked Mar 17 at 21:04

BenBen

264

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have a mistake in the Karnaugh map. The term A'B'C is misplaced.
  $endgroup$
  – Jens
  Mar 17 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Whoops, thank you. I had B'C and BC' reversed from my scratch paper. Fixed now.
  $endgroup$
  – Ben
  Mar 17 at 22:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Karnaugh now agrees with the first simplification.
  $endgroup$
  – Jens
  Mar 17 at 22:18
  

  
  
  
  

add a comment | 

0

$begingroup$

I was given a question, simplify the expression represented by the sum of minterms 0,1,3 and 7 for the 3 parameter function f. Writing this out I got $f = A'B'C' + A'B'C + A'BC + ABC = A'B' + BC$. However, using a karnaugh map

BC 00 10 11 01
A
0 1 0 1 1
1 0 0 1 0

I found 3 groups of 2 ones, which gave me the expression $A'B' + A'C + BC$. I know the two expressions are equivalent, but I was under the impression that a karnaugh map would produce the simplest possible sum of products, which it demonstrably did not do here. Did I make a mistake somewhere or are my assumptions wrong?

Note: the second answer was also the answer given.

boolean-algebra

share|cite|improve this question

edited Mar 17 at 22:12

Ben

asked Mar 17 at 21:04

BenBen

264

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have a mistake in the Karnaugh map. The term A'B'C is misplaced.
  $endgroup$
  – Jens
  Mar 17 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Whoops, thank you. I had B'C and BC' reversed from my scratch paper. Fixed now.
  $endgroup$
  – Ben
  Mar 17 at 22:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Karnaugh now agrees with the first simplification.
  $endgroup$
  – Jens
  Mar 17 at 22:18
  

  
  
  
  

add a comment | 

$begingroup$

I was given a question, simplify the expression represented by the sum of minterms 0,1,3 and 7 for the 3 parameter function f. Writing this out I got $f = A'B'C' + A'B'C + A'BC + ABC = A'B' + BC$. However, using a karnaugh map

BC 00 10 11 01
A
0 1 0 1 1
1 0 0 1 0

I found 3 groups of 2 ones, which gave me the expression $A'B' + A'C + BC$. I know the two expressions are equivalent, but I was under the impression that a karnaugh map would produce the simplest possible sum of products, which it demonstrably did not do here. Did I make a mistake somewhere or are my assumptions wrong?

Note: the second answer was also the answer given.

boolean-algebra

share|cite|improve this question

edited Mar 17 at 22:12

Ben

asked Mar 17 at 21:04

BenBen

264

$endgroup$

share|cite|improve this question

edited Mar 17 at 22:12

Ben

asked Mar 17 at 21:04

BenBen

264

share|cite|improve this question

edited Mar 17 at 22:12

Ben

asked Mar 17 at 21:04

BenBen

264

asked Mar 17 at 21:04

BenBen

264

asked Mar 17 at 21:04

BenBen

264

1 Answer
1

active

oldest

votes

0

$begingroup$

I looked up k-maps and saw that you edited your to reflect the proper Karnaugh map.
$$
beginarrayc
& B'C' & B'C & BC & BC' \
hline
A' & 1 & 1 & 1 & 0 \
A & 0 & 0 & 1 & 0
endarray
$$

I noticed that whenever neither $A$ nor $B$ the result is $1$. Ergo:
$$A'B'$$

Next, I also noticed that whenever $B$ and $C$ are both true, the result is $1$. Ergo:
$$A'B' + BC$$

share|cite|improve this answer

edited Mar 20 at 21:57

mwt

953416

answered Mar 20 at 20:34

NeilNeil

1013

$endgroup$

add a comment | 

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0

$begingroup$

I looked up k-maps and saw that you edited your to reflect the proper Karnaugh map.
$$
beginarrayc
& B'C' & B'C & BC & BC' \
hline
A' & 1 & 1 & 1 & 0 \
A & 0 & 0 & 1 & 0
endarray
$$

I noticed that whenever neither $A$ nor $B$ the result is $1$. Ergo:
$$A'B'$$

Next, I also noticed that whenever $B$ and $C$ are both true, the result is $1$. Ergo:
$$A'B' + BC$$

share|cite|improve this answer

edited Mar 20 at 21:57

mwt

953416

answered Mar 20 at 20:34

NeilNeil

1013

$endgroup$

add a comment | 

0

$begingroup$

I looked up k-maps and saw that you edited your to reflect the proper Karnaugh map.
$$
beginarrayc
& B'C' & B'C & BC & BC' \
hline
A' & 1 & 1 & 1 & 0 \
A & 0 & 0 & 1 & 0
endarray
$$

I noticed that whenever neither $A$ nor $B$ the result is $1$. Ergo:
$$A'B'$$

Next, I also noticed that whenever $B$ and $C$ are both true, the result is $1$. Ergo:
$$A'B' + BC$$

share|cite|improve this answer

edited Mar 20 at 21:57

mwt

953416

answered Mar 20 at 20:34

NeilNeil

1013

$endgroup$

add a comment | 

$begingroup$

I looked up k-maps and saw that you edited your to reflect the proper Karnaugh map.
$$
beginarrayc
& B'C' & B'C & BC & BC' \
hline
A' & 1 & 1 & 1 & 0 \
A & 0 & 0 & 1 & 0
endarray
$$

I noticed that whenever neither $A$ nor $B$ the result is $1$. Ergo:
$$A'B'$$

Next, I also noticed that whenever $B$ and $C$ are both true, the result is $1$. Ergo:
$$A'B' + BC$$

share|cite|improve this answer

edited Mar 20 at 21:57

mwt

953416

answered Mar 20 at 20:34

NeilNeil

1013

$endgroup$

share|cite|improve this answer

edited Mar 20 at 21:57

mwt

953416

answered Mar 20 at 20:34

NeilNeil

1013

edited Mar 20 at 21:57

mwt

953416

edited Mar 20 at 21:57

mwt

953416

answered Mar 20 at 20:34

NeilNeil

1013

answered Mar 20 at 20:34

NeilNeil

1013