How to Simplify a Boolean Reduction Problem?

  • Thread starter DethRose
  • Start date
In summary: The boolean reduction is (ab)(\overline c+\overline d)+(a+b)(\overline c\overline d). The revised formula is (ab)(\overline c+\overline d)+(a+b)(\overline c\overline d) (the plus sign means the input is inverted). Andrew, you are a bit confused by your symbology. An AB means A and B are combined, while an AB+ means A and B are combined and the bar on top means the input is inverted. The star is just to keep the bars over top of the correct letter so you know which ones are inverted. And Andrew, to learn how
  • #1
DethRose
101
0
Ive been working on this problem for hours and its due on monday i really need help :bugeye: !

heres the boolean reduction i have to do
[tex](ab)[/tex][tex]([/tex][tex]\overline c[/tex][tex]+[/tex][tex]\overline d[/tex][tex])[/tex][tex]+[/tex][tex](a+b)[/tex][tex]([/tex][tex]\overline c[/tex][tex]\overline d[/tex][tex])[/tex]

heres what I've done
***_******_ _**_ _
AB C+ABD+AC D+BC D
***_***_ _
AB(C+D)+C D (A+B)

i have no idea where to go from there...thanks for you help

the stars are just to keep the bars over top of the correct letter so you know which ones are inverted

andrew
 
Last edited:
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  • #2
I am a bit confused by your symbology.

I assume + equal to AND , the bar should be NOT but what is meant by AB?

look here to learn how to do this

[tex] AB \overline {C} [/tex]
 
  • #3
Correction:

AB = A and B
A+B = A or B.

That seems to be the archaic form according to various websites. But that is the form that I was familiar with along time ago.
 
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  • #4
when the numbers are right beside each other they are anded together..and when there is a plus sign it means they are ord...thats the way I've been taught to represent them haha...the bar on top means the input is inverted
 
  • #5
heres the revised formula haha

[tex](ab)[/tex][tex]([/tex][tex]\overline c[/tex][tex]+[/tex][tex]\overline d[/tex][tex])[/tex][tex]+[/tex][tex](a+b)[/tex][tex]([/tex][tex]\overline c[/tex][tex]\overline d[/tex][tex])[/tex]
 
  • #6
DethRose said:
heres the revised formula haha

[tex](ab)[/tex][tex]([/tex][tex]\overline c[/tex][tex]+[/tex][tex]\overline d[/tex][tex])[/tex][tex]+[/tex][tex](a+b)[/tex][tex]([/tex][tex]\overline c[/tex][tex]\overline d[/tex][tex])[/tex]


[tex]\overline{C} + \overline{D}=\overline{CD}[/tex]

[tex](A + B)\overline{CD} + (AB)\overline {CD} = \overline{CD}(AB + A +B ) [/tex]

[tex] A + B + AB = A + B [/tex]

[tex]AB(\overline{C} + \overline{D}) + (A+B)(\overline{C}\overline{D}) = (A+B)\overline{CD}[/tex]


ehild
 
  • #7
that doesn't work because the cd in brackets doesn't have a whole bar on top of it...it is split in half
 
  • #8
Correctly written, original problem should be:
[tex]AB(\overline{C} + \overline{D}) + (A+B)(\overline{C}\,\overline{D})[/tex]

What if one lets:

[tex]W=AB[/tex]

[tex]X=\overline{C}+\overline{D}[/tex]

[tex]Y=A+B[/tex]

[tex]Z=\overline{C}\,\overline{D}[/tex]


Then the above original expression becomes

(WX) + (YZ)

((WX)+Y)((WX)+Z)

(W+Y)(X+Y)(W+Z)(X+Z)

or

[tex]\Big({(AB)+(A+B)}\Big) \Big({\overline{C}+\overline{D}+A+B}\Big)\Big({AB\overline{C}\,\overline{D}}\Big)\Big(({\overline{C}+\overline{D})\overline{C}\,\overline{D}}\Big)[/tex]

then make use of commutative and associative properties and

a(a+b) = a+(ab) = a

so for example

(AB)+(A+B) = ((AB)+A)+B = A+B

You'll need to do a sanity check on the intermedate relationships (W, X, Y, Z). It's been awhile since I have done this.
 
  • #9
i have no idea what you are doing there...there has to be a way to do it without having to replace the variables and just using the boolean rules i think
 

1. What is a Boolean Reductions problem?

A Boolean Reductions problem is a type of computational problem that involves finding the most efficient way to reduce a given Boolean expression to its simplest form.

2. Why are Boolean Reductions problems important?

Boolean Reductions problems are important because they are used to solve a wide range of real-world problems, such as circuit design, database optimization, and algorithm development. They also have applications in fields like computer science, mathematics, and engineering.

3. What are some common methods for solving Boolean Reductions problems?

Some common methods for solving Boolean Reductions problems include Boolean algebra, truth tables, Karnaugh maps, and Quine-McCluskey method. These methods use logical operations and rules to simplify Boolean expressions and find the most efficient solution.

4. What is the difference between a Boolean Reductions problem and a Boolean satisfiability problem?

A Boolean Reductions problem involves simplifying a given Boolean expression to its simplest form, while a Boolean satisfiability problem involves determining whether a given Boolean expression can be satisfied by assigning true or false values to its variables. In other words, Boolean satisfiability problems focus on finding a solution, while Boolean Reductions problems focus on finding the most efficient solution.

5. How can Boolean Reductions problems be applied to real-world scenarios?

Boolean Reductions problems can be applied to real-world scenarios in various ways, such as optimizing circuit design to reduce power consumption, simplifying logical conditions in computer programs to improve performance, and minimizing the size of databases for efficient storage and retrieval. They can also be used in decision-making processes, such as determining the most efficient route for a delivery truck.

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