Why Am I Getting an Extra Area in My Karnaugh Map Simplification?

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Discussion Overview

The discussion revolves around the simplification of a Boolean expression using Karnaugh maps (K-maps). Participants explore the discrepancies in the areas represented on the K-map and the resulting simplified equations derived from it. The context includes homework-related problem-solving and technical reasoning regarding Boolean algebra.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a Boolean expression and expresses confusion over obtaining an extra area in their K-map simplification.
  • Another participant requests clarification on the initial problem statement and the approach taken to solve it.
  • A third participant provides a simplified equation derived from the K-map and suggests that there is further simplification possible using the Consensus Theorem.
  • A later reply indicates that the advice received was helpful and inquires about software options for visual modeling of the function.

Areas of Agreement / Disagreement

There is no clear consensus on the correct simplification or the presence of the extra area in the K-map, as participants present differing equations and approaches without resolving the discrepancies.

Contextual Notes

Participants do not fully agree on the simplification steps or the interpretation of the K-map results, and there are unresolved aspects regarding the application of the Consensus Theorem and its implications for the original equation.

Who May Find This Useful

Students and individuals interested in Boolean algebra, K-map simplification techniques, and those seeking software tools for visual modeling of logical functions.

Koopan
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Hi,
I'm new to this forum.
My problem led me here.

Homework Statement


I need to solve this, using simplifying rules.
∫ = Ʃ(0,5,6,7,13)

But I always get one extra area when checking with Karnaugh map, which isn't necessary.

Homework Equations


∫() = a' b' c' d' + a' b c' d + a' b c d' + a' b c d + a b c' d


The Attempt at a Solution


Well I get always this as a solution, no matter the order i simplify

∫() = a' b' c' d' + b c' d + a' b c + a' b d


Now when applied to Karnaugh map the areas (equation parts from left to right 1-4) 1, 2, 3 are the solution I get when only using the map. But 4 overlaps with 2 and 3.

http://s8.postimage.org/dbo6zqrab/img.jpg
img.jpg


If anyone has time to help me out, then please do.

Thanks
 
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Koopan said:
I need to solve this, using simplifying rules.
\int = \sum (0,5,6,7,13)
Can you explain what this means, and how you went about solving it? I might catch on to what you are doing.
 
The k-map produces the right result: The simplified equation is:

a' b c +b c' d + a' b' c' d'

There is more you can do with your equation

a' b' c' d' + b c' d + a' b c + a' b d

Hint: Consensus Theorem :)
 
Thank you Ivan92 your advice helped me solve this.

Now as I'm here does anyone know any free or opensource software for modelling this function visually?
 

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