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I've tried for many times to simplify the five variable boolean algebra but can't get it done.

Question Here is the question

$$F = \bar{A}\bar{B}\bar{C}\bar{D}E + \bar{A}\bar{B}C\bar{D}E + \bar{A}B\bar{C}\bar{D}E + \bar{A}BC\bar{D}\bar{E} + \bar{A}BC\bar{D}E + A\bar{B}\bar{C}DE + AB\bar{C}D\bar{E} + ABC\bar{D}\bar{E} + ABCD\bar{E} + ABCDE$$

Solution Here is the solution I've tried

I have seen simplified form using k-map. But it isn't matching with mine solution. Here is the k-map simplification ,y = A'D'E + BCD'E' + ABDE' + ABCD + AB'C'DE . I've tried many times but in vain . Have also tried many online simplifiers but they doesn't seem to work with 5 variables.

Dan
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Asir Shahriar Roudra
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1 Answers1

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If transcribed your question correctly, you want a function of 5 Boolean variables $F(A, B, C, D, E)$ that matches precisely these ten $ABCDE$ combinations: $\{00001, 00101, 01001, 01100, 01101, 10011, 11010, 11100, 11110, 11111\}$.

There are $3^5 = 243$ possible AND expressions using these 5 variables: Each one can be either true, false, or ignored. (Including one combinations with all 5 variables ignored, which is useless.)

Brute-forcing these by computer, I found 20 combinations that are true only if $F$ is true (but not necessarily vice-versa). These are the 10 five-variable terms from the original question, plus these 10:

  • $\bar{A}\bar{D}E$
  • $ABCD$
  • $\bar{A}BC\bar{D}$
  • $ABC\bar{E}$
  • $ABD\bar{E}$
  • $\bar{A}B\bar{D}E$
  • $\bar{A}\bar{B}\bar{D}E$
  • $\bar{A}C\bar{D}E$
  • $\bar{A}\bar{C}\bar{D}E$
  • $BC\bar{D}\bar{E}$

We'd prefer not to use all of them, though, so let's see how few we can get away with.

Brute-forcing all $2^{20}$ = 1,048,576 possible ORed combinations of presence/absence of the 20 matching ANDed terms, we get 187,245 valid ways of writing $F$, which I obviously won't list individually. But it turns out that there is one unique combination that minimizes the total number of AND+OR logic gates, with 19 of them:

$$ABCD + A\bar{B}\bar{C}DE + ABD\bar{E} + \bar{A}\bar{D}E + BC\bar{D}\bar{E}$$

Note that this is the same as “the k-map simplification” at the end of your post.

You seem to have made an algebra error somewhere. I evaluated the truth tables for all 32 possible $ABCDE$ combinations, and found that your solution incorrectly matches $AB\bar{C}DE$ while failing to match $AB\bar{C}D\bar{E}$.

Dan
  • 14,978