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I am new to K-maps. I am a bit confused on this. I have this K-Map that gives different min SoP and min PoS for the same map.

I have attached an image containing the K-Map and my calculations.

I know minSoP and minPoS should be the same, but I am not able to prove algebraically that min SoP of this K-Map = min PoS of this K-Map. I have checked my calculations many times but still I am at a loss. Please help.

The question is that we first have to find SoP and PoS independently and then prove they are same so we cannot use de-Morgan to find PoS from SoP, instead we need to group 0's and find PoS.enter image description here

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    \$\begingroup\$ You'll probably need to put variable names on that kmap \$\endgroup\$
    – Voltage Spike
    Aug 25, 2021 at 15:54
  • \$\begingroup\$ Please draw a proper K map, with labels for the inputs and their values for each row and column. Show us how you circled implicants. If you are trying to use Boolean algebra to demonstrate equivalence, show us all of your work. \$\endgroup\$ Aug 25, 2021 at 15:56
  • \$\begingroup\$ You probably aren't ordering the inputs on the rows and columns properly. It needs to follow gray code (i.e. only one bit changes at a time 00 01 11 10, that way you can actually group them to make sense, NOT 00 01 10 11). Hard to say what you did though when we can't see what you did. \$\endgroup\$
    – DKNguyen
    Aug 25, 2021 at 15:57
  • \$\begingroup\$ You need to expand and deduct PoS to SoP, not using Karnaugh as your prof said. The first expansion becomes: (a'c+ a'd+a' + bc+bd+a'b).(a+b+d').(c'+d+b'). If you don't get the same result as SoP, your SoP or PoS, or both were incorrect. \$\endgroup\$
    – jay
    Aug 25, 2021 at 16:17
  • \$\begingroup\$ Oh I thought it's a standard so I didn't include my working. I am attaching an image of the kmap. \$\endgroup\$
    – Alex
    Aug 25, 2021 at 16:56

1 Answer 1

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I've circled things up a bit, with red and blue:

enter image description here

For SOP (red), you have \$bd + \overline{a\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{c\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}d + \overline{a\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{b\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{c\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}} + \overline{b\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{c\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{d\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\$. But this can be reduced in two ways (easily seen in the above image):

  1. \$bd + \overline{a\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{b\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{c\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}} + \overline{b\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{c\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{d\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\$
  2. \$bd + \overline{a\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{c\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}d + \overline{b\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{c\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\overline{d\vphantom{bd + \overline{a}\overline{c}d + \overline{a}\overline{b}\overline{c} + \overline{b}\overline{c}\overline{d}}}\$

For POS (blue), you have \$\overline{b\overline{d} + \overline{b}c + a\overline{b}d+c\overline{d}+\overline{b}cd}\$ but the term, \$\overline{b}cd\$, is completely unnecessary. But if you look closely there, you also can see that the term, \$c\overline{d}\$, is also unnecessary. So all you need is:

  1. \$\overline{b\overline{d} + \overline{b}c + a\overline{b}d}\$

Is this sufficient help?

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    \$\begingroup\$ yes yes thanks a lot \$\endgroup\$
    – Alex
    Aug 26, 2021 at 2:31
  • \$\begingroup\$ One doubt. if I expand 3. will I get exp-1 or exp-2? I am getting something different :( \$\endgroup\$
    – Alex
    Aug 26, 2021 at 21:32
  • \$\begingroup\$ I am getting an extra term of bc'd + a'c'd after simplifying exp-3 \$\endgroup\$
    – Alex
    Aug 26, 2021 at 21:40
  • \$\begingroup\$ @Alex You should get (b' + d)(b + c')(a' + b + d'), yes? \$\endgroup\$
    – jonk
    Aug 27, 2021 at 3:36
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    \$\begingroup\$ ahhh got it now, I was actually simplifying blindly \$\endgroup\$
    – Alex
    Aug 27, 2021 at 5:37

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