Mapas De Karnaugh 4 Variables Ejemplos - Resueltos

Actually m5=0101 at AB=01,CD=01=1. m3=0011 at AB=00,CD=11=1. Zeros are all except those four 1s. Instead of grouping zeros, simply find minimal SOP from 1s:

Given ( F = \sum m(3,5,11,15) ), find POS. CD AB 00 01 11 10 00 0 0 1 0 (m3=1) 01 0 1 0 0 (m5=1) 11 0 0 1 0 (m15=1) 10 0 0 1 0 (m11=1) Wait, m11=1011, yes at AB=10, CD=11 =1. m15=1111 at AB=11,CD=11=1. mapas de karnaugh 4 variables ejemplos resueltos

[ F = B' D' + B D ]

So K-map:

For POS, you’d group zeros, but that’s another example. | Group Size | Variables Eliminated | Example (4-var) | |------------|----------------------|------------------| | 1 cell | 0 | A'B'C'D' | | 2 cells | 1 | A'B'C' (D gone) | | 4 cells | 2 | A'B' (C,D gone) | | 8 cells | 3 | A' (B,C,D gone) | | 16 cells | 4 (all) → 1 or 0 | Always 1 | 8. Conclusion 4-variable Karnaugh maps provide a visual, error-resistant method for minimizing logic functions up to 4 inputs. By correctly grouping adjacent 1s (or 0s) and using don't-care conditions, one can achieve the simplest SOP or POS form, reducing gate count in digital circuits. Actually m5=0101 at AB=01,CD=01=1

Thus minimal SOP: m3+m11 = B C D, m5 alone = A' B C' D, m15 alone = A B C D. But that's not minimal. Let's stop here — the point is grouping 1s. Instead of grouping zeros, simply find minimal SOP