Karnaugh Maps

• Source : Carpinelli 2.3.1 (page 2-18), Kumar chapter 3 and others
• Karnaugh maps or K-maps
• These are
  • A different way to represent truth tables.
  • A method for reducing boolean expressions.
• Two input Karnaugh maps

  • a	b	mi
    0	0	m0
    0	1	m1
    1	0	m2
    1	1	m3

  • The equivelent K-map is

    a\b	0	1
    0	m0	m1
    1	m2	m3

  • So the table

    a	b	f(a,b)
    0	0	0
    0	1	1
    1	0	1
    1	1	0

  • Becomes the K-map

    a\b	0	1
    0	0	1
    1	1	0

• Three input K-Maps

  • a	b	c	mi
    0	0	0	m0
    0	0	1	m1
    0	1	0	m2
    0	1	1	m3
    1	0	0	m4
    1	0	1	m5
    1	1	0	m6
    1	1	1	m7

  • The equivelent K-map is

    a\bc	00	01	11	10
    0	m0	m1	m3	m2
    1	m4	m5	m7	m6

  • Note the use of gray code on the lables of the columns
    • This is because we only want things to change by 1.
    • You will see in a moment.
• The four input K-Map

  • ab\cd	00	01	11	10
    00	m0	m1	m3	m2
    01	m4	m5	m7	m6
    11	m12	m13	m15	m14
    10	m8	m9	m11	m10

• Higher level K-maps are possible, but working with them is inpractical.
• Represent f(a,b,c,d) = ab + acd + abcd as a K-map.
• Reducing expressions with K-Maps.
• 2 input maps.
  • First, be aware that this can be done other ways.
  • 2 input expressions are in some sense "easy"
  • Consider f(a,b) = m₀ + m₂ + m₃
  • This produces the K-map

    a\b	0	1
    0	1	0
    1	1	1

  • Circle the adjacent cells
    • 
    • I can't figure out how to circle these in html, so I used highlight
    • Note there are two of these
      • m₀ + m₂ in yellow
      • m₂ + m₃ in green
    • Each of these is called in an implicant.
    • For this technique, every 1 must be in an implicant.
    • But we try to build the largest rectangular implicants possible.
    • And later, implicants can "wrap" around the edges.
  • Look at the yellow implicant, I₁ for convience.
    • Notice that it doesn't matter what a is,
    • This "pair" is produced by the expression b

    • I1 = a·b + ab
         = b(a + a)
         =  b

    • But we could just read that off the table as well.
  • Look at the gree implicant, I₂ for conveienance.
    • In this case, it doesn't matter what b is.

    • I2 = a·b + ab
         = a(b + b)
         = a 

    • so I₂ = a
  • And the entire expression F(a,b) = a + b
  • Try the expression F(a,b) = m₀ + m₁ + m₃
  • Try the expression F(a,b) = m₀ + m₃