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CENG 342 Digital Systems Tabular Minimization Larry Pyeatt SDSM&T Tabular Minimization Karnaugh Maps are good for up to six input variables, but cannot be extended beyond that. Karnaugh Maps are not easily implemented in a computer

  1. CENG 342 – Digital Systems Tabular Minimization Larry Pyeatt SDSM&T

  2. Tabular Minimization Karnaugh Maps are good for up to six input variables, but cannot be extended beyond that. Karnaugh Maps are not easily implemented in a computer program. The tabular method 1 overcomes both of these limitations. 1 Also known as the Quine-McCluskey method, because it was developed by Willard V. Quine and extended by Edward J. McCluskey between 1950 and 1956.

  3. From Truth Table to Minterms A B C D f 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 � F ( A , B , C , D ) = m ( 4 , 8 , 10 , 11 , 12 , 15 ) + d ( 9 , 14 ) 0 1 1 0 0 0 1 1 1 0 1 0 0 0 1 Recall that d stands for don’t-care. The output does 1 0 0 1 d not matter in those cases. 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 0 1 1 1 0 d 1 1 1 1 1

  4. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Group the minterms by how many bits are set to 1. Number of 1s Term Binary Size 2 m 4 0100 1 m 8 1000 ( m 9 ) 1001 2 m 10 1010 m 12 1100 m 11 1011 3 ( m 14 ) 1110 4 m 15 1111

  5. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Compare each term in group 1 to each term in group 2. Combine terms m 12 and m 4. (Only one bit differs) Number of 1s Term Binary Size 2 m ( 4 , 12 ) m 4 0100 � -100 1 m 8 1000 ( m 9 ) 1001 2 m 10 1010 m 12 1100 � m 11 1011 3 ( m 14 ) 1110 4 m 15 1111

  6. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Combine terms m 8 and m 9. (Only one bit differs) Number of 1s Term Binary Size 2 m ( 4 , 12 ) m 4 0100 � -100 1 m ( 8 , 9 ) m 8 1000 � 100- ( m 9 ) 1001 � 2 m 10 1010 m 12 1100 � m 11 1011 3 ( m 14 ) 1110 4 m 15 1111

  7. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Combine terms m 8 and m 10. (Only one bit differs) Number of 1s Term Binary Size 2 m ( 4 , 12 ) m 4 0100 � -100 1 m ( 8 , 9 ) m 8 1000 � 100- m ( 8 , 10 ) 10-0 ( m 9 ) 1001 � 2 m 10 1010 � m 12 1100 � m 11 1011 3 ( m 14 ) 1110 4 m 15 1111

  8. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Combine terms m 8 and m 12. (Only one bit differs) Number of 1s Term Binary Size 2 m ( 4 , 12 ) m 4 0100 � -100 1 m ( 8 , 9 ) m 8 1000 � 100- m ( 8 , 10 ) 10-0 ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 2 m 10 1010 � m 12 1100 � m 11 1011 3 ( m 14 ) 1110 4 m 15 1111

  9. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Compare each term in group 2 to each term in group 3. Combine terms m 9 and m 11. (Only one bit differs) Number of 1s Term Binary Size 2 m ( 4 , 12 ) m 4 0100 � -100 1 m ( 8 , 9 ) m 8 1000 � 100- m ( 8 , 10 ) 10-0 ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 m 11 1011 � 3 ( m 14 ) 1110 4 m 15 1111

  10. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Combine terms m 10 and m 11. (Only one bit differs) Number of 1s Term Binary Size 2 m ( 4 , 12 ) m 4 0100 � -100 1 m ( 8 , 9 ) m 8 1000 � 100- m ( 8 , 10 ) 10-0 ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 m ( 10 , 11 ) 101- m 11 1011 � 3 ( m 14 ) 1110 4 m 15 1111

  11. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Combine terms m 10 and m 14. (Only one bit differs) Number of 1s Term Binary Size 2 m ( 4 , 12 ) m 4 0100 � -100 1 m ( 8 , 9 ) m 8 1000 � 100- m ( 8 , 10 ) 10-0 ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 m ( 10 , 11 ) 101- m 11 m ( 10 , 14 ) 1011 � 1-10 3 ( m 14 ) 1110 � 4 m 15 1111

  12. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Combine terms m 12 and m 14. (Only one bit differs) Number of 1s Term Binary Size 2 m ( 4 , 12 ) m 4 0100 � -100 1 m ( 8 , 9 ) m 8 1000 � 100- m ( 8 , 10 ) 10-0 ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 m ( 10 , 11 ) 101- m 11 m ( 10 , 14 ) 1011 � 1-10 3 ( m 14 ) m ( 12 , 14 ) 1110 � 11-0 4 m 15 1111

  13. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Compare each term in group 3 to each term in group 4. Combine terms m 11 and m 15. (Only one bit differs) Number of 1s Term Binary Size 2 m ( 4 , 12 ) m 4 0100 � -100 1 m ( 8 , 9 ) m 8 1000 � 100- m ( 8 , 10 ) 10-0 ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 m ( 10 , 11 ) 101- m 11 m ( 10 , 14 ) 1011 � 1-10 3 ( m 14 ) m ( 12 , 14 ) 1110 � 11-0 4 m 15 � m ( 11 , 15 ) 1111 1-11

  14. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Combine terms m 14 and m 15. (Only one bit differs.) All groups in this column have been combined. Start a new column. Number of 1s Term Binary Size 2 Size 4 m ( 4 , 12 ) m 4 0100 � -100 1 m ( 8 , 9 ) m 8 1000 � 100- m ( 8 , 10 ) 10-0 ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 m ( 10 , 11 ) 101- m 11 m ( 10 , 14 ) 1011 � 1-10 3 ( m 14 ) m ( 12 , 14 ) 1110 � 11-0 4 m 15 � m ( 11 , 15 ) 1111 1-11 m ( 14 , 15 ) 111-

  15. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Compare each term in group 1 to each term in group 2. Combine terms m ( 8 , 9 ) and m ( 10 , 11 ) . (Dashes match and only one bit differs) Number of 1s Term Binary Size 2 Size 4 m ( 4 , 12 ) m ( 8 , 9 , 10 , 11 ) m 4 0100 � -100 10-- 1 m ( 8 , 9 ) m 8 1000 � 100- � m ( 8 , 10 ) 10-0 ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 m ( 10 , 11 ) 101- � m 11 m ( 10 , 14 ) 1011 � 1-10 3 ( m 14 ) m ( 12 , 14 ) 1110 � 11-0 4 m 15 � m ( 11 , 15 ) 1111 1-11 m ( 14 , 15 ) 111-

  16. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Combine terms m ( 8 , 10 ) and m ( 9 , 11 ) . (Dashes match and only one bit differs) The term m ( 8 , 9 , 10 , 11 ) is already listed, so don’t add it again, but do put a checkmark by the terms m ( 8 , 10 ) and m ( 9 , 11 ) Number of 1s Term Binary Size 2 Size 4 m ( 4 , 12 ) m ( 8 , 9 , 10 , 11 ) m 4 0100 � -100 10-- 1 m ( 8 , 9 ) m 8 1000 � 100- � m ( 8 , 10 ) 10-0 � ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 � m ( 10 , 11 ) 101- � m 11 m ( 10 , 14 ) 1011 � 1-10 3 ( m 14 ) m ( 12 , 14 ) 1110 � 11-0 4 m 15 � m ( 11 , 15 ) 1111 1-11 m ( 14 , 15 ) 111-

  17. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Combine terms m ( 8 , 10 ) and m ( 12 , 14 ) . (Dashes match and only one bit differs) Number of 1s Term Binary Size 2 Size 4 m ( 4 , 12 ) m ( 8 , 9 , 10 , 11 ) m 4 0100 � -100 10-- 1 m ( 8 , 9 ) m ( 8 , 10 , 12 , 14 ) m 8 1000 � 100- � 1--0 m ( 8 , 10 ) 10-0 � ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 � m ( 10 , 11 ) 101- � m 11 m ( 10 , 14 ) 1011 � 1-10 3 ( m 14 ) m ( 12 , 14 ) 1110 � 11-0 � 4 m 15 � m ( 11 , 15 ) 1111 1-11 m ( 14 , 15 ) 111-

  18. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Combine terms m ( 8 , 12 ) and m ( 10 , 14 ) . (Dashes match and only one bit differs) The term m ( 8 , 10 , 12 , 14 ) is already listed, so don’t add it again, but do put a checkmark by the terms m ( 8 , 12 ) and m ( 10 , 14 ) Number of 1s Term Binary Size 2 Size 4 m ( 4 , 12 ) m ( 8 , 9 , 10 , 11 ) m 4 0100 � -100 10-- 1 m ( 8 , 9 ) m ( 8 , 10 , 12 , 14 ) m 8 1000 � 100- � 1--0 m ( 8 , 10 ) 10-0 � ( m 9 ) m ( 8 , 12 ) 1001 � 1-00 � 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 � m ( 10 , 11 ) 101- � m 11 m ( 10 , 14 ) 1011 � 1-10 � 3 ( m 14 ) m ( 12 , 14 ) 1110 � 11-0 � 4 m 15 � m ( 11 , 15 ) 1111 1-11 m ( 14 , 15 ) 111-

  19. Phase 1: Finding Prime Implicants � F ( A , B , C , D ) = ( m ( 4 , 8 , 10 , 11 , 12 , 15 )) + d ( 9 , 14 ) Compare each term in group 2 to each term in group 3. Combine terms m ( 10 , 14 ) and m ( 11 , 15 ) . (Dashes match and only one bit differs) Number of 1s Term Binary Size 2 Size 4 m ( 4 , 12 ) m ( 8 , 9 , 10 , 11 ) m 4 0100 � -100 10-- 1 m ( 8 , 9 ) m ( 8 , 10 , 12 , 14 ) m 8 1000 � 100- � 1--0 m ( 8 , 10 ) 10-0 � ( m 9 ) m ( 8 , 12 ) m ( 10 , 11 , 14 , 15 ) 1001 � 1-00 � 1-1- 2 m 10 1010 � m 12 m ( 9 , 11 ) 1100 � 10-1 � m ( 10 , 11 ) 101- � m 11 m ( 10 , 14 ) 1011 � 1-10 � 3 ( m 14 ) m ( 12 , 14 ) 1110 � 11-0 � 4 m 15 � m ( 11 , 15 ) � 1111 1-11 m ( 14 , 15 ) 111-

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