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Lecture 5

Lecture 5. Converting to use NAND and NOR Minimizing functions using Boolean cubes. X Y X nor Y 0 0 1 1 1 0. X Y X nand Y 0 0 1 1 1 0. Minimal set. Can implement any logic function from NOT, NOR, and NAND In fact, can do it with only NORs and NANDs

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Lecture 5

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  1. Lecture 5 • Converting to use NAND and NOR • Minimizing functions using Boolean cubes

  2. X Y X nor Y0 0 11 1 0 X Y X nand Y0 0 11 1 0 Minimal set • Can implement any logic function from NOT, NOR, and NAND • In fact, can do it with only NORs and NANDs • NOT is just NAND or NOR with two identical inputs

  3. Why NAND/NOR? • NAND/NOR preferred for real hardware implementation • More efficient (less switches per gate) • But how do we convert from the canonical forms that are expressed in AND/OR?

  4. X Y X' Y' (X + Y)' X' • Y'0 0 1 1 1 10 1 1 0 0 01 0 0 1 0 0 1 1 0 0 0 0 (X + Y)' = X' • Y'NOR is equivalent to AND with inputs complemented X Y X' Y' (X • Y)' X' + Y'0 0 1 1 1 10 1 1 0 1 11 0 0 1 1 1 1 1 0 0 0 0 (X • Y)' = X' + Y'NAND is equivalent to OR with inputs complemented NAND/NOR truth tables

  5. NOR NAND AND OR Pushing the bubble • Apply DeMorgan's A'B' = (A + B)' A' + B' = (AB)' A + B = (A'B')' (AB) = (A' + B')'

  6. Converting to NAND/NOR • Introduce inversions ("bubbles") • Introduce bubbles in pairs • Conserve inversions • Do not alter logic function

  7. A NAND A A B Z NAND NAND B B C Z Z NAND C C D NAND D D AND/OR to NAND/NAND Z = AB + CD = (A'+B')'+(C'+D')' = [(A'+B')(C'+D')]' = [(AB)'(CD)']'

  8. A A’ A’ A B B’ Z Z B’ Z NOR NOR NOR NOR NOR NOR NOR B C C’ Z C’ C D D’ D’ D AND/OR to NOR/NOR Z = AB + CD = (A'+B')'+(C'+D')’ = [(A'+B')'+(C'+D')’]’’ = {[(A'+B')'+(C'+D')']'}'

  9. A’ A A’ B B’ Z Z B’ Z NAND NAND C’ C NAND NAND C’ D D’ D’ A B Z NAND NAND NAND C D OR/AND to NAND/NAND

  10. A NOR A B Z C B Z NOR A C D NOR B D Z NOR C NOR D OR/AND to NOR/NOR

  11. Goal: Logic minimization • Algebraic simplification • Not a systematic procedure • Hard to know when we reached the minimum • Computer-aided design tools • Require very long computation times (NP hard) • Heuristic methods employed–"educated guesses”

  12. Goal: Logic minimization • Visualization methods are useful • Our brain is good at figuring "simple" things out • Many real-world problems are solvable by hand

  13. Key tool: Uniting Theorem • Uniting theorem: A(B’+B) = A • The approach: • Find where some variables don’t change (the A’s above) and others do (the B’s above) • Eliminate the changing variables (the B’s) A B F 0 0 1 0 1 1 1 0 0 1 1 0 A has the same value in both “on-set” rows keep A B has a different value in the both rows eliminate B F = A'B'+A'B = A'(B+B') = A'

  14. Boolean cubes • Visualization tool for the uniting theorem • n input variables = n-dimensional "cube" 01 11 0 1 2-cube 1-cube Y X 00 10 X 1011 1111 0111 0011 011 111 4-cube 0010 3-cube 101 Y Y Z Z W 1100 1000 000 100 X X 0000 0100

  15. Mapping to Boolean cubes • ON set = solid nodes • OFF set = empty nodes Look for on-set adjacent to each other Sub-cube (a line) comprises two nodes A varies within the sub-cube; B does not This sub-cube represents B' A B F 0 0 1 0 1 0 1 0 1 1 1 0 F 01 11 B 00 10 A

  16. (A'+A)BCin A(B+B')Cin AB(Cin'+Cin) Example: Full adder carry-out • Binary full-adder carry-out logic • On-set is covered by the OR of three 1-D subcubes A B Cin Cout 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 111 B 101 Cin 000 A Cout = BCin+AB+ACin

  17. (A'+A)BCin A(B+B')(Cin+Cin‘) Example • Changed one bit from the previous function • On-set is covered by the OR of one 2-D subcube and one 3-D subcube A B Cin D 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 111 101 B Cin 1 000 A D = BCin+A

  18. M-D cubes in N-D space • In a 3-cube (three variables): • A 0-cube (a single node) yields a term in 3 literals • A 1-cube (a line of two nodes) yields a term in 2 literals • A 2-cube (a plane of four nodes) yields a term in 1 literal • A 3-cube (a cube of eight nodes) yields a constant term "1" 011 111 F(A,B,C) = m(4,5,6,7) On-set forms a square (a 2-D cube) A is asserted (true) and unchanging B and C vary This sub-cube represents the literal A 110 010 B 101 C 000 100 A

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