HW1 solution. 2.2 Prove the above two equations using Shannon s expansion.

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1 HW1 solution Please notice that just the fact that you get the points for homework does not mean your solution is correct, since some of them are just checked for completion. Rubrics ( =20 points in total) 2.1.A 1 point for completion 2.1.B 1 point for completion 2.2.A 1 point for completion 2.2.B 1 point for completion 3.i. 1 point for completion 3.ii. 1 points for correctness 3.iii. 1 point for correctness (both expressions or using min/max term index are OK) 4.A.i 1 point for completion + 1 point for simplified form (13 literals) + 1 point for #terms 4.A.ii 1 point for completion 4.A.iii 1 point for completion 4.B.i 1 point for completion + 1 point for simplified form (7 literals) + 1 points for #terms 4.B.ii 1 point for completion 4.B.iii 1 point for completion 5. 3 points for 3 blanks check for correctness 2 Application of Boolean Algebra Theorems 2.1 Prove the following equations using Boolean algebra: A. ad + bc d +abc = ad + bc d B. (a+d)(b+c +d )(a+b+c )=(a+d)(b+c +d ) Are the above equations related to the consensus theorem? 2.1 A. ad + bc d + abc = ad + abc (d+d )+bc d (complement) = ad + abc d+ abc d +bc d (distributed) = ad(1+bc )+(a+1)bc d (distributed) = ad+bc d B. (a+d)(b+c +d )(a+b+c ) = (a+d)(b+c +d )( a+b+c +d d) (complement) = (a+d)(a+b+c +d)(a+b+c +d )(b+c +d ) (distributed) = (a+d)(1+b+c )(b+c +d )(a+1) (distributed) = (a+d)(b+c +d ) 2.2 Prove the above two equations using Shannon s expansion.

2 2.2 A. LHS = f(a,b,c,d) = d.f(a,b,c,1)+ d f(a,b,c,0) = d(a+abc ) +d (bc +abc ) = ad+bc d = RHS B. LHS=f(a,b,c,d) =(d +f(a,b,c,1)) (d+f(a,b,c,0)) =(d + (b+c )(a+b+c ))(d+a(a+b+c ) ) =(b+c +d )(a+d) =RHS 3. From Problem to Boolean Expression A majority voting machine inputs four binary bits (x3, x2, x1, x0), and outputs y = 1 when the more than half of the input bits are 1. Otherwise, the output is y = 0. i. Write the truth table. ii. Write the functions in sum-of-products canonical form. iii. Write the functions in product-of-sums canonical form. 3. i. index x3 x2 x1 x0 y

3 ii. F= m(7, 11, 13, 14, 15) = x 3 x2 x1 x0 + x3 x2 x1 x0 + x 3 x2 x1 x0 + x 3 x2 x1 x0 + x 3 x2 x1 x0 iii.f= M(0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12) = ( x3 + x 2 + x 1 + x 0)(x3 + x 2 + x 1 + x 0 )(x3 + x 2 + x 1 + x 0) ( x3 + x 2 + x 1 + x 0 )(x3 + x 2 + x 1 + x 0)(x3 + x 2 + x 1 + x 0 )(x3 + x 2 + x 1 + x 0) ( x3 + x 2 + x 1 + x 0)(x3 + x 2 + x 1 + x 0 )(x3 + x 2 + x 1 + x 0)(x3 + x 2 + x 1 + x 0) 4. Boolean Algebra and Implementation i. Simplify each of the following two Boolean equations. ii. Sketch a reasonably simple combinational circuit implementing the simplified equation. iii. Compare the numbers of literals and operators versus the numbers of gates, nets, and pins in the schematic diagrams A. a bc + a b d + a b cd + bc d + acd + abcd + ac d B. (a+b+c+d )(a+b+c+d)(a+b+c )(a +b+c+d)(a +c+d )(a +b+c +d) 4. A.i. Solution 1: aʹbcʹ + aʹbʹd + aʹbʹcdʹ + bcʹd + acd + abcdʹ + acʹd = a bc +a b d+a b cd +bc d+acd+abcd +ac d (commutative) =(a b+bd+ad)c +ab(d+cd )+ac(d+bd ) (distributive) =(a b+ad)c +a b (d+c)+ac(d+b) (consensus) = a bc +ac d+a b c+a b d+acd+abc (distributive) = a bc +ac d+a b c+a(c d+cd+bc) (distributive) = a bc +ac d+a b c+a(d+bc) (complement) = a bc +a b c+a b d+ad+abc (distributive) = a b c+ ad +b d+a bc +a b c+abc (consensus) =(a +1)b d+ad+a bc +a b c+abc (distributive) =b d+ad+a bc +a b c+abc

4 Shorter and more obvious approach: Solution 2: a bc + a b d + a b cd + bc d + acd + abcd + ac d = a bc + a b d + a b cd + acd + abcd + ac d --bc d is the consensus term of a bc and ac d = a bc + a b d + a b cd + ad(c + c ) + abcd --distributive consensus = a bc + a b d + a b cd + ad + abcd --complement = a bc + a b d + a b cd + ad + abcd + a b c + abc --consensus of ab d & a b cd, ad & abcd = a bc + a b d + a b c(1 + d ) + ad + abc(1+d ) --distributive = a bc + a b d + a b c + ad + abc --1+ a =1 = a bc + a b d + a b c + ad + abc + b d --consensus of a b d and ad = a bc + b d(a + 1) + a b c + ad + abc --distributive = a bc + b d + a b c + ad + abc --1+ a =1 Solution 3: a bc +a b c+a b cd +bc d+acd+abcd +ac d Let s say T1=a bc, T2= a b c, T3 =a b cd, T4 = bc d, T5=acd, T6=abcd and T7=ac d Then, Original expression =T1+T2+T3+T4+T5+T6+T7 =T1+T2+T3+T4+T5+T6+T7 + a c d + abc + ad (T2, T3 consensus a b c, T5, T6 consensus abc, T5, T7 consensus ad, T2, T4 consensus a c d) =T1+T2+T3+T4+T5+T6+T7 + a c d + abc + ad + c d (a c d and T7 consensus c d) =T1+a b c+abc+ad+c d (c d covers T7; abc covers T6; ad covers T5; c d covers a c d; c d covers T4; T2 covers T3) =a bc +a b c+abc+ad+c d Other possible solution: ad + c d + abc + a bc + a b c and b d + c d + abc + a bc + a b c ii)

5 iii)literals : 13 (ad + b d + a bc + a b c + abc) Operators: 6 operators ( 5 AND, 1 OR) or 12 for two input gates. Gates: - 6 gates (multiple-input gates or 12 gates for two-input gates), Nets: 10 nets ( 16 nets for two-input gates) Pins: =24 pins (36 pins for two-input gates) We use multiple-input gates then #variables+#operators = #nets #literals+#gates*2-1 = #pins B i) Solution 1: (a + b + c + dʹ)(a + b + c + d)(a + b + cʹ)(aʹ + b + c + d)(aʹ + c + dʹ)(aʹ + b + cʹ + d) = (a + b + c)(a + b + cʹ)(aʹ + b + d)(aʹ + c + dʹ) (distributive) = (a + b)(aʹ + b + d)(aʹ + c + dʹ) = (a+b)(a + b +d ) (b +d) (a +c +d ) (consensus) //(a+b)(a +b+d ) = (a+b)(a +b+d)(b+d) = (a+b)(b + d +a.0) (a +c +d ) (distributive) = (a+b)(b + d )(a +c +d ) = ( ad +b) (a +c +d ) = (a+b)(d+b)(a +c+d ) --simplest POS = acd + a b + bc + bd = acd + a b + bc(d +d ) + bd = acd + a b + bcd + bcd + bd = acd + a b + bd (c+1) (consensus) = acd + a b + bd --simplest SOP Both SOP and POS have 7 literals.

6 Solution 2: Let s say T1= a + b + c +d T2 = a + b + c+ d T3 = a + b + c T4 = a + b + c + d T5 = a + c + d T6 = a + b + c + d Then the original expression =T1.T2.T3.T4.T5.T6 = T3.T5.(a+b +c)(a +b+d) ( T1, T2 consensus a+b+c; T4, T6 consensus a +b+d ) = T5(a +b+d)(a + b) ( a+b+c and T3 consensus a + b ) = T5(a+b)(b+d) ( a +b+d and a+b consensus b +d ) =(a +c+d )(a+b)(b+d) ii) iii) acd + a b + bd Literals :- 7 ( acd + a b + bd ) Operators:- 4 operators (3 AND, 1 OR) or 6 operators if use two input gates Gates: - 4 gates (or 6 gates if use two input gates) Nets: 8 nets (10 nets if use two input gates) Pins: =14 pins (18 pins if use two input gates)

7 We use multiple-input gates then #variables+#operators = #nets #literals+#gates*2-1 = #pins 5. Bluespec Adder Complete the following implementation of a full adder using &,,! and bit selection []. Remember that for Bit(n) x, x[0] represents the Least Significant Bit (LSB) of x. Please refer to BSV by Example for any syntax, semantics of Bluespec. function Bit#(2) fa(bit#(1) a, Bit#(1) b, Bit#(1) c_in); Bit#(2) ab = ha(a, b); Bit#(2) abc = ha(, ); Bit#(1) c_out = ; return {c_out, }; endfunction 5. function Bit#(2) fa(bit#(1) a, Bit#(1) b, Bit#(1) c_in); Bit#(2) ab = ha(a, b); Bit#(2) abc = ha( ab[0], c_in ); Bit#(1) c_out = ab[1] abc[1]_ ; return {c_out, abc[0] }; endfunction