Daniel Mihai Problem 1.1
The parse tree Pe
The equivalent tree to Pe
The parse tree of e is Ge=<E,V,f> V=cfw_v1,v2,v3,v4,x1,x2,x3,x4, E=cfw_<v1,v2>,<v2,x1>,<v1,v3>,<v3,v4>,<v3,x4>,<v4,x2>,<v4,x3> f(v1)=+,f(v2)=-,f(v3)=*,f(v4)=+ The Graph represented in th
Daniel Mihai Problem 1.2 P(n):"Any acyclic digraph with non-empty set of nodes has at least one node with in-degree 0". Base case: P(1): The graph contains just one node and no edges (due to the fact that is acyclic), therefore the indegree is 0. Step cas
Daniel Mihai Problem 1.3 1. The minimum number of bits necessary to represent 7 and 5 in two's complement is 4,because 710=1112 and 510=1012, and in two's complement 7=<0111>32s and 5=<0101>32s 2. The depth of a CCA is 3n for adding n-bit numbers. In this
Daniel Mihai Problem 1.4 Binary comparator for 3 1-bit numbers
Binary comparator for 3 n-bit numbers a1 b1 c1 . . . . an bn cn
an .an . . . . a2 . a1 2 a1
n-ary AND
x1 . . . . xn
The depth of the circuit is 7+2*log2 n because a a n-ary AND made from AND g
Daniel Mihai Problem 2.1 The Quine McCluskey algorithm for f1:
X1 0 0 0 0
X2 0 1 0 1
X3 0 0 1 1
f1 1 1 1 1
X1 0 0 0 0
X2 X 0 1 X
X3 0 X X 1
X1 0 0 0 0
X2 X 0 1 X
X3 0 X X 1
X1 0
X2 X
X3 X
Since we have only one prime implicant, it is essential. f1= The Qu
Daniel Mihai Problem 2.2
We can design the upper circuit in order to determine if the player lost. If we set the D input of the D flip-flop to 1 and maintain it constant, AND the signals from all the cells of a battleship and set the Q output to be 0 init