Summer 2013
CENG 242
Solution 4
1.
The system is stable if and only if
1 =
1 + 1 4c
1 1 4c
< 1 and 2 =
<1.
2
2
Case 1: Real eigenvalues, i.e., 1 4c 0 c 1 / 4 .
1 + 1 4c < 2
1 1 4c < 2
We must have
(1 + 1 4c ) < 2
together:
for 1 and
(1 1 4c ) < 2
for

Summer 2013
CENG 242
Solution 1
1. Note that P = U \ P = cfw_0,2,4,6 = Q P Q = P P = U and P Q = P P = cfw_.
Then,
( P Q ) ( P R ) = U ( P R ) = P R = Q R,
( P Q ) ( P R ) (Q R ) = U (Q R ) (Q R ) = Q R.
Similarly,
( P Q ) ( P R ) = P R = Q R,
( P Q ) ( P

Summer 2013
CENG 242
Solution 6
1. For an NM image with given probabilities pjk for pixels mik and given row and
column intensities rj and ck, our 0-1 ILP problem is formulated below.
N
Maximize
M
p
j =1 k =1
Subject to
jk
m jk
N
m jk = ck , k cfw_1,2,.

Summer 2013
CENG 242
Solution 3
1. Since R1 = R2 = 1 , our second-order HLR becomes vn+2 3vn+1 + vn = 0. We
need two initial conditions, one of which is given: v0 = 1 V. The other initial
condition, the value of v1, can be calculated using v1 = v0 R1i0, b

Summer 2013
CENG 242
Solution 5
1.
Let the edge ordering be [e1 e2 e3 | e4 e5 e6]. We can construct Bf = [BfT | I] first
(three circuits due to e4, e5, e6), and then derive Qf = [I | QfT*], using QfT* = BfTT.
We have:
Circuit due to e4: e4 (e3) (e1) e2;
C