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Computer Science 290, Spring, 2009
Final Exam
Name________________________________
1. Let S = {w | w in {a,b,c}*, w is a substring of ‘aabcbb} and let R be the relation “is a substring of.” For
example, abc R abcb, but not aa R abcb.
(15) (a) Prove that (S, R) is a partially ordered set.
1. For any w in S, w is a substring of itself, so R is reflexive.
2. Suppose v, w in S and vRw and wRv. This means v is a substring of w and w is a substring of v. This implies
that the lengths of v and w are the same, and therefore they are the same string: v = w. So R is antisymmetric.
3. Suppose for u, v, w in S that uRv and vRw. The u is a substring of v and v is a substring of w. This means
that u appears in the substring v within w, so u is a substring of w, uRw. Thus R is transitive.
R is a partial ordering by 1,2,3, so (S, R) is a partially ordered set.
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Computer Science 290, Spring, 2009
Final Exam
Name________________________________
(10) (b) Draw the Hasse diagram for (S, R).
aabcbb
aabcb
aabc
aab
abcbb
abcb
abc
aa
ab
bcbb
bcb
cbb
bc
cb
b
c
a
bb



(10) (c) Is (S, R) a lattice? Why or why not? (A formal proof is not required, just an explanation of why or
why not, based on the diagram.)
No. b and c have no LUB since bc and cb are both upper bounds for b and c and bc, cb are not related under R.
Similarly, bc and cb have no GLB, since b and c are both lower bounds for bc, cb.
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Computer Science 290, Spring, 2009
Final Exam
Name________________________________
2. Consider the set F = {(x, y) | x, y in Z, not y = 0}. (x could be 0.)
Define the operation + on F as follows: (r, s) + (x, y) = (ry + sx, sy)
(10) (a) Prove that (F, +) is a commutative monoid.
1. Consider (r,s), (u,v), and (x,y) all in F:
(r,s) + ((u,v) + (x,y)) = (r,s) + (uy+vx, vy)
def
= (r(vy) + s(uy + vx), s(vy))
def
= (rvy + suy + svx, svy)
associativity of multiplication and addition on Z
= ((rv + su)y + (sv)x, (sv)y)
associativity of multiplication and addition on Z
= (rv + su, sv) + (x,y)
def
= ((r,s) + (u,v)) + (x,y)
def
So + on F is associative.
2. (r,s) + (x,y) = (ry + sx, sy)
def
= (xs + yr, ys)
addition and multiplication on Z commutative
= (x,y) + (r,s)
def
So + on F is commutative.
3. Consider (0,1) in F and any (r,s) in F
(0,1) + (r,s) = (0s + 1r, 1s)
= (r, s)
def
properties of 0, 1 in Z
So (0,1) is the identity on (F, +)
By 1, 2, 3, (F, +) is a commutative monoid.
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Computer Science 290, Spring, 2009
Final Exam
Name________________________________
(10) (b) Define the relation ~ on F as follows: (a, b) ~ (c, d) if and only if ad = bc.
Show that ~ is an equivalence relation.
1. For any (x,y) in F, xy = yx (since mult on Z is commutative), so (x,y) ~ (x,y) and ~ is reflexive.
2. Assume (r,s) ~ (x,y), for (r,s) and (x,y) in F. Then ry = sx (def of ~). So xs = yr (mult on Z is commutative
and = is symmetric) and (x,y) ~ (r,s) (def of ~). So ~ is symmetric.
3. Assume that (r,s) ~ (u,v) and (u,v) ~ (x,y), for (r,s), (u,v), (x,y) in F. Then rv = su and uy = vx (def of ~).
Since rv = su, we have rvy = suy, multiplying both sides by y.
Then rvy = svx, by substitution.
Since v is not 0 (def of F), we can conclude that ry = sx.
Then (r,s) ~ (x,y) (def of ~).
Thus ~ is transitive. (Note: you must cite that v is not 0 to come to this conclusion – this relation depends on
that part of the definition of F.)
By 1, 2, and 3, ~ is an equivalence relation.
(10) (c) Show that ~ is a congruence relation for (F, +).
Suppose (a,b) ~ (c,d) and (r,s) ~ (x,y), all in F. Then ad = bc and ry = sx (def of ~).
Consider (a,b) + (r,s) = (as + br, bs) and (c,d) + (x,y) = (cy + dx, dy).
To show these are equivalent by ~, we need to show that
(as + br)dy = bs(cy + dx).
Start with (as + br)dy = asdy + brdy
distributive
= adsy + bdry
mult commutative
= bcsy + bdsx
substitution from above
= bscy + bsdx
mult commutative
= bs(cy + dx)
distributive
So if (a,b) ~ (c,d) and (r,s) ~ (x,y), then (a,b) + (r,s) ~ (c,d) + (x,y)
~ is a congruence.
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Computer Science 290, Spring, 2009
Final Exam
Name________________________________
(5) (d) Consider (F/~, [+]), where [+] is the operation + applied to the congruence classes.
Give a partial list of some of the congruence classes (five classes with four or five representative elements
each).
What well-known mathematical structure is this?
{ (0,1), (0,2), (0,3), (0,4), (0,5), …}
{ (1,1), (2,2), (3,3), (4,4), (5,5), …}
{(1,2), (2,4), (3,6), (4,8), (5,10), …}
{(5,3), (10,6), (15,9), (20,12), (25,15),…}
{(-3,4), (3,-4), (-6,8), (6,-8), (-9, 12),….}
The rational numbers. Each congruence class is the set of fractions that are equal.
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Computer Science 290, Spring, 2009
Final Exam
Name________________________________
3. Consider the Moore Machine, M, specified by the following diagram with T = {S2, S3}.
0
0
1
S0
S1
S2
1
0
1
1
0
1
S3
1
S4
0
S5
0
(5) (a) Fill out the table that specifies the function fw for w = 01101
s
fw (s)
s0
s1
s2
s3
s4
s5
s5
s0
s5
s0
s5
s0
(10) (b) Find the special congruence R so that L(M/R) = L(M). Do this by refining the partitions that correspond
to the sequence of relations R0, R1, …, where P0 = {{S0, S1, S4, S5} {S2, S3}}.
P0 = {{S0, S1, S4, S5} {S2, S3}} input 0 splits S0 C S1 and S0 and S4. S0 and s5 stay together as do S1 and S4
S2, S3 stay together under both inputs.
P1 = {{S0, S5} {S1, S4} {S2, S3}}
P2 = P1
no further splitting occurs.
So R is the relation defined by the partition P1.
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Computer Science 290, Spring, 2009
Final Exam
Name________________________________
(10) (c) Draw M/R.
1
0
1
[S0]
[S1]
0
1
[S3]
0
(5) (d) Give a regular expression for L(M/R).
First build the r.e. that is generated by going around the loop and ending at [S0]:
0*11*00*1
This loop can be repeated many times (or none), then must be followed by the sequence that put the machine
into the accepting state:
(0*11*00*1)*0*11*00*
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