Mathematics Masters Examination
OPTION 4
March 30, 2004
COMPUTER SCIENCE
2–5 PM
NOTE: Any student whose answers require clarification may be required to submit to an
oral examination.
Each of the fourteen numbered questions is worth 20 points. All questions will be graded,
but your score for the examination will be the sum of your scores on your eight best questions.
Please observe the following:
•
DO NOT answer two or more questions on the same sheet
(not even on both sides of the same sheet).
•
DO NOT write your name on any of your answer sheets.
You will be given separate instructions on the use of these answer sheets. When you have
completed a question, place it in the large envelope provided.
Spring 04 CS
1/4
Computer Algorithms
1. Given an array A of n elements, we wish to find the k largest in sorted order using the
following 3 algorithms. Find their (worst-case) running-times in terms of n and k.
(a) MergeSort A and list k largest.
(b) Build a max-heap and call EXTRACT-MAX k times.
(c) SELECT the k largest number, partition around that number and sort the k largest
numbers.
2. Let G = (V, E) be a directed graph with |V | = n and |E| = m. Suppose that each vertex
u ∈ V is labelled with a unique integer L(u) from the set {1, 2, . . . , n}. For each u ∈ V , let
R(u) = {v ∈ V : u
v} be the set of the vertices that can be reached by a directed path
starting from u. Define min(u) to be the vertex in R(u) whole label is minimum. Give an
O(n + m)-time algorithm that computes min(u) for all vertices u ∈ V .
Programming Language Design
3. C++ uses pointers to elements of type (say) E as a common alternative to indexes when
accessing elements of arrays.
Give at least one reason in favor of this scheme and one reason against it (observe that the
scheme has been abandoned in Java).
C++ (at least most versions) does not guard against a pointer going “out of bounds” of
an array it is indexing. Why not? Could such a check be put into the language as a
modification? What would be the “cost” of doing so?
Combinatorics
4. Consider a rectangular box with distinct dimensions. Color the 12 edges B, W, or R.
Find the number of inequivalent colorings with 4B, 4W, and 4R edges. Colorings are
equivalent with respect to rotations of the box. (Note that since the dimensions are distinct,
the rotation must use an angle of 180 degrees and the axis must join the middle of opposite
faces.)
5. Find the number of permutations of 1, 2, . . . , 6 in which 1 and 2 are in positions 3,4,5,6; 3
and 4 are in positions 1,2,5,6; and 5 and 6 are in positions 1,2,3,4.
Graph Theory
6. All graphs are simple and undirected.
Let H be a subgraph of Kn,n with more than (k − 1)n edges. Prove that H has a matching
of size k.
Spring 04 CS
2/4
Theory of Computation
7. Let Σ = {0, 1}. Construct a DFA for the following language:
L = {w | w does not contain the substring 00, and |w| is even}.
Explain and justify your construction.
Notes: |w| is the length of w, and substring means a sequence of consecutive symbols.
8. Let Σ = {0, 1, #}. Construct a CF grammar generating the following language:
L = {w # x#wR | w, x ∈ {0, 1}∗ }.
Explain and justify your construction.
Symbolic Computation
9. Explain the difference between Int and int in Maple.
Give an example to illustrate why we need a command like Int.
Numerical Analysis


1.5906E +00 −1.1367E −01 −8.8133E −01
10. Consider the matrix A =  3.8693E −01 −2.7644E −02 −2.1440E −01 
1.0159E +00 −7.2596E −02 −5.6290E −01


−3.8816E +09 −4.5674E +09 7.8171E +09
with inverse A−1 =  −1.5738E +08 −1.8505E +08 3.1690E +08 .
−6.9849E +09 −8.2190E +09 1.4067E +10
1. Estimate the condition number of A using ||.||1 (the maximum column sum) as norm.
2. For this given matrix A, we now consider the linear system Ax = b, for any general
right-hand side vector b, and we forget about the inverse A−1 .
What is the relation between the relative error on the solution x to Ax = b and the
relative error on A? Use cond(A) in your answer.
If we want to know the first eight decimal places of x, how accurately must we know A?
Justify your answer.
Spring 04 CS
3/4
11. Suppose we want to approximate the first derivative f 0 (x0 ) of some function f (x) at the
point x0 .
1. Show that the forward-difference approximation using step h > 0 is only a first-order
approximation in h for f 0 (x0 ).
2. Develop extrapolation formulas to compute a third-order approximation in h for f 0 (x0 ),
cutting the step h in half each time.
Computational Geometry
12. A Star-shaped polygon (SSP) is a simple polygon all of whose interior is visible from at
least one point in the polygon’s interior. We shall call such a point a point of total visibility
(PTV). It is clear that a star-shaped polygon need not be convex.
The Kernel of a SSP is a region of its interior containing all of its PTV’s. Observe that if
the SSP happens to be convex, its kernel will span the entire interior of the SSP.
Outline a proof sketch that the kernel of a SSP is a convex polygon inside the SSP. (Hint:
Start with a 4-sided polygon with vertex angles 3 of which are convex, turning to the right
in a clockwise traversal of the polygon (just like a convex polygon) and 1 of which is reflex
(turning to the left). Draw the kernel of this polygon. Now consider how the kernel is
constrained when there are several reflex angles. For instance, draw the kernel when there
are two reflex angles.)
Error-Correcting Codes and Cryptography


1 1 1 0 1 0 0
13. A binary code C is described by its parity-check matrix H =  1 1 0 1 0 1 0 .
1 0 1 1 0 0 1
(a) How would you encode the messages m1 = 0000 and m2 = 1101?
(b) How would you decode the received vectors v1 = 0000111 and v2 = 0001110?
(c) Find the parameters n, k, and d for C.
14. In a public-key cryptosystem using RSA, you intercept the cryptogram C = 10 sent to a
receiver whose public key is e = 5 and n = 35. Find the integer M that was sent.
Spring 04 CS
4/4
Solutions to the
Mathematics Masters Examination
OPTION 4
March 30, 2004
COMPUTER SCIENCE
2–5 PM
NOTE: Any student whose answers require clarification may be required to submit to an
oral examination.
Each of the fourteen numbered questions is worth 20 points. All questions will be graded,
but your score for the examination will be the sum of your scores on your eight best questions.
Please observe the following:
•
DO NOT answer two or more questions on the same sheet
(not even on both sides of the same sheet).
•
DO NOT write your name on any of your answer sheets.
You will be given separate instructions on the use of these answer sheets. When you have
completed a question, place it in the large envelope provided.
Spring 04 CS
1/9
Computer Algorithms
1. Given an array A of n elements, we wish to find the k largest in sorted order using the
following 3 algorithms. Find their (worst-case) running-times in terms of n and k.
(a) MergeSort A and list k largest.
(b) Build a max-heap and call EXTRACT-MAX k times.
(c) SELECT the k largest number, partition around that number and sort the k largest
numbers.
Solution:
(a) Sorting costs n log n and listing costs k. Total O(n log n).
(b) Building a max-heap cost n and extracting k times costs k log n. Total O(n + k log n).
(c) Selecting and partitioning costs n and sorting costs k log k. Total is O(n + k log k).
2. Let G = (V, E) be a directed graph with |V | = n and |E| = m. Suppose that each vertex
u ∈ V is labelled with a unique integer L(u) from the set {1, 2, . . . , n}. For each u ∈ V , let
R(u) = {v ∈ V : u
v} be the set of the vertices that can be reached by a directed path
starting from u. Define min(u) to be the vertex in R(u) whole label is minimum. Give an
O(n + m)-time algorithm that computes min(u) for all vertices u ∈ V .
Solution: Compute GT in the usual way, so that GT is G with its edges reversed. Then do
a depth-first search on GT , but in the main loop of DFS, consider the vertices in order of
increasing values of L(v). If vertex u is in the depth-first tree with root v, then min(u) = v.
Clearly, this algorithm takes O(V + E) time.
Spring 04 CS
2/9
Programming Language Design
3. C++ uses pointers to elements of type (say) E as a common alternative to indexes when
accessing elements of arrays.
Give at least one reason in favor of this scheme and one reason against it (observe that the
scheme has been abandoned in Java).
C++ (at least most versions) does not guard against a pointer going “out of bounds” of
an array it is indexing. Why not? Could such a check be put into the language as a
modification? What would be the “cost” of doing so?
Solution: The indexing can be done, at the target level, with no checking and no conversion
of an integer index when the elements of the array occupy other than 1 word. It is thus
extremely efficient. There is no conversion from an integer to an index, as would be the case
for an integer used as an index on an array whose elements are types occupying more than
one word.
It is easy to implement in the compiler the generation of target code that will check each
pointer applied to an array as an index, in order to ensure that the pointer is in bounds.
This requires a run-time check each time the pointer is used as an index on an array, and
obliges the use of a separate pointer for each array accessed, which is annoying when (say)
two arrays of identical size are being accessed by the same index value. The result is a
classic trade-off between efficiency and security. With today’s fast processors, the “right”
way is almost always to opt for security, unless every extra cycle needs to be squeezed from
the computer.
Spring 04 CS
3/9
Combinatorics
4. Consider a rectangular box with distinct dimensions. Color the 12 edges B, W, or R.
Find the number of inequivalent colorings with 4B, 4W, and 4R edges. Colorings are
equivalent with respect to rotations of the box. (Note that since the dimensions are distinct,
the rotation must use an angle of 180 degrees and the axis must join the middle of opposite
faces.)
Solution: P6 = 14 (z112 + 3z26 ), with z1 = B + W + R and z2 = B 2 + W 2 + R2 .
The coefficient of B 4 W 4 R4 in 14 ((B + W + R)12 + 3(B 2 + W 2 + R2 )6 )
12
6
1
+3
= 8730.
is 4
4, 4, 4
2, 2, 2
5. Find the number of permutations of 1, 2, . . . , 6 in which 1 and 2 are in positions 3,4,5,6; 3
and 4 are in positions 1,2,5,6; and 5 and 6 are in positions 1,2,3,4.
Solution: (1 + 4z + 2z 2 )3 = 1 + 12z + 54z 2 + 112z 3 + 108z 4 + 48z 5 + 8z 6 .
Then the answer is 6! − 12 · 5! + 54 · 4! − 112 · 3! + 108 · 2! − 48 · 1! + 8 · 0! = 80.
Graph Theory
6. All graphs are simple and undirected.
Let H be a subgraph of Kn,n with more than (k − 1)n edges. Prove that H has a matching
of size k.
Solution: Each vertex covers at most n edges, hence the minimum number of vertices
needed to cover all the edges of H is at least k. By the König-Egerváry theorem, the
maximum size of a matching in H is equal to the minimum size of a vertex cover. From
this the result follows.
Spring 04 CS
4/9
Theory of Computation
7. Let Σ = {0, 1}. Construct a DFA for the following language:
L = {w | w does not contain the substring 00, and |w| is even}.
Explain and justify your construction.
Notes: |w| is the length of w, and substring means a sequence of consecutive symbols.
Solution:
- i
f
1
f
i
i
I@ 0
@0
6
6@
@@
@
1
@@
@
@@
@
1
0,1 0,1
@@
@
@@
@
0
1@@
0
@
? @@
R
R ?
@
i
i
i
8. Let Σ = {0, 1, #}. Construct a CF grammar generating the following language:
L = {w # x#wR | w, x ∈ {0, 1}∗ }.
Explain and justify your construction.
Solution:
S → 0 S 0 | 1 S 1 | S0
S0 → # A #
A→A0|A1|
Symbolic Computation
9. Explain the difference between Int and int in Maple.
Give an example to illustrate why we need a command like Int.
Solution:
1. Int is the ”inert” Maple command: it just displays the integral, but does not attempt
to evaluate the integral, this is done by int.
2. Quite often, a symbolic antiderivative does not exist, and then we must resort to
numerical approximation. The command evalf(Int(f(x),x=a..b)) immediately computes such a numerical approximation, without first trying to compute a symbolic
antiderivative.
Spring 04 CS
5/9
Numerical Analysis


1.5906E +00 −1.1367E −01 −8.8133E −01
10. Consider the matrix A =  3.8693E −01 −2.7644E −02 −2.1440E −01 
1.0159E +00 −7.2596E −02 −5.6290E−01

−3.8816E +09 −4.5674E +09 7.8171E +09
with inverse A−1 =  −1.5738E +08 −1.8505E +08 3.1690E +08 .
−6.9849E +09 −8.2190E +09 1.4067E +10
1. Estimate the condition number of A using ||.||1 (the maximum column sum) as norm.
2. For this given matrix A, we now consider the linear system Ax = b, for any general
right-hand side vector b, and we forget about the inverse A−1 .
What is the relation between the relative error on the solution x to Ax = b and the
relative error on A? Use cond(A) in your answer.
If we want to know the first eight decimal places of x, how accurately must we know A?
Justify your answer.
Solution: Condition number of A:
1. cond(A) = ||A||1 × ||A−1 ||1 . Since ||A||1 ≈ 1 and ||A−1 ||1 ≈ 1010 , cond(A) ≈ 1010 .
2. Let x̄ be the system Āx = b, where Ā is the given A, known with limited precision.
Then the relative error on x is estimated by
||A − Ā||
||x − x̄||
≤ cond(A)
.
||x||
||A||
Ā||
For ||x−x̄||
to be less than 10−8 , ||A−
must be less than 10−18 , because the condition
||x||
||A||
number amplifies errors on A with a factor of 1010 . Thus we must know A with at
least 18 decimal places accurately to be sure of the first eight decimal places in x.
Spring 04 CS
6/9
11. Suppose we want to approximate the first derivative f 0 (x0 ) of some function f (x) at the
point x0 .
1. Show that the forward-difference approximation using step h > 0 is only a first-order
approximation in h for f 0 (x0 ).
2. Develop extrapolation formulas to compute a third-order approximation in h for f 0 (x0 ),
cutting the step h in half each time.
Solution: Approximation of f 0 (x0 ) with forward differences:
00
(x0 ) 2
1. Using f (x0 + h) = f (x0 ) + f 0 (x0 )h + f 2!
h + O(h3 ), we find the error for the forward
difference difference approximation ∆f (x0 , h) as
∆f (x0 , h) =
f (x0 + h) − f (x0 )
f 00 (x0 )
= f 0 (x0 ) +
h + O(h2 ),
h
2!
which shows ∆f (x0 , h) is a first-order approximation in h for f 0 (x0 ).
2. Cutting h in half each time, we derive
∆f (x0 , h) = f 0 (x0 ) + C1 h + C2 h2 + O(h3 )
∆f (x0 , h/2) = f 0 (x0 ) + C1 h/2 + C2 h2 /4 + O(h3 )
∆f (x0 , h/4) = f 0 (x0 ) + C1 h/4 + C2 h2 /16 + O(h3 )
for some constants C1 and C2 . Elimination of C1 (or equivalently, the term in h) leads
to a second-order approximation:
∆f (x0 , h, h/2) =
∆f (x0 , h/2, h/4) =
2∆f (x0 ,h/2)−∆f (x0 ,h)
2−1
= f 0 (x0 ) − C2 h2 /2 + O(h3 )
2∆f (x0 ,h/4)−∆f (x0 ,h/2)
2−1
= f 0 (x0 ) − C2 h2 /8 + O(h3 )
(x0 ,h,h/2)
We now see that 4∆f (x0 ,h/2,h/4)−∆f
eliminates the term in h2 and gives us the
4−1
desired third-order approximation for f 0 (x0 ).
Spring 04 CS
7/9
Computational Geometry
12. A Star-shaped polygon (SSP) is a simple polygon all of whose interior is visible from at
least one point in the polygon’s interior. We shall call such a point a point of total visibility
(PTV). It is clear that a star-shaped polygon need not be convex.
The Kernel of a SSP is a region of its interior containing all of its PTV’s. Observe that if
the SSP happens to be convex, its kernel will span the entire interior of the SSP.
Outline a proof sketch that the kernel of a SSP is a convex polygon inside the SSP. (Hint:
Start with a 4-sided polygon with vertex angles 3 of which are convex, turning to the right
in a clockwise traversal of the polygon (just like a convex polygon) and 1 of which is reflex
(turning to the left). Draw the kernel of this polygon. Now consider how the kernel is
constrained when there are several reflex angles. For instance, draw the kernel when there
are two reflex angles.)
Solution: Suppose it is already known that the polygon P is star-shaped.
Consider every reflex vertex v and the two vertices u and w on P. From v extend two
half-lines continuing uv and vw. These form a wedge. From any point in the wedge, the
inside of both uv and vw are visible; from outside the wedge, they are not both visible.
Now reshape the polygon by connectiing w and u, forgetting v. It can be shown easily
that P is still star-shaped, and that uw will still not pass through any edge of P.
Continue this process until there are no more reflex vertices. The remaining polygon Q
is convex, and forms the convex hull of P. The kernel must be the intersection of all the
half-planes on either side of the wedges (two per wedge), which is convex, with Q, which is
convex, and so finally the kernel is convex.
Spring 04 CS
8/9
Error-Correcting Codes and Cryptography


1 1 1 0 1 0 0
13. A binary code C is described by its parity-check matrix H =  1 1 0 1 0 1 0 .
1 0 1 1 0 0 1
(a) How would you encode the messages m1 = 0000 and m2 = 1101?
(b) How would you decode the received vectors v1 = 0000111 and v2 = 0001110?
(c) Find the parameters n, k, and d for C.
Solution:
(a) 00000000 and 1101010
(b) 1000111 and 0011110
(c) [7, 4, 3]
14. In a public-key cryptosystem using RSA, you intercept the cryptogram C = 10 sent to a
receiver whose public key is e = 5 and n = 35. Find the integer M that was sent.
Solution: p = 5, q = 7
Hence d is the inverse of 5 mod 24. This is 5 again. So M = 105 (mod 35) = 5.
Spring 04 CS
9/9