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Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.1 What is an algorithm?
14.1.1
The following layout describes the usual algorithm for multiplying an n-digit
integer x by a single-digit integer y, in base 10 notation.
xn−1 xn−2 . . . x2 x1 x0 x
× y y
cn cn−1 . . . c2 c1 (carrying digits)
pn pn−1 pn−2 . . . p2 p1 p0 xy
Write down the formulae which determine the digits of xy and the carrying
digits.
Solution Using t and u to denote the tens-digit and the units-digit, as in the
text, we have
c1 = t(x0 y), p0 = u(x0 y);
ci+1 = t(xi y + ci ), pi = u(xi y + ci ) (1 ≤ i ≤ n − 1);
pn = cn .
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.1.2
Explain carefully how the algorithm described in Ex. 14.1.1 can be extended to
yield the usual ‘long multiplication’ algorithm for multiplying an n-digit integer
by an m-digit integer (n ≥ m ≥ 1).
Solution In the usual method of long multiplication, the multiplier y, in base
10 notation, is
y = ym−1 ym−2 · · · y1 y0 .
The products
x × ym−1 , x × ym−2 , . . . , x × y0
are computed by the method given in Ex. 14.1.1, and 0s are inserted to give the
products
x × 10m−1 ym−1 , x × 10m−2 ym−2 , . . . , x × y0 .
These are then added, using the method explained in the Example, to give
x × y.
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.2 The language of programs
14.2.1
Suppose that initially the values of the identifiers x, y, z are 46, 23, 78, respec-
tively. Write down the values of the identifiers after the execution of the follow-
ing commands.
(a) x:= y + z
(b) x:= x + 12
(c) z:= z - 17
Solution
(a) x y z
101 23 78
(b) x y z
58 23 78
(c) x y z
46 23 61
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.2.2
Use a table such as Table 14.2.1 to work out the values of x, y, z after the
execution of the following commands, when the initial values in each case are
as in Ex. 14.2.1.
(a) begin x:= 27; y:= 13 end
(b) begin z:= 25; x:= y - 12; y:= z + 6 end
(c) begin x:= x + 28; y:= x; x:= y - 14; y:= x + z end
Solution The tables are as follows.
(a) x y z
46 23 78
27 23 78
27 13 78
(b) x y z
46 23 78
46 23 25
11 23 25
11 31 25
(c) x y z
46 23 78
74 23 78
74 74 78
60 74 78
60 138 78
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.3 Algorithms and programs
14.3.1
Tabulate the values of x, y, and z in the version of the Euclidean algorithm given
above, when the initial values of s and t are 725 and 441.
Solution
x y z
725 441
441 284 441
284 157 284
157 127 157
127 30 127
30 7 30
7 2 7
2 1 2
1 0 1
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.3.2
What value of ans is computed by the following algorithm?
sum:= 0;
for i:= 1 to 17 do
sum:= sum + i;
ans:= sum
Solution The algorithm computes the sum
17
i,
i=1
which is equal to 153.
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.4 Proving that an algorithm is correct
14.4.1
The following algorithm is intended to calculate the product of two positive
integers.
x:= m; y:= n; z:= 0;
while y > 0 do
begin z:= z + x; y:= y - 1 end
p:= z
Show that
(a) the algorithm terminates,
(b) the value of z + x ∗ y is invariant under the command begin z := z + x;
y := y − 1 end, and
(c) the value assigned to p is the product of the initial values m and n.
Solution
(a) The sequence of values of y is altered only by the command y := y − 1 and
so it forms a strictly decreasing sequence. By the fundamental property
of integers, the sequence eventually becomes negative, and the algorithm
terminates.
(b) Suppose that the values of x, y, z before and after one execution of the
begin end command are x, y, z; x , y , z . Then
z + x y = (z + x) + x(y − 1) = z + xy.
(c) The initial value of z + x ∗ y is m ∗ n, so this is also the final value. But
the final value of y is 0, so the final value of z is m ∗ n, which is the value
assigned to p.
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.4.2
Tabulate the values of a, b and c during the course of the following algorithm
when the given values of m and n are (a) 31 and 12, (b) 15 and 77.
a:= 0; b:= n; c:= m;
while b > 0 do
begin while b mod 2 = 0 do
begin b:= b div 2; c:= 2 * c end;
a:= a + c; b:= b - 1
end;
q:= a
Solution
(a)
a b c
0 12 31
6 62
3 124
124 2
1 248
q = 372 0
(b)
a b c
0 77 15
15 76
38 30
19 60
75 18
9 120
195 8
4 240
2 480
1 960
q = 1155 0
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.5 Efficiency of algorithms
14.5.1
Suppose we wish to find the least member of a finite set of integers {x1 , x2 , . . . , xn }
using the algorithm indicated in Ex. 14.3.3.
(i) Suggest a good measure of the size of an instance.
(ii) How many comparisons are needed?
(iii) What is the worst case with respect to the number of assignment com-
mands required?
Solution
(i) The obvious choice is n, the number of elements in the set.
(ii) Each one of x2 , x3 , . . . , xn is compared with the current minimum, so
n − 1 comparisons are required.
(iii) The worst case is when
x1 > x2 > · · · > xn ,
because then each comparison much be followed by the assignment of a
new value to the current minimum.
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.6 Growth rates: the O notation
14.6.1
Find a function g(n) (of the form AB ) such that f (n) is O(g(n)), in each of the
cases:
n
(i) f (n) = ;
3
5n3 + 6
(ii) f (n) = ;
n+2
3f (n − 1) (n > 1),
(iii) f (n) =
2 (n = 1);
(iv) f (n) = n!.
Solution
(i) We have
n 1 1
= n(n − 1)(n − 2) ≤ n3 .
3 6 6
Hence the expression is O(n3 ).
(ii) Here we have
5n3 + 6 5n3 + 6
≤ = 5n2 + 6/n ≤ 6n2 ,
n+2 n
for all n ≥ 2. Hence the expression is O(n2 ).
(iii) 3n .
(iv) nn .
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.6.2
Show that for any positive constants C0 , C1 , . . . , Ck , the expression
C0 + C1 n + C2 n2 + · · · + Ck nk
is O(nk ). Prove that the expression is not O(nk−1 ).
Solution Since ni ≤ nk for all natural numbers n and 0 ≤ i ≤ k, it follows
that
C0 + C1 n + · · · + Ck nk ≤ (C0 + C1 + · · · + Ck )nk ,
and so the left-hand side is O(nk ).
Suppose the left-hand side is O(nk−1 ), that is, for some constant A,
C0 + C1 n + · · · + Ck nk ≤ Ank−1 for all n ≥ n0 .
Then, since all the terms are positive,
A
Ck nk ≤ Ank−1 , that is Ck ≤ .
n
But this is clearly only true for a finite number of n, specifically n ≤ A/Ck .
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.7 Comparison of algorithms
14.7.1
Estimate the time taken to compute n100 000 using Algorithms A and B, on the
assumption that we can carry out one multiplication per second.
Solution Using Algorithm A requires about 100 000 seconds. A rough esti-
mate can be found using the fact that there are 3600 seconds in one hour, which
implies that 100 000 seconds is roughly 30 hours.
Using Algorithm B requires about
2(log2 100 000) = 2 × log2 (105 ) = 10 × log2 (10)
seconds. Since log2 (10) is about 3.6, this is roughly 36 seconds.
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.8 Introduction to sorting algorithms
14.8.1
Describe the result of each ‘pass’ of the bubble sort algorithm when the given
list is 4, 3, 9, 5, 1, 2, 7, 8, 6.
Solution
After first pass : 3 4 5 1 2 7 8 6 9
After second pass : 3 4 1 2 5 7 6 8 9
After third pass : 3 1 2 4 5 6 7 8 9
After fourth pass : 1 2 3 4 5 6 7 8 9
The remaining passes result in no changes.
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.8.2
Use insertion sort (by hand!) to arrange the following list in increasing order:
516, 207, 321, 581, 762, 163, 921, 105, 721, 813, 316, 188, 733, 909, 281, 312,
871, 950, 135, 888, 417.
Solution The list is built up as follows:
516
207, 516
207, 321, 516
207, 321, 516, 581
and so on.
Solutions to Chapter 14 Exercises
in Discrete Mathematics by Norman L. Biggs;
2nd Edition 2002
14.8.6
An alternative sorting algorithm is known as selection sorting. In this method,
at the jth pass we select the smallest of xj , xj+1 , . . . , xn and (if it is not xj ) we
switch it with xj . Describe how this method operates when the given list is as
in Ex. 14.8.1. How many comparisons and how many switches are needed, in
general?
Solution
After first pass : 1 3 9 5 4 2 7 8 6
After second pass : 1 2 9 5 4 3 7 8 6
After third pass : 1 2 3 5 4 9 7 8 6
and so on.
There are n − j comparisons in the jth pass, and so the total number is
1
n(n − 1).
(n − 1) + (n − 2) + · · · + 2 + 1 =
2
There is (at most) one switch for each pass, so the total number is at most
n − 1.
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