Algorithms Homework – Fall 2000
2.1-1 Let f(n) and g(n) be asymptotically nonnegative functions. Using the basic
definition of  notation, prove that max(f(n), g(n)) = ( f(n) + g(n)).
Show: 0  c1(f(n) + g(n))  max(f(n), g(n))  c2(f(n) + g(n))
max(f(n), g(n))  c2(f(n) + g(n)), since f(n) and g(n) are asymptotically non-negative.
Therefore, c2 = 1 and n0 = 1
max(f(n), g(n))  ½ (f(n) + g(n)),
since the maximum value of ½ (f(n) + g(n)) = max(f(n), g(n))
Therefore, c1 = ½
2.1-2 Show that for any real constants a and b, where b > 0,
(n + a)b = (nb)
Note: (a ch b) means “a choose b” for integers a and b
(n + a)b = nb + (b ch 1)nb–1a + (b ch 2)nb–2a2 + … + ab
So nb is the leading term in the polynomial.
By the Sum Rule (proven in Exercise 2.1-1) the leading term sets the  value.
Therefore, (n + a)b = (nb)
2.2-1 Show that if f(n) and g(n) are monotonically increasing functions, then so are
the functions f(n) + g(n) and f(g(n)), and if f(n) and g(n) are in addition nonnegative,
then f(n) * g(n) is monotonically increasing.
If two monotonically increasing functions are added together, the sum must also be
monotonically increasing since there is nothing in the sum that decreases either of them.
Since g(n) is monotonically increasing, once it is evaluated at any given point and a value
is produced and this value is fed into another monotonically increasing function, f(n), the
result must be monotonically increasing by the definition of a monotonically increasing
function.
If the results of two functions are the same sign, then the product of these results will be
positive. Since both f(n) and g(n) are non-negative and monotonically increasing, the
product of these two must also be monotonically increasing because there is nothing to
decrease them.
PROBLEMS
2-1
Asymptotic behavior of polynomials
Let p(n) = di=0 aini
where ad  0, be a degree-d polynomial in n, and let k be a constant.
Use the definitions of the asymptotic notations to prove the following properties.
a)
If k  d, then p(n) = O(nk).
p(n) = a1nd + a2nd-1 + …
Since k  d, k – d  0
Therefore, nk grows as fast or faster than nd
Therefore, p(n) = O(nk)
Alternatively:
p(n)  d(adnd), for some large n
Therefore, by setting c = d
d(adnd) = O(nd)
Therefore, since k  d
d(adnd) = O(nk)
Therefore, since p(n)  d(adnd)
p(n) = O(nk)
d)
If k  d, then p(n) = o(nk).
Let f(n) = d(adnd)  p(n)
limn f(n)/nk = limn d(adnd)/nk = limn d(adnd-k) = limn dad/nk-d = 0
Therefore, p(n) = o(nk)
2-4
Asymptotic notation properties
Let f(n) and g(n) be asymptotically positive functions. Prove of disprove each of the
following conjectures.
a)
f(n) = O(g(n)) implies g(n) = O(f(n)).
FALSE.
Let f(n) = n2 and g(n) = n3
Then, f(n) = O(g(n)), but g(n)  O(f(n))
b)
f(n) + g(n) = (min(f(n), g(n))).
FALSE.
Let f(n) = n2 and g(n) = n3
Then min(f(n), g(n)) = n2
If f(n) + g(n) = n2 + n3 = ( n2) = (min(f(n), g(n))), then
 c1 and c2 0 and n0  0 such that 0  c1n2  n2 + n3  c2n2, for all n  n0
But, 1 + n  c2, for sufficiently large n
So, this is a contradiction.
c)
f(n) = O(g(n)) implies lg(f(n)) = O(lg(g(n))), where lg(g(n))  0 and
f(n)  1 for all sufficiently large n.
TRUE.
Given: f(n)  cg(n)
Show: lg(f(n))  clg(g(n))
lg(f(n))  clg(g(n)) = lg(f(n))  lg(g(n))c
Since we are given that lg(g(n))  0, then c must be positive.
Therefore, lg(g(n))c must grow faster than lg(f(n)) as n goes to , and is therefore an
asymptotic upper bound.
d)
f(n) = O(g(n)) implies 2f(n) = O(2g(n)).
FALSE.
Let f(n) = n and g(n) = n/2,
then f(n) = O(g(n)), with c  2
However, limn 2n/2/2n  1
Therefore, 2f(n)  O(2g(n))
e)
f(n) = O((f(n))2).
FALSE.
Let f(n) = 1/n2
Since, 1/n2  c(1/n2)2 = c/n4 for sufficiently large n,
f(n)  O((f(n))2)
f)
f(n) = O(g(n)) implies g(n) = (f(n)).
TRUE.
By transpose symmetry.
f(n) = O(g(n)) implies g(n) = (f(n)) means that
f(n)  c(g(n)) implies g(n)  c(f(n))
If, f(n)  c(g(n)), then, 1/c f(n)  g(n), which means that g(n)  c(f(n))
g)
f(n) = (f(n/2)).
FALSE.
f(n) = (f(n/2)) means that
 c1 and c2 0 and n0  0 such that 0  c1f(n/2)  f(n)  c2f(n/2), for all n  n0
Let f(n) = 2n, then
 c1 and c2 0 and n0  0 such that 0  c12n/2  2n  c22n/2, for all n  n0
However, 2n grows faster than 2n/2 for sufficiently large n.
Therefore, the constants c1 and c2 do not exist.