MCA 202: Discrete
Mathematics
Instructor
Neelima Gupta
ngupta@cs.du.ac.in
Table Of Contents
Solving Recurrences
The Master Theorem
Recurrence Relations
Recurrences
• The expression:
c



T (n)  
n

 2 T    cn
  2 
n 1
n 1
is a recurrence.
– Recurrence: an equation that describes a function in
terms of its value on smaller functions
Recurrence Examples
0

s(n)  
 c  s ( n  1)
c



T (n)  
n

 2T    c
  2 
n0
n0
n 1
n 1
0

s(n)  
 n  s ( n  1)


c

T (n)  

n
 aT    cn
b

n0
n0
n 1
n 1
Solving Recurrences
• Substitution method
• Iteration method
• Master method
Solving Recurrences
• The substitution method (CLR 4.1)
– “Making a good guess” method
– Guess the form of the answer, then use
induction to find the constants and show that
solution works
– Examples:
• T(n) = 2T(n/2) + (n)  T(n) = (n lg n)
• T(n) = 2T(n/2) + n  ???
Solving Recurrences
• The substitution method (CLR 4.1)
– “Making a good guess” method
– Guess the form of the answer, then use
induction to find the constants and show that
solution works
– Examples:
• T(n) = 2T(n/2) + (n)  T(n) = (n lg n)
• T(n) = 2T(n/2) + n  T(n) = (n lg n)
• T(n) = 2T(n/2 )+ 17) + n  ???
Substitution method
• Guess the form of the solution .
• Use mathematical induction to find the constants and
show that the solution works .
The substitution method can be used to establish either upper
or lower bounds on a recurrence.
An example (Substitution method )
• T(n) = 2T(floor(n/2) ) +n
We guess that the solution is T(n)=0(n lg n).
i.e. to show that T(n) ≤ cn lg n , for some constant c> 0 and n ≥ m.
Assume that this bound holds for [n/2]. So , we get
T(n) ≤ 2(c floor (n/2) lg(floor(n/2))) + n
≤ cn lg(n/2) + n
= cn lg n – cn lg 2 + n
= cn lg n – cn + n
≤ cn lg n
where , the last step holds as long as c≥ 1.
• Boundary conditions :
Suppose , T(1)=1 is the sole boundary condition of the
recurrence .
then , for n=1 , the bound T(n)≤ c n lg n yields
T(1)≤ c lg1=0 , which is at odds with T(1)=1.
Thus ,the base case of our inductive proof fails to hold.
To overcome this difficulty , we can take advantage of the
asymptotic notation which only requires us to prove
T(n)≤c n lg n for n≥ m.
The idea is to remove the difficult boundary condition T(1)= 1
from consideration.
Thus , we can replace T(1) by T(2) as the base cases in the
inductive proof , letting m=2.
Contd....
From the recurrence , with T(1) = 1, we get
T(2)=4
We require T(2)≤ c 2 lg 2
It is clear that , any choice of c≥2 suffices for
the base cases
Are we done?
• Have we proved the case for n =3.
• Have we proved that T(3) ≤ c 3 lg 3.
• No. Since floor(3/2) = 1 and for n =1, it does not hold. So
induction does not apply on n= 3.
• From the recurrence, with T(1) = 1, we get T(3) = 5.
The inductive proof that T(n) ≤ c n lg n for some constant c≥1
can now be completed by choosing c large enough that
T(3)≤c 3 lg 3 also holds.
It is clear that , any choice of c ≥ 2 is sufficient for this to hold.
Thus we can conclude that T(n) ≤ c n lg n for any c ≥ 2 and n ≥ 2.
Wrong Application of induction
Given recurrence: T(n) = 2T(n/2) + 1
Guess: T(n) = O(n)
Claim : Ǝ some constant c and n0 , such that T(n)
<= cn , Ɏ n >= n0 .
Proof: Suppose the claim is true for all values <= n/2
then
T(n)=2T(n/2)+1 (given)
<= 2.c.(n/2) +1 (by induction
hypothesis)
= cn+1
<= (c+1) n , Ɏ n>=1
Why is it Wrong?
Note that T(n/2)<=cn/2 => T(n)<=(c+1)n (this
statement is true but does not help us in
establishing a solution to the problem)
T(1)<=c (when n=1)
T(2)<=(c+1).2
T(2^2)<=(c+2).2^2
.
.
.
T(2^i)<=(c+i).2^I
So, T(n)= T(2^logn)= (c+ log n)n
= Ɵ (n log n)
.
What if we have extra lower order
terms?
• So, does that mean that the claim we initially
made that T(n)=O(n) was wrong ?
• No.
• Recall:
• T(n)=2T(n/2)+1 (given)
<= 2.c.(n/2) +1 (by induction hypothesis)
= cn+1
• Note that in the proof we have an extra lower
order term in our inductive proof.
Given Recurrence: T(n) = 2T(n/2) + 1
Guess: T(n) = O(n)
Claim : Ǝ some constant c and n0 , such that T(n) <= cn - b , Ɏ n >= n0
Proof: Suppose the claim is true for all values <= n/2
then
T(n)=2T(n/2)+1
<=2[c(n/2)-b]+1
<= cn-2b+1
<= cn-b+(1-b)
<= cn-b , Ɏ b>=1
Thus,
T(n/2) <= cn/2 – b
=> T(n) <= cn – b, Ɏ b>=1
Hence, by induction T(n) = O(n), i.e. Our claim was true.
Hence proved.
Another Example (Substitution method )
• T(n) = 2T( n/2 + 17 ) + n,
T(34) < c.34.lg 34 for an appropriate constant c.
•
Since it looks similar to T(n) = T(n/2) + n, We guess that the
solution is T(n) = O(n lg n).
Thanks to Manoj Patial, Roll No. - 20 (MCA 2012)
Let us assume that it holds ∀ m ≤  n/2 + 17
꞊꞊> T(m) ≤ cm lg m ∀ m ≤  n/2 + 17
Hence we can write
T(n) = 2T( n/2 + 17 ) + n
≤ 2c (n/2+17)lg (n/2 +17) + n
≤ 2c (n/2+17)lg(n/2 +17) + n
≤ 2c(n/2+17) lg(n/2+n/4) + n
∀ n/4 ≥ 17
= (cn + 34c) lg(3n/4) + n
= cn lgn – cn lg 4/3 + 34c lg n – 34c lg 4/3 + n
≤ cn lgn – cn lg 4/3 + 34c lg n + n
As lg n = 0(n) therefore 34 c lgn ≤ kn ∀ n ≥ no hence
T(n) ≤ cn lg n + n(1- c lg 4/3) + kn ∀ n ≥ max {no , 68}
= cn lg n + n(k+1-c lg 4/3) …………..(1)
now we can always choose c such that
k+1 – c lg 4/3 ≤ 0
Thanks to Manoj Patial, Roll No. - 20 (MCA 2012)
꞊꞊> k+1 ≤ c lg 4/3
꞊꞊> c ≥ (k+1)/ lg(4/3) = k
꞊꞊ T(n) ≤ cn lg n
for some constant c ≥ k and n ≥ max {no , 68}
…………….(Hence proved)
Boundary Conditions:
n = 35, it holds since it holds for  35/2 + 17 =34, therefore it holds for 35 by
induction.
n = 36, it holds since it holds for  36/2 + 17 =35, therefore it holds for 36 by
induction.
.
.
.
And so on.
Thanks to Manoj Patial, Roll No. - 20 (MCA 2012)
Solving Recurrences using Recursion Tree Method
• Here while solving recurrences, we divide the problem into
subproblems of equal size.
For e.g., T(n) = a T(n/b) + f(n) where a > 1 ,b > 1 and f(n) is a given
function .
F(n) is the cost of splitting or combining the sub problems.
n
T
n/b
T
n/b
Thanks : MCA 2012 Anurag
Aggarwal and Aniruddh Jarial
1) T(n) = 2T(n/2) + n
The recursion tree for this recurrence is :
n
n
T
n/2
T
n/2
n/2
log2 n
n/22
n/22
n
n/2
n/22
n/22
:
:
:
1
1
1
1
1
1
Thanks : MCA 2012 Anurag
Aggarwal and Aniruddh Jarial
1
When we add the values across the levels of the recursion tree, we get a
value of n for every level.
We have
n + n + n + ……
log n times
= n (1 + 1 + 1 + …… log n times)
= n (log2 n)
= Ɵ (n log n)
T(n) = Ɵ (n log n)
Thanks : MCA 2012 Anurag
Aggarwal and Aniruddh Jarial
II.
Given : T(n) = 2T(n/2) + 1
Solution : The recursion tree for the above recurrence is
1
1
1
log n
2
4
:
:
:
Thanks : MCA 2012 Anurag
Aggarwal and Aniruddh Jarial
Now we add up the costs over all levels of the recursion tree, to determine
the cost for the entire tree :
We get series like
1 + 2 + 22 + 23 + …… log n times
which is a G.P.
[ So, using the formula for sum of terms in a G.P. :
a + ar + ar2 + ar3 + …… + ar n – 1 = a( r n – 1 )
r – 1
]
= 1 (2log n – 1)
2–1
= n–1
= Ɵ (n – 1)
(neglecting the lower order terms)
= Ɵ (n)
Thanks : MCA 2012 Anurag
Aggarwal and Aniruddh Jarial
III.
Given : T(n) = T(n/3) + T(2n/3) + n
Solution : The recursion tree for the above recurrence is
n
n
2n/3
n/3
n/3
n
32
log3 n
1
2 n
3 3
2n/3
1 2n
3 3
n/3i
:
:
:
Thanks : MCA 2012 Anurag
Aggarwal and Aniruddh Jarial
n
n .
(3/2)2
log3/2 n
n
When we add the values across the levels of the recursion tree , we get a
value of n for every level.
Since the shortest path from the root to the leaf is
n → n → n → n → …. 1
3
32
33
we have 1 when n = 1
3i
=> n = 3i
Taking log₃ on both the sides
=> log₃ n = i
Thus the height of the shorter tree is log₃ n
T(n) > n log₃ n … A
Thanks : MCA 2012 Anurag
Aggarwal and Aniruddh Jarial
Similarly, the longest path from root to the leaf is
n → 2 n → 2 2n → …1
3
3
So rightmost will be the longest
when
2 k n =1
3
or
n = 1
(3/2)k
=>
k = log3/2 n
T(n) < n log3/2 n
… B
Since base does not matter in asymptotic notation , we guess
from A and B
T(n) = Ɵ (n log2 n)
Thanks : MCA 2012 Anurag
Aggarwal and Aniruddh Jarial
Solving Recurrences
• Another option is “iteration method”
– Expand the recurrence
– Work some algebra to express as a
summation
– Evaluate the summation
• We will show some examples
Assignment 4
• Solve the following recurrence:
T(n) = T(αn) + T(βn) +
where 0 < α ≤ β < 1
n,
Assume suitable initial conditions.
The Master Theorem
• Given: a divide and conquer algorithm
– An algorithm that divides the problem of size
n into a subproblems, each of size n/b
– Let the cost of each stage (i.e., the work to
divide the problem + combine solved
subproblems) be described by the function
f(n)
• Then, the Master Theorem gives us a
cookbook for the algorithm’s running time:
The Master Theorem
• if T(n) = aT(n/b) + f(n) then


log a
 n b



log b a
T (n )   n
log n


  f (n ) 






log a  

f (n)  O n b


  0
log b a
f (n)   n

 c 1

log a  

f (n)   n b
AND

af ( n / b )  cf ( n ) for large n 







Using The Master Method
• T(n) = 9T(n/3) + n
– a=9, b=3, f(n) = n
– nlog a = nlog 9 = (n2)
– Since f(n) = O(nlog 9 - ), where =1, case 1
applies:
b
3
3

T (n)   n
log b a
 when

f (n)  O n
log b a  
– Thus the solution is T(n) = (n2)

More Examples of Master’s
Theorem
•
•
•
•
•
T(n) = 3T(n/5) + n
T(n) = 2T(n/2) + n
T(n) = 2T(n/2) + 1
T(n) = T(n/2) + n
T(n) = T(n/2) + 1
When Master’s Theorem
cannot be applied
• T(n) = 2T(n/2) + n logn
• T(n) = 2T(n/2) + n/ logn
Up Next
Proving the correctness of Algorithms