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CS 350 Algorithms and Complexity
Winter 2015
Lecture 4: Analyzing Recursive Algorithms
Andrew P. Black
Department of Computer Science
Portland State University
Tuesday, 20 January 2015
General Plan for Analysis of Recursive algorithms
✦
✦
✦
✦
✦
Decide on parameter n indicating input size
Identify algorithm’s basic operation
Determine worst, average, and best cases
for input of size n
Set up a recurrence relation, with initial
condition, for the number of times the basic
operation is executed
Solve the recurrence, or at least ascertain
the order of growth of the solution (see
Levitin Appendix B)
2
Tuesday, 20 January 2015
Ex 2.4, Problem 1(a)
!
Use a piece of paper and do this now,
individually.
"
Solve this recurrence relation:
x(n) = x(n
1) + 5
for n > 1
x(1) = 0
3
Tuesday, 20 January 2015
Individual Problem (Q1):
Solve the recurrence x(n) = x(n﹣1) + 5 for n > 1
x(1) = 0
What’s the answer?
4
Tuesday, 20 January 2015
Individual Problem (Q1):
Solve the recurrence x(n) = x(n﹣1) + 5 for n > 1
x(1) = 0
What’s the answer?
A.
B.
C.
D.
x(n) = n﹣1
x(n) = 5n
x(n) = 5n﹣5
None of the above
5
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
1) = x(n
2) + 5
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
1) = x(n
2) + 5
substitute for x(n−1):
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
substitute for x(n−1):
x(n) = x(n
1) = x(n
2) + 5
2) + 5 + 5
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
substitute for x(n−1):
x(n) = x(n
1) = x(n
2) + 5
2) + 5 + 5
substitute for x(n−2):
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
substitute for x(n−1):
x(n) = x(n
substitute for x(n−2):
= x(n
1) = x(n
2) + 5
2) + 5 + 5
3) + 5 + 5 + 5
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
substitute for x(n−1):
x(n) = x(n
substitute for x(n−2):
= x(n
1) = x(n
2) + 5
2) + 5 + 5
3) + 5 + 5 + 5
generalize:
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
substitute for x(n−1):
x(n) = x(n
substitute for x(n−2):
= x(n
3) + 5 + 5 + 5
generalize:
= x(n
i) + 5i
1) = x(n
2) + 5
2) + 5 + 5
8i < n
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
substitute for x(n−1):
x(n) = x(n
substitute for x(n−2):
= x(n
3) + 5 + 5 + 5
generalize:
= x(n
i) + 5i
1) = x(n
2) + 5
2) + 5 + 5
8i < n
put i = (n−1) :
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
substitute for x(n−1):
x(n) = x(n
substitute for x(n−2):
= x(n
3) + 5 + 5 + 5
generalize:
= x(n
i) + 5i
put i = (n−1) :
= x(n
(n
1) = x(n
2) + 5
2) + 5 + 5
8i < n
1)) + 5(n
1)
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
substitute for x(n−1):
x(n) = x(n
substitute for x(n−2):
= x(n
3) + 5 + 5 + 5
generalize:
= x(n
i) + 5i
put i = (n−1) :
= x(n
(n
1) = x(n
2) + 5
2) + 5 + 5
8i < n
1)) + 5(n
1)
substitute for x(1):
6
Tuesday, 20 January 2015
My Solution
x(n) = x(n
1) + 5 for all n > 1
x(1) = 0
replace n by n−1:
x(n
substitute for x(n−1):
x(n) = x(n
substitute for x(n−2):
= x(n
3) + 5 + 5 + 5
generalize:
= x(n
i) + 5i
put i = (n−1) :
= x(n
(n
substitute for x(1):
= 5(n
1)
1) = x(n
2) + 5
2) + 5 + 5
8i < n
1)) + 5(n
1)
6
Tuesday, 20 January 2015
Ex 2.4, Problem 1(c)
!
Use a piece of paper and do this now,
individually.
"
Solve this recurrence relation:
x(n) = x(n
1) + n
for n > 0
x(0) = 0
7
Tuesday, 20 January 2015
Ex 2.4, Problem 1(d)
!
Use a piece of paper and do this now,
individually.
k
" Solve this recurrence relation for n = 2 :
x(n) = x(n/2) + n for n > 1
x(1) = 1
8
Tuesday, 20 January 2015
What does that mean?
9
Tuesday, 20 January 2015
What does that mean?
!
“Solving a Recurrence relation” means:
9
Tuesday, 20 January 2015
What does that mean?
!
“Solving a Recurrence relation” means:
"
find an explicit (non-recursive) formula that
satisfies the relation and the initial condition.
9
Tuesday, 20 January 2015
What does that mean?
!
“Solving a Recurrence relation” means:
"
"
find an explicit (non-recursive) formula that
satisfies the relation and the initial condition.
For example, for the relation
x(n) = 3x(n
the solution is
x(n) = 4 ⇥ 3n
1) for n > 1, x(1) = 4
1
9
Tuesday, 20 January 2015
What does that mean?
!
“Solving a Recurrence relation” means:
"
"
find an explicit (non-recursive) formula that
satisfies the relation and the initial condition.
For example, for the relation
x(n) = 3x(n
the solution is
"
x(n) = 4 ⇥ 3n
1) for n > 1, x(1) = 4
1
Check:
9
Tuesday, 20 January 2015
What does that mean?
!
“Solving a Recurrence relation” means:
"
"
find an explicit (non-recursive) formula that
satisfies the relation and the initial condition.
For example, for the relation
x(n) = 3x(n
the solution is
"
x(n) = 4 ⇥ 3n
1) for n > 1, x(1) = 4
1
Check:
x(1) = 4 ⇥ 30 = 4 ⇥ 1 = 4
9
Tuesday, 20 January 2015
What does that mean?
!
“Solving a Recurrence relation” means:
"
"
find an explicit (non-recursive) formula that
satisfies the relation and the initial condition.
For example, for the relation
x(n) = 3x(n
the solution is
"
x(n) = 4 ⇥ 3n
1) for n > 1, x(1) = 4
1
Check:
x(1) = 4 ⇥ 30 = 4 ⇥ 1 = 4
x(n) = 3x(n 1)
definition of recurrence
=
=
Tuesday, 20 January 2015
3 ⇥ [4 ⇥ 3(n
4 ⇥ 3n
1
1) 1
]
substitute solution
= x(n)
9
What does that mean?
!
“Solving a Recurrence relation” means:
"
"
find an explicit (non-recursive) formula that
satisfies the relation and the initial condition.
For example, for the relation
x(n) = 3x(n
the solution is
"
x(n) = 4 ⇥ 3n
1) for n > 1, x(1) = 4
1
Check:
x(1) = 4 ⇥ 30 = 4 ⇥ 1 = 4
x(n) = 3x(n 1)
definition of recurrence
=
=
Tuesday, 20 January 2015
3 ⇥ [4 ⇥ 3(n
4 ⇥ 3n
1
1) 1
]
substitute solution
= x(n)
9
What does that mean?
!
“Solving a Recurrence relation” means:
"
"
find an explicit (non-recursive) formula that
satisfies the relation and the initial condition.
For example, for the relation
x(n) = 3x(n
the solution is
"
x(n) = 4 ⇥ 3n
1) for n > 1, x(1) = 4
1
Check:
x(1) = 4 ⇥ 30 = 4 ⇥ 1 = 4
x(n) = 3x(n 1)
definition of recurrence
=
=
Tuesday, 20 January 2015
3 ⇥ [4 ⇥ 3(n
4 ⇥ 3n
1
1) 1
]
substitute solution
= x(n)
9
c. x(n) = x(n − 1) + n for n > 0,
x(0) = 0
Ex
2.4,
Problem
2
e. x(n) = x(n/3) + 1 for n > 1, x(1) = 1
d. x(n) = x(n/2) + n for n > 1,
x(1) = 1 (solve for n = 2k )
(solve for n = 3k )
2. Set up and solve a recurrence relation for the number of calls made by
F (n), the recursive algorithm for computing n!.
3. Consider the following recursive algorithm for computing the sum of the
F(n)n ≝ if
= 0 = 13 + 23 + ... + n3 .
first
cubes:n S(n)
then return 1
Algorithm S(n)
else return F(n – 1) ⨉ n
//Input: A positive integer n
//Output: The sum of the first n cubes
if n = 1 return 1
else return S(n − 1) + n ∗ n ∗ n
a. Set up and solve a recurrence relation for the number of times the
algorithm’s basic operation is executed.
b. How does this algorithm compare with the straightforward nonrecursive
algorithm for computing this function?
10
4. Consider the following recursive algorithm.
Tuesday, 20 January 2015
my solution
F(n) ≝ if n = 0
then return 1
else return F(n – 1) ⨉ n
11
Tuesday, 20 January 2015
my solution
F(n) ≝ if n = 0
then return 1
else return F(n – 1) ⨉ n
Let C(n) be the number of calls made in computing F (n)
11
Tuesday, 20 January 2015
my solution
F(n) ≝ if n = 0
then return 1
else return F(n – 1) ⨉ n
Let C(n) be the number of calls made in computing F (n)
C(n) = C(n
C(0) = 1
1) + 1
(when n = 0, there is 1 call)
11
Tuesday, 20 January 2015
my solution
F(n) ≝ if n = 0
then return 1
else return F(n – 1) ⨉ n
Let C(n) be the number of calls made in computing F (n)
C(n) = C(n
1) + 1
C(0) = 1
(when n = 0, there is 1 call)
C(n) = C(n
= [C(n
1)
+1
2) + 1] + 1
11
Tuesday, 20 January 2015
my solution
F(n) ≝ if n = 0
then return 1
else return F(n – 1) ⨉ n
Let C(n) be the number of calls made in computing F (n)
C(n) = C(n
1) + 1
C(0) = 1
(when n = 0, there is 1 call)
C(n) = C(n
1)
+1
= [C(n
2) + 1] + 1
= C(n
2) + 2
11
Tuesday, 20 January 2015
my solution
F(n) ≝ if n = 0
then return 1
else return F(n – 1) ⨉ n
Let C(n) be the number of calls made in computing F (n)
C(n) = C(n
1) + 1
C(0) = 1
(when n = 0, there is 1 call)
C(n) = C(n
1)
+1
= [C(n
2) + 1] + 1
= C(n
= C(n
2) + 2
i) + i 8i < n
(generalize)
11
Tuesday, 20 January 2015
my solution
F(n) ≝ if n = 0
then return 1
else return F(n – 1) ⨉ n
Let C(n) be the number of calls made in computing F (n)
C(n) = C(n
1) + 1
C(0) = 1
(when n = 0, there is 1 call)
C(n) = C(n
= [C(n
Put i = n:
1)
+1
2) + 1] + 1
= C(n 2) + 2
= C(n i) + i 8i < n
= C(0) + n
(generalize)
11
Tuesday, 20 January 2015
my solution
F(n) ≝ if n = 0
then return 1
else return F(n – 1) ⨉ n
Let C(n) be the number of calls made in computing F (n)
C(n) = C(n
1) + 1
C(0) = 1
(when n = 0, there is 1 call)
C(n) = C(n
= [C(n
Put i = n:
1)
+1
2) + 1] + 1
= C(n 2) + 2
= C(n i) + i 8i < n
= C(0) + n
(generalize)
= 1+n
11
Tuesday, 20 January 2015
my solution
F(n) ≝ if n = 0
then return 1
else return F(n – 1) ⨉ n
Let C(n) be the number of calls made in computing F (n)
C(n) = C(n
1) + 1
C(0) = 1
(when n = 0, there is 1 call)
C(n) = C(n
= [C(n
Put i = n:
1)
+1
2) + 1] + 1
= C(n 2) + 2
= C(n i) + i 8i < n
= C(0) + n
(generalize)
= 1+n
Now check!
Tuesday, 20 January 2015
11
e. x(n) = x(n/3) + 1 for n > 1,
x(1) = 1 (solve for n = 3k )
Ex 2.4, Problem 3
2. Set up and solve a recurrence relation for the number of calls made by
F (n), the recursive algorithm for computing n!.
3. Consider the following recursive algorithm for computing the sum of the
first n cubes: S(n) = 13 + 23 + ... + n3 .
Algorithm S(n)
//Input: A positive integer n
//Output: The sum of the first n cubes
if n = 1 return 1
else return S(n − 1) + n ∗ n ∗ n
a. Set up and solve a recurrence relation for the number of times the
algorithm’s basic operation is executed.
b. How does this algorithm compare with the straightforward nonrecursive
algorithm for computing this function?
4. Consider the following recursive algorithm.
Algorithm Q(n)
//Input: A positive integer n
if n = 1 return 1
else return Q(n − 1) + 2 ∗ n − 1
Tuesday, 20 January 2015
12
e. x(n) = x(n/3) + 1 for n > 1, x(1) = 1 (solve for n = 3kk )
e. x(n) = x(n/3) + 1 for n > 1, x(1) = 1 (solve for n = 3 )
Set up and solve a recurrence relation for the number of calls made by
Set up and solve a recurrence relation for the number of calls made by
F (n), the recursive algorithm for computing n!.
F (n), the recursive algorithm for computing n!.
Consider the following recursive algorithm for computing the sum of the
Consider the following recursive algorithm for computing the sum of the
first n cubes: S(n) = 133 + 233 + ... + n33 .
first n cubes: S(n) = 1 + 2 + ... + n .
Ex 2.4, Problem 3
2.
2.
3.
3.
Algorithm S(n)
Algorithm S(n)
//Input: A positive integer n
//Input: A positive integer n
//Output: The sum of the first n cubes
//Output: The sum of the first n cubes
if n = 1 return 1
if n = 1 return 1
else return S(n − 1) + n ∗ n ∗ n
else return S(n − 1) + n ∗ n ∗ n
a.
and solve
solve aa recurrence
recurrence relation
relation for
for the
the number
number of
of times
times the
the
a. Set
Set up
up and
algorithm’s
basic operation
operation is
is executed.
executed.
algorithm’s basic
b.
this algorithm
algorithm compare
compare with
with the
the straightforward
straightforwardnonrecursive
nonrecursive
b. How
How does
does this
algorithm
for computing
computing this
this function?
function?
algorithm for
S
1 the
4.
the following
following recursive
recursive algorithm.
algorithm.
4. Consider
Consider
for i
2 to n do
Algorithm
Q(n)
S
S + Q(n)
i⇥i⇥i
Algorithm
//Input:
A positive
positive integer
integer nn
return SA
//Input:
if
return 11
if n
n=
=1
1 return
else
Q(n −
− 1)
1) +
+ 22 ∗∗ nn −
− 11
else return
return Q(n
Tuesday, 20 January 2015
12
algorithm’s basic operation is executed.
Ex 2.4, Problem 4(a)
b. How does this algorithm compare with the straightforward nonrecursive
algorithm for computing this function?
4. Consider the following recursive algorithm.
Algorithm Q(n)
//Input: A positive integer n
if n = 1 return 1
else return Q(n − 1) + 2 ∗ n − 1
a. Set up a recurrence relation for this function’s values and solve it
to determine what this algorithm computes.
b. Set up a recurrence relation for the number of multiplications made by
this algorithm and solve it.
c. Set up a recurrence relation for the number of additions/subtractions
made by this algorithm and solve it.
27
Tuesday, 20 January 2015
13
algorithm’s basic operation is executed.
Ex 2.4, Problem 4(a)
b. How does this algorithm compare with the straightforward nonrecursive
algorithm for computing this function?
4. Consider the following recursive algorithm.
Algorithm Q(n)
//Input: A positive integer n
if n = 1 return 1
else return Q(n − 1) + 2 ∗ n − 1
a. Set up a recurrence relation for this function’s values and solve it
to determine what this algorithm computes.
b. Set up a recurrence relation for the number of multiplications made by
this algorithm and solve it.
c. Set up a recurrence relation for the number of additions/subtractions
made by this algorithm and solve it.
27
Tuesday, 20 January 2015
13
algorithm’s basic operation is executed.
Ex 2.4, Problem 4(a)
b. How does this algorithm compare with the straightforward nonrecursive
algorithm for computing this function?
4. Consider the following recursive algorithm.
Algorithm Q(n)
//Input: A positive integer n
if n = 1 return 1
else return Q(n − 1) + 2 ∗ n − 1
a. Set up a recurrence relation for this function’s values and solve it
to determine what this algorithm computes.
b. Set up a recurrence relation for the number of multiplications made by
this algorithm and solve it.
c. Set up a recurrence relation for the number of additions/subtractions
made by this algorithm and solve it.
27
Tuesday, 20 January 2015
13
Ex 2.4, Problem 8
b. Set up a recurrence relation for the number of additions made by
the nonrecursive version of this algorithm (see Section 2.3, Example 4)
and solve it.
7. a. Design a recursive algorithm for computing 2n for any nonnegative
integer n that is based on the formula: 2n = 2n−1 + 2n−1 .
b. Set up a recurrence relation for the number of additions made by
the algorithm and solve it.
c. Draw a tree of recursive calls for this algorithm and count the number
of calls made by the algorithm.
d. Is it a good algorithm for solving this problem?
8. Consider the following recursive algorithm.
Algorithm Min1 (A[0..n − 1])
//Input: An array A[0..n − 1] of real numbers
if n = 1 return A[0]
else temp ← Min1 (A[0..n − 2])
if temp ≤ A[n − 1] return temp
else return A[n − 1]
Tuesday, 20 January 2015
14
the one for the recursive version:
Solution to Problem 8
A(n) = A(⌊n/2⌋) + 1 for n > 1,
A(1) = 0,
with the solution A(n) = ⌊log2 n⌋ + 1.
7. a. Algorithm Power (n)
//Computes 2n recursively by the formula 2n = 2n−1 + 2n−1
//Input: A nonnegative integer n
//Output: Returns 2n
if n = 0 return 1
else return P ower(n − 1) + P ower(n − 1)
We didn’t
simplify!
36
15
Tuesday, 20 January 2015
the one for the recursive version:
Solution to Problem 8
A(n) = A(⌊n/2⌋) + 1 for n > 1,
A(1) = 0,
with the solution A(n) = ⌊log2 n⌋ + 1.
7. a. Algorithm Power (n)
//Computes 2n recursively by the formula 2n = 2n−1 + 2n−1
//Input: A nonnegative integer n
//Output: Returns 2n
if n = 0 return 1
else return P ower(n − 1) + P ower(n − 1)
We didn’t
simplify!
b. A(n) = 2A(n − 1) + 1, A(0) = 0.
A(n) =
=
=
=
=
=
=
36
2A(n − 1) + 1
2[2A(n − 2) + 1] + 1 = 22 A(n − 2) + 2 + 1
22 [2A(n − 3) + 1] + 2 + 1 = 23 A(n − 3) + 22 + 2 + 1
...
2i A(n − i) + 2i−1 + 2i−2 + ... + 1
...
2n A(0) + 2n−1 + 2n−2 + ... + 1 = 2n−1 + 2n−2 + ... + 1 = 2n − 1.
c. The tree of recursive calls for this algorithm looks as follows:
n
Tuesday, 20 January 2015
n-2
15
n-1
n-1
n-2
n-2
n-2
(n) = 2A(n − 1) + 1, A(0) = 0.
Solution to Problem 8
=
=
=
=
=
=
=
2A(n − 1) + 1
2[2A(n − 2) + 1] + 1 = 22 A(n − 2) + 2 + 1
22 [2A(n − 3) + 1] + 2 + 1 = 23 A(n − 3) + 22 + 2 + 1
...
2i A(n − i) + 2i−1 + 2i−2 + ... + 1
...
2n A(0) + 2n−1 + 2n−2 + ... + 1 = 2n−1 + 2n−2 + ... + 1 = 2n − 1.
he tree of recursive calls for this algorithm looks as follows:
n
n-1
n-1
1
0
...
n-2
0
n-2
1
0
0
...
n-2
n-2
...
1
0
0
0
1
0
16
that it has one extra level compared to the similar tree for the Tower
Tuesday, 20 January 2015
b. Which of the algorithms Min1 or Min2 is faster? Can you suggest an algorithm for the problem they solve that would be more efficient
than either of them?
Ex 2.4, Problem 11
10. The determinant of an n-by-n matrix
⎡
a11
⎢ a21
A=⎢
⎣
an1
⎤
a1n
a2n ⎥
⎥,
⎦
ann
denoted det A, can be defined as a11 for n = 1 and, for n > 1, by the
recursive formula
n
'
det A =
sj a1j det Aj ,
j=1
where sj is +1 if j is odd and -1 if j is even, a1j is the element in row
1 and column j, and Aj is the (n − 1)-by-(n − 1) matrix obtained from
matrix A by deleting its row 1 and column j.
a.◃ Set up a recurrence relation for the number of multiplications made
by the algorithm implementing this recursive definition. (Ignore multiplications by sj .)
b.◃ Without solving the recurrence, what can you say about the solution’s order of growth as compared to n! ?
17
9. von Neumann neighborhood revisited Find the number of cells in the von
Neumann
neighborhood of range n (see Problem 11 in Exercises 2.3) by
Tuesday,
20 January 2015
b. Which of the algorithms Min1 or Min2 is faster? Can you suggest an algorithm for the problem they solve that would be more efficient
than either of them?
Ex 2.4, Problem 11
10. The determinant of an n-by-n matrix
⎡
a11
⎢ a21
A=⎢
⎣
an1
⎤
a1n
a2n ⎥
⎥,
⎦
ann
denoted det A, can be defined as a11 for n = 1 and, for n > 1, by the
recursive formula
n
'
det A =
sj a1j det Aj ,
j=1
where sj is +1 if j is odd and -1 if j is even, a1j is the element in row
1 and column j, and Aj is the (n − 1)-by-(n − 1) matrix obtained from
matrix A by deleting its row 1 and column j.
a.◃ Set up a recurrence relation for the number of multiplications made
by the algorithm implementing this recursive definition. (Ignore multiplications by sj .)
b.◃ Without solving the recurrence, what can you say about the solution’s order of growth as compared to n! ?
17
9. von Neumann neighborhood revisited Find the number of cells in the von
Neumann
neighborhood of range n (see Problem 11 in Exercises 2.3) by
Tuesday,
20 January 2015
my solution
18
Tuesday, 20 January 2015
my solution
Let M (n) be the number of multiplications made in
computing the determinant of an n ⇥ n matrix.
M (1) = 0
n
X
M (n) =
(1 + M (n
j=1
1))
8n > 1
18
Tuesday, 20 January 2015
my solution
Let M (n) be the number of multiplications made in
computing the determinant of an n ⇥ n matrix.
M (1) = 0
n
X
M (n) =
(1 + M (n
1))
8n > 1
M (n) = n ⇥ (1 + M (n
1))
8n > 1
j=1
M (n) = n + nM (n
1)
18
Tuesday, 20 January 2015
my solution
Let M (n) be the number of multiplications made in
computing the determinant of an n ⇥ n matrix.
M (1) = 0
n
X
M (n) =
(1 + M (n
1))
8n > 1
M (n) = n ⇥ (1 + M (n
1))
8n > 1
j=1
M (n) = n + nM (n
1)
Without solving this relation, what can you say about M’s
order of growth, compared to n! ?
18
Tuesday, 20 January 2015