Solving Recurrences
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Transcript Solving Recurrences
Homogeneous Linear
Recurrences
To solve such recurrences we must first know how
to solve an easier type of recurrence relation:
DEF: A linear recurrence relation is said to be
homogeneous if it is a linear combination of
the previous terms of the recurrence without
an additional function of n or a constant.
an = 2an-1 is homogeneous
an = 2an-1 + 2n-3 - an-3 is not.
bk = 2bk-1 - bk-2 + 4bk-3 + 2 is not.
Homogeneous Linear
Recurrences
To solve such recurrences we must first know how
to solve an easier type of recurrence relation:
DEF: A linear recurrence relation is said to be
homogeneous if it is a linear combination of
the previous terms of the recurrence without
an additional function of n or a constant.
an = 2an-1 is homogeneous
an = 2an-1 + 2n-3 - an-3 is not.
bk = 2bk-1 - bk-2 + 4bk-3 + 2 is not.
First Order Homogenous Recurrences
Quiz tomorrow from Solving Recurrence Relations
b0 2
bk 3bk 1k 0
Characteristic equation
D3
General Solution:
bk A(root 1) k
bk A(3) k
Using initial conditions:
b0 A(3) 0
2 A
Closed Form Solution:
bk 2(3) k
2
6
18
54
162
486
1458
4374
0
1
2
3
4
5
6
7
2
6
18
54
162
486
1458
4374
Solving Homogenous Recurrences
Finding a closed form solution to a recurrence
relation.
Multiply both sides by
a0 1
r n 2
a1 8
r n 2 r n r n 2 r n1 2r n 2 r n2 0
an an 1 2an 2n 2
r n 2 n r n 2 n1 2r n 2 n2 0
an an 1 2an 2
Let
an r n
r n r n 1 2r n 2
r n r n 1 2r n 2 0
r 2 r 2 0 Characteristic equation
r 2 2r r 2 0
r (r 2) (r 2) 0
(r 2)(r 1) 0
r2
roots
r 1
Example continued…
The closed form solution is given by:
Let
an A(root1) n B(root2) n
an A(2) n B(1) n
be the solution.
1
Where constants A and B are to
determined by employing initial
conditions. a0 1, a1 8
Plug in n = 0 in equation 1 to obtain:
a0 A(2)0 B(1)0 1 A B
Plug in n = 1 in equation 1 to obtain:
a1 A(2)1 B(1)1 8 2 A B
Adding the two equation, we get
A B 1
2A B 8
3A 9
A 3, B 2
Example continued…
By substituting A = 3 and B = -2 in equation
1, we get the final solution in the closed
form:
an 3(2) 2(1)
n
Verification
an 3(2) n 2(1) n
Test Data
a0 3(2) 0 2(1) 0
a0 1
a0 3 2 1
a1 3(2) 2(1)
1
1
a1 3(2) 2(1) 6 2 8
a2 3(2) 2(1) 3 * 4 2 *1
2
2
12 2 10
a3 3(2) 3 2(1) 3 24 2 26
a4 3(2) 4 2(1) 4 48 2 46
a5 3(2) 5 2(1) 5 96 2 98
a1 8
a2 10
a3 26
a4 46
a5 98
n
Example 2
b0 2
b1 2
bk 7bk 1 10bk 2k 2
D 2 7 D 10
D 2 7 D 10 0
( D 2)(D 5) 0
D 2, D 5
The closed form
solution is given by:
bk A(root1) k B(root2) k
bk A(2) k B(5) k
Using initial conditions:
bk A(2) k B(5) k
b0 A(2) 0 B(5) 0
2 A B
b1 A(2)1 B(5)1
2 2 A 5B
2 A B
2 2 A 5B
2
8
B
,A
3
3
8
2
bk (2) k (5) k
3
3
Solution of Fibonacci Recurrence
a0 1
a1 1
an an1 an2n 2
Solve the quadratic equation:
r 2= r + 1
r 2 - r - 1 = 0 Characteristic equation
to obtain
r1 = (1+5)/2
r2 = (1-5)/2
General solution:
an = A [(1+5)/2]n +B [(1-5)/2]n
Fibonacci Recurrence
Use initial conditions a0 = 0, a1 = 1 to find A,B and obtain
specific solution.
0=a0 = A [(1+5)/2]0 +B [(1-5)/2]0 = A +B
That is A +B = 0 --- (1)
1=a1 = A [(1+5)/2]1 +B [(1-5)/2]1 = A(1+5)/2 +B (1-5)/2
= (A+B )/2 + (A-B )5/2
That is (A+B )/2 + (A-B )5/2 = 1 or
(A+B ) + (A-B )5 = 2 -- (2)
First equation give B = -A. Plug into 2nd:
A = 1/5, B = -1/5
n
Final answer:
1 1 5
an
5
2
1 1 5
2
5
n
Case where the roots are identical
Finding a closed form solution to a recurrence
relation when the roots are identical.
t0 1
t1 3
tk 6tk 1 9tk 2k 2
D 6D 9
2
D 2 6D 9 0
( D 3) 2 0
D3
Test Data
The closed form
solution is given by:
t k A( root 1) k Bk ( root 2) k
t k A(3) k Bk (3) k
t0 1
t1 3
t2 9
t3 27
t 4 81
t5 243
t6 729
Where constants A and B are to determined by
employing initial conditions.
t k A(3) k Bk (3) k
t0 1
I
t1 3
Plug in k = 0 in equation 1 to obtain:
Plug in k= 1 in equation 1 to obtain:
t0 A(3) 0 B * 0 * (3)0 1 A
A 1.
t1 A(3)1 B *1* (3)1 3 3 A 3B
A B 1
A 1
B0
tk (3)
k
A variation of the previous Example
Finding a closed form solution to a recurrence
relation when the roots are identical.
t0 0
t1 3
tk 6tk 1 9tk 2k 2
D 6D 9
2
D 2 6D 9 0
( D 3) 2 0
D3
Test Data
The closed form
solution is given by:
t k A( root 1) k Bk ( root 2) k
t k A(3) k Bk (3) k
t0 0
t1 3
t 2 18
t3 81
t 4 324
Where constants A and B are to determined by
employing initial conditions.
t k A(3) k Bk (3) k
t0 0
t0 0
t1 3
I
t 2 18
t1 3
t3 81
t 4 324
Plug in k = 0 in equation 1 to obtain:
Plug in k= 1 in equation 1 to obtain:
t0 A(3) 0 B * 0 * (3) 0 0 A
A 0.
t1 A(3)1 B *1* (3)1 3 3 A 3B
A B 1
A0
B 1
tk k (3)
k
Relationship between Recurrence Relation and
Sequences
Finding sequences that satisfy a recurrence
relation.
bk 7bk 1 10bk 2k 2
Characteristic equation
D 2 7 D 10
b0, b1, b2, ….
D 7 D 10 0
( D 2)(D 5) 0
D 2, D 5
2
1, 2, 4, 8,
The given recurrence is
satisfied by the following
sequences:
Or
…
1, 5, 25, 125,
1, (2)1, (2)2, (2)3 , …
1, (5)1, (5)2, (5)3 , …
…
Verification
bk 7bk 1 10bk 2k 2
1, 2, 4, 8,
…
b2 7b1 10b0
b2 7 * 2 10*1 4
b3 7b2 10b1
7 * 4 10* 2 28 20 8
1, 5, 25, 125,
…
b2 7b1 10b0
b2 7 * 5 10*1 25
b3 7b2 10b1
7 * 25 10* 5 175 50 125
Practice Problem
a0 2
a1 3
a k 3a k 1 2a k 2 k 2
List first 5 terms of sequence generated by the recurrence.
Solve the recurrence in the close form.
Step 1: Write characteristic equation.
Step 2: Find roots of the characteristic equation.
Step 3: Write the general solution.
ak A(root1) k B(root2) k
Step 4: Find constants A and B using boundary conditions.
Step 5: Substitute constants in the general solution.
Solving Recurrences using
Spreadsheet
a0 2
a1 3
a k 3a k 1 2a k 2 k 2
2
3
5
9
17
33
65
129
257
513
2
3
=3*A2-2*A1
=3*A3-2*A2
=3*A4-2*A3
=3*A5-2*A4
=3*A6-2*A5
=3*A7-2*A6
=3*A8-2*A7
=3*A9-2*A8
3rd order Homogenous Recurrences
a 0 2,
Where A, B, and C are
constants to be
determined by using
boundary conditions.
Let r = 0
a1 1,
a2 1
ar 6ar 1 11ar 2 6ar 3
Characteristic equation
a0 A(1) 0 B(2) 0 C (3) 0
D 3 6D 2 11D 6
2 A B C
A B C 2
D 6D 11D 6 0
3
2
Roots D = 1, D = 2, and D = 3
General Solution:
ar A(1) B(2) C(3)
r
r
r
(I)
Let r = 1
a1 A(1)1 B(2)1 C (3)1
1 A 2 B 3C
A 2 B 3C 1
(II)
3rd order Homogenous Recurrences
Let r = 2
The solution is not complete.
a 2 A(1) 2 B(2) 2 C (3) 2
1 A 4 B 9C
A 4 B 9C 1
(III)
We have to solve (I), (II), and
(III) simultaneously for A, B,
and C.
A BC 2
A 2 B 3C 1
A 4 B 9C 1