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CS 361, Lecture 10
Jared Saia University of New Mexico
Outline
Annihilator Operators
We define three basic operations we can perform on this se- quence:
- Multiply the sequence by a constant: cA = 〈ca 0 , ca 1 , ca 2 , · · · 〉
- Shift the sequence to the left: LA = 〈a 1 , a 2 , a 3 , · · · 〉
- Add two sequences: if A = 〈a 0 , a 1 , a 2 , · · · 〉 and B = 〈b 0 , b 1 , b 2 , · · · 〉, then A + B = 〈a 0 + b 0 , a 1 + b 1 , a 2 + b 2 , · · · 〉
Annihilator Description
- We first express our recurrence as a sequence T
- We use these three operators to “annihilate” T , i.e. make it all 0 ’s
- Key rule: can’t multiply by the constant 0
- We can then determine the solution to the recurrence from the sequence of operations performed to annihilate T
Example
- Consider the recurrence T (n) = 2T (n − 1), T (0) = 1
- If we solve for the first few terms of this sequence, we can see they are 〈 20 , 21 , 22 , 23 , · · · 〉
- Thus this recurrence becomes the sequence:
T = 〈 20 , 21 , 22 , 23 , · · · 〉
Example (II)
Let’s annihilate T = 〈 20 , 21 , 22 , 23 , · · · 〉
- Multiplying by a constant c = 2 gets: 2 T = 〈 2 ∗ 20 , 2 ∗ 21 , 2 ∗ 22 , 2 ∗ 23 , · · · 〉 = 〈 21 , 22 , 23 , 24 , · · · 〉
- Shifting one place to the left gets LT = 〈 21 , 22 , 23 , 24 , · · · 〉
- Adding the sequence LT and − 2 T gives: LT − 2 T = 〈 21 − 21 , 22 − 22 , 23 − 23 , · · · 〉 = 〈 0 , 0 , 0 , · · · 〉
Distributive Property
- The distributive property holds for these three operators
- Thus can rewrite LT − 2 T as (L − 2)T
- The operator (L − 2) annihilates T (makes it the sequence of all 0’s)
- Thus (L − 2) is called the annihilator of T
0, the “Forbidden Annihilator”
- Multiplication by 0 will annihilate any sequence
- Thus we disallow multiplication by 0 as an operation
- In particular, we disallow (c−c) = 0 for any c as an annihilator
- Must always have at least one L operator in any annihilator!
Uniqueness
- An annihilator annihilates exactly one type of sequence
- In general, the annihilator L − c annihilates any sequence of the form 〈a 0 cn〉
- If we find the annihilator, we can find the type of sequence, and thus solve the recurrence
- We will need to use the base case for the recurrence to solve for the constant a 0
Example
If we apply operator (L − 3) to sequence T above, it fails to annihilate T
(L − 3)T = LT + (−3)T = 〈 21 , 22 , 23 , · · · 〉 + 〈− 3 × 20 , − 3 × 21 , − 3 × 22 , · · · 〉 = 〈(2 − 3) × 20 , (2 − 3) × 21 , (2 − 3) × 22 , · · · 〉 = (2 − 3)T = −T
Example (II)
What does (L−c) do to other sequences A = 〈a 0 dn〉 when d 6 = c?:
(L − c)A = (L − c)〈a 0 , a 0 d, a 0 d^2 , a 0 d^3 , · · · 〉 = L〈a 0 , a 0 d, a 0 d^2 , a 0 d^3 , · · · 〉 − c〈a 0 , a 0 d, a 0 d^2 , a 0 d^3 , · · · 〉 = 〈a 0 d, a 0 d^2 , a 0 d^3 , · · · 〉 − 〈ca 0 , ca 0 d, ca 0 d^2 , ca 0 d^3 , · · · 〉 = 〈a 0 d − ca 0 , a 0 d^2 − ca 0 d, a 0 d^3 − ca 0 d^2 , · · · 〉 = 〈(d − c)a 0 , (d − c)a 0 d, (d − c)a 0 d^2 , · · · 〉 = (d − c)〈a 0 , a 0 d, a 0 d^2 , · · · 〉 = (d − c)A
Uniqueness
- The last example implies that an annihilator annihilates one type of sequence, but does not annihilate other types of sequences
- Thus Annihilators can help us classify sequences, and thereby solve recurrences
The Details
- Consider: T = 〈a^0 + b^0 , a^1 + b^1 , a^2 + b^2 , · · · 〉
- LT = 〈a^1 + b^1 , a^2 + b^2 , a^3 + b^3 , · · · 〉
- aT = 〈a^1 + a ∗ b^0 , a^2 + a ∗ b^1 , a^3 + a ∗ b^2 , · · · 〉
- LT − aT = 〈(b − a)b^0 , (b − a)b^1 , (b − a)b^2 , · · · 〉
- We know that (L − a)T annihilates the a terms and multiplies the b terms by b − a
- Thus (L − a)T = 〈(b − a)b^0 , (b − a)b^1 , (b − a)b^2 , · · · 〉
- And so the sequence (L − a)T is annihilated by (L − b)
- Thus the annihilator of T is (L − b)(L − a)
Key Point
- In general, the annihilator (L − a)(L − b) (where a 6 = b) will anihilate only all sequences of the form 〈c 1 an^ + c 2 bn〉
- We will often multiply out (L − a)(L − b) to L^2 − (a + b)L + ab
- Left as an exercise to show that (L − a)(L − b)T is the same as (L^2 − (a + b)L + ab)T
Lookup Table
- The annihilator L−a annihilates sequences of the form 〈c 1 an〉
- The annihilator (L − a)(L − b) (where a 6 = b) anihilates se- quences of the form 〈c 1 an^ + c 2 bn〉
Fibonnaci Sequence
- We now know enough to solve the Fibonnaci sequence
- Recall the Fibonnaci recurrence is T (0) = 0, T (1) = 1, and T (n) = T (n − 1) + T (n − 2)
- Let Tn be the n-th element in the sequence
- Then we’ve got:
T = 〈T 0 , T 1 , T 2 , T 3 , · · · 〉 (1) LT = 〈T 1 , T 2 , T 3 , T 4 , · · · 〉 (2) L^2 T = 〈T 2 , T 3 , T 4 , T 5 , · · · 〉 (3)
- Thus L^2 T − LT − T = 〈 0 , 0 , 0 , · · · 〉
- In other words, L^2 − L − 1 is an annihilator for T
Factoring
- L^2 − L − 1 is an annihilator that is not in our lookup table
- However, we can factor this annihilator (using the quadratic formula) to get something similar to what’s in the lookup table
- L^2 − L − 1 = (L − φ)(L − φˆ), where φ = 1+
√ 5 2 and ˆφ^ =^
1 −√ 5
Quadratic Equation
“Me fail English? That’s Unpossible!” - Ralph, the Simpsons
High School Algebra Review:
- To factor something of the form ax^2 + bx + c, we use the Quadratic Formula:
- ax^2 + bx + c factors into (x − φ)(x − ˆφ), where:
φ = b^ +
√ b^2 − 4 ac 2 a
φˆ = b^ −
√ b^2 − 4 ac 2 a
Example
- To factor: L^2 − L − 1
- Rewrite: 1 ∗ L^2 − 1 ∗ L − 1, a = 1, b = −1, c = − 1
- From Quadratic Formula: φ = 1+
√ 5 2 and ˆφ^ =^
1 −
√ 5 2
- So L^2 − L − 1 factors to (L − φ)(L − φˆ)
Back to Fibonnaci
- Recall the Fibonnaci recurrence is T (0) = 0, T (1) = 1, and T (n) = T (n − 1) + T (n − 2)
- We’ve shown the annihilator for T is (L − φ)(L − φˆ), where φ = 1+
√ 5 2 and ˆφ^ =^
1 −
√ 5 2
- If we look this up in the “Lookup Table”, we see that the sequence T must be of the form 〈c 1 φn^ + c 2 ˆφn〉
- All we have left to do is solve for the constants c 1 and c 2
- Can use the base cases to solve for these
Finding the Constants
- We know T = 〈c 1 φn^ + c 2 ˆφn〉, where φ = 1+
√ 5 2 and ˆφ^ =^
1 −√ 5 2
T (0) = c 1 + c 2 = 0 (6) T (1) = c 1 φ + c 2 φˆ = 1 (7)
- We’ve got two equations and two unknowns
- Can solve to get c 1 = √^15 and c 2 = − √^15 ,
The Punchline
- Recall Fibonnaci recurrence: T (0) = 0, T (1) = 1, and T (n) = T (n − 1) + T (n − 2)
- The final explicit formula for T (n) is thus:
T (n) = √^1 5
( 1 +
)n − √^1 5
( 1 −
)n
(Amazingly, T (n) is always an integer, in spite of all of the square roots in its formula.)
Todo
