Intro to Annihilators. CS 362, Lecture 3. Today s Outline. Annihilator Operators

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1 Intro to Annihilators CS 36, Lecture 3 Jared Saia University of New Mexico Suppose we are given a sequence of numbers A = a 0, a 1, a, This might be a sequence like the Fibonacci numbers I.e. A = a 0, a 1, a,...) = (T(1), T(), T(3), Today s Outline Annihilator Operators Listen and Understand! That terminator is out there. It can t be bargained with, it can t be reasoned with! It doesn t feel pity, remorse, or fear. And it absolutely will not stop, ever, until you are dead! - The Terminator Solving Recurrences using Annihilators We define three basic operations we can perform on this sequence: 1. Multiply the sequence by a constant: ca = ca 0, ca 1, ca,. Shift the sequence to the left: LA = a 1, a, a 3, 3. Add two sequences: if A = a 0, a 1, a, and B = b 0, b 1, b,, then A + B = a 0 + b 0, a 1 + b 1, a + b, 1 3

2 Annihilator Description Example (II) Let s annihilate T = 0, 1,, 3, 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 Multiplying by a constant c = gets: T = 0, 1,, 3, = 1,, 3, 4, Shifting one place to the left gets LT = 1,, 3, 4, Adding the sequence LT and T gives: LT T = 1 1,, 3 3, = 0,0,0, The annihilator of T is thus L 4 6 Example Distributive Property Consider the recurrence T(n) = T(n 1), T(0) = 1 If we solve for the first few terms of this sequence, we can see they are 0, 1,, 3, Thus this recurrence becomes the sequence: T = 0, 1,, 3, The distributive property holds for these three operators Thus can rewrite LT T as (L )T The operator (L ) annihilates T (makes it the sequence of all 0 s) Thus (L ) is called the annihilator of T 5 7

3 0, the Forbidden Annihilator Example 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! If we apply operator (L 3) to sequence T above, it fails to annihilate T (L 3)T = LT + ( 3)T = 1,, 3, + 3 0, 3 1, 3, = ( 3) 0,( 3) 1,( 3), = ( 3)T = T 8 10 Uniqueness Example (II) What does (L c) do to other sequences A = a 0 d n when d c?: An annihilator annihilates exactly one type of sequence In general, the annihilator L c annihilates any sequence of the form a 0 c n 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 (L c)a = (L c) a 0, a 0 d, a 0 d, a 0 d 3, = L a 0, a 0 d, a 0 d, a 0 d 3, c a 0, a 0 d, a 0 d, a 0 d 3, = a 0 d, a 0 d, a 0 d 3, ca 0, ca 0 d, ca 0 d, ca 0 d 3, = a 0 d ca 0, a 0 d ca 0 d, a 0 d 3 ca 0 d, = (d c)a 0,(d c)a 0 d,(d c)a 0 d, = (d c) a 0, a 0 d, a 0 d, = (d c)a 9 11

4 Uniqueness Example First calculate the annihilator: 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 Recurrence: T(n) = 4 T(n 1), T(0) = Sequence: T =, 4, 4, 4 3, Calulate the annihilator: LT = 4, 4, 4 3, 4 4, 4T = 4, 4, 4 3, 4 4, Thus LT 4T = 0,0,0, And so L 4 is the annihilator 1 14 Lookup Table Example (II) Now use the annihilator to solve the recurrence The annihilator L a annihilates any sequence of the form c 1 a n Look up the annihilator in the Lookup Table It says: The annihilator L 4 annihilates any sequence of the form c 1 4 n Thus T(n) = c 1 4 n, but what is c 1? We know T(0) =, so T(0) = c = and so c 1 = Thus T(n) = 4 n 13 15

5 In Class Exercise Multiple Operators Consider the recurrence T(n) = 3 T(n 1), T(0) = 3, Q1: Calculate T(0),T(1),T() and T(3) and write out the sequence T Q: Calculate LT, and use it to compute the annihilator of T Q3: Look up this annihilator in the lookup table to get the general solution of the recurrence for T(n) Q4: Now use the base case T(0) = 3 to solve for the constants in the general solution We can string operators together to annihilate more complicated sequences Consider: T = , , + 3, We know that (L ) annihilates the powers of while leaving the powers of 3 essentially untouched Similarly, (L 3) annihilates the powers of 3 while leaving the powers of essentially untouched Thus if we apply both operators, we ll see that (L )(L 3) annihilates the sequence T Multiple Operators The Details We can apply multiple operators to a sequence For example, we can multiply by the constant c and then by the constant d to get the operator cd We can also multiply by c and then shift left to get clt which is the same as LcT We can also shift the sequence twice to the left to get LLT which we ll write in shorthand as L T Consider: T = a 0 + b 0, a 1 + b 1, a + b, LT = a 1 + b 1, a + b, a 3 + b 3, at = a 1 + a b 0, a + a b 1, a 3 + a b, LT at = (b a)b 0,(b a)b 1,(b a)b, We know that (L a)t annihilates the a terms and multiplies the b terms by b a (a constant) Thus (L a)t = (b a)b 0,(b a)b 1,(b a)b, And so the sequence (L a)t is annihilated by (L b) Thus the annihilator of T is (L b)(l a) 17 19

6 Key Point Fibonnaci Sequence In general, the annihilator (L a)(l b) (where a b) will anihilate only all sequences of the form c 1 a n + c b n We will often multiply out (L a)(l b) to L (a+b)l+ab Left as an exercise to show that (L a)(l b)t is the same as (L (a + b)l + ab)t 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 ) Let T n be the n-th element in the sequence Then we ve got: T = T 0, T 1, T, T 3, (1) LT = T 1, T, T 3, T 4, () L T = T, T 3, T 4, T 5, (3) Thus L T LT T = 0,0,0, In other words, L L 1 is an annihilator for T 0 Lookup Table Factoring The annihilator L a annihilates sequences of the form c 1 a n The annihilator (L a)(l b) (where a b) anihilates sequences of the form c 1 a n + c b n L 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 L 1 = (L φ)(l ˆφ), where φ = 1+ 5 and ˆφ =

7 Quadratic Formula Back to Fibonnaci Me fail English? That s Unpossible! - Ralph, the Simpsons High School Algebra Review: To factor something of the form ax + bx + c, we use the Quadratic Formula: ax + bx + c factors into (x φ)(x ˆφ), where: φ = b + b 4ac (4) a ˆφ = b b 4ac (5) a Recall the Fibonnaci recurrence is T(0) = 0, T(1) = 1, and T(n) = T(n 1) + T(n ) We ve shown the annihilator for T is (L φ)(l ˆφ), where φ = 1+ 5 and ˆφ = 1 5 If we look this up in the Lookup Table, we see that the sequence T must be of the form c 1 φ n + c ˆφ n All we have left to do is solve for the constants c 1 and c Can use the base cases to solve for these 4 6 Example Finding the Constants To factor: L L 1 Rewrite: 1 L 1 L 1, a = 1, b = 1, c = 1 From Quadratic Formula: φ = 1+ 5 and ˆφ = 1 5 So L L 1 factors to (L φ)(l ˆφ) We know T = c 1 φ n + c ˆφ n, where φ = 1+ 5 and ˆφ = 1 5 We know T(0) = c 1 + c = 0 (6) T(1) = c 1 φ + c ˆφ = 1 (7) We ve got two equations and two unknowns Can solve to get c 1 = 1 and c 5 = 1, 5 5 7

8 The Punchline Example Recall Fibonnaci recurrence: T(0) = 0, T(1) = 1, and T(n) = T(n 1) + T(n ) The final explicit formula for T(n) is thus: T(n) = 1 5 ( ) n 1 5 ( ) n 1 5 (Amazingly, T(n) is always an integer, in spite of all of the square roots in its formula.) (L a) na n = (n + 1)a n+1 (a)na n = (n + 1)a n+1 na n+1 = (n + 1 n)a n+1 = a n+1 (L a) na n = (L a) a n+1 = A Problem Generalization Our lookup table has a big gap: What does (L a)(l a) annihilate? It turns out it annihilates sequences such as na n It turns out that (L a) d annihilates sequences of the form p(n)a n where p(n) is any polynomial of degree d 1 Example: (L 1) 3 annihilates the sequence n 1 n = 1,4,9,16,5 since p(n) = n is a polynomial of degree d 1 = 9 31

9 Lookup Table Examples (L a) annihilates only all sequences of the form c 0 a n (L a)(l b) annihilates only all sequences of the form c 0 a n + c 1 b n (L a 0 )(L a 1 )...(L a k ) annihilates only sequences of the form c 0 a n 0 + c 1a n c ka n k, here a i a j, when i j (L a) annihilates only sequences of the form (c 0 n+c 1 )a n (L a) k annihilates only sequences of the form p(n)a n, degree(p(n)) = k 1 Q: What does (L 3)(L )(L 1) annihilate? A: c 0 1 n + c 1 n + c 3 n Q: What does (L 3) (L )(L 1) annihilate? A: c 0 1 n + c 1 n + (c n + c 3 )3 n Q: What does (L 1) 4 annihilate? A: (c 0 n 3 + c 1 n + c n + c 3 )1 n Q: What does (L 1) 3 (L ) annihilate? A: (c 0 n + c 1 n + c )1 n + (c 3 n + c 4 ) n 3 34 Lookup Table Annihilator Method (L a 0 ) b 0(L a 1 ) b 1...(L a k ) b k annihilates only sequences of the form: p 1 (n)a n 0 + p (n)a n p k(n)a n k where p i (n) is a polynomial of degree b i 1 (and a i a j, when i j) Write down the annihilator for the recurrence Factor the annihilator Look up the factored annihilator in the Lookup Table to get general solution Solve for constants of the general solution by using initial conditions 33 35

10 Todo HW 1 36