Using Operators - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

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CS 173:

Discrete Mathematical Structures

Recurrences

Example: solve an = 5an-1 - 6an-2 + 4

Rewrite: an - 5an-1 + 6an-2 = 4 Rewrite again so n is smallest index: an+2 - 5an+1 + 6an = 4 Rewrite again as a sequence: 〈an+2 - 5an+1 + 6an〉 = 〈 4 〉 Rewrite again using operators: (E 2 - 5E + 6)〈an〉 = 〈 4 〉 (E-1) “annihilates” 〈c〉

Recurrences

Example: solve an - 6an-1 + 11an-2 - 6an-3 = 0

Characteristic equation: (r^3 - 6r^2 + 11r - 6) = 0 (r-1)(r-2)(r-3) = 0

General solution: an = A 1 + A 22 n^ + A 33 n

Homogeneous!!!

Not a coincidence that characteristic equation looks the same as the annihilating operator: f(r)=0 corresponds exactly to g(E)〈an〉 = 〈 0 〉

Recurrences

Technique:

Rewrite recurrence in sequence notation. Rewrite left side as OP〈an〉. Find an annihilator for sequence on right side and apply to both sides Read characteristic equation off left side. Solve homogeneous equation as before.

How?

Recurrences

Linear NONhomogeneous recurrence relations with constant coefficients.

c 0 an + c 1 an-1 + c 2 an-2 + … + c (^) k an-k = f(n),

Where f(n) is constant polynomial in n cn^ for some constant c cn^ · polynomial(n)

Group challenge: find the annihilators for the remaining function types.

Use operator (E - 1) “Discrete derivative”

(E - 1)〈a (^) n〉 = E〈a (^) n〉 - 〈a (^) n〉 = 〈a (^) n+1〉 - 〈a (^) n〉 = 〈a (^) n+1 - a (^) n〉

Recurrences

Linear NONhomogeneous recurrence relations with constant coefficients.

c 0 an + c 1 an-1 + c 2 an-2 + … + c (^) k an-k = f(n),

Where f(n) is constant polynomial in n of degree k

Group challenge: find the annihilators for the remaining function types.

Use operator (E - 1) “Discrete derivative”

(E - 1)〈a (^) n〉 = E〈a (^) n〉 - 〈a (^) n〉 = 〈a (^) n+1〉 - 〈a (^) n〉 = 〈a (^) n+1 - a (^) n〉

Annihilator: (E - 1) k+

Recurrences

Table of handy annihilators:

〈c〉 E-

〈polynomial(n) of degree ≤ k〉 (E-1) k+

〈c n〉 E-c 〈c n·polynomial(n) of degree ≤ k〉 (E-c) k+

Sequence Annihilator

Recurrences

Another little helpful fact:

Suppose operator A annihilates 〈an〉

And B annihilates 〈b (^) n〉

Then AB annihilates 〈an + bn〉

Recurrences

Solve an = 2an-1 + 2n^ - 1, a 0 = 0.

General solution: an = (A 1 + A 2 n)2n^ + A 3

Need 3 initial values because we have 3 unknowns to solve for.

A 1 = 2(0) + 2^1 - 1 = 1, A 2 = 2(1) + 2 2 - 1 = 5

A 1 + A 3 = 0

2A 1 + 2A 2 + A 3 = 1

4A 1 + 8A 2 + A 3 = 1

A 1 = -1, A 2 = A 3 = 1

an = (n - 1)2 n^ + 1

Solve 3 equations in 3 unknowns (using the general solution):

Recurrences

Solve an = 2an/2 + n-1, a 1 = 0.

Mergesort

of comparisons