Tutorial 2 WANG PENG. Department of Systems Engineering and Engineering Management, The Chinese University of Hong Kong. September 28, 2017

Transcription

1 Tutorial 2 WANG PENG Department of Systems Engineering and Engineering Management, The Chinese University of Hong Kong September 28, 2017 WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

2 Outline 1 Multiple Sums 2 Recurrences WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

3 Two Basic Rules for Summations Two simple rules can help you simplify your computation of sums. The Distributive Law ( a i ) m j=1 b j = j=1 m a i b j Change of Variable a i = j=1 a j WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

4 Mutiple Sums Multiple Sums Consider a m n matrix as follows: a 11 a 12 a 1n a 21 a 22 a 2n..... a m1 a m2 a mn Then we want to sum up all the elements in the matrix, i.e., S = a 11 + a a 1n + a a m1 + + a mn And we can use the double sum to express above summation compactly as follows: m S = j=1 a ij WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

5 Multiple Sums Interchange the Order of Summation When inner and outer indices are unrelated, we can change the order of summation, that is, m m a ij = j=1 j=1 Note: Summing the rows and summing the columns should give the same result. When inner and outer indices are related, we need to take care of the relationship between them, that is, j a ij = Note: j=1 a ij j=i j=i a ij j=1 WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17 a ij a ij

6 Multiple Sums Example 1 for Multiple Summation Problem: Evaluate the sum i 1 (i j) j=1 in closed form. Show all your work. You are allowed to use without proof the fact that i 2 = n(n + 1)(2n + 1), 6 = n(n + 1) 2 WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

7 Multiple Sums Example 1 for Multiple Summation Solution: i 1 (i j) = j=1 = i 1 i 1 i j j=1 j=1 [i(i 1) 12 ] i(i 1) = 1 (i 2 i) 2 ( = 1 n ) i 2 i 2 = 1 [ n(n + 1)(2n + 1) 2 6 ] n(n + 1) 2 = 1 n(n 1)(n + 1) 6 WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

8 How to find the Recurrence Definition 1 A recurrence describes a sequence of numbers. Early terms are specified explicitly and later terms are expressed as a function of their predecessors. Example 2 How many different ways are there to climb n stairs, if you can either step up (S) one stair or hop up two (H)? After making an S move, there are n 1 stairs to climb. After making an H move, there are n 2 stairs to climb. Let T (n) be the number of ways to climb n stairs. Then, since the first move must either S or H, so we get the recurrence as follows: T (n) = T (n 1) + T (n 2) Impose the initial conditions on the recurrences WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

9 How to Solve the Recurrences Finally, we can get the following recurrence { T (n) = T (n 1) + T (n 2), n 3 ( ) T (1) = 1, T (2) = 2 Then we need to use the tools from Linear Homogeneous Recurrences to solve it. Definition 3 Recurrence has the following form is the Linear Homogeneous Recurrences: T (n) = a 1 T (n 1) + a 2 T (n 2) + + a d T (n d) ( ) for some given constants a 1, a 2,, a d and d 1. WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

10 Linear Homogeneous Recurrences Then we develop the following steps to solve the Linear Homogeneous Recurrences: 1 Guess and Verify Method to try T (n) = x n, then plugs it into (*): x n = a 1 x n 1 + a 2 x n a d x n d ( ) 2 Divide (**) by x n d, we get its characteristic equation as follows: x d = a 1 x d 1 + a 2 x d a d ( ) 3 Solve ( ) to get the solutions to ( ). Theorem 4 Let r be the root of ( ), then T (n) = r n solves ( ). WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

11 Linear Homogeneous Recurrences Note: If we apply the above steps to solve ( ), we found that the roots of the characteristic equation of ( ) didn t satisfy the initial condition of ( ). Theorem 5 Let r 1, r 2,, r d be the roots of ( ), then from Theorem 4, we get that T 1 (n) = r1 n, T 2 (n) = r2 n,, T d (n) = rd n all solve ( ). Moreover, for any numbers α 1,, α d, the linear combination T 0 (n) of T 1 (n),, T d (n) also solves (*). T 0 (n) = α 1 T 1 (n) + α 2 T 2 (n) + + α d T d (n) WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

12 Proof of Theorem 5 Recurrences Proof. Since T 1 (n) = r n 1, T 2 (n) = r n 2,, T d (n) = r n d all solve ( ), we get that T i (n) = a 1 T i (n 1) + a 2 T i (n 2) + + a d T i (n d), i = 1,, d Then since T 0 (n) = α 1 T 1 (n) + α 2 T 2 (n) + + α d T d (n), plugs in above equation for i = 1,, d, T 0 (n) = d α i (a 1 T i (n 1) + a 2 T i (n 2) + + a d T i (n d)) d = a 1 α i T i (n 1) + + a d d α i T i (n d) = a 1 T 0 (n 1) + + a d T d (n d) (By definition of T 0 (n)) WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

13 Proceed the Roots with k Counting Multiplicity, k 1 Theorem 6 If r is a root of the characteristic equation of the multiplicity k 1, then all solve the Recurrence (*). r n, nr n,, n k 1 r n Goal: Solve the Linear Homogeneous Recurrence as follows: { T (n) = a 1 T (n 1) + a 2 T (n 2) + + a d T (n d) (Main Body) T (0) = t 0, T (1) = t 1,, T (d 1) = t d 1 (Initial Condition) And we can use the following framework to solve the above problem. WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

14 Framework to Solve the Linear Homogeneous Recurrences 1 Find the Characteristic Equation x d = a 1 x d 1 + a 2 x d a d 2 Find the roots of Characteristic Equation By Theorem 6, if r is a root with multiplicity k 1, then r n, nr n,, n k 1 r n will solve the main body of the recurrence. 3 Form the Linear Combination By Theorem 5, if T 1 (n), T 2 (n),, T d (n) are the solutions obtained in Step 2, then for any α 1, α 2,, α d, T 0 (n) = α 1 T 1 (n) + α 2 T 2 (n) + + α d T d (n) is also a solution to the main body of the recurrence. 4 Impose the initial condition Use the initial condition for T 0 (n) to fix α 1, α 2,, α d WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

15 Example 2 for Linear Homogeneous Recurrence Problem: What is the solution of the recurrence relation with a 0 = 2 and a 1 = 7 a n = a n 1 + 2a n 2 WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

16 Example 2 for Linear Homogeneous Recurrence Problem: What is the solution of the recurrence relation a n = a n 1 + 2a n 2 with a 0 = 2 and a 1 = 7 Solution: Follow the framework above, 1 Find the Characteristic Equation and roots: 2 Form the Linear Combination 3 Impose the initial condition r 2 = r + 2 r 1 = 2, r 2 = 1 a n = α 1 2 n + α 2 ( 1) n a 0 = α 1 + α 2 = 2, a 1 = 2α 1 α 2 = 7 α 1 = 3, α 2 = 1 Based on above steps, the solution to the recurrence is as follows: a n = 3 2 n ( 1) n WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

17 Example 3 for Linear Homogeneous Recurrence Problem: What is the solution of the recurrence relation with a 0 = 1 and a 1 = 6 a n = 6a n 1 9a n 2 WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

18 Example 3 for Linear Homogeneous Recurrence Problem: What is the solution of the recurrence relation a n = 6a n 1 9a n 2 with a 0 = 1 and a 1 = 6 Solution: Follow the framework above, 1 Find the Characteristic Equation and roots: 2 Form the Linear Combination 3 Impose the initial condition r 2 = 6r 9 r 1 = r 2 = 3 a n = (α 1 + α 2 n) 3 n a 0 = α 1 = 1, a 1 = 3α 1 + 3α 2 = 6 α 1 = α 2 = 3 Based on above steps, the solution to the recurrence is as follows: a n = 3 n + n 3 n WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17

19 Thank You! If you have any questions, feel free to contact me. WANG PENG (ENGG 2440B) Tutorial 2 September 28, / 17