Notes for Recitation 14

Transcription

1 6.04/18.06J Mathematics for Computer Science October 7, 006 Tom Leighton and Ronitt Rubinfeld Notes for Recitation 14 Guessing a Particular Solution A general linear recurrence has the form: fn) = b 1 fn 1) + b fn ) b d fn d) + gn) One step in solving this recurrence is finding a particular solution. This is a function fn) that satisfies the recurrence equation, but may not be consistent with the boundary conditions. Here s a recipe to help you guess a particular solution: If gn) is a constant, guess that fn) is some constant c. Plug this into the recurrence equation and see if any constant actually works. If not, try fn) = bn + c, then fn) = an + bn + c, etc. More generally, if gn) is a polynomial, try a polynomial of the same degree. If that fails, try a polynomial of degree one higher, then two higher, etc. For example, if gn) = n, then try fn) = bn + c and then fn) = an + bn + c. If gn) is an exponential, such as 3 n, then first guess that fn) = c3 n. Failing that, try fn) = bn3 n + c3 n and then an 3 n + bn3 n + c3 n, etc. In practice, your first or second guess will almost always work.

2 Recitation 14 Mini-Tetris Problem 1. A winning configuration in the game of Mini-Tetris is a complete tiling of a n board using only the three shapes shown below: For example, here are several possible winning configurations on a board: a) Let T n denote the number of different winning configurations on a n board. Determine the values of T 1, T, and T 3. Solution. T 1 = 1, T = 3, and T 3 =. b) Find a recurrence equation that expresses T n in terms of T n 1 and T n. Solution. Every winning configuration on a n board is of one three types, distinguished by the arrangment of pieces at the top of the board. n 1 n n There are T n 1 winning configurations of the first type, and there are T n winning configurations of each of the second and third types. Overall, the number of winning configurations on a n board is: T n = T n 1 + T n

3 Recitation 14 3 c) Find a closed-form expression for the number of winning configurations on a n Mini-Tetris board. Solution. The characteristic polynomial is r r = r )r + 1), so the solution is of the form A n + B 1) n. Setting n = 1, we have 1 = T 1 = A B. Setting n =, we have 3 = T = A + B 1) = 4A + B. Solving these two equations, we conclude A = /3 and B = 1/3. That is, T n = 3 n )n = n+1 + 1) n. 3 Problem. Find closed-form solutions to the following linear recurrences. a) T 0 = 0 T 1 = 1 T n = T n 1 + T n + 1 *) Solution. First, we find the general solution to the homogenous recurrence. The characteristic equation is r r 1 = 0. The roots of this equation are: r 1 = 1 + r = 1 Therefore, the solution to the homogenous recurrence is of the form 1 + ) n 1 ) n T n = A + B. Next, we need a particular solution to the inhomogenous recurrence. Since the inhomogenous term is constant, we guess a constant solution, c. So replacing the T terms in *) by c, we require c = c + c + 1, namely, c = 1. That is, T n 1 is a particular solution to *). The complete solution to the recurrence is the homogenous solution plus the particular solution: T n = A 1 + ) n 1 ) n + B 1 All that remains is to find the constants A and B. Substituting the initial conditions gives a system of linear equations.

4 Recitation 14 4 The solution to this linear system is: 0 = A + B ) 1 ) 1 = A + B 1 A = + 3 B = 3 Therefore, the complete solution to the recurrence is + 3 ) 1 + ) n T n = + b) S 0 = 0 S 1 = 1 S n = 6S n 1 9S n 3 ) B 1 ) n 1. Solution. The characteristic polynomial is r 6r + 9 = r 3), so the solution is of the form A3 n + Bn3 n for some constants A and B. Setting n = 0, we have 0 = S 0 = A3 0 + B = A. Setting n = 1, we have 1 = S 1 = A3 1 + B = 3B, so B = 1/3. That is, S n = 0 3 n n3n = n3 n 1.

5 Recitation 14 Short Guide to Solving Linear Recurrences A linear recurrence is an equation fn) = a 1 fn 1) + a fn ) a d fn d) homogeneous part together with boundary conditions such as f0) = b 0, f1) = b 1, etc. 1. Find the roots of the characteristic equation: x n = a 1 x n 1 + a x n a k +gn) inhomogeneous part. Write down the homogeneous solution. Each root generates one term and the homogeneous solution is the sum of these terms. A nonrepeated root r generates the term c r r n, where c r is a constant to be determined later. A root r with multiplicity k generates the terms: c r1 r n, c r nr n, c r3 n r n,..., c rk n k 1 r n where c r1,..., c rk are constants to be determined later. 3. Find a particular solution. This is a solution to the full recurrence that need not be consistent with the boundary conditions. Use guess and verify. If gn) is a polynomial, try a polynomial of the same degree, then a polynomial of degree one higher, then two higher, etc. For example, if gn) = n, then try fn) = bn+c and then fn) = an +bn+c. If gn) is an exponential, such as 3 n, then first guess that fn) = c3 n. Failing that, try fn) = bn3 n + c3 n and then an 3 n + bn3 n + c3 n, etc. 4. Form the general solution, which is the sum of the homogeneous solution and the particular solution. Here is a typical general solution: fn) = c n + d 1) n + 3n }{{ + 1 } homogeneous solution particular solution. Substitute the boundary conditions into the general solution. Each boundary condition gives a linear equation in the unknown constants. For example, substituting f1) = into the general solution above gives: = c 1 + d 1) = c d Determine the values of these constants by solving the resulting system of linear equations.