a03a - MAD 6206 Combinatorics I, Fall 2011 CRN: 87401...

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Unformatted text preview: MAD 6206 Combinatorics I, Fall 2011 CRN: 87401 Assignment 3 – suggested solution Question 1 . (Binary strings and Fibonacci numbers) Suppose that the digit “0” takes one unit of time to transmit through a communication channel while the digit “1” takes two units of time. (Then the time needed to transmit a binary string with i “0”s and j “1”s equals i + 2 j units.) Let f k,n be the number of binary strings of length n that can be transmitted in k units of time. (a) Find a closed form for F ( x, y ) = summationdisplay k ≥ summationdisplay n ≥ f k,n x k y n . Find G ( x, y ) = ∂ ∂x F ( x, y ) and the number 2 − n [ y n ] G (1 , y ). What is the meaning of this number? (b) Find the number of strings g k that can be transmitted using exactly k units of time (for each string). (c) Find the number of strings that can be transmitted using at most k units of time each. Solution . (a) The enumerator for each digit is ( x + x 2 ) y . Thus, the generating function for p k,n is summationdisplay k ≥ summationdisplay n ≥ f k,n x k y n = summationdisplay n ≥ ( x + x 2 ) n y n = 1 1 − xy − x 2 y . Then G ( x, y ) = ∂ ∂x F ( x, y ) = y + 2 xy (1 − xy − x 2 y ) 2 , and G (1 , y ) = 3 y (1 − 2 y ) 2 . Then [ y n ] G (1 , y ) = 3(2 n − 1 n ), and this gives the result 2 − n [ y n ] G (1 , y ) = 3 n 2 . Note that 2 − n [ y n ] G (1 , y ) = summationdisplay k ≥ kf k,n 2 n , and that there are 2 n strings of length n . Hence the number 2 − n [ y n ] G (1 , y ) = 3 n 2 gives the average time required to transmit a string of length n . (Note: The average time required to transmit one bit is 3/2, and so it takes an average of 3 n/ 2 units of time to transmit n bits – this verifies our result. Our generating function can be used to find more complicated properties.) (b) Note that g k equals the coefficient of x k in F ( x, 1) = 1 / (1 − x − x 2 ). Recall that the generating function of Fibonacci numbers a n is x/ (1 − x − x 2 ). Hence, the number of strings that can be transmitted using exactly k units of time equals a k +1 , where we showed in class that a n = [ x n ] x 1 − x − x 2 = 1 √ 5 parenleftBiggparenleftBigg 1 + √ 5 2 parenrightBigg n − parenleftBigg 1 − √ 5 2 parenrightBigg n parenrightBigg . (c) By a lemma that we showed in class, we have k summationdisplay n =0 g n = [ x k ] 1 1 − x F ( x, 1) = [ x k ] 1 (1 − x )(1 − x − x 2 ) = [ x k ] 1 x 2 parenleftbigg 1 1 − x − x 2 − 1 1 − x parenrightbigg = [ x k +3 ] x 1 − x − x 2 − 1 = a k +3 − 1 . In terms of Fibonacci numbers, we have used generating functions to show that ∑ k n =0 a n = a k +2 − 1. The identity can also be proved using a combinatorial argument number of strings (in the context of this problem) with at least one 1, needing exactly k + 2 units of time to transmit. Question 2 . (Binary strings with no 3 adjacent 0’s) Find a formula for the number f n of strings, with n bits, containing no 3 consecutive 0‘s. Two adjacent 0’s are allowed. Note. Using a combinatorial argument,bits, containing no 3 consecutive 0‘s....
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This note was uploaded on 01/02/2012 for the course MAD 6206 taught by Professor Suen during the Spring '11 term at University of South Florida - Tampa.

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a03a - MAD 6206 Combinatorics I, Fall 2011 CRN: 87401...

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