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Unformatted text preview: Solution to HW 5 Ning SUN August 16, 2009 7.1: 2, 10; 7.2: 8 (a,b); 8.1: 10, 12, 20, 26, 36; 8.2: 2, 6, 8 7.1#2 (a) a n = a n 1 + a n 2 + a n 4 ; a = 1 , a 1 = 1 . ( a k = 0 for k &lt; .) (b) a 4 = a 3 + a 2 + a = ( a 2 + a 1 ) + a 2 + a = ( a 1 + a + a 1 ) + ( a 1 + a ) + a = 6 . So there are 6 ways to climb four stairs. 7.1#10 # of female in the n th previous generation=(# of female + # of male) in the ( n 1) th previous generation = a n 1 , so # of male in the n th previous generation=# of female in the ( n 1) th previous generation = a n 2 . So a n = (# of female + # of male) in the n th previous generation = a n 1 + a n 2 ; a 1 = 1 , a 2 = 2 . Find a recurrence relation for the number of ndigit binary sequences with (a) no &quot;11&quot;; (b) no &quot;10&quot;. (a) There are 2 choices for the rst digit: 1 or 0. If it's 0... , you can simplify the subcase to an ( n 1) digit sequence restarting from the second digit. The corresponding number is a n 1 . If it's 1... , then the second digit can only be 0 to aviod 11... . And then you can simplify the subcase to an ( n 2) digit sequence restarting from the third digit. The corresponding number is a n 2 . In total, a n = a n 1 + a n 2 , a = 1 , a 1 = 2 . (b) There are 2 choices for the rst digit: 1 or 0. If it's 0... , you can simplify the subcase to an ( n 1) digit sequence restarting from the second digit. The corresponding number is a n 1 . If it's 1... , then the second digit can only be 1 to aviod 10... . However you cannot restart the sequence after this 11... since the next digit can be 0 which makes 110... . So we subtract the undesired cases 10... (we can restart the sequence after the 0 . So the number is a n 2 ) from the total 1... ( a n 1 ). In total, a n = 2 a n 1 a n 2 , a = 1 , a 1 = 2 . 1 Find a recurrence relation for the number of ndigit ternary sequences with (a) no &quot;10&quot;; (b) any 1 not in the last position is followed by a 0. (a) There are 3 choices for the rst digit: 0, 1 or 2. If it's 0... or 2... (2 cases), you can simplify the subcase to an ( n 1) digit sequence restarting from the second digit. The corresponding number is 2 a n 1 . If it's 1... , then the second digit must be 1 or 2 to avoid 10... . However, if the second digit is 1, then the sequence cannot restart from the third digit since if the third digit digit is 0, we get 110... which violates the constraint. So we need to subtract the undesired cases 10... (we can restart the sequence after the 0 . So the number is a n 2 ) from the total 1... ( a n 1 )....
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 Spring '08
 ARKIN

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