MATH 243E Test #3 Solutions
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1 MATH 4E Test # Solutions () Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s. You do not need to solve this recurrence relation. (Hint: Consider the following cases: the last bit is a, the last two bits are 0, or the last two bits are 00. Don t forget the initial conditions.) (5 points) Solution: Let a n be the number of strings of length n containing a pair of consecutive 0s. Using the hint, we know that every string of length n either has its last bit being a, its last two bits being 0, or its last two digits being 00. We consider each case in turn. If the last bit is a, then the string with this removed must have a consecutive pair of 0s. Hence exactly a n of the a n strings we are considering end in a. Similarly, if the last two bits are 0, then the string with these bits removed still contains a consecutive pair of 0s. Hence exactly a n of the a n strings we are considering end in 0. Something different happens if the last two bits of the string we are considering are 00; the smaller string can be anything. Hence, exactly n of the a n strings we are considering end in 00. Because these cases are mutually exclusive, we have a n = a n + a n + n, for n (or for n if you feel squeamish about dealing with strings of length zero). We still need to supply initial conditions; counting shows that a = 0 and a =. Hence the recurrence relation for a n is a n = a n + a n + n, for n, a = 0, a =. Grading: 5 points for using a n instead of n ; 5 points for forgetting the initial conditions.
2 () Solve the following recurrence relation. (5 points) d n 6 d n 6 d n = n, n, d 0 =, d = 5. Solution: This is a linear, nonhomogeneous recurrence relation, so we have three steps to perform: () Solve the homogeneous version: h n 6 h n 6 h n = 0. The auxiliary equation for this (linear, homogeneous) recurrence is r 6r 6 = 0 or (r 8)(r + ) = 0. Hence r = 8 or, and the solution to the homogeneous equation is for some constants c and c. h n = c ( ) n + c 8 n, () Find one solution p n satisfying the original recurrence. The right-hand side of the original recurrence is n, a constant polynomial times n. Because is not a root of the auxiliary equation of the homogeneous recurrence, our deep theorem tells us that there is a solution of the form p n = C n, where C depends on the recurrence. Because p n must satisfy the recurrence, we must have C n 6C n 6C n = n. Dividing both sides by n, we have C 6 C 6 4 C =, or C = 6. Hence our particular solution is p n = 6 n. () Add them together and plug in the initial conditions. Hence d n = c ( ) n + c 8 n 6 n. Plugging in the initial conditions, we have = d 0 = c + c 6 (and hence c + c = 7 6 ) and 5 = d = c +8c, or c +8c = 6. Solving these equation yields c = 5 and c = (more work than I had anticipated). Thus 0 d n = 5 ( )n + 0 8n 6 n. Grading: 0 points for no particular solution; 5 points for not finding c and c.
3 () Recall that an onto function is a function f : A B such that for every element b B, there exists an element a A with f(a) = b. In this problem we will count the number of onto functions from A to B, where A = {a, b, c, d, e} and B = {,, }. (a) How many functions g : S T are there, if S has m elements and T has n elements? (5 points) Answer: n m. Grading: + points for anything else, as long as it was used for (b). Let the set A i consist of all functions f : A B such that i is not in the range of f, where i is,, or. (For instance, A is the set of all functions f from A to B which do not have in the range of f; i.e., f is then a function from A to B \ {} = {, }.) (b) What is A A A? (5 points) Solution: Inclusion-exclusion tells us that A A A = A + A + A A A A A A A + A A A. Since A is the set of all functions with domain A and range {, }, A is just the number of such functions, or 5 =. Similarly, A = A = as well. Now, what is A A? It is a subset of the set of functions with domain A, but what is the range of a function f A A? Neither nor can be in the range of f, so the range of f is only {}. Hence A A is the set of all functions with domain A and range {}, and thus A A = 5 =. Similarly, A A = A A = as well. Now what about A A A? Thinking along the lines above, it contains precisely the functions from A to B, whose range does not include,, or. There are no such functions, since B = {,, }, so A A A =, and it has size 0. Plugging these values into our inclusion-exclusion equation, we have A A A = = 9. Grading: full credit if the number of functions computed was consistent with the answer to (a). (c) What does the set A A A have to do with onto functions from A to B? How many onto functions are there from A to B? (5 points) Solution: The set A A A consists of all functions from A to B which are not onto, since at least one of,, or is missing from the range of a function in this union. Hence the number of onto functions from A to B is the total number of functions from A to B minus the size of the set A A A (the bad functions ), or 5 9 = 60. Grading: + points for saying that A A A is the set of all onto functions.
4 (4) How many integers between 0 and 9,999,999 (inclusive) are there whose digits add up to? (5 points) Solution: The question can be viewed as the following problem: How many solutions are there to the equation ( ) x + x + + x 7 =, where 0 x i 9, for all i. ( ) ( ) Without the upper bounds, the answer would be =. This answer includes solutions where at least one of the x i s is at least 0. Let us call such a solution bad ; the question is now: How many bad solutions are there? Let A i consist of all solutions to ( ) with x i 0. Then the number of bad solutions to ( ) is A A A 7, which is equal to i 7 A i i<j 7 A i A j + i<j<k n A i A j A k, where the last term is A A A 7. Since all the bounds on the x i s are the same, the above expression simplifies to ( ) ( ) ( ) A A A + A A A. So how big is A? A consists of all solutions of ( ) with x 0. If we let x = x 0, we have x + x + x + + x 7 = 0 =, where x, x, x,..., x 7 0. The ( number of ) solutions ( ) to this equation is equal to the size of A ; hence A = =. Similarly, for a solution to belong to A A, it must have ( x, x ) 0. If ( we ) let x = x 0 and x = x 0, we see that A A has = elements. Clearly, A A A must be empty, because otherwise the first three digits would add up to at least 0. Hence the number of bad solutions is ( ) 9 and the number of good solutions to ( ) is ( ) 9 ( ) 9 + ( ) 9, ( ) 9 = 86,860. Grading: +5 points for just setting up A i ; +5 points for using 6 instead of 7, the rest being okay. 4
5 (5) Suppose that C n satisfies the homogeneous linear recurrence relation C n+5 = 5C n+4 7C n+ C n+ + 8C n+ 4C n, for n 5. What is the general form for the solution to this recurrence? (Hint: The auxiliary equation for this recurrence is (r + )(r ) (r ) = 0.) (5 points) Solution: The roots of the auxiliary equation are (with multiplicity ), (with multiplicity ), and (with multiplicity ), so the general form of the solution to the recurrence is C n = α ( ) n + α n + α n n + α 4 n + α 5 n n, where α,..., α 5 are constants (yet to be determined). Grading: +5 points if the roots were interpreted to be, ±, ± (and the rest of the problem done correctly); +0 points if α n + α n showed up instead of α + α n; +0 points if the α n term and/or the α 4 n term was omitted. MAT 4 Website: checkman/4/ 5
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