hard a problem o hard help e help problem, p help o problem a f hard g e f help help g please g o f problem g

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1 MATH2440 / SOLUTIONS FOR MIDTERM 2 (1) Let L := {a, b, d, e, p, r, o} and A := {hard, help, please, problem}. Consider the map f : L A sending a letter to the alphabetically first word in A in which it appears. Is f surjective? Let g : A L be the map sending a word to the first vowel occurring in it. The map f is given by a hard b problem d hard e help p help r hard o problem Is f g the identity map? Since please is not an output, f is not surjective. The map g is given by hard a help e please e problem o and hence for the composition f g : A A, we have hard g a f hard help g e f help please g e f help problem g o f problem, which is not the identity map, since it sends please to help. (2) The seven colors of the rainbow in order are Red, Orange, Yellow, Green, Blue, Indigo, Violet. What is the code of the subset of colors having an o in their name? What colors are represented by the code ? What set does the code represent if the colors are listed instead alphabetically? Red Orange Yellow Green Blue Indigo Violet with o: decode:

2 2 From the last line, we see that represents {Orange, Yellow, Blue}. To decode in the alphabetical listing, we have Blue Green Indigo Orange Red Violet Yellow decode: so that it represents {Orange, Violet}. (3) In the elevator, of the nine students that take Calculus, three also take Discrete Math. Math, three also take World History. Moreover, of the seven students that take Discrete How many students in this elevator are taking a math class? Let C and D be the subsets of students taking respectively Calculus and Discrete Math. We get the following information C = 9, C D = 3 and D = 7 (we do not need the information on World History). By the inclusion/exclusion principle, we get C D = = 13 students taking a math class. (4) Let D be the set of all words in the sentence A leaf is falling off a branch. Consider the map f : D N sending a word to its length. Is f injective? What is f 1 ({3, 5})? The map f is given by a 1 leaf 4 is 2 falling 7 off 3 branch 6 (note that the word a only appears once in the set, although it appears twice in the sentence!) Since all outputs are different, we get an injective map. Since no word has length 5, we get f 1 ({3, 5}) = {off}. (5) Let H be the set of letters in the word head, E those in endear, and D those in deer. Draw a Venn diagram containing these three sets. Note that there are six letters: h, e, a, d, n, r. Each of these letters has to be put in the right region: for instance, h lies in H only, whereas e lies in H E D, etc. Filling in, we get

3 3 H E h a e d r n D (6) Show that (B A) (C A) = (B C) A, using a membership table. A B C B A C A (B A) (C A) B C (B C) A Both columns are indeed equal. (7) Calculate 2 3 j=1 i2 j. We make a table with the first two columns the values of the counters and the last the cumulative sum i j i 2 j Sum (8) Calculate 11 i=5 i2 using the table below.

4 4 We calculate the sum of squares up to n = 11 and subtract the sum up to n = 4 to get the given sum: 4 i=5 i 2 = i 2 = i 2 = i 2 = 30 = i 2 = = 476 (9) Describe an algorithm that checks if two given lists are disjoint (meaning that they have no element in common); write the pseudo-code for your algorithm by calling it disjoint(a 1,..., a n, b 1,..., b m : elements) (note that the lists can have different size). Description of the algorithm: take the first element of the first list, and compare it with each element in the second list; if no match is found, do the same with the second element of the first list, etc. If there was never a match, output answer=true. Here is the pseudo-code: Procedure disjoint(a 1,..., a n, b 1,..., b m : elements) answer:=true for i := 1 to n for j := 1 to m if a i = b j, then answer:=false return answer. This is not the most efficient program, as it makes all possible checks, even if a match was already found, at which point we could have exited the loops. This can be remedied by using while loops instead, but this is not important at this point. (10) Use the bubble sort to sort the list (W, O, R, D); do the same for the insertion sort (show all lists as you go through the algorithms). bubble: In the first run, we get WORD, OWRD, ORWD, ORDW note that we now know that the last letter is on the right spot. Starting anew, for the second run, we get ORDW, ODRW at this point, we are sure that the last two letters are RW; for the third, we get ODRW, DORW so that ORW is in order whence so is the whole list DORW. insertion: we start with the list split as W ORD and insert O in the right spot, giving OW RD; next we insert R as ORW D and finally we put D on its proper place, yielding DORW.

5 5 (11) Let f : R R: x 3x+1 and g : R R: x 2x 2. Calculate g f and f g and decide whether there equal or not. Since (g f)(x) = g(f(x)) = 2(f(x)) 2 = 2(3x + 1) 2 = 6x (f g)(x) = f(g(x)) = 3(g(x)) + 1 = 3(2x 2) + 1 = 6x 5 these compositions are not the same.