Computer Fundamentals.ASSIGNMENT 1..Click here to have a similar paper done for you by one of our writers within the set deadline at a discountedApplications for extension:Applications for extension (on medical or other grounds) should be made in writing to the unit coordinator before the due date, and should be accompanied byappropriate evidence (eg medical certificate) to support your application.Presentation:You may use a computer package to prepare your submission, but it is not essential. In some cases it may be preferable to put in diagrams and symbolic expressions byhand if you are not competent to produce an acceptable result using a computer package. You should use one side of each sheet only. Plastic covers are not required; infact they usually complicate the processing of submissions.2Question 1(a) Consider the following argument. If bears are brown then giraffes are not green. If bears are not brown then rabbits are red. Rabbits are red. Therefore giraffesare not green. Define the following propositions: b: bears are brown g: giraffes are green r: rabbits are red (i) (ii) (iii) (iv) State the argument in symbols.Rewrite the symbolic argument as a proposition Construct a truth table for the proposition. State whether the argument is valid.(b)The propositions below can be arranged in three groups so that each member of a group is logically equivalent to the other two. Find the three groups. 1 2 3 4 5 6 7 89 Either the light is not on or the system is ready. If the system is ready then the light is on. If the light is on then the system is not ready. If the light is noton then the system is not ready. If the system is not ready then the light is not on. Either the system is not ready or the light is on. If the light is on then thesystem is ready. If the system is ready then the light is not on. Either the system is not ready or the light is not on.Give your answer by listing the three groups, for example 2, 5 and 9? (which is not necessarily a correct answer) [ Hint: It may help to realise that the first threeof the expressions p q, ~q ~p, ~p q, q p are logically equivalent to each other but they are not logically equivalent to the fourth (as you could verify using a truthtable ). ][5 + 3 = 8 marks]ENS1161_A1_ECU_v12_2_2014.docPage 2 of 63.Click here to have a similar paper done for you by one of our writers within the set deadline at a discountedQuestion 2A group of 100 students is polled to see how many watched three TV shows, Action, Buzz and Calypso. The results showed that 59 watched Action, 68 watched Buzz, 52watched Calypso, 41 watched Action and Buzz, 34 watched Action and Calypso, 38 watched Buzz and Calypso, and 11 did not watch any of the three. Let A denote the set ofstudents who watched Action, and similarly define sets B and C. (a) Calculate the number of students in each of the eight subsets shown in the Venn diagram. Copy theVenn diagram and enter the number of students in each subset.A ? ? ? ? ? ? B ? C . .?(b)Hence find how many students watched: (i) Action and Calypso, but not Buzz; (ii) Buzz only; (iii) only two of the three shows; (iv) at least two of the shows. [8 + 2 =10 marks]Question 3Suppose that P is a set of people and M a set of movies. Define the predicate S(p, m) to mean person p has seen movie m. Consider the following list of sentences. 12 3 4 5 6 7 8 9 10 11 12 There is at least one movie that some person has seen. For each movie there is somebody who has not seen it. There is some movie that nobodyhas seen. There is at least one movie that everybody has seen. For each movie there is somebody who has seen it. There is some person who has not seen any of themovies. There is at least one movie that some person has not seen. Nobody has seen any of the movies. For each person there is some movie that they have not seen.Everybody has seen at least one movie. There is some person who has seen all the movies. Everybody has seen all of the movies.ENS1161_A1_ECU_v12_2_2014.docPage 3 of 64 For each sentence, the negation is also in the list. Your task is to match each sentence with its negation. Give your answer by listing the number of each sentenceand its negation, for example 4 and 10? (which is not necessarily a correct answer.) [6 marks]Question 4Introduction: There is a simple way of representing sets using so-called bit strings. A bit string is simply a string of 0ýs and 1ýs. Suppose that U is the universalset. Then the rule for the bit string for a particular set S is: If the kth element of U is in S, then the kth bit of the bit string is 1. If the kth element of U isnot in S, then the kth bit of the bit string is 0.Example 1: Suppose the universal set is U = {1, 2, 3, 4, , 10}. Then the set A = {1, 3, 5, 9} is represented by the bit string 1010100010 because: the 1st element ofU is in A, so the 1st element of the bit string is 1; the 2nd element of U is not in A, so the 2nd element of the bit string is 0; and so on. Similarly, the set {5, 6,.Click here to have a similar paper done for you by one of our writers within the set deadline at a discounted7} is represented by the bit string 0000111000. Conversely, the bit string 0110010001 represents the set C = {2, 3, 6, 10} because: the 1st element of the bit stringis 0, so the 1st element of U, that is 1, is not in C; the 2nd element of the bit string is 1, so the 2nd element of U, that is 2, is in C; and so on. Similarly, thebit string 0000000110 represents {8, 9}, and the bit string 0101010101 represents {2, 4, 6, 8, 10}. The operations of union, intersection and complement may be carriedout on the bit strings. The simplest way is to use a table and to work bitwise, that is carry out the operations on the first bits, then the second bits, and so on.(This is just like constructing a truth table that is turned on its side.) Example 2: Let the universal set be U = {4, 6, 9, 13, 18, 25}, and consider the sets A = {6, 13, 18 } and B = {4, 13, 18, 25}. Suppose that we want to find A B, A B and A. Using a table and performing the operations bitwise: U A B A A A Then, from thetable, B B 4 0 1 1 0 1 6 1 0 1 0 0 9 0 0 0 0 1 13 1 1 1 1 0 18 1 1 1 1 0 25 0 1 1 0 1ENS1161_A1_ECU_v12_2_2014.docPage 4 of 65 A B is represented by: A B is represented by: A is represented by: 110111, so A B = {4, 6,13, 18, 25} 000110, so A B = {13, 18} 101001, so A = {4, 9, 25}Your task: Let the universal set U be the following set of 16 countries: U = {Argentina, Australia, Belgium, Brazil, Cameroon, Chile, Cuba, Denmark, Egypt, Ethiopia,Fiji, France, Germany, Ghana, Greece, Haiti} (i) (ii) (iii) Find the bit string that represents the set L = { Brazil, Fiji, Ghana, Greece } Find the set M representedby the bit string 0000 0110 1001 0001 Let P, Q, R and S be sets represented by the following four bit strings respectively:P Q R S 1 0 0 1 1 1 1 1 0 1 0 0 1 1 1 1 0 1 1 0 1 1 1 0 0 0 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 0 0 1 0 0 1 0 1 1 1 1 1 0 0 0 1 1 0 0 0 1 1 1By using bitwise operations on the four bit strings, find the bit string that represents the set T = (P Q) (R S). (Set out your working in a table, with a row foreach of Q, P Q, R, etc.) (iv) List the countries in the set T defined in (iii). [1+ 1 + 3 + 1 = 6 marks]Question 5.Click here to have a similar paper done for you by one of our writers within the set deadline at a discountedIn the lecture notes there is an application of Karnaugh maps to a 7-segment display that is used in some calculators and similar digital devices. The display issometimes extended to include all hexadecimal digits. A typical extended display is:Notice the display for the hexadecimal digits for 10 through 15. Some are shown as upper case and some lower case: A b c d E F. As in the lecture notes we label eachof the seven segments as shown:a f g eENS1161_A1_ECU_v12_2_2014.docbc dPage 5 of 66 When any one of the keys 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E or F is pressed, a 4-bit binary signal wxyz is generated, as shown in the tableKey w 0 1 2 3 4 5 6 7 8 9 A B C D E F 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1Binary representation x 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1Segment Fundamental productFor each of the seven segments there is a corresponding Boolean function and a circuit that has output 0 or 1 as the various keys are pressed. Denote these functionsby a(w, x, y, z), b(w, x, y, z), , g(w, x, y, z).Your task concerns the two segments d and g, and is as follows. For each of the two functions d(w, x, y, z) and g(w, x, y, z): (i) (ii) (iii) construct a truth tablefor the function (like the one shown above); construct the corresponding Karnaugh map; find a minimal sum of products for the function.NOTE: Use the simple labelling on the Karnaugh maps and show your groups clearly. [10 marks].Click here to have a similar paper done for you by one of our writers within the set deadline at a discounted
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