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# Discrete Structures Comp Sci CSE 260

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This 5 page Class Notes was uploaded by Donnell Kertzmann on Saturday September 19, 2015. The Class Notes belongs to CSE 260 at Michigan State University taught by Sakti Pramanik in Fall. Since its upload, it has received 37 views. For similar materials see /class/207409/cse-260-michigan-state-university in Computer Science and Engineering at Michigan State University.

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Date Created: 09/19/15

CSE 260 Automated Theorem Prover p q A q r gt q r is a tautology Truth Table p V q p V r pvq pvr qu 3915 q I HHHanJa TJ TJ TJ TJHHHH HH HH quotUH H H H HHHHHH TJ Tli li J TJH TJH quotUHHH HHH HHHHHHHH F T Another way of justifying the tautology q 7 true makes 1 q A q r gt q 7 true q 7 false also makes 1 q A q r gt q 7 true because of the following q 7 false implies q and r both false Then if p true 1 q A q r is false and if p false then also 1 q A q r is false Above tautology gives the following Inference Rule p V q q r Therefore7 q r The above is called resolution rule and q r is called the resolvent Method of proof by Resolution principle Hypothesis and conclusion are expressed as clauses A clause is a disjunction of variables or negation of variables It can be a single variable or it s negation We can convert propositions into clauses Here are some examples Example 1 p gt q q q Example 2 p q ltgt q A q Ip7 Iq List the clauses corresponding to the hypothesis and the negation of the conclusion Show a contradiction by applying resolution inference rule on this list7 recursively The recursive algorithm is as follows If S is a set of clauses then R650S S CSE 260 Discrete Structures An Application of Relations S Pramanik Relations and Relational Databases 0 A relational database consists of a collection of n ary relations Each relation is called a table 0 n ary Relation Let 141142 An be sets n ary relation R Q A1 gtlt A2 gtlt X A A1 A2 A are the domains of the relation and n is called the degree of the relation 0 Example Sid 513253 A set of student id s for the stu dents Courses cl 62 set of courses that the students are taking Grades A B C D a set of possible grades We de ne the following 3 ary relation SCG indicat ing which students have taken which courses receiv ing what grades SCG 51 cl A 31 02 D 32 cl B 33 cl B 33 02 A Operations on n ary Relations 0 selection Select some particular elements from a re lation example Get all the elements of the relation SGG for the student with SIDs3 53 cl B 33 02 A operator notation Selectmdqg SCG Note the above expression represents a 3 ary relation Projection project on some particular components of all the elements of the relation example Get the rst and the second components of all the elements of SGG 8161 5102 5201 5301 5302 operator notation project12SCG Note the above expression represents a 2 ary binary relation Get only the courses of the student s3 prOjeCtzltS l Ct3 d33SCG will return 01 62 CSE 260 Computing Functions in Turing Machines Computing functions in TM is somewhat di erent from accepting a language by a TM ln accepting a string of the language the TM has to go to Halt state Thus going to a halt state is same as saying 77yes ie the string is accepted If it does not go to a halt state it means 77no ie the string is not accepted It does not matter what is left on the tape at the time the TM halts ln computing a function however you have to come up with a mechanism to present the result output of the function The output for a TM is left on the same tape replacing what ever was there before Representing the input is also di erent from a TM accepting a string of a language because you may have more than one values that you need for input For example for the function fxy xy you have to have a way of representing the two numbers X and y on the input Tape before the TM starts and a convention for leaving the result ie xy on the Tape when the TM halts after the computation Following is a convention for handling the problem 1 will expalin it with an example for computing fxyxy We will assume that the numbers X and y are unsigned integers We will represent an integer by the number of 1 bits corresponding to the value of the integer Thus the integer zero is represented by zero number of 17s ie no 17s the integer 1 by one 1 bit 2 by two 1 bits ie 11 3 by three 1 bits ie 111 and so on You could have used a di erent convension where integer 0 will be represented by one 1 integer 1 by two 17s integer 2 by three 17s and so on For the input to the TM we will place the two integers X and y on the tape separated by a A Thus the separating A is needed by the TM to distinguish the rst number from the second If we want to add X and y where X2 and y3 the input tape will be as follows note that the output will be left on the same tape replacing this input A11A111AAA After the TM has nished computation on this input the result 5 will be left on the tape left justi ed as follows Also note that we did not save the input on the tape A11111AAA Actual computation by this TM is rather very simple Only thing the

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