The exercises and answer for programming language class

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1 The exercises and answer for programming language class Neversay: March 14, 2006 Abstract There are exercises and its answer in detail. Please read it carefully. This exercise practice includes: 2.4: Determine whether or not the given statement therminateing by apply the rule or axiom you learn in natrual semantics. 2.7: Define the natrual semantics(ns) of repeat-construct in language While, and prove the equivalence of two sentences. 2.8: Define the natrual semantics(ns) of for-construct in language While, and evaluate some statement by using derivation tree. 2.17: Define the structural operational semantics(sos) of repeatconstruct in language While. 2.18: Define the structural operational semantics(sos) of for-construct in language While. Homework 2: Define the NS and SOS of (for)-construct in language While. If you have any question about this article, please call TA for help. 1

2 [ass ns ] [skip ns ] [comp ns ] [if tt [if ff ns ] [while tt [while ff < x := a, s > s[x A[ α ]s] < skip, s > s < S 1, s > s, < S 2, s > s < S 1 ; S 2, s > s < S 1, s > s < if b then S 1 else S 2, s > s if B[ b ]s = tt < S 2, s > s < if b then S 1 else S 2, s > s if B[ b ]s = ff < S, s > s, < while b do S, s > s < while b do S, s > s if B[ b ]s = tt < while b do S, s > s if B[ b ]s = ff 1 Exercise Question Consider the following statements: Table 1: Natural semantics for While 1. while (x = 1) do (y := y x; x := x 1) 2. while 1 x do (y := y x; x := x 1) 3. while true do skip For each statement determine whether or not it always terminates and whether or not it always loops. Try to argue for your answers using the axioms and rules of Table Answer 1. while (x = 1) do (y := y x; x := x 1) It terminates when only x 1 otherwise it loops. 2. while 1 x do (y := y x; x := x 1) It always terminates. 3. while true do skip It always loops because the derivation tree show that:... < skip, s > s, < while true do skip, s > s < skip, s > s, < while true do skip, s > s < while true do skip, s > s [while tt because B[ true ]s == tt 2

3 2 Exercise Question 1.Extend the language While with the statement repeat S until b and define the relation for it. (The semantic of the repeat-construct is not allowed to rely on the existence of a while-construct in the language.) 2. Prove that repeat S until b and S; if b then skip else (repeat S until b) are semantically equivalent. 2.2 Answer 1. [repeat tt < S, s > s < repeat S until b, s > s if B[ b ]s = tt [repeat ff < S, s > s, < repeat S until b, s > s < repeat S until b, s > s if B[ b ]s = ff 2. We must prove that repeat S until b S; if b then skip else repeat S until b a. First we prove part 1 : repeat S until b S; if b then skip else (repeat S until b) 1. case [repeat tt < S, s > s < repeat S until b, s > s if B[ b ]s = tt by axiom < skip, s > s, because B[ b ]s = tt, then < skip, s > s < if b then skip else (repeat S until b), s > s if B[ b ]s = tt by applying to [if tt, and then < S, s > s, < if b then skip else (repeat S until b), s > s < S; if b then skip else (repeat S until b), s > s by applying to [comp ns ] we got the prove. 2. case [repeat ff < S, s > s, < repeat S until b, s > s < repeat S until b, s > s if B[ b ]s = ff 3

4 < (repeat S until b), s > s < if b then skip else (repeat S until b), s > s if B[ b ]s = ff by applying [if ff ns ] because B[ b ]s = ff < S, s > s, < if b then skip else (repeat S until b), s > s < S; if b then skip else (repeat S until b), s > s by applying [comp ns ] and then we got the prove. By combining of case 1 and 2 we complete the part 1 of this prove. b. Now we prove part 2 : repeat S until b S; if b then skip else (repeat S until b) 1. case : [if tt. we assume s where B[ b ]s = tt First, we extract sentence we want to prove by applying [comp ns ]: < S, s > s, < if b then skip else (repeat S until b), s > s < S; if b then skip else (repeat S until b), s > s And we are going to extract < skip, s > s from if b then skip else (repeat S until b): < skip, s > s < if b then skip else (repeat S until b), s > s if B[ b ]s = tt and by axiom < skip, s > s, we know s = s. (equation 1) And then we prove by combining equation 1 and definition of [repeat tt : < S, s > s < repeat S until b, s > s = s because B[ b ]s = tt 2. case : [if ff ns ]. we assume s where B[ b ]s = ff First, we extract sentence we want to prove by applying [comp ns ]: < S, s > s, < if b then skip else (repeat S until b), s > s < S; if b then skip else (repeat S until b), s > s And we are going to extract < repeat S until b, s > s from if b then skip else (repeat S until b): < (repeat S until b), s > s < if b then skip else (repeat S until b), s > s if B[ b ]s = ff and we have: 4

5 [repeat ff < S, s > s, < repeat S until b, s > s < repeat S until b, s > s because B[ b ]s = ff, so we complete this prove. 3 Exercise Question Another iterative construct is : for x := a 1 to a 2 do S 1. Extend the language While with this statement and define the relation for it. 2. Evaluate the statement y := 1; for z := 1 to x do (y := y x; x := x 1) from a state where x has the value 5. Hint: You may need to assume that you have an inverse to N, so that there is a numeral for each number that may arise during the computation. (The semantics for the for-construct is not allowed to rely on the existence of a while-coonstruct.) 3.2 Answer 1. [for tt < x := a 1, s 0 > s 1, < S, s 1 > s 2, < for x := a 3 to a 2 do S, s 2 > s 3 < for x := a 1 to a 2 do S, s 0 > s 3 where a 3 = N 1 [ A[ a 1 ]s ] if B[ (x = a 2 ) ]s 1 = tt [forns ff < x := a 1, s 0 > s 1 ] < for x := a 1 to a 2 do S, s 0 > s 1 if B[ (x = a 2 ) ]s 1 = ff Please note that the state of x always increases by 1 after evaluating S. 2. We evaluate this by using derivation tree: We encode the state variable s x,y,z which corresponding to the value of x, y, and z in s. For example, s 1,2,3 means x=1,y=2,z=3 in this state enviorment. For saving space, we use S replace (y := y x; x := x 1). < y := 1, s 5,0,0 > s 5,1,0, < for z := 1 to x do S, s 5,1,0 > s 3,20,3 < y := 1; for z := 1 to x do S, s 5,0,0 > s 3,20,3 by [comp ns ]. And now we derivate the tree by [for tt : < z := 1, s 5,1,0 > s 5,1,1, < S, s 5,1,1 > s 4,5,1, < for z := 2 to x do S, s 4,5,1 > s 3,20,3 < for z := 1 to x do S, s 5,1,0 > s 3,20,3 where 2 = N 1 [ A[ 1 ]s 4,5,1 + 1 ] if B[ (z = x) ]s 5,1,1 = tt 5

6 [ass sos ] [skip sos ] [comp 1 sos] [comp 2 sos] [if tt sos] [if ff sos] < x := a, s > s[x A[ α ]s] < skip, s > s < S 1, s > < S 1, s > < S 1 ; S 2, s > < S 1 ; S 2, s > < S 1, s > s < S 1 ; S 2, s > < S 2, s > < if b then S 1 else S 2, s > < S 1, s > if B[ b ]s = tt < if b then S 1 else S 2, s > < S 2, s > if B[ b ]s = ff [while sos ] < while b do S, s > < if b then (S; while b do S) else skip, s > Table 2: Structural Operational Semantics for While For now we know how to extract < (y := y x; x := x 1), s 5,1,1 > s 4,5,1 by [comp ns ]: < y := y x, s 5,1,1 > s 5,5,1, < x := x 1, s 5,5,1 > s 4,5,1 < y := y x; x := x 1, s 5,1,1 > s 4,5,1 For simplizing our derivation process, we will not explain tree like this any more. And now we focus on the tree of [for]: < z := 2, s 4,5,1 > s 4,5,2, < S, s 4,5,2 > s 3,20,2, < for z := 3 to x do S, s 3,20,2 > s 3,20,3 < for z := 2 to x do S, s 4,5,1 > s 3,20,3 where 3 = N 1 [ A[ 2 ]s 3,20,2 + 1 ] if B[ (z = x) ]s 4,5,2 = tt < z := 3, s 3,20,2 > s 3,20,3 < for z := 3 to x do S, s 3,20,2 > s 3,20,3 if B[ (z = x) ]s 3,20,3 = ff 4 Exercise Question Extend While with the construct repeat S until b and specify a structural operational semantics for it. (The semantics for the repeat-construct is not allowed to rely on the existence of a while-construct.) 4.2 Answer It is a piece of cake: [repeat sos ] < repeat S until b, s > < S; if b then skip else (repeat S until b), s > 6

7 5 Exercise Question 1. Extend While with the construct for x := a 1 to a 2 do S and specify the structural operational semantics for it. Hint: You may need to assume that you have an inverse to N, so that there is a numeral for each number that may arise during the computation. (The semantics for the for-construct is not allowed to rely on the existence of a while-coonstruct.) 2. Evaluate the statement y := 1; for z := 1 to x do (y := y x; x := x 1) from a state where x has the value Answer 1. The answer for for-construct is simple: [for sos ] < x := a 1, s > s < for x := a 1 to a 2 do S, s > < if (x = a 2 ) then (S; for x := a 3 to a 2 do S) else skip, s > where a 3 = N 1 [ A[ a 1 ]s + 1 ] 2. We place S with (y := y x; x := x 1) for shorting the shape of evaluation. < y := 1; for z := 1 to x do S, s 5,0,0 > < for z := 1 to x do S, s 5,1,0 > < if (z = x) then (S; for z := 2 to x do S) else skip, s 5,1,1 > < (y := y x; x := x 1); for z := 2 to x do S, s 5,1,1 > < x := x 1; for z := 2 to x do S, s 5,5,1 > < for z := 2 to x do S, s 4,5,1 >... < for z := 3 to x do S, s 3,20,2 > < if (z = x) then (S; for z := 1 to x do S) else skip, s 3,20,3 > < skip, s 3,20,3 > s 3,20,3 6 Homework 2 Extend the language While with the statement for x := a 1 to a 2 step S s do S where S s is the step statement and always executes after S. And define the relation and for it. Hint: You may need to assume that you have an inverse to N, so that there is a numeral for each number that may arise during the computation. (The semantics for the for-construct is not allowed to rely on the existence of a while-coonstruct.) 7