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Expressions Interpretation in BSL Format Using Dr Racket Assignment Solution

July 02, 2024
Dr. Madison Davidson
Dr. Madison
🇺🇸 United States
Programming
Dr. Madison Davidson, an accomplished programming expert, earned her Ph.D. from the University of York, United Kingdom. With 15 years of experience, she excels in solving intricate programming assignments with finesse and expertise.
Key Topics
  • Instructions
    • Objective
  • Requirements and Specification
Tip of the day
When working on OCaml assignments, make use of pattern matching to simplify your code. It provides a clean and efficient way to handle different cases, especially for recursive functions and data structures like lists and trees. This can reduce complexity and improve readability.
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Instructions

Objective

Write a Racket assignment program that allows users to to Interpret expressions in BSL format.

Requirements and Specification

It should be with "BSL" with an amateur language. We run the Programming on Dr. Racket with the beginner setting.

Use functions like "(check-expect...)" I am also adding the script for the programming language. I added the translated ones and the original one. "Important: This time, the homework will only be processed on paper. Give your solution to your tutor at the beginning of your next tutorial, or send your tutor a digitised version (photo / scan) via the forum.

Source Code

Aufgabe 1: Grammatiken und Ableitungsbäume -------------------------------------------------------------------------------- 1) Example 1: (+ (* 3 4) (- 9 2)) Derivation tree: (+ (* 3 4 ) (- 9 2 ) ) Example 2: (/ (* (+ 1 2) (- 4 3)) 2) (/ (* (+ 1 2 ) (- 4 3 ) ) 2 ) 2) Example 1: (+ (- 2 3) (* 4 "a")) This expression doesn't conform to the grammar since the second expression in "(* )" must be either another expression or a and "a" is neither. Example 2: (- (+ (/ 24 2) (* 4 1)) (- 8)) This expression doesn't conform to the grammar since the expression that starts with "(-" must be followed by two expressions and (- 8) has only one: the 8. Aufgabe 2: Struktur von Ausdrücken -------------------------------------------------------------------------------- 1) a) x Is syntactically correct from the derivation tree: x b) (= (= y z) 0) Is syntactically correct from the derivation tree: ( = ( = y z ) 0 ) 2) a) (3 + 4) Is not syntactically correct because the first symbol in a expression with parentheses must be one of: , and, or, cond, if, but 3 is a . b) number?(1) Is not syntactically correct. The text doesn't start with parenthesis so it must be an , but it's not a , , , or so it could only be a , however a must not contain parentheses so the text is not syntactically correct. 3) a) (define (f x) x) Is syntactically correct from the derivation tree: ( define ( f x ) x ) b) (define (f x y z) y) Is syntactically correct from the derivation tree: ( define ( f x y z ) y ) 4) a) (define (f "x") "x") Is not syntactically correct since the second symbol inside the parentheses in a define must be a +, that is one or more symbols, however "x" is a which can't be derived from . b) (define (f) 10) Is not syntactically correct since the parentheses after define must have an initial and after that there must be one or more and (f) has only the initial . 5) a) (+ 1 (not x)) Using the derivation tree we get: ( + 1 ( not x ) ) So the text is syntactically correct. b) ("x") Using the derivation tree we get: ( "x" <- this is not a name It's not syntactically correct since "x" is not a . Aufgabe 3: Bedeutung von Ausdrücken -------------------------------------------------------------------------------- (* 1 (- 11 4) (+ 190 1)) Since the first expression is a value 1, we use the next position for the hole. Thus, we start with the evaluation context: e = (* 1 (- 11 4) (+ 190 1)) E = (* 1 [] (+ 190 1)) e_1 = (- 11 4) e_1 reduces as: (- 11 4) -> 7 with (KONG) we get: E[e_1] -> E[7] so e -> (* 1 7 (+ 190 1)) Now 1 and 7 are values so we use next position for the hole. Thus the new evaluation context is: e = (* 1 7 (+ 190 1)) E = (* 1 7 []) e_1 = (+ 190 1) e_1 reduces as: (+ 190 1) -> 191 with (KONG) we get: E[e_1] -> E[191] so e -> (* 1 7 191) Since the expression has only values after the name, it can be evaluated: (* 1 7 191) -> 1337 Aufgabe 4: Bedeutung von Programmen mit algebraischen Datentypen -------------------------------------------------------------------------------- Start in (PROG) with env = empty The first element of the program is a structure definition so it's added to env: env = (define-struct order (item quantity)) Tne next element is a function definition so it's added to the env: env = (define-struct order (item quantity)) (define (item-price item) (cond ([string=? "sugar" item] 1) ([string=? "milk" item] 2))) Tne next element is a function definition so it's added to the env: env = (define-struct order (item quantity)) (define (item-price item) (cond ([string=? "sugar" item] 1) ([string=? "milk" item] 2))) (define (order-price order) (* (item-price (order-item order)) (order-quantity order))) The next element is a constant definition, according to (PROG) we need to evaluate first (make-order "milk" 2): e = (make-order "milk" 2), the expression has only values so it can be evaluated: e -> according to (STRUCT-make) The env is now: env = (define-struct order (item quantity)) (define (item-price item) (cond ([string=? "sugar" item] 1) ([string=? "milk" item] 2))) (define (order-price order) (* (item-price (order-item order)) (order-quantity order))) (define MILK-ORDER ) The next program element is an expression so we evaluate it according to (PROG) using the current environment: e = (order-price MILK-ORDER) breaks down to: E = (order-price []) and e_1 = MILK-ORDER. According to (CONST) it follows that MILK-ORDER -> and following (KONG) we get: e -> (order-price ) Following (FUN) we get: (order-price ) -> (* (item-price (order-item )) (order-quantity )) Breaking down the expression: e = (* (item-price (order-item )) (order-quantity )) E = (* [] (order-quantity )) and e_1 = (item-price (order-item )) Using again the evaluation context with e_1 we get: e = (item-price (order-item )) E = (item-price []) and e_1 = (order-item ) Following (STRUCT-select) we get: e_1 -> "milk" Then using (KONG) we get e -> (item-price "milk") Using (FUN) we get: (item-price "milk") -> (cond ([string=? "sugar" "milk"] 1) ([string=? "milk" "milk"] 2)) Using the eval context: e = (cond ([string=? "sugar" "milk"] 1) ([string=? "milk" "milk"] 2)) E = (cond ([] 1) ([string=? "milk" "milk"] 2)) with e_1 = (string=? "sugar" "milk") Using (PRIM) we get e_1 -> #false Using (KONG) we get e -> (cond (#false 1) ([string=? "milk" "milk"] 2)) Using (COND-False) we get (cond (#false 1) ([string=? "milk" "milk"] 2)) -> (cond ([string=? "milk" "milk"] 2)) Using the eval context: e = (cond ([string=? "milk" "milk"] 2)) E = (cond ([] 2)) e_1 -> (string=? "milk" "milk") Using (PRIM) we get e_1 -> #true Using (KONG) we get e -> (cond (#true 2)) Using (COND-True) we get (cond (#true 2)) -> 2 Using (PRIM) (item-price "milk") -> 2 Using (KONG) we get e -> (* 2 (order-quantity )) Evaluating the expression again using the evaluation context: e = (* 2 (order-quantity )) E = (* 2 []) and e_1 = (order-quantity ) Following (STRUCT-select) we get: e_1 -> 2 Using (KONG) we get: e -> (* 2 2) Using (PRIM) we get: (* 2 2) -> 4 Using (PRIM) we get (order-price ) -> 4 The environment at the end is: env = (define-struct order (item quantity)) (define (item-price item) (cond ([string=? "sugar" item] 1) ([string=? "milk" item] 2))) (define (order-price order) (* (item-price (order-item order)) (order-quantity order))) (define MILK-ORDER ) Aufgabe 5: Bedeutung von Datendefinitionen -------------------------------------------------------------------------------- 1) "yes" "no" 2) (make-pair "yes" "yes") (make-pair "yes" "no") (make-pair "no" "yes") (make-pair "no" "no") 3) "yes" "No" #false 4) (make-pair "yes" "yes") (make-pair "yes" "no") (make-pair "yes" #false) (make-pair "no" "yes") (make-pair "no" "no") (make-pair "no" #false) (make-pair #false "yes") (make-pair #false "no") (make-pair #false #false) 5) (make-pair "yes" "yes") (make-pair "yes" "no") (make-pair "no" "yes") (make-pair "no" "no") #false

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