Exam 2 Study Guide

It will likely contain questions on:

• Programming with mutation, environment model of evaluation
• Object-oriented programming
• Measuring work using orders of growth
• Computability
• Databases
• Static Type Checking
• Memoization
• Networks

Mutation and Objects

1. For Problem Set 6, Question 4, two common answers were given:

Answer A:

(define (make-professor name)
  (let ((super (make-lecturer name)))
    (lambda (message)
      (if (eq? message 'profess)
	    (lambda (self message)
	          (ask self 'say (list "It is intuitively obvious that"))
		  (ask self 'lecture message))
	      (get-method super message)))))

Answer B:

(define (make-professor name)
  (let ((super (make-lecturer name)))
    (lambda (message)
      (if (eq? message 'profess)
	    (lambda (self message)
	          (ask self 'say (list "It is intuitively obvious that"))
		  (ask self 'say message)
		  (ask self 'say (list "you should be taking notes")))
	    (get-method super message)))))

2. Consider the following definitions taken from the PS6 code (without modification except for removing some methods that are not needed for this question):

(define (make-object name)
  (lambda (message)
    (case message
      ((class) (lambda (self) 'object))
      ((name) (lambda (self) name))
      (else no-method))))

(define (make-physical-object name)
  (let ((super (make-object name))
	(location null-place)) ; location set when we install
    (lambda (message) ; Normal actions
      (case message
	((class) (lambda (self) 'physical-object))
	((location) (lambda (self) location))
	(else (get-method super message))))))

(define (make-mobile-object name)
  (let ((super (make-physical-object name)))
    (lambda (message)
      (case message
	((class) (lambda (self) 'mobile-object))
	(else (get-method super message))))))

(define (make-person name)
  (let ((super (make-mobile-object name))
        (possessions '())  
	(restlessness 0.0))
    (lambda (message)
      (case message
	((class) (lambda (self) 'person))
	(else (get-method super message))))))

3. How many frames would be created when (make-mobile-object "bike") is evaluated? (Explain why each frame is generated. Recall that let is syntactic sugar for ((lambda (...) )).)

Objects and Environments

(From Spring 2003 Exam 2)

Consider the classes defined below: sail-boat is a subclass of boat which is a subclass of object.

(define (ask object message . args)
  (apply (get-method object message) object args))

(define (get-method object message)
  (object message))

(define make-object 
  (lambda (name)
    (lambda (message)
      (case message
	((class) (lambda (self) 'object))
	((object?) (lambda (self) #t))
	((name) (lambda (self) name))
	(else (error "No method"))))))

(define make-boat
  (lambda (name)
    (let ((super (make-object name))
	    (floating #t))
      (lambda (message)
	(case message
	    ((class) (lambda (self) 'boat))
	      ((boat?) (lambda (self) #t))
	        ((is-floating?) (lambda (self) floating))
		  ((sink) (lambda (self) (set! floating #f)))
		    (else (get-method super message)))))))

(define make-sail-boat 
  (lambda (name)
    (let ((super (make-boat name))
	    (sailing #f))
      (lambda (message)
	(case message
	    ((class) (lambda (self) 'sail-boat))
	      ((raise-sail) (lambda (self) (set! sailing #t)))
	        ((lower-sail) (lambda (self) (set! sailing #f)))
		  (else (get-method super message)))))))

2. There are three frames in the diagram (labeled F1, F2 and F3) that contain a place named name with the value "discovery". If we evaluate (ask discovery 'name) in the global environment, in which frame with the value of name be found?

Objects and Environments

Cy D. Fect defines a procedure reverse! that uses mutation to reverse a list:

(define (next-to-last-pair lst)
  (if (null? lst) (error "Null list!")
      (if (null? (cdr lst)) (error "Single length list!")
          (if (null? (cdr (cdr lst))) lst
              (next-to-last-pair (cdr lst))))))

(define (reverse! lst)
    (if (or (null? lst) (eq? 1 (length lst)))
        #f ;; Nothing to do for empty or length 1 list (value doesn't matter)
	(let* ((nexttolast (next-to-last-pair lst))
	       (lastpair (cdr nexttolast))
	       (first (car lst)))
	  ;;; Point 0
	  (set-cdr! nexttolast null)
	  ;;; Point 1
	  (reverse! (cdr lst))
	  (set-cdr! nexttolast lastpair)
	  (set-car! lst (car lastpair))
	  ;;; Point 2
	  (set-car! lastpair first))))

> (define ilist (list 1 2 3))

> (reverse! ilist)

> ilist

(3 2 1)

At Point 0 on the first call to reverse! the environment looks like:

The E1 environment is created by the application (reverse! ilist). The E2 environment is created by the let inside reverse! (which desugars to an application of a lambda). (The let* actually desugars into more than one lambda, and more than one environment, but for this question we combine them into one environment.)

4. Show what the environment looks like at Point 1 (after evaluating (set-cdr! nexttolast null)). To make your drawing easier, here is a template drawing. You can answer this question by only adding to this template.

5. Show what the environment looks like at Point 2. You can answer this question by only adding to this template.

Computability

Input: A description of a Turing Machine and input tape.

Output: If executing the Turing Machine on the given input would ever repeat a machine configuration (the state of the finite state machine and the contents of the tape), output true. Otherwise, output false. That is, the output should be true if and only if executing the Turing Machine on the given input would encounter the same exact machine configuration more than once.

Input: A description of a Turing Machine and input tape.

Output: If executing the Turing Machine on the given input would ever repeat a finite state machine state, output true. Otherwise, output false. That is, the output should be true if and only if executing the Turing Machine on the given input would enter the same finite state machine state more than once.

Computability

Is the Greater than Zero Problem described below decidable?

Input: A program P and input n
Output: If evaluating P applied to n would produce a value greater than 0, true. If evaluating P applied to n would produce a value that is not a number, or is a number less than or equal to 0, or would not terminate, output is false.

Universal Language

Consider the preSQL language described below:

Commands ::= Command ; Commands
Commands ::= Command
Command ::= CreateTableCommand | InsertCommand | CountCommand | DeleteCommand

CreateTable ::= CREATE TABLE Name
    Create a new table named Name with no entries.
InsertCommand ::= INSERT INTO Table Value
    Insert a value (TRUE or FALSE) as the last element in table Table.
Value ::= TRUE
Value ::= FALSE
CountCommand ::= COUNT Table
    Evaluates to the number of entries in Table.
DeleteCommand ::= DELETE FROM Table
    Remove the first (earliest inserted) element from the table Table.

• Entries in a table have one value (TRUE or FALSE). There are no fields, since each table entry is only one value. (Note that there is no way to get the actual values of entries.)
• The DELETE command just deletes the first entry in the table.
• There is no SELECT command and no UPDATE command.
• The COUNT command can be used to obtain the number of entries in a table.

> CREATE TABLE tab

> INSERT INTO tab TRUE

> INSERT INTO tab FALSE

> COUNT tab

2

> DELETE FROM tab

> COUNT tab

1

PreSQLsql adds one additional command to PreSQL:

Command ::= UNTILEMPTY Table { Commands }

For example, UNTILEMPTY tab { DELETE FROM tab } would delete all entries from that table tab.

10. Show how you could do addition using PreSQLsql. Your addition code should assume the input is in the form of two tables, tableA and tableB. After addition, a new table named output should have the result. Provide code that could be used for the missing code below that would correctly add any two natural numbers (your code may modify the input tables):

> COUNT tableA n > COUNT tableB m > insert your code here > COUNT output Value of n + m

Models of Computation

Phine Knight suggests the modelling computation using a Phine Machine consisting of a list of numbered instructions (the Instructions), a pointer to the current instructions (the InstructionPointer), and a store that associates name and values (the Store).

A program executes by executing the instruction numbered by the InstructionPointer, and then increasing the InstructionPointer by 1. This continues unless the Instruction is HALT.

Instructions can do one of four things:

• Set the value associated with a name to zero (e.g., X := 0)
• Set the value associated with a name to the value of a name plus 1 (e.g., Y := X - 1)
• Set the value associated with a name to the value of a name minus 1 (e.g., X := X - 1)
• Halt (e.g., HALT)

   ( <1: X := 0
      2: X := X + 1
      3: Y := X - 1
      4: HALT >,
     1,                     
     { } )

   ( <1: X := 0
      2: X := X + 1
      3: Y := X - 1
      4: HALT >,
     2,                     
     { <X, 0> } )

   ( <1: X := 0
      2: X := X + 1
      3: Y := X - 1 
      4: HALT>,
     4,                     
     { <X, 1>, <Y, 0> } )

9. Is this Phine Machine model of computation equivalent to a Turing Machine? Argue convincingly why it is or isn't.

10. Condy Shonal suggests adding one new instruction to Knight's model:

    Instruction ::= IF Name = 0 GOTO Number

For example,

    ( <1: IF A = 0 GOTO 5
       2: B := B + 1
       3: A := A - 1
       4: GOTO 1
       5: HALT>,
      1,
      { <A, n>, <B, m> } )

Condy claims her new model of computation is as powerful as a Turing Machine, but Fine Knight does not believe her. Write an informal but convincing argument that this model is as powerful as a Turing Machine.