1. Lecture 2
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2. Review: syntax vs. semantics
Syntax Semantics
Primitives 23 + *
Proc for adding
Proc for multiplying
Means of Combination
( )
Application of proc to arguments
Result = 25
Means of Abstraction
(define score 23)
Associates score with 23 in environment table
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3. An example ==> (define score 23)* + 5 6 - 23 * 3 2 11 11 12
121
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4. Substitution model
To Apply a compound procedure to a list of arguments:
Replace the formal parameters with the corresponding actual values.
Evaluate the body of the procedure with these values.
(define square (lambda (x) (* x x)))
1. (square 4)
2. (* 4 4)
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5. Review: evaluation of an expression
The value of a numeral: number
The value of any name: the associated object in the environment
The value of a built-in operator: machine instructions to execute
To Evaluate a combination (other than special form) or a compound procedure
Evaluate all of the sub-expressions in any order
Apply the procedure that is the value of the leftmost sub-expression to the arguments (the values of the other sub-expressions)
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6. IF special form
(if <predicate> <consequent> <alternative>)
(if (< 2 3) 2 3) ==> 2
(if (< 2 3) 2 (/ 1 0)) ==>
ERROR 2
If the value of <predicate> is #t,
Evaluate <consequent> and return it
Otherwise
Evaluate <alternative> and return it
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7. IF is a special form
In a general form, we first evaluate all arguments and then apply the function
(if <predicate> <consequent> <alternative>) is different:
<predicate> determines whether we evaluate <consequent> or <alternative>. We evaluate only one of them
Can you see why it has to be that way?
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8. Lambda special form lambda syntax (lambda (x y) (+ x y x 2))
1st operand position: the parameter list (x y)
a list of names (perhaps empty)
determines the number of operands required
2nd operand position: the body (+ x y x 2)
may be any expression
not evaluated when the lambda is evaluated
evaluated when the procedure is applied
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9. Naming procedures An application of an unnamed procedure:
((lambda (x) (* x x)) 4) ==> 16
2. Naming the procedure:
(define square (lambda (x) (* x x)))
(square 3)
3. Syntactic Sugar:
(define (square x) (* x x))

10. Some examples: (define twice ) (twice 2) ==> 4 (twice 3) ==> 6
(define second )
(second ) ==> 15 (second ) ==> -5
(lambda (x) (* 2 x))
(lambda (x y z) y)
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11. An example for the substitution model
(define square (lambda (x) (* x x))) (define average (lambda (x y) (/ (+ x y) 2)))
(average 5 (square 3)) (average 5 (* 3 3)) (average 5 9) first evaluate operands, then substitute (applicative order)
(/ (+ 5 9) 2) (/ 14 2) if operator is a primitive procedure, replace by result of operation
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12. Sum of squares S(n) = 02 + 12 + 22 ………. …… (n-1)2 + n2 Notice that:
S(n) = S(n-1) + n2
S(0) = 0
These two properties completely define the function

13. An algorithm for sum of squares
(define sum-squares (lambda (n) (if (= n 0) (+ (sum-squares (- n 1)) (square n))))

14. Evaluating (sum-squares 3)
(define (sum-squares n) (if (= n 0) 0 (+ (sum-squares (- n 1)) (square n))))
(sum-squares 3)
(if (= 3 0) 0 (+ (sum-squares (- 3 1)) (square 3)))
(+ (sum-squares (- 3 1)) (square 3))
(+ (sum-squares (- 3 1)) (* 3 3))
(+ (sum-squares (- 3 1)) 9)
(+ (sum-squares 2) 9)
(+ (if (= 2 0) 0 (+ (sum-squares (- 2 1)) (square 2))) 9) …
(+ (+ (sum-squares 1) 4) 9)
(+ (+ (+ (sum-squares 0) 1) 4) 9)
(+ (+ (+ (if (= 0 0) 0(+ (sum-squares (- 0 1)) (square 0))) 1) 4) 9)
(+ (+ (+ 0 1) 4) 9) 14
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15. How to design recursive algorithms
Show how to solve big instances, if you know the solution to
smaller instances.
Wishful thinking: if I could only solve the smaller instance …
Solve the “base cases”.
For example, for sum of squares:
S(n) = S(n-1) + n (induction rule)
S(0)= (base case)
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16. General form of recursive algorithms
test, base case, recursive case
(define sum-sq (lambda (n)
(if (= n 0) ; test for base case
; base case
(+ (sum-sq (- n 1)) (square n))
; recursive case
)))
base case: non-decomposable problem
recursive case: larger (decomposable) problem
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17. Evaluating (sum-squares 3) in the substitution model, with IF a regular form….
(define (sum-squares n) (if (= n 0) 0 (+ (sum-squares (- n 1)) (square n))))
(sum-squares 3)
(if (= 3 0) 0 (+ (sum-squares (- 3 1)) (square 3)))
(if #f 0 (+ (sum-squares 2) 9))
(if #f 0 (+ (if #f 0 (+ (sum-squares 1) 4))) 9)) ..
calling (sum-squares 0) ….
calling (sum-squares -1)
We evaluate all operands.
We always call (sum-squares) again.
We get an infinite loop…….. OOPS
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18. Another example of a recursive algorithm
even?
(define even? (lambda (n)
(not (even? (- n 1))) ; recursive case
)))
(if (= n 0) ; test for base case
#t ; base case
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19. Short summary Design a recursive algorithm by
1. Solving big instances using the solution to smaller instances.
2. Solving directly the base cases.
Recursive algorithms have
1. test
2. recursive case
3. base case
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20. SQRT To find an approximation of square root of x: Make a guess G
Improve the guess by averaging G and x/G
Keep improving the guess until it is good enough
G = 1
X = 2
X/G = 2
G = ½ (1+ 2) = 1.5
X/G = 4/3
G = ½ (3/2 + 4/3) = 17/12 =
X/G = 24/17
G = ½ (17/ /17) = 577/408 =
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21. (define initial-guess 1.0) (define precision 0.0001)
(define (sqrt-iter guess x)
(if (good-enough? guess x)
guess
(sqrt-iter (improve guess x) x)))
(define (good-enough? guess x)
(< (abs (- (square guess) x)) precision))
(define (improve guess x)
(average guess (/ x guess)))
(define (sqrt x)
(sqrt-iter initial-guess x))
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22. Good programming 1. Divide the task to well-defined, natural, and
simple sub-tasks.
E.g: good-enough? and improve .
Thumb of rule: If you can easily name it, it does a well-defined task.
2. Use parameters. E.g.: precision, initial-guess.
3. Use meaningful names.
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23. Procedural abstraction
It is better to:
Export only what is needed
Hide internal details.
The procedure SQRT is of interest for the user.
The procedure improve-guess is an internal detail.
Exporting only what is needed leads to:
A clear interface
Avoids confusion
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24. Rewriting SQRT (Block structure)
(define (sqrt x)
(define (sqrt-iter guess x)
(if (good-enough? guess x)
guess
(sqrt-iter (improve guess x) x)))
(define (good-enough? guess x)
(< (abs (- (square guess) x)) precision))
(define (improve guess x)
(average guess (/ x guess)))
(define initial-guess 1.0)
(define precision )
(sqrt-iter initial-guess x))
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25. Lexical Scoping Every variable is:
Recognized within the procedure where it is defined (its scope).
Not recognized outside it.
Allows different procedures to use the same variable name. E.g., different procedures can use the variable name i as a counter.
A variable can be used globally within its scope. No need to pass it from one procedure to another.
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26. SQRT again, x is used globally.
(define (sqrt x)
(define (good-enough? guess)
(< (abs (- (square guess) x)) precision))
(define (improve guess)
(average guess (/ x guess)))
(define (sqrt-iter guess)
(if (good-enough? guess)
guess
(sqrt-iter (improve guess))))
(define initial-guess 1.0)
(define precision )
(sqrt-iter initial-guess))
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27. Variables scope (define (sqrt x) (define (good-enough? guess)
(< (abs (- (square guess) x)) precision))
(define (improve guess)
(average guess (/ x guess)))
(define (sqrt-iter guess)
(if (good-enough? guess)
guess
(sqrt-iter (improve guess))))
(define initial-guess 1.0)
(define precision )
(sqrt-iter initial-guess))
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28. An example (define (proc1 x) (define (proc2 y) (+ x y))
(define (proc3 x) (proc2 x))
(proc3 (* 2 x)))
Proc1.x
Proc3.x
(proc1 4) proc1.x = 4
(proc3 8) proc3.x = 8
(proc2 8) proc2.y = 8 proc2.x=proc1.x=4 12
Local bindings override more global bindings.
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