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Understanding Procedures and Evaluations in Computer Science

Explore procedural patterns, naming abstractions, and the Substitution Model. Learn applicative and normal order evaluation rules, with examples like Factorial. Compare iterative and recursive approaches to computing, discussing processes and space complexities.

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Understanding Procedures and Evaluations in Computer Science

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  1. Lecture 3 Material in the textbook on Pages 32-46 of 2nd Edition Sections 1.2.1 to 1.2.4 מבוא מורחב

  2. Review • Procedures capture common patterns • Naming abstractions allows us to use them as primitive objects • עוץ-לי-גוץ-לי Principle: “If you capture a process (in a procedure), and give it a name, you have power over that process” • The Substitution Model מבוא מורחב

  3. Today • Refine our model • Understand how it captures the nature of processes • Use it to quantify various properties of processes: • Number of steps a process takes (Time Complexity) • Amount of Space a process uses (Space Complexity) מבוא מורחב

  4. Applicative order evaluation rules Combination... (<operator> <operand1> …… <operand n>) • Evaluate <operator> to get the procedure and evaluate <operands> to get the arguments • If <operator> is primitive: do whatever magic it does • If <operator> is compound: evaluate body with formal parameters replaced by arguments מבוא מורחב

  5. Example: Factorial • wishful thinking : • base case: n! =n * (n-1)! n = 1 (define fact (lambda (n) (if (= n 1) 1 (* n (fact (- n 1)))))) מבוא מורחב

  6. Example: Factorial • ==>(fact 3) • ([[n](if(= n 1) 1 (* n (fact (- n 1))))] 3) • (if (= 3 1) 1 (* 3 (fact (- 3 1)))) • (if #f 1 (* 3 (fact (- 3 1)))) • (* 3 (fact (- 3 1))) • (* 3 ([[n](if(= n 1) 1 (* n (fact (- n 1))))] (- 3 1))) • (* 3 (if (= 2 1) 1 (* 2 (fact (- 2 1))))) • (* 3 (if #f 1 (* 2 (fact (- 2 1))))) • (* 3 (* 2 (fact (-2 1)))) • (* 3 (* 2 (([[n](if(= n 1) 1 (* n (fact (- n 1))))] (- 2 1)))) • (* 3 (* 2 (if (= 1 1) 1 (* (fact (- 1 1))))))) • (* 3 (* 2 (if #t 1 (* 1 (fact (- 1 1)))))) • (* 3 (* 2 1)) • (* 3 2) • 6 מבוא מורחב

  7. Normal order evaluation Combination … (<operator> <operand1> …… <operand n>) • Evaluate <operator> to get the procedure and evaluate <operands> to get the arguments • If <operator> is primitive: do whatever magic it does • If <operator> is compound: evaluate body with formal parameters replaced by arguments מבוא מורחב

  8. The Difference Normal ((lambda (x) (+ x x)) (* 3 4)) (+ (* 3 4) (* 3 4)) (+ 12 12) 24 Applicative ((lambda (x) (+ x x)) (* 3 4)) ((lambda (x) (+ x x)) 12) (+ 12 12) 24 This may matter in some cases: ((lambda (x y) (+ x 2)) 3 (/ 1 0)) Scheme is an Applicative Order Language! מבוא מורחב

  9. Compute ab (Recursive Approach) • wishful thinking : • base case: ab=a * a(b-1) a0 = 1 • (define exp-1 • (lambda (a b) • (if (= b 0) • 1 • (* a (exp-1 a (- b 1)))))) מבוא מורחב

  10. ab=a * a * a*…*a • Which is: b ab = a2 *a*…*a= a3 *…*a • Operationally: • Halting condition: result  result * a counter  counter - 1 counter = 0 Compute ab (Iterative Approach) • Another approach: מבוא מורחב

  11. Syntactic Recursion Compute ab (Iterative Approach) (define (exp-2 a b) (define (exp-iter a b product) (if (= b 0) product (exp-iter a (- b 1) (* a product)))) (exp-iter a b 1) How then, do the two procedures differ? They give rise to different processes – lets use our model to understand how. מבוא מורחב

  12. Recursive Process • (define exp-1 • (lambda (a b) • (if (= b 0) • 1 • (* a (exp-1 a (- b 1)))))) • (exp-1 3 4) • (* 3 (exp-1 3 3)) • (* 3 (* 3 (exp-1 3 2))) • (* 3 (* 3 (* 3 (exp-1 3 1)))) • (* 3 (* 3 (* 3 (* 3 (exp-1 3 0))))) • (* 3 (* 3 (* 3 (* 3 1)))) • (* 3 (* 3 (* 3 3))) • (* 3 (* 3 9)) • (* 3 27) • 81 מבוא מורחב

  13. Iterative Process (define (exp-2 a b) (define (exp-iter a b product) (if (= b 0) product (exp-iter a (- b 1) (* a product)))) (exp-iter a b 1) (exp-2 3 4) (exp-iter 3 4 1) (exp-iter 3 3 3) (exp-iter 3 2 9) (exp-iter 3 1 27) (exp-iter 3 0 81) 81 מבוא מורחב

  14. (exp-1 3 4) • (* 3 (exp-1 3 3)) • (* 3 (* 3 (exp-1 3 2))) • (* 3 (* 3 (* 3 (exp-1 3 1)))) • (* 3 (* 3 (* 3 (* 3 (exp-1 3 0))))) • (* 3 (* 3 (* 3 (* 3 1)))) • (* 3 (* 3 (* 3 3))) • (* 3 (* 3 9)) • (* 3 27) • 81 (exp-2 3 4) Growing amount of space (exp-iter 3 4 1) (exp-iter 3 3 3) (exp-iter 3 2 9) Constant amount of space (exp-iter 3 1 27) (exp-iter 3 0 81) 81 The Difference מבוא מורחב

  15. operation pending no pending operations Why More Space? • Recursive exponentiation: (define exp-1 (lambda (a b) (if (= b 0) 1 (* a (exp-1 a (- b 1))))) • Iterative exponentiation: • (define (exp-2 a b) • (define (exp-iter a b product) • (if (= b 0) • product • (exp-iter a (- b 1) (* a product)))) • (exp-iter a b 1))

  16. Summary • Recursive process num of deferred operations “grows proportional to b” • Iterative process num of deferred operations stays “constant” (actually its zero) Can we better quantify these observations? מבוא מורחב

  17. Orders of Growth • Suppose n is a parameter that measures the size of a problem (the size of its input) • R(n)measures the amount of resources needed to compute a solution procedure of size n. • Two common resources are space, measured by the number of deferred operations, and time, measured by the number of primitive steps. מבוא מורחב

  18. Orders of Growth • Want to estimate the “order of growth” of R(n): R1(n)=100n2 R2(n)=2n2+10n+2 R3(n) = n2 Are all the same in the sense that if we multiply the input by a factor of 2, the resource consumption increase by a factor of 4 Order of growth is proportional to n2 מבוא מורחב

  19. Orders of Growth • We say R(n)has order of growth Q(f(n))if there are constants c1 0and c2 0such that for all n>c0 c1f(n)<= R(n)<= c2f(n) • R(n)O(f(n)) if there is a constant c  0such that for all n • R(n) <= cf(n) • R(n)(f(n)) if there is a constant c  0such that for all n • cf(n)<= R(n) מבוא מורחב

  20. Orders of Growth True or False? f 100n2 O(n) t 100n2 (n) 100n2 Q(n) f 100n2 Q(n2) t True or False? f 2100 (n) t 2100n  O(n2) 2n  Q(n) f 210 Q(1) t מבוא מורחב

  21. (exp-1 3 4) “n”=b=4 • (* 3 (exp-1 3 3)) • (* 3 (* 3 (exp-1 3 2))) • (* 3 (* 3 (* 3 (exp-1 3 1)))) • (* 3 (* 3 (* 3 (* 3 (exp-1 3 0))))) • (* 3 (* 3 (* 3 (* 3 1)))) • (* 3 (* 3 (* 3 3))) • (* 3 (* 3 9)) • (* 3 27) • 81 Resources Consumed by EXP-1 • Space b <= R(b) <= b which is Q(b) • Time b <= R(b) <= 2b which is Q(b) Linear Recursive Process מבוא מורחב

  22. (exp-2 3 4) “n”=b=4 (exp-iter 3 4 1) (exp-iter 3 3 3) (exp-iter 3 2 9) (exp-iter 3 1 27) (exp-iter 3 0 81) 81 Resources Consumed by EXP-2 • Space Q(1) • Time Q(b) Linear Iterative Process מבוא מורחב

  23. Summary • Refine our model • captures the nature of processes • Use it to quantify various properties of processes: • Number of steps a process takes (Time Complexity) • Amount of Space a process uses (Space Complexity) מבוא מורחב

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