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6.001 SICP – September 22

6.001 SICP – September 22. 6001-Introduction Trevor Darrell trevor@csail.mit.edu 32-D512 6.001 web page: http://sicp.csail.mit.edu/ section web page: http://www.csail.mit.edu/trevor/6001/ Office Hours W11, 32-D512 Orders of Growth Linear Exponential Logarithmic Let. 6.001 SICP.

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6.001 SICP – September 22

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  1. 6.001 SICP – September 22 6001-Introduction Trevor Darrell trevor@csail.mit.edu 32-D512 6.001 web page: http://sicp.csail.mit.edu/ section web page: http://www.csail.mit.edu/trevor/6001/ Office Hours W11, 32-D512 • Orders of Growth • Linear • Exponential • Logarithmic • Let 6.001 SICP

  2. 6.001 SICP • Orders of Growth • Linear • Exponential • Logarithmic • let 6.001 SICP

  3. Orders of growth of processes • Suppose n is a parameter that measures the size of a problem • Let R(n) be the amount of resources needed to compute a procedure of size n. • We say R(n) has order of growth Q(f(n)) if there are constants k1 and k2 such that k1f(n)<= R(n)<= k2f(n)for large n • Two common resources are space, measured by the number of deferred operations, and time, measured by the number of primitive steps. 6.001 SICP

  4. Orders of Growth N=2 N=10 N=100 • Constant Q(1) 1 1 1 • Logarithmic Q(log N) 1 3.33 6.66 • Linear Q(n) 2 10 100 • Quadratic Q(n2) 4 100 10000 • Exponential Q(2n) 4 1024 1.26x1030 • At 1 billion operations per second (current state of the art), if you were to run an exponential time algorithm in the lab on a data set of size n=100, you would be waiting for approximately 4x1011 centuries for the code to finish running! 6.001 SICP

  5. Order of growth examples For each, find simplest and slowest growing f for which R(n)=Q(f(n)) • R(n)=6 • R(n)=n2 + 3 • R(n)=6*n3 + 3*n2 + 7n + 100 • R(n)=23n+7 6.001 SICP

  6. Order of growth examples For each, find simplest and slowest growing f for which R(n)=Q(f(n)) • R(n)=6 Q(1) 1 * 1 <= 6 <= 6 * 1 for all n • R(n)=n2 + 3 Q(n2) 1*n2 <= n2 + 3 <= 2*n2 for all n > 2 • R(n)=6*n3 + 3*n2 + 7n + 100 Q(n3) 1*n3 <= 6*n3 + 3*n2 + 7n + 100 <= 7*n3 for all n>100 • R(n)=23n+7 Q(8n) 1*8n <= 23n+7 <= 28*8n for all n>0 6.001 SICP

  7. Examples (define (factorial n) (if (= n 1) 1 (* n (factorial (- n 1))))) (factorial 5) (* 5 (factorial 4)) (* 5 (* 4 (factorial 3))) (* 5 (* 4 (* 3 (factorial 2)))) (* 5 (* 4 (* 3 (* 2 (factorial 1))))) (* 5 (* 4 (* 3 (* 2 1)))) (* 5 (* 4 (* 3 2))) (* 5 (* 4 6)) 120 Time = Q(n) Space = Q(n) 6.001 SICP

  8. Examples (define (fact2 n) (define (helper cur k) (if (= k 1) cur (helper (* cur k) (- k 1)))) (helper 1 n)) (fact2 5) (helper 1 5) (helper 5 4) (helper 20 3) (helper 60 2) (helper 120 1) Time = Q(n) Space = Q(1) 6.001 SICP

  9. Towers of Hanoi • Three posts, and a set of different size disks • any stack must be sorted in decreasing order from bottom to top • the goal is to move the disks one at a time, while preserving these conditions, until the entire stack has moved from one post to another 6.001 SICP

  10. Towers of Hanoi (define move-tower (lambda (size from to extra) (cond ((= size 0) true) (else (move-tower (- size 1) from extra to) (print-move from to) (move-tower (- size 1) extra to from))))) (define print-move (lambda (from to) (write-line ``Move top disk from ``) (write-line from) (write-line `` to ``) (write-line to))) 6.001 SICP

  11. Move 4 Move 3 Move 3 Move 2 Move 2 Move 2 Move 2 Move 1 Move 1 Move 1 Move 1 Move 1 Move 1 Move 1 Move 1 A tree recursion 6.001 SICP

  12. Orders of growth for towers of Hanoi • Let tn be the number of steps that we need to take to solve the case for n disks. Then •  tn = 2tn-1 + 1 = 2(2tn-2 +1) + 1 ….= 2n -1 •  So in time we have Q(2n) -- exponential • In space, we have one deferred operation for each increment of the stack of disks -- Q(n) -- linear 6.001 SICP

  13. Quiz What is simplest expression for the order of growth of running time of procedure mul1 & mul2? (define (mul1 n m) (if (= n 0) 0 (+ m (mul1 (- n 1) m)))) (define (mul2 n m) (define (help count ans) (if (= count 0) ans (help (- count 1) (+ m ans)))) (help n 0)) 1. Theta(1) 2. Theta(2n) 3. Theta(n) 4. Theta(n/2) 6.001 SICP

  14. Mul Now write a procedure mul4 that computes m*n in Q(log n) time Hint: Compare with a^b = (a^2)^(b/2) if b is even, and a^b = a*a^(b-1) if b is odd, n*m = 2 * ( (n / 2) * m) if n even, n*m = m + ((n-1) * m) if n odd. 6.001 SICP

  15. Local Variables • Suppose we want to compute F(x,y)=x(1+xy)2 + y(1-y) + (1+xy)(1-y) • Useful to name intermediate quantities: A = 1 + xy B = 1 - y F(x,y) = xA2 + yB + AB • How can we do this? Using a “helper function?” Using a nested lambda? Using Let… 6.001 SICP

  16. Local variables with helper fcn. / lambda A = 1 + xy; B = 1 - y; F(x,y) = xA2 + yB + AB • Can use an auxiliary procedure to bind the intermediate variables: • (define (f x y) • (define (f-helper a b) • (+ (* x (square a)) • (* y b) • (* a b))) • (f-helper (+ 1 (* x y)) • (- 1 y))) 6.001 SICP

  17. Local variables with helper fcn. / lambda A = 1 + xy; B = 1 - y; F(x,y) = xA2 + yB + AB Or use lambda directly…. • (define (f x y) • ((lambda (a b) • (+ (* x (square a)) • (* y b) • (* a b))) • (+ 1 (* x y)) • (- 1 y))) • (define (f x y) • (define (f-helper a b) • (+ (* x (square a)) • (* y b) • (* a b))) • (f-helper (+ 1 (* x y) ) • (- 1 y)) ) 6.001 SICP

  18. Local variables with let • Let is a special form which is syntactic sugar for this use of lambda to create local variables (let (( <var1> <exp1> ) ( <var2> <exp2> ) … ( <varn> <expn> )) <body> ) Which means: let <var1> be<exp1>, <var2> be <exp2>, <var3> be <exp3>in <body> , and return the value of the last expression in <body> 6.001 SICP

  19. Local variables with let A = 1 + xy; B = 1 - y; F(x,y) = xA2 + yB + AB • (define (f x y) • ((lambda (a b) • (+ (* x (square a)) • (* y b) • (* a b))) • (+ 1 (* x y)) • (- 1 y))) • (define (f x y) • (let ((a (+ 1 (* x y)) • (b (- 1 y)) ) • (+ (* x (square a)) • (* y b) • (* a b)))) 6.001 SICP

  20. Let vs. nested define Use of nested defines for local variable definition is considered bad form, but is technically legal: (define (f x) (define k 2) (+ x k)) Let can be used inside another expression: (define x 5) (+ (let ((x 3)) (+ x (* x 10))) 5) • (define (f x) • (let ((k 2)) • (+ x k))) 6.001 SICP

  21. Let* • Let first evaluates all of the values, then binds each to the variable names. (define x 0) (define y 0) (let ((x 1) (y 2) (z (+ x y))) z) === 0! • Let* evaluates and binds each name/value pair in sequence, so that each definition is available to the next (let* ((x 1) (y 2) (z (+ x y))) z) === 3! 6.001 SICP

  22. Let/Let* Write a single function using let or let* to solve (x- 5)^3 + 2(x-5)^2 where x = y – 1 and y = K*z and K = 39.2, where z is given. (let* ((k 39.2) (y (* k z)) (x (- y 1)) (a (- x 5))) (+ (* a a a) (* 2 a a))) 6.001 SICP

  23. Remember… • calendar… 6.001 SICP

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