</executions>
<configuration>
<sourceDirectories>
- <sourceDirectory>src/main/clojure</sourceDirectory>
+ <sourceDirectory>src</sourceDirectory>
</sourceDirectories>
</configuration>
</plugin>
+++ /dev/null
-(ns sicp.ch1-1)
-
-(defn square [x] (* x x))
-
-(defn sum-of-squares [x y]
- (+ (square x) (square y)))
-
-(defn f [a]
- (sum-of-squares (+ a 1) (* a 2)))
-
-(defn abs
- "find absolute value of x"
- [x]
- (if (< x 0) (- x) x))
-
-(defn abs1 [x]
- (cond (< x 0) (- x)
- :else x))
-
-;; exercise 1.1: What is the result printed by the interpreter in response
-;; to each expression (in order) ?
-
-(def a 3)
-(def b (+ a 1))
-
-(+ a b (* a b)) ; 19
-(= a b) ; false
-
-(if (and (> b a) (< b (* a b)))
- b
- a) ; 4
-
-(cond (= a 4) 6
- (= b 4) (+ 6 7 a)
- :else 25) ; 16
-
-(+ 2 (if (> b a) b a)) ; 6
-
-(* (cond (> a b) a
- (< a b) b
- :else -1)
- (+ a 1)) ; 16
-
-;; exercise 1.2: Translate the given expression into prefix form
-(/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 1 5)))))
- (* 3 (- 6 2) (- 2 7))) ; -71/300
-
-;; exercise 1.3: Define a procedure that takes three numbers as
-;; arguments and returns the sum of the squares of
-;; the two larger numbers.
-(defn sort3 [a b c]
- (cond (> b a) (sort3 b a c)
- (< b c) (sort3 a c b)
- :else [a b c]))
-
-(defn sum-of-sq-of-two-largest [a b c]
- (apply sum-of-squares (take 2 (sort3 a b c))))
-
-;; well, I cheated above. Let me use only the constructs introduced
-;; so far in the book. (follows after the sicp meetup #2 on 28/mar/2010.
-(defn sum-of-square-of-two-largest [a b c]
- (if (> a b)
- (if (> b c)
- (sum-of-squares a b) ; a > b > c
- (sum-of-squares a c))
- (if (> a c)
- (sum-of-squares b a)
- (sum-of-squares b c))))
-
-;; exercise 1.4: Observe that our model of evaluation allows for
-;; combinations whose operators are compound
-;; expressions. Use this observation to describe the
-;; behavior of the following procedure:
-;; (defn a-plus-abs-b [a b]
-;; ((if (> b 0) + -) a b))
-(comment
- If b is positive, we do (+ a b) and if it is negative, we do (- a b).
- This makes use of the fact that the first element in a list is an
- operand. Here, the operand is chosen based on other operators.
- )
-
-;; exercise 1.5: Ben Bitdiddle has invented a test to determine
-;; whether the interpreter he is faced with is
-;; using applicative-order evaluation or normal-order
-;; evaluation. He defines the following two procedures:
-;; (defn p [] (p))
-;; (defn test [x y]
-;; (if (= x 0)
-;; 0
-;; y))
-;;
-;; Then he evaluates the expression
-;;
-;; (test 0 (p))
-;;
-;; What behavior will Ben observe with an interpreter that uses
-;; applicative-order evaluation?
-(comment
- In the case of applicative order evaluation, the test gets into
- and infinite loop (eventually using all the stack), as the parameters
- are evaluated before they are actualy used in the function.
- )
-;; What behavior will he observe with an interpreter that uses
-;; normal-order evaluation? Explain your answer.
-(comment
- It will print 0, as (p) is not evaluated in this case.
- )
-
-;; 1.1.7 Square root finding using Newton's method
-(defn average [x y]
- (/ (+ x y) 2))
-
-(defn improve [guess x]
- (average guess (/ x guess)))
-
-(defn good-enough? [guess x]
- (< (abs (- (square guess) x)) 0.001))
-
-(defn sqrt-iter [guess x]
- (if (good-enough? guess x)
- guess
- (sqrt-iter (improve guess x)
- x)))
-
-(defn sqrt [x]
- (sqrt-iter 1.0 x))
-
-;; exercise 1.6
-;; Alyssa P. Hacker doesn't see why if needs to be provided as a special form.
-;; ``Why can't I just define it as an ordinary procedure in terms of cond?''
-(defn new-if [predicate then-clause else-clause]
- (cond predicate then-clause
- :else else-clause))
-
-(new-if (= 3 2) 0 5) ; 5
-(new-if (= 1 1) 0 5) ; 0
-
-;; Delighted, Alyssa uses new-if to rewrite the square-root program:
-
-(defn sqrt-iter [guess x]
- (new-if (good-enough? guess x)
- guess
- (sqrt-iter (improve guess x)
- x)))
-
-;; what happens when Alyssa attempts to use this to compute square roots? Explain.
-(comment
- Since `new-if' is a function, when it is called from sqrt-iter, the parameters
- are evaluated before it gets called. good-enough? will return a false unless
- the guess and x are almost the same. guess evaluated to the initial value of
- guess. sqrt-iter gets evaluated, but gets into an infinite loop. The predicate
- will have no effect.)
-
-;; Exercise 1.7: The good-enough? test used in computing square roots will not
-;; be very effective for finding the square roots of very small numbers. Also,
-;; in real computers, arithmetic operations are almost always performed with
-;; limited precision. This makes our test inadequate for very large numbers.
-;; Explain these statements, with examples showing how the test fails for small
-;; and large numbers.
-(comment
- user> (sqrt (square 0.001))
- 0.031260655525445276
- user> (sqrt (square 0.2))
- 0.20060990407779591
- user> (sqrt (square 0.01))
- 0.03230844833048122
- user> (sqrt (square 0.02))
- 0.0354008825558513
- user> (sqrt (square 10))
- 10.000000000139897
- user> (sqrt (square 100))
- 100.00000025490743
- user> (sqrt (square 200))
- 200.000000510076
- user> (sqrt (square 2))
- 2.0000000929222947
- user> (sqrt (square 0.1))
- 0.10032578510960607
- user> (sqrt (square 0.01))
- 0.03230844833048122
- user> (sqrt (square 10000))
- 10000.0
- user> (sqrt (square 20000))
- 20000.0
- user> (sqrt (square 200000))
- 200000.0
- user> (sqrt (square 20000000))
- 2.0E7
- user> (sqrt (square 20000000000))
- 2.0E10
- user> (sqrt (square 200000.012))
- 200000.012
- user> (sqrt (square 2000000.123))
- 2000000.123
- user> (sqrt (square 200000000.123))
- 2.00000000123E8
- user> (sqrt (square 2000000000.123))
- 2.000000000123E9
- user> (sqrt (square 20000000000.123))
- 2.0000000000123E10
- user> (sqrt (square 2000000000000.123))
- 2.000000000000123E12
- )
-;; An alternative strategy for implementing good-enough? is to watch how guess
-;; changes from one iteration to the next and to stop when the change is a very
-;; small fraction of the guess.
-(defn sqrt-iter [old-guess new-guess x]
- (if (good-enough? old-guess new-guess x)
- new-guess
- (sqrt-iter new-guess (improve new-guess x)
- x)))
-
-(defn improve [guess x]
- (average guess (/ x guess)))
-
-(defn average [x y]
- (/ (+ x y) 2))
-
-(defn good-enough? [old-guess new-guess x]
- (< (/ (abs (- new-guess old-guess)) new-guess) 0.0001))
-
-(defn sqrt [x]
- (sqrt-iter x 1.0 x))
-(comment
-user> (sqrt (square 0.01))
-0.010000000025490743
-user> (sqrt (square 0.001))
-0.0010000000000000117
-user> (sqrt (square 0.0001))
-1.0000000000082464E-4
-user> (sqrt (square 0.02))
-0.020000000050877154
-user> (sqrt (square 0.002))
-0.0020000000000000235
-user> (sqrt (square 4))
-4.000000000000051
-user> (sqrt (square 20))
-20.000000000298428
-user> (sqrt (square 25))
-25.000000063076968
-user> (sqrt 5)
-2.236067977499978
-user> (sqrt 25)
-5.000000000053722
-user> (sqrt 9)
-3.000000001396984
-user> (sqrt 81)
-9.000000000007091
-)
-;; exercise 1.8: cube root
-(defn cube [x]
- (* x x x))
-
-(defn improve [guess x]
- (/ (+ (/ x (square guess)) (* 2 guess)) 3))
-
-(defn cubert-iter [old-guess new-guess x]
- (if (good-enough? old-guess new-guess x)
- new-guess
- (cubert-iter new-guess (improve new-guess x)
- x)))
-
-(defn cuberoot [x]
- (cubert-iter x 1.0 x))
-
-(comment
-user> (cuberoot (cube 2))
-2.000000000012062
-user> (cuberoot (cube 10))
-10.000000000000002
-user> (cuberoot (cube 9))
-9.000000000053902
-user> (cuberoot (cube 0.001))
-0.001000000000000962
-user> (cuberoot (cube 0.0001))
-1.000000000000001E-4
-user>
-)
-;; section 1.1.8
-;; hiding the non-public procedure definitions
-(defn- sqrt-iter [guess x]
- (if (good-enough? guess x)
- guess
- (sqrt-iter (improve guess x)
- x)))
-
-(defn- improve [guess x]
- (average guess (/ x guess)))
-
-(defn- average [x y]
- (/ (+ x y) 2))
-
-(defn- good-enough? [guess x]
- (< (abs (- (square guess) x)) 0.001))
-
-(defn sqrt [x]
- (sqrt-iter 1.0 x))
+++ /dev/null
-(ns sicp.ch1-2
- (:use (clojure.contrib trace math)))
-
-
-(defn factorial [n]
- (if (= n 1)
- 1
- (* n (factorial (- n 1)))))
-
-;; stack friendly version
-(defn factorial2 [n]
- (loop [x n acc 1]
- (if (= x 1)
- acc
- (recur (- x 1) (* acc x)))))
-
-;; even better
-(defn factorial3 [n]
- (reduce * (range 1 (inc n))))
-
-;; (comment
-;; user> (dotrace [factorial] (factorial 3))
-;; TRACE t2701: (factorial 3)
-;; TRACE t2702: | (factorial 2)
-;; TRACE t2703: | | (factorial 1)
-;; TRACE t2703: | | => 1
-;; TRACE t2702: | => 2
-;; TRACE t2701: => 6
-;; 6
-;; )
-
-;; (comment
-;; sicp.chapter1.ch1_2> (dotrace [factorial] (factorial 6))
-;; TRACE t2778: (factorial 6)
-;; TRACE t2779: | (factorial 5)
-;; TRACE t2780: | | (factorial 4)
-;; TRACE t2781: | | | (factorial 3)
-;; TRACE t2782: | | | | (factorial 2)
-;; TRACE t2783: | | | | | (factorial 1)
-;; TRACE t2783: | | | | | => 1
-;; TRACE t2782: | | | | => 2
-;; TRACE t2781: | | | => 6
-;; TRACE t2780: | | => 24
-;; TRACE t2779: | => 120
-;; TRACE t2778: => 720
-;; 720
-;; )
-
-;; (comment
-;; sicp.chapter1.ch1_2> (dotrace [factorial2] (factorial2 6))
-;; TRACE t2806: (factorial2 6)
-;; TRACE t2806: => 720
-;; 720
-;; )
-
-(defn fact-iter [product counter max-count]
- (if (> counter max-count)
- product
- (fact-iter (* product counter)
- (inc counter)
- max-count)))
-
-(defn factorial [n]
- (fact-iter 1 1 n))
-
-;; (comment
-;; user> (dotrace [factorial fact-iter] (factorial 6))
-;; TRACE t2378: (factorial 6)
-;; TRACE t2379: | (fact-iter 1 1 6)
-;; TRACE t2380: | | (fact-iter 1 2 6)
-;; TRACE t2381: | | | (fact-iter 2 3 6)
-;; TRACE t2382: | | | | (fact-iter 6 4 6)
-;; TRACE t2383: | | | | | (fact-iter 24 5 6)
-;; TRACE t2384: | | | | | | (fact-iter 120 6 6)
-;; TRACE t2385: | | | | | | | (fact-iter 720 7 6)
-;; TRACE t2385: | | | | | | | => 720
-;; TRACE t2384: | | | | | | => 720
-;; TRACE t2383: | | | | | => 720
-;; TRACE t2382: | | | | => 720
-;; TRACE t2381: | | | => 720
-;; TRACE t2380: | | => 720
-;; TRACE t2379: | => 720
-;; TRACE t2378: => 720
-;; 720
-;; )
-
-;; observation: clojure loop-recur construct is essentially the same as
-;; the iterative version.
-
-;; exercise 1.9
-(defn ++ [a b]
- (if (= a 0)
- b
- (inc (++ (dec a) b))))
-
-;; (comment
-;; This version is a recursive process, where the previous call increments
-;; the sum by 1 and each call decrement the first operand by 1.
-
-;; user> (dotrace [++] (++ 4 5))
-;; TRACE t3745: (++ 4 5)
-;; TRACE t3746: | (++ 3 5)
-;; TRACE t3747: | | (++ 2 5)
-;; TRACE t3748: | | | (++ 1 5)
-;; TRACE t3749: | | | | (++ 0 5)
-;; TRACE t3749: | | | | => 5
-;; TRACE t3748: | | | => 6
-;; TRACE t3747: | | => 7
-;; TRACE t3746: | => 8
-;; TRACE t3745: => 9
-;; 9
-;; )
-
-(defn ++ [a b]
- (if (= a 0)
- b
- (++ (dec a) (inc b))))
-
-;; (comment
-
-;; user> (dotrace [++] (++ 4 5))
-;; TRACE t3766: (++ 4 5)
-;; TRACE t3767: | (++ 3 6)
-;; TRACE t3768: | | (++ 2 7)
-;; TRACE t3769: | | | (++ 1 8)
-;; TRACE t3770: | | | | (++ 0 9)
-;; TRACE t3770: | | | | => 9
-;; TRACE t3769: | | | => 9
-;; TRACE t3768: | | => 9
-;; TRACE t3767: | => 9
-;; TRACE t3766: => 9
-;; 9
-;; )
-
-;; exercise 1.10
-;; ackerman functions
-(defn A [x y]
- (cond (= y 0) 0
- (= x 0) (* 2 y)
- (= y 1) 2
- :else (A (- x 1)
- (A x (- y 1)))))
-
-(comment
-user> (A 1 10)
-1024
-user> (A 2 4)
-65536
-user> (A 3 3)
-65536
-)
-
-(defn f [n] (A 0 n)) ; f(n) = 2n
-(defn g [n] (A 1 n)) ; g(n) = 2^n
-;; (comment
-;; g (n) = A (1,n)
-;; = A (0, A (1, n-1)) = f (A(1,n-1))
-;; = f (f (1,n-2))
-;; .....
-;; = f (f (f ... f (1,(n- (n-1)))))
-;; = f (f (f ... 2))
-;; = 2 * (2^(n-1))
-;; = 2^n
-;; )
-
-(defn h [n] (A 2 n)) ; h(n) = 2^(n^2)
-
-;; section 1.2.2: Tree Recursion
-(defn fib [n]
- (cond (= n 0) 0
- (= n 1) 1
- :else (+ (fib (- n 1))
- (fib (- n 2)))))
-
-;; iterative version
-(defn fib-iter [a b count]
- (if (= count 0)
- b
- (fib-iter (+ a b) a (dec count))))
-
-(defn fib [n]
- (fib-iter 1 0 n))
-
-;; example: counting changes
-(defn first-denomination [kinds-of-coins]
- (cond (= kinds-of-coins 1) 1
- (= kinds-of-coins 2) 5
- (= kinds-of-coins 3) 10
- (= kinds-of-coins 4) 25
- (= kinds-of-coins 5) 50))
-
-
-(defn first-denomination [kinds-of-coins]
- ({1 1 2 5 3 10 4 25 5 50} kinds-of-coins))
-
-(defn cc [amount kinds-of-coins]
- (cond (= amount 0) 1
- (or (< amount 0) (= kinds-of-coins 0)) 0
- :else (+ (cc amount (- kinds-of-coins 1))
- (cc (- amount
- (first-denomination kinds-of-coins))
- kinds-of-coins))))
-
-(defn count-change [amount]
- (cc amount 5))
-
-;; exercise 1.11: A function f is defined by the rule that f(n) = n if n < 3
-;; and f(n) = f(n - 1) + 2f(n - 2) + 3f(n - 3) if n> 3.
-;; Write a procedure that computes f by means of a recursive
-;; process. Write a procedure that computes f by means of an
-;; iterative process.
-(defn f [n]
- (if (< n 3)
- n
- (+ (f (- n 1))
- (* 2 (f (- n 2)))
- (* 3 (f (- n 3))))))
-
-(comment
-user> (map f (range 10))
-(0 1 2 4 11 25 59 142 335 796)
-)
-
-;; ex 1.11: iterative version
-(defn f-iter [count prev0 prev1 prev2]
- (if (= count 3)
- (+ prev0
- (* 2 prev1)
- (* 3 prev2))
- (f-iter (dec count)
- (+ prev0
- (* 2 prev1)
- (* 3 prev2))
- prev0
- prev1)))
-
-(defn f [n]
- (if (< n 3)
- n
- (f-iter n 2 1 0)))
-
-;; ex 1.11: iterative version with let
-(defn f-iter [count prev0 prev1 prev2]
- (let [res (+ prev0 (* 2 prev1) (* 3 prev2))]
- (if (= count 3)
- res
- (f-iter (dec count)
- res
- prev0
- prev1))))
-
-;; exercise 1.12. The following pattern of numbers is called Pascal's triangle.
-;; 1
-;; 1 1
-;; 1 2 1
-;; 1 3 3 1
-;; 1 4 6 4 1
-;; ...................
-;;
-;; The numbers at the edge of the triangle are all 1, and each
-;; number inside the triangle is the sum of the two numbers above
-;; it. Write a procedure that computes elements of Pascal's triangle
-;; by means of a recursive process.
-(defn pascal [row col]
- (when (<= col row)
- (if (or (= col 0) (= row col))
- 1
- (+ (pascal (dec row) col)
- (pascal (dec row) (dec col))))))
-
-;; exercise 1.13:
-(comment
-See the pdfs in the directory for the answers.
-)
-
-;; ex 1.13 (contd)
-(defn fib-approx [n]
- (let [phi (/ (+ 1 (sqrt 5)) 2)]
- (/ (expt phi n) (sqrt 5))))
-
-;; (comment
-;; user> (map fib-approx (range 10))
-;; (0.4472135954999579 0.7236067977499789 1.1708203932499368 1.8944271909999157 3.065247584249853 4.959674775249769 8.024922359499623 12.984597134749393 21.009519494249016 33.99411662899841)
-;; )
-
-;; exercise 1.14: tree of (count-changes 11)
-(comment
- See the pdf for the tree representation.
-)
-
-
-;; order of size and computation
-;; see PDF, but below is the trace tree.
-;; (comment
-;; user> (use 'clojure.contrib.trace)
-;; nil
-;; user> (dotrace [count-change cc] (count-change 11))
-;; TRACE t2609: (count-change 11)
-;; TRACE t2610: | (cc 11 5)
-;; TRACE t2611: | | (cc 11 4)
-;; TRACE t2612: | | | (cc 11 3)
-;; TRACE t2613: | | | | (cc 11 2)
-;; TRACE t2614: | | | | | (cc 11 1)
-;; TRACE t2615: | | | | | | (cc 11 0)
-;; TRACE t2615: | | | | | | => 0
-;; TRACE t2616: | | | | | | (cc 10 1)
-;; TRACE t2617: | | | | | | | (cc 10 0)
-;; TRACE t2617: | | | | | | | => 0
-;; TRACE t2618: | | | | | | | (cc 9 1)
-;; TRACE t2619: | | | | | | | | (cc 9 0)
-;; TRACE t2619: | | | | | | | | => 0
-;; TRACE t2620: | | | | | | | | (cc 8 1)
-;; TRACE t2621: | | | | | | | | | (cc 8 0)
-;; TRACE t2621: | | | | | | | | | => 0
-;; TRACE t2622: | | | | | | | | | (cc 7 1)
-;; TRACE t2623: | | | | | | | | | | (cc 7 0)
-;; TRACE t2623: | | | | | | | | | | => 0
-;; TRACE t2624: | | | | | | | | | | (cc 6 1)
-;; TRACE t2625: | | | | | | | | | | | (cc 6 0)
-;; TRACE t2625: | | | | | | | | | | | => 0
-;; TRACE t2626: | | | | | | | | | | | (cc 5 1)
-;; TRACE t2627: | | | | | | | | | | | | (cc 5 0)
-;; TRACE t2627: | | | | | | | | | | | | => 0
-;; TRACE t2628: | | | | | | | | | | | | (cc 4 1)
-;; TRACE t2629: | | | | | | | | | | | | | (cc 4 0)
-;; TRACE t2629: | | | | | | | | | | | | | => 0
-;; TRACE t2630: | | | | | | | | | | | | | (cc 3 1)
-;; TRACE t2631: | | | | | | | | | | | | | | (cc 3 0)
-;; TRACE t2631: | | | | | | | | | | | | | | => 0
-;; TRACE t2632: | | | | | | | | | | | | | | (cc 2 1)
-;; TRACE t2633: | | | | | | | | | | | | | | | (cc 2 0)
-;; TRACE t2633: | | | | | | | | | | | | | | | => 0
-;; TRACE t2634: | | | | | | | | | | | | | | | (cc 1 1)
-;; TRACE t2635: | | | | | | | | | | | | | | | | (cc 1 0)
-;; TRACE t2635: | | | | | | | | | | | | | | | | => 0
-;; TRACE t2636: | | | | | | | | | | | | | | | | (cc 0 1)
-;; TRACE t2636: | | | | | | | | | | | | | | | | => 1
-;; TRACE t2634: | | | | | | | | | | | | | | | => 1
-;; TRACE t2632: | | | | | | | | | | | | | | => 1
-;; TRACE t2630: | | | | | | | | | | | | | => 1
-;; TRACE t2628: | | | | | | | | | | | | => 1
-;; TRACE t2626: | | | | | | | | | | | => 1
-;; TRACE t2624: | | | | | | | | | | => 1
-;; TRACE t2622: | | | | | | | | | => 1
-;; TRACE t2620: | | | | | | | | => 1
-;; TRACE t2618: | | | | | | | => 1
-;; TRACE t2616: | | | | | | => 1
-;; TRACE t2614: | | | | | => 1
-;; TRACE t2637: | | | | | (cc 6 2)
-;; TRACE t2638: | | | | | | (cc 6 1)
-;; TRACE t2639: | | | | | | | (cc 6 0)
-;; TRACE t2639: | | | | | | | => 0
-;; TRACE t2640: | | | | | | | (cc 5 1)
-;; TRACE t2641: | | | | | | | | (cc 5 0)
-;; TRACE t2641: | | | | | | | | => 0
-;; TRACE t2642: | | | | | | | | (cc 4 1)
-;; TRACE t2643: | | | | | | | | | (cc 4 0)
-;; TRACE t2643: | | | | | | | | | => 0
-;; TRACE t2644: | | | | | | | | | (cc 3 1)
-;; TRACE t2645: | | | | | | | | | | (cc 3 0)
-;; TRACE t2645: | | | | | | | | | | => 0
-;; TRACE t2646: | | | | | | | | | | (cc 2 1)
-;; TRACE t2647: | | | | | | | | | | | (cc 2 0)
-;; TRACE t2647: | | | | | | | | | | | => 0
-;; TRACE t2648: | | | | | | | | | | | (cc 1 1)
-;; TRACE t2649: | | | | | | | | | | | | (cc 1 0)
-;; TRACE t2649: | | | | | | | | | | | | => 0
-;; TRACE t2650: | | | | | | | | | | | | (cc 0 1)
-;; TRACE t2650: | | | | | | | | | | | | => 1
-;; TRACE t2648: | | | | | | | | | | | => 1
-;; TRACE t2646: | | | | | | | | | | => 1
-;; TRACE t2644: | | | | | | | | | => 1
-;; TRACE t2642: | | | | | | | | => 1
-;; TRACE t2640: | | | | | | | => 1
-;; TRACE t2638: | | | | | | => 1
-;; TRACE t2651: | | | | | | (cc 1 2)
-;; TRACE t2652: | | | | | | | (cc 1 1)
-;; TRACE t2653: | | | | | | | | (cc 1 0)
-;; TRACE t2653: | | | | | | | | => 0
-;; TRACE t2654: | | | | | | | | (cc 0 1)
-;; TRACE t2654: | | | | | | | | => 1
-;; TRACE t2652: | | | | | | | => 1
-;; TRACE t2655: | | | | | | | (cc -4 2)
-;; TRACE t2655: | | | | | | | => 0
-;; TRACE t2651: | | | | | | => 1
-;; TRACE t2637: | | | | | => 2
-;; TRACE t2613: | | | | => 3
-;; TRACE t2656: | | | | (cc 1 3)
-;; TRACE t2657: | | | | | (cc 1 2)
-;; TRACE t2658: | | | | | | (cc 1 1)
-;; TRACE t2659: | | | | | | | (cc 1 0)
-;; TRACE t2659: | | | | | | | => 0
-;; TRACE t2660: | | | | | | | (cc 0 1)
-;; TRACE t2660: | | | | | | | => 1
-;; TRACE t2658: | | | | | | => 1
-;; TRACE t2661: | | | | | | (cc -4 2)
-;; TRACE t2661: | | | | | | => 0
-;; TRACE t2657: | | | | | => 1
-;; TRACE t2662: | | | | | (cc -9 3)
-;; TRACE t2662: | | | | | => 0
-;; TRACE t2656: | | | | => 1
-;; TRACE t2612: | | | => 4
-;; TRACE t2663: | | | (cc -14 4)
-;; TRACE t2663: | | | => 0
-;; TRACE t2611: | | => 4
-;; TRACE t2664: | | (cc -39 5)
-;; TRACE t2664: | | => 0
-;; TRACE t2610: | => 4
-;; TRACE t2609: => 4
-;; 4
-;; )
-
-
-;; TODO: orders of growth in space and number of steps.
-
-;; exercise 1.15: sin (x) calculation
-;; a. How many times is the procedure p applied when (sine 12.15)
-;; is evaluated?
-;; b. What is the order of growth in space and number of steps (as
-;; a function of a) used by the process generated by the sine
-;; procedure when (sine a) is evaluated?
-(defn cube [x] (* x x x))
-
-(defn p [x] (- (* 3 x) (* 4 (cube x))))
-
-(defn sine [angle]
- (if (not (> (abs angle) 0.1))
- angle
- (p (sine (/ angle 3.0)))))
-
-;; solution to (a) => 5
-;; (comment
-;; user> (dotrace [p] (sine 12.15))
-;; TRACE t2490: (p 0.049999999999999996)
-;; TRACE t2490: => 0.1495
-;; TRACE t2491: (p 0.1495)
-;; TRACE t2491: => 0.4351345505
-;; TRACE t2492: (p 0.4351345505)
-;; TRACE t2492: => 0.9758465331678772
-;; TRACE t2493: (p 0.9758465331678772)
-;; TRACE t2493: => -0.7895631144708228
-;; TRACE t2494: (p -0.7895631144708228)
-;; TRACE t2494: => -0.39980345741334
-;; -0.39980345741334
-;; )
-
-;; solution to b
-;; both space and number of steps grows as log3(a) -> log a to the base 3.
-;;
-;; proof:
-;; a * (1/3)^n <= 0.1
-;; => take log to the base 3 on both the sides.
-
-;; Note: Finding the order of space in a recursive process is sort of, equiv
-;; to finding the number of deferred operations. Which is in-turn the
-;; same as the depth of the evaluation tree.
-
-;; 1.2.4: exponentiation
-;; computing b^n
-
-;; iterative
-(defn expt-iter [base counter product]
- (if (= counter 0)
- product
- (expt-iter base
- (dec counter)
- (* product base))))
-
-(defn myexpt [b n]
- (expt-iter b n 1))
-
-(defn myexpt [b n]
- (if (= n 0)
- 1
- (* b (myexpt b (dec n)))))
-
-;; fast version
-(defn square [x]
- (* x x))
-
-(defn fast-expt [b n]
- (cond (= n 0) 1
- (even? n) (square (fast-expt b (/ n 2)))
- :else (* b (fast-expt b (dec n)))))
-
-;; exercise 1.16:
-(defn myexpt [b n]
- (expt-iter b n 1))
-
-(defn expt-iter [b n a]
- (cond (= n 0) a
- (even? n) (expt-iter (square b) (/ n 2) a)
- :else (expt-iter b (- n 1) (* a b))))
-
-;; exercise 1.17:
-(defn mult [a b]
- (if (= b 0)
- 0
- (+ a (mult a (- b 1)))))
-
-;; double
-(defn twice [x]
- (* 2 x))
-
-(defn half [x]
- (/ x 2))
-
-;; product = 2 * (a * (b/2)) for even b
-;; = a + (a * (b - 1)) for odd b
-(defn fast-mult [a b]
- (cond (= b 0) 0
- (= b 1) a
- (even? b) (twice (fast-mult a (half b)))
- :else (+ a (fast-mult a (- b 1)))))
-
-;; exercise 1.18: iterative multiply thru addition
-;; the idea is to keep a state variable.
-(defn fast-mult-iter [a b k]
- (cond (= b 0) k
- (even? b) (fast-mult-iter (twice a) (half b) k)
- :else (fast-mult-iter a (- b 1) (+ k a))))
-
-(defn fast-mult-new [a b]
- (fast-mult-iter a b 0))
-
-;; (comment
-;; user> (dotrace [fast-mult-new fast-mult-iter] (fast-mult-new 2 3))
-;; TRACE t2915: (fast-mult-new 2 3)
-;; TRACE t2916: | (fast-mult-iter 2 3 0)
-;; TRACE t2917: | | (fast-mult-iter 2 2 2)
-;; TRACE t2918: | | | (fast-mult-iter 4 1 2)
-;; TRACE t2919: | | | | (fast-mult-iter 4 0 6)
-;; TRACE t2919: | | | | => 6
-;; TRACE t2918: | | | => 6
-;; TRACE t2917: | | => 6
-;; TRACE t2916: | => 6
-;; TRACE t2915: => 6
-;; 6
-;; )
-
-;; exercise 1.19: fast fibonacci
-;; see the pdf of the notebook scan for the derivation of p' and q'
-(defn ffib-iter [a b p q count]
- (cond (= count 0) b
- (even? count)
- (ffib-iter a
- b
- (+ (* p p) (* q q))
- (+ (* 2 p q) (* q q))
- (/ count 2))
- :else (ffib-iter (+ (* b q) (* a q) (* a p))
- (+ (* b p) (* a q))
- p
- q
- (- count 1))))
-
-(defn ffib [n]
- (ffib-iter 1 0 0 1 n))
-
--- /dev/null
+(ns sicp.ch1-1)
+
+(defn square [x] (* x x))
+
+(defn sum-of-squares [x y]
+ (+ (square x) (square y)))
+
+(defn f [a]
+ (sum-of-squares (+ a 1) (* a 2)))
+
+(defn abs
+ "find absolute value of x"
+ [x]
+ (if (< x 0) (- x) x))
+
+(defn abs1 [x]
+ (cond (< x 0) (- x)
+ :else x))
+
+;; exercise 1.1: What is the result printed by the interpreter in response
+;; to each expression (in order) ?
+
+(def a 3)
+(def b (+ a 1))
+
+(+ a b (* a b)) ; 19
+(= a b) ; false
+
+(if (and (> b a) (< b (* a b)))
+ b
+ a) ; 4
+
+(cond (= a 4) 6
+ (= b 4) (+ 6 7 a)
+ :else 25) ; 16
+
+(+ 2 (if (> b a) b a)) ; 6
+
+(* (cond (> a b) a
+ (< a b) b
+ :else -1)
+ (+ a 1)) ; 16
+
+;; exercise 1.2: Translate the given expression into prefix form
+(/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 1 5)))))
+ (* 3 (- 6 2) (- 2 7))) ; -71/300
+
+;; exercise 1.3: Define a procedure that takes three numbers as
+;; arguments and returns the sum of the squares of
+;; the two larger numbers.
+(defn sort3 [a b c]
+ (cond (> b a) (sort3 b a c)
+ (< b c) (sort3 a c b)
+ :else [a b c]))
+
+(defn sum-of-sq-of-two-largest [a b c]
+ (apply sum-of-squares (take 2 (sort3 a b c))))
+
+;; well, I cheated above. Let me use only the constructs introduced
+;; so far in the book. (follows after the sicp meetup #2 on 28/mar/2010.
+(defn sum-of-square-of-two-largest [a b c]
+ (if (> a b)
+ (if (> b c)
+ (sum-of-squares a b) ; a > b > c
+ (sum-of-squares a c))
+ (if (> a c)
+ (sum-of-squares b a)
+ (sum-of-squares b c))))
+
+;; exercise 1.4: Observe that our model of evaluation allows for
+;; combinations whose operators are compound
+;; expressions. Use this observation to describe the
+;; behavior of the following procedure:
+;; (defn a-plus-abs-b [a b]
+;; ((if (> b 0) + -) a b))
+(comment
+ If b is positive, we do (+ a b) and if it is negative, we do (- a b).
+ This makes use of the fact that the first element in a list is an
+ operand. Here, the operand is chosen based on other operators.
+ )
+
+;; exercise 1.5: Ben Bitdiddle has invented a test to determine
+;; whether the interpreter he is faced with is
+;; using applicative-order evaluation or normal-order
+;; evaluation. He defines the following two procedures:
+;; (defn p [] (p))
+;; (defn test [x y]
+;; (if (= x 0)
+;; 0
+;; y))
+;;
+;; Then he evaluates the expression
+;;
+;; (test 0 (p))
+;;
+;; What behavior will Ben observe with an interpreter that uses
+;; applicative-order evaluation?
+(comment
+ In the case of applicative order evaluation, the test gets into
+ and infinite loop (eventually using all the stack), as the parameters
+ are evaluated before they are actualy used in the function.
+ )
+;; What behavior will he observe with an interpreter that uses
+;; normal-order evaluation? Explain your answer.
+(comment
+ It will print 0, as (p) is not evaluated in this case.
+ )
+
+;; 1.1.7 Square root finding using Newton's method
+(defn average [x y]
+ (/ (+ x y) 2))
+
+(defn improve [guess x]
+ (average guess (/ x guess)))
+
+(defn good-enough? [guess x]
+ (< (abs (- (square guess) x)) 0.001))
+
+(defn sqrt-iter [guess x]
+ (if (good-enough? guess x)
+ guess
+ (sqrt-iter (improve guess x)
+ x)))
+
+(defn sqrt [x]
+ (sqrt-iter 1.0 x))
+
+;; exercise 1.6
+;; Alyssa P. Hacker doesn't see why if needs to be provided as a special form.
+;; ``Why can't I just define it as an ordinary procedure in terms of cond?''
+(defn new-if [predicate then-clause else-clause]
+ (cond predicate then-clause
+ :else else-clause))
+
+(new-if (= 3 2) 0 5) ; 5
+(new-if (= 1 1) 0 5) ; 0
+
+;; Delighted, Alyssa uses new-if to rewrite the square-root program:
+
+(defn sqrt-iter [guess x]
+ (new-if (good-enough? guess x)
+ guess
+ (sqrt-iter (improve guess x)
+ x)))
+
+;; what happens when Alyssa attempts to use this to compute square roots? Explain.
+(comment
+ Since `new-if' is a function, when it is called from sqrt-iter, the parameters
+ are evaluated before it gets called. good-enough? will return a false unless
+ the guess and x are almost the same. guess evaluated to the initial value of
+ guess. sqrt-iter gets evaluated, but gets into an infinite loop. The predicate
+ will have no effect.)
+
+;; Exercise 1.7: The good-enough? test used in computing square roots will not
+;; be very effective for finding the square roots of very small numbers. Also,
+;; in real computers, arithmetic operations are almost always performed with
+;; limited precision. This makes our test inadequate for very large numbers.
+;; Explain these statements, with examples showing how the test fails for small
+;; and large numbers.
+(comment
+ user> (sqrt (square 0.001))
+ 0.031260655525445276
+ user> (sqrt (square 0.2))
+ 0.20060990407779591
+ user> (sqrt (square 0.01))
+ 0.03230844833048122
+ user> (sqrt (square 0.02))
+ 0.0354008825558513
+ user> (sqrt (square 10))
+ 10.000000000139897
+ user> (sqrt (square 100))
+ 100.00000025490743
+ user> (sqrt (square 200))
+ 200.000000510076
+ user> (sqrt (square 2))
+ 2.0000000929222947
+ user> (sqrt (square 0.1))
+ 0.10032578510960607
+ user> (sqrt (square 0.01))
+ 0.03230844833048122
+ user> (sqrt (square 10000))
+ 10000.0
+ user> (sqrt (square 20000))
+ 20000.0
+ user> (sqrt (square 200000))
+ 200000.0
+ user> (sqrt (square 20000000))
+ 2.0E7
+ user> (sqrt (square 20000000000))
+ 2.0E10
+ user> (sqrt (square 200000.012))
+ 200000.012
+ user> (sqrt (square 2000000.123))
+ 2000000.123
+ user> (sqrt (square 200000000.123))
+ 2.00000000123E8
+ user> (sqrt (square 2000000000.123))
+ 2.000000000123E9
+ user> (sqrt (square 20000000000.123))
+ 2.0000000000123E10
+ user> (sqrt (square 2000000000000.123))
+ 2.000000000000123E12
+ )
+;; An alternative strategy for implementing good-enough? is to watch how guess
+;; changes from one iteration to the next and to stop when the change is a very
+;; small fraction of the guess.
+(defn sqrt-iter [old-guess new-guess x]
+ (if (good-enough? old-guess new-guess x)
+ new-guess
+ (sqrt-iter new-guess (improve new-guess x)
+ x)))
+
+(defn improve [guess x]
+ (average guess (/ x guess)))
+
+(defn average [x y]
+ (/ (+ x y) 2))
+
+(defn good-enough? [old-guess new-guess x]
+ (< (/ (abs (- new-guess old-guess)) new-guess) 0.0001))
+
+(defn sqrt [x]
+ (sqrt-iter x 1.0 x))
+(comment
+user> (sqrt (square 0.01))
+0.010000000025490743
+user> (sqrt (square 0.001))
+0.0010000000000000117
+user> (sqrt (square 0.0001))
+1.0000000000082464E-4
+user> (sqrt (square 0.02))
+0.020000000050877154
+user> (sqrt (square 0.002))
+0.0020000000000000235
+user> (sqrt (square 4))
+4.000000000000051
+user> (sqrt (square 20))
+20.000000000298428
+user> (sqrt (square 25))
+25.000000063076968
+user> (sqrt 5)
+2.236067977499978
+user> (sqrt 25)
+5.000000000053722
+user> (sqrt 9)
+3.000000001396984
+user> (sqrt 81)
+9.000000000007091
+)
+;; exercise 1.8: cube root
+(defn cube [x]
+ (* x x x))
+
+(defn improve [guess x]
+ (/ (+ (/ x (square guess)) (* 2 guess)) 3))
+
+(defn cubert-iter [old-guess new-guess x]
+ (if (good-enough? old-guess new-guess x)
+ new-guess
+ (cubert-iter new-guess (improve new-guess x)
+ x)))
+
+(defn cuberoot [x]
+ (cubert-iter x 1.0 x))
+
+(comment
+user> (cuberoot (cube 2))
+2.000000000012062
+user> (cuberoot (cube 10))
+10.000000000000002
+user> (cuberoot (cube 9))
+9.000000000053902
+user> (cuberoot (cube 0.001))
+0.001000000000000962
+user> (cuberoot (cube 0.0001))
+1.000000000000001E-4
+user>
+)
+;; section 1.1.8
+;; hiding the non-public procedure definitions
+(defn- sqrt-iter [guess x]
+ (if (good-enough? guess x)
+ guess
+ (sqrt-iter (improve guess x)
+ x)))
+
+(defn- improve [guess x]
+ (average guess (/ x guess)))
+
+(defn- average [x y]
+ (/ (+ x y) 2))
+
+(defn- good-enough? [guess x]
+ (< (abs (- (square guess) x)) 0.001))
+
+(defn sqrt [x]
+ (sqrt-iter 1.0 x))
--- /dev/null
+(ns sicp.ch1-2
+ (:use (clojure.contrib trace math)))
+
+
+(defn factorial [n]
+ (if (= n 1)
+ 1
+ (* n (factorial (- n 1)))))
+
+;; stack friendly version
+(defn factorial2 [n]
+ (loop [x n acc 1]
+ (if (= x 1)
+ acc
+ (recur (- x 1) (* acc x)))))
+
+;; even better
+(defn factorial3 [n]
+ (reduce * (range 1 (inc n))))
+
+;; (comment
+;; user> (dotrace [factorial] (factorial 3))
+;; TRACE t2701: (factorial 3)
+;; TRACE t2702: | (factorial 2)
+;; TRACE t2703: | | (factorial 1)
+;; TRACE t2703: | | => 1
+;; TRACE t2702: | => 2
+;; TRACE t2701: => 6
+;; 6
+;; )
+
+;; (comment
+;; sicp.chapter1.ch1_2> (dotrace [factorial] (factorial 6))
+;; TRACE t2778: (factorial 6)
+;; TRACE t2779: | (factorial 5)
+;; TRACE t2780: | | (factorial 4)
+;; TRACE t2781: | | | (factorial 3)
+;; TRACE t2782: | | | | (factorial 2)
+;; TRACE t2783: | | | | | (factorial 1)
+;; TRACE t2783: | | | | | => 1
+;; TRACE t2782: | | | | => 2
+;; TRACE t2781: | | | => 6
+;; TRACE t2780: | | => 24
+;; TRACE t2779: | => 120
+;; TRACE t2778: => 720
+;; 720
+;; )
+
+;; (comment
+;; sicp.chapter1.ch1_2> (dotrace [factorial2] (factorial2 6))
+;; TRACE t2806: (factorial2 6)
+;; TRACE t2806: => 720
+;; 720
+;; )
+
+(defn fact-iter [product counter max-count]
+ (if (> counter max-count)
+ product
+ (fact-iter (* product counter)
+ (inc counter)
+ max-count)))
+
+(defn factorial [n]
+ (fact-iter 1 1 n))
+
+;; (comment
+;; user> (dotrace [factorial fact-iter] (factorial 6))
+;; TRACE t2378: (factorial 6)
+;; TRACE t2379: | (fact-iter 1 1 6)
+;; TRACE t2380: | | (fact-iter 1 2 6)
+;; TRACE t2381: | | | (fact-iter 2 3 6)
+;; TRACE t2382: | | | | (fact-iter 6 4 6)
+;; TRACE t2383: | | | | | (fact-iter 24 5 6)
+;; TRACE t2384: | | | | | | (fact-iter 120 6 6)
+;; TRACE t2385: | | | | | | | (fact-iter 720 7 6)
+;; TRACE t2385: | | | | | | | => 720
+;; TRACE t2384: | | | | | | => 720
+;; TRACE t2383: | | | | | => 720
+;; TRACE t2382: | | | | => 720
+;; TRACE t2381: | | | => 720
+;; TRACE t2380: | | => 720
+;; TRACE t2379: | => 720
+;; TRACE t2378: => 720
+;; 720
+;; )
+
+;; observation: clojure loop-recur construct is essentially the same as
+;; the iterative version.
+
+;; exercise 1.9
+(defn ++ [a b]
+ (if (= a 0)
+ b
+ (inc (++ (dec a) b))))
+
+;; (comment
+;; This version is a recursive process, where the previous call increments
+;; the sum by 1 and each call decrement the first operand by 1.
+
+;; user> (dotrace [++] (++ 4 5))
+;; TRACE t3745: (++ 4 5)
+;; TRACE t3746: | (++ 3 5)
+;; TRACE t3747: | | (++ 2 5)
+;; TRACE t3748: | | | (++ 1 5)
+;; TRACE t3749: | | | | (++ 0 5)
+;; TRACE t3749: | | | | => 5
+;; TRACE t3748: | | | => 6
+;; TRACE t3747: | | => 7
+;; TRACE t3746: | => 8
+;; TRACE t3745: => 9
+;; 9
+;; )
+
+(defn ++ [a b]
+ (if (= a 0)
+ b
+ (++ (dec a) (inc b))))
+
+;; (comment
+
+;; user> (dotrace [++] (++ 4 5))
+;; TRACE t3766: (++ 4 5)
+;; TRACE t3767: | (++ 3 6)
+;; TRACE t3768: | | (++ 2 7)
+;; TRACE t3769: | | | (++ 1 8)
+;; TRACE t3770: | | | | (++ 0 9)
+;; TRACE t3770: | | | | => 9
+;; TRACE t3769: | | | => 9
+;; TRACE t3768: | | => 9
+;; TRACE t3767: | => 9
+;; TRACE t3766: => 9
+;; 9
+;; )
+
+;; exercise 1.10
+;; ackerman functions
+(defn A [x y]
+ (cond (= y 0) 0
+ (= x 0) (* 2 y)
+ (= y 1) 2
+ :else (A (- x 1)
+ (A x (- y 1)))))
+
+(comment
+user> (A 1 10)
+1024
+user> (A 2 4)
+65536
+user> (A 3 3)
+65536
+)
+
+(defn f [n] (A 0 n)) ; f(n) = 2n
+(defn g [n] (A 1 n)) ; g(n) = 2^n
+;; (comment
+;; g (n) = A (1,n)
+;; = A (0, A (1, n-1)) = f (A(1,n-1))
+;; = f (f (1,n-2))
+;; .....
+;; = f (f (f ... f (1,(n- (n-1)))))
+;; = f (f (f ... 2))
+;; = 2 * (2^(n-1))
+;; = 2^n
+;; )
+
+(defn h [n] (A 2 n)) ; h(n) = 2^(n^2)
+
+;; section 1.2.2: Tree Recursion
+(defn fib [n]
+ (cond (= n 0) 0
+ (= n 1) 1
+ :else (+ (fib (- n 1))
+ (fib (- n 2)))))
+
+;; iterative version
+(defn fib-iter [a b count]
+ (if (= count 0)
+ b
+ (fib-iter (+ a b) a (dec count))))
+
+(defn fib [n]
+ (fib-iter 1 0 n))
+
+;; example: counting changes
+(defn first-denomination [kinds-of-coins]
+ (cond (= kinds-of-coins 1) 1
+ (= kinds-of-coins 2) 5
+ (= kinds-of-coins 3) 10
+ (= kinds-of-coins 4) 25
+ (= kinds-of-coins 5) 50))
+
+
+(defn first-denomination [kinds-of-coins]
+ ({1 1 2 5 3 10 4 25 5 50} kinds-of-coins))
+
+(defn cc [amount kinds-of-coins]
+ (cond (= amount 0) 1
+ (or (< amount 0) (= kinds-of-coins 0)) 0
+ :else (+ (cc amount (- kinds-of-coins 1))
+ (cc (- amount
+ (first-denomination kinds-of-coins))
+ kinds-of-coins))))
+
+(defn count-change [amount]
+ (cc amount 5))
+
+;; exercise 1.11: A function f is defined by the rule that f(n) = n if n < 3
+;; and f(n) = f(n - 1) + 2f(n - 2) + 3f(n - 3) if n> 3.
+;; Write a procedure that computes f by means of a recursive
+;; process. Write a procedure that computes f by means of an
+;; iterative process.
+(defn f [n]
+ (if (< n 3)
+ n
+ (+ (f (- n 1))
+ (* 2 (f (- n 2)))
+ (* 3 (f (- n 3))))))
+
+(comment
+user> (map f (range 10))
+(0 1 2 4 11 25 59 142 335 796)
+)
+
+;; ex 1.11: iterative version
+(defn f-iter [count prev0 prev1 prev2]
+ (if (= count 3)
+ (+ prev0
+ (* 2 prev1)
+ (* 3 prev2))
+ (f-iter (dec count)
+ (+ prev0
+ (* 2 prev1)
+ (* 3 prev2))
+ prev0
+ prev1)))
+
+(defn f [n]
+ (if (< n 3)
+ n
+ (f-iter n 2 1 0)))
+
+;; ex 1.11: iterative version with let
+(defn f-iter [count prev0 prev1 prev2]
+ (let [res (+ prev0 (* 2 prev1) (* 3 prev2))]
+ (if (= count 3)
+ res
+ (f-iter (dec count)
+ res
+ prev0
+ prev1))))
+
+;; exercise 1.12. The following pattern of numbers is called Pascal's triangle.
+;; 1
+;; 1 1
+;; 1 2 1
+;; 1 3 3 1
+;; 1 4 6 4 1
+;; ...................
+;;
+;; The numbers at the edge of the triangle are all 1, and each
+;; number inside the triangle is the sum of the two numbers above
+;; it. Write a procedure that computes elements of Pascal's triangle
+;; by means of a recursive process.
+(defn pascal [row col]
+ (when (<= col row)
+ (if (or (= col 0) (= row col))
+ 1
+ (+ (pascal (dec row) col)
+ (pascal (dec row) (dec col))))))
+
+;; exercise 1.13:
+(comment
+See the pdfs in the directory for the answers.
+)
+
+;; ex 1.13 (contd)
+(defn fib-approx [n]
+ (let [phi (/ (+ 1 (sqrt 5)) 2)]
+ (/ (expt phi n) (sqrt 5))))
+
+;; (comment
+;; user> (map fib-approx (range 10))
+;; (0.4472135954999579 0.7236067977499789 1.1708203932499368 1.8944271909999157 3.065247584249853 4.959674775249769 8.024922359499623 12.984597134749393 21.009519494249016 33.99411662899841)
+;; )
+
+;; exercise 1.14: tree of (count-changes 11)
+(comment
+ See the pdf for the tree representation.
+)
+
+
+;; order of size and computation
+;; see PDF, but below is the trace tree.
+;; (comment
+;; user> (use 'clojure.contrib.trace)
+;; nil
+;; user> (dotrace [count-change cc] (count-change 11))
+;; TRACE t2609: (count-change 11)
+;; TRACE t2610: | (cc 11 5)
+;; TRACE t2611: | | (cc 11 4)
+;; TRACE t2612: | | | (cc 11 3)
+;; TRACE t2613: | | | | (cc 11 2)
+;; TRACE t2614: | | | | | (cc 11 1)
+;; TRACE t2615: | | | | | | (cc 11 0)
+;; TRACE t2615: | | | | | | => 0
+;; TRACE t2616: | | | | | | (cc 10 1)
+;; TRACE t2617: | | | | | | | (cc 10 0)
+;; TRACE t2617: | | | | | | | => 0
+;; TRACE t2618: | | | | | | | (cc 9 1)
+;; TRACE t2619: | | | | | | | | (cc 9 0)
+;; TRACE t2619: | | | | | | | | => 0
+;; TRACE t2620: | | | | | | | | (cc 8 1)
+;; TRACE t2621: | | | | | | | | | (cc 8 0)
+;; TRACE t2621: | | | | | | | | | => 0
+;; TRACE t2622: | | | | | | | | | (cc 7 1)
+;; TRACE t2623: | | | | | | | | | | (cc 7 0)
+;; TRACE t2623: | | | | | | | | | | => 0
+;; TRACE t2624: | | | | | | | | | | (cc 6 1)
+;; TRACE t2625: | | | | | | | | | | | (cc 6 0)
+;; TRACE t2625: | | | | | | | | | | | => 0
+;; TRACE t2626: | | | | | | | | | | | (cc 5 1)
+;; TRACE t2627: | | | | | | | | | | | | (cc 5 0)
+;; TRACE t2627: | | | | | | | | | | | | => 0
+;; TRACE t2628: | | | | | | | | | | | | (cc 4 1)
+;; TRACE t2629: | | | | | | | | | | | | | (cc 4 0)
+;; TRACE t2629: | | | | | | | | | | | | | => 0
+;; TRACE t2630: | | | | | | | | | | | | | (cc 3 1)
+;; TRACE t2631: | | | | | | | | | | | | | | (cc 3 0)
+;; TRACE t2631: | | | | | | | | | | | | | | => 0
+;; TRACE t2632: | | | | | | | | | | | | | | (cc 2 1)
+;; TRACE t2633: | | | | | | | | | | | | | | | (cc 2 0)
+;; TRACE t2633: | | | | | | | | | | | | | | | => 0
+;; TRACE t2634: | | | | | | | | | | | | | | | (cc 1 1)
+;; TRACE t2635: | | | | | | | | | | | | | | | | (cc 1 0)
+;; TRACE t2635: | | | | | | | | | | | | | | | | => 0
+;; TRACE t2636: | | | | | | | | | | | | | | | | (cc 0 1)
+;; TRACE t2636: | | | | | | | | | | | | | | | | => 1
+;; TRACE t2634: | | | | | | | | | | | | | | | => 1
+;; TRACE t2632: | | | | | | | | | | | | | | => 1
+;; TRACE t2630: | | | | | | | | | | | | | => 1
+;; TRACE t2628: | | | | | | | | | | | | => 1
+;; TRACE t2626: | | | | | | | | | | | => 1
+;; TRACE t2624: | | | | | | | | | | => 1
+;; TRACE t2622: | | | | | | | | | => 1
+;; TRACE t2620: | | | | | | | | => 1
+;; TRACE t2618: | | | | | | | => 1
+;; TRACE t2616: | | | | | | => 1
+;; TRACE t2614: | | | | | => 1
+;; TRACE t2637: | | | | | (cc 6 2)
+;; TRACE t2638: | | | | | | (cc 6 1)
+;; TRACE t2639: | | | | | | | (cc 6 0)
+;; TRACE t2639: | | | | | | | => 0
+;; TRACE t2640: | | | | | | | (cc 5 1)
+;; TRACE t2641: | | | | | | | | (cc 5 0)
+;; TRACE t2641: | | | | | | | | => 0
+;; TRACE t2642: | | | | | | | | (cc 4 1)
+;; TRACE t2643: | | | | | | | | | (cc 4 0)
+;; TRACE t2643: | | | | | | | | | => 0
+;; TRACE t2644: | | | | | | | | | (cc 3 1)
+;; TRACE t2645: | | | | | | | | | | (cc 3 0)
+;; TRACE t2645: | | | | | | | | | | => 0
+;; TRACE t2646: | | | | | | | | | | (cc 2 1)
+;; TRACE t2647: | | | | | | | | | | | (cc 2 0)
+;; TRACE t2647: | | | | | | | | | | | => 0
+;; TRACE t2648: | | | | | | | | | | | (cc 1 1)
+;; TRACE t2649: | | | | | | | | | | | | (cc 1 0)
+;; TRACE t2649: | | | | | | | | | | | | => 0
+;; TRACE t2650: | | | | | | | | | | | | (cc 0 1)
+;; TRACE t2650: | | | | | | | | | | | | => 1
+;; TRACE t2648: | | | | | | | | | | | => 1
+;; TRACE t2646: | | | | | | | | | | => 1
+;; TRACE t2644: | | | | | | | | | => 1
+;; TRACE t2642: | | | | | | | | => 1
+;; TRACE t2640: | | | | | | | => 1
+;; TRACE t2638: | | | | | | => 1
+;; TRACE t2651: | | | | | | (cc 1 2)
+;; TRACE t2652: | | | | | | | (cc 1 1)
+;; TRACE t2653: | | | | | | | | (cc 1 0)
+;; TRACE t2653: | | | | | | | | => 0
+;; TRACE t2654: | | | | | | | | (cc 0 1)
+;; TRACE t2654: | | | | | | | | => 1
+;; TRACE t2652: | | | | | | | => 1
+;; TRACE t2655: | | | | | | | (cc -4 2)
+;; TRACE t2655: | | | | | | | => 0
+;; TRACE t2651: | | | | | | => 1
+;; TRACE t2637: | | | | | => 2
+;; TRACE t2613: | | | | => 3
+;; TRACE t2656: | | | | (cc 1 3)
+;; TRACE t2657: | | | | | (cc 1 2)
+;; TRACE t2658: | | | | | | (cc 1 1)
+;; TRACE t2659: | | | | | | | (cc 1 0)
+;; TRACE t2659: | | | | | | | => 0
+;; TRACE t2660: | | | | | | | (cc 0 1)
+;; TRACE t2660: | | | | | | | => 1
+;; TRACE t2658: | | | | | | => 1
+;; TRACE t2661: | | | | | | (cc -4 2)
+;; TRACE t2661: | | | | | | => 0
+;; TRACE t2657: | | | | | => 1
+;; TRACE t2662: | | | | | (cc -9 3)
+;; TRACE t2662: | | | | | => 0
+;; TRACE t2656: | | | | => 1
+;; TRACE t2612: | | | => 4
+;; TRACE t2663: | | | (cc -14 4)
+;; TRACE t2663: | | | => 0
+;; TRACE t2611: | | => 4
+;; TRACE t2664: | | (cc -39 5)
+;; TRACE t2664: | | => 0
+;; TRACE t2610: | => 4
+;; TRACE t2609: => 4
+;; 4
+;; )
+
+
+;; TODO: orders of growth in space and number of steps.
+
+;; exercise 1.15: sin (x) calculation
+;; a. How many times is the procedure p applied when (sine 12.15)
+;; is evaluated?
+;; b. What is the order of growth in space and number of steps (as
+;; a function of a) used by the process generated by the sine
+;; procedure when (sine a) is evaluated?
+(defn cube [x] (* x x x))
+
+(defn p [x] (- (* 3 x) (* 4 (cube x))))
+
+(defn sine [angle]
+ (if (not (> (abs angle) 0.1))
+ angle
+ (p (sine (/ angle 3.0)))))
+
+;; solution to (a) => 5
+;; (comment
+;; user> (dotrace [p] (sine 12.15))
+;; TRACE t2490: (p 0.049999999999999996)
+;; TRACE t2490: => 0.1495
+;; TRACE t2491: (p 0.1495)
+;; TRACE t2491: => 0.4351345505
+;; TRACE t2492: (p 0.4351345505)
+;; TRACE t2492: => 0.9758465331678772
+;; TRACE t2493: (p 0.9758465331678772)
+;; TRACE t2493: => -0.7895631144708228
+;; TRACE t2494: (p -0.7895631144708228)
+;; TRACE t2494: => -0.39980345741334
+;; -0.39980345741334
+;; )
+
+;; solution to b
+;; both space and number of steps grows as log3(a) -> log a to the base 3.
+;;
+;; proof:
+;; a * (1/3)^n <= 0.1
+;; => take log to the base 3 on both the sides.
+
+;; Note: Finding the order of space in a recursive process is sort of, equiv
+;; to finding the number of deferred operations. Which is in-turn the
+;; same as the depth of the evaluation tree.
+
+;; 1.2.4: exponentiation
+;; computing b^n
+
+;; iterative
+(defn expt-iter [base counter product]
+ (if (= counter 0)
+ product
+ (expt-iter base
+ (dec counter)
+ (* product base))))
+
+(defn myexpt [b n]
+ (expt-iter b n 1))
+
+(defn myexpt [b n]
+ (if (= n 0)
+ 1
+ (* b (myexpt b (dec n)))))
+
+;; fast version
+(defn square [x]
+ (* x x))
+
+(defn fast-expt [b n]
+ (cond (= n 0) 1
+ (even? n) (square (fast-expt b (/ n 2)))
+ :else (* b (fast-expt b (dec n)))))
+
+;; exercise 1.16:
+(defn myexpt [b n]
+ (expt-iter b n 1))
+
+(defn expt-iter [b n a]
+ (cond (= n 0) a
+ (even? n) (expt-iter (square b) (/ n 2) a)
+ :else (expt-iter b (- n 1) (* a b))))
+
+;; exercise 1.17:
+(defn mult [a b]
+ (if (= b 0)
+ 0
+ (+ a (mult a (- b 1)))))
+
+;; double
+(defn twice [x]
+ (* 2 x))
+
+(defn half [x]
+ (/ x 2))
+
+;; product = 2 * (a * (b/2)) for even b
+;; = a + (a * (b - 1)) for odd b
+(defn fast-mult [a b]
+ (cond (= b 0) 0
+ (= b 1) a
+ (even? b) (twice (fast-mult a (half b)))
+ :else (+ a (fast-mult a (- b 1)))))
+
+;; exercise 1.18: iterative multiply thru addition
+;; the idea is to keep a state variable.
+(defn fast-mult-iter [a b k]
+ (cond (= b 0) k
+ (even? b) (fast-mult-iter (twice a) (half b) k)
+ :else (fast-mult-iter a (- b 1) (+ k a))))
+
+(defn fast-mult-new [a b]
+ (fast-mult-iter a b 0))
+
+;; (comment
+;; user> (dotrace [fast-mult-new fast-mult-iter] (fast-mult-new 2 3))
+;; TRACE t2915: (fast-mult-new 2 3)
+;; TRACE t2916: | (fast-mult-iter 2 3 0)
+;; TRACE t2917: | | (fast-mult-iter 2 2 2)
+;; TRACE t2918: | | | (fast-mult-iter 4 1 2)
+;; TRACE t2919: | | | | (fast-mult-iter 4 0 6)
+;; TRACE t2919: | | | | => 6
+;; TRACE t2918: | | | => 6
+;; TRACE t2917: | | => 6
+;; TRACE t2916: | => 6
+;; TRACE t2915: => 6
+;; 6
+;; )
+
+;; exercise 1.19: fast fibonacci
+;; see the pdf of the notebook scan for the derivation of p' and q'
+(defn ffib-iter [a b p q count]
+ (cond (= count 0) b
+ (even? count)
+ (ffib-iter a
+ b
+ (+ (* p p) (* q q))
+ (+ (* 2 p q) (* q q))
+ (/ count 2))
+ :else (ffib-iter (+ (* b q) (* a q) (* a p))
+ (+ (* b p) (* a q))
+ p
+ q
+ (- count 1))))
+
+(defn ffib [n]
+ (ffib-iter 1 0 0 1 n))
+
+;;; Section 1.2.5: GCD
+(defn mygcd [a b]
+ (if (= b 0)
+ a
+ (mygcd b (rem a b))))
\ No newline at end of file
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