One thing that has made going through this book super fun so far is REPL-driven development. My current workflow uses VSCode’s Calva plugin and it is truly powerful. For instance, I can evaluate parts of an expression easily and also instrument code and do step-through debugging. I also took some time to learn about Paredit, which has made working with all those S-expressions much easier! Anyway, on to the meat of the article:
One key takeaway from §1.2 is that we must not confuse Recursive Procedures with Recursive Processes. The following exercises are an application of this concept:
;; 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 ex111-r [n] (cond (< n 3) n (>= n 3) (+ (ex111-r (- n 1)) (* 2 (ex111-r (- n 2))) (* 3 (ex111-r (- n 3)))) :else (throw (new AssertionError "invalid n")))) (ex111-r 10) (defn ex111-i [n] (let [iter (fn [a b c count] (if (zero? count) c (recur b c (+ (* 3 a) (* 2 b) c) (dec count))))] (iter 0 1 2 (- n 2)))) (ex111-i 10)
This one uses a trick I learnt from doing 4Clojure exercises -
partition 2 1 gives overlapping pairs in a list which work well for Pascal’s triangle.
;; Exercise 1.12. Write a procedure that computes elements of Pascal's triangle by means of a recursive process. (defn pascal [n] (let [pp (fn [i prev] (cond (< n 0) (throw (new AssertionError "invalid n")) (= n 1)  (= n 2) [1 1] (= i n) prev :else (let [pairs (partition 2 1 prev) new-inner (map #(apply + %) pairs)] (recur (inc i) (concat  new-inner )))))] (pp 1 ))) (assert (= (pascal 5) [1 4 6 4 1]))
Ah yes; invariants! Designing a loop with the right invariants can often assure you of the correctness of the loop. Exercises 1.16 and 1.18 are a great example of this idea.
;; Exercise 1.16 Design a procedure that evolves an iterative exponentiation process that uses successive squaring and uses a logarithmic number of steps, as does fast-expt. (defn fast-expt [b n] (cond (zero? n) 1 (even? n) (* (fast-expt b (/ n 2)) (fast-expt b (/ n 2))) :else (* b (fast-expt b (- n 1))))) (fast-expt 2 5) (defn fast-expt-iter [b n] ;; a﹒b^n is invariant (let [f (fn [b n a] (cond (= n 0) a (even? n) (recur (* b b) (/ n 2) a) :else (recur b (dec n) (* b a))))] (f b n 1))) (fast-expt-iter 2 5)
;; Exercise 1.18. Using the results of exercises 1.16 and 1.17, devise a procedure that generates an iterative process for multiplying two integers in terms of adding, doubling, and halving and uses a logarithmic number of steps.40 ;; After thinking of the invariant, it comes naturally. (defn fast-mult-iter [a b n] ;; a﹒b + n is invariant (let [double (fn [a] (* a 2)) halve (fn [a] (/ a 2))] (cond (zero? b) n (even? b) (recur (double a) (halve b) n) :else (recur a (dec b) (+ n a))))) (fast-mult-iter 11 3 0)
A logarithmic algorithm for computing Fibonacci numbers, super cool! My derivation of the transform is here: https://notability.com/n/~S4C78hC359sd2EDdHWkW.
;; Exercise 1.19 (defn fib-iter [a b p q count] (cond (= count 0) b (even? count) (recur a b (+ (* q q) (* p p)) (+ (* q q) (* 2 p q)) (/ count 2)) :else (recur (+ (* b q) (* a q) (* a p)) (+ (* b p) (* a q)) p q (dec count)))) (fib-iter 1 0 0 1 10)
For the following exercises, we use Clojure’s built-in
time function, which prints out the runtime of the expression as a side-effect.
;; Exercise 1.21. Use the smallest-divisor procedure to find the smallest divisor of each of the following numbers: 199, 1999, 19999. (defn smallest-divisor-slow [n] (first (filter #(zero? (mod n %)) (range 2 (inc n))))) (defn smallest-divisor [n] (first (filter #(zero? (mod n %)) (concat '(2) (range 3 (inc n) 2))))) (smallest-divisor 199) ;; 199 (smallest-divisor 1999) ;; 1999 (smallest-divisor 19999) ;; 7 ;; Ex 1.23 (defn prime-slow? [n] (= n (smallest-divisor-slow n))) (defn prime? [n] (= n (smallest-divisor n))) (time (dotimes [n 10] (prime? 1000037))) ;; 555 ms (time (dotimes [n 10] (prime-slow? 1000037))) ;; 1073 ms ;; 2x speedup, as expected!
Here we apply another trick to measure the runtime of the code - we use
doall since Clojure’s sequences are lazy and we want to force evaluation.
;; Ex 1.21 (defn fermat-test [n] (let [expmod (fn expmod [base exp m] (cond (zero? exp) 1 (even? exp) (let [x (expmod base (/ exp 2) m)] (mod (* x x) m)) :else (mod (* base (expmod base (dec exp) m)) m))) try-it (fn [a] (= (expmod a n n) a))] (try-it (+ 1 (rand-int (dec n)))))) ;; Uses the mod property (fermat-test 19999) (defn fast-prime? [n times] (cond (zero? times) true (fermat-test n) (fast-prime? n (- times 1)) :else false)) ;; Ex 1.22 (defn search-for-primes [lower-bound num-primes-wanted] (take num-primes-wanted (filter #(fast-prime? % 8) (filter odd? (range lower-bound (java.lang.Integer/MAX_VALUE)))))) (defn search-for-primes-slow [lower-bound num-primes-wanted] (take num-primes-wanted (filter prime? (filter odd? (range lower-bound (java.lang.Integer/MAX_VALUE)))))) ;; This is approximately O(sqrt(n)) growth. (time (doall (search-for-primes-slow 1000 3))) ;;/ (1009 1013 1019) 0.6 msecs" (time (doall (search-for-primes-slow 10000 3))) ;; (10007 10009 10037) 4 msecs (time (doall (search-for-primes-slow 100000 3))) ;; (100003 100019 100043) 30 msecs" (time (doall (search-for-primes-slow 1000000 3))) ;; (1000003 1000033 1000037) 224 ms ;; Ex 1.24 ;; This is approximately O(log(n)) growth and demonstrates the power of logarithmic runtimes - a multiplication of the input's cardinality results in an additive runtime increase. (trace/trace (time (doall (search-for-primes 1000 3)))) ;; (1009 1013 1019) 0.24 msecs" (time (doall (search-for-primes 10000 3))) ;; (10007 10009 10037) 0.50 msecs (time (doall (search-for-primes 100000 3))) ;; (100003 100019 100043) 0.57 msecs" (time (doall (search-for-primes 1000000 3))) ;; (1000003 1000033 1000037) 0.6 msecs"
;; Exercise 1.25. Alyssa P. Hacker complains that we went to a lot of extra work in writing expmod. After all, she says, since we already know how to compute exponentials, we could have simply written: ;; (define (expmod base exp m) ;; (remainder (fast-expt base exp) m)) ;; Theoretically (in terms of math), it would work. Computationally it fails since the numbers (without the intermediate mods) are too large and either overflow or take too long.
;; Exercise 1.26 ;; Previously, the process was linear recursive and you halved the search space on each level (the /2). Now, every time you halve the seach space, you make 2 recursive calls at the same level. So the runtime is now O(N) instead of O(log N).s
;; Exercise 1.28 ;; Proof that if there is nontrivial square root of n it is not prime ;; https://crypto.stanford.edu/pbc/notes/numbertheory/poly.html (defn miller-rabin [n] (let [expmod (fn expmod [base exp m] (cond (zero? exp) 1 (even? exp) (let [x (expmod base (/ exp 2) m) sq (mod (* x x) m)] (if (and (not= x 1) (not= x (dec m)) (= sq 1)) 0 sq)) :else (mod (* base (expmod base (dec exp) m)) m))) try-it (fn [a] (= (expmod a (dec n) n) 1))] (try-it 3) ;; (every? true? (map try-it (range 2 n))) ))