#lang scheme
;; 1.9
;; 1. primer: rekurziven ;; (+ 4 5) = inc((inc(inc(inc(5)))) = inc((inc(inc(6))) = inc((inc(7)) = inc(8) = 9 ;; 2. primer: iterativen ;; (+ 4 5): ;; (+ 3 6) ;; (+ 2 7) ;; (+ 1 8) ;; (+ 0 9) ;; 9
;; 1.10 ;; Ackermannova funkcija ;; https://en.wikipedia.org/wiki/Ackermann_function - tukaj malo drugacna definicija, mislim, da se indeksiranje premakne za 3
( define (A x y) ( cond ((= y 0 ) 0 ) (( = x 0 ) ( * 2 y)) ((= y 1 ) 2 ) ( else (A ( - x 1 ) ( A x ( - y 1 ) ) ) ) ) )
(A 1 10 ) (A 2 4) (A 3 3)
;; (f n) = 2*n ;; (g n) = 2n ;; (h n) =
;;;; iz knjige ;;;; (define (fib n) ( cond (( = n 0 ) 0) (( = n 1) 1 ) ( else ( + ( fib ( - n 1 ) ) ( fib ( - n 2 ) ) ) ) ) )
(fib 10) ;;;; ;;;;
;; 1.11
;; rekurzivno (define (f-rec n) ( cond (( < n 3 ) n) ( else ( + (f-rec ( - n 1 )) (* 2 (f-rec (- n 2))) (* 3 (f-rec (- n 3)))))))
(f-rec 6)
;;;; iz knjige ;;;; ( define (fib2 n) ( fib-iter 1 0 n) )
( define ( fib-iter a b count ) ( if (= count 0 ) b ( fib-iter (+ a b) a ( - count 1 ) ) ) )
(fib2 10) ;;;; ;;;;
;; iterativno ( define (f2 n) ( f-iter n (if (< n 2) n 2)) )
( define ( f-iter max-count sum) ( if (< max-count 3 ) max-count (+ (f-iter (- max-count 1) sum) (* 2 (f-iter (- max-count 2) sum )) (* 3 (f-iter (- max-count 3) sum )))))
(f2 6)
;; 1.12 (define (pascal r c) (if (or (= r 0) (= c 1) (= c r)) 1 (+ (pascal (- r 1) c) (pascal (- r 1) (- c 1)))))
(pascal 5 3)
;; 1.13 na papirju
;; 1.14 na papirju
;;;; iz knjige ;;;; ( define ( count-change amount ) ( cc amount 5 ) )
( define ( 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 ) ) ) ) ) (define ( 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) ) )
( count-change 11) ;;;; ;;;;
;;;; iz knjige ;;;; (define (square x) (* x x))
( define ( even? n) (= (remainder n 2 ) 0 ) )
( define ( fast-expt b n) ( cond ( (= n 0 ) 1 ) ( ( even? n ) ( square ( fast-expt b ( / n 2 ) ) ) ) ( else ( * b ( fast-expt b ( - n 1 ) ) ) ) ) )
(fast-expt 2 5) ;;;; ;;;;
;; 1.15 ;; a) 5 ;; b) O(log n) (?)
;; 1.16 ( define ( fast-exp-iter n b a ) ( if (= n 0 ) a (if (even? n) (fast-exp-iter ( / n 2 ) (square b) a ) ( * b ( fast-exp-iter ( - n 1 ) (* a b) a ) ) ) ) )
(fast-exp-iter 10 2 1 )
;; 1.17
;;;; iz knjige ;;;; (define (mult a b) (if (= b 0) 0 (+ a (mult a (- b 1 ) ) ) ) )
(mult 3 4) ;;;; ;;;;
;; naloga (define (double x) (+ x x)) (define (halve x) (/ x 2)) ;; a bi to moralo biti kako drugace; zakaj uporabljam deljenje pri implementaciji mnozenja s sestevanjem
( define ( fast-mult a b) ( cond ( (= b 0 ) 0 ) ( ( even? b ) ( double ( fast-mult a (halve b) ) ) ) ( else ( + a ( fast-mult a ( - b 1 ) ) ) ) ) )
(fast-mult 8 7)
;; 1.18
( define ( fast-mult-iter b a c) ( if (= b 0 ) 0 (if (even? b) (fast-mult-iter (halve b) (double a) c ) ( + a ( fast-mult-iter ( - b 1 ) a c ) ) ) ) )
(fast-mult-iter 18 10 0 )
;; 1.21
;;;; iz knjige ;;;; ( define ( smallest-divisor n) ( find-divisor n 2 ) ) ( define (find-divisor n test-divisor) ( cond (( > ( square test-divisor) n) n) (( divides? test-divisor n) test-divisor) (else ( find-divisor n (+ test-divisor 1 ) ) ) ) ) ( define (divides? a b ) (= (remainder b a) 0 ) )
(smallest-divisor 19999)
( define (prime? n) (= n ( smallest-divisor n) ) ) ;;;; ;;;;
;; 1.22
(define (runtime) (current-milliseconds)) ;; brez tega error: runtime: unbound identifier in: runtime
;;;; iz knjige ;;;; (define (timed-prime-test n) (newline) (display n) (start-prime-test n (runtime))) (define (start-prime-test n start-time) (if (prime? n) (report-prime (- (runtime) start-time)) -1)) (define (report-prime elapsed-time) (display "*" ) (display elapsed-time)) ;;;; ;;;;
(timed-prime-test 87178291199) ;; 35742549198872617291353508656626642567
;; https://en.wikipedia.org/wiki/List_of_prime_numbers
;; 1.26 ;; ker se s klicanjem (/ exp 2) v navadnem mnozenju parameter exp ne razpolovi v naslednjem koraku ?