((* P (Puissance1 P (- N 1))))))
(println "Puissance1")
(println (Puissance1 5 5))
+(println (Puissance1 2 12))
;(trace true)
;O(log N)
(define (Puissance2 P N)
(cond
- ((= N 0) 1)
((= N 1) P)
- ((> N 1)
+ ((= N 2) (* P P))
+ ((> N 2)
(cond
((= (mod N 2) 0) (Puissance2 (Puissance2 P 2) (/ N 2)))
- ((* P (Puissance2 (Puissance2 P 2) (/ (- 1 N) 2))))))))
+ ((* P (Puissance2 (Puissance2 P 2) (/ (- N 1) 2))))))))
(println "Puissance2")
-;(println (Puissance2 5 5))
+(println (Puissance2 5 5))
+(println (Puissance2 2 12))
;(trace nil)
;(trace true)
; https://fr.wikipedia.org/wiki/Coefficient_binomial
-; relation de pascal
+; relation de pascal commenté
(define (comb N P)
(cond
((= P 0) 1)
((= N P) 1)
- ((+ (comb (- N 1) P) (comb (- N 1) (- P 1))))))
+ ;((+ (comb (- N 1) P) (comb (- N 1) (- P 1))))))
+ ((/ (* N (comb (- N 1) (- P 1))) P))))
(println "comb")
(println (comb 5 4))
(println (comb 60 4))
+(println "(comb 12 8) = "(comb 12 8))
;(trace nil)
+;(trace true)
(setq L '(3 7 + 4 2 + *))
-(setq P '())
+(setq M '(4 3 7 + * 2 -))
+(setq N '(10 10 5 / +))
(define (calculExp P L)
(cond
- ((null? L) 0)
- ((= (first L) '+) (+ (first P) (calculExp (rest P) (rest L))))
- ((= (first L) '-) (- (first P) (calculExp (rest P) (rest L))))
- ((= (first L) '*) (* (first P) (calculExp (rest P) (rest L))))
+ ((null? L) (first P))
+ ; all these conditions could probably be simplified
+ ((= (first L) '+) (calculExp (cons (+ (P 1) (first P)) (rest (rest P))) (rest L)))
+ ((= (first L) '-) (calculExp (cons (- (P 1) (first P)) (rest (rest P))) (rest L)))
+ ((= (first L) '*) (calculExp (cons (* (P 1) (first P)) (rest (rest P))) (rest L)))
;FIXME: test for divide by zero
- ((= (first L) '/) (/ (first P) (calculExp (rest P) (rest L))))
- ((cons (first L) (calculExp P (rest L))))))
-;(println (calculExp P L))
+ ((= (first L) '/) (calculExp (cons (/ (P 1) (first P)) (rest (rest P))) (rest L)))
+ ((number? (first L)) (calculExp (cons (first L) P) (rest L)))))
+(println "calculExp")
+(println (calculExp '() L))
+;(trace true)
+(println (calculExp '() M))
+(println (calculExp '() N))
+
+;(trace nil)
+
+(setq Q '(+ (* x 0) (* 10 (+ y 0))))
+(define (algsimplificator L)
+ (cond
+ ((null? L) '())
+ ; I'm having hard time to find a way of escaping the '(' and ')' characters
+ ((= (first L) ) (rest L))
+ ;here is the idea: detect the lower well formed expression: begin with (op and finish with ) where op = + - * / and have only two parameters that are atoms.
+ ;then if it match a know pattern, simplify it by following the matching rule.
+ ;do it again on the upper layer recursively until we only have (op A B) that just match no known simplication rules.
+
+ ))
+(println "algsimplificator")
+;(println algsimplificator(Q))
+
+(define (fibonacci N)
+ (cond
+ ((= N 0) 0)
+ ((= N 1) 1)
+ ((> N 1) (+ (fibonacci (- N 1)) (fibonacci (- N 2))))))
+(println "fibonacci")
+;(println (fibonacci 21))
+;(println (fibonacci 14))
+(println (fibonacci 20))
+(println (time (fibonacci 20)))
+
+;(trace true)
+
+(define (fibo:fibo n)
+ (if (not fibo:mem) (set 'fibo:mem '(0 1)))
+ (dotimes (i (- n 1))
+ ;this create a LIFO (or stack) of all previous fibonnaci serie result values
+ (push (+ (fibo:mem -1) (fibo:mem -2)) fibo:mem -1))
+ (last fibo:mem))
+(println "fibo")
+(println (fibo 20))
+(println (time (fibo 20)))
+
+;(trace nil)
+;(trace true)
+
+(setq S '())
+(define (Somme3ou5 N)
+ (dolist (i (sequence 0 N))
+ (cond
+ ((= (mod i 3) 0) (setq S (cons i S)))
+ ((= (mod i 5) 0) (setq S (cons i S)))))
+ (apply + S))
+(println "Somme3ou5")
+(println (Somme3ou5 100))
+
+;(trace nil)
(exit)