1 files changed, 86 insertions, 0 deletions
diff --git a/Computer_Science/SICP/ex2-4.scm b/Computer_Science/SICP/ex2-4.scm
new file mode 100644
index 0000000..3ad6d15
--- /dev/null
+++ b/Computer_Science/SICP/ex2-4.scm
@@ -0,0 +1,86 @@
+(define (cons x y)
+  (define (dispatch m)
+    (cond ((= m 0) x)
+          ((= m 1) y)
+          (else
+            (error "Argument not 0 or 1:
+                   CONS" m))))
+  dispatch)
+
+(define (car z) (z 0))
+(define (cdr z) (z 1))
+
+(define pair (cons 111 2))
+
+(car pair)
+
+;Exercise 2.4
+(define (cons x y)
+  (lambda (m) (m x y)))
+
+(define (car z)
+  (z (lambda (p q) p)))
+(define (cdr z)
+  (z (lambda (p q) q)))
+
+(cdr (cons 1 2))
+
+;Exercise 2.5
+(define (cons x y) (* (pow 2 x) (pow 3 y)))
+
+(define (pow x y)
+  (if (= y 0) 
+    1
+  (* x (pow x (- y 1)))))
+
+
+(define (car z)
+ (if (not (= (modulo z 2) 0))
+    0
+    (+ (car (/ z 2)) 1)))
+(define (cdr z)
+  (if (not (= (modulo z 3) 0))
+    0
+    (+ (cdr (/ z 3)) 1)))
+
+(car (cons 10 2))
+(cdr (cons 1 119))
+
+;Exercise 2.6
+
+(define zero (lambda (f) (lambda (x) x)))
+
+(define (add-1 n)
+  (lambda (f) (lambda (x) (f ((n f) x)))))
+;; (add-1 zero)
+;; (lambda (f) (lambda (x) (f ((zero f) x))))
+;; (lambda (f) (lambda (x) (f (f x))))
+
+(define (add a b)
+  (lambda (f)
+    (lambda (x)
+      ((a f) ((b f) x)))))
+(define one (lambda(f) (lambda (x) (f x))))
+(define two (lambda(f) (lambda (x) (f (f x)))))
+(define three (lambda(f) (lambda (x) (f (f (f x))))))
+
+(define (add a b)
+  (lambda (f)
+    (lambda (x)
+      ((a f) ((b f) x)))))
+
+;; Excellent answer for add a b, actually it's the
+;; same compared with above imp.
+(define (add a b)
+  ((a add-1) b))
+
+(zero 1)
+
+(define (square x) (* x x))
+
+((two square) 2)
+((three square) 2)
+
+(((add two one) square) 2)
+
+