SICP: 1.1 through 1.5

Author: shyba

sicp/chap1.md

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+1.1
+```scheme
+10
+> 10
+(+ 5 3 4)
+> 12
+(- 9 1)
+> 8
+(/ 6 2)
+> 3
+(+ (* 2 4) (- 4 6))
+> 6
+(define a 3)
+>
+(define b (+ a 1))
+>
+(+ a b (* a b))
+> 19
+(= a b)
+> #f
+(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
+```
+
+1.2
+```scheme
+(/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5)))))
+   (* 3 (- 6 2) (- 2 7)))
+```
+
+1.3
+```scheme
+(define (sum-squares x y z)
+    (define (squares p q) (+ (* p p) (* q q)))
+    (cond ((> x y z) (squares x y))
+          ((> x z y) (squares x z))
+          ((> y z x) (squares y z))
+          ((> y x z) (squares y x))
+          ((> z x y) (squares z x))
+          (else (squares z y))))
+```
+
+1.4
+```
+This function will change from a+b to a-b if b is less than 0, ensuring b as absolute.
+```
+
+1.5
+```
+In normal order the interpreter will "fully expand and then reduce", using values as needed. so:
+(test 0 (p)) fully expands to:
+(if (= 0 0) 0 (p))
+Since its true, 0 is returned.
+
+On applicative order, however, the interpreter will evaluate the arguments then apply. So it will evaluate (p) before applying test, which will make it loop forever.
+```