* status Excercise 1.1 - 1.5 * notes ** 1 *** 1 **** 5 #+BEGIN_QUOTE In general, when modeling phenomena in science and engineering, we begin with simplified, incomplete models. As we examine things in greater detail, these simple models become inadequate and must be replaced by more refined models. #+END_QUOTE * exercises ** 1 *** 4 #+BEGIN_SRC scheme (define (a-plus-abs-b a b) ((if (> b 0) + -) a b)) #+END_SRC If ~b~ is greater than 0, do ~a + b~; otherwise do ~a - b~. *** 5 code at [[./one/five.scm]] If the interpreter uses *applicative order* to evaluate the expression: #+BEGIN_SRC scheme (test 0 (p)) #+END_SRC The parameters are evaluated before applying the compound procedure ~test~; 0 evaluates to 0, ~(p)~ never finishes evaluating as the compound procedure ~p~ recursively calls itself again and again infinitely. If same expression is evaluated by the interpreter using *normal order*, the expression will be expanded to #+BEGIN_SRC scheme (if (= 0 0) 0 (p))) #+END_SRC and will evaluate to ~0~. *** 6 If I've understood it correctly, scheme uses applicative-order evaluation, meaning, it evaluates the operands before appling the procedure. In the case when ~new-if~ used in the ~sqrt-iter~ procedure, the operands/arguments for the ~new-if~ -- ~(good-enough? guess x)~, ~guess~, ~(sqrt-iter (improve guess x) x)~ -- are evaluated. Due to the last operand, which is a call to the ~sqrt-iter~ procedure, we get into infinite loop of evaluating the ~sqrt-iter~ procedure again and again.