From 8db66297da582005ceae5e333a12aa1f4e53df28 Mon Sep 17 00:00:00 2001
From: Correl Roush <correl@gmail.com>
Date: Mon, 19 May 2014 00:46:41 -0400
Subject: [PATCH] 1.1 in org format

---
 1-1.org | 279 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1-1.scm | 243 ------------------------------------------------
 2 files changed, 279 insertions(+), 243 deletions(-)
 create mode 100644 1-1.org
 delete mode 100644 1-1.scm

diff --git a/1-1.org b/1-1.org
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+++ b/1-1.org
@@ -0,0 +1,279 @@
+#+BEGIN_HTML
+---
+title: 1.1 - The Elements of Programming
+layout: page
+---
+#+END_HTML
+
+* Code
+  #+BEGIN_SRC scheme :tangle yes
+    (define (square x)
+      (* x x))
+
+
+    (define (sum-of-squares a b)
+      (+ (square a) (square b)))
+
+    ;; (define (abs x)
+    ;;   (cond ((< x 0) (- x))
+    ;;         ((= x 0) 0)
+    ;;         ((> x 0) x)))
+
+    (define (abs x)
+      (if (< x 0) (- x)
+          x))
+
+    (define (average x y)
+      (/ (+ x y) 2))
+
+    (define (sqrt x)
+      (define (sqrt-iter guess x)
+        (if (good-enough? guess x)
+            guess
+            (sqrt-iter (improve guess x)
+                       x)))
+
+      (define (improve guess x)
+        (average guess (/ x guess)))
+
+      (define (good-enough? guess x)
+        (< (abs (- (square guess) x)) 0.001))
+
+      (sqrt-iter 1.0 x))
+
+  #+END_SRC
+* Exercises
+** Exercise 1.1
+   Below is a sequence of expressions.  What is the result printed by
+   the interpreter in response to each expression?  Assume that the
+   sequence is to be evaluated in the order in whichit is presented.
+
+   #+BEGIN_SRC scheme
+     ;; -------------------------------------------------------------------
+     ;; Exercise 1.1
+     ;; -------------------------------------------------------------------
+
+     10
+     ;; 10
+
+     (+ 5 3 4)
+     ;; 12
+
+     (- 9 1)
+     ;; 8
+
+     (/ 6 2)
+     ;; 3
+
+     (+ (* 2 4) (- 4 6))
+     ;; 10
+
+     (define a 3)
+     ;; 3
+
+     (define b (+ a 1))
+     ;; 4
+
+     (+ 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
+   #+END_SRC
+** Exercise 1.2
+   Translate the following expression into prefix form.
+
+   #+BEGIN_EXAMPLE
+     5 + 4 + (2 - (3 - (6 + 4/5)))
+     -----------------------------
+            3(6 - 2)(2 - 7)
+   #+END_EXAMPLE
+
+   #+BEGIN_SRC scheme :tangle yes
+     ;; -------------------------------------------------------------------
+     ;; Exercise 1.2
+     ;; -------------------------------------------------------------------
+
+     (define ex1.2
+       (/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5)))))
+          (* 3 (- 6 2) (- 2 7))))
+   #+END_SRC
+** Exercise 1.3
+   Define a procedure that takes three numbers as arguments and
+   returns the sum of the squares of the two larger numbers.
+
+   #+BEGIN_SRC scheme :tangle yes
+     ;; -------------------------------------------------------------------
+     ;; Exercise 1.3
+     ;; -------------------------------------------------------------------
+
+     (define (ex1.3 a b c)
+       (cond ((and (> b a) (> c a)) (sum-of-squares b c))
+             ((and (> a b) (> c b)) (sum-of-squares a c))
+             (#t (sum-of-squares a b))))
+
+     (define (ex1.3-fancy a b c)
+       (let* ((sorted (sort (list a b c) >))
+              (x (car sorted))
+              (y (cadr sorted)))
+         (+ (square x) (square y))))
+   #+END_SRC
+
+** Exercise 1.4
+   Observe that our model of evaluation allows for
+   combinations whose operators are compound expressions.  Use this
+   observation to describe the behavior of the following procedure:
+
+   #+BEGIN_SRC scheme
+     (define (a-plus-abs-b a b)
+     ((if (> b 0) + -) a b))
+   #+END_SRC
+
+   -------------------------------------------------------------------
+   
+   When the function is called, the following will happen:
+
+   * The first expression in the list, (if (> b 0) + -) will be
+     evaluated. Within it, (> b 0) will be evaluated first, and based
+     on the value of b, the result of the evaluation will be + or -.
+   * The remaining expressions (a and b) will be evaluated to their
+     passed-in values.
+   * The resulting expression will be evaluated, e.g. (+ 3 2)
+   * The final result will be the result of applying the + or -
+     operator to the operands a and b
+
+** Exercise 1.5
+   Ben Bitdiddle has invented a test to determine
+   whether the interpreter he is faced with is using
+   applicative-order evaluation or normal-order evaluation.  He
+   defines the following two procedures:
+
+   #+BEGIN_SRC scheme
+     (define (p) (p))
+
+     (define (test x y)
+       (if (= x 0)
+           0
+           y))
+   #+END_SRC
+
+   Then he evaluates the expression
+
+   #+BEGIN_SRC scheme
+     (test 0 (p))
+   #+END_SRC
+
+   What behavior will Ben observe with an interpreter that uses
+   applicative-order evaluation?  What behavior will he observe with
+   an interpreter that uses normal-order evaluation?  Explain your
+   answer.  (Assume that the evaluation rule for the special form `if'
+   is the same whether the interpreter is using normal or applicative
+   order: The predicate expression is evaluated first, and the result
+   determines whether to evaluate the consequent or the alternative
+   expression.)
+   
+   -------------------------------------------------------------------
+
+   Applicative order evaluation will evaluate test, 0 and (p), then
+   evaluate the application of the operator test on its
+   operands. However, attempting to evaluate (p) will hang, as it is a
+   recursive function that never exits.
+   
+   Normal order evaluation will first apply the operator test on its
+   operands, which will then evaluate 0 in the if statment. The
+   conditional expression will succeed, and so the function will
+   return 0, never evaluating (p).
+
+** Exercise 1.6
+   Alyssa P. Hacker doesn't see why `if' needs to be
+   provided as a special form.  "Why can't I just define it as an
+   ordinary procedure in terms of `cond'?" she asks.  Alyssa's friend
+   Eva Lu Ator claims this can indeed be done, and she defines a new
+   version of `if':
+
+   #+BEGIN_SRC scheme
+     (define (new-if predicate then-clause else-clause)
+       (cond (predicate then-clause)
+             (else else-clause)))
+   #+END_SRC
+   
+   Eva demonstrates the program for Alyssa:
+
+   #+BEGIN_SRC scheme
+     (new-if (= 2 3) 0 5)
+     5
+       
+     (new-if (= 1 1) 0 5)
+     0
+   #+END_SRC
+   
+   Delighted, Alyssa uses `new-if' to rewrite the square-root program:
+
+   #+BEGIN_SRC scheme
+     (define (sqrt-iter guess x)
+       (new-if (good-enough? guess x)
+               guess
+               (sqrt-iter (improve guess x)
+                          x)))
+   #+END_SRC
+   
+   What happens when Alyssa attempts to use this to compute square
+   roots?  Explain.
+  
+   -------------------------------------------------------------------
+ 
+   Calls to sqrt-iter will recurse indefinitely. This is because both
+   the then-clause and the else-clause passed to new-if will be
+   evaluated before the function new-if is applied.
+ 
+** Exercise 1.7
+   The `good-enough?' test used in computing square
+   roots will not be very effective for finding the square roots of
+   very small numbers.  Also, in real computers, arithmetic operations
+   are almost always performed with limited precision.  This makes
+   our test inadequate for very large numbers.  Explain these
+   statements, with examples showing how the test fails for small and
+   large numbers.  An alternative strategy for implementing
+   `good-enough?' is to watch how `guess' changes from one iteration
+   to the next and to stop when the change is a very small fraction
+   of the guess.  Design a square-root procedure that uses this kind
+   of end test.  Does this work better for small and large numbers?
+
+   -------------------------------------------------------------------
+ 
+** Exercise 1.8
+   Newton's method for cube roots is based on the
+   fact that if y is an approximation to the cube root of x, then a
+   better approximation is given by the value
+
+   #+BEGIN_EXAMPLE
+       x/y^2 + 2y
+       ----------
+           3
+   #+END_EXAMPLE
+   
+   Use this formula to implement a cube-root procedure analogous to
+   the square-root procedure.  (In section *Note 1-3-4:: we will see
+   how to implement Newton's method in general as an abstraction of
+   these square-root and cube-root procedures.)
+
+   -------------------------------------------------------------------
diff --git a/1-1.scm b/1-1.scm
deleted file mode 100644
index 9b6df6e..0000000
--- a/1-1.scm
+++ /dev/null
@@ -1,243 +0,0 @@
-;; ====================================================================
-;; SICP - 1.1: The Elements of Programming
-;; ====================================================================
-
-(define (square x)
-  (* x x))
-
-
-(define (sum-of-squares a b)
-  (+ (square a) (square b)))
-
-;; (define (abs x)
-;;   (cond ((< x 0) (- x))
-;;         ((= x 0) 0)
-;;         ((> x 0) x)))
-
-(define (abs x)
-  (if (< x 0) (- x)
-      x))
-
-(define (average x y)
-  (/ (+ x y) 2))
-
-(define (sqrt x)
-  (define (sqrt-iter guess x)
-    (if (good-enough? guess x)
-        guess
-        (sqrt-iter (improve guess x)
-                   x)))
-
-  (define (improve guess x)
-    (average guess (/ x guess)))
-
-  (define (good-enough? guess x)
-    (< (abs (- (square guess) x)) 0.001))
-
-  (sqrt-iter 1.0 x))
-
-
-;; ====================================================================
-;; *Exercise 1.1:* Below is a sequence of expressions.  What is the
-;; result printed by the interpreter in response to each expression?
-;; Assume that the sequence is to be evaluated in the order in which
-;; it is presented.
-;; -------------------------------------------------------------------
-
-10
-;; 10
-
-(+ 5 3 4)
-;; 12
-
-(- 9 1)
-;; 8
-
-(/ 6 2)
-;; 3
-
-(+ (* 2 4) (- 4 6))
-;; 10
-
-(define a 3)
-;; 3
-
-(define b (+ a 1))
-;; 4
-
-(+ 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
-
-;; ===================================================================
-;; *Exercise 1.2:* Translate the following expression into prefix
-;; form.
-;;
-;;      5 + 4 + (2 - (3 - (6 + 4/5)))
-;;      -----------------------------
-;;             3(6 - 2)(2 - 7)
-;; -------------------------------------------------------------------
-
-(define ex1.2
-  (/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5)))))
-     (* 3 (- 6 2) (- 2 7))))
-
-;; ===================================================================
-;; *Exercise 1.3:* Define a procedure that takes three numbers as
-;; arguments and returns the sum of the squares of the two larger
-;; numbers.
-;; -------------------------------------------------------------------
-
-(define (ex1.3 a b c)
-  (cond ((and (> b a) (> c a)) (sum-of-squares b c))
-        ((and (> a b) (> c b)) (sum-of-squares a c))
-        (#t (sum-of-squares a b))))
-
-(define (ex1.3-fancy a b c)
-  (let* ((sorted (sort (list a b c) >))
-         (x (car sorted))
-         (y (cadr sorted)))
-    (+ (square x) (square y))))
-
-;; ===================================================================
-;; *Exercise 1.4:* Observe that our model of evaluation allows for
-;; combinations whose operators are compound expressions.  Use this
-;; observation to describe the behavior of the following procedure:
-;;
-     (define (a-plus-abs-b a b)
-       ((if (> b 0) + -) a b))
-;; -------------------------------------------------------------------
-
-;; When the function is called, the following will happen:
-;;
-;;   * The first expression in the list, (if (> b 0) + -) will be
-;;     evaluated. Within it, (> b 0) will be evaluated first, and
-;;     based on the value of b, the result of the evaluation will be +
-;;     or -.
-;;   * The remaining expressions (a and b) will be evaluated to their
-;;     passed-in values.
-;;   * The resulting expression will be evaluated, e.g. (+ 3 2)
-;;   * The final result will be the result of applying the + or -
-;;     operator to the operands a and b
-
-;; ===================================================================
-;; *Exercise 1.5:* Ben Bitdiddle has invented a test to determine
-;; whether the interpreter he is faced with is using
-;; applicative-order evaluation or normal-order evaluation.  He
-;; defines the following two procedures:
-;;
-     (define (p) (p))
-
-     (define (test x y)
-       (if (= x 0)
-           0
-           y))
-;;
-;; Then he evaluates the expression
-;;
-;;      (test 0 (p))
-;;
-;; What behavior will Ben observe with an interpreter that uses
-;; applicative-order evaluation?  What behavior will he observe with
-;; an interpreter that uses normal-order evaluation?  Explain your
-;; answer.  (Assume that the evaluation rule for the special form
-;; `if' is the same whether the interpreter is using normal or
-;; applicative order: The predicate expression is evaluated first,
-;; and the result determines whether to evaluate the consequent or
-;; the alternative expression.)
-;; -------------------------------------------------------------------
-
-;; Applicative order evaluation will evaluate test, 0 and (p), then
-;; evaluate the application of the operator test on its
-;; operands. However, attempting to evaluate (p) will hang, as it is a
-;; recursive function that never exits.
-;;
-;; Normal order evaluation will first apply the operator test on its
-;; operands, which will then evaluate 0 in the if statment. The
-;; conditional expression will succeed, and so the function will
-;; return 0, never evaluating (p).
-
-;; ===================================================================
-;; *Exercise 1.6:* Alyssa P. Hacker doesn't see why `if' needs to be
-;; provided as a special form.  "Why can't I just define it as an
-;; ordinary procedure in terms of `cond'?" she asks.  Alyssa's friend
-;; Eva Lu Ator claims this can indeed be done, and she defines a new
-;; version of `if':
-;; 
-;;      (define (new-if predicate then-clause else-clause)
-;;        (cond (predicate then-clause)
-;;              (else else-clause)))
-;; 
-;; Eva demonstrates the program for Alyssa:
-;; 
-;;      (new-if (= 2 3) 0 5)
-;;      5
-;; 
-;;      (new-if (= 1 1) 0 5)
-;;      0
-;; 
-;; Delighted, Alyssa uses `new-if' to rewrite the square-root program:
-;; 
-;;      (define (sqrt-iter guess x)
-;;        (new-if (good-enough? guess x)
-;;                guess
-;;                (sqrt-iter (improve guess x)
-;;                           x)))
-;; 
-;; What happens when Alyssa attempts to use this to compute square
-;; roots?  Explain.
-;; -------------------------------------------------------------------
-
-;; Calls to sqrt-iter will recurse indefinitely. This is because both
-;; the then-clause and the else-clause passed to new-if will be
-;; evaluated before the function new-if is applied.
-
-;; ===================================================================
-;; *Exercise 1.7:* The `good-enough?' test used in computing square
-;; roots will not be very effective for finding the square roots of
-;; very small numbers.  Also, in real computers, arithmetic operations
-;; are almost always performed with limited precision.  This makes
-;; our test inadequate for very large numbers.  Explain these
-;; statements, with examples showing how the test fails for small and
-;; large numbers.  An alternative strategy for implementing
-;; `good-enough?' is to watch how `guess' changes from one iteration
-;; to the next and to stop when the change is a very small fraction
-;; of the guess.  Design a square-root procedure that uses this kind
-;; of end test.  Does this work better for small and large numbers?
-;; -------------------------------------------------------------------
-
-;; ===================================================================
-;; *Exercise 1.8:* Newton's method for cube roots is based on the
-;; fact that if y is an approximation to the cube root of x, then a
-;; better approximation is given by the value
-;; 
-;;      x/y^2 + 2y
-;;      ----------
-;;          3
-;; 
-;; Use this formula to implement a cube-root procedure analogous to
-;; the square-root procedure.  (In section *Note 1-3-4:: we will see
-;; how to implement Newton's method in general as an abstraction of
-;; these square-root and cube-root procedures.)
-;; -------------------------------------------------------------------