Learn:

1. Review 1.1 if necessary.
2. Watch lecture 2. Feel free to skip to 6:50 – the beginning is administrative.

Practice:

The class syllabus provides exercises for the first week. I copied them here for convenience. After this week, practice will more or less be in the form of "do as many of the exercises from the book as you can."

Do exercise 1.6, page 25. If you had trouble understanding the square root program in the book, explain instead what will happen if you use new-if instead of if in the pigl Pig Latin procedure.

Write a procedure squares that takes a sentence of numbers as its argument and returns a sentence of the squares of the numbers:

> (squares ’(2 3 4 5)) (4 9 16 25)

Write a procedure switch that takes a sentence as its argument and returns a sentence in which every instance of the words I or me is replaced by you, while every instance of you is replaced by me except at the beginning of the sentence, where it’s replaced by I. (Don’t worry about capitalization of letters.) Example:

> (switch ’(You told me that I should wake you up)) (i told you that you should wake me up)

Write a predicate ordered? that takes a sentence of numbers as its argument and returns a true value if the numbers are in ascending order, or a false value otherwise.

Write a procedure ends-e that takes a sentence as its argument and returns a sentence containing only those words of the argument whose last letter is E:
> (ends-e ’(please put the salami above the blue elephant)) (please the above the blue)

Most versions of Lisp provide and and or procedures like the ones on page 19. In principle there is no reason why these can’t be ordinary procedures, but some versions of Lisp make them special forms. Suppose, for example, we evaluate

(or (= x 0) (= y 0) (= z 0))

If or is an ordinary procedure, all three argument expressions will be evaluated before or is invoked. But if the variable x has the value 0, we know that the entire expression has to be true regardless of the values of y and z. A Lisp interpreter in which or is a special form can evaluate the arguments one by one until either a true one is found or it runs out of arguments.
Your mission is to devise a test that will tell you whether Scheme’s and and or are special forms or ordinary functions. This is a somewhat tricky problem, but it’ll get you thinking about the evaluation process more deeply than you otherwise might.

Why might it be advantageous for an interpreter to treat or as a special form and evaluate its arguments one at a time? Can you think of reasons why it might be advantageous to treat or as an ordinary function?