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## cs150: Notes 7

#### Assignments Due

- (extended from 29 January) Before Monday, 5 February: Read GEB,
*Little Harmonic Labyrinth* and GEB, Chapter 5
- Friday, 2 February (beginning of class): Problem Set 2. (Note: Problem
Set 2 will be accepted without penalty or any permission required until
the beginning of class on Monday, 5 February, as long as you promise to
still finish reading GEB Chapter 5 before Monday's class.)

#### List Procedures Practice

- Be
*very* optimistic! Since lists themselves are recursive data
structures, most problems involving lists can be solved with recursive procedures.
- Think of the simplest version of the problem, something you can
already solve. This is the base case. For lists, this is usually when
the list is
`null`.
- Consider how you would solve a big version of the problem by using
the result for a slightly smaller version of the problem. This is the
recursive case. For lists, the smaller version of the problem is the rest (
`cdr`) of
the list.
- Combine the base case and the recursive case to solve the problem.

#### List?

Define a procedure `list?` that takes one operand and
evaluates to `#t` if the operand is a list, and `#f` otherwise.

- What should we do when the input is
`null`?

- When the input is a
`cons` pair (the built-in procedure
`pair?` evaluates to `#t` if and only if its input is a
cons pair), what should we do?

- When the input is not
`null` and not a `cons` what
should we do?

(define (list? p)
(if (null? p)
___________________________
(if (pair? p)
______________________________________
________
)))

#### Sumlist

Define a procedure `sumlist` that takes a list as its operand and
evaluates to the sum of elements in the list. For example, `(sumlist
(list 1 2 3))` should evaluate to 6, and `(sumlist null)`
should evaluate to 0.
(define (sumlist p)
)

#### Map

Define a procedure `map` that takes a procedure as its first
operand and a list as its second operand and evaluates to the list
containing the values resulting from applying the procedure to each
element of the input list.
Examples:

> (map car (list (cons 1 2) (cons 2 3)))

(1 2)

> (map + null)

()

> (map (lambda (x) (* x x)) (list 1 2 3))

(1 4 9)

> (map (lambda (el) (> el 3)) (list 2 4 6))

__________________

> (map (lambda (el) (+ 1 el)) (list 1 2 3))

__________________

(define (map f lst)
)

The `analyze-flop-situation` proceudre from ps2 uses `map`:
(map (lambda (turn-card)
(analyze-turn-situation hole1 hole2 (cons turn-card community)))
current-deck)))

What is this expression doing?

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