# Laboratory: Recursion Basics

Summary: In this laboratory, you will explore some basic concepts in recursing over lists.

## Preparation

In this laboratory, we will not be working with images (just with colors and with lists), so you need not create an image.

a. Make a copy of `recursion-basics-lab.scm`, which contains most of the code from the reading.

b. Review the procedures in the file so that you understand their purpose (if not necessarilly the process by which they acheive their purpose).

c. Create a list of a dozen or so RGB colors (red, black, green, blue, yellow, orange, magenta, white, black, etc.). Name it `my-colors`. You may find it easiest to create a list of color names and convert it to RGB colors using `map`.

```(define my-colors
(map color->rgb
(list "red" "orange" "yellow" "green" "blue" "indigo" "violet"
"black" "white" "grey" "magenta")))
```

## Exercises

### Exercise 1: Testing Sum

a. Read through `sum` so that you have a sense of its approach to accomplishing its purpose

b. Verify that `sum` produces the same results as in the corresponding reading.

c. What value do you expect `sum` to produce for the empty list?

e. What value do you expect `sum` to produce for a singleton list? (A “singleton list” is a list with only one value.)

g. Try `sum` for a few other lists, too.

h. What do you expect the following to compute?

````>` `(sum 1 2 3)`
```

### Exercise 2: Removing Dark Colors

a. Read through the definition of `rgb-filter-out-dark` to try to understand what it does.

b. Determine which colors in `my-colors` are dark with `(map rgb-dark? my-colors)`.

c. Create a list of non-dark colors with ```(rgb-filter-out-dark my-colors)```.

d. Verify that all the resulting colors are not dark, using a technique similar to the one that you used in step b.

e. Find out the names of the non-dark colors with

````>` `(map rgb->color-name (rgb-filter-out-dark my-colors))`
```

### Exercise 3: Counting Values

Suppose the `length` procedure, which computes the length of a list, were not defined. We could define it by recursing through the list, counting 1 for each value in the list. In some sense, this is much like the definition of `sum`, except that we use the value 1 rather than the value of each element.

a. Using this idea, write a recursive procedure, `(list-length lst)` that finds the length of a list. You may not use `length` in defining `list-length`.

b. Check your answer on a few examples: the empty list, the list of colors you created, and a few more lists of your choice.

### Exercise 4: Product

Write a recursive procedure, ```(product nums)```, that computes the product of a list of numbers. You should feel free to use `sum` as a template for `product`. However, you should think carefully about the base case.

### Exercise 5: Counting Special Values

The `length` procedure counts the number of values in a list. What if we don't want to count every value in a list? For example, what if we only want to count the dark values in a list of colors? In this case, we still recurse through the list, but we sometimes count 1 (when the color is dark) and sometimes count 0 (when the color is not dark).

a. Using this idea, write a procedure, ```(rgb-tally-dark colors)```, that, given a list of colors, counts how many are dark.

### Exercise 6: Summing Components

In the past, we've found it useful to find the average of two colors. Let's consider how we might find the average of a list of colors. First, we would need to find the number of colors in the list. That's easy, we just use the `length` procedure. Next, we need to find the sum of each component. That's a bit harder, but let's suppose we can do it. We next divide each sum by the length, and get the average of that component. Finally, we put it all together with `rgb-new`.

That is, we might write

```(define rgb-list-average
(lambda (colors)
(let ((count (length colors)))
(rgb-new (/ (sum-red colors) count)
(/ (sum-green colors) count)
(/ (sum-blue colors) count)))))
```

Of course, for this to work, we need to write `sum-red`, `sum-blue`, and `sum-green`. For now, we'll write one of the three. (One we've written that one, the other two should be obvious.)

a. Write a procedure, ```(sum-red colors)```, that computes the sum of the red components in a list of colors. You should use direct recursion in your definition `sum-red`. (That is, you should use recursion, and not take advantage of the already-written `sum` procedure, other than as a template for your code.) You may want to base your definition on the definition of `sum`.

b. Test your procedure on a list of a single color.

c. Test your procedure on the `my-colors` list you wrote earlier.

d. It is possible to write `sum-red` without using direct recursion. How? An appropriate combination of `map` and `sum`. Try doing so.

If you can't find a solution, look at the notes on this problem.

### Exercise 7: Filtering Out Reds

Write a procedure, ```(filter-out-reds colors)```, that filters out all elements of `colors` with a red component of at least 128.

## For Those With Extra Time

### Extra 1: The Darkest Color

Write a procedure, `(rgb-darkest colors)`, that, given a nonempty list of colors, finds the darkest of those colors.

### Extra 2: Closest to Zero

Write a procedure, ```(closest-to-zero values)```, that, given a list of real numbers (including both positive and negative numbers), returns the value closest to zero in the list.

Hint: Think about how, given two numbers, you determine which is closer to zero.

Hint: Think about how this problem is similar to a problem or problems we've solved before.

### Extra 3: Averaging Colors

We've seen how to average two colors and a list of colors. But what if we want to do something different: Given a list of colors, we want averages, but only of neighboring elements in the list.

Write a procedure, ```(rgb-averages colors)```, that, given a list of colors, computes a new list of colors, by averaging subsequent pairs of colors. For example, if the input list is the standard seven rainbow colors (red, orange, yellow, green, blue, indigo, and violet), the output list will consist of a red-orange average, an orange-yellow average, a yellow-green average, a green-blue average, a blue-indigo average, and an indigo-violet average.

Once again, the length of the result list is one less than the length of the input list.

## Notes

### Notes on Problem 6: Summing Components

We can use `map` to extract the red component of each color.

```(map rgb-red colors)
```

That gives us a list of numbers, which we can sum with `sum`.

```(sum (map rgb-red colors))
```

Putting it all together in a procedure, we get

```(define sum-red
(lambda (colors)
(sum (map rgb-red colors))))
```

Samuel A. Rebelsky, rebelsky@grinnell.edu

Copyright (c) 2007-10 Janet Davis, Matthew Kluber, Samuel A. Rebelsky, and Jerod Weinman. (Selected materials copyright by John David Stone and Henry Walker and used by permission.)

This material is based upon work partially supported by the National Science Foundation under Grant No. CCLI-0633090. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License. To view a copy of this license, visit `http://creativecommons.org/licenses/by-nc/2.5/` or send a letter to Creative Commons, 543 Howard Street, 5th Floor, San Francisco, California, 94105, USA.