CSci 1100 Lab 12 — Recursion


Category: You will Instantly receive a download link for .zip solution file upon Payment


5/5 - (2 votes)

Checkpoint 1
Attempts to define the formal foundations of mathematics at the start of the 20th century
depended heavily on the mathematical notion of recursion. While ultimately completing
this formalization was shown to be impossible, the theory that was developed contributed
heavily to the formal basis of computer science.
The idea is to start with primitive, axiomatic definitions and build from there. We will build
basic arithmetic starting with just one value — 0 — and three operations: (1) adding 1 to a
value (the successor operation), (2) subtracing 1 from a value (the predecessor operation),
and (3) testing a value to see if it is 0. In this case, the value 1 is the successor of 0, the
value 2 is the successor of the successfor of 0, etc.
Using just these operations, we can implement addition of positive integers using a Python
recursive function:
def add(m,n):
if n == 0:
return m
return add(m,n-1) + 1
To make sure you understand this, show the sequence of calls made by
print add(5,3)
You may add print statements to show the results.
Building on this idea, please write a recursive function to multiply two non-negative integers
using only the add function we’ve just defined, together with +1, -1 and the equality with 0
test. Call this function mult. Demonstrate the result by multiplying 8 and 3. As a recursive
function, mult call itself of course.
Now, define the integer power function, power(x, n) = x
, in terms of the mult function
you just wrote, together with +1, -1, and equality. Demonstrate the result by computing
To complete Checkpoint 1: Show:
1. the calls from add(5,3),
2. a working version of mult, and
3. a working version of power.
Checkpoint 2
In this section, you will write a recursive function to draw a self repeating plus sign. You
are given the code — — to draw the first plus sign in the middle
of the canvas. First, remember that (0,0) is the upper left corner, and (600,600) is the
lower right corner.
At each iteration, your code will draw the same pattern at the four end points of the current
plus sign and then reduce the length of the sign to half of its current value. The expected
figure at iterations 0, 1, 2 and a much higher level is shown below.
When you start, your origin is at location (300,300) and original length is given as 150 on
each side. You first draw a plus sign by drawing two lines, between:
(150,300)-(450,300) (horizontal line, total length 300)
(300,150)-(300,450) (vertical line, total length 300)
Now, you must start from the end points of this plus sign and draw four new plus signs
(recursively) of length 75 at on each side (the center of these four new plus signs will be:
(150,300), (450,300), (300,150), and (300,450).)
Each time you call the function recursively, the length will decrease by half. Think of a
good stopping condition for your recursion. May be a lower limit on the length would be a
good choice.
To solve this, you are going to follow the same pattern we have used for the Sierpinski
triangle in class.
To complete Checkpoint 2: Show your working program.
Congratulations! You have completed all the labs in the course!
Some extra things to try for fun!
You can change a few things in your program for Checkpoint 2 and get different designs.
First, try drawing the lines not at the end of the plus sign, but 1/4 way in from the end
points. For example, the second lines will have center locations at follows:
(300,375), (375,300), (225,300), (300,225)
(basically length/2 away from the center in all four dimensions.) If you repeat this process,
you will get a different pattern, as you can see below:
You can use the same code for the original checkpoint, but simply change the center locations. You may want to keep two versions of the code or include an if statement to toggle
between the two. We will call this new version the rectange version.
For both the rectangle and the plus version, the new lines don’t overlap with the old ones
because we keep dividing the length by half and moving the center by the same amount
at least. You can change this as well and change the length by a different fraction at each
step. Try setting the length at each iteration to:
length = 2*length/3
length = 3*length/5
instead of length/2. You will see that as patterns overlap, you start seeing some very
interesting Moir´e patterns:
Here are some example patterns you might get (for 2*length/3 for plus and rectangle versions):
Now, try changing the length to 3/2 of its current value every third iteration and halving it
at other iterations. Try shifting the starting values a little every time and see new patterns