In [1]:
function b(z)
prod = a(z, z)
println(z, " ", prod)
prod
end
function a(x, y)
x = x + 1
x * y
end
function c(x, y, z)
total = x + y + z
square = b(total)^2
square
end
x = 1
y = x + 1
println(c(x, y+3, x+y))
A palindrome is a word that is spelled the same backward and forward, like “noon” and “redivider”. Recursively, a word is a palindrome if the first and last letters are the same and the middle is a palindrome.
The following are functions that take a string argument and return the first, last, and middle letters:
In [1]:
function first(word)
word[1]
end
function last(word)
word[end]
end
function middle(word)
word[nextind(word, 1) : prevind(word, sizeof(word))]
end
Out[1]:
We’ll see how they work in lecture 10
Test theses functions out. What happens if you call middle with a string with two letters? One letter? What about the empty string, which is written "" and contains no letters?
Write a function called is_palindrome that takes a string argument and returns true if it is a palindrome and false otherwise. Remember that you can use the built-in function length to check the length of a string.
The Ackermann function, $A(m, n)$, is defined:
$$
A(m, n) =
\begin{cases}
n+1& \textrm{if } m = 0 \\
A(m−1, 1)& \textrm{if } m > 0 \textrm{ and } n = 0 \\
A(m−1, A(m, n−1))& \textrm{if } m > 0 \textrm{ and } n > 0.
\end{cases}
$$
See http://en.wikipedia.org/wiki/Ackermann_function. Write a function named ack that evaluates the Ackermann function. Use your function to evaluate ack(3, 4), which should be 125. What happens for larger values of m and n?
The greatest common divisor (GCD) of $a$ and $b$ is the largest number that divides both of them with no remainder.
One way to find the GCD of two numbers is based on the observation that if $r$ is the remainder when $a$ is divided by $b$, then gcd(a, b) = gcd(b, r). As a base case, we can use gcd(a, 0) = a.
Write a function called gcd that takes parameters a and b and returns their greatest common divisor.
Copy the loop from the square root calculation of the lecture and encapsulate it in a function called mysqrt that takes a as a parameter, chooses a reasonable value of x, and returns an estimate of the square root of a.
To test it, write a function named test_square_root that prints a table like this:
a mysqrt(a) sqrt(a) diff
- --------- ------- ----
1.0 1.0 1.0 0.0
2.0 1.41421356237 1.41421356237 2.22044604925e-16
3.0 1.73205080757 1.73205080757 0.0
4.0 2.0 2.0 0.0
5.0 2.2360679775 2.2360679775 0.0
6.0 2.44948974278 2.44948974278 0.0
7.0 2.64575131106 2.64575131106 0.0
8.0 2.82842712475 2.82842712475 4.4408920985e-16
9.0 3.0 3.0 0.0The first column is a number, a; the second column is the square root of a computed with mysqrt; the third column is the square root computed by sqrt; the fourth column is the absolute value of the difference between the two estimates.
In [3]:
expr = parse("1+2*3")
eval(expr)
Out[3]:
In [4]:
expr = parse("sqrt(π)")
eval(expr)
Out[4]:
Write a function called eval_loop that iteratively prompts the user, takes the resulting input and evaluates it using eval, and prints the result. It should continue until the user enters "done", and then return the value of the last expression it evaluated.
The mathematician Srinivasa Ramanujan found an infinite series that can be used to generate a numerical approximation of $\frac{1}{\pi}$:
$$
\frac{1}{\pi}=\frac{2\sqrt2}{9801}\sum_{k=0}^\infty\frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}
$$
Write a function called estimate_pi that uses this formula to compute and return an estimate of π. It should use a while loop to compute terms of the summation until the last term is smaller than 1e-15 (which is Julia notation for 10−15). You can check the result by comparing it to pi.