Exercises

Type the following code into the REPL

var1 = "a"

var2 = 'b'

What does the code do? Which of the following lines of code are valid and which will give you an error? First think what you expect to happen and then check.

var1+var2

var1*var2

var1+1

var2+1

var1*1

var2*1

var1^3

var2^3

var1 < "c"

var1 < 'c'

var2 > 'c'

var1*"1+$var2"

var1*"$(1+var2)"

"var1*$(var2)"

"$(var1)*$(var2)"

length(var1)+2

length(var2)+3

[i for i in 1:10]

string("0",join([i for i in 1:9]))

join("0",[i for i in 1:9])

string("0",[i for i in 1:9])

[var1+i for i in 1:5]

[string(var1,i) for i in 1:5]

join([var2+i for i in 1:5])

string(2,pad=4)

Using list comprehension create an array that contains all the lower case characters of the alphabet and bind this array to the variable name alphabet. (Hint: Recall that 'a'+0='a', 'a'+1='b' etc.)

Then create a single string that contains all the letters of the alphabet and bind it to the name alphabetstring. Also create another string variable alphabetcomma that contains all letters of the alphabet separated by comma.

Next create an array that contains the string "letter number 1 in the alphabet is a" as first element, the string "letter number 2 in the alphabet is b" as its second element and so on until "letter number 26 in the alphabet is z".

Out of how many code blocks does the following code block consist and what does the function move1 do?

var = "xyz"
function move1(input)
    output = string(input[2:end],input[1])
    return output
end
move1(var)

If we add the following to the code of the previous exercise

function move1(input::Number)
    output = input + 1
    return output
end

What output do you expect for the following function calls? Think first before trying!

move1(3)
move1(3.0)
move1("3")
move1("3.0")
move1('3')
move1([1, 2, 3])

Write a function evaluatef that takes two inputs: (i) another function (that I call f in the following), (ii) a range (e.g. 1:5). The function evaluatef should evaluate f at each point of the range and print the output. For example, if \(f(x)=x^2\), then \(evaluatef(f,1:3)\) should print:

f(1) equals 1

f(2) equals 4

f(3) equals 9

(Hint: to get the text printed line by line you have to recall what we did last week, namely list comprehension and escape sequences.)

In the following we write a function that solves the equation \(f(x)=0\) for some function \(f\). (As an example, suppose we want to know for which value of \(x\) the equation \(3^x=2\) holds, i.e. for which \(x\) we have \(3^x-2=0\). Then the function is \(f(x)=3^x-2\).)

We will use a method called "bisection". Our function bisect(f,a,b,eps) takes as inputs: (i) the function \(f\) (\(3^x-2\) in our example), (ii) values \(a\) and \(b\) for which we know that \(f(a)\) and \(f(b)\) have opposite signs (e.g. we could use 0 and 1 in our example as we know that \(3^0-2=-1<0\) and \(3^1-2=1>0\)), (iii) a small positive number \(eps\) which is the precision of our solution method, i.e. we consider \(f(x)\) as being zero whenever \(|f(x)|\leq eps\) (for example we could use \(0.0001\) and be satisfied with every \(x\) that yields \(|f(x)|\leq0.0001\) as a solution to our equation).

The method works as follows: We know that there is an \(x\) between \(a\) and \(b\) such that \(f(x)=0\). Next we check the function value at the midpoint, i.e. \(f((a+b)/2)\). If \(f((a+b)/2)\) has the same sign as \(f(a)\), then \((a+b)/2\) becomes our "new \(a\)" and we know that the \(x\) we are looking for is between \((a+b)/2\) and \(b\). If, however, \(f((a+b)/2)\) has the same sign as \(f(b)\), then \((a+b)/2\) is our "new \(b\)" and we know that the \(x\) we are looking for is between \(a\) and \((a+b)/2\). (In our example, \(3^{1/2}-2=-0.2679...<0\) and therefore we know that the \(x\) is between \(1/2\) and \(1\).) Then we iterate this procedure: In each step we compute \(f((a+b)/2)\) with the current \(a\) and \(b\) and get a new smaller interval in which our \(x\) lies. We stop iterating if \(|f((a+b)/2)|\leq eps\) and return \((a+b)/2\) as the result then.