# Programming in JuliaIteration

We have already seen one way of doing something to each element in a collection: the *array comprehension*

smallest_factor = Dict(2 => 2, 3 => 3, 4 => 2, 5 => 5, 6 => 2, 7 => 7, 8 => 2, 9 => 3) [v for (k,v) in pairs(smallest_factor)]

In this array comprehension, we **iterate** over the pairs of the **for loop**.

*For* statements

The code above could also be rewritten as follows:

smallest_factor = Dict(2 => 2, 3 => 3, 4 => 2, 5 => 5, 6 => 2, 7 => 7, 8 => 2, 9 => 3) A = [] for (k,v) in pairs(smallest_factor) push!(A,v) end A

The statement `for item in collection:`

works as follows: the first element of `collection`

is assigned to `item`

, and the block indented below the `for`

statement is executed. Then, the second element of `collection`

is assigned to `item`

, the indented block is executed again, etc., until the end of the collection is reached.

We can nest `for`

statements. For example, suppose we have a matrix represented as an array of arrays, and we want to sum all of the matrix entries. We can do that by iterating over the rows and then iterating over each row:

""" Return the sum of the entries of M """ function sum_matrix_entries(M) s = 0 for row in M for entry in row s = s + entry end end s end using Test M = [[1,2,3],[4,5,6],[7,8,9]] @test sum_matrix_entries(M) == 45

**Exercise**

Suppose you have imported a function `file_bug_report`

with two parameters: `id`

and `description`

. Suppose also that you have a `Dict`

called `bugs`

whose keys are ids and whose values are strings representing descriptions. Write a loop which performs the action of filing each bug report in the dictionary.

"A dummy function which represents filing a bug report" function file_bug_report(id, description) println("bug $id ($description) successfully filed") end bugs = Dict( "07cc242a" => "`trackShipment` hangs if `trackingNumber` is missing", "100b359a" => "customers not receiving text alerts" )

*Solution.* We loop over the pairs of the dictionary:

for (id, desc) in pairs(bugs) file_bug_report(id, desc) end

**Exercise**

Write a `sumorial`

which takes a positive integer `n`

as an argument and sums of the integers 1 to `n`

using a loop.

"Return the sumorial of a positive integer n" function sumorial(n) # add code here end using Test @test sumorial(3) == 6 @test sumorial(8) == 36 @test sumorial(200) == 20100

*Solution.* We loop through `1:n`

and add as we go.

function sumorial(n) total = 0 for k in 1:n total = total + k end total end using Test @test sumorial(3) == 6 @test sumorial(8) == 36 @test sumorial(200) == 20100

*While* statements

The **Collatz conjecture** is one of the easiest-to-state unsolved problems in mathematics. Starting from any given positive integer, we halve it if it's even and triple it and add one if it's odd. The Collatz conjecture states that repeatedly applying this rule always gets us to the number 1 eventually. For example, the *Collatz sequence* starting from 17 is

17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1

If we want to write a Julia function which returns the Collatz sequence for any given starting number, we face a problem: we don't know from the start how many steps it will take to reach 1, so it isn't clear how we could use a *for loop*. What we want to do is execute a block of code until a given condition is met. Julia provides the `while`

loop for this purpose.

"Return the Collatz sequence starting from n" function collatz_sequence(n) sequence = [n] while n > 1 if n % 2 == 0 n = n ÷ 2 else n = 3n + 1 end push!(sequence,n) end sequence end using Test @test collatz_sequence(17) == [17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]

The expression which appears immediately following the `while`

keyword is called the **condition**, and the block indented below the `while`

statement is the **body** of the loop. The rules of the language stipulate the following execution sequence for a `while`

statement: the condition is evaluated, and if it's true, then the body is executed, then condition is evaluated again, and so on. When the condition returns `false`

, the loop is exited. An exit can also be forced from within the body of the while loop with the keyword `break`

.

**Exercise**

Newton's algorithm for finding the square root of a number `n`

starts from 1 and repeatedly applies the function . For example, applying this algorithm to approximate , we get

1, 3/2, 17/12, 577/408, ...

This algorithm converges very fast: 577/408 approximates with a relative error of about 0.00015%.

Write a function `newtonsqrt`

which takes as an argument the value `n`

to square root and applies Newton's algorithm until the relative difference between consecutive iterates drops below .

Note that can be represented in Julia using scientific notation `1e-8`

.

function newtonsqrt(n) """Use Newton's algorithm to approximate √n""" # add code here end using Test @test abs(newtonsqrt(2) - 1.4142135623730951) < 1e-6 @test abs(newtonsqrt(9) - 3) < 1e-6

*Solution.* We keep up with two separate variables, which we call `x`

and `old_x`

, to compare the most recent two iterates:

"""Use Newton's algorithm to approximate √n""" function newtonsqrt(n) x = 1 while true old_x = x x = 1/2 * (x + n/x) if abs(x - old_x)/old_x < 1e-8 return x end end end

## Exercises

**Exercise**

Write a function which prints an checkerboard pattern of `x`

's and `o`

's.

*Note*: `\n`

in a string literal represents the "newline" character. You'll need to print this character after each row you've printed.

""" Prints an n × n checkerboard, like: xoxo oxox xoxo oxox """ function checkerboard(n) # add code here end

*Solution.* We loop through the rows and use an `if`

statement to print a different output depending on whether the row is even-numbered or odd-numbered.

"Prints an n × n checkerboard" function checkerboard(n) for i in 1:n if iseven(i) print("xo" ^ (n÷2)) else print("ox" ^ (n÷2)) end print("\n") end end

**Exercise**

Write a function which prints Pascal's triangle up to the $n$th row, where the top row counts as row zero. You might want to use a helper function `print_row(n,row)`

to manage the responsibility of printing each row, as well as a helper function `next_row(row)`

to calculate each row from the previous one.

Example output, for `n = 4`

:

```
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
```

*Note*: there's no solution to this one, but you can do it on your own!

""" Prints the nth row (`row`) of Pascal's triangle with appropriate spacing. """ function print_row(n,row) # add code here end """ Returns the next row in Pascal's triangle. Example: next_row([1,3,3,1]) == [1,4,6,4,1] """ function next_row(row) # add code here end """ Print the first n rows of Pascal's triangle """ function pascals_triangle(n) # add code here end