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# Programming in JuliaIteration

Час читання: ~25 min

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 dictionary to produce a new list. Although list comprehensions are very useful, they are not flexible enough to cover all our iteration needs. A much more flexible tool is 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" =>
)

Solution. We loop over the pairs of the dictionary:

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

Exercise
Write a function called 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)
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"""
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)
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)
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)
end  Bruno