Step by Step solution about How to resolve the algorithm Magic constant step by step in the Julia programming language

This code demonstrates a few concepts:

1. Using the Lazy package: This package provides lazy evaluation, which means that values are computed only when needed. This enables efficient handling of infinite sequences.

2. Defining a function magic: It calculates a sequence of numbers using a mathematical expression.

3. Creating a Lazy sequence magics: Using @>> to create a lazy sequence, which applies the magic function to a range of inputs. The resulting sequence is lazy, meaning it's not computed until it's actually used.

4. Filtering the sequence magics: Using the filter function to remove values less than 10 from the lazy sequence magics.

5. Printing the first 20 elements of the filtered sequence magics: Using take to obtain the first 20 values and converting them to integers (Int) for printing.

6. Getting the thousandth element of the filtered sequence magics: Using collect to materialize the thousandth element of the lazy sequence and then accessing the last element of the resulting array.

7. Finding the smallest magic square with a magic constant greater than a given value:

   • for expo in 1:20 iterates through a range of exponents (1 to 20).
   • goal = big"10"^expo sets the target magic constant as a big number using the big literal.
   • ordr = Int(floor((2 * goal)^(1/3))) + 1 calculates the order of the magic square required to achieve the desired magic constant. This formula is based on the mathematical property of magic squares.
   • println prints the exponent and the corresponding order of the magic square.

This code showcases the efficiency and power of lazy evaluation in handling infinite sequences, filtering out unwanted values, and performing mathematical calculations. It demonstrates how to use the Lazy package to create lazy sequences and manipulate them effectively in Julia.

using Lazy

magic(x) = (1 + x^2) * x ÷ 2
magics = @>> Lazy.range() map(magic) filter(x -> x > 10) # first 2 values are filtered out
println("First 20 magic constants: ", Int.(take(20, magics)))
println("Thousandth magic constant is: ", collect(take(1000, magics))[end])

println("Smallest magic square with constant greater than:")
for expo in 1:20
    goal = big"10"^expo
    ordr = Int(floor((2 * goal)^(1/3))) + 1
    println("10^", string(expo, pad=2), "    ", ordr)
end