6 - Sum Square Difference

Problem

The sum of the squares of the first ten natural numbers is, $$1^2 + 2^2 + ... + 10^2 = 385.$$ The square of the sum of the first ten natural numbers is, $$(1 + 2 + ... + 10)^2 = 55^2 = 3025.$$ Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is $3025 - 385 = 2640$. Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Other programming languages

I don't really know what to say about this problem. I think this is simple for computers but very tedious for humans.

Certainly it is a trivial problem for Uiua and easy for many other programming languages too.

I'll just demonstrate quickly with a few programs before we solve it in Uiua.

Ruby

Ruby has a way of instantiating ranges with (from..to). With the sum and map methods on ranges/arrays, and raising to the second power with num ** 2, we can solve this problem. A program might look like this.

sum_of_squares = (1..10).map{|v|v**2}.sum
square_of_sums = (1..10).sum**2
puts square_of_sums - sum_of_squares

Python

Very similar to Ruby but uses list comprehensions because why not?

sum_of_squares = sum([i**2 for i in range(1, 101)])
square_of_sums = sum(range(1, 101))**2
print(square_of_sums - sum_of_squares)

Lua

I claim this is trivial for most programming languages. To back that up I searched for a random programming language and got handed Lua. I have no experience with Lua.

I read a little on the language but didn't find any obvious mappers or magic methods like sum.

Here's how I decided to tackle this problem more directly.

local range_sum = 0
local squares_sum = 0
for i=1,100 do
    range_sum = range_sum + i
    squares_sum = squares_sum + i^2
end

print(range_sum^2 - squares_sum)

A very optional exercise 1

Pick a programming language, other than Uiua, and try solving this problem.

Solving it with Uiua

Let's solve first for the example range of 1-10.

    range,1 10
    ⇡₁10
[1 2 3 4 5 6 7 8 9 10]

Raising all numbers to the power of 2

    pow,2 ⇡₁10
    ⁿ₂ ⇡₁10
[1 4 9 16 25 36 49 64 81 100]

and summing them up.

    /+ ⁿ₂ ⇡₁10
385

Tacking on another range, but this time summing right away

    /+ ⇡₁10 /+ ⁿ₂ ⇡₁10
385
55

raising to the second power

    ⁿ₂/+ ⇡₁10 /+ ⁿ₂ ⇡₁10
385
3025

and subbing.

    - ⁿ₂/+ ⇡₁10 /+ ⁿ₂ ⇡₁10
¯2640

Whoops, nothing backward won't fix.

    ˜- ⁿ₂/+ ⇡₁10 /+ ⁿ₂ ⇡₁10
2640

And that's the answer.

Solution

    ˜- ⁿ₂/+ ⇡₁100 /+ ⁿ₂ ⇡₁100
25164150

How fast?

Uiua is optimized for array operations and the math we're doing isn't crazy. We should expect this to be blazing fast.

    ⊙◌⍜now(˜- ⁿ₂/+ ⇡₁100 /+ ⁿ₂ ⇡₁100)
0.000020265579223632813

Yup. Near instant.

Make it pretty

The solution we have is straightforward. However, one thing we can take note of is that the same range is created twice. When we see that happen, often we can use fork to run two operations on the same input.

    fork(/add pow,2|pow,2 /add) range,1 10
    ⊃(/+ⁿ₂|ⁿ₂/+) ⇡₁10
3025
385

The fork outputs two arguments. We simply sub them.

    -⊃(/+ⁿ₂|ⁿ₂/+) ⇡₁100
25164150