Advent of Code 2023 Day 6

Another breather day!

This puzzle describes a boat race. To win, you must get the boat at least \(d\) distance within \(t\) seconds. You power the boat by pressing a button for \(b\) seconds; the boat then travels at \(b\) units per second for the remaining \(b-t\) seconds.

I saw immediately that this could be represented in a quadratic equation \(-b^2 + bt - d = 0\), where \(t\) is time, \(d\) is distance, and \(b\) is the length of the button press. All I had to do was find the roots of this equation, compute the lowest and highest integers in that interval, and subtract.

Brute force probably would have worked, but math is so much more satisfying.

local quadratic = function(a, b, c)
    local discrim = math.sqrt((b ^ 2) - (4 * a * c))
    local denom = 2 * a
    return { (-b + discrim) / denom, (-b - discrim) / denom }
end

local intercepts = function(time, dist)
    return quadratic(-1, time, -dist)
end

local num_bool = function(e)
    return (e and 1) or 0
end

local solve = function(data)
    local result = 1
    for i, _ in ipairs(data.time) do
        local points = intercepts(data.time[i], data.distance[i])
        -- Need to handle case where intercepts are integers and therefore invalid
        local left_i = math.min(points[1], points[2])
        local left = math.max(1, math.ceil(left_i + num_bool(left_i % 1 == 0)))
        local right_i = math.max(points[1], points[2])
        local right = math.min(
            math.floor(right_i - num_bool(right_i % 1 == 0)),
            data.time[i] - 1
        )
        result = result * (right - left + 1)
    end
    return result