Project Euler – Problem 112 Solution

Yan Cui

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Problem

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a “bouncy” number; for example, 155349.

Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.

Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.

Find the least number for which the proportion of bouncy numbers is exactly 99%.

Solution

// define function f which compares the sorted and original digits of the supplied
// number n and returns if they’re the same
let f (sort:int[] -> int[]) (n:int) =
let digits = n.ToString().ToCharArray() |> Array.map (fun c -> int(c.ToString()))
not(digits |> Seq.exists2 (fun e e’ -> e <> e’) (sort digits))

// curry the function f to form the isIncreasing and isDecreasing functions
let isIncreasing = f (fun l -> Array.sort l)
let isDecreasing = f (fun l -> Array.sortBy (fun x -> 1 – x) l)

let isBouncy n = not(isIncreasing n) && not(isDecreasing n)

let answer =
let mutable bouncyCount, bouncyRatio, n = 0, 0.0, 100
while bouncyRatio <> 0.99 do
n <- n+1 if isBouncy n then bouncyCount <- bouncyCount+1 bouncyRatio <- double(bouncyCount)/double(n) n [/code]

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