Daily Challenge #186 - Jumping Frog

Setup

There is a lonely frog that lives in a pond. Lily-pads are laid out on a coordinate axis atop the pond. The frog can only jump one unit more than the length of the last jump.

With a starting point of 0, reach the target point of n using the frog's jumping ability. You can choose to jump forward to backward. Reach the target with the minimal amount of steps.

Examples

For n = 2, the output should be 3.

step 1:  0 ->  1  (Jump forward, distance 1)
step 2:  1 -> -1  (Jump backward, distance 2)
step 3: -1 ->  2  (Jump forward, distance 3)

For n = 6, the output should be 3.

step 1: 0 -> 1  (Jump forward, distance 1)
step 2: 1 -> 3  (Jump forward, distance 2)
step 3: 3 -> 6  (Jump forward, distance 3)

Tests

n = 1

n = 10

n = 25

n = 100

n = 1000

Good luck!

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This challenge comes from myjiinxin2015 on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

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Comments

[SavagePixie]

Some Elixir using recursion:

def jump(n), do: jump(n, 1, 1)
def jump(n, step, value)
  when rem(value - n, 2) == 0
  and value >= n, do: step
def jump(n, step, value), do: jump(n, step + 1, value + step + 1)

[Craig McIlwrath]

My O(1) solution in Haskell:

jump :: Integral a => a -> a
jump v = let n = ceiling $ (-1 + sqrt (1 + 8 * (fromIntegral v))) / 2
             s = n * (n + 1) `div` 2
             s2 = (n + 1) * (n + 2) `div` 2
         in  if even (s - v) then n
             else if even (s2 - v) then n + 1
                  else n + 2

I tested this against SavagePixie's solution from 1-10,000 and handwritten tests from 1-36. It passed all those tests. I'll try my best to explain how this works.

The n value that is calculated is the 1-indexed index of the first value greater than v (the target value) in the recursive sequence a{n} = a{n-1} + n. For example, when v = 13, n = 5 because a{5} = 15 and a{4} = 10.

s is the term a{n}. s2 is the term a{n+1}.

I determined that there are 3 patterns that will construct a shortest path 3 in different situations, which are represented in the if statements.

1. If s - v is even, then the target is some even number below a value in the sequence. This is an easy path to construct: to construct s, it's simply 1 + 2 + ... + n, and if s - v = 4, then v = 1 + 2 + ... + n - 4 = 1 - 2 + ... + n. The length of this path is equal to n. This works for any difference, not just 4. The summand that is negated is (s - v) / 2.

2. If s2 - v is even, it's the same as (1) except with s2 and n + 1, so the path length is equal to n + 1. This is longer than (1) when both conditions are met, but shorter than (3) when both conditions are met.

3. In all other cases, a shortest path is to construct a path to v + 1 and then subtract 1. Subtracting 1 requires 2 operations (ex. 6 - 7 after going to the 5th term). Using 12 as an example, construct the path to 13: -1 + 2 + 3 + 4 + 5, then subtract 1: -1 + 2 + 3 + 4 + 5 + (6 - 7). The length is n + 2.

I have an intuition of why these are the only three cases, but I can't seem to put it into words right now.