## Lowest-cost Search

### Problem 328

We are trying to find a hidden number selected from the set of integers {1, 2, ..., `n`} by asking questions.
Each number (question) we ask, has a __cost equal to the number asked__ and we get one of three possible answers:

- "Your guess is lower than the hidden number", or
- "Yes, that's it!", or
- "Your guess is higher than the hidden number".

Given the value of `n`, an *optimal strategy* minimizes the total cost (i.e. the sum of all the questions asked) __for the worst possible case__. E.g.

If `n`=3, the best we can do is obviously to ask the number "**2**". The answer will immediately lead us to find the hidden number (at a total cost = 2).

If `n`=8, we might decide to use a "binary search" type of strategy: Our first question would be "**4**" and if the hidden number is higher than 4 we will need one or two additional questions.

Let our second question be "**6**". If the hidden number is still higher than 6, we will need a third question in order to discriminate between 7 and 8.

Thus, our third question will be "**7**" and the total cost for this worst-case scenario will be 4+6+7=**17**.

We can improve considerably the worst-case cost for `n`=8, by asking "**5**" as our first question.

If we are told that the hidden number is higher than 5, our second question will be "**7**", then we'll know for certain what the hidden number is (for a total cost of 5+7=**12**).

If we are told that the hidden number is lower than 5, our second question will be "**3**" and if the hidden number is lower than 3 our third question will be "**1**", giving a total cost of 5+3+1=**9**.

Since **12**>**9**, the worst-case cost for this strategy is **12**. That's better than what we achieved previously with the "binary search" strategy; it is also better than or equal to any other strategy.

So, in fact, we have just described an optimal strategy for `n`=8.

Let C(`n`) be the worst-case cost achieved by an optimal strategy for `n`, as described above.

Thus C(1) = 0, C(2) = 1, C(3) = 2 and C(8) = 12.

Similarly, C(100) = 400 and $\sum \limits_{n = 1}^{100} {C(n)} = 17575$.

Find $\sum \limits_{n = 1}^{200000} {C(n)}$ .