### Bekh's blog

By Bekh, history, 5 days ago, ,

Hello,

I know and already got AC with other neater solutions, but I was trying various dp states/approaches for practice. I wanted to know what I am missing in this dp approach since I'm getting a bit lower value than the sample output (0.5012511 instead of 0.5002286)

approach:

• i is zero based

Dp definition:

• dp[i][rem] is probability to get a final even 'usedCandy' count in range [i, m) using 'rem' candies such that usedCandy = totalCandies — rem

base case:

• at i == m, return isEven(usedCandy)

transition:

• at each [i][rem], if 'tk' is amount given to man 'i'. 'tk' is in [0, rem]

• answer = Summation for all 'tk' in [0, rem] -> P(man 'i' getting exactly tk candies) * dp[i + 1][rem — tk]

• P(a man getting exactly 'tk' candies) having 'rem' candies in total can be found through binomial theorom C[rem][tk] * p^tk q^(rem — tk) such that p is probability for

• a person to get a single candy = 1/(m+w) and q = 1 — p

If something is not clear with my solution, please ask. Where am I going wrong with this dp?

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• +5
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 » 5 days ago, # |   0 Auto comment: topic has been updated by Bekh (previous revision, new revision, compare).
 » 5 days ago, # |   0 Auto comment: topic has been updated by Bekh (previous revision, new revision, compare).
 » 5 days ago, # | ← Rev. 6 →   +3 Here's the error in your code: Consider that we are at the i-th man with rem candies left. Since the first i men have already been distributed the candies and we will not give them any more candies so, the probability of the first i men getting a candy now is 0. So, we are effectively left now with (m - i) men and w women and the next candies can be distributed to them with equal probability. So, in the state dp[i][rem], the probability of success becomes p = 1.0 / ((m - i) + w). Make this change and you'll get AC.BTW, I still don't get why are you doing if(ret == ret) return ret;. So, I used a visited array instead. Please explain that line. Here's the updated code.
•  » » 5 days ago, # ^ |   +6 Oh, thank you. I've completely missed that!for the (ret == ret) part. I initially memset the dp array with (-1) this fills that memory chunk with all 1's (because the 2's complement representation of (-1) is all 1's). In double values, all 1's is defined as NaN (Not a Number). NaN's has some unusual properties, one of them is that NaN is not equal to any other value including itself. Checking (ret == ret) should return false for the initial state of the dp, otherwise for any other value it should return true.for more about NaN's: http://www.cs.technion.ac.il/users/yechiel/c++-faq/nan.html
•  » » » 5 days ago, # ^ |   +2 Well that's something interesting. Thanks :)
•  » » » » 5 days ago, # ^ |   +3 No, problem. Good luck!