There are 1000 buckets, one and only one of them contains poison, the rest are filled with water. They all look the same. If a pig drinks that poison it will die within 15 minutes. What is the minimum amount of pigs you need to figure out which bucket contains the poison within one hour.

Answer this question, and write an algorithm for the follow-up general case.

Follow-up:

If there are n buckets and a pig drinking poison will die within m minutes, how many pigs (x) you need to figure out the "poison" bucket within p minutes? There is exact one bucket with poison.

 

我草，说是简单题目，简单你妹啊！出题人脑壳有问题！！！明显是数学题，太尼玛打击老子了想了一个晚上没有搞定。

摘录别人的解答：

如果只有一只猪，考虑到间隔时间为15分钟，而测试时间是60分钟，所以一维的情况下，一只猪可以测出最多五只桶(为什么不是60/15, 4只呢，因为如果前面四个测试猪都没有死，则证明第五只桶有毒，但是不能六只桶，因为四次测试时间用完，会剩下两只，无法确定哪只有毒)。也就是说如果只有一只猪，可以最多检查(测试时间/毒发时间 + 1)个桶。

 

解答的思路：

1. 一只猪在一小时内最多能验多少桶？

一次喝一个桶的，15分钟后没挂再喝第二桶，一小时60分钟内可以喝 60/15 = 4 次，如果有5桶水，那个只要喝前4桶就只能第5桶是否有毒。

因此一只小猪在一小时可以验5桶水

 

2. 两只呢？

既然一只能验5桶，那么用二维的思路，2只猪应该可以验5*5桶：

猪A负责行，猪B负责列，每15分钟试喝一行/一列的所有5桶水，通过2只猪上天的时间能推断出毒水在几行几列。

1   2   3   4   5

6   7   8   9  10

11 12 13 14 15

16 17 18 19 20

21 22 23 24 25

 


With 2 pigs, poison killing in 15 minutes, and having 60 minutes, we can find the poison in up to 25 buckets in the following way. Arrange the buckets in a 5×5 square:

1  2  3  4  5
 6  7  8  9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

Now use one pig to find the row (make it drink from buckets 1, 2, 3, 4, 5, wait 15 minutes, make it drink from buckets 6, 7, 8, 9, 10, wait 15 minutes, etc). Use the second pig to find the column (make it drink 1, 6, 11, 16, 21, then 2, 7, 12, 17, 22, etc).

 

Having 60 minutes and tests taking 15 minutes means we can run four tests. If the row pig dies in the third test, the poison is in the third row. If the column pig doesn't die at all, the poison is in the fifth column (this is why we can cover five rows/columns even though we can only run four tests).

With 3 pigs, we can similarly use a 5×5×5 cube instead of a 5×5 square and again use one pig to determine the coordinate of one dimension (one pig drinks layers from top to bottom, one drinks layers from left to right, one drinks layers from front to back). So 3 pigs can solve up to 125 buckets.

In general, we can solve up to (⌊minutesToTest / minutesToDie⌋ + 1)pigs buckets this way, so just find the smallest sufficient number of pigs for example like this:

def poorPigs(self, buckets, minutesToDie, minutesToTest):
    pigs = 0
    while (minutesToTest / minutesToDie + 1) ** pigs < buckets:
        pigs += 1
    return pigs

Or with logarithm like I've seen other people do it. That's also where I got the idea from (I didn't really try solving this problem on my own because the judge's solution originally was wrong and I was more interested in possibly helping to make it right quickly).

 

 

还是觉得 minutesToTest / minutesToDie + 1 中+1有问题？因为60分钟，只能测试4次，也就是最后的25个桶实际上是喝不到的。

1  2  3  4  5
 6  7  8  9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

如果所有轮次后，两只猪都没有死，其实是可以确定有毒的在25号桶上！也就是说，虽然25号桶没有喝但是依然可以判断有毒是否是它！

参考：https://leetcode.com/problems/poor-pigs/discuss/94298/2-pigs-for-25-buckets.

 

数学解法，化简后：

class Solution(object):
    def poorPigs(self, buckets, minutesToDie, minutesToTest):
        """
        :type buckets: int
        :type minutesToDie: int
        :type minutesToTest: int
        :rtype: int
        """
        return int(math.ceil(math.log(buckets, minutesToTest / minutesToDie + 1)))