## Thursday, September 21, 2017

### Test To Detect The Poison

Here binary number system can come to rescue. Just for a  moment, let's assume there are 15 bottles. Now let's number the bottles from 1 to 15. To test these 15 bottles we need 4 prisoners as below. Let's number the prisoners from in descending 4 to 1.

Wherever 1 is written for the particular bottle number, that bottle should be given to particular prisoner. Otherwise should not.

So for the specific bottle with unique number a specific combination of prisoners (they are bits here) would be formed.

For example, if bottle labeled as 11 has a poison then prisoner no. 4,2,1 would die. In other words, if prisoner 4 & 2 die then the bottle no. 10 had poison.

For 16th bottle we would have needed 1 more prisoner.

In similar way, to test 1000 bottles, we need 10 prisoners (2^10=1024). Depending on what combination of prisoner die we can determine which bottle had poison. If prisoners numbered from 10 to 1 & if prisoner 10,8,6,3 & 2 die then bottle no.678 (binary -
1010100110) must had poison. Since the poison takes some time to take effect, even if prisoners taste this bottle, we still would have time to test rest of all bottles in given binary pattern.

 Poisonous Bottle

In case there were 1025 bottle, we would have needed 11 prisoners.