Question:
There are 8 bottles, one has poison. What's the minimum number of rats you need to find the poison bottle in time T, and how? (You get the rats you need all at once, feed them all at the same time, and poison kills them after time T.)
Solution:
so , now if we take 4 rats , and suppose 0 represents that the rat does not drink from that bottle and 1 means the rat drinks from that bottle.
so solution 1 is fed to the 4th rat only , solutin 2 is fed only to the 3rd rat , soln 3 is fed to the 3rd and 4th rat .. so on , at the end , we can make out which one has poison , seeing which rat dies/rats die .....
POST YOUR OPINION IF YOU HAVE BETTER SOLUTION
4 comments
Dude srsly??
Represent the poisons in 3 digit binary.
You need 3 rats only. Bottles b1, b2, ....b8. Rats, rat1, rat2, rat3
feed the Mixture of b1,b3,b5,b7 to rat1
Feed the mixture of b1,b2,b5,b6 to rat2
Feed the mixture of b1,b2,b3,b4 to rat3
If rat1 dies r1 =0 else r1=1, rat2 dies, r2 =0 else r2=1, rat3 dies r3 =0 else r3 =1
Now we have
r3 r2 r3 Poison bottle
0 0 0 b1
0 0 1 b2
0 1 0 b3
0 1 1 b4
1 0 0 b5
1 0 1 b6
1 1 0 b7
1 1 1 b8
hey that`s not the case u told...for eight bottles you have to consider 4 bits only not 3 bits...because to find poison in bottle...u have to give poison to atleast one rat,so your 0 0 0 b1 case is eliminated.I hope ur doubt is cleared.:)
no only 3 rats are needed.. becoz when after giving d dose from bottles by 7 ways if no rat dies that means d bottle from which d dose is not given is poisoned !!! :) so u need 3
when r1=0,r2=0,r3=0-this means dat b1 is poisioned from which no rat is feeded
I'll try to make it more clear:
You have 8 bottles A, B, C, D, E, F, G and H. You take 3 rats R1, R2 and R3. You feed R1 from A+D+E+F, R2 from B+D+E+G and R3 from C+D+F+G. It's easy to tell the poisoned bottle by which rat(s) die(s).
(If not obvious, here's how: If R1 dies, A is poisoned bottle, R2 dies -> B, R3 dies -> C, R1+R2 die -> E, R1+R3 die -> F, R2+R3 die -> G, R1+R2+R3 die -> D, no rat dies -> H (which was left out)
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