Dgf Test: Fill & Download for Free

n bottles, x hours.Option 1 : straight from the bottle1 hours : use n-1 rats.2 hours : use n/2 rats with half the bottles and the same rats again if need be for the second half3 hours : n/3 ratsin general : n/x rats would suffice.Option 2 : mixing the bottles…1 hour,1 bottle : 1 rat2 bottles : 1 rat3 bottles : 2 rats4 bottles : 2 rats receiving mixes of ab and bc, d left untouched (if R1 dies, a is poison, if R1 and R2 dies, b is poison, if R2 dies, c is poison, no death is poison in bottle d)5 bottles : 3 rats, mixes of 2 bottles each (a left alone, bc, de, ce)6 bottles : 3 rats, (a left alone, mixes bcd, cef, df)7 bottles : 3 rats (a left alone, bcd, cef, dgf)10 bottles : 4 rats (one bottle left alone, each rat with a different bottle in common with every other rat and one to itself, each bottle used by 0,1or 2 rats )we need for each bottle a unique combinaison of a few rats or no rats..the old .factorial something / factorial other calcul…thingy that I am not able to write in mathematical notation in a comment. Try and calculate there or some such :Calcul de Combinaison - K Parmi N - Générateur en LigneNOW with 2 hours left, you can make ‘m’ mixes of as many bottles as possible to separate at the last step, ‘b’so hour 2 : m-1 rats take mixes, and hour 1 : the survivors take b-1 of the b bottles used to make the deadly batch (or the left over batch)m*b >= n;m>b;m-1 rats neededchecking with some cases :say 2 hours, and 10 bottles3 rats gets 3 mixed samples and one bottle lefteither no rats dies, you get the intact bottle as poisonedor 1 rat dies, you give sample from one of his bottles to one of the survivor and one bottle unusedthen :1 dead: that bottle is poison, no dead, the leftover bottle is poison.2 hours and 110 bottles :10 batches of 10 bottles; 5 rats to try mixes of those batches, the killer batch’s bottles checked on simpler mixes on 5 rats (might need a new one here)so 6 rats.so 2 hours and n bottles : root(n) rats work, or “r rats, with r^2=n” but you can get quite a lot lowerin 3 hours you can do 3 trials , so you need r rats with r*r*r = n“3 hours :r rats with r^3=n”(check : 3 hours, 70 bottles : 4 rats not enough, 5 rats,14 bottles per rat first batch , then 5 batch of 3 bottles, then last round 3 rats to 3 bottlesor 6 batches of 12, one unused; second round 12 to 4-5 rats so 3 bottle per mix, then 1 each to 3-4 ratsor …last batch 2 rats : can proof 4 bottlesprevious 3 rats : have to separate 4-bottle mixes, can proof 7 batchesfirst 4 rats : can proof up to 11 assemblagesaltogether, in 3 hours, 4 rats can clear up to 11*7*4 = 308 bottlesbut I am not quite sure you have the time to produce all the mixes before time’s upin 4 hours you can do 4 rounds and so onI say you need less than r rats with r^x=nOPTION 3 : the poison is really really precise, and you have a timerSo you feed the same rat samples every 5 minutes, when he dies, it was the bottle from one hour ago.1 hour allows to check one bottle per rat1 hour 5 min lets you check 2 bottles per rat1 hour 10 min : 3 bottles per rat1 hour 15 min : 4 bottles pre rat…2 hours : 13 bottles a ratand every hour after that : 12 more bottlesso..n bottles, x hours => x=1 : n-1 rat; x>1 : n/((12*(x-1))+1) ratsOption 4: both option 2 and 3, feeding the poor critters successive mixes.Except you do not know the result of a particular mix until 1 hour later so you might try a lot of useless batch that you have to mix…1 hour si still r=n-1 rats2 hours, 20 bottles : one rat get 10 successive mix of pairs until death, and at the same time one of the bottle of the mix is given to a second rat. 2 rats does it.2 hours, 100 bottles : one rat allows for 13 checks, we need 8 packs of bottles checked4 rats gets pairs of bottles R1:a and b; R2 : c and d ;R3 : e and f; R4 : g and h2 rats gets pairs of bottles , one from each of the first rats’ mixesR5: a and c; R6: e and gAt each batch, no death until the bad bottle get used, and then one or two rats dies, depending on if the bottle was used once or twice.2 hours, 100 bottles : 6 rats , a bit better than the two previous options3 hours, 100 bottles :bottles in 24 batches of 4 (a b c d), 3 rats dozed with batches ab, bd, acIf R1 and R2 dies it is bottle b, if R1 and R3 dies it is bottle a, if R3 dies alone it is c, if R2 dies alone it is d.If no rats dies at all, the last batch of 4 can be tested on our 3 rats at the last hour3 hours, 100 bottles : is 3 ratsBut my brain is tiring…