26 Aug 2022 7:46 pm
I am with you for the first step.Cannonpointer » 26 Aug 2022, 4:12 pm » wrote: ↑ Okay, I'm back. I have solved for the initial 8 being control, and I have solved for the sequestered 4 being control, but only when sequestering 2 from a and 1 from b and getting an even weight. An odd weight defeated me.
An odd weight left us with 5 suspects in 2 groups of 4 - a and b, light and heavy, respectively. Two are in one group, and three are in the other. We have identified neither the suspect nor the weight problem, and we have only one weighing left.
I need an applied math course - I should be able to set this up. I don't now whether I need to back out of my second solution (Replacing 2&1 of the suspects from the control group of four), or think harder about the present problem in the event of an unequal weighing (Leaving us with 3 suspects on one side and 2 on the other - and 9 control marbles identified). Hmmm...
Musing...
I simply cannot see any way to work out the 3v2 dead end in a single weighing... so I would have to go back and try something OTHER than replacing 1 from a and 2 from b...
That is another dead end, for me. I am going to don my softest leather loafers - **** that, my cloth slippers - and prepare to kick my own ***. @RunningWithScissors, @Jantje_Smit, I surrender.
4x4 gets you a control group (or two if they balance).
The remaining odd four are easy to sort in two weighings.
If the first 4x4 do not balance, isn't this where the "solution" and actually the problem really is?
How do you find 1 odd ball out of 8 in just two weighings?
Please seat yourself.
I like the very things you hate.
I like the very things you hate.