Waste less time on Facebook — follow Brilliant.

Battery Optimization

Hi all,

Suppose we have \(n\) batteries, \(g\) good ones, and a flashlight that takes \(c\) batteries. And you don't know which are the bad ones.

And, let \(f(n,g,c) =\) The minimum number of attempts (insertions of \(c\) batteries into the flashlight) you need in order to get your flashlight working.

And assume that the only way the flashlight will work is if all \(c\) batteries you put in the flashlight are good. Otherwise, it won't work, and there is no way to distinguish between any combination that contains one or more bad batteries.

What I am wondering is how to show/prove the general solution for \(f(n,g,c)\).

We have a few trivial cases:

  • \(f(n,g,1) = n - g +1\)
  • \(f(n,g,g) = \binom{n}{g}\)
  • \(f(n,n,c) = 1\)
  • \(f(4,3,2) = 2\)

So, I think the first non-trivial case becomes:

  • \(f(5,3,2)\)

This one can be done in four moves, (1 vs 2, 3 vs 4, 4 vs 5, 3 vs 5) but can we prove that it can't be done in fewer?

And, does anyone have any idea about how we can solve this problem for the general case?



Note by Geoff Pilling
7 months, 1 week ago

No vote yet
1 vote


Sort by:

Top Newest

@Brian Charlesworth In a previous thread you had suggested that for \(f(n,3,2)\) you might want to split the group into two, one of which is guaranteed to have two good batteries, and run through each of the combinations in each group, giving:

\(f(2n,3,2) = 2\dbinom{n}{2}\) and \(f(2n + 1,3,2) = \dbinom{n}{2} + \dbinom{n + 1}{2}\)

One generalization of that for \(g>3\) might be to split up the batteries into the greatest number of groups (\(g-1\)?) such that at least one is guaranteed to have at least two good batteries and then go through all the combinations in each group. Without showing that this is the best approach, it certainly seems like a promising one.

So, for example if we take \(g=4\), then we would split into 3 groups, yielding:

\(f(8,4,2) = \binom{2}{2} + \binom{3}{2} + \binom{3}{2} = 1 + 3 + 3 = 7\)

So for \(f(100,50,2)\) we would have:

\(f(100,50,2) = 2\binom{3}{2} + 47\binom{2}{2} = 53\)

Recognize that answer? :0)

A couple more examples using the above algorithm gives:

  • \(f(100,10,2) = 8\binom{11}{2} + \binom{12}{2} = 506\)
  • \(f(100,25,2) = 4\binom{5}{2} + 20\binom{4}{2} = 160\)

@Calvin Lin Thoughts on how one might go about proving that these could be optimal approaches (outside of using a computer)? Geoff Pilling · 7 months, 1 week ago

Log in to reply

@Geoff Pilling Once you summarize the approach it does look quite promising. :)

As far as I can tell, with \(f(n,g,2)\) your goal is to form the sum

\(a\dbinom{\lfloor\frac{n}{g - 1}\rfloor}{2} + b\dbinom{\lfloor\frac{n}{g - 1}\rfloor + 1}{2}\)

so as to maximize \(a\) given \(a + b = g - 1\).

I can see one other approach to get \(f(100,10,2) = 506\), where we have \(9\) groupings of \(11\) and one singleton left over. Then if after \(9\binom{11}{2} = 495\) tests with no success we can conclude that the singleton is good and that each of the \(9\) groups has one good battery as well. So we then choose one of these groups and pair all \(11\) in the group with the good singleton to finally get a working pair after a grand total of \(495 + 11 = 506\) trials. This isn't an improvement but it does suggest that the \(g - 1\) paradigm may not be exclusively optimal. Brian Charlesworth · 7 months, 1 week ago

Log in to reply

@Brian Charlesworth Without a proof, I suspect that the general solution for \(f(n,g,c)\) might be given by this:

1) Divide the \(n\) batteries up into the maximum number of groups you can (minimizing the maximum number for any group) so that at least one group contains at least \(c\) batteries. I think this number is given by:

  • Number of groups \(= \lfloor {\frac{g-1}{c-1}} \rfloor \)

2) Perform all possible iterations within each group.

So, for example, for \(f(100,50,3)\) you would split it up into \(20\) groups of \(4\) and \(4\) groups of \(5\), guaranteeing that at least one group has \(3\).

\(\implies f(100,50,3) = 20\binom{4}{3} + 4\binom{5}{3}\)

@Brian Charlesworth @Calvin Lin Geoff Pilling · 7 months, 1 week ago

Log in to reply

@Geoff Pilling That looks good. The values ramp up quite quickly with \(c\), with

\(f(100,50,4) = 12\binom{6}{4} + 4\binom{7}{4} = 320, f(100,50,5) = 8\binom{8}{5} + 4\binom{9}{5} = 952\)

and \(f(100,50,6) = 8\binom{11}{6} + \binom{12}{6} = 4620\).

We could probably write out a messy general formula now but I think the description of the process suffices. The optimization proof remains elusive, however ... Brian Charlesworth · 7 months, 1 week ago

Log in to reply

@Brian Charlesworth Yes. Looks like a good generalization/summary for \(c=2\)... Now we just gotta figure out a proof and/or generalize it for \(c>2\)... Hmmmm.... Geoff Pilling · 7 months, 1 week ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...