Radioactive Rods

Let's get a baseline answer:

Split the rods into groups of 2, test them (4).

If only one group is hot, you're good.
Otherwise, you have 2 groups of 2, 1 each which is radioactive. Test one of each. (2)
If hot, found; otherwise the other one is hot.

Edit: For a more generalized answer, I think the worst case is $2 * log_2{N}$

Split the group in half, check both groups.
If one each, do binary search in each.
Otherwise, split the hot half, checking both groups.
Repeat.

Here's another way to get

First, split the rods into groups of 4 and test them (2)

If both groups have a radioactive rod, do the following for both groups: take two of the rods and test them. If the test is positive, check one of those two. If the test of two rods was negative, test one of the other two rods. If the check of the single rod is negative, it was the other rod in that pair. If it is positive, obviously you've found the radioactive rod. This takes 4 tests to check both groups.

If only one group has radioactive rods, split the group in two. If the first pair you check is not radioactive, it is the remaining two rods. Otherwise, test one of the two by itself - if it is not radioactive, you know the other one is and the second radioactive rod is in the other pair and can determine which with one more test. If the one you check is radioactive, test the other one. If it is not radioactive, the radioactive rod is in the other pair and can be found with one more test. The worst case scenario is 4 tests.

I guess I'll just add another answer for:

But for a lowest average number of tests over many scenarios:

Label your rods A,B,C,D,E,F,G,H
To make use of all possible outcomes, test two equal groups, but leave two behind, so...

1. Test A,B,C and D,E,F (2 tests)
   
   • if neither --> G and H are radioactive (2 total and done)
     
   • if only one group --> Go to Step 2. (Thanks, Aggie Kidd)
     
   • if both --> Go to Step 3.
     

P(done after step 1) = 1/28

2. Test A,B,C (or D,E,F) and G,H one at a time
   • Keep testing (starting with G,H until two have been found or all but one is tested (4-6 tests)

P(done at step 2) = P(both in same trio) + P(trio and G,H) = 3/14 + 6/14 = 9/14

3. Test A and D (2 tests) (this works testing A,B and D,E too)
   
   • if both --> A and D (4 total)
     
   • if A, but not D --> Test E --> then A and E or F (5 total)
     
   • if D, but not A --> Test B --> then D and B or C (5 total)
     
   • if neither --> Test B and E and deduce (6 total)
     

P(done at step 3) = P(one in each trio) = 9/28

I haven't worked through the probabilities/combinatorics completely, but this seems to have a lower average number of tests. If you had to stick with an algorithm from the beginning, this technically works better than some other algorithms, though the worst-case-scenario will still be 6 tests.

I'd be really interested to see someone prove that their answer is correct, which would involve calculating the probabilities of each test scenario to find the lowest average test number via Monte Carlo. If someone wants to do that, I'll owe them a lifetime's reputation.

Just because a test is successful doesn't mean we can eliminate the unchosen rods. Therefore, we always have to cover every group not yet confirmed to fail the test (if there aren't already 2 "successful" groups with no common rods), and their rods. The larger the groups that can be confirmed to fail are, the better. If the rods aren't divided equally, the smallest group might fail and the biggest ones succeed, so it doesn't give the best worst-case scenario.

After doing these, you'll know which quarter(s) has/have the radioactive rods. Even if they're two different quarters, you'll only need 2 more tests, which makes 6:

• Take 4.

• Take the other 4.

• Take half of both groups.

• Take the other halves.

Since I don't see the non-adaptive answer posted yet... You can also derive the answer of "six tests" following the strategy described here in "Wolves and sheep":

A: Test rods 1, 4, and 5 together.  
B: Test rods 1, 6, and 7 together.  
C: Test rods 2, 4, and 6 together.  
D: Test rods 2, 5, and 7 together.  
E: Test rods 3, 5, and 6 together.  
F: Test rods 3, 4, and 7 together.  

Then look up the set of positive results in this table:

ABCD: 12    ACDF: 24    BDEF:  37    AB:  18
ABEF: 13    ACDE: 25    ACDEF: 45    CD:  28
ABCF: 14    BCDE: 26    ABCEF: 46    EF:  38
ABDE: 15    BCDF: 27    ABCDF: 47    ACF: 48
ABCE: 16    ACEF: 34    ABCDE: 56    ADE: 58
ABDF: 17    ADEF: 35    ABDEF: 57    BCE: 68
CDEF: 23    BCEF: 36    BCDEF: 67    BDF: 78

Julian Rosen's answer gives a proof (which I have not checked rigorously, but it sounds good) that it cannot be done in fewer than 6 tests, not even if you permit "adaptive" strategies.