• coming soon

Part 4: An actual solution:

• likely to come later

When using the weighing scale we initially (after 1 or 2 measurements) cannot conclude much, while the balance scale can immediately discard around 2/3rd of the possibilities. It seems attractive to at least get some answers fast. For example weigh one half and then the other half, then we can at least know if the fakes are distributed 1-1 or 2-0. I could not get that to work however. And it is not needed!

With the weighing scale we can always reason backwards after we have enough data. Then however we must make sure that no big problem area's can emerge. To make sure of that taking orthogonal slices of half the coins seems best. We can do that 3 times, and then are down to groups of 3. Taking 2 of each 3 seems a useful 4th measurement (better than 1 of each , since it leaves only one coin completely unmeasured)

So my (initial) strategy:
Divide the coins in 8 groups of 3: {a,b,c,d,e,f,g,h}.
Measure orthogonal slices; abcd,abef, aceh, nr1+2 of each group.

Part 4a Testing the strategy with a (difficult?) case:

Case 1: all measurements suggest the same weight:
the first 3 measurements showed 12Q and the last one 16Q for some weight Q

- Q is the genuine weight.
Then the fakes must be canceling each other and that is only possible if they are the number 1 and 2 of a group.
- Q is not the genuine weight. That means that the fourth group has at least one fake.
If that one is in b..g, the 1 or 2 groups that do not measure those must have an identical fake. Then the offset of the 4th group can only be 1 or 2 times that offset, but it needs to be 16/12 times the offset. -> contradiction, thus the fakes are in a+h.

Possibilities if the genuine weight is Q:
nr 1+2 of 1 of the 8 groups.
Possibilities if the genuine weight is Q+x:
a1 or a2 Q-11x and h1 or h2 Q-3x
a1 or a2 Q-15x and a3 Q+5x
a3 Q-11x and h1 or h2 Q-15x

The 5th measurement: If we end up with canceling fakes after this measurement, only 5 (having 10 possibilities) can be distinguished enough with the remaining 2 measurements. (weigh 3 and then balance or weigh 1 new group). So we weigh b1..d1.

- if 3Q, they are genuine: we can weigh e1..g1 and the either balance them or measure h1 to distinguish the 5 possibilities.

- if 3Q+y the remaining possibilities are:
The true weight is Q
- b1 or c1 or d1 weigh Q+y the corresponding nr2 Q-y
The true weight is Q+y/3
- a1 or a2 Q-11y/3 and h1 or h2 Q-y (and the rest Q+y/3)
- a1 or a2 Q-15y/3 and a3 Q-5y/3 (and the rest Q+y/3)
- a3 Q-11y/3 and h1 or h2 Q-15x

6) We measure a combo that discerns the different true weight cases and splits up the true weight is Q case.
Note that now the power of the measuring scale really shows.
Weigh h1h2a3b1f1
- 5Q: c1+c2 or d1 + s2 do not weigh Q ( 7) weigh c1)
- 5Q+y: b1+ b2 do not weigh Q (already done)
- 5Q+y/3: a1 or a2 weighs Q-11y/3; h1 or h2 weighs Q-y; (the others Q+y/3) ( 7) weigh a1+h1 to distinguish the 4 cases)
- 5Q-y/3: a1 or a2 weighs Q-15y/3; a3 is the other fake ( 7) weigh a1)
- 5Q-23y/3: a3 weighs Q-11y/3; h1 or h2 Q-15x ( 7) weigh h1)

So the presumed most difficult case of equal average weights is solvable in 7 measurements when using this strategy.

Part 4: An actual solution:

• other branches likely to come later

An attempt to solve it with 7 measurements

Part 1 : possibility analysis

A strict lower bound on possibilies:

We only need to find the two fakes, that is 24x23/2 = 276 possibilies

A reasonlable approximate lower bound:

However, it is impossible to do that without finding out the actual offset of the fakes for most distributions of the fakes (if they are heavy/light compared to the genuine coins and other fake). Also solving the relative weight of the coins will therefor be almost the same problem in practice and has a lot more possibilities:

- 2 equal fakes: 24x23/2 = 276 possibilies
- 2 cancelling fakes: 24x23 = 552 possibilies
- a heavy and less heavy fake: 24x23 = 552 possibilies
- a heavy and less light fake: 24x23 = 552 possibilies
- a light and less heavy fake: 24x23 = 552 possibilies
- a light and less light fake: 24x23 = 552 possibilies

Part 2 : an (approximate) lower bound

When using the balance scale we can distinguish 3 states.
When using the weighing scale we can distinguish:
- 3 states for case 1: 0,1 or 2 fakes
- 3 states for case 2: the light fake, the heavy fake or 0/2 fakes
- 4 states in the other cases: no fake, lesser fake, greater fake, both fakes
So the weighing scale is superior , and should be used often!

Now, in theory, we can with the first measurement split the possibilities in 3:
- half of case1+2 solvable in 3log(414) = 5.48 tries
- half of case1+2 solvable in 3log(414) = 5.48 tries
- all of case 3..6; solvable in 4log(2208) = 5.55 tries

So the lower bound is around 6.5

Although it is still not proved that 6 is impossible if we only have to determine the position of the fakes, it is extremely unlikely.
On the other hand 6.5 is 'far' from 7. Making it likely, but not certain, that a solution that only requires 7 measurements exists.

Part 3: A strategy to (attempt to) get an actual solution with 7 measurements.

• coming soon

Part 4: An actual solution:

• likely to come later

An attempt to solve it with 7 measurements

Part 1 : possibility analysis

A strict lower bound on possibilies:

We only need to find the two fakes, that is 24x23/2 = 276 possibilies

A reasonlable approximate lower bound:

However, it is impossible to do that without finding out the actual offset of the fakes for most distributions of the fakes (if they are heavy/light compared to the genuine coins and other fake). Also solving the relative weight of the coins will therefor be almost the same problem in practice and has a lot more possibilities:

- 2 equal fakes: 24x23/2 = 276 possibilies
- 2 cancelling fakes: 24x23 = 552 possibilies
- a heavy and less heavy fake: 24x23 = 552 possibilies
- a heavy and less light fake: 24x23 = 552 possibilies
- a light and less heavy fake: 24x23 = 552 possibilies
- a light and less light fake: 24x23 = 552 possibilies

Part 2 : an (approximate) lower bound

When using the balance scale we can distinguish 3 states.
When using the weighing scale we can distinguish:
- 3 states for case 1: 0,1 or 2 fakes
- 3 states for case 2: the light fake, the heavy fake or 0/2 fakes
- 4 states in the other cases: no fake, lesser fake, greater fake, both fakes
So the weighing scale is superior , and should be used often!

Now, in theory, we can with the first measurement split the possibilities in 3:
- half of case1+2 solvable in 3log(414) = 5.48 tries
- half of case1+2 solvable in 3log(414) = 5.48 tries
- all of case 3..6; solvable in 4log(2208) = 5.55 tries

So the lower bound is around 6.5

Although it is still not proved that 6 is impossible if we only have to determine the position of the fakes, it is extremely unlikely.
On the other hand 6.5 is 'far' from 7. Making it likely, but not certain, that a solution that only requires 7 measurements exists.

added note: If weighing is used, we have the additional complication that the genuine weight is not known. That is likely to resolve itself however, since as long as we do not know it, we do not have 4, but infinite possible results of the next weighing. Thus, the acquired lower bound does not need a significant increase for that reason.

Part 3: A strategy to (attempt to) get an actual solution with 7 measurements.

• coming soon

Part 4: An actual solution:

• likely to come later