Odd One Out

Note that there are $19 \cdot 2 = 38$ possible cases. As every weighing gives $\log_2 3$ bits of information (there are three possibilities: left is heavier, balanced, right is heavier), we need $\left\lceil \dfrac{\log_2 38}{\log_2 3} \right\rceil = 4$ weighings in the worst case.

This is also easily achievable, mostly because of having only $38$ cases to distinguish while we can distinguish $3^4 = 81$ cases.

Divide the eggs into groups of 6, 6, and 7, and weigh the two groups of 6 against each other.

• If they balance, then the group of 7 has the odd egg and the two groups of 6 have normal weights. Weigh 7 normal eggs against the offending group of 7; this will tell whether the odd egg is heavier or lighter than normal.

• If they don't balance, then the group of 7 is clear. Weigh 6 normal eggs against the heavier side of the previous weighing. If they balance, then the other group of 6 has the odd egg and it's lighter; if they don't, then the current group of 6 has the odd egg and it's heavier.

So after two weighings, we manage to figure out a group of at most 7 eggs containing the odd egg, and whether the odd egg is heavier or lighter. To make things easier, add normal eggs to this offending group so there are 9 suspected eggs, and assume that the egg is heavier; if it's lighter, reverse all meanings of heavier and lighter.

Now we have 9 eggs, one of them is heavier than the rest. This is a simple ternary search: divide the eggs into groups of 3 and compare any two of them. This will tell which group of 3 has the odd egg. Next, take any two eggs from the offending group of 3; this will tell the odd egg in 4 weighings.

I Think 3 weighing is sufficient. Justification :

Divide the eggs in the group of 6,6,7. Weigh the 6,6 group.

Case 1: The two group weigh equal (1 weigh done). It implies the remaining 7 has the heavy egg. Now take the seven and divide it into 3,3,1. Now Weigh the 3,3 set (2 weigh done) . If they settle equally=> implies the left out is the odd man (This is the minimum set i.e 2 weighs ). Now Consider one of the 3,3 set is unequal. Take the heavier set and pick any 2 and weigh them (3 weigh done). you would get the heavy egg depending on if both settle equal or any one pulls the balance done. Hence max iteration is 3.

Case 2: If any one of the 6 set eggs weigh down (1 weigh done ). Implies => Discard the remaining seven and other lighter set of 6 eggs. Take this 6 eggs. Divide into 3,3. Weigh them (2 weigh done). One will definitely pull the balance down. Take this heavy set of 3 eggs and weigh any 2 egg (as earlier case: 3 weighs done) and you will get the odd man.

Hence either way : Max is 3

Separate the eggs into groups of 5, 5, 5, and 4. Call them A, B, C, D.

First, weigh A and B (weighing 1). Then weigh A and C (weighing 2). There are scenarios that can arise from here:

• Both weighings are equal.

This means that the odd egg must be in group D.

• Weighing 1 is equal while weighing 2 is different.

This means that the odd egg must be in group C.

• Weighing 1 is different while weighing 2 is equal.

This means that the odd egg must be in group B.

• Both weighings are different.

This means that the odd egg is in group A.

If the egg is in group A, B, or C, we can use a single example to describe all of them because they are all groups of 5. But first, if the egg is in group D:

Group D

Label the four eggs w, x, y, and z. First, weigh w and x (weighing 3). Then, weigh w and y (weighing 4).

• Both weighings are equal.

This means that the odd egg is egg z. Notice that we don't have to identify whether the egg is heavy or light.

• Weighing 3 is equal while weighing 4 is different.

This means that the odd egg is egg y.

• Weighing 3 is different while weighing 4 is equal.

This means that the odd egg is egg x.

• Both weighings are different.

This means that the odd egg is odd w.

For all scenarios in which the egg is in group D, we have $\boxed{4}$ weighings.

Group A, B, or C

If the egg is in any of these groups, we know whether the egg is light or heavy, since we know how each group weighed in comparison to the others. The egg is in a group with 4 other eggs. Label these eggs as v, w, x, y, and z. First, weigh eggs v and w against eggs x and y (weighing 3).

• Weighing 3 is equal.

This means that the egg is egg z.

• Weighing 3 is not equal.

You know whether the egg is heavy or light. Thus, pick the group of two with the different egg. Weigh these two eggs against each other (weighing 4).

• Weighing 4 is equal.

Wait, no, this doesn't happen :P

• Weighing 4 is not equal.

This weighing should never be equal. You can now identify which egg is lighter or heavier than the other, and thus you can identify which egg is the odd egg.

We did this again in $\boxed{4}$ weighings.

Thus, this algorithm can identify the egg in a maximum of $\boxed{4}$ weighings.

Take out 1 egg out of 19. 18 Left. 9 9 (1st weighing) if = that 1st egg we took out was heavier else take the heavy pile of 9 eggs & take out 1 egg again (8 Left) 4 4 (2nd weighing) if = last egg which we took out was heavier else take the heavy pile of 4 eggs & divide into 2 2 2 (3rd weighing) cant be equal take heavier pile of 2 eggs & divide into 1 1 1 (4th weighing) REST YOU CAN :P Thank You

Not clear, depends if this is a balance or a scale.

Shouldnt it work out with 3? 1) 6-6-7 compare groups containing 6 eggs 2) 2-2-3 (assuming that in the first trial both groups were equal; otherwise the groups would be 2-2-2) compare the groups containing 2 eggs 3) we are left with 2 eggs and compare them --> so just 3 weighings are needed

i only need 1 weighing.... i will weight 1 and see how heavy it is,then i'll add another one without taking off the first one and see if it is double the weight....then keep adding eggs one by one until i see the weight did not increase as it should..this way i will only use the weighing machine once. so 1 weighing needed!