Photo by Elena Mozhvilo on Unsplash

Puzzle Statement

You arrive at an island in which there are 12 residents. You are told that all but one of the islanders weighs the same, but that one of the twelve weighs more or less than the others. You only have a pan-scale, like that of the background, and you have a limit of three measurements.

1. How do you determine the islander with differing weight?
2. Is the differing islander heaver or lighter than the others?

I first heard of this puzzle on the show Brooklyn-99, and then in various places on the internet. Largely the solutions either:

• Were incorrect. Usually this was because they were only solving the first numbered item
• Were correct, but they only partially explained how they generated the solution.

Analysis

If we only consider the first question above, then we just have to select the subsets of the twelve into three sets (on the left-pan, on the right-pan, and sitting-this-measurement-out). I won’t solve this part, but rest-assured, that we have three measurements, (each of which gives a reading of “left-heavy”, “right-heavy”, or “balanced”), of which we can get 3³=27 outcomes. So we’re probably okay.

If we tackle the second question, there are 24 possible outcomes, so that tells us that we (potentially) have enough information, but since 24 is close to 27, we likely have to be very careful with the way we combine the information for each measurement. Hey, there’s a reason there is some amount of buzz about this puzzle!

Solution

First, let’s number the islanders 1 though 12, and use the superscript X⁺ denote that X is heavier, and X^- that X is lighter.

Secondly, if we let the symbol L denote the scale falls to the left, R that the scale falls to the right, and C denote the case where a balanced scale.

Aside: Splitting up the Islanders

My solution, and every other one that I have see involves weighing three groups of four: one on the left, one on the right, and one sitting-out the measurement. I will use the middle of the scale to visualize this in the tables. I only have intuition to support why this is necessary. Clearly weighing six on each side is a waste of time since you only deduce that it is imbalanced, not that the imbalance arises from a heavier or lighter islander. Perhaps the puzzle would admit a solution in which 5 islanders are on the scale for at least one of the measurements? Perhaps.

So, we start with two tables, and start to fill things out. At the start they look like this:

Sequence Table:

Sequence	Assignment	Sequence	Assignment	Sequence	Assignment
CCC		LCC		RCC
CCL		LCL		RCL
CCR		LCR		RCR
CLC		LLC		RLC
CLL		LLL		RLL
CLR		LLR		RLR
CRC		LRC		RRC
CRL		LRL		RRL
CRR		LRR		RRR

Person allocation table:

No.	Left	Sitting	Right
		Out
1
2
3

Now we will start filling in entries of the table, starting with 1 and ending with 12. In the person allocation table, the person must be allocated in one of the three columns for each of the three measurements. After the placement, we record where in the sequence table what each of the two outcomes are by considering the heavy and light version of that person.

There are some constraints:

• Discard “CCC” as it does not tell us anything useful.
• When placing a heavy person on one of the sequence entry, its lighter counterpart will be placed in the sequence entry by reversing the Ls and Rs, and leaving the Cs. (e.g. LCR becomes RCL).
• Make sure no person matches another’s allocation for each of the three measurements.
• Four people on each of the three columns.

I started by thinking in terms of the sequence table, and filling that up, which determines the corresponding allocation table. These steps happen to be the order I picked, I am sure there are a large number of permutations.

Person 1

I decided to assign LLL=1⁺, which requires RRR=1^- This gives the following entries:

Sequence Table:

Sequence	Assignment	Sequence	Assignment	Sequence	Assignment
CCC	XXX	LCC		RCC
CCL		LCL		RCL
CCR		LCR		RCR
CLC		LLC		RLC
CLL		LLL	1+	RLL
CLR		LLR		RLR
CRC		LRC		RRC
CRL		LRL		RRL
CRR		LRR		RRR	1-

Person allocation table:

No.	Left	Sitting	Right
		Out
1	1
2	1
3	1

Person 2

Assign LLC=2⁺, which requires RRC=2^-

Sequence Table:

Sequence	Assignment	Sequence	Assignment	Sequence	Assignment
CCC	XXX	LCC		RCC
CCL		LCL		RCL
CCR		LCR		RCR
CLC		LLC	2+	RLC
CLL		LLL	1+	RLL
CLR		LLR		RLR
CRC		LRC		RRC	2+
CRL		LRL		RRL
CRR		LRR		RRR	1-

Person allocation table:

No.	Left	Sitting	Right
		Out
1	1,2
2	1,2
3	1	2

Person 3

Assign LLR=3⁺, which requires RRL=3^-

Sequence Table:

Sequence	Assignment	Sequence	Assignment	Sequence	Assignment
CCC	XXX	LCC		RCC
CCL		LCL		RCL
CCR		LCR		RCR
CLC		LLC	2+	RLC
CLL		LLL	1+	RLL
CLR		LLR	3+	RLR
CRC		LRC		RRC	2+
CRL		LRL		RRL	3-
CRR		LRR		RRR	1-

Person allocation table:

No.	Left	Sitting	Right
		Out
1	1,2,3
2	1,2,3
3	1	2	3

Person 4

Assign CCL=4⁺, which requires CCR=4^-

Person 5

Assign RLC=5⁺, which requires LRC=5^-

Person 6

Assign CRL=6⁺, which requires CLR=6^-

After filling in half the entries it looks like this:

Sequence	Assignment	Sequence	Assignment	Sequence	Assignment
CCC	XXX	LCC		RCC
CCL	4+	LCL		RCL
CCR	4-	LCR		RCR
CLC		LLC	2+	RLC	5+
CLL		LLL	1+	RLL
CLR	6-	LLR	3+	RLR
CRC		LRC	5-	RRC	2+
CRL	6+	LRL		RRL	3-
CRR		LRR		RRR	1-

Person allocation table:

No.	Left	Sitting	Right
		Out
1	1,2,3	4,6	5
2	1,2,3,5	4	6
3	1,4,6	2,5	3

Persons 7 through 12

Similarly, the assignments can be chosen as:

CRR = 7⁺, CLL=7^-

RCR = 8⁺, LCL=8^-

RCC = 9⁺, LCC=9^-

RCL = 10⁺, LCR=10^-

LRR = 11⁺, RLL=11^-, and

CRC = 12⁺, CLC=12^-

The tricky bit is to ensure that you don’t over-fill any of the allocations, and you don’t repeat any prior allocation. At this stage it sometimes helped to think of the allocations first, and then hypothesize a heavy- or light-person and see if it would fit into the sequence table. In fact, I got almost all the way to the end, and I realized that I had an allocation that was repeated. To solve this I had to think in terms of swapping people in one of the measurements, and change the corresponding sequence table entries.

After filling in all the entries:

Sequence	Assignment	Sequence	Assignment	Sequence	Assignment
CCC	XXX	LCC	9-	RCC	9+
CCL	4+	LCL	8-	RCL	10+
CCR	4-	LCR	10-	RCR	8+
CLC	12-	LLC	2+	RLC	5+
CLL	7-	LLL	1+	RLL	11-
CLR	6-	LLR	3+	RLR	unused
CRC	12+	LRC	5-	RRC	2-
CRL	6+	LRL	unused	RRL	3-
CRR	7+	LRR	11+	RRR	1-

Person allocation table:

No.	Left	Sitting	Right
		Out
1	1,2,3,11	4,6,7,12	5,8,9,10
2	1,2,3,5	4,8,9,10	6,7,11,12
3	1,4,6,10	2,5,9,12	3,7,8,11

Checking to see if our answer is correct, consider the 3⁺ case. You can see in the allocation table that we would get the result “LLR”, which is the one indicated in the sequence table.

---