12 balls puzzle: You have 12 balls identical in looks with one fake
Find the fake among 12 balls in 3 weighs. All 12 balls look same and you have a pan balance with no weights.
Recommended time to solve: 40 minutes.
Give it a try before going ahead.
Can you find all possible ways to solve the puzzle?
No time limit for this one and no solution from us. It’s only for you, if you are curious.
Solution to Find the fake among 12 balls in 3 weighs puzzle
What should be the best plan for first weighing?
Should you first weigh 6 against 6 balls? That will cause all 12 balls suspect after first weighing—the number of suspect balls not reduced at all except that the balls are split into two groups of 6 containing the lighter or the heavier ball.
Dealing with 12 suspect balls even if divided into two groups with only two remaining weighing attempts should be impossible.
So divide the 12 balls into groups of smaller size.
Better will be to divide 12 balls into three groups: two equal sized sets of balls weighed against each other and a third set of balls left aside.
Maximum number of suspect balls will be at most 10 with this strategy of first weighing.
Next question arises at once,
How best can the 12 balls be divided into 3 such groups?
Let us list out all possible ways 12 as a sum can be split into 3 numbers in this weighing case. Two of the numbers should be equal to each other and number of balls kept aside should be as small as possible.
12 can be split in three unique ways that satisfy the conditions,
12 = 5 + 5 + 2,
12 = 4 + 4 + 4, or,
12 = 3 + 3 + 6.
The other two combinations, 12 = 2 + 2 + 8 or 12 = 1 + 1 + 10 will have the left aside group having 8 or 10 balls—too large to work on. More so because even the nature of the balls is not known for any of these two left aside groups of balls.
Which among the three should be chosen for the first weigh?
For 5 + 5 + 2 combination, in the worst case of one pan going down, 10 balls become suspect. Even with the knowledge of lighter or heavier 5-ball groups, it will be too complex to work on.
On the other hand, for combination 3 + 3 + 6, when the two pans are balanced, 6 left aside balls will be suspect. With no knowledge of nature of balls, finding the fake ball among 6 balls in two weighs will be impossible. Verify it.
Best choice will then be to keep the sizes of both weighed groups and left aside group as small as possible.
In this situation, the most promising course of action will be to follow a general principle of problem solving drawn from experience—how best to divide and conquer,
Divide the enemy in the smallest sized groups with group sizes as same to each other as possible.
Following this principle, for the first weighing, 12 is split into three groups each with 4 balls.
It also satisfies principle of symmetry,
If you increase the symmetry in the problem by an action, that should be your MOST PROMISING ACTION.
Most promising course of action for first weighing is then,
4 balls weighed against 4 others, with 4 balls kept aside.
The figure below shows the first weighing combination.
Three outcomes of the first weighing are possible.
First weighing Result 1 of 12 balls puzzle: The right pan descends and the left pan goes up.
Conclusion: Four kept aside balls are good and all eight balls weighed are suspect. Specifically,
1.1. Either the left side up-going four balls contain the fake lighter ball, or
1.2. The right side descending pan has the fake heavier ball.
The picture below shows the situation.
First weighing Result 2 of 12 balls puzzle: The left pan descends and right pan goes up.
Conclusion: Four kept aside balls are good and all 8 balls weighed are suspect. Specifically,
2.1. Either the left side descending 4 balls contain the fake heavier ball, or
2.2. The right side up-going pan has the fake lighter ball.
These two results will need similar actions as the outcomes 1.1 and 1.2.
Analysis of the outcomes 1.1 and 1.2 will be enough to know how to find the fake ball for outcomes 2.1 and 2.2. So you won’t analyze these two results further.
First weighing Result 3 of 12 balls puzzle: The pans are equally balanced:
The 4 left aside balls must contain the fake ball, lighter or heavier and,
All 8 balls weighed are good balls.
Let’s solve for the third result first.
Solving for Result 3 of 12 balls puzzle: In 4 + 4 first weigh, the two pans are balanced: Finding the fake ball among 4 left aside suspect balls in 2 weighing
First conclusion on analysis,
In 2nd weighing, you cannot weigh two suspect balls against the two other suspect balls. You will still be left with 4 suspect balls, though divided into two groups of opposite nature. With only 1 weighing quota left, solution will be impossible.
Verify the conclusion.
Taking the cue from decision of first weighing combination, apply again how best to divide and conquer strategy. Split the 4 balls into three small sized groups.
But how best to split the suspect group of 4 balls into three groups? Can you leave aside 2 balls and weigh 1 ball against 1 in the second weighing? In the worst case, when the pans are balanced, you will still have 2 balls of unknown nature of weight. Solution will be impossible.
only one way to go.
You will leave aside only 1 ball in second weighing.
To make the number balls being weighed EVEN, you will add 1 good ball from the group of 8 good balls. You will weigh then 2 suspect balls against 1 good ball and 1 suspect ball in the second weighing.
The following schematic shows this second weighing combination. The suspect balls are shaded, left pan suspect ball marked L and right pan suspect ball marked R. The suspect ball left aside is also shaded but not marked as its nature will stay unknown if the pans balance in this weighing.
The possible outcomes of this weighing are,
3.1. Pans are unbalanced: One pan goes up and the other descends in the second weigh after the pans balanced in 1st weigh. This has two possibilities,
3.1.1. Left Pan with good ball descends and right pan with 2 suspect ball goes up.
3.1.2. Left pan with good ball goes up and right pan with 2 suspect balls descends.
3.2. Pans are again balanced in second weigh after the pans balanced in the first weigh.
Conclusion and action for result 3.2 is simple—the left aside single suspect ball must be the fake. Just weigh it against a good ball and you will know whether it is the lighter or the heavier ball.
Conclusion for Result 3.1.1. When the left pan with 1 good ball descends: Either the L- ball in left pan is heavier or the two R-balls in the right pan are lighter. If the pan with 1 good ball goes up, the conclusion will be the reverse and the actions will be similar.
So we will analyze only the result 3.1.1 when the left pan with 1 good ball descends in 2nd weighing.
Solving for Result 3.1.1 of 12 balls puzzle: When pans are equally balanced in 1st weighing and pan with 1 good ball descends in 2nd weighing
Possible conclusions are—if the left pan with 1 good ball descends,
Either the single suspect ball (shaded and marked L) with the good ball is heavier,
Or, the two right pan balls (shaded and marked R) contain the fake lighter one.
You are left with 3 suspect balls split into two groups of balls numbering 1 and 2.
You are not totally sure yet, but,
One weighing is still left with you and number of suspect balls is 3—1 possibly heavier fake or one of the other two possibly lighter fake.
Again, carry out worst case consequence analysis.
In 3rd try, if you weigh the suspect heavier ball against a suspect lighter ball from right pan, and the pan with the suspect lighter ball goes up, you cannot be sure which one of the two is the fake ball. So this weighing combination is rejected.
The only possible 3rd weighing combination is then to weigh 1 suspect lighter fake ball in right pan against the other right pan ball of the same nature leaving aside 1 suspect heavier ball from left pan. Whichever pan goes up has the fake lighter ball (as both balls weighed are suspect lighter).
The weighing combination is shown.
Final Conclusions from 3rd weighing for Result 3.1.1 of 12 balls puzzle:
There can only be two possible conclusions,
- The pan going up has the fake lighter ball.
- If the pans balance, the single suspect ball in left pan earlier is the fake heavier ball.
First is shown in the figure left and the second in figure right.
You have not finished solving the puzzle yet—you’ll have to analyze the possibilities for the Result 1 of first weighing.
Solution for Result 1 of 12 balls puzzle: Exploring 1st result of 4 + 4 first weighing when left pan with 4 balls went up
The schematic of Result 1 of first weighing is shown below for convenience.
Conclusion: All 8 balls are suspect. Specifically,
1.1. Either the left side up-going 4 ball pan contains the fake lighter ball,
1.2. Or, The right side descending 4 ball pan has the fake heavier ball.
All 8 suspected fake balls are shaded with up-going left pan balls marked L and descending right pan balls marked R for ease of later reference.
Analysis and weighing decision for the second weighing for two groups of 4 suspect balls each
For second weighing, without hesitation you will apply the strategy of best way to divide and conquer by dividing 8 balls into 3 groups,
Divide 8 balls into 3 groups numbering as near to each other as possible. The groups will have 3, 3, and 2 balls.
An immediate question arises, will you simply weigh 3 left pan L balls against 3 right pan R balls?
This is infeasible as in the worst outcome you will still be left with 6 suspect balls with 1 weighing quota left.
What should you do?
Answer is clear.
You must improve the worst possible outcome of the 2nd weighing.
Again, answer is not difficult to reach,
Instead of weighing 3 against 3, you will keep aside 3 balls and add 1 good ball to the rest 5 to weigh 3 against 3—number of suspect balls in worst case will come down to 5.
That makes good sense, and you make a firm decision—you will keep aside 3 suspect balls.
Which ones to keep aside? Will you keep aside 3 L balls (or 3 R balls) of the same nature?
Being already experienced in finding the answer in 3rd weighing with 1 L ball and 2 R balls turning out to be finally suspect (in third weighing scheme for result 3.1.1), you take the next decision with confidence,
You will keep aside 1 L ball and 2 R balls.
3 L balls and 2 R balls will be left.
Add 1 good ball to make the number for weighing even—3+3.
This will further simplify the worst outcome of last weigh, because, you may still have 5 suspect balls, but these will have a more balanced combination of nature of balls—3 L balls and 2 R balls (not 4 L balls and 1 R ball).
This is the powerful general technique of increasing symmetry or decreasing asymmetry in nature of balls being weighed coming into play.
Drawn from experience, we have found the truth to be,
If you can increase the symmetry in a problem by an action, that action will be your most promising action for solving the problem.
Note: In solving a hard puzzle and many difficult math problems this abstract technique is used resulting in assured quick solution.
Okay, coming back to track with our puzzle, after deciding about mix of balls to be left aside, you have 3 L balls and 2 R balls.
Think for a moment that if you simply weigh 3 L in left pan with 2 R plus 1 good ball in right pan, the worst outcome will still be these 5 suspect balls remaining suspect. No progress at all.
How to improve this situation then?
STOP and think of a way out.
Yes, of course. You could again,
Apply the same technique of increasing symmetry in nature of balls in the two groups weighed simply by EXCHANGING 1 L ball with 1 R ball.
This is the maximum that you can do. The weighing combination for 2nd weighing will then be,
Left pan: 2 L balls + 1 R ball. Right pan: 1 R ball + 1 L ball + 1 good ball.
The 2nd weighing combination for Result 1 of first weighing is shown.
Three possible outcomes, actions and conclusions for solving the puzzle are now easy to visualize,
1.1 Left pan goes up: The fake ball is in the two suspected lighter L balls in left pan or the single suspected heavier R ball in right pan. You will now easily find the fake ball by using the method in solution for result 3.1.1.
1.2 Right pan goes up: The fake ball will either be the suspect lighter L ball in right pan or the suspect heavier R ball in the left pan. Weigh any of these two remaining suspects against a good ball and you will know the answer.
1.3 The two pans balance perfectly: The fake ball must be in the 3 kept aside balls. It will be one of the two suspected heavier R balls, or the suspected lighter L ball. Solve as before using method of solution for result 3.1.1.
No need to elaborate any further. Right?
You may think so at first. But the steps you followed never deviated from a systematic and logical path.
For key breakthroughs, you have used powerful general problem solving techniques as well.
With such a step by step reason-bound solution method, if you just remember a few key ideas, you could easily repeat the process yourself without referring to this long solution, made longer by detailed explanation.
Actual solution using the techniques should take maximum of 30 minutes.
You may discover new ways to solve this hard puzzle and should understand the reasons for the actions taken if you try to solve the puzzle on your own.
Know how to solve difficult problems easily without wasting time on random attempts
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