Find the fake ball in 3 attempts—12 balls weighing puzzle
12 balls puzzle with 1 fake ball
You have 12 balls and a two-pan balance with no weights.
All the balls look same and each ball weighs same except for one, which is the fake ball, heavier or lighter than the other 11 balls.
Find the fake ball and its nature, lighter or heavier, by weighing the balls 3 times using the balance scale.
Recommended time: 45 minutes.
Give it a good try before going ahead.
Can you find all possible ways to solve the puzzle?
No time limit for this one as well as no solution from us. It's only for you, if you are curious.
Solution to the 12 balls puzzle with 1 fake ball
To decide the first weigh combination, apply how best to divide and conquer strategy and use principle of symmetry
What should be the best plan for first weighing?
Should you first weigh 6 against 6 balls? That would result in 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 seems to be too complex.
Leaving this more complicated path of action with more future complications, adopt the action that would have lesser number of suspect balls after the weighing,
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 would be at most 10 after this weighing.
Next question arises immediately,
How best can the 12 balls be split up into 3 groups?
The task is then,
Two groups of equal number of balls are to be weighed against each other with a third group of balls kept aside.
First enumerate all possible ways 12 can be split into 3 numbers in this weighing case. This is exhaustive enumeration technique. You are using number combinations possible with the condition of two numbers equal to each other.
The ways to split 12 into three such groups would be,
You stop at the combination of 3rd left aside group size of 6, because the next combination $12=2+2+8$ would have left aside group size as 8 balls, too large a number to handle in 2 weighing with not even the nature of fake ball nature known. This would be more difficult than having 12 suspect balls split into two groups of 6 after first weighing.
There is a trade-off here. Let us be clear about the trade-off,
If the number of balls selected for 1st weighing is larger, the left aside group of balls will be fewer, and vice versa.
Why do we call it a trade-off?
It is because of the nature of results of 1st weigh,
- If the two pans perfectly balance, the fake ball must be in 3rd group of left aside balls with not even the fake ball lighter or heavier known. Answer would be harder to find with this number of left aside suspect balls larger.
- If one of the pans goes down, all the balls weighed are suspect, but divided into two groups containing lighter or heavier fake ball. Larger this total number of balls, again answer would be more difficult to reach.
Easiest option would 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 would be to follow a general principle of problem solving drawn from experience—how best to divide and conquer,
Divide the enemy in smallest sized groups with group sizes as nearly equal to each other as possible.
Following this principle, all the three group sizes to be kept as small as possible, that is—12 is to be split into three groups each with 4 balls,
This strategy also follows principle of symmetry,
If you increase the symmetry in the problem by an action, that would be your MOST PROMISING ACTION.
Thus you decide the most promising course of action for quicker solution as to,
Weigh first, 4 balls against 4 others, with 4 balls kept aside.
The figure below shows the first weighing combination.
There can be three results or outcomes of your weighing.
Result 1. First result of 1st weighing—the right pan goes down and left pan goes up
Conclusion: All eight balls are suspect. Specifically,
1.1. Either the left side up-going four balls contain the fake lighter ball, or
1.2. The right side down-going pan contains the fake heavier ball.
The picture below shows the situation.
Result 2. Second possibility—the left pan goes down:
Conclusion: All 8 balls are suspect. Specifically,
2.1. Either the left side down-going 4 balls contain the fake heavier ball, or
2.2. The right side up-going pan contains the fake lighter ball.
Essentially, these two results would need very similar actions as the results 1.1 and 1.2.
So you won't analyze these two results further—analyzing results 1.1 and 1.2 should be enough for reaching the solution for this result also.
Result 3. Third possibility—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 1st 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 because after this second weighing, you would still be left with 4 suspect balls though divided into two groups of opposite nature, with only 1 weighing attempt left. Solution would be impossible in this way.
You may verify whether this conclusion is correct.
So to increase the number of identified good balls and decrease the number of suspect balls in this second step,
You have to split the problematic group of 4 balls into three groups.
Again you are applying how best to divide and conquer strategy.
Look ahead and think forward to imagine what would happen in the worst case when you leave aside 2 balls and weigh 1 against 1. You are doing worst case analysis coupled with consequence analysis that are general management and problem solving techniques.
If the scale is balanced you would be left with 10 good balls but 2 balls of unknown type. Conclusion,
It would be impossible to find the fake ball and its nature among 2 untested balls in 1 weighing with this approach also.
You can leave aside only 1 ball in second weighing.
And to make the number balls being weighed EVEN,
Simply add 1 good ball from the group of 8 good ones and weigh 2 suspect against 1 good ball and 1 suspect ball.
Final decision: Second weighing combination for 3rd result:
Add 1 good ball from 8 to the 4 suspect balls, keep aside 1 suspect ball and weigh 1 good ball plus 1 suspect ball against 2 suspect balls. Out of 8, seven good balls are not used.
The following schematic shows this weighing combination. The suspect balls are shaded orange, left pan suspect ball marked L and right pan suspect ball marked R. The suspect ball left aside is also shaded orange but not marked as its nature would still remain unknown if the pans balance in this weighing.
The possible results are,
3.1. Pans are unbalanced: One pan goes up and the other goes down in the second weigh after the pans balanced in 1st weigh. This has two possibilities,
3.1.1. Left Pan with good ball goes down 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 goes down.
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 would know whether it the lighter or the heavier ball.
Conclusion for Result 3.1.1. If the left pan with 1 good ball goes down: 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 would be the just the opposite and the actions would be very similar.
So we would analyze only the result 3.1.1 that the left pan with 1 good ball went down in 2nd weigh.
Solving for Result 3.1.1. When pans are equally balanced in 1st weighing AND pan with 1 good ball goes down in 2nd weighing
Possible conclusions are—if the left pan with 1 good ball goes down,
Either the single suspect ball with the good ball (orange shaded and marked L) is heavier,
Or, the two right pan balls (orange shaded and marked R) contain the fake lighter one.
You are left with 3 suspect balls split into two nature-identified groups of number of balls numbering 1 and 2.
This is the best possible outcome after second weighing. You have reached this outcome by using systematic analytical method and techniques—not randomly at all.
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 do worst case consequence analysis.
In 3rd attempt, 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 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).
If the two pans balance, the suspect heavier left out ball is indeed the fake heavier ball.
Observe how in this case also we have used the how best to divide and conquer strategy along with gradually narrowing down reasoning by series of question answer and analysis or QAA technique.
Let us see the this third weighing combination visually.
Solution for Result 3.1.1: Third weighing scheme for result 3.1.1
Weigh two suspect balls in right pan against each other—Orange shaded R against the other orange shaded R. One of these is the fake lighter ball if one side goes up. If the two pans balance, the left out suspect ball is the fake heavier ball.
This is shown in the following schematic.
Final Conclusions from 3rd weighing for Result 3.1.1:
There can only be two possible conclusions,
1. The pan going up contains the fake lighter ball.
2. 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 as below.
You have not finished solving the puzzle yet—you'll have to analyze the possibilities for the first result 1 of first weighing.
Solving for Result 1 of 12 balls puzzle: Exploring 1st result of 4 + 4 first weigh when left pan with 4 balls went up
The schematic of Result 1 of first weighing is shown below for convenience.
And the CONCLUSION: All 8 balls are suspect. Specifically,
1.1. Either the left side up-going 4 ball pan contain the fake lighter ball,
1.2. Or, The right side down-going 4 ball pan contains the fake heavier ball.
All 8 suspected fake balls are shaded orange with up-going left pan balls marked L and down-coming 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—a total of 8 suspect balls
For second weighing, without hesitation you would apply the strategy of best way to divide and conquer by dividing 8 into 3 groups,
Divide 8 balls into 3 groups of number strengths as near to each other as possible—3+3+2.
An immediate question arises, would you simply weigh 3 left pan L balls against 3 right pan R balls?
It is a clearly infeasible combination as in the worst outcome of this 2nd weigh plan, if the pan with 3 L balls goes up again, you would still be left with 6 suspect balls and 1 remaining weighing attempt. This plan won't give you the solution.
What should you do?
Answer is clear.
You must improve the worst possible outcome of the 2nd weighing.
Again, answer is not difficult to arrive at,
Instead of weighing 3 against 3, you would 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 would keep aside 3 suspect balls.
Which ones to be kept aside? Would you keep aside 3 L balls (or 3 R balls) all of 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 would keep aside 1 L ball and 2 R balls.
3 L balls and 2 R balls would be left.
Add 1 good ball to make the number for weighing even—3+3 (the suspect ball breakup will be 2+3+3, instead of 3+3+2).
This would further simplify last weighing worst possible outcome because, you may still have 5 suspect balls, but these would have a more balanced combination of nature balls—3 L balls and 2 R balls (instead of 4 L balls and 1 R ball that would have more complex worst outcome.
This is the very 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 would be your most promising action for solving the problem.
Note: In solving a hard puzzle and many difficult math problems this abstract technique has been used resulting in assured quick solution.
Okay, coming back to track with our puzzle, after taking the decision 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 in right pan, the worst outcome would still be these 5 suspect balls remaining suspect. No progress at all.
How to improve this situation then?
If you are still with us this long, 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 at this point, you feel. The weighing combination for 2nd weighing would 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 below.
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 would 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 would 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 would be one of the two suspect heavier R balls or the suspect lighter L ball. Solve as before using method of solution for result 3.1.1.
No need to elaborate any further. Right?
May seem so at first. But the steps you followed never deviated from a systematic, logical and the approach of selecting most promising 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 would easily be able to repeat the process yourself without referring to this long solution, made longer for detailed explanation.
Actual solution using the techniques should take maximum of 30 minutes.
Lastly, if you seriously explore to find the answer yourself, you may discover new ways to solve this hard puzzle, and also you would understand the reasons and the needs of the actions taken and the techniques applied.
Puzzles you may enjoy
Logic analysis puzzles
River crossing puzzles
Ball weighing puzzles
Find the fake ball among 12 identical balls in 3 weighing hard puzzle with solution