Find the heavier ball in 2 weighs—8 balls weight puzzle
In this first 8 balls weight puzzle, find the single heavier ball in 2 weighs. All 8 balls look alike and you are given a pan balance with no weights.
How many ways can you solve the puzzle?
Total time to solve: 12 minutes.
Try before going through the solution.
Solution to the 8 balls weight puzzle: find the heavier ball in 2 weighs
With no weights, you have the only choice of weighing a few balls in one pan against same number of balls on the other pan.
In simple words, the only way is to compare weights of equal number of balls.
Question is: How many balls should be in each pan for the first weighing?
Let’s see the consequence of weighing all 8 split into two groups of 4 balls each in the first weighing.
Whichever pan descends, that group of 4 balls must have the heavier ball.
You have only one weighing left. Will you again take the simple path of splitting the 4 balls into 2 groups of 2 balls each and weigh? What will be the result?
The heavier pan with 2 balls will have the ball you are trying to find, but you won’t find which one is heavier. You have finished your quota of weighing. The simple way won’t work.
You decide to avoid this difficult approach.
First stage analysis: Strategic approach of working backwards
Taking a more promising approach, you decide to check whether you can find the heavier among 3 balls identified as suspect in the first weighing itself.
You have reduced the number of suspect balls by 1. Don’t think how you will isolate a set of 3 suspect balls in the first weighing. Assume you have done it.
You are analyzing the end state of the second weighing. If you find that you can spot the heavier ball among the 3 suspect in 1 weighing, you’ll work back to analyze how to isolate the 3 suspect balls in the first weighing.
Okay, with 3 balls containing the heavier ball, naturally you will weigh 1 ball against 1 other from the three, keeping aside the 3rd ball.
Whichever pan descends has the heavier ball, and if both sides are balanced equally, the 3rd ball kept aside must be the heavier ball.
You could then identify the heavier ball if after the first weighing you could identify 3 balls having the heavier ball.
Now think of the first weighing. Your goal is to isolate 3 balls with the heavier ball.
Second stage: How to isolate 3 balls having the heavier ball from 8 balls in the first weighing
Now you know what you have to do. You will split 6 balls into two groups of 3 each. 2 balls are kept aside. The two groups of 3 balls are weighed against each other.
The technique of reducing number of balls in the active set of balls applied.
The possibilities are,
- Whichever pan descends has the heavier ball, and, you already know how to find the heavier ball among suspect 3 balls in 1 weighing.
- If the pans are equally balanced, the 2 balls left out must have the heavier ball. You know by now, finding the heavier ball among 2 balls will be dead easy.
All the rest six balls are good balls.
You will weigh one of the two balls kept aside against one good ball. If the pans are balanced, the single ball kept aside this time is the heavier one. Otherwise, the heavier ball is the one in the pan that descends.
Okay, first part of the puzzle is solved. Now you’ll have to think of the second part.
How many ways can the puzzle be solved?
Can we consider any other weighing combination?
Remember when you first met with the only other possible starting combination? That was when you attempted to weigh all 8 balls split into two groups of 4 balls each.
You ended up with the challenge of identifying the heavier ball among 4 suspect balls in 1 weighing. The suspect set will either be the 4 balls in the pan that descends or the 4 balls set aside when the pans are balanced.
How to identify 1 heavier ball among 4 suspect balls in 1 weighing
You have seen, weighing 2 balls against 2 in the second weighing fails. This time, you take a new approach of keeping 1 ball aside. Number of balls weighed reduces from 4 to 3, an improvement.
How to weigh 3 balls? Simple. Add 1 of the good balls to the pan with 1 ball and weigh 2 against 2 balls.
- The single ball with the good ball must be the heavier if the pan with the two descends.
- The left out fourth ball must be the heavier if the two pans are equally balanced.
- But, you will still have 2 suspect balls left with weighing quota exhausted if the pan with 2 suspect balls descends.
This won’t finally work.
With no other approach possible, you are sure now it is impossible to find 1 heavier ball among 4 in 1 weighing.
You have already found the only way to solve the puzzle.
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