Abnormal Ball, I found you die: Easy Algorithms in pseudo-code

1. Description:

The question this article focuses on is a very old intellectual question. However, the thoughts and algorithms solving it is still worth thinking.

2. Question:

There are 12 identical balls in the desk except for one abnormal ball inside having different weight. Besides, apart from the 12 balls, there is scale on the desk. But the scale is without counterweight. So, this scale can only be used for telling which one is heavier or lighter between two items. The question is how to locate the abnormal ball by 3 times weighing for those balls.

Remark: in this question, we do not know whether the abnormal ball is lighter or heavier than the normal.

3. Solution:

Notation: we denote WR as the weighing relationship between any two different items

In this place,

if A is equal to B, we denote WR(A,B) = equal;

if A is heavier than B, we denote WR(A,B) = down;

if A is lighter than B, we denote WR(A,B) = up;

First we need to divide those 12 balls into 3 groups: A = {a1,a2,a3,a4}; B = {b1,b2,b3,b4}; C = {c1,c2,c3,c4};
1st Weighing: A B
	if WR(A,B) = equal
		the abnormal ball should be in C
		2st Weighing {a1,a2,a3},{c1,c2,c3}
			if WR({a1,a2,a3}, {c1,c2,c3}) = equal
				c4 is the abnormal ball
			else
				the abnormal ball should be among {c1,c2,c3}
				note down the WR of {a1,a2,a3} and {c1,c2,c3}
				Here, we consider the case: the abnormal ball is heavier,i.e. WR({a1,a2,a3},{c1,c2,c3}) = up
				3st Weighing c1 c2
					if WR(c1,c2) = equal
						c3 is the abnormal ball
					else
						whatever the heavier one between c1 and c2 is the abnormal
					end
			end
	else
	the abnormal ball should be among {a1,a2,a3,a4,b1,b2,b3,b4}
	note down the WR of A = {a1,a2,a3,a4} and B = {b1,b2,b3,b4}
	in this place we denote the case is WR(A,B) = up for convenient statement
	Reconstruct: 
	divide those balls into three groups including X = {a1,a2,b1}, Y = {a3,b2,b3}, and Z = {b4,a4,c1}; 
	c1 is a normal ball as checked before
	2st Weighing: X Z
		if WR(X,Z) = equal
			the abnormal ball is in Z
			3st Weighing: c1 a4
				if WR(c1,a4) = equal
					the abnormal ball is b4
				else
					the abnormal ball is a4
				end
		end
		if WR(X,Z) = up
			the abnormal ball should be among a1, a2 and b4
			3st Weighing: a1 a2
				if WR(a1,a2) = up
					a1 is the abnormal ball
				end
				if WR(a1,a2) = down
					a2 is the abnormal ball
				end
				if WR(a1,a2) = equal
					b4 is the abnormal ball
				end
		end
		if WR(X,Z)=down
			the abnormal ball should be between b1 and a4
			3st Weighing b1 c1
				if WR(b1,c1) = equal
					a4 is the abnormal ball
				else
					b1 is the abnormal ball
				end
		end
	end
Finally, we can obtain the abnormal ball within 3 moves.

Tips: 

There is unique abnormal ball among the 12 balls. 

The weight relationship among balls does not change during procedure. 

(Applied Math is dreaming of coding... ...)

Author: Clarence