TADM2E 4.46 - Revision history

Matt: Undo revision 833 by Ufsuraifs (talk)

2020-07-23T14:46:17Z

Undo revision 833 by Ufsuraifs (talk)

Ufsuraifs: Blanked the page

2020-07-17T02:50:05Z

Blanked the page

HappyEngineer at 23:20, 13 June 2016

2016-06-13T23:20:27Z

Bogdandumitrescu at 22:17, 8 September 2015

2015-09-08T22:17:13Z

Dentin: Remove unnecessary header

2015-04-17T00:40:23Z

Remove unnecessary header

Neonstalwart at 05:23, 19 February 2015

2015-02-19T05:23:25Z

Neonstalwart: added an alternative solution

2015-02-19T05:21:29Z

added an alternative solution

Algowikiadmin: Recovering wiki

2014-09-11T18:23:55Z

Recovering wiki

Algowikiadmin: Recovering wiki

2014-09-11T18:14:10Z

Recovering wiki

New page

The first solution was so wrong I was compelled to answer this question correctly.
The original poster needs to have his posting privileges SEVERELY limited for the sake of limiting stupid activites whenever possible.

Question: You are given 12 coins. One of them is heavier or lighter than the rest. Identify
this coin in just three weighings

Solution:
Number the coins 1 through 12 and divide them coins into 4 sets of 3...

There are multiple comparison sets possible. This is an acceptable template to find a few of them.
(This template is NOT definitive, there are other solutions that don't follow this template)

Compare (set 1 & 1st coin from set 4) against (set 2 + 2nd coin from set 4)
Compare (set 1 & 2nd coin from set 4) against (set 3 + 1st coin from set 4)
Compare (1st coin from each set) against (3rd coin from each set)

A more concise example:

Compare 1 2 3 10 against 4 5 6 11
Compare 1 2 3 11 against 7 8 9 10
Compare 1 4 7 10 against 3 6 9 12

Each weighing can have 3 possible outcomes: Left Heavy, Right Heavy, or Balanced (L,R or B)

Build a truth table to interpret outcomes...many outcomes are not possible.
Note: THE TABLE VALUES ARE DERIVED FROM THE CHOSEN COMPARISON SETS!

outcome: fake coin:
l l l 1 is heavy
r r r 1 is light
l l b 2 is heavy
r r b 2 is light
l l r 3 is heavy
r r l 3 is light
r b l 4 is heavy
l b r 4 is light
r b b 5 is heavy
l b b 5 is light
r b r 6 is heavy
l b l 6 is light
b r l 7 is heavy
b l r 7 is light
b r b 8 is heavy
b l b 8 is light
b r r 9 is heavy
b l l 9 is light
r l r 10 is heavy
l r l 10 is light
l r b 11 is heavy
r l b 11 is light
b b r 12 is heavy
b b l 12 is light

There are multiple comparison set possibilities, each with their own comparison table solution.

← Older revision		Revision as of 02:50, 17 July 2020
Line 1:		Line 1:
−	~~Question: You are given 12 coins. One of them is heavier or lighter than the rest. Identify~~
−	~~this coin in just three weighings~~

−	~~Solution:~~
−	~~Number the coins 1 through 12 and divide them coins into 4 sets of 3...~~
−
−	~~There are multiple comparison sets possible. This is an acceptable template to find a few of them.~~
−	~~(This template is NOT definitive, there are other solutions that don't follow this template)~~
−
−	~~Compare (set 1 & 1st coin from set 4) against (set 2 + 2nd coin from set 4)~~
−	~~Compare (set 1 & 2nd coin from set 4) against (set 3 + 1st coin from set 4)~~
−	~~Compare (1st coin from each set) against (3rd coin from each set)~~
−
−	~~A more concise example:~~
−
−	~~Compare 1 2 3 10 against 4 5 6 11~~
−	~~Compare 1 2 3 11 against 7 8 9 10~~
−	~~Compare 1 4 7 10 against 3 6 9 12~~
−
−	~~Each weighing can have 3 possible outcomes: Left Heavy, Right Heavy, or Balanced (L,R or B)~~
−
−	~~Build a truth table to interpret outcomes...many outcomes are not possible.~~
−	~~Note: THE TABLE VALUES ARE DERIVED FROM THE CHOSEN COMPARISON SETS!~~
−
−	~~outcome: fake coin:~~
−	~~l l l 1 is heavy~~
−	~~r r r 1 is light~~
−	~~l l b 2 is heavy~~
−	~~r r b 2 is light~~
−	~~l l r 3 is heavy~~
−	~~r r l 3 is light~~
−	~~r b l 4 is heavy~~
−	~~l b r 4 is light~~
−	~~r b b 5 is heavy~~
−	~~l b b 5 is light~~
−	~~r b r 6 is heavy~~
−	~~l b l 6 is light~~
−	~~b r l 7 is heavy~~
−	~~b l r 7 is light~~
−	~~b r b 8 is heavy~~
−	~~b l b 8 is light~~
−	~~b r r 9 is heavy~~
−	~~b l l 9 is light~~
−	~~r l r 10 is heavy~~
−	~~l r l 10 is light~~
−	~~l r b 11 is heavy~~
−	~~r l b 11 is light~~
−	~~b b r 12 is heavy~~
−	~~b b l 12 is light~~
−
−	~~There are multiple comparison set possibilities, each with their own comparison table solution.~~
−
−	~~----~~
−
−	~~A simpler solution #2:~~
−
−	* put 6 coins on each side of the scale, one side will be heavier.
−	* use the heavier side from the first weighing and put 3 coins on each side of the scale.
−	* using the heavier side from the 2nd weighing, pick 2 coins and put 1 on each side of the scale.
−
−	~~If the scale is balanced then the coin you didn't weigh is the heavier one. Otherwise, the scale will show which one of the other 2 is the heavy coin.~~
−
−	~~----~~
−
−	~~Since we do not know if the faulty coin is heavier or lighter , the soluition #2 is not correct~~
−
−	~~The only solution is solution nr 1, for which we can also use a binary tree~~
−
−	~~----~~
−
−	I found the explanation at mathforum.org/library/drmath/view/55618.html to be helpful. The key is to build a truth table, but how that table was built is a little tricky because you can't just write out all possible combinations of results and use that. I tried writing a truth table and found that I had to swap some values around to make sure that each measurement step had exactly 4 on each side.

← Older revision		Revision as of 23:20, 13 June 2016
Line 66:		Line 66:

	The only solution is solution nr 1, for which we can also use a binary tree		The only solution is solution nr 1, for which we can also use a binary tree
		+
		+	----
		+
		+	I found the explanation at mathforum.org/library/drmath/view/55618.html to be helpful. The key is to build a truth table, but how that table was built is a little tricky because you can't just write out all possible combinations of results and use that. I tried writing a truth table and found that I had to swap some values around to make sure that each measurement step had exactly 4 on each side.

@@ Line 53: / Line 53: @@
 ----
-A simpler solution:
+A simpler solution #2:
 * put 6 coins on each side of the scale, one side will be heavier.
@@ Line 60: / Line 60: @@
 If the scale is balanced then the coin you didn't weigh is the heavier one.  Otherwise, the scale will show which one of the other 2 is the heavy coin.

← Older revision		Revision as of 00:40, 17 April 2015
Line 1:		Line 1:
−	~~The first solution was so wrong I was compelled to answer this question correctly.~~
−	~~The original poster needs to have his posting privileges SEVERELY limited for the sake of limiting stupid activites whenever possible.~~
−
	Question: You are given 12 coins. One of them is heavier or lighter than the rest. Identify		Question: You are given 12 coins. One of them is heavier or lighter than the rest. Identify
	this coin in just three weighings		this coin in just three weighings

← Older revision		Revision as of 05:21, 19 February 2015
Line 1:		Line 1:
		+	A simple solution:
		+
		+	* put 6 coins on each side of the scale, one side will be heavier.
		+	* use the heavier side from the first weighing and put 3 coins on each side of the scale.
		+	* using the heavier side from the 2nd weighing, pick 2 coins and put 1 on each side of the scale.
		+
		+	If the scale is balanced then the coin you didn't weigh is the heavier one. Otherwise, the scale will show which one of the other 2 is the heavy coin.
		+
		+	----
		+
		+
	The first solution was so wrong I was compelled to answer this question correctly.		The first solution was so wrong I was compelled to answer this question correctly.
	The original poster needs to have his posting privileges SEVERELY limited for the sake of limiting stupid activites whenever possible.		The original poster needs to have his posting privileges SEVERELY limited for the sake of limiting stupid activites whenever possible.

← Older revision	Revision as of 14:46, 23 July 2020
Line 1:	Line 1:
	+	Question: You are given 12 coins. One of them is heavier or lighter than the rest. Identify
	+	this coin in just three weighings

	+	Solution:
	+	Number the coins 1 through 12 and divide them coins into 4 sets of 3...
	+
	+	There are multiple comparison sets possible. This is an acceptable template to find a few of them.
	+	(This template is NOT definitive, there are other solutions that don't follow this template)
	+
	+	Compare (set 1 & 1st coin from set 4) against (set 2 + 2nd coin from set 4)
	+	Compare (set 1 & 2nd coin from set 4) against (set 3 + 1st coin from set 4)
	+	Compare (1st coin from each set) against (3rd coin from each set)
	+
	+	A more concise example:
	+
	+	Compare 1 2 3 10 against 4 5 6 11
	+	Compare 1 2 3 11 against 7 8 9 10
	+	Compare 1 4 7 10 against 3 6 9 12
	+
	+	Each weighing can have 3 possible outcomes: Left Heavy, Right Heavy, or Balanced (L,R or B)
	+
	+	Build a truth table to interpret outcomes...many outcomes are not possible.
	+	Note: THE TABLE VALUES ARE DERIVED FROM THE CHOSEN COMPARISON SETS!
	+
	+	outcome: fake coin:
	+	l l l 1 is heavy
	+	r r r 1 is light
	+	l l b 2 is heavy
	+	r r b 2 is light
	+	l l r 3 is heavy
	+	r r l 3 is light
	+	r b l 4 is heavy
	+	l b r 4 is light
	+	r b b 5 is heavy
	+	l b b 5 is light
	+	r b r 6 is heavy
	+	l b l 6 is light
	+	b r l 7 is heavy
	+	b l r 7 is light
	+	b r b 8 is heavy
	+	b l b 8 is light
	+	b r r 9 is heavy
	+	b l l 9 is light
	+	r l r 10 is heavy
	+	l r l 10 is light
	+	l r b 11 is heavy
	+	r l b 11 is light
	+	b b r 12 is heavy
	+	b b l 12 is light
	+
	+	There are multiple comparison set possibilities, each with their own comparison table solution.
	+
	+	----
	+
	+	A simpler solution #2:
	+
	+	* put 6 coins on each side of the scale, one side will be heavier.
	+	* use the heavier side from the first weighing and put 3 coins on each side of the scale.
	+	* using the heavier side from the 2nd weighing, pick 2 coins and put 1 on each side of the scale.
	+
	+	If the scale is balanced then the coin you didn't weigh is the heavier one. Otherwise, the scale will show which one of the other 2 is the heavy coin.
	+
	+	----
	+
	+	Since we do not know if the faulty coin is heavier or lighter , the soluition #2 is not correct
	+
	+	The only solution is solution nr 1, for which we can also use a binary tree
	+
	+	----
	+
	+	I found the explanation at mathforum.org/library/drmath/view/55618.html to be helpful. The key is to build a truth table, but how that table was built is a little tricky because you can't just write out all possible combinations of results and use that. I tried writing a truth table and found that I had to swap some values around to make sure that each measurement step had exactly 4 on each side.

@@ Line 1: / Line 1: @@
 The first solution was so wrong I was compelled to answer this question correctly.
   The original poster needs to have his posting privileges SEVERELY limited for the sake of limiting stupid activites whenever possible.
@@ Line 64: / Line 53: @@
 There are multiple comparison set possibilities, each with their own comparison table solution.

@@ Line 11: / Line 11: @@
 (This template is NOT definitive, there are other solutions that don't follow this template)
-  Compare   (set 1 &amp; 1st coin from set 4)  against  (set 2 + 2nd coin from set 4)
+  Compare   (set 1 & 1st coin from set 4)  against  (set 2 + 2nd coin from set 4)
-  Compare   (set 1 &amp; 2nd coin from set 4)  against  (set 3 + 1st coin from set 4)
+  Compare   (set 1 & 2nd coin from set 4)  against  (set 3 + 1st coin from set 4)
   Compare   (1st coin from each set)       against  (3rd coin from each set)