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Guest Message by DevFuse

Weighing IV.

35 replies to this topic

#11 amw8181

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Posted 20 September 2007 - 04:30 PM

To clarify:

The solution of weighing 6 and 6, 3 and 3, and then finding the oddball doesn't work. When you get one group to be heavier and one group to be lighter, you still don't know whether the oddball is heavier or lighter-- if it's heavier, you want to start weighing the heavy group, if it's lighter, you want to start weighing the lighter 6.
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#12 osoriocj

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Posted 23 October 2007 - 06:39 AM

Separate the balls in three groups of four and place a letter on them, from A to L.
The known balls will be called "X".

So, here we go:

ABCD EFGH IJKL

The 1st weighing.
Use the first two groups.
There will be three results:

1) ABCD EFGH (I) => Group IJKL contains "The Ball"; ABCD EFGH will be called XXXX XXXX

or
2) ABCD ____ (I) => IJKL will be called XXXX
_______ EFGH
or
3) ____ EFGH (I)
_ ABCD
Obviously, the solution for case 3 is the same for case 2.

Solution for Case 1.

First result:
1) ABCD EFGH (I) => Group IJKL contains "The Ball"

4) IJK XXX (II) => L<>X

5) L _ (III) => L is the lightest ball
___ X
or
6) _X (III) => L is the heaviest ball
__L

Second result:
1) ABCD EFGH (I) => Group IJKL contains "The Ball"

7) IJK ___ (II) => IJK is lighter than the others
_____ XXX

8) I J (III) => K is the lightest ball

or
9) I __ (III) => K is the lightest ball
___ J
or
10) _J (III) => K is the lightest ball
___I

Third result:
1) ABCD EFGH (I) => Group IJKL contains "The Ball"

11) ___ XXX (II) => IJK is heavier than the others
___ IJK

12) Use solutions 8,9 and 10 changing "lightest" for "heaviest".

Solution for Case 2.

First result:
2) ABCD ____ (I) => IJKL will be called XXXX
_______ EFGH

13) AFGH EXXX (II) => BCD is lighter than the others

14) Use solutions 8,9 and 10 to find "The Ball".

Second result:
2) ABCD ____ (I) => IJKL will be called XXXX
_______ EFGH

15) AFGH ____ (II) => A<>E
________ EXXX

16) Use solutions 5 and 6 to find "The Ball".

Third result:
2) ABCD _____ (I) => IJKL will be called XXXX
_______ EFGH

17) _____ EXXX (II) => FGH is heavier than the others
___ AFGH

18) Use solution 12 to find "The Ball".

That's all.
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#13 erandaa

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Posted 21 November 2007 - 09:44 PM

How about you just put all the balls in one side then balance the other side with weights. Then take each ball and put it back till the “ball side of the scale goes up? :-)
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#14 al_rawi

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Posted 22 November 2007 - 04:18 PM

I have another solution with three steps that follows similar logic to the author/admin:
using the same terminology of x>, x<, x? and x:
the first step is the same, but I will repeat for completeness!
put 1? 2? 3? 4? on the left, 5? 6? 7? 8? on the left,
if there is equilibrium
____ we know that the odd ball is 9?, 10?, 11?, 12?
____ we now put 5, 6, 7 on the left and 9?, 10?, 11? on the right
____ if there is equilibrium
________ we know that 12? is the odd ball, we weigh it against, say 5
________ if it is lighter, then answer is 12<
________ else if it is heavier, then the answer is 12>
____ else if the left is heavier
________ we know that the odd ball is 9<, 10< or 11<
________ we now put 9< on the left and 10< on the right
________ if there is equilibrium then the answer is 11<
________ else if the left is heavier, then the answer is 10<
________ else if the right is heavier then the answer 9<
____ else if the right is heavier
________ (same as previous case with > substituting ________ we know that the odd ball is 9>, 10> or 11>
________ we now put 9> on the left and 10> on the right
________ if there is equilibrium then the answer is 11>
________ else if the left is heavier, then the answer is 9>
________ else if the right is heavier then the answer 10>
________ (this concludes the part that is the same as the author)
else if left is heavier
____ we know that 9,10,11,12 are not odd and that we have 1>, 2>,3>,4> and 5<,6<,7<,8<
____ we now put 1>,2>,5< on the left and 3>,4>,9 on the right
____ if there is equilibrium
________ we know that 1,2,3,4,5 are not odd so we are left with 6<,7<,8<
________ we now put 6< on the left and 7< on the right
________ if there is equilibrium then the answer is 8<
________ if left is heavier then the answer is 6<
________ if the right is heavier then the answer is 7<
____ else if the left is heavier then
________ we know that 3,4,5,6,7,8 are not odd, and we are left with 1>,2>
________ we now put 1> on the left and 2> on the right
________ if the left is heavier then the answer is 1>
________ if the right is heavier then the answer is 2>
____ else if the right is heavier then
________ we know that 1,2,6,7,8 are not odd, so we are left with 3>,4>,5<
________ we now put 3> on the left and 4> on the right
________ if we have equilibrium then the answer is 5<
________ if we left is lighter then the answer is 3>
________ if the right is lighter then the answer is 4>
else if right is heavier
____ the same as previous (wide) condition as left and right in this case are interchangeable (i.e. exchange 1,2,3,4 with 5,6,7,8 and follow same logic!)

I think this is correct, can anyone disqualify this?
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#15 firestar413

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Posted 30 November 2007 - 11:09 PM

Why Can't I delete my comment.
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#16 rookie1ja

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Posted 30 November 2007 - 11:22 PM

Why Can't I delete my comment.

only I can do that
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#17 BrainMagMo

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Posted 09 January 2008 - 07:36 AM

Weigh 4 balls on left, 4 balls on right, rest left off.

Gives three possibilities.

1) Left side is heavier (down).
2) Left side is lighter (up).
3) Equilibrium.
-
For 2), turn the scale around so we only have two possibilities.
-
For 1), either the left contains a heavy ball or the right contains a light ball, but the rest is definitely normal.
Replace the right side with the remainder. Weigh.
Two possibilities:
A) Left side is still heavier; it contains a heavy ball.
B)Equilibrium, right side contained light ball.

Therefore, we know there is a group or four and whether the ball is heavy or light.
Put two balls on the scale, weigh.
Same three possibilities, as in the beginning.
For 1, right ball is the light ball or left is the heavy ball, as determined. (3 weighs)
For 2, vice versa. (3 weighs)
For 3, weigh the remaining two balls, and you'll get 1) or 2). (4 weighs)
-
For 3) on first weigh, place two balls from the left-off group and weigh.
1. Left ball heavier
2. Right ball heavier, turn it around and make 1)
3. equilibrium.

For 1., replace right ball with a left-off ball and weigh.
A) Left still heavier, left ball is the heavy ball. (3 weighs)
B) Equilibrium. The taken off ball is the light ball. (3 weighs)

For 3., switch the other two and weigh, the proceed as 1. or 2. (4 weighs)

max: 4 weighs.

to osoriocj, 8), 9), 10) maken't sense.
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#18 grottylittlewanker

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Posted 25 January 2008 - 09:16 AM

All the solutions I read were too complex to understand or simplified and didn't sound legit.

Here is my solution:

1: Three groups of Four Balls (A,B,C)

2:
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#19 grottylittlewanker

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Posted 25 January 2008 - 09:33 AM

All the solutions I read were too complex to understand or simplified and didn't sound legit.

Here is my solution, in which, you need 4 tests to find the solution in any case, I wrote it out as simple as possible:

Three groups of Four Balls (1-4, 5-8, 9-12)

Test 1) weigh 1-4 vs 5-8

Test 2) weigh (either 1-4 or 5-8) vs 9-12

with these two tests we will know which group has the odd ball and whether it is heavy or light

Spoiler for Explanation

now the group that has the odd one, the balls are a, b, c, and d

Test 3) weigh a,b vs c,d

Test 4) weigh (a or b) vs (c or d)

this will let us know which ball it is

Spoiler for Explanation

hope this is an easier read than all the others
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#20 Icedragon_of_Solitude

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Posted 03 February 2008 - 08:44 AM

While my solution is a max of 4 steps, i think it is much simpler than some of the previously mentioned solutions.
(Answers are denoted with a * )

First you break the balls up into 4 groups of 3: A(1-3), B(4-6), C(7-9), D(10-12)

1: Test A and B

_1[1](A=B): The oddball is in C or D
__2: Test A and C

__2[1](A=C):The oddball is in D.
___3:Test 10 and 11
*__3[1](10=11): The oddball is 12
___3[2](10<>11):10 or 11 is the oddball
____4:Test 10 and 12
*___4[1](10=12): The oddball is 11
*___4[2](10<>12): The oddball is 10

__2[2](A<>C): The oddball is in C
___3: Test 7 and 8
*__3[1](7=8): The oddball is 9
___3[2](7<>8):7 or 8 is the oddball
____4:Test 7 and 9
*___4[1](7=9): The oddball is 8
*___4[2](7<>9): The oddball is 7

_1[2](A<>B): The oddball is in A or B
__2:Test A and C

__2[1](A=C):The oddball is in B
___3:Test 4 and 5
*__3[1](4=5): The oddball is 6
___3[2](4<>5):4 or 5 is the oddball
____4:Test 4 and 6
*___4[1](4=6): The oddball is 5
*___4[2](4<>6): The oddball is 4

__2[2](A<>C):The oddball is in A
___3:Test 1 and 2
*__3[1](1=2): The oddball is 3
___3[2](1<>2):1 or 2 is the oddball
____4:Test 1 and 3
*___4[1](1=3): The oddball is 2
*___4[2](1<>3): The oddball is 1
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