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Weighing VII.


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28 replies to this topic

#21 MayLover24

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Posted 20 February 2008 - 01:04 AM

You have to do it 26 times otherwise you don't know which one is the heavy.
Example-
Say i said the answer was 1 time...9 groups which i weigh. Then if one is heavier then that is the group.
If there is equlibrum then the group unweighted is the heavier one.
BUT-this means i only know the group, not the specific ball.
So I have to weigh 26 times.
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#22 hendle

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Posted 28 February 2008 - 02:41 AM

You could actually accomplish this in just one weighing using the simplest scale possible. Imagine a teeter-totter perfectly centered on its fulcrum(simple balance scale)....you align the balls however you like on the scale . plugging a few numbers into a physics equation will tell you exactly which ball is the heaviest.Thinking Equation would look something like this B1 D1+B2 D2+B3 D3 +B4 D4= B5 D5+ B6 D6....etc...etc... Where B is the weight(or ball) and D is the distance it is from the fulcrum.
I could be wrong....sounded good to me though ::shrugs::

Edit: I'm really stoned, that wouldn't work...maybe a more advanced equation adding variable/constant for gravity, and the amount the scale traveled. Perhaps someone out there is experienced at physics and gimme advice on this method.

Edited by hendle, 28 February 2008 - 02:44 AM.

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#23 Dynamic

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Posted 07 March 2008 - 08:58 PM

You know the scale has three outcomes, L = left heavy, B = balanced, R = right heavy
Given that we know the odd ball is heavier than the rest, with a good choice of n comparisons, you can guarantee to identify the heavier ball out of 3^n balls. (3^n means three raised to the power of n, so 3^3 = 3 x 3 x 3 = 27 meaning only three weighings.

That's actually all you need to answer the question of how many trials to always be sure of the correct answer.

Rather than use the result of each weighing to choose which balls to weigh next, you can decide to always weigh one third of the total number of balls on each side of the scales and construct a truth table with all 27 unique combinations of three weighing trial results, meaning you can predetermine the sequence of trials you'll run rather than having to wait for the result of the first trial to determine the balls you weigh on each side at the second trial etc.

Here's an example truth table, working systematically through all 27 permutations and assigning a meaning, which determines whether the ball in question goes on the left side, the right side or is omitted for each trial.

Results in order:

RRR = ball 1 is heaviest
RRB = ball 2
RRL = ball 3
RBR = 4
RBB = 5
RBL = 6
RLR = 7
RLB = 8
RLL = 9
BRR = 10
BRB = 11
BRL = 12
BBR = 13
BBB = 14
BBL = 15
BLR = 16
BLB = 17
BLL = 18
LRR = 19
LRB = 20
LRL = 21
LBR = 22
LBB = 23
LBL = 24
LLR = 25
LLB = 26
LLL = 27


Other allocations of numbers to each outcome are possible of course.

Using this truth table, you can set up the three trials by looking down the list for L and R in the first column for the first trial, in the second column for the second trial, and the third for the third trial, which will have 9 balls (assuming balls are numbered 1 to 27) on each side and produce those results as follows:

ON LEFT SIDE OF SCALES ON RIGHT SIDE OF SCALES
FIRST TRIAL 19 20 21 22 23 24 25 26 27 1 2 3 4 5 6 7 8 9
SECOND TRIAL 7 8 9 16 17 18 25 26 27 1 2 3 10 11 12 19 20 21
THIRD TRIAL 3 6 9 12 15 18 21 24 27 1 4 7 10 13 16 19 22 25

_linenums:0'> ON LEFT SIDE OF SCALES ON RIGHT SIDE OF SCALESFIRST TRIAL 19 20 21 22 23 24 25 26 27 1 2 3 4 5 6 7 8 9 SECOND TRIAL 7 8 9 16 17 18 25 26 27 1 2 3 10 11 12 19 20 21THIRD TRIAL 3 6 9 12 15 18 21 24 27 1 4 7 10 13 16 19 22 25

If you had come up with a different truth table you'd have a different set of trials of 9 balls. The ball with result BBB never gets put on the scales. In my truth table, that's ball 14.

Pick a number from 1 to 27 and see what the results would be in sequence, then look that up in the truth table. E.g. ball 23 will give the results Left-side heavy, Balanced, Balanced. The sequence LBB in the truth table is next to the number 23.

You can also apply the same technique to the 12 balls, only one of which differs but you don't know if it's heavier or lighter than the normal balls. You then have 24 truth table items ('ball is light' and 'ball is heavy' outcomes for each of the 12 balls, which must be unique). Certain truth table results out of the 27 possible won't give you the full answer (e.g. BBB is useless because it doesn't tell you if the ball is heavier or lighter, but you only need 24 outcomes). You then need to ensure that you assign the numbers to ensure you have 4 balls on each side for each trial. You also know that if ball 1 being heavy is LRB then ball 1 being light must have result RLB (i.e. L and R swap over, but B stays the same because that ball wasn't weighed).

You could envisage balls with a 1 gram difference on scales with a 0.5 gram blob of glue stuck to the right-hand side of the scales and unable to be removed or counterbalanced. This eliminates the possibility of balancing evenly leaving only 2 possible outcomes of each trial weighing (L or R) and makes the problem into a 2^n problem, e.g. 4 trials then allows you to identify which one of 16 balls is heavier than the rest. This is a binary truth table (e.g. 1 = L, 0 = R) more familiar to computer programmers.
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#24 RiddleRookie

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Posted 12 March 2008 - 05:51 PM

I think if you split it in half 13/13 and 1 out. if scales equal its odd one out (so minimum weighs)... (First weigh)
next - take heavier side and repeat 6/6/ 1out.... (Second time)
next take heavier side and repeat 3/3 (third time)
next take heavier side and you'll have 1/1/1out... so if the scale is equal - its odd one out, otherwise its the heavier of the two..
SO - Four Times Max needed.
B))

******
OK - darn... its three by splitting into 3 groups versus two you get there one time faster. Great quick puzzle.

Edited by RiddleRookie, 12 March 2008 - 05:56 PM.

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#25 Oblivion

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Posted 15 March 2008 - 08:43 AM

0. Set fire to all balls and the one that finishes burning last is the heavier one
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#26 ALFRED

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Posted 20 March 2008 - 06:04 PM

Now that everyone has tried this puzzle and seen the answer, let's try it with a little twist:

What if you didn't know if the one ball that weighed different was heavier or lighter? Then how many times on the scales would it take?

Spoiler for Solution

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#27 rookie1ja

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Posted 20 March 2008 - 07:42 PM

Now that everyone has tried this puzzle and seen the answer, let's try it with a little twist:

What if you didn't know if the one ball that weighed different was heavier or lighter? Then how many times on the scales would it take?

Spoiler for Solution

the variation is already posted with good discussion ... check this

I have dedicated one whole page to such Weighing Puzzles
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#28 vinoth

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Posted 28 April 2008 - 12:15 PM

9 9 9 is my first choice.. if somene is interested in alternative answer here u go.. (but it takes 6 times)

14 13

take 1 ball from any 14 (lhs)

if you take the heavier ball by first chance from lhs, scale equals

if not ,remove the first ball taken aside

13 13 (assuming that normal ball is taken, the heavier side will slide as both sides are having equal balls)

7 6

if you take the heavier ball by chance from lhs, scale equals

if not, remove the ball taken aside..so

6 6

now look which side is heavier..n split those in to two halves

3 3

now look which side is heavier

2 1

now remove one ball from lhs n see if it equals..

1 1

if the heavier ball is still left out, it slides and thats it .
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#29 Presence

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Posted 04 July 2008 - 07:12 AM

Heres how I would of solved it.

First you split the groups to three groups, 13, 13, 1.
Second Weigh the two Thirteen groups to eachother, Should one be heavier you eliminated most of them, Should they be the same then the last ball you have is the heavier one.
If second step does not work then sepereate the balls again into three groups, 6, 6, 1.
Weigh them again following step 2
Weigh the heavier side, 3, 3.
Weigh the heavier side, 1, 1, 1.
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