# Solve this if you can !

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I meant BN to mean the number of Balls which can be solved in N Weighings.

For example I can solve 120 Balls in 5 Weighings which means B5 = 120 !

That means if the given number of Balls are 120 instead of 39 the answer will be : 5 Weighings and the procedure to find that Culprit Ball in those 5 Weighings !!

That can be taken as a second Puzzle !!!

I don't believe you can do it in 5 weighings!

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I meant BN to mean the number of Balls which can be solved in N Weighings.

OK. Thnx.

So when you say:

I think I have one which gives BNwhich means the number of Balls that can be decided upon [spot and Fix the Direction of Variation Lighter or Heavier]as a Function of BN-1

Does this mean you can do the 39 balls in 4 weighings?

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OK. Thnx.

So when you say:Does this mean you can do the 39 balls in 4 weighings?

Yes, its a Clue I can give that it can be done in 4 Weighings !

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Imatfaal,

Again I am not finding where to mark a section of text as spoiler, so don't read any further.

My overall scheme is two-fold. Increase the number of known good balls with each wheighing or narrow the field of balls containing the odd ball. If at any point you have known good balls on the left then the group on the right can only be all good, all good but one heavy, or all good but one light.

Once you make your first measurement you have either identified a whole stable of good balls for use on the left side of the scale, because they have balanced, or you have selected two groups which contain an odd ball, which means the OTHER balls, left unwheighed are GOOD, and become your stable of good balls. They don't need to be whieghed because you have already identified their character.

At that point, once you know the unwheighed group is all good, you place 10 of those on the left platform and select 10 from the unknown group. Any 10. The right hand platform will either go down, stay balanced or go up. At that point you know the character of the selected 10, as well as the character of the 9 or 10 you did not select. And this is after 2 wheighings.

One wheighing will identify 19 or 20 good balls, which leaves 19 or 20 to figure.

Two trials leaves 10

Three trials leaves 5 unknown max.

Take just those 5 unknown and balance two against 2. If they balance...

In anycase after 3 trials you have a new puzzle. You have 5 balls, one of which is heavier or lighter than the other 4, plus you have 34 balls that are the same whieght as the four good balls in the group.

There seems to me to be several strategies you could take from here, depending on the results of that fourth measurement, but in all cases you have 2,2 and 1 with certain things known and unknown about each group and 34 good balls to test any unknown ball against.

Perhaps I have not yet designed the test for each of the possible results for each of the possible trials available once you are down to 5 suspects, but in any case its a new puzzle where you must find the oddball, and whether it is heavy or light, it two tries. And this seems very likely to have not only one successful solution, but many. All depending on the results of the forth trial, which would set the stage for the design of the final fifth test which would determine the identity and character of the oddball.

Regards, TAR

edit - by imatfaal

Tar

Highlight text to be hidden - hit button 3rd along from left hand side - top row and a box saying special bb code appears - scroll down to spoiler.

Edited by imatfaal

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I don't believe you can do it in 5 weighings!

Me neither.

If you are using three groups as Sensei and I did the max in 5 weighings is 81. You can think in reverse order

5th 1 v 2 , 3 spare

4th 123 v 456, 789 spare

3rd 1-9 v 10-18, 19-27 spare

2 check over or under

1st 1-27 v 28-54, 55-81 spare

But as I said above - I never tend to be very efficient at these tests

Imatfaal,

Again I am not finding where to mark a section of text as spoiler, so don't read any further.

Tar - I am afraid I still think you have it wrong

One wheighing will identify 19 or 20 good balls, which leaves 19 or 20 to figure.

check

Two trials leaves 10

Almost check.

You can end up with 9 balls that you know contain the odd but you dont know if heavy or light.

or 10 balls where you know that one of those balls is heavy

or 10 balls where you know that one of those balls is light

the 10 ball scenerios are easy to polish off in 3 weightings - but 9 with an unknown odd is not

Three trials leaves 5 unknown max.

But you still don't ALWAYS know if you have heavy or light

Take just those 5 unknown and balance two against 2. If they balance...

Then the odd one is in the remainder - but you dont know heavy or light.

But if they don't balance you don't know if it is because LH is heavy or RH is light !

In anycase after 3 trials you have a new puzzle. You have 5 balls, one of which is heavier or lighter than the other 4, plus you have 34 balls that are the same whieght as the four good balls in the group.

There seems to me to be several strategies you could take from here, depending on the results of that fourth measurement, but in all cases you have 2,2 and 1 with certain things known and unknown about each group and 34 good balls to test any unknown ball against.

Perhaps I have not yet designed the test for each of the possible results for each of the possible trials available once you are down to 5 suspects, but in any case its a new puzzle where you must find the oddball, and whether it is heavy or light, it two tries. And this seems very likely to have not only one successful solution, but many. All depending on the results of the forth trial, which would set the stage for the design of the final fifth test which would determine the identity and character of the oddball.

After two trials you COULD have 9 balls which includes an odd one - you don't know heavier or lighter. You claim that you can sort that in 3 weighings - BUT you haven't said how.

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tar, simply write [spoiler ] ..... [/spoiler ]

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But you still don't ALWAYS know if you have heavy or light

Does it matter as long as there are an even number of balls and there is an imbalance

So start with 2 balls one on each scale are they the same?

if they are balanced place them both on the one pan and weigh 2 more.

if they are balanced place them both on the one pan and weigh 4 more.

if they are balanced place them both on the one pan and weigh 8 more. Total 16.

if they are balanced place them both on the one pan and weigh 16 more. Total 32.

At some point you can't keep on doubling it. so then you just split what is remaining in two and weight them.

Plus a bit more mucking around.

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Does it matter as long as there are an even number of balls and there is an imbalance

Weighing #1 So start with 2 balls one on each scale are they the same?

if they are balanced place them both on the one pan and weigh 2 more.Weighing #2

Weighing #3 if they are balanced place them both on the one pan and weigh 4 more.

Weighing #4if they are balanced place them both on the one pan and weigh 8 more. Total 16.

Weighing #5if they are balanced place them both on the one pan and weigh 16 more. Total 32.

At some point you can't keep on doubling it. so then you just split what is remaining in two and weight them.

Plus a bit more mucking around.

If you check Sensei and my method you will note we got to 39 in 5 weighings - and from my note above we could have got to 81. And my challenge to Tar to prove his worked was to do 9 balls in 3 weighings - yours weighs 8 balls in 3 weighings (note if your 4good v 4unknown in #3 was equal you would have identified the odd ball but would not know if it was heavy or light)

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If you check Sensei and my method you will note we got to 39 in 5 weighings - and from my note above we could have got to 81. And my challenge to Tar to prove his worked was to do 9 balls in 3 weighings - yours weighs 8 balls in 3 weighings (note if your 4good v 4unknown in #3 was equal you would have identified the odd ball but would not know if it was heavy or light)

Yes I have lost track of which puzzle we were working on, but I think I'd always start with 2 balls and when you notice the odd one you only need to compare it to any other one to see if it is heavy or light.

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I have not read the spoilers.

IIRC the all (entire) problem is in the definition of "weighing".

Some of these Balls can by weighed in the scale against each other and any Inference taken counts for one Weighing !

The underlined part in the above.

One weighing

For example, put a single ball on each side, then add simultaneitly a second ball and it still counts for one weighing.

Continuing adding a ball left & right till the balance acts.

You need only one weighing.

Oops, sorry, if you are very unlucky, two weighings.

Edited by michel123456

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I have not read the spoilers.

IIRC the all problem is in the definition of "weighing".

Did you mean 'ball' problem? And by all means please expound on the issue of definition. If you think it's necessary, then put your explication in spoilers.

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Did you mean 'ball' problem? And by all means please expound on the issue of definition. If you think it's necessary, then put your explication in spoilers.

Sorry, i ment "entire".(i corrected my post)

The definition is :what is ment by "weighing". What kind of things can you do in "one weighing".

it is an old problem, I cannot recall where I have encountered it. Maybe from school. 40 years ago or so.

Edited by michel123456

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Yes, its a Clue I can give that it can be done in 4 Weighings !

Roger.

Is it possible fellas that we are overlooking an inference that we can take from an earlier weighing in your algorithms that would tell us heavier/lighter? I'm thinking this because Commander said something about paying attention to Left-up [pan], Balanced [pans], and Right-up [pan]. I got a late start today and am still wiping the bleary from my eyes and will go over the solutions again when I get my java on.

Sorry, i ment "entire".

The definition is :what is ment by "weighing". What kind of things can you do in "one weighing".

it is an old problem, I cannot recall where I have encountered it.

Acknowledged. So possibly some element of 'trickery' aside from the math. The thought crossed my mind when I was asking Commander about the illustration he gave. Mmmmmm......

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Maybe the problem is thousands of years old.

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Maybe the problem is thousands of years old.

You piqued my interest in history and I committed the unpardonable sin of consulting the oracle, i.e. I did a web search. (I just finished reading The Archimedes Codex by Netz & Noel and they revealed that Archimedes used combinatorics to solve the puzzle of the stomachion. The authors also put the problem to modern mathematicians who solved it a couple different ways. The Wiki article on the puzzle that I just linked to is not up to these current efforts, but it does have an illustration of the puzzle.)

While I can't say how old the weighing problems are, they are not new. I can say that the solution is not a matter of the definition of 'weighing'. While Commander says this is a puzzle of his own invention I can only see that may be so in his choice of the number of balls and/or his solution.

In light of the weight of my sin I must of course withdraw from further discussion here. Good luck!

Edited by Acme

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You piqued my interest in history and I committed the unpardonable sin of consulting the oracle, i.e. I did a web search. (I just finished reading Bachet, Diophantus , Metrodorus. I think one can go to the mesopotamians & the chinese.

anyway I am not sure it was originally a mathematical puzzle, rather a cleverness exercise.

Edited by michel123456
• 1

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i was googling too.

Finding Bachet, Diophantus , Metrodorus. I think one can go to the mesopotamians & the chinese.

anyway I am not sure it was originally a mathematical puzzle, rather a cleverness exercise.

Roger & thanks. By-the-by, your Bachet link is wrong and goes back to this thread. At any rate, cleverness is no anathema to mathematics.

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Imatfaal,

You are right, I can not solve 5 completely unknown in 2 tries, all the time. Sometimes yes, but I have not come up with a foolproof stategy. So I need 6 with my method. Your 5 method looks good though.

Regards, TAR

just testing my ability to mark text as spoiler

Edited by tar

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I don't believe you can do it in 5 weighings!

Well, I already said I can do it in 4 Weighings !

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Yes, its a Clue I can give that it can be done in 4 Weighings !

39 in 4 weighings

1-19 compared 20 -38 +1

they will be either the same or out of balance but you won't know which side is heavy or light. - fail

put 3 ball in each side if balanced all the same weight

put 6 in one pan another 6 to counterbalance

3 possible results even, heavy, light

if even put 12 in one pan and 12 to counter balance,

3 possible results even, heavy, light

if even put 15 in one pan and 15 to counter balance, fail

put 4 balls in each side if balanced all the same weight (35 unweighed)

put 8 in one pan another 8 to counterbalance (23 unweighed)

3 possible results even, heavy, light

if even put 16 in one pan and 16 to counter balance, (7 unweighed)

3 possible results even, heavy, light

if even put 5 in one pan and 5 to counter balance, (2 unweighed)

fail.

put 13 balls in each side if balanced all the same weight you know the problem ball in last 13 (13 unweighed)

3 possible results even, heavy, light

if uneven you know the problem ball is not in the last 13, leave the heavy 13 in one pan place last 13 to counterbalance (0 unweighed)

2 possible results either even, or heavy,

if "even" problem ball in the 13 put aside and you know it is light.

Can you sort the last 13 in 2 steps?

Edit by mod

Please use spoilers - there are two lots of info about usage above

Edited by imatfaal
spoiler

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You piqued my interest in history and I committed the unpardonable sin of consulting the oracle, i.e. I did a web search. (I just finished reading The Archimedes Codex by Netz & Noel and they revealed that Archimedes used combinatorics to solve the puzzle of the stomachion. The authors also put the problem to modern mathematicians who solved it a couple different ways. The Wiki article on the puzzle that I just linked to is not up to these current efforts, but it does have an illustration of the puzzle.)

While I can't say how old the weighing problems are, they are not new. I can say that the solution is not a matter of the definition of 'weighing'. While Commander says this is a puzzle of his own invention I can only see that may be so in his choice of the number of balls and/or his solution.

In light of the weight of my sin I must of course withdraw from further discussion here. Good luck!

Yes,

The Balls, Shapes , Coins or the Balance are not my Original Inventions !

I had already said there are similar problems I had solved earlier and they were not my own. ie with different number of balls.

This problem of solving with 39 balls is my own proposal.

If it was proposed by anyone already I am not aware of it.

I too have solved a similar problem as early as 1973 and therefore I know similar problems have a long history of existence !

I have proposed the Puzzle with 39 Balls and another with 120 balls here which are MY OWN !

I also have a generic algorithm to calculate the maximum number of such balls that can be decided upon in N number of weighings ! WHICH IS ALSO MY OWN DERIVATION !!

More Later ........

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Yes,

The Balls, Shapes , Coins or the Balance are not my Original Inventions !

I had already said there are similar problems I had solved earlier and they were not my own. ie with different number of balls.

This problem of solving with 39 balls is my own proposal.

If it was proposed by anyone already I am not aware of it.

I too have solved a similar problem as early as 1973 and therefore I know similar problems have a long history of existence !

I have proposed the Puzzle with 39 Balls and another with 120 balls here which are MY OWN !

I also have a generic algorithm to calculate the maximum number of such balls that can be decided upon in N number of weighings ! WHICH IS ALSO MY OWN DERIVATION !!

More Later ........

Using upper case is considered shouting and there is no need for that. Changing the number of balls from the 'traditional' puzzle is mildly clever, but only insomuch as to make it not easily recognized. Having a general algorithm is rather more interesting and I -and I'm sure the others- look forward to your posting it.

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Can someone create a TEST PLATFORM for this Puzzle.

You can pick and load Balls 1 to 39 on either side of the Balance.

On the press of a button 'Weigh' the balance tilts or balances

Of course everytime the Trial Starts one of the Balls is defined as Culprit randomly and oddity of Lightness or Heaviness too decided randomly.

Just a kind of Demo / Work Bench

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Well, I already said I can do it in 4 Weighings !

But you never showed it in spoiler.. Or I am missing something?

I also have a generic algorithm to calculate the maximum number of such balls that can be decided upon in N number of weighings ! WHICH IS ALSO MY OWN DERIVATION !!

I have already showed you algorithm in post #24, when you asked.

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But you never showed it in spoiler.. Or I am missing something?

I have already showed you algorithm in post #24, when you asked.

Yes, of course it is in my head only right now !

If you give up and I have a consensus of demand for it I can reveal it !!

Maybe next time you want me to pose a puzzle and give its solution in the Spoiler !!!

Use your algorithm and give the solution here.

In how many weighings are you solving it and what is the procedure ?