# Solve this if you can !

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Use your algorithm and give the solution here.

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

??

Everything is in post #17.

Average is 5.3 weightings.

Edited by Sensei

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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.

Bachet

in this case, as in many others, it is not a matter of cleverness, I was wrong on that. It is a matter of memory. I mean i remembered the puzzle, ithat's all.

Also many of that kind of puzzle have a trick in the first statements. Many have been made by "clever" philosophs who still laugh from their graves at the mathematicians who try (stupidly) to solve their "problem" the straight way.

Ah, and I was wrong, by the "clever way of weighing" in can be done in ...

One weighing

The "clever weighing" consists into "pouring" a ball one by one one simultaneously on both sides of the balance. All the procedure will count as one weighing. at some time, you will "pour" the wrong ball and the balance will act. If you go through all the balls till number 37-38, it means the last ball is the wrong one. But you don't know if ball 39 is heavier or lighter. So in order to avoid a second weighing, what you have to do is to take back balls 37 and 38 simultaneously from the balance, and put balls 38 and 39. then the balance will tilt. One weighing.

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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.

Acme :

I have this habit of emphasizing with Capitals now and then and please don't take offence to it.

I know in internet parlance it also means Shouting and therefore I'll avoid it amap !

Regards

??

Everything is in post #17.

Average is 5.3 weightings.

Sensei :

Yes, you have given the Solution in #17 to solve in 5 Steps.

It is a nice Solution and I think you solve it always in 5 Steps and I can not see why you need a sixth step and in what sequence.

Therefore you and imatfaal have solved it in 5 Steps !

Correct me and point out if I have missed out anyone else's Solution in 5 Steps.

I am still to view Michel 123456's links and will comment on it if he/she wants.

Regards

Regards to all

i am a he.

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i am a he.

TY

Best regards

Thomas

PS : May not bother to correct old post as Moderators prefer not to encourage it !

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 !!!

Hi all,

If BN is the number of balls which can be solved in this particular context of given problem [ie just one odd ball of different weight among B] in N number of weighings

then, BN+1 = 3 ( BN + 1 )

If N = 4 and B4 = 39 then , B5 = 3 ( 39 + 1 ) = 120

The Solution for 39 Balls I mean the procedure I will give if there is a consensus !

I am sure imatfaal or sensei will find the solution on their own soon !!

In the meantime I am looking for a demo Platform [ I have one for lesser number of balls but may not have the right to put it up here as I have not coded it myself ]

Best Regards

Thomas

Edited by Commander

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Thomas - Hold off on posting the answer for 4 weighings max please - I think I might have it but it will be a few hour before I can check.

It would be cool if you could wait - I would like to get it right without help

Thanks

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Thomas - Hold off on posting the answer for 4 weighings max please - I think I might have it but it will be a few hour before I can check.

It would be cool if you could wait - I would like to get it right without help

Thanks

imatfaal :

Sure, I'll wait and take your time.

I know it is tough and those who persists with puzzle solving have great mental and analytical Energy !

Good Show and Keep it up !

Regards

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Setting up post

Since it is not too late to try! I will continue on from where I left off before http://www.scienceforums.net/topic/86946-solve-this-if-you-can/page-3#entry842811

Attempt 3

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?

So after 2 steps you know the problem ball is heavy or light and in which 13 it is.

So split the 13 up into 3 groups of 4 Plus one.

Put 4 balls in each side if balanced all the same weight you know the problem ball in last 4 +1 (5 unweighed).
3 possible results even, heavy, light (but we already know if the problem ball is heavy or light)
If uneven you know the problem ball is not in the last 5, leave the 4 balls that are normal in one pan and place last 4 to counterbalance (0 unweighed)
2 possible results either even, or uneven,
if "even" problem ball is the plus 1 ball.

Now we still could have 4 balls to sort in 1 remaining move so it can't always be done in 5 moves.

Edited by Robittybob1

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First though, I have to ask what shape other than spherical are balls?

Lots of shapes if you consider the following balls:

Football, soccer ball, golf ball, rugby ball, basketball, etc.

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Yes, you have given the Solution in #17 to solve in 5 Steps.

It is a nice Solution and I think you solve it always in 5 Steps and I can not see why you need a sixth step and in what sequence.

Therefore you and imatfaal have solved it in 5 Steps !

Correct me and point out if I have missed out anyone else's Solution in 5 Steps.

Because you're not counting correctly needed weightings.

If you have 3 unknown balls. You need 1 weighting if you pick up two normal balls. If you pick up different and normal balls, then you need yet another weighting to make sure which is which.

• 1
• 1

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Because you're not counting correctly needed weightings.

If you have 3 unknown balls. You need 1 weighting if you pick up two normal balls. If you pick up different and normal balls, then you need yet another weighting to make sure which is which.

Correct That is if you want to know which one it is and its sign, is it heavier or lighter.

Edited by Robittybob1

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Setting up post

Since it is not too late to try! I will continue on from where I left off before http://www.scienceforums.net/topic/86946-solve-this-if-you-can/page-3#entry842811

Attempt 3

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?

So after 2 steps you know the problem ball is heavy or light and in which 13 it is.

So split the 13 up into 3 groups of 4 Plus one.

Put 4 balls in each side if balanced all the same weight you know the problem ball in last 4 +1 (5 unweighed).

3 possible results even, heavy, light (but we already know if the problem ball is heavy or light)

If uneven you know the problem ball is not in the last 5, leave the 4 balls that are normal in one pan and place last 4 to counterbalance (0 unweighed)

2 possible results either even, or uneven,

if "even" problem ball is the plus 1 ball.

Now we still could have 4 balls to sort in 1 remaining move so it can't always be done in 5 moves.

If the above approach is taken :

After the first two weighings we have B1 , B2, B3, B4, B5, B6, B7, B8, B9, B10, B11, B12, B13

within which the odd ball lies and we know whether it is lighter or heavier.

sensei has solved it further in 3 more weighings by Binary Search Algo.

Weigh B1 , B2, B3, B4, B5, B6 VS B7, B8, B9, B10, B11, B12 and if this balances it is B13

Else if it unbalances take the six in the correct pan [going up or down according to known condition of the oddball] and name it

B1 , B2, B3, B4, B5, B6

On the 4th weighing B1 , B2, B3 VS B4, B5, B6 gets the oddball within three; name it B1 , B2, B3

On the 5th and last weighing B1 VS B2 and if this balances it is B3 or else B1 or B2 which tilts the scale in the known direction.

imatfaal has pointed out that he can handle more number of balls at the 3rd weighing onwards using a different algorithm.

Even if starting with 81 use the first two weighings to compare 27 balls instead of 13 and decide on 27 balls group within which lies the oddball with known deviational direction.

Thereafter 9 VS 9 and if it balances the rest 9 or else among the 9 going the right direction.

Its a kind of trinary algorithm.

Then the 4th weighing reduces the oddball within 3 and thereafter it is the same procedure !

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ok done it.

4 weighings is possible. Cannot explain it so have to post pdf.

Good - means a known good ball which you have managed to exclude from enquiries. I haven't included the actual weighing - just the three results - but it is obvious what the weighing was

Thanks Thomas - that was a good puzzle with an unexpected answer. Although I did spend 4 hours of my weekend pouring over my laptop and msexcel Glad I got it though - I do hope i have made a typo. My handwritten notes definitely work

Book1.pdf

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Because you're not counting correctly needed weightings.

If you have 3 unknown balls. You need 1 weighting if you pick up two normal balls. If you pick up different and normal balls, then you need yet another weighting to make sure which is which.

But you already know that those 3 are either heavier or lighter so just 1 weighing will do !

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But you already know that those 3 are either heavier or lighter so just 1 weighing will do !

That is a point as long as it is prior knowledge.

ok done it.

4 weighings is possible. Cannot explain it so have to post pdf.

Good - means a known good ball which you have managed to exclude from enquiries. I haven't included the actual weighing - just the three results - but it is obvious what the weighing was

Thanks Thomas - that was a good puzzle with an unexpected answer. Although I did spend 4 hours of my weekend pouring over my laptop and msexcel Glad I got it though - I do hope i have made a typo. My handwritten notes definitely work

I couldn't understand that.

If the above approach is taken :

After the first two weighings we have B1 , B2, B3, B4, B5, B6, B7, B8, B9, B10, B11, B12, B13

within which the odd ball lies and we know whether it is lighter or heavier.

sensei has solved it further in 3 more weighings by Binary Search Algo.

Weigh B1 , B2, B3, B4, B5, B6 VS B7, B8, B9, B10, B11, B12 and if this balances it is B13

Else if it unbalances take the six in the correct pan [going up or down according to known condition of the oddball] and name it

B1 , B2, B3, B4, B5, B6

On the 4th weighing B1 , B2, B3 VS B4, B5, B6 gets the oddball within three; name it B1 , B2, B3

On the 5th and last weighing B1 VS B2 and if this balances it is B3 or else B1 or B2 which tilts the scale in the known direction.

imatfaal has pointed out that he can handle more number of balls at the 3rd weighing onwards using a different algorithm.

Even if starting with 81 use the first two weighings to compare 27 balls instead of 13 and decide on 27 balls group within which lies the oddball with known deviational direction.

Thereafter 9 VS 9 and if it balances the rest 9 or else among the 9 going the right direction.

Its a kind of trinary algorithm.

Then the 4th weighing reduces the oddball within 3 and thereafter it is the same procedure !

Now for the perfected run!

Attempt 3

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?

So after 2 steps you know the problem ball is heavy or light and in which 13 it is.

So split the last 13 up into 2 groups of 5 Plus 3.

Put 5 balls in each side if balanced all the same weight you know the problem ball in last 3 (3 unweighed).

3 possible results even, heavy, light (but we already know if the problem ball is heavy or light)

If uneven you know the problem ball is not in the last 3,

Now we still could have 5 balls to sort in 2 remaining moves so it can always be done in 5 moves.

So split the last 5 up into 2 groups of 2 Plus 1.

Put 2 balls in each side if balanced all the same weight you know the problem ball in last 1 (1 unweighed).

3 possible results even, heavy, light (but we already know if the problem ball is heavy or light)

If uneven you know the problem ball is not in the last 1,

Split the correct uneven side.

Choose the problem ball on the 5th move.

,

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ok done it.

4 weighings is possible. Cannot explain it so have to post pdf.

Good - means a known good ball which you have managed to exclude from enquiries. I haven't included the actual weighing - just the three results - but it is obvious what the weighing was

Thanks Thomas - that was a good puzzle with an unexpected answer. Although I did spend 4 hours of my weekend pouring over my laptop and msexcel Glad I got it though - I do hope i have made a typo. My handwritten notes definitely work

Hi imatfaal,

Good you have found solution in 4 weighings !

I will study it and get back.

Regards

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Hi imatfaal,

Good you have found solution in 4 weighings !

I will study it and get back.

Regards

It looks wrong to me.

• 1

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Imatfaal, try solving your weightings when ball 28 is heavy. I have feeling it's not covered case on the list. Am I blind? Have not sleep whole night, so I might be..

Is this PDF covering the all cases, or just general algorithm?

Another thing. If 1-14 is equal 15-28, why would 28 being heavier?

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So even when you have been told it is a trick you continue.

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So even when you have been told it is a trick you continue.

It is not a trick - there is a 4 weighing.

Imatfaal, try solving your weightings when ball 28 is heavy. I have feeling it's not covered case on the list. Am I blind? Have not sleep whole night, so I might be..

Is this PDF covering the all cases, or just general algorithm?

Another thing. If 1-14 is equal 15-28, why would 28 being heavier?

Brilliantly spotted - damn damn damn. I know this can be done. The thought that clicked it for me - and it is in opposition to the five level that both you and I worked out - is that you MUST NOT get any balls ruled out in less than four. You have a total of 81 branches - and a total of 78 potenial unique outcomes; so your utilization must be perfect (or at least very close - you cannot have a third of test finishing after 3 weighings as you have lost too much "investigatory phase space".

Will revert with corrected - even if it kill me

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Damn I hope this is right

Second Attempt cleaned up.pdf

There is no Capital N - that was the problem last time I miscorrected the capital N and that lead to hopelessness.

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It is not a trick - there is a 4 weighing.

The trick is to do it with a single weighing. No maths required.

-------------------

As i explained before, the stupid mathematician jumps into calculations. The smart philosopher re-read the question and find the solution wiithout calculating anything.

IOW it is not a mathematical problem, it is a bad joke against mathematicians.

---------------

It is a trap and you jumped into it with both feets.

• 1

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for those that

The trick is to do it with a single weighing. No maths required.

-------------------

As i explained before, the stupid mathematician jumps into calculations. The smart philosopher re-read the question and find the solution wiithout calculating anything.

IOW it is not a mathematical problem, it is a bad joke against mathematicians.

---------------

It is a trap and you jumped into it with both feets.

Dont be ridiculous - I don't care about your silly trick weighing in one go. There is a proper mathematical brain-teaser here that you seem determined to avoid. Your trick method in one go doesn't even work - if you are gonna cheat at least get it right; you put the first two balls on simultaneously and it tips to the left hand down. You either have a heavy on the left or a light on the right - and you cannot do anything else (like slipping a replacement in to check) cos to do that you must do something different from your predetermined schema ie you have already used your one inference.

The links you gave were all to the sort of mathematicians who enjoyed puzzles - not to the sort of chop-logicians who try and skirt the rules in an attempt to get an easy solution. Puzzles like the above always have loopholes - you could spend hours writing a new puzzle and yet some wannabe lawyer would find a flaw and claim that the loophole lead to the best solution; but it is pretty obvious what the parameters of the puzzle are - so I try to stick to them

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Damn I hope this is right

There is no Capital N - that was the problem last time I miscorrected the capital N and that lead to hopelessness.

So you have labeled your balls with the letters of the alphabet, was that allowed?

Do they need to be labeled? Or could you work with the groups as I did "just take any 5" not a specific 5?

As the OP said "We have Only a SIMPLE BALANCE and nothing else to use", so we don't have a pen either.

Edited by Robittybob1

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So you have labeled your balls with the letters of the alphabet, was that allowed?

Not so much labelled as called them a name to allow you to keep track of them - there is no need to write anything on them, just put them down on the table after weighing in a recallable manner. But you do need to keep track of them as some weighings will lead to an inference that a previously weighed ball was heavy

For instance - take the fourth line down

kl > m,good k = l m is light

You know that the odd ball is one of k, l or m, but you still don't know if the odd is light or heavy. you weigh k against l - and it is even - therefore m must be light.

This weighing DOES NOT tell you if m is heavy - because if m were heavy then the first expression k,l > m, good would not be true

...

Do they need to be labeled? Or could you work with the groups as I did "just take any 5" not a specific 5?

You need to know which are in heavier group and which are in lighter group, which are good and which have not yet been weighed. That is all you need to keep track off - so as i say - four piles on a table, or front and back pockets

..

As the OP said "We have Only a SIMPLE BALANCE and nothing else to use", so we don't have a pen either.

No need for a pen - thank God - otherwise I would be back to the drawing board. I could write it as a set of consecutive instructions as you did with no letters - but that would require 81 lines and it would be hellishly complicated. The letters are more to make it clear what is happening

AND I am still praying that Sensei doesn't find another mistake

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