Commander

Solve this if you can !

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imatfaal :

 

I had a look into the Book1.pdf and the second attempt.

 

I have not yet convinced myself that there is a sure solution.

 

I need to study more.

 

However it was a brilliant attempt at it !

 

Thanks, yes it is indeed a very difficult problem based on very simple definition.

 

I consider this as the best ever mechanical problem I have encountered and also a challenge to logic.

 

I remember the days some years back I took a week's leave from Office to sit with the Rubic Cube and evolve a set of formulas and solution to Rubic cube without looking at the other published solutions and methods. It was a thrilling experience to derive a solution.

 

I think it is now my duty to give the solution , ie my solution and I know there are some variations which can be also employed with balls arrangements to solve in a similar way.

 

Can I now give my solution ?

 

Regards

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for those that

 

 

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

Oh, you are correct.

Two weighings.

The cheater wins anyway.

I am pretty sure this particular puzzle has been made for ridiculizing the mathematicians. A long long time ago.

Mathematicians are so stupid that they will continue calculating for centuries upon this problem, waisting their time.

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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

 

 

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 :)

 

 

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

I still see lots of errors in your tables, sorry.

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I still see lots of errors in your tables, sorry.

Then point them out in a spoiler. Your hit & run one-liners are unhelpful.

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Then point them out in a spoiler. Your hit & run one-liners are unhelpful.

In the second round what does "good" mean? is it a ball named "Good" for no reason?

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In the second round what does "good" mean? is it a ball named "Good" for no reason?

 

It's any ball that was confirmed to have normal weight.

f.e. if A=B then either A and B is good.

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In the second round what does "good" mean? is it a ball named "Good" for no reason?

 

 

 

It's any ball that was confirmed to have normal weight.

f.e. if A=B then either A and B is good.

 

thnks Boss

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thnks Boss

I have also done it in 4 weighings but only if it is possible to track individual ball's identity as you have done.

So you win.

 

Nominate ball 1 as the odd one out.

do modular screen

light

1

3

5

7

9

11

13

15

17

19

21

23

25

27

29

31

33

35

37

then

light

1

8

15

22

29

36

4

11

18

25

32

0

7

14

21

28

35

3

10

 

then

light

1

5

9

13

17

21

25

29

33

37

2

6

10

14

18

22

26

30

34

 

then

light

1

6

11

16

21

26

31

36

2

7

12

17

22

27

32

37

3

8

13

 

By this stage only 1 number is in the light tray 3 times or there would 1 number in the heavy side so test whether 1 is light or the other is heavy, By default.

 

Edited by Robittybob1

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I have also done it in 4 weighings but only if it is possible to track individual ball's identity as you have done.

So you win.

 

Nominate ball 1 as the odd one out.

do modular screen

light

1

3

5

7

9

11

13

15

17

19

21

23

25

27

29

31

33

35

37

then

light

1

8

15

22

29

36

4

11

18

25

32

0

7

14

21

28

35

3

10

 

then

light

1

5

9

13

17

21

25

29

33

37

2

6

10

14

18

22

26

30

34

 

then

light

1

6

11

16

21

26

31

36

2

7

12

17

22

27

32

37

3

8

13

 

By this stage only 1 number is in the light tray 3 times or there would 1 number in the heavy side so test whether 1 is light or the other is heavy, By default.

 

 

I will have to think about that one.

 

BTW I do not track the balls identity - the numbering was for the readers benefit - I just need to know if it was heavy, light, untested or good.

Bob - you don't weigh balls (by your numbering) 12,16,20 or 24. And multiple balls have similar weigh counts - you weigh 25, 21, 29 etc as many times as 1; how can you tell them apart?

 

I love the idea of three large weighings - but unless I have the idea mixed up in and around my head then your method doesnt work yet

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It doesn't matter if you mark the balls or not. That is, it's neither 'against the rules' or necessary. As Imatfaal pointed out, you can 'identify' individual balls by how you place them when handling them. For example you could line them up in a row and in your minds eye 'number' them and then just keep track of which 'numbers' you use by counting and reordering them in rows during and after handling. While Commander said his illustration wasn't a clue when I asked about it, my thought was prompted by the idea of numbering. Some of the solutions that I read to the classic 12-ball problem used such 'numbering', some did not. (Note that I asked the question before I read solutions.)

Edited by Acme

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I will have to think about that one.

 

BTW I do not track the balls identity - the numbering was for the readers benefit - I just need to know if it was heavy, light, untested or good.

Bob - you don't weigh balls (by your numbering) 12,16,20 or 24. And multiple balls have similar weigh counts - you weigh 25, 21, 29 etc as many times as 1; how can you tell them apart?

 

I love the idea of three large weighings - but unless I have the idea mixed up in and around my head then your method doesnt work yet

Second check it's failing - sorry

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It doesn't matter if you mark the balls or not. That is, it's neither 'against the rules' or necessary. As Imatfaal pointed out, you can 'identify' individual balls by how you place them when handling them. For example you could line them up in a row and in your minds eye 'number' them and then just keep track of which 'numbers' you use by counting and reordering them in rows during and after handling. While Commander said his illustration wasn't a clue when I asked about it, my thought was prompted by the idea of numbering. Some of the solutions that I read to the classic 12-ball problem used such 'numbering', some did not. (Note that I asked the question before I read solutions.)

 

Yep - agree entirely. Although my method relies on stacks/pockets/piles rather than numbering

I have also done it in 4 weighings but only if it is possible to track individual ball's identity as you have done.

So you win.

 

Nominate ball 1 as the odd one out.

do modular screen

light

1

3

5

7

9

11

13

15

17

19

21

23

25

27

29

31

33

35

37

then

light

1

8

15

22

29

36

4

11

18

25

32

0

7

14

21

28

35

3

10

 

then

light

1

5

9

13

17

21

25

29

33

37

2

6

10

14

18

22

26

30

34

 

then

light

1

6

11

16

21

26

31

36

2

7

12

17

22

27

32

37

3

8

13

 

By this stage only 1 number is in the light tray 3 times or there would 1 number in the heavy side so test whether 1 is light or the other is heavy, By default.

 

 

Bob -

 

unless I have your method completely misread it boils down to having unique patterns of weighing - ie left heavy, left heavy, right heavy, left heavy. And each unique pattern will have a heavy or light ball associated

 

I cannot see how you can get close to 39 balls. If I prearrange the weightings I can only manage 16 balls for the simple reason that is the number of combos of unique patterns

 

Heavy left on

1,2,3,4

1,2,3

1,3,4

1,2,4

1,2

1,3

1,4

1

 

And Heavy Right similarly. 16 Patterns means 16 balls max. If I can use a stack of known good balls IN ADDITION to the test balls I can extend your method to 30 balls - cos you can now add

 

Heavy Left on

2,

3,

4,

2,3

2,4

3,4

2,3,4

 

 

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Yep - agree entirely. Although my method relies on stacks/pockets/piles rather than numbering

 

Bob -

 

unless I have your method completely misread it boils down to having unique patterns of weighing - ie left heavy, left heavy, right heavy, left heavy. And each unique pattern will have a heavy or light ball associated

 

I cannot see how you can get close to 39 balls. If I prearrange the weightings I can only manage 16 balls for the simple reason that is the number of combos of unique patterns

 

Heavy left on

1,2,3,4

1,2,3

1,3,4

1,2,4

1,2

1,3

1,4

1

 

And Heavy Right similarly. 16 Patterns means 16 balls max. If I can use a stack of known good balls IN ADDITION to the test balls I can extend your method to 30 balls - cos you can now add

 

Heavy Left on

2,

3,

4,

2,3

2,4

3,4

2,3,4

 

 

I'm not sure,

 

what I was doing was to arrange 38 of the balls in various patterns ( firstly even with odds) then every 3rd number modulated by 39 (maybe I should have brought that down to 38 for the last ball is known from the beginning whether it is heavy or light if the first 38 balance).

So how many columns do I have to compare till I get only one ball to be in the always heavy or always light combinations?

It appeared to me to be 4 and then I had to compare the last two to see if it was the heavy one or light variety.

 

I have run it again lowering the modulator to 38 and changing the increment steps, and it is looking rather interesting.

 

 

Is this logic right if a number ever ended up on a heavy side and on a light side it can't be the problem ball?

 

If that is the case after 3 weighings I have only 2 balls to check which is heavy or light so that is 4 steps to a solution. -

I'll double check it later.

 

If the balls are split into 2 groups provided the 39th ball isn't "it" the scale will be out balance, but whatever number is the faulty one it will always be in the group (always consistently up or consistently down but never up and down within the same set of weighings, so it is just a matter of working off some regular pattern, hence my incrementing number-line and modulated by N-1 or N balls depending on which number was even.

If the ball put aside from a prior weighing is put back in and the modulator decreased accordingly, I won't be surprised if it comes down to a solution after just 3 weighings.

 

Edited by Robittybob1

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I'm not sure,

 

what I was doing was to arrange 38 of the balls in various patterns ( firstly even with odds) then every 3rd number modulated by 39 (maybe I should have brought that down to 38 for the last ball is known from the beginning whether it is heavy or light if the first 38 balance).

So how many columns do I have to compare till I get only one ball to be in the always heavy or always light combinations?

It appeared to me to be 4 and then I had to compare the last two to see if it was the heavy one or light variety.

 

I have run it again lowering the modulator to 38 and changing the increment steps, and it is looking rather interesting.

 

 

Is this logic right if a number ever ended up on a heavy side and on a light side it can't be the problem ball?

 

If that is the case after 3 weighings I have only 2 balls to check which is heavy or light so that is 4 steps to a solution. -

I'll double check it later.

 

If the balls are split into 2 groups provided the 39th ball isn't "it" the scale will be out balance, but whatever number is the faulty one it will always be in the group (always consistently up or consistently down but never up and down within the same set of weighings, so it is just a matter of working off some regular pattern, hence my incrementing number-line and modulated by N-1 or N balls depending on which number was even.

 

Failed again!

Then some more success!

 

Then seemed to have some success when I put the 39th ball back in for we know that one is OK after the first weighing.

 

Edited by Robittybob1

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Another method with one weighing. No cheating.

 

 

 

Dismantle the balance. In fact not really needed, that is only for not be suspected of weighing.

Hang freely one disk of the balance with your left hand, take one ball at a time with your right hand and make it fall on the disk from the exactly same height (use the strings that hang the disk to take the each time same distance. You will hear something like dang dang dang ding**, or dang dang dang *dong*. If you hear ding** then the ball is lighter, if the sound is **dong, then the ball is heavier. Remount the balance and make a weighing for verification.

 

 

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Another method with one weighing. No cheating.

 

 

 

Dismantle the balance. In fact not really needed, that is only for not be suspected of weighing.

Hang freely one disk of the balance with your left hand, take one ball at a time with your right hand and make it fall on the disk from the exactly same height (use the strings that hang the disk to take the each time same distance. You will hear something like dang dang dang ding**, or dang dang dang *dong*. If you hear ding** then the ball is lighter, if the sound is **dong, then the ball is heavier. Remount the balance and make a weighing for verification.

 

 

No not if you are tone deaf.

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imatfaal :

 

Yes, your solution is correct and I have verified.

 

I wish I had enlarged the text and read your second file earlier.

 

I was busy creating the file / graphics to indicate my Solution.

 

As I said earlier and as I see it now, there can be a few variations in the arrangement of weighings / balls.

 

I CREDIT YOU WITH THE SOLUTION !

 

Well done !!

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attachicon.gif39 ball solution.jpg

 

imatfaal :

 

Yes, your solution is correct and I have verified.

 

I wish I had enlarged the text and read your second file earlier.

 

I was busy creating the file / graphics to indicate my Solution.

 

As I said earlier and as I see it now, there can be a few variations in the arrangement of weighings / balls.

 

I CREDIT YOU WITH THE SOLUTION !

 

Well done !!

 

 

Thank You Thomas. And Congratulations on your solution and thank you for bringing a good little brainteaser to my attention. From first glance it seems that your solution and mine are different - I had assumed that any minimal solution would basically be the same (ignoring minor differences). But yours and mine involve different logic as far as I can tell

Another method with one weighing. No cheating.

 

 

 

Dismantle the balance. In fact not really needed, that is only for not be suspected of weighing.

Hang freely one disk of the balance with your left hand, take one ball at a time with your right hand and make it fall on the disk from the exactly same height (use the strings that hang the disk to take the each time same distance. You will hear something like dang dang dang ding**, or dang dang dang *dong*. If you hear ding** then the ball is lighter, if the sound is **dong, then the ball is heavier. Remount the balance and make a weighing for verification.

 

 

 

And surely the pitch depends on the size of the resonant surface not on the weight of the strike.

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Thank You Thomas. And Congratulations on your solution and thank you for bringing a good little brainteaser to my attention. From first glance it seems that your solution and mine are different - I had assumed that any minimal solution would basically be the same (ignoring minor differences). But yours and mine involve different logic as far as I can tell

 

And surely the pitch depends on the size of the resonant surface not on the weight of the strike.

 

Thank you imatfaal !

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If they are lighter than water, float them. One will float higher or lower than the others.
If they are heavier, drop them in and see which one takes more or less time to hit the bottom than the rest.

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I found the solution .jpg file posted by me is missing

 

Might have been deleted by mistake.

 

Please let me know if someone needs it !

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