Brain Twisters

[Dr.T.N.Jambunathan]

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gadda gadda guddu guddu naduvulay pallam :drum:

சிகுபுப சிகுபுக ரயிலேலே
கலகுதுபாரு உன் styleஎ styleஎ
Jambu:roll:

 

[Raghy]

No. One ball is odd it may be heavier or lighter It is not one ball is heavier as you have assumed. In that situation your observation Chance 1 may be due to one ball being heavier is correct. It may be that one ball may be lighter . In that case the pan which goes up contains the lighter ball This is really a complicated affair Only 3 weighing are allowed. I am doing it for the past 3 or 4 days spending about 2 to 3 hours a day No where near a solution Let us try. Jambu

Sri.Jambunathar,

Since one ball is either heavier or lighter than the others, the problem can not be solved in 3 weighings only. However,It can be solved by 4 weighings.
However if we know that the oddity before hand (heavy/light), then the problem can be solved in 3 weighings.

Cheers!

 

[Dr.T.N.Jambunathan]

Dear This problem was given to me some years ago by a co passenger, to while away time when we were traveling from Madras to Delhi by Train He assured me that it has an answer But may take lot of time and patience to solve like Rubik Cube But I have forgotten about it. When I saw this thread I picked it up தூசி தட்டி I have posted. I am spending at least 2 hours on this for the past 3 or 4 days Now I have got a strategy and working on that line. But if I hit I will post it at once My strategy works on the following understanding of mine

I think in third weighing you can have only 3 or less balls and the rouge ball should be one of them and your suggestion on the oddity before hand(heavy/light) could make it much easier Jambu

 

[Raghy]

I think in third weighing you can have only 3 or less balls and the rouge ball should be one of them and if you could identify say it it weighs more or less it will be much easier

I have only 3 balls at the start of 3rd weighing; rogue ball is one of them; but I can't identify it weighs more or less. That's why I need one more weighing, 4 in total.

Cheers!

 

[Dr.T.N.Jambunathan]

Have you got the oddity of the ball in the 2 weighing? If not you wont crack it in the 3rd weighing Jambu:loco:

 

[Raghy]

Have you got the oddity of the ball in the 2 weighing? If not you wont crack it in the 3rd weighing Jambu:loco:

Sri.Jambunathar,

My solution comprise of different combinations and various possibilities. I can solve the problem in 3 weighings for some possiblities; but, there is atleast one possibility that demands more than 3 weighings; such a possibility can not be ruled out. That's my progress, so far.

Cheers!

 

[Dr.T.N.Jambunathan]

I think I am nearer to the solution I am sure you must follow what I have suggested.
3 rd weighing can have 3 or less balls and the rouge should be among them and you should have fixed the oddity of the ball in 2nd weighing itself What is the time you spend on this. I spend at least 2hrs on this working with paper and pencil! Frustating and interesting Jambu:clap2:

 

[Dr.T.N.Jambunathan]

I am slightly modifying the line on which you work on the problem YOU SHOULD BE ABLE TO FIX THE ODDITY IN THE 2 WEIGHING. Then automatically you will be left with 3 only for the last weighing including the rouge. Is it frustrating or interesting? Jambu:angel:

 

[Raghy]

I am slightly modifying the line on which you work on the problem YOU SHOULD BE ABLE TO FIX THE ODDITY IN THE 2 WEIGHING. Then automatically you will be left with 3 only for the last weighing including the rouge. Is it frustrating or interesting? Jambu:angel:

Sri.Jambunathar,

I am not able to spend on this problem. But, like I said, although I satisfy in certain possiblities, I am out in other possibilities. In short, I am not finding the solution yet. It is very challenging. Thanks for the challenge. (Once we done this, hopefully, I have a challenge which may be equally interesting!).

Cheers!

 

[Dr.T.N.Jambunathan]

I think I have got the answer.But before I post I need to check and counter check it May take about 2 hours Then I have to put it in paper edit and finally post it In the meanwhile if you have a solution it let me know Jambu:cheer2:

 

[Dr.T.N.Jambunathan]

Dear

Here is my solution for that Rouge Ball puzzle It looks OK to me Your feed back is expected

Divide the 12 balls into 3 groups say A B & C

Group A ( A1 A2 A3 A4) and similarly Group B & C

Put Group A in Left Pan (A1 to A4 ) & Group B (B1 to B4) in Right Pan

Possibility 1 They balance Conclusion all these 8 are standard The rouge is in the left behind group C

2 weighing. Take 3 balls from C (C1 to C3) Put in one pan take 3 from the established standard in the other say (A1 to A3)
Possibility 1. They balance. Conclusion the 4th left over of C is the rouge

Possibility 2 They do not balance . Then the rouge is one among the 3 from (C1 C2 C3). Since the is is compared with standard balls you arrive at the oddity by the tilting of pan.

3 Weighing. Put one ball in the let and the other in the right (say C1 C2)

Possibility 1 They balance Conclusion the left behind C3 is the rouge
Possibility 2 they don"t Balance since you know the oddity already established in weighing 2 you pick the rouge easily
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The problem gets complicated when in the first weighing they don"t balance

If they do not Balance The rouge is here one among the 8(A1 A2 A3 A4 & B1 B2 B3 B4)
The left behind C1 C2 C3 C4 are all standard

2nd weighing Take A1A2A3 leaving A4 in the pan and put them in the other pan replacing B1B2B3 leaving B4in its place Place C1C2C3 in the ban replacing A1A2A3 This will be like this

C1C2C3 A4 A1A2A3B4

Possibility 1. They Balance. Conclusion The rouge has been taken away from the balance
It is among B1B2orB3 you should fix the oddity by the tilt of the pan from which B1B2b3(before they are taken away)

3rd weighing is similar to the 3rd weighing already see (Above the dividing line)

Possibility 2 Status Co maintained . conclusion This is possible only when the rogue is among the status co ball( 2 balls not disturbed The rouge A4 or B4)
3rd weighing Weigh A4 or B4 with any standard in the other pan and you can identify the rouge

Possibility 3 Tilt of the pan is reversed conclusion the The rouge is among the 3 which have been shifted from one pan to the other they carry the tilt also along with them (A1A2A3) Original tilt of A1 to A4 will decide the oddity of Rogue.

3rd weighing you know now will get you the rouge Jambu:whoo:
If you have difficulty in any step let me know I will try my best to clarify

EUREKA