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Thursday, January 13, 2011

Marble Riddle and Solution!

I hope you guys enjoyed the puzzle I posted a few days ago!  I'll reiterate it here and give the solution!

You have 12 marbles.  They are identical in size and shape.  One of the marbles is of a different weight than the other 11, either lighter or heavier.  Using a simple scale ONLY THREE TIMES and with only these marbles, isolate the unique marble AND determine whether it is light or heavy.


Here's my solution:


Step 1) Weigh 1:Amongst the 12, weigh 4 against 4.

Step 1) Condition1) If they weigh even, the candidate is among the 4 you didn't weigh.

Step 2) Weigh 2:Remove one of the 4; put it aside.  Weigh your three candidates with 3 you have discarded.  This will let you know for sure if the odd one out is lighter or heavier.

Step 2) Condition 1) If the scales are even
Step 3) Weigh 3:Weigh your isolated 1 vs a discarded standard one.  DONE

If Step 2) Condition 2) is an uneven scale, you know for sure now whether the odd one out is lighter or heavier.
At this point your Step 3) Weigh 3 involves weighing 2 of your remaining 3.  If the scales are even, the odd one out is the one you most recently discarded.  If the scale is uneven, you know which is the odd one out because you discovered whether the odd one out is lighter or heavier in the last step. DONE

-----

Step 1) Condition2)Assuming your first weigh yielded an uneven result in your 4 vs 4, discard the remaining 4.  This leaves you with 8 candidates, and only two weighs available.  Remember the result (/ or \)

Step 2) Weigh 2: You have four on your left side and four on your right side, with a certain sign (/ or \).  Remove one candidate from the left half and put it aside.  Remove two candidates from the right side and put them aside.  Add a standard discarded marble to the right side.  Finally, swap one of your candidates from the right side with one of your candidates on the left side.  To clear this up: this gives you three candidates on the left side, one which used to reside on the right side.  On the right side is two candidates and a standard marble, with one of those 2 candidates having used to reside on the left side.

Step 2) Condition 1) If the scale is even, you know those 6 you just weighed are all standard, which leaves you simply with the three you have removed in this step as candidates.

Step 3) Weigh 3: On the left side of the scale now, weigh the one marble you had removed from the left side along with one of the marbles you had removed from the right side against 2 standard marbles on the right side.

Step 3) Condition 1) If Step 1) Condition2) was / and Step 3) Condition 1) was an even weigh, then the one on the right side on weigh 2 is now determined to be light. DONE

Step 3) Condition 2) If Step 1) Condition 2) was / and step 3 condition 2 was /, then the one on the left side on weigh 2 is now determined to be heavy. DONE

Step 3) Condition 3)If Step 1) Condition 2) was / and step 3 condition 3 was \, then the candidate that was on the right side on weigh 2 is now determined to be light. DONE

The reasoning is symmetric during Step 3) if Step 1) Condition 3) had been \.

Refer to Step 2) Weigh 2, we continue with condition 2 here:
Step 2) Condition 2) If Weigh 1 was / and Weigh 2 yields /, then the 3 you had removed in Weigh 2) are standard, and don't need to be considered.  Additionally, since the results kept the same sign (/), then the two candidates that you swapped at Weigh 2 were obviously standards, and can be discarded as well. Now either a candidate on the left is heavy, or one of the candidates on the right is light.
Step 3) Weigh 3:  At this point you're left with two candidates on the left and one on the right.  Of these 3, take the one from the left and one from the right and weigh them against two standards.
Step 3) Condition 1) If the scale is even, then the leftover one from the left (that you didn't weigh in Weigh 3) is the odd one out, and heavy.  DONE

Step 3) Condition 2) If the scale is heavy on the side of the candidates, then the one from the right was standard, so the one from the left that you just weighed is the odd one out, and heavy.  DONE

Step 3) Condition 3) If the scale is light on the side of the candidates, then the one from the left was standard, so the one from the right right is the odd one out, and light. DONE

The reasoning is symmetric  if step 1) condition 3) had been \.


Step 2) Condition 3) If Weigh 1 yielded / and then changed to \ in Weigh 2) [or in \ then / inversely], then you know the two you swapped in Weigh 2) are the only two remaining candidates.  At this point you know that the candidate A on the left (remember weigh 2 yielded \) is either light or standard, or candidate B either heavy or standard.

Step 3) Weigh 3: the one to the left [arbitrarily] against a standard.

Step 3) Condition 1) If the scales are even, then A is a standard, so B is heavy.  DONE

Step 3) condition 2) If the scale shows A to be light, then there you go. DONE

Hurrah for logic puzzles!


Took me 7 hours, but it's the hardest puzzle I've ever solved, and satisfying!





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Wednesday, January 12, 2011

Saturday, January 8, 2011

MARBLE RIDDLE

My friend Kurt gave me a logic based riddle he found.  I've been focusing on it for many hours, and I finally solved it.  This is actually why I've missed a couple days of blogging.  Here it is, the Marble Riddle:

You have 12 marbles.  They are identical in size and shape.  One of the marbles is of a different weight than the other 11, either lighter or heavier.  Using a simple scale ONLY THREE TIMES and with only these marbles, isolate the unique marble AND determine whether it is light or heavy.

I wrote out my answer and I will post it tomorrow!  It's a time-eater, I warn you!  Probably the most fun logic riddle I've ever seen.  I'm very proud of myself for figuring out the answer.

A Tip: After developing a strategy, go through it step by step, on paper, to double check and triple check your strategy.  Expect several dead ends.

GOOD LUCK!

Here's the answer post on my blog.
http://astronasty.blogspot.com/2011/01/marble-riddle-and-solution.html


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