My brain isn't functioning properly enough to create an original puzzle at the moment, so instead, here's a classic:
/begin puzzle
You have 12 weights, identical in mass except for one of them. You don't know which is the odd weight, nor do you know whether it is heavier or lighter than the other 11. You also have a balance scale.
With only three weighings, how do you determine which of the 12 weights is the odd one and whether it is heavier or lighter?
Fadeblue: EDIT: now an actual answer: DAMN THAT WAS HARD.
Okay.
Take eight weights. Put four on each side of the scale (Weighing 1).
The ones currently on the left are 1-4. The ones on the right are 5-8. The ones sitting out are 9-12.
Now, in order to find a weight that is off, and whether it is heavier or lighter, it must be weighed on both sides of the scale, with the same results (going down/going up) on both sides, and each time that weight appears on the scales in which the scales are tipped, it must have an example of the same type of scale-tipping without any "buddies" from the other weighing, on the same side or opposite.
Let's see if I can create a set of weighings in which each weight either:
1) Appears only once, with all the weights it was weighed against also later appearing, in which case if that weighing is tipped, and level the other times then it will be revealed as the heavier/lighter.
2) Appears twice, with all weights that were with/against it switching places for the second appearance with the same results, OR those weights being triple-appearances to prove their regular weight (if they were not regular, the third appearance would also be slanted).
3) Appears all three times, with all weights spending a weighing both with this weight and against it.
Since it will probably work best without deviating from the 8-weights-per-weighing standard, there are 24 weight slots to fill. That is 2 appearances per weight, with one single appearance for each triple appearance.
Since the weights appearing with a single-appearance weight must be weighed later again, there cannot be two single-appearances in one weighing. So, there can be no more than 3 single-appearances and thus no more than 3 triple-appearances.
I'll try it with 3 and hope that works :slant:
Let's make #1 easy, since it doesn't have to be anything else, and work from there:
Left-----------Right
1) 1,2,3,4---------5,6,7,8
Let's decide which was the single for this time (3, for simplicity, and 9 and 10).
'Kay. For #2, let's choose our three triples (1, 2, and 5 for simplicity. Can't be 1,2,3 b/c there isn't enough time for them to be all on the same side, as per the diagram). One of the triples must switch sides, and again for #3.
Left-----------Right
1) 1,2,3,4---------5,6,7,8
2) 1,5-----2 (9)
3) 1----5,2 (10)
That means 4, 6, 7, and 8 are doubles. Since 4 cannot oppose 6,7,or 8 again, and 6,7, and 8 cannot be on the same side again, it is impossible for all of them to appear in the same weighing again. If 6, 7, and 8 are all left to the same, two will be on the same side. That means that one must be used now. (6). Since 7 and 8 will be opposed in #3, 4 must also be repeated now, or else it would be opposite either 7 or 8.
Left-----------Right
1) 1,2,3,4---------5,6,7,8
2) 1,5-----2 (9,4+6)
3) 1----5,2 (10,7<>8)
Plugging in numbers:
Left-----------Right
1) 1,2,3,4---------5,6,7,8
2) 1,5,9,-----2,4,6,
3) 1,10,7,----5,2,8,
That leaves two doubles, 11 and 12. In one of the levels they must be on the same side, and on the other they must be opposing. I don't think it matters which. However, there must be room for this. One of the singletons can move.
Plugging in numbers:
Left-----------Right
1) 1,2,3,4----5,6,7,8
2) 1,5,9,11-----2,4,6,12
3) 1,7,11,12----5,2,8,10
Correct me if I'm wrong, but recording the results after those weighings should tell you which weight is different and which way it is if you deduct correctly afterward.
Wow. That solution works, and it's damn nice, too. The solution I came up with does it by cases, meaning that the results of each weighing determines what you do next. However, yours gives one sequence of weighings which, from the looks of it, does indeed provide unique results that directly tells you which weight is the odd one, and it's just a matter of seeing which way the scales tipped to determine whether it's heavier or lighter.
Now that I've seen your solution, I realize that that's perhaps the most logical approach. It turns out, in fact, that with 3 weighings, there are exactly 12 distinct result sequences, so it only makes sense to assign each of those results to one of the weights. It's only a matter of filling in the slots for the weighings after that.
You certainly solved it more quickly than I did. Using my approach of conditional weighings, it took me a full two days before I came up with my solution, but it isn't as elegant as yours. Brilliant! *applause*
purple polka dots...cause you must be tripping. west then south then east is not where you originally were, it is south of there....oh.
is it
white...polar?
-Alpha
Yep!
/begin puzzle
1. A king has given a convict one last chance at life. Two boxes are placed in front of the convict, one containing 100 black marbles, the other containing 100 white marbles. The boxes are too large for the marbles cannot sit ontop of each other in the boxes; they are too long, wide and high. The convict is instructed to mix up the marbles in the boxes. There are only two conditions:
1. There must be at least one marble inside each box.
2. All marbles must be accounted for.
After the marbles have been thoroughly mixed, a judge will come in, blindfolded, first picking a box, then opening it and picking one marble. If the marble is black, the man dies. If it is white, he lives. Is there any way to mix the marbles that gives the convict better than a fifty-fifty chance at living?
Put 49 white marbles and 50 black marbles into one box, and put the last white marble in the other box. The man now has a 50% * 100% + 50% * 49.5%, or 74.75% chance of living.
Put 49 white marbles and 50 black marbles into one box, and put the last white marble in the other box. The man now has a 50% * 100% + 50% * 49.5%, or 74.75% chance of living.
/begin puzzle
You have 12 weights, identical in mass except for one of them. You don't know which is the odd weight, nor do you know whether it is heavier or lighter than the other 11. You also have a balance scale.
With only three weighings, how do you determine which of the 12 weights is the odd one and whether it is heavier or lighter?
Fadeblue: EDIT: now an actual answer: DAMN THAT WAS HARD.
Okay.
Take eight weights. Put four on each side of the scale (Weighing 1).
The ones currently on the left are 1-4. The ones on the right are 5-8. The ones sitting out are 9-12.
Now, in order to find a weight that is off, and whether it is heavier or lighter, it must be weighed on both sides of the scale, with the same results (going down/going up) on both sides, and each time that weight appears on the scales in which the scales are tipped, it must have an example of the same type of scale-tipping without any "buddies" from the other weighing, on the same side or opposite.
Let's see if I can create a set of weighings in which each weight either:
1) Appears only once, with all the weights it was weighed against also later appearing, in which case if that weighing is tipped, and level the other times then it will be revealed as the heavier/lighter.
2) Appears twice, with all weights that were with/against it switching places for the second appearance with the same results, OR those weights being triple-appearances to prove their regular weight (if they were not regular, the third appearance would also be slanted).
3) Appears all three times, with all weights spending a weighing both with this weight and against it.
Since it will probably work best without deviating from the 8-weights-per-weighing standard, there are 24 weight slots to fill. That is 2 appearances per weight, with one single appearance for each triple appearance.
Since the weights appearing with a single-appearance weight must be weighed later again, there cannot be two single-appearances in one weighing. So, there can be no more than 3 single-appearances and thus no more than 3 triple-appearances.
I'll try it with 3 and hope that works :slant:
Let's make #1 easy, since it doesn't have to be anything else, and work from there:
Left-----------Right
1) 1,2,3,4---------5,6,7,8
Let's decide which was the single for this time (3, for simplicity, and 9 and 10).
'Kay. For #2, let's choose our three triples (1, 2, and 5 for simplicity. Can't be 1,2,3 b/c there isn't enough time for them to be all on the same side, as per the diagram). One of the triples must switch sides, and again for #3.
Left-----------Right
1) 1,2,3,4---------5,6,7,8
2) 1,5-----2 (9)
3) 1----5,2 (10)
That means 4, 6, 7, and 8 are doubles. Since 4 cannot oppose 6,7,or 8 again, and 6,7, and 8 cannot be on the same side again, it is impossible for all of them to appear in the same weighing again. If 6, 7, and 8 are all left to the same, two will be on the same side. That means that one must be used now. (6). Since 7 and 8 will be opposed in #3, 4 must also be repeated now, or else it would be opposite either 7 or 8.
Left-----------Right
1) 1,2,3,4---------5,6,7,8
2) 1,5-----2 (9,4+6)
3) 1----5,2 (10,7<>8)
Plugging in numbers:
Left-----------Right
1) 1,2,3,4---------5,6,7,8
2) 1,5,9,-----2,4,6,
3) 1,10,7,----5,2,8,
That leaves two doubles, 11 and 12. In one of the levels they must be on the same side, and on the other they must be opposing. I don't think it matters which. However, there must be room for this. One of the singletons can move.
Plugging in numbers:
Left-----------Right
1) 1,2,3,4----5,6,7,8
2) 1,5,9,11-----2,4,6,12
3) 1,7,11,12----5,2,8,10
Correct me if I'm wrong, but recording the results after those weighings should tell you which weight is different and which way it is if you deduct correctly afterward.
-Goblinboy
Now that I've seen your solution, I realize that that's perhaps the most logical approach. It turns out, in fact, that with 3 weighings, there are exactly 12 distinct result sequences, so it only makes sense to assign each of those results to one of the weights. It's only a matter of filling in the slots for the weighings after that.
You certainly solved it more quickly than I did. Using my approach of conditional weighings, it took me a full two days before I came up with my solution, but it isn't as elegant as yours. Brilliant! *applause*
Yep!
/begin puzzle
1. A king has given a convict one last chance at life. Two boxes are placed in front of the convict, one containing 100 black marbles, the other containing 100 white marbles. The boxes are too large for the marbles cannot sit ontop of each other in the boxes; they are too long, wide and high. The convict is instructed to mix up the marbles in the boxes. There are only two conditions:
1. There must be at least one marble inside each box.
2. All marbles must be accounted for.
After the marbles have been thoroughly mixed, a judge will come in, blindfolded, first picking a box, then opening it and picking one marble. If the marble is black, the man dies. If it is white, he lives. Is there any way to mix the marbles that gives the convict better than a fifty-fifty chance at living?
(You're gonna have to know some math for this.)
Bucket holds Roman equities. Viking leaders triumph. (7)
Invert real chaos; it's extraneous. (10)
Soft money. (6)
You're missing a whole lotta marbles.
What Yax probably meant to type:
Put 100 black and 99 white marbles in one box. Put the last white marble in the other box.
You have 100%*50% + 49.5%*50% = 74.75% of picking a white.
-Goblinboy