Conceptual problems

I remember this one. No, I should say I vaguely remember doing this one in high school, and I'm going to need all my fingers and toes to count how long ago that was.

Something about splitting the 12 into 3 groups of 4, and balancing them against each other. If the first two groups balance, the odd one is in the third group. After that, I forget, but I think this is the right way to start.

 

This method requires a long explanation. :blink:

I don't remember the exact order, but you compare four coins against four coins. On the first weighing, you will know if four coins are potentially light AND for coins are potentially heavy, OR if all eight coins are equal.

First weighing: 1,2,3,4 on the left, and 5,6,7,8 on the right.

Which ever side the scale drops to, you mark those coins with a plus sign.
If the side goes up, you mark those coins with a minus sign.
If they balance, you mark all the coins with an equal sign. (Those eight equal coins are now known not to be the odd coin, but you don't throw them out. Some of them will be used in the next two measurements.

For simplicity sake, let's say coin one is the odd coin and is light.

On the first weighing, the left side went up, so the numbers became

1-
2-
3-
4-

5+
6+
7+
8+

On the second weighing, you swap around some coins and place three unweighed coins on the right (but always comparing four coins against four coins). If the first weighing wasn't equal, this weighing will start the process of disproving some coins as contenders. . .

Second weighing: 1,2,7,8 on the left, and 4,9,10,11 on the right. If this second weighing turns out equal, then 3,5,6 have been proven suspect.
Since we cheated and already know coin 1 is the light coin, the results will be

1-
2-
7-
8-

4+
9+
10+
11+

So far with two measurements we can eliminate 4,7,8 because their plus and minus signs cancel out. You can't be on both the heavy side and the light side. We can also cancel out 3 because it got a minus sign, but wasn't involved in the second minus weigh in.

If fact, so far the experiment has proven either 1 or 2 as the light coin, so we can eliminate all other coins from contention as the odd coin, but we still need to prove 1 or 2 as the light coin.

This is where I get a little fuzzy with the exact coin placement, but either coin 1 or coin 2 is weighed on the third weighing. The exact order causes some coins to be weighed one, two, or three times . . . but the exact order will narrow down the odd coin and its positive or negative weight.

For this example, let's say coin 2 was the one scheduled to be measured for the third time. If on this third weighing coin two does anything except be on the side which weighs light, then it is eliminated . . . but in this example we know the scale will measure equal, and coin 2, and all seven of the other coins will be proven equal.

With all coins measured, coin 1 will be the only one measured with two negative signs, while all other coins either measured equal or positive and negative.

This method provides a total of 24 coin measurements (4x2x3=24), and a way to find the odd coin and its positive or negative.

 

<div class='quotetop'>QUOTE(eagle33199 @ May 2 2007, 09:36 AM) [snapback]433875[/snapback]</div>

Can the queens attack each other?

 

<div class='quotetop'>QUOTE(eagle33199 @ May 2 2007, 07:03 AM) [snapback]433900[/snapback]</div>

One of two ways are possible:
1. Place the 1st queen at any corner. Place the 2nd queen any knight's move distant. Place the 3rd the only knight's move distant that doesn't attack any of the other queens. Repeat that last step for queens 4 and 5.

2. Use all white or all black queens and stick 'em anywhere on the board - pieces cannot attack another piece of the same color!

MB

 

<div class='quotetop'>QUOTE(eagle33199 @ May 2 2007, 07:38 AM) [snapback]433923[/snapback]</div>

Put 2 queens one atop the other in the leftmost column at rows 3 and 4 (counting up from the bottom).
Put 1 queen next to the top queen in row 4 in the 2nd column.
Put the last 2 queens side by side in the bottom row in columns 2 and 3.

2 of the 3 pawns will find sanctuary side by side in the top row in columns 4 and 5.
The 3rd pawn will find sanctuary in the 2nd row in column 5.

Took awhile, but having unlimited sheets of graph paper marked in 5ths of an inch helped greatly.

Mark Baird
Alameda CA

 

<div class='quotetop'>QUOTE(eagle33199 @ May 2 2007, 01:28 PM) [snapback]434211[/snapback]</div>

To steal from one of the great lines from Douglas Adams' hapless hero, Arthur Dent, you are obviously using a meaning for "easier" I was not previously aware of. That's an ancient problem, cooked up by a couple of bored brontosauruses before the asteroid got them, but it has not gotten softer with age. The original dialogue was terse and elegant (brontosauruses were creatures of few words):

P: Hell if I know.

S: I knew that.

P: Well now I know.

S: Me too.

Solving this gnarly bit of logic is an exercise in long tedious lists of elimination - hours of work. I did it years ago (just after the asteroid incident - I'm giving away my age) - and I ain't gonna do all that again. I'll leave this to those who haven't yet encountered it. It is NOT a problem any of MY meanings for "easier" would describe.

MB

 

I'm gonna go with 2 and 6.

 

<div class='quotetop'>QUOTE(eagle33199 @ May 2 2007, 02:17 PM) [snapback]434245[/snapback]</div>

Not so fast, old friend, you better go take a proper look at Number 3 - Two Logicians
The problem is CONSIDERABLY more complex than you think. I'm not sure what the flaw is in your example yet, but the correct answer is something other than 3 and 4. And I'll try to identify the flaw in your example (I hope it doesn't take as much work as solving the damn thing)!

Mark

 

<div class='quotetop'>QUOTE(eagle33199 @ May 2 2007, 01:28 PM) [snapback]434211[/snapback]</div>

Guessing that the pawns go in a corner and putting 2 there makes it easy to see where the Queens can't go. It leaves 6 places for 5 queens and One place for a Pawn that would only remove one of those 6 options.

<div class='quotetop'>QUOTE(jeneric @ May 2 2007, 02:07 PM) [snapback]434240[/snapback]</div>

S=8 is (4,4), (5,3), (2,6). (5,3) doesn't work, but 16 and 12 both have two two factor sets. So, yeah, I was wrong.

 

<div class='quotetop'>QUOTE(eagle33199 @ May 2 2007, 02:17 PM) [snapback]434245[/snapback]</div>

Unless I'm missing something, this one is super easy. For the first two guys, the possible combinations are:

2 1
B B
B W
W B
W W

If the third guys sees BB, he knows he must have W, and since he didn't say anything, it must be one of the other three. The second guy knows this. If he sees B on the guy in front of him, he knows he is wearing W, but he doesn't say anything so that can't be it. That leaves BW and WW, in either case that means a white hat for the first guy. So after the other two guys say nothing, he deduces he is wearing W.

 

How about this one?

There are 3 light switches outside a room, each one connected to a different light inside a room. You can't see in the room from outside. What do you do to the switches, so that when you go into the room, you will know which switch is hooked to which light.

 

<div class='quotetop'>QUOTE(jeneric @ May 2 2007, 04:40 PM) [snapback]434339[/snapback]</div>

Turn two switches on (or leave two on); turn off the third switch. Go have a sandwich. Come back, and turn off one of the two switches that was on. Enter the room, take your Swiss Army collapsible 12 foot stepladder out of your pocket, climb up to the lights and feel the two that are unlit. The hot one of the two will be connected to the switch you just turned off, the cold one to the switch you turned off earlier, and the still burning light is connected to the switch that is turned on.

This is why one must never leave home without a 12 foot collapsible step ladder, so that if this sort of thing comes up, you won't be helpless and have to depend on customer service via a tech supprt hotline to Zanzibar.

MB