# Logic games

Just realized my rather simple and cool solution has one shortcoming. In ONE of a bunch of possible outcomes, I am unable to determine if the Buddha is heavier or lighter.

A B C D (each a set of 3 Buddhas, e.g. A is made up of A1 A2 A3)

1. compare A & B
2. if equal, then compare A & C
3. if equal, then compare D1 vs D2. if equal, then D3 is the correct Buddha

But I dont know if D3 is heavier or lighter!

Back to the drawing board :x

Yep, it’s tough. And you also have to solve it for all possibilities, eg, if A and B are not equal.

Brian

(Please excuse my thought processes as I type this…)

If A and B are not equal, then weigh A1 + A2 against B1 and B2. Which ever side is heavier, weigh them against each other the heavier (or lighter) one is the real deal. If A1&2 and B1&2 are the same then weigh A3 and B3 against each other.

Don’t weigh A again, Neo…you know it’s not A because it weighs the same as B and there’s only one heavy Buddha.

Maybe if you divide them into three groups instead of 4 you can eliminate them more quickly…
A1234, B1234, and C1234. Step 1: Weigh A and B. If they are the same, weigh C12 and C34 (Step 2). Because you know the real gold has to be in that group since it’s not in the first two. Which ever side is heavier…or lighter (Step 3) weigh it against each other (c1 vs. c2 or c3 vs. c4). That should tell you which one is the real gold.
If when weighing A and B, one side is heavier…or lighter, do the same steps for the heavier…or lighter sides respectively.

That took longer to type than to solve it.

[quote]Maybe if you divide them into three groups instead of 4 you can eliminate them more quickly.
A1234, B1234, and C1234. Weigh A and B. If they are the same, weigh C12 and C34. Because you know the real gold has to be in that group since it’s not in the first two. Which ever side is heavier weigh it against each other(c1 vs. c2 or c3 vs. c4). That should tell you which one is the real gold.

Phew. It took longer to type it than to solve it.
If when weighing A and B, one side is heavier, do the same.
[/quote]

Won’t work.

If A and B are the same it DOES mean the real gold is in C, but remember we don’t know if real gold is heavier or lighter than fool’s gold.

So when you weigh C1&C2 vs C3&C4 and C1&C2 turns out heavier, your possibilities are C1H (C1 is heavy), C2H, C3L (C3 is light) or C4L. You’ve already used up 2 of your 3 weighs, and you’re not going to find out which of those 4 possibilities is right with just one more weigh. So using your example you figure C1 or C2 is heavy, so you weigh them against each other and, oops, they’re the same. It must have been that either C3 or C4 is lighter, and you don’t know which one.

Sorry

Brian

Simple, throw them all in the river, the fake ones will float to the surface, while the real one will sink to the bottom and get lost in the mud, oh… that won’t work either… :?

First weighing: 1,2,3,4 vs. 5,6,7,8

If they balance, weigh 9,10 vs. 11, X (X represents any buddha you know to be normal.)

If they also balance, then 12 is the odd buddha. Weigh 12 vs. X to determine if it’s heavy or light.

If they don’t balance, weigh 9 vs. 10. That and the previous weighing will give you enough information to solve the puzzle.

If the first weighing doesn’t balance, weigh 1,2,3,5 vs. 4,X,X,X

If they balance, then the odd buddha must be 6, 7 or 8. Weigh 6 vs. 7.

If they are unbalanced in the same direction as the first weighing, then the odd buddha must be 1, 2 or 3. Weigh 1 vs. 2

If they are unbalanced in the opposite direction of the first weighing, the odd buddha must be 4 or 5. Weigh 4 vs. X

[quote=“Bu Lai En”]Yep, it’s tough. And you also have to solve it for all possibilities, eg, if A and B are not equal.

Brian[/quote]

My former solution worked for all combinations except the one exception I mentioned above.

1. Let’s say A and B are unequal, then it must be in either A or B.
2. Weigh A against C. If A is heavier, then B and C must be the same and you know that the Buddha is heavier and in A. If A is lighter, you know that the Budda is lighter and in A. If A and C weigh the same, you know that the Buddha is lighter (since A is heavier than B) and in B.
3. So if A was heavier in step 2, weigh A1 vs A2. If they’re the same, you know it’s A3 and it’s heavier. If A1 is heavier, you know it’s A1. If A2 is heavier, you know it’s A2.
If A was lighter in step 2, then step 3 is the same but you know that the Budda is lighter.
If A was the same in step 2, then step 3 is the same but test B1 vs B2. The Buddha is lighter (since A is heavier than B)

So this solutions works when things dont weigh the same. The one failure is when everything keeps weighing the same. Even then, you get the right Buddha, you just dont know if it’s heavier or not.

Alien- yours seems to work. Not sure I would have come up wit that. Congrats!

[quote=“Bu Lai En”]Just once (I won’t write the answer though. I don’t want to spoil it).

Brian[/quote]

HA! WRONG! You have to cross twice – once to open the doors and once to return to the room with the switches.
Of course, its a bad question because it also forces you to make the assumption that you can see into one room from the other.
So Neo, I presume that the answer your friend’s interviewers were looking for was “it depends.”

Okay…Well, say the fake Buddhas weigh 2 kg and the real one weighs 1 kg. two fakes versus a fake and a real one is 4kg vs. 3kg. You can tell it’s lighter because the fake ones all weigh the same. Even if it’s heavier…fake=2kg, real=3kg, you get fake+fake=4kg vs. fake+real=5kg. Ergo, it works.

Let’s do this with numbers:

f=2kg. r=1kg.

Scenario 1: A and B are the same weight then the scales would be (if they could give you numbers) 8kg vs. 8kg…Well, those can’t be it, you think to yourself, because the real one doesn’t weigh the same as the others so it must be in the C group. C12 weighs less (3kg)than C34(4kg). Eureka, you think…it must be either C1 or C2. You weigh C1 …oh shit. Now I see where the problem comes in about not knowing if it’s heavier or lighter. But wait…if C34 weighs half as much as A1234 or B1234, then you know. Okay. Wait a minute. Okay…Skip weighing C1 against C2. You just switch off the pairing. Let’s say C1 is the real deal and weighs less. Okay nevermind that, weigh C123 with A123. If they are the same then you know C4 is the one and you just weigh it against a fake one. If they are different, you know what ever that difference is, lighter or heavier, and can weigh C1 against C2. If they balance, then it’s C3 (and you know whether or not it’s heavier than the fakes from the A1234 vs. C1234). If they don’t, well, you already know that the real one is heavier or lighter from weighing three unknowns against the three known fakes and so you know which one is the real one.

Better?

[quote=“sandman”][quote=“Bu Lai En”]Just once (I won’t write the answer though. I don’t want to spoil it).

Brian[/quote]

HA! WRONG! You have to cross twice – once to open the doors and once to return to the room with the switches.
Of course, its a bad question because it also forces you to make the assumption that you can see into one room from the other.
So Neo, I presume that the answer your friend’s interviewers were looking for was “it depends.”[/quote]
Sorry, Brian was right, only once…but try to figure out why

[quote=“Neo”][quote=“Bu Lai En”]Yep, it’s tough. And you also have to solve it for all possibilities, eg, if A and B are not equal.

Brian[/quote]

My former solution worked for all combinations except the one exception I mentioned above.

1. Let’s say A and B are unequal, then it must be in either A or B.
2. Weigh A against C. If A is heavier, then B and C must be the same and you know that the Buddha is heavier and in A. If A is lighter, you know that the Budda is lighter and in A. If A and C weigh the same, you know that the Buddha is lighter (since A is heavier than B) and in B.
3. So if A was heavier in step 2, weigh A1 vs A2. If they’re the same, you know it’s A3 and it’s heavier. If A1 is heavier, you know it’s A1. If A2 is heavier, you know it’s A2.
If A was lighter in step 2, then step 3 is the same but you know that the Budda is lighter.
If A was the same in step 2, then step 3 is the same but test B1 vs B2. The Buddha is lighter (since A is heavier than B)

So this solutions works when things dont weigh the same. The one failure is when everything keeps weighing the same. Even then, you get the right Buddha, you just dont know if it’s heavier or not.[/quote]

But Alien, how do you know it’s not A4, B4, or C4? There are 12 of them so three groups of 4.

Highlight this with your mouse if you don’t mind a possible spoiler…

[color=yellow]
For the light switch question, it doesn’t say which side of the hall you start on.
[/color]

Alien got the Buddha question right. The other solutions still don’t work. Her solution was a little different than mine, but the key is to start with balancing 4 against 4, then you’ll need to have both putting a buddha that you know to be normal back in, and also some switching of a buddha from one group to another.

Here’s my solution:

W1 (Weigh One)
1,2,3,4 vs 5,6,7,8
If it’s equal then possibilities are 9L (9 is light), 9H, 10L, 10H, 11L, 11H, 12L, 12H go to W2.1
If 1,2,3,4 > 5,6,7,8 then 1H, 2H, 3H, 4H, 5L, 6L, 7L or 8L go to W2.2
If 1,2,3,4 < 5,6,7,8 it’s the same but opposite

W2.1
9,10 vs, 11, X (X is a buddha you know to be normal)
If equal then 12H or 12L (just weigh one against X for 3rd weigh)
If 9,10 > 11,X, then 9H, 10H, or 11L (weigh 9 vs 10 to find out)

W2.2
1,2,6 v 5,3,X
If same then 4H, 7L, 8L (weigh 7 vs 8 to find out)
If 1,2,6 > 5,3,9 then 1H, 2H, 5L (weigh 1 vs 2 to find out)

Congrats to Alien

Brian

[quote]HA! WRONG! You have to cross twice – once to open the doors and once to return to the room with the switches.
Of course, its a bad question because it also forces you to make the assumption that you can see into one room from the other.
So Neo, I presume that the answer your friend’s interviewers were looking for was “it depends.”[/quote]

You only need to walk once from the lightswitches to the room with the lights (which was downstairs and definitely out of sight in the version I heard). You should probably specify that these are lightswitches of the kind where up is off (ie you knwo whether it’s on or off).

Brian

Actually, this isnt a trick question. You start in the room with the switches and they are your normal run-of-the-mill switches. The two rooms are totally separate from each other.

I’m lost. Totally lost. Neo, you owe me several million brain synapses.

heres the solution to the light switches question. highlight the below if you want to know the solution

spoilers spoilers spoilers spoilers spoilers spoilers spoilers spoilers spoilers spoilers spoilers spoilers

[color=yellow]Turn on light switches A and B. After 5-10 minutes, turn B off. Then walk into the next room. Bulb D is on. Touch the two dark bulbs (E & F). The light bulb that is still warm is connected to light switch B. The cold one connects to light switch C.[/color]

[quote=“Neo”]heres the solution to the light switches question. highlight the below if you want to know the solution

spoilers spoilers spoilers spoilers spoilers spoilers spoilers spoilers spoilers spoilers spoilers spoilers

[color=yellow]Turn on light switches A and B. After 5-10 minutes, turn B off. Then walk into the next room. Bulb D is on. Touch the two dark bulbs (E & F). The light bulb that is still warm is connected to light switch B. The cold one connects to light switch C.[/color][/quote]

Then you need to reword the question, which said the two rooms are acroos the hall. I thought they were both across the hall from me, not from each other.

thanks, its hard to think of how a question can be interpreted sometimes…i edited my original post to be more precise