Todays riddle

roflmao!!! That second one is great.

In a world without <font color=red>walls </font color=red>or <font color=green>fences </font color=green>, what use have we for <font color=red>Windows </font color=red>or <font color=green>Gates.</font color=green>

Well done.

Did you figure that out, or did you get some help from a <font color=green> lizard<font color=black>?

Wingding - the guy your mother warned you about

....*sigh of admiration*....<font color=green>Lizards<font color=black>....is there anything they CAN'T do?

Wingding - the guy your mother warned you about

That's fair. They are still amazing.

This post will elevate me to Ancient Poster. I dedicate this post to the might of your <font color=green>lizards<font color=black>.

Wingding - the guy your mother warned you about

Every time I masturbate, I enter a post. That's the kind of sad existence I have. My computer monitor and keyboard are a right mess.

Wingding - the guy your mother warned you about

I score that many in an afternoon.

Wingding - the guy your mother warned you about

That would be 3 groups of 4. The first weigh would leave 4 balls (shush Wingding!). Then you weigh 2 and 2, giving you the 'heavy pair'. Then you weigh the pair to get the heavy one.

That one was easy!

<b><font color=blue>~ <A HREF="http://www.btvillarin.com/phpBB/viewtopic.php?t=324" target="_new">My System Specs</A> ~<font color=blue></b>

Yes, I told you. So here is the REAL challenge. Whoever solves this will be rewarded with extreeme respect and WELL DONE!

The challenge is basically the same. You have 12 golf balls, and are allowed to use the scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether its heavier or lighter than the rest.

Is this at all possible? And how do you do that?

<i><b>"I don't understand what it is! Let me kill it!" -- Worf</b></i>

Right I've had a look at this...

I don't think it is possible.

Take 3 groups of 4 and do the last one. At the end of the third weigh, you have two balls, but you need a 4th weigh to tell which is the deviant.

Take 6 and 6 as your first weigh. You have no way of knowing which 6 is the deviant 6. Needs I think 5 weighs to do this.

Am I right?

<b><font color=blue>~ <A HREF="http://www.btvillarin.com/phpBB/viewtopic.php?t=324" target="_new">My System Specs</A> ~<font color=blue></b>

If you divide them into three groups of four you can determine whether you have a heavy or a light ball by weighing to and observing and then weighing the third set with one of the other two. If the first two are the same then the odd ball is in the third group but you do not know if it is heavy or light. So you weigh it with one of the other groups. From that you can determine if it is heavy or light. After that you take a the set that has the heavy/light ball and weigh those in pairs of two. From that you can determine which of two balls is heavier/lighter. After that, since you are out of scale uses, you take the two with different masses, mount them on a quartz strand with two large masses around the mass. Measure the force on the balls due to the gravitational constant and from that you can determine which ball is heavier/lighter. And best of all you only used the scale three times.

In a world without <font color=red>walls </font color=red>or <font color=green>fences </font color=green>, what use have we for <font color=red>Windows </font color=red>or <font color=green>Gates.</font color=green>

4 groups of 3. Weigh two of them to eliminate 6 balls.

Weigh one of the eliminated groups of 3 with one of the suspect groups of 3, to detemine which is the rogue group.

Finally taking into account the movement of the scales on the first and second weighings, weigh 2 of the rogue 3 to get your results.

I get the feeling it's not that easy.

<b><font color=blue>~ <A HREF="http://www.btvillarin.com/phpBB/viewtopic.php?t=324" target="_new">My System Specs</A> ~<font color=blue></b>

You then weigh 3 of the suspect 6 with 3 of the equilibriem 6. That will give you the 3.

<b><font color=blue>~ <A HREF="http://www.btvillarin.com/phpBB/viewtopic.php?t=324" target="_new">My System Specs</A> ~<font color=blue></b>

Yes. With that approach you sometimes will need more than 3 weighings (this words sounds suspeciously wrong).

I will give you a hint: When you weigh some balls, you not only gain information about the balls on the scale, but also on the balls you don't weigh. But I sense from your attempt that you already figured that out?

<i><b>"I don't understand what it is! Let me kill it!" -- Worf</b></i>

U'r allowed to weigh 3 times, right? 4 groups of 3.
Weigh first 2 groups. If they give equilibrium, weigh next 2 groups. Then on the 3rd weighing, whichever group has the heavier, take the 2 balls and weigh em against each other. Thus one will either be heavier, or the reamining one not weighed will be the heavier. With this approch, it can be done in even 2 steps sometimes.....
1*** 2*** 3*** 4***
weigh 1*** vs 2***
if 1***weighs same as 2***
weigh 3*** vs 4***
whichever is heavier, take 2 balls from and weigh against each other X* vs X*
if both r at equilibrium, X* u left out is it, otherwise heavier is it.
If 1***!=2***
do same approch where u weigh 2 balls.

What if you had admin rights to life?

O, I see.

What if you had admin rights to life?

Damn, that 12 ball one is tough. I think I have it though.

Divide them up into 3 groups of 4. Call them Group A, B, and C.

Weigh Group A vs Group B.



Case 1: Group A = Group B.

In this case, you know the deviant ball is in group C. Weigh c1, c2 and c3 against a1, a2, and a3.



case 1.1: They are the same.

You know the deviant ball is c4. Weigh it against a1 to find out if it is light or heavy.

case 1.2: c1, c2 and c3 are heavier.

Weigh c1 vs c2. If they are the same, you know the ball is c3 and it is heavier. If they differ, you know that the heavier ball is the one.

case 1.3: c1, c2, and c3 are lighter.

Same as above. Weigh c1 vs c2. If they are the same, you know c3 is the light ball. If they differ, the ball that is lighter is the one you're after.




Case 2: A is heavier than B.

Take a1, a2, and b1, and weigh them against a3, b2, and c1.



case 2.1: a1, a2, and b1 are heavier.

Since you know the heavy ball must be in A or the light ball is in B, you can tell that the ball you want is either a1, a2, or b2. Weigh a1 vs a2.


case 2.1.1: They differ.

You know that the heavier ball is your ball.

case 2.1.2: They are the same.

You know that b2 is your ball and it is lighter.



Case 2.2: a1, a2, and b1 are lighter.

You know either b1 is your ball and it is lighter, or a3 is your ball and it is heavier. Weigh a3 vs c1.


case 2.2.1: a3 is heavier than c1.

You know a3 is your ball and it is heavy.

case 2.2.2: a3 is the same as c1.

You know b1 is your ball and it is lighter.



Case 2.3: a1, a2, and b1 are the same weight as a3, b2, and c1.

Basically, at this point, you have a4, b3, and b4 left. From your first measurement, you also know that a4 cannot be the lighter ball, nor can b3 nor b4 be the heavier ball. Take a4 and b3, and weigh them against c1 and c2.


case 2.3.1: a4 and b3 are heavier.

Since you know that b3 cannot be the heavier ball, a4 must be the ball you want, and it is heavy.

case 2.3.2: a4 and b3 are lighter.

Since you know that a4 cannot be the lighter ball, b3 must be the ball you want and it is lighter.

case 2.3.3: a4 and b3 are the same as c1 and c2.

You know that b4 is the deviant ball, and from your very first measurement, you know that it must be light.


Case 3: B is heavier than A:

Just repeat case 2 over, but switch all the balls labelled a for b.

DONE


Phew, that was hard.

Some day I'll be rich and famous for inventing a device that allows you to stab people in the face over the internet.

My hamster would just mollest your leg until you weighed each one!

<b>Cogposto tomsa, ergo sum - <i>Descartes</i>
</b>Translation:<b> I post at Toms Hardware, Therefor I am. </b>

Ah!! it didn't format it correctly. Oh, well, it is still legible, but the indentation made it a lot easier to read.

In a world without <font color=red>walls </font color=red>or <font color=green>fences </font color=green>, what use have we for <font color=red>Windows </font color=red>or <font color=green>Gates.</font color=green>

Sorry, I didnt get back to you earlier. I do need to sleep once in a while (bloody waste of time)

But yes. <b>RESPECT and WELL DONE !!!</b>

My solution differs slightly though. But its basically the same. Ill include it here, just for completion:

<pre>1) The first measurement always starts out with 4 balls on each side of the scale. Three

outcomes are possible:
1.1) Equal, all balls on the scale are normal and the deviating ball is among the

remaining 4. [Proceed with step 2.1]
1.2) Left side is heaviest. The deviating ball is among the eight on the scale, and the

remaining 4 are normal. But we also know that one of the balls on the left side is

heavier than normal OR one of the ball on the right side is lighter than normal. [Proceed

with step 2.2]
1.3) Right side is heaviest. The deviating ball is among the eight on the scale, and the

remaining 4 are normal. But we also know that one of the balls on the right side is

heavier than normal OR one of the ball on the left side is lighter than normal. [Proceed

with step 2.3]
</pre><p><pre>2) Second measurement
2.1) Put 3 normal balls one the left side, and 3 of the unknown on the right side. Three

outcomes are possible:
2.1.1) Equal. The selected 3 unknown are all normal, and the last unknown must be the

deviating. [Proceed with step 3.1]
2.1.2) Left side is heaviest. One of the 3 unknown is the deviating ball and it must be

lighter than normal. Further the last unknown ball must me normal. [Proceed with step

3.2]
2.1.3) Right side is heaviest. One of the 3 unknown is the deviating ball and it must be

heavier than normal. Further the last unknown ball must be normal. [Proceed with step

3.3]

2.2) We now have three groups: A) 4 normal balls. B) 4 balls where one of them may be

heavier. C) 4 balls where one of them may be lighter. Put 2 balls from B and 2 balls from

C on the left side. Put 2 balls from A, 1 ball from B and 1 from C on the right side.

Three outcomes are possible:
2.2.1) Equal. All balls on the scale are normal balls. We are left with two unknowns,

where 1 may be heavier and 1 may be lighter. [Proceed with step 3.4]
2.2.2) Left side heaviest. The two balls on the left side that could have been lighter

are now known to be normal. Further the 1 ball on the right side that could have been

heavier is now known to be normal. We are left with 3 unknowns, where 2 may be heavier

and 1 may be lighter [Proceed with step 3.5]
2.2.3) Right side heaviest. The two balls on the left side that could have been heavier

are now known to be normal. Further the 1 ball on the right side that could have been

lighter is now known to be normal. We are left with 3 unknowns, where 2 may be lighter

and 1 may be heavier [Proceed with step 3.6]

2.3) This is practically identical to 2.2, so i leave it out.
</pre><p><pre>3) Third measurement
3.1) The last deviating ball is measured against a normal ball to figure out whether is

lighter or heavier than normal.
3.2) We are left with three balls, and we know the deviating is lighter than normal. Put

one on each side of the scale. If equal the remain is the deviating, otherwise the scale

shows which is the lighter.
3.3) We are left with three balls, and we know the deviating is heavier than normal. Put

one on each side of the scale. If equal the remain is the deviating, otherwise the scale

shows which is the heavier.
3.4) Put the two unknown balls on the left side and two normal on the right side. If the

left side is heavier, the ball that could be lighter is normal and the other is the

deviating (heavier) one. If left side is lighter, the ball that could be heavier is

normal and the other is the deviating (lighter) one. This measurement cant turn out equal

since that would mean that all 12 balls are equal in weight.
3.5) Put one of the balls that may be heavier and the ball that may be lighter on the

left side. Put 2 normal balls on the right side. If equal the, all balls on the weight

are normal and the remaining unknown is the deviating (heavier) one. If not equal the

deviating ball is identified as in step 3.4
3.6) Put one of the balls that may be lighter and the ball that may be heavier on the

left side. Put 2 normal balls on the right side. If equal the, all balls on the weight

are normal and the remaining unknown is the deviating (lighter) one. If not equal the

deviating ball is identified as in step 3.4
</pre><p><pre>All this can be put in a graphical way, which may make more sense:
</pre><p><pre>Notation:

____|____ : The scale with some balls on the left and right side


Ball weight:

X: Unknown
-: Perhaps lighter than normal
+: Perhaps heavier than normal
H: Heavier than normal
L: Lighter than normal
O: Normal

Weigh result:
=: Left and right side are equal
/: Left side is heavier than right side
\: Left side is lighter than right side

!: All 12 balls are equal in weight (never occurs)

Initial condition:

abcdefghijkl
XXXXXXXXXXXX


1. weighing and result 2. weighing and result 3. weighing and result

abcd efgh abcdefghijkl abc ijk abcdefghijkl a l abcdefghijkl
XXXX|XXXX = OOOOOOOOXXXX OOO|XXX = OOOOOOOOOOOX O|X = OOOOOOOOOOOO <- !
/ OOOOOOOOOOOL
\ OOOOOOOOOOOH

i j abcdefghijkl
/ OOOOOOOO---O -|- = OOOOOOOOOOLO
/ OOOOOOOOOLOO
\ OOOOOOOOLOOO

i j abcdefghijkl
\ OOOOOOOO+++O +|+ = OOOOOOOOOOHO
/ OOOOOOOOOHOO
\ OOOOOOOOHOOO



abcdefghijkl aebf cgij abcdefghijkl dh kl abcdefghijkl
/ ++++----OOOO +-+-|+-OO = OOO+OOO-OOOO +-|OO = OOOOOOOOOOOO <- !
/ OOOHOOOOOOOO
\ OOOOOOOLOOOO

abcdefghijkl bg kl abcdefghijkl
/ ++OOOO-OOOOO +-|OO = HOOOOOOOOOOO
/ OHOOOOOOOOOO
\ OOOOOOLOOOOO

abcdefghijkl ce kl abcdefghijkl
\ OO+O--OOOOOO +-|OO = OOOOOLOOOOOO
/ OOHOOOOOOOOO
\ OOOOLOOOOOOO



abcdefghijkl aebf cgij abcdefghijkl dh kl abcdefghijkl
\ ----++++OOOO -+-+|-+OO = OOO-OOO+OOOO -+|OO = OOOOOOOOOOOO <- !
/ OOOLOOOOOOOO
\ OOOOOOOHOOOO

abcdefghijkl ce kl abcdefghijkl
/ OO-O++OOOOOO -+|OO = OOOOOHOOOOOO
/ OOOOHOOOOOOO
\ OOLOOOOOOOOO

abcdefghijkl bg kl abcdefghijkl
\ --OOOO+OOOOO -+|OO = LOOOOOOOOOOO
/ OOOOOOHOOOOO
\ OLOOOOOOOOOO</pre><p><i><b>"I don't understand what it is! Let me kill it!" -- Worf</b></i>

<b>That's</b> why I couldn't solve it. I hate maths!



<b><font color=blue>~ <A HREF="http://www.btvillarin.com/phpBB/viewtopic.php?t=324" target="_new">My System Specs</A> ~<font color=blue></b>

I just went over your solution very quickly (have to go to work soon), but it seems as though our solutions are exactly the same, with the exception of formatting. The actual method is the same though

Some day I'll be rich and famous for inventing a device that allows you to stab people in the face over the internet.

All these riddles are driving me crazy. I have VACATION... so I don't want to think to much.

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NO... please NO!

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I'm up for that. I feel like i get dumber everyday.

What good fortune for those in power that people
do not think. - <b>Adolf Hitler</b>

lol.. good way to put it.

Hey does anyone here play Red Alert 2 Yuri's revenge? Wanna go at it in an online game.

What good fortune for those in power that people
do not think. - <b>Adolf Hitler</b>