Condom Logic #2

27 Nov 2004

Following the raid on the Munch (Condom Logic #1), the leader of the condom collectors has decided to go to the Swingingland condom dealer and trade in some of his more valuable condoms for an extremely rare and collectable Chateauneuf de Pape 69 flavoured condom. After some negotiation, a price of one dozen strawberry Pavlova condoms is agreed.
However, the condom dealer knows from experience that one in every twelve strawberry Pavlova condoms is fake, and tells his customer he wants to examine the proffered condoms. A visual inspection is no good because the fakes are so good the only way they can be distinguished from the genuine article is by their relative weight, so the dealer announces that he will check the condoms by weighing them against each other in pairs on his balance scales. This will enable him to identify a fake condom after a maximum of seven trials, but the collector is in a hurry and demands that the dealer completes his investigation after no more than three weighings. The dealer thinks for a minute and agrees to to the collector's condition.
Sure enough, after three weighings, the dealer announces that one of the 12 condoms is fake, he knows which one it is, and he knows whether it is lighter or heavier than the real thing.
How was he able to guarantee identifying the fake and determining its weight in only three weighings?

27 Nov 2004

U got me there!!
Having said that i am half pissed so it wont be very hard!!

27 Nov 2004

Oh no no no no no!!!!!!!!!
I've gotta take the kids to see the Incredibles now!!!!!!!! I'm gonna be about 3 hours behind!!!!!
OK nobody look at this thread until after nine o'clock ok? biggrin

27 Nov 2004

Quote by MISSCHIEF
Oh no no no no no!!!!!!!!!
I've gotta take the kids to see the Incredibles now!!!!!!!! I'm gonna be about 3 hours behind!!!!!
OK nobody look at this thread until after nine o'clock ok?

OK, I won't look. rotflmao

27 Nov 2004

I'm guessing this is wrong since it took me several hours to work out last nights when I saw it today. But I have to go back to work again so I'm going to have a quick guess at
3 and 3
3 and 3
From the 3 which weighed lightest one and one.
confused

27 Nov 2004

:scared:
Not looking

27 Nov 2004

does he first put six condoms in each balance dish ( weigh 1)
if the dishes balance then there are no fakes however on this occasion the dishes do not
he takes the six condoms from the dish that weighed the heaviest (keeping the other 6 as genuine) and splits them into 2 piles of three each and weighs again with three in each weigh dish. (weigh 2)
he saves the three condoms from the dish that weighs the lightest as genuine.
he then puts one condom from the three in the heavier dish onto the table and weighs off the remainig two against each other (weigh three)
if the condoms balance each other out then the one on the table is the fake. If they dont then the heaviest of the condoms in the balance scale is the fake

27 Nov 2004

he takes the six condoms from the dish (saving all those in the lighter dish as genuine) that weighed the heaviest

You don't know whether the fake is heavy or light; working that out is part of the problem.

27 Nov 2004

Quote by davej
then the heaviest of the condoms in the balance scale is the fake

But what if the fakes are lighter than the originals? confused

27 Nov 2004

The closest I can come up with is this:
1. He weighs six against six. One set of six will weigh more than the other. (because one condom weighs more or less than the others)
2. He takes one set of six, doesn't matter which but we'll say the heaviest, and weighs them 3 against 3. If they balance, the odd one isn't in those 6. He also now knows that the bad one is lighter than the good ones.
3. He now takes the other 6, the lighter ones, and weighs them 3 against 3. Whichever side goes up, he knows the bad one is amongst those 3.
This is where it all goes pear shaped. If he had one more go on the scales, he could put any two, one against one. If they balance, it's the third one. If they don't balance, it's the lighter of the two on the scales.
Nearly, but not quite!! I'll check late tomorrow, as I will be otherwise engaged tonight and wont be fit for anything in the morning!!
Dr Mal
cool

27 Nov 2004

Quote by celticq
I'm guessing this is wrong since it took me several hours to work out last nights when I saw it today. But I have to go back to work again so I'm going to have a quick guess at
3 and 3
3 and 3
From the 3 which weighed lightest one and one.

Same mistake as davej... you're assuming that the weight of the fake is known in advance. It isn't. :twisted:

27 Nov 2004

Quote by Ice Pie
does he first put six condoms in each balance dish ( weigh 1)

That tells him nothing he doesn't already know, i.e. the scales won't balance.

he takes the six condoms from the dish (saving all those in the lighter dish as genuine) that weighed the heaviest

You don't know whether the fake is heavy or light; working that out is part of the problem.
bollocks and damn icepie......davej is now sitting in amongst a heap of screwed up bits of paper scribbling, adding and subtracting like a loony tune.......

27 Nov 2004

Weighing six against six, one side will be heavier than the other. As he removes them one at a time from each side, ready for the second weigh (does that still count as one weigh?) Eventually the scales will balance, giving him a choice of two. Knowing all the others still on the scale are good, he can use the following two weighs to compare a good one with each of the suspect pair. Whichever is lighter/heavier than the good one is the fake!

27 Nov 2004

PoloLady I hope your right cos this has been doing my head in ever since icepie posted it......now I know thats not a difficult task but I need to know the answer...
Ice pie, is poloLady right....is she....is she.... :bounce:

27 Nov 2004

Quote by PoloLady
Weighing six against six, one side will be heavier than the other. As he removes them one at a time from each side, ready for the second weigh (does that still count as one weigh?)

By removing a pair and noting whether the scales are affected, he's making a measurement, so that's not going to work unless the first pair he removes happens to include the dud.

Eventually the scales will balance, giving him a choice of two.

Eventually yes, but he is only allowed to make 3 weighings. This method requires up to 7 weighings.

27 Nov 2004

Quote by davej
PoloLady I hope your right cos this has been doing my head in ever since icepie posted it......now I know thats not a difficult task but I need to know the answer...
Ice pie, is poloLady right....is she....is she.... :bounce:

No, sorry.
Removing one from each side and then noting whether the scales balance or stay unbalanced counts as a weighing. If they stay unbalanced, you would still have ten suspects and only one weighing left.

27 Nov 2004

Quote by Ice Pie
PoloLady I hope your right cos this has been doing my head in ever since icepie posted it......now I know thats not a difficult task but I need to know the answer...
Ice pie, is poloLady right....is she....is she.... :bounce:

No, sorry.
Removing one from each side and then noting whether the scales balance or stay unbalanced counts as a weighing. If they stay unbalanced, you would still have ten suspects and only one weighing left.
Double bollocks :doh:

27 Nov 2004

Quote by davej
Double bollocks :doh:

Polyorchidism they call it, which is strange cos orchids actually remind me of the female nether bits. dunno

27 Nov 2004

Quote by davej
PoloLady I hope your right cos this has been doing my head in ever since icepie posted it......now I know thats not a difficult task but I need to know the answer...
Ice pie, is poloLady right....is she....is she.... :bounce:

Sorry davej - I tried :cry:

27 Nov 2004

Quote by Ice Pie
Double bollocks :doh:

Polyorchidism they call it, which is strange cos orchids actually remind me of the female nether bits. dunno

It's a good thing that I'm not well educated cos there's probably a piss take tucked up in there somewhere, fortunately I'm thick and therefore blissfully unaware of it ......
...now then is there going to be any help with this thing or not...or have I got to wait for misschief to get back from the pictures........

27 Nov 2004

ARRRRRGGGGGGHHHHHH!!!!!!! I'm sat waiting to go out and Iain won't stop thinkin about which friggin condom is the fake!! lol

Ice pie i could kill ya!! evil

27 Nov 2004

Quote by PoloLady
PoloLady I hope your right cos this has been doing my head in ever since icepie posted it......now I know thats not a difficult task but I need to know the answer...
Ice pie, is poloLady right....is she....is she.... :bounce:

Sorry davej - I tried :cry:
you and me both poloLady, never mind at least we had a go, I don't mind getting it wrong whats the old saying...something about the only people that don't make mistakes are those that never try......
must be right, every time I fLuck up mrs davej says something about me being very trying.

27 Nov 2004

Quote by red_931
ARRRRRGGGGGGHHHHHH!!!!!!! I'm sat waiting to go out and Iain won't stop thinkin about which friggin condom is the fake!! Ice pie i could kill ya!!

Don't worry honey we can go out now coz I think I got it all worked out.
How's this theory Icepie, does it stand up to examination??
Divide condoms into groups of 9 and 3
Further divide group of 9 into 3 groups of 3 and mark them A B C
Divide group of 3 into single condoms marked a b c
Weigh Aa against Bb then Aa against Cb
If Aa and Bb and Aa and Cb balance then fake condom is c and it's weight will be known when it's measured against either a or b
If Aa and Bb balance but Aa and Cb imbalance then fake is in group C and because we already know that Aa and Bb balance then the weight of the fake can be determined by the direction of the imbalance so if Cb outweighs Aa then fake is heavier so by weighing 2 of the 3 in C against each other the fake can be determined by either the 2 condoms balancing in which case the fake is the condom left over or in the case of imbalance then the heavier condom is the fake.
TA-DAAAAAAAAA

27 Nov 2004

Ok then
First divide the 12 condoms in to 4 groups of three.
weigh the first two groups,
two outcomes.
1the dud is in the first six
2 the dud is in the unweighed 6
eat the end of the day your left with 6 condoms to play with
redface

split the remaining 6 condoms into 3 groups of 2
weigh the first two groups.
2 scenarios
1. the dud is in the 4 weighed condoms
2. the dud is in the 2 remaining condoms. (pretty straight forward from there on)
so under scenario 1 you'd then - bollox that won't work - you need to somehow build in a "test" group that you know are genuine to determine the dud - i'd need 4 weighs to be absolutly sure
back to the drawing board.
p.s didn't want to delete all my ramblings - may spark any idea off with someone

27 Nov 2004

Red take Iain out FFS!!!!!....his answer is making me brain hurt more than the bloody question..
:silly:

27 Nov 2004

I'm gonna do now......he thinks his answer is right so i might just be able to tear him away lol

27 Nov 2004

Just rushed in on my tea break to see if the answer was up yet.
I like Iain's idea but am wondering if it will still work if Aa does not balance with either Bb or Cb meaning you then have to find the fake in a group of 4 Aa with only one weigh?

27 Nov 2004

Quote by iain_nw
ARRRRRGGGGGGHHHHHH!!!!!!! I'm sat waiting to go out and Iain won't stop thinkin about which friggin condom is the fake!! Ice pie i could kill ya!!

Don't worry honey we can go out now coz I think I got it all worked out.
How's this theory Icepie, does it stand up to examination??
Divide condoms into groups of 9 and 3
Further divide group of 9 into 3 groups of 3 and mark them A B C
Divide group of 3 into single condoms marked a b c
Weigh Aa against Bb then Aa against Cb
If Aa and Bb and Aa and Cb balance then fake condom is c and it's weight will be known when it's measured against either a or b
If Aa and Bb balance but Aa and Cb imbalance then fake is in group C and because we already know that Aa and Bb balance then the weight of the fake can be determined by the direction of the imbalance so if Cb outweighs Aa then fake is heavier so by weighing 2 of the 3 in C against each other the fake can be determined by either the 2 condoms balancing in which case the fake is the condom left over or in the case of imbalance then the heavier condom is the fake.
TA-DAAAAAAAAA
Do you mind if I rewrite your solution a bit?

Divide condoms into groups of 9 and 3
Further divide group of 9 into 3 groups of 3 and mark them (1,2,3) - (5,6,7) - (9,10,11)
Divide group of 3 into single condoms marked 4, 8, 12
Weigh (1,2,3,4 against 5,6,7,8) then (1,2,3,4 against 9,10,11,8)
If (1,2,3,4 and 5,6,7,8) and (1,2,3,4 and 9,10,11,8) balance then fake condom is 12 and its weight will be known when it's measured against either 9 or 10

So far so good. - What you actually did in effect was divide them into three groups of four, with one condom moving from one group to another. I changed the lettered groups to individual numbered condoms to make that a bit clearer.

If (1,2,3,4 and 5,6,7,8) balance but (1,2,3,4 and 9,10,11,8) imbalance then fake is in group (9,10,11) and because we already know that(1,2,3,4) and (5,6,7,8) balance then the weight of the fake can be determined by the direction of the imbalance so if (9,10,11,8) outweighs (1,2,3,4) then fake is heavier so by weighing 2 of the 3 in (9,10,11) against each other the fake can be determined by either the 2 condoms balancing in which case the fake is the condom left over or in the case of imbalance then the heavier condom is the fake.
TA-DAAAAAAAAA

What if your first weighing, (1,2,3,4) against (5,6,7,8) does not balance?

27 Nov 2004

OK, here I go.
Divide the 12 into 3 groups of 4.
Weigh these against each other you will know which 4 have the Fake one and if it is heavier or lighter than the real one.
You now have 4 condoms, split into two, take the two with the fake in and weigh against each other.
I think!!

27 Nov 2004

Quote by cplonamission
OK, here I go.
Divide the 12 into 3 groups of 4.
Weigh these against each other you will know which 4 have the Fake one and if it is heavier or lighter than the real one.

This requires two weighings. There is only a one in three chance that one weighing will tell you which group contains the fake, and it won't tell you its weight.

You now have 4 condoms, split into two, take the two with the fake in

How do you know which two contains the fake? You need to weigh two suspects against two known good ones. That uses up your third weighing.

and weigh against each other.

Nope, that's four weighings. :twisted: