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Condom Logic #2
Sex God
No clues on this one cos Iain is really close. ;)
Okay - assuming
(1) 1,2,3,4 > 5, 6,7 ,8 Then 9,10,11,12 are all = G
(2) a 1,6,7,8 > 5,10,11,12 Then F = 1,5 or 9
b 1,6,7,8 = 5,10,11,12 Then F = 2, 3 or 4
c 1,6,7,8 < 5,10,11,12 Then F = 6,7or 8
Right I've got this far now - I can now get the fake from result a but not from result b or c
Oh if Ice has gone to bed - I'm going to start crying now.
(1) 1,2,3,4 > 5, 6,7 ,8 Then 9,10,11,12 are all = G
(2) a 1,6,7,8 > 5,10,11,12 Then F = 1,5 or 9
b 1,6,7,8 = 5,10,11,12 Then F = 2, 3 or 4
c 1,6,7,8 < 5,10,11,12 Then F = 6,7or 8
Right I've got this far now - I can now get the fake from result a but not from result b or c
Oh if Ice has gone to bed - I'm going to start crying now.
Sex God
Quote by celticq
Okay - assuming
(1) 1,2,3,4 > 5, 6,7 ,8 Then 9,10,11,12 are all = G
(2) a 1,6,7,8 > 5,10,11,12 Then F = 1,5 or 9 (not 9, you've already identified it as good)]
b 1,6,7,8 = 5,10,11,12 Then F = 2, 3 or 4
c 1,6,7,8 < 5,10,11,12 Then F = 6,7or 8
Right I've got this far now - I can now get the fake from result a
Yes...
but not from result b or c
Why not?
Okay by substituting the condoms for 11 tealights and a stapler (which I have pretended looks like a tealight) I think I have now got it.
If fake is 2, 3 or 4 then its heavier
If its 6, 7 or 8 then its lighter
Hurray! Either way one weigh decides the fake.
TF you had not gone to bed Ice I had visions of sitting here all night waiting for the answer.
If fake is 2, 3 or 4 then its heavier
If its 6, 7 or 8 then its lighter
Hurray! Either way one weigh decides the fake.
TF you had not gone to bed Ice I had visions of sitting here all night waiting for the answer.
Sex God
Quote by celticq
Okay by substituting the condoms for 11 tealights and a stapler (which I have pretended looks like a tealight) I think I have now got it.
If fake is 2, 3 or 4 then its heavier
If its 6, 7 or 8 then its lighter
Hurray! Either way one weigh decides the fake.
You know that and I know that, but explain it to the audience - how do you identify the fake from three suspects in one weighing?
If 1 = 9 then 5 = F or if 1 < or > 9 then 1 = F
If 2 > 3 then 2 = F, if 2 = 3 then 4 = F, if 2 < 3 then 3 = F
If they balance the unweighed is the fake
Removing 2,3 and 4 made the weigh less so 2,3 or 4 will be heavier than G
6,7 and 8 are always on the lightest side so they will be lighter than G
All these numbers are confusing me now - I'm starting to doubt myself. It is very much easier to demostrate than explain
If 2 > 3 then 2 = F, if 2 = 3 then 4 = F, if 2 < 3 then 3 = F
If they balance the unweighed is the fake
Removing 2,3 and 4 made the weigh less so 2,3 or 4 will be heavier than G
6,7 and 8 are always on the lightest side so they will be lighter than G
All these numbers are confusing me now - I'm starting to doubt myself. It is very much easier to demostrate than explain
OK, this better be right!
Number the Condoms 1 to 12.
Weigh 1-4 against 5-8 - if the same go to A if different go to B
A - get a condom you know is not fake (IE 1-8 - call it 0) weigh 0 and 9 against 10 and 11
-if the same go to (1) if 0and 9 heavier go to (2) if 10 and 11 heavier go to (3)
B - ok condoms 9-12 are real. Take one real one ( called 0) weigh 0, 6 and 3 against 5,7 and 2.
- if same go to (4). If 0,6,3 heavier go to (5) if 5,7,2 heavier go to (6)
(1) - we know 12 is the fake and weigh against any other condom.
(2) - weigh 10 against 11. If balanced 9 is fake and heavy, if one is heavier then that is the fake.
(3) - weigh 10 against 11, if balanced 9 is fake and light, if one is heavier that is the fake.
(4) Weigh 1 against 4, if the same 8 is fake and light. If one if heavier it is the fake.
(5) Weigh 5 against 7, if the same 3 is fake and heavy. If one is lighter then it is fake.
(6) Either 2 is heavy or 6 is light, weight one and see.
Nor that better be right after I typed it, I am sure there is a simpler way by giving all the condoms a 3 digit no and having weighting combinations but couldn't be arsed to figure it out.
Number the Condoms 1 to 12.
Weigh 1-4 against 5-8 - if the same go to A if different go to B
A - get a condom you know is not fake (IE 1-8 - call it 0) weigh 0 and 9 against 10 and 11
-if the same go to (1) if 0and 9 heavier go to (2) if 10 and 11 heavier go to (3)
B - ok condoms 9-12 are real. Take one real one ( called 0) weigh 0, 6 and 3 against 5,7 and 2.
- if same go to (4). If 0,6,3 heavier go to (5) if 5,7,2 heavier go to (6)
(1) - we know 12 is the fake and weigh against any other condom.
(2) - weigh 10 against 11. If balanced 9 is fake and heavy, if one is heavier then that is the fake.
(3) - weigh 10 against 11, if balanced 9 is fake and light, if one is heavier that is the fake.
(4) Weigh 1 against 4, if the same 8 is fake and light. If one if heavier it is the fake.
(5) Weigh 5 against 7, if the same 3 is fake and heavy. If one is lighter then it is fake.
(6) Either 2 is heavy or 6 is light, weight one and see.
Nor that better be right after I typed it, I am sure there is a simpler way by giving all the condoms a 3 digit no and having weighting combinations but couldn't be arsed to figure it out.
Sex God
Quote by cplonamission
OK, this better be right!
Number the Condoms 1 to 12.
Weigh 1-4 against 5-8 - if the same go to A if different go to B
A - get a condom you know is not fake (IE 1-8 - call it 0) weigh 0 and 9 against 10 and 11
-if the same go to (1) if 0and 9 heavier go to (2) if 10 and 11 heavier go to (3)
B - ok condoms 9-12 are real. Take one real one ( called 0) weigh 0, 6 and 3 against 5,7 and 2.
- if same go to (4). If 0,6,3 heavier go to (5) if 5,7,2 heavier go to (6)
(1) - we know 12 is the fake and weigh against any other condom. Yes.
(2) - weigh 10 against 11. If balanced 9 is fake and heavy, Yes. if 10 is heavier then that is the fake. No. According to condition A, (0,9) is heavier than (10,11), therefore if 10 and 11 don't balance, the fake is the lighter of those two.
(3) - weigh 10 against 11, if balanced 9 is fake and light, if one is heavier that is the
(4) Weigh 1 against 4, if the same 8 is fake and light. If one if heavier it is the is only true if on your first weighing (1,2,3,4) is heavier than (5,6,7,8). If number 1 was fake and light, you would not get the right answer, you would get the answer that number 4 was fake and heavy.
(5) Weigh 5 against 7, if the same 3 is fake and Number 2 could be fake and light.. If one is lighter then it is
(6) Either 2 is heavy or 6 is light, weight one and Number 3 could be light.
Nor that better be right after I typed it, I am sure there is a simpler way by giving all the condoms a 3 digit no and having weighting combinations but couldn't be arsed to figure it out.
The main reason statements 4-6 aren't guaranteed correct is because you made an assumption about which group was heavier on your first weighing and didn't allow for the scales tipping the other way.
There is another reason, a strategic one on the second weighing - you're on the right lines though.
CQ has actually solved it I think, she's just having trouble explaining the details of the third weighing. ;)
Iain has solved the (1,2,3,4) = (5,6,7,8) scenarios, and his strategy for the third weighing will adapt to the outcomes of (1,2,3,4) > (5,6,7,8) and (1,2,3,4) < (5,6,7,8).
I'm expecting a quiet day at work, so I'll be composing Condom Logic #3. I'm sure all the pieces of this puzzle will have been fitted together by the time I get back tonight ;)
Later dudes
OH dear have just come into my office. Hundreds of bits of paper with numbers and arrows on, steel ruler, lemon meringue pie plate, pound coins and condoms all over the floor. Yikes.
Having read cplonamissions theory I actually thought it was better, as mine has the flaw that in deciding between 2 fakes and a genuine you can always find the fake but not always say if it is heavy or lighter.
Assuming the fake is in the group of 8 not the 4 (since we can all see who to do that bit)
You are left with 4 heavies, 4 lights and 4 genuine and then I spent a long time using their theory to balance Heavy+Light+Genuine V's Heavy+Light+Light checking this out only to read this bit.......
Closely followed by this bit
Can you not release #3 until late afternoon or nobody will be getting any lunch
Having read cplonamissions theory I actually thought it was better, as mine has the flaw that in deciding between 2 fakes and a genuine you can always find the fake but not always say if it is heavy or lighter.
Assuming the fake is in the group of 8 not the 4 (since we can all see who to do that bit)
You are left with 4 heavies, 4 lights and 4 genuine and then I spent a long time using their theory to balance Heavy+Light+Genuine V's Heavy+Light+Light checking this out only to read this bit.......
Quote by Ice Pie:bounce:
CQ has actually solved it
Closely followed by this bit
Quote by Ice Pie
I think
Can you not release #3 until late afternoon or nobody will be getting any lunch
Well done CQ :bounce: :bounce: :bounce:
I got absolutely nowhere as also thought we had to say whether the condom would actually be heavier or lighter - phone up Ice to say 'Ice hunny, it's impossible to tell if it's heavier or lighter' or words to that effect anyway
Half an hour of "discussions" later
:fuckinghell: :fuckinghell:
:fuckinghell: :fuckinghell: :banghead:
and he finally realised that I was trying to work out if it was actually heavier or lighter. By that time I had lost it totally (stuff happening to distract me - ie pissed parents ringing me up to ask who they were watching on telly!! )
I think it was pure fluke that I got the first one - only by working backwards (much more my style) and got the final bit only after being directed to the priority rule
So, come on Ice gis another ............. actually, can you wait until Thursday, when I've got time to concentrate :?
I got absolutely nowhere as also thought we had to say whether the condom would actually be heavier or lighter - phone up Ice to say 'Ice hunny, it's impossible to tell if it's heavier or lighter' or words to that effect anyway
Half an hour of "discussions" later
:fuckinghell: :fuckinghell:
:fuckinghell: :fuckinghell: :banghead: and he finally realised that I was trying to work out if it was actually heavier or lighter. By that time I had lost it totally (stuff happening to distract me - ie pissed parents ringing me up to ask who they were watching on telly!!
)I think it was pure fluke that I got the first one - only by working backwards (much more my style) and got the final bit only after being directed to the priority rule
So, come on Ice gis another ............. actually, can you wait until Thursday, when I've got time to concentrate :?
Quote by Ice Pie
OK, this better be right!
Number the Condoms 1 to 12.
Weigh 1-4 against 5-8 - if the same go to A if different go to B
A - get a condom you know is not fake (IE 1-8 - call it 0) weigh 0 and 9 against 10 and 11
-if the same go to (1) if 0and 9 heavier go to (2) if 10 and 11 heavier go to (3)
B - ok condoms 9-12 are real. Take one real one ( called 0) weigh 0, 6 and 3 against 5,7 and 2.
- if same go to (4). If 0,6,3 heavier go to (5) if 5,7,2 heavier go to (6)
(1) - we know 12 is the fake and weigh against any other condom. Yes.
(2) - weigh 10 against 11. If balanced 9 is fake and heavy, Yes. if 10 is heavier then that is the fake. No. According to condition A, (0,9) is heavier than (10,11), therefore if 10 and 11 don't balance, the fake is the lighter of those two.
(3) - weigh 10 against 11, if balanced 9 is fake and light, if one is heavier that is the
(4) Weigh 1 against 4, if the same 8 is fake and light. If one if heavier it is the is only true if on your first weighing (1,2,3,4) is heavier than (5,6,7,8). If number 1 was fake and light, you would not get the right answer, you would get the answer that number 4 was fake and heavy.
(5) Weigh 5 against 7, if the same 3 is fake and Number 2 could be fake and light.. If one is lighter then it is
(6) Either 2 is heavy or 6 is light, weight one and Number 3 could be light.
Nor that better be right after I typed it, I am sure there is a simpler way by giving all the condoms a 3 digit no and having weighting combinations but couldn't be arsed to figure it out.
The main reason statements 4-6 aren't guaranteed correct is because you made an assumption about which group was heavier on your first weighing and didn't allow for the scales tipping the other way.
There is another reason, a strategic one on the second weighing - you're on the right lines though.
CQ has actually solved it I think, she's just having trouble explaining the details of the third weighing. ;)
Iain has solved the (1,2,3,4) = (5,6,7,8) scenarios, and his strategy for the third weighing will adapt to the outcomes of (1,2,3,4) > (5,6,7,8) and (1,2,3,4) < (5,6,7,8).
I'm expecting a quiet day at work, so I'll be composing Condom Logic #3. I'm sure all the pieces of this puzzle will have been fitted together by the time I get back tonight ;)
Later dudes
OKOk, the strategy is right if you assume on the first weigh in if they are different that condoms 5-8 are heavier.
If condoms 1-4 are heavier then you reverse the procedure, that will work, but I am not going to type it up.
I should have seen that last night but it was in the morning!!