Discussion
Just joined in the discussion about cakes https://www.pistonheads.com/gassing/topic.asp?h=0&... and that made me think of a logic problem that I solved many years ago ... so here is the problem which I hope the PH massive should enjoy solving :
You have 12 objects that all look and feel exactly the same; lets use matchsticks as an example. One of the matchsticks is slightly different to the 11 others in that it weighs more or could be less.
You have a chemical balance but no standard weights
You can only use the chemical balance three times to discover which matchstick is different and whether it's heavier or lighter taking into account all possibilities
Good luck if you care to have a go
You have 12 objects that all look and feel exactly the same; lets use matchsticks as an example. One of the matchsticks is slightly different to the 11 others in that it weighs more or could be less.
You have a chemical balance but no standard weights
You can only use the chemical balance three times to discover which matchstick is different and whether it's heavier or lighter taking into account all possibilities
Good luck if you care to have a go
Edited by RATATTAK on Thursday 11th January 21:08
Edited by RATATTAK on Thursday 11th January 21:36
Clue
That first weighing tells you that it could be one of the four lighter ones on the LHS or one of the four heavier ones on the RHS but ...
more importantly, it tells you that it's NOT one of the four that were not on the scales, so they become 'standards'
so graphically you have:
^^^^ it could be one of these and if it is, it's got to be lighter
vvvv it could be one of these and if it is, it's got to be heavier
ssss all of these are "standards'
That first weighing tells you that it could be one of the four lighter ones on the LHS or one of the four heavier ones on the RHS but ...
more importantly, it tells you that it's NOT one of the four that were not on the scales, so they become 'standards'
so graphically you have:
^^^^ it could be one of these and if it is, it's got to be lighter
vvvv it could be one of these and if it is, it's got to be heavier
ssss all of these are "standards'
Edited by RATATTAK on Friday 12th January 21:49
Phud said:
Next two weighs are to find which of the vvvv or ^^^^ has the mismatched stick.
Then as you will have two sticks in each pan, remove one stick from each pan, this will tell you if the stick remains in a pan or in your hand, if mismatched stick is in the pan it will rise or fall, if pans are level then you will know form which pan you too the stick from.
is this what you mean ?Then as you will have two sticks in each pan, remove one stick from each pan, this will tell you if the stick remains in a pan or in your hand, if mismatched stick is in the pan it will rise or fall, if pans are level then you will know form which pan you too the stick from.
second weighing
vv l vv
which will give you vv and ssssssssss
third weighing
v l v
which will give you v and ssssssssssss
but what if your second weighing balances ?
you've then got ^^^^ and ssssssss and you will need a third and a fourth weighing
so FAIL I'm afraid
Edited by RATATTAK on Friday 12th January 22:27
ash73 said:
Matches 123456ABCDEF, let's say 2 is lighter (or E could be heavier)
123456 / ABCDEF, left side up so 1-6 must be lighter (or ABCDEF must be heavier)
123ABC / 456DEF, left side up so 1-3 must be lighter ( or DEF must be heavier)
1 / 2, right side up so 2 lighter (if level 3 would be lighter) (but what about DEF ?)
123456 / ABCDEF, left side up so 1-6 must be lighter (or ABCDEF must be heavier)
123ABC / 456DEF, left side up so 1-3 must be lighter ( or DEF must be heavier)
1 / 2, right side up so 2 lighter (if level 3 would be lighter) (but what about DEF ?)
ash73 said:
True!
Maybe need to do three combinations, e.g.
123456 / ABCDEF
123ABC / 456DEF
135ACE / 246BDF
If left up, left up, right up 2 must be lighter
If left up, left up, left up D must be heavier
etc
Devil's advocate says:Maybe need to do three combinations, e.g.
123456 / ABCDEF
123ABC / 456DEF
135ACE / 246BDF
If left up, left up, right up 2 must be lighter
If left up, left up, left up D must be heavier
etc
How does that work if 3 is lighter or for that matter if F is heavier ?
Phud said:
second weighing
vv/vv if nothing third weighing is ^^/^^ as you have already seen SSSS can be ignored
if you your second or third weighing one side dips then remove a single from each side so you would end up with v/v or ^/^ and then either v and v in a hand or ^ and ^
the result of you taking a single stick out will show you which one of the 4 was the odd one
That's four weighings ... you can only have threevv/vv if nothing third weighing is ^^/^^ as you have already seen SSSS can be ignored
if you your second or third weighing one side dips then remove a single from each side so you would end up with v/v or ^/^ and then either v and v in a hand or ^ and ^
the result of you taking a single stick out will show you which one of the 4 was the odd one
Clue:
Do NOT ignore your standards
V8LM said:
2nd weighing is take three from the left and put on the right, take three from the right and put on the table, and take three from the table on put on the left.
Well done, you've got it down to one of three and you know if it's heavier or lighter and the third weighing is easysecond weighing
sss^ l ^^^v
RHS goes up it's one of ^^^ and it's lighter
Scale balances it's one of vvv (on the table) and it's heavier
lj04 said:
5 on both sides if the same heavier is matchstick leftover. If not take the heavier group of 5 , 2 on each scale if the same, heavier matchstick is one leftover. If not take the heavier group of 2 I think you can guess the rest.
Have you taken into account that it could be lighter ?Gassing Station | Science! | Top of Page | What's New | My Stuff