Rotten apples

anna_gg · 2016-02-11 01:19:32

We have a basket with 8 apples of which 2 are rotten; one is lighter than the good apples and the second, strangely enough, is heavier; maybe because of the worms that have penetrated it All the good apples are of the same weight.
How can we tell if the two rotten apples together weight more or equal or less than two good ones - in only 3 weightings with a two-arm balance scale? Obviously there are no visible signs of the rotten apples.

Last edited by anna_gg (2016-02-20 03:05:42)

anna_gg · 2016-02-18 07:12:02

No chance to make apple pie...

bobbym · 2016-02-18 18:59:50

I have heard of 6 weighings and 4 weighings for this problem. I did not know there is a 3 weighing.

anna_gg · 2016-02-19 06:16:57

Me neither and I'm struggling to find it!

bobbym · 2016-02-19 13:28:55

You must be aware that for two defective coins some of the questions are still open areas of research.

anna_gg · 2016-02-19 19:32:44

Yes but we are not looking for the exact rotten apples, only if the two of them together weight more or less than two of the good ones.

bobbym · 2016-02-20 02:20:28

Are we given that the sum of the two bad apples ≠ sum of two good apples?

anna_gg · 2016-02-20 03:05:19

No, it may also be equal. I will update the wording. Thanks for noticing!!

bobbym wrote:

Are we given that the sum of the two bad apples ≠ sum of two good apples?

Grantingriver · 2016-02-22 00:43:36

Divide the 8 apples into two groups each consists of four apples (call them group A and group B) and divide each goup into subgroup each contains two apples. Now put on each the scale's arm one subgroup of group A , and do the same for group B. So two of the weighing attempts will be used and you will get one of the following possibilities:

1) The subgroups of each group are balanced
2) One group has its subgroups balanced and the othe has not.
3) Both groups have their subgroups unbalanced.

In the first possibility you will know of course that the two rotten apples are equal to two good ones. If you get the second possibility you will know that the group which has its subgroups balanced are all consists of good apples, so put each of its subgroup together and each of the subgroup of the other group together and put each collection in on arm of the scale (using the third attempt) then you will have one of the following results:

1) both of them palanced (in this case the two rotten apples equal in weight to a two good apples)
2) the collection of the group of the unbalanced subgroup weighs more then the collection of the other group (in this case the two rotten apples weigh more then the two good apples)
3) the collection of the group of the balanced subgroup weighs more then the collection of the group of the unbalanced subgroup (in this case the two rotten apples weighs less two good apples)

Finally, if you get the third possibility, put the subgroup of the group A which weighs less with the subgroup of group B which weighs more and vice versa, and put on each arm of the scale one collection (using the third attempt), then you will get the same results that we get in the second prossibility hence you know either the two rotten apples are equal to or more or less in weight then two good apple.

Q.E.F

anna_gg · 2016-02-22 07:50:09

In the third possibility, I understand that if the two sides balance, then the two rotten apples weight the same as two good ones.
However, if they do not balance, and say, A1+B2>A2+B1 - how do we know in which side we have the rotten apples? So, in essence, how do we deduce if they are lighter or heavier?

Grantingriver wrote:

Finally, if you get the third possibility, put the subgroup of the group A which weighs less with the subgroup of group B which weighs more and vice versa, and put on each arm of the scale one collection (using the third attempt), then you will get the same results that we get in the second prossibility hence you know either the two rotten apples are equal to or more or less in weight then two good apple.

Grantingriver · 2016-02-22 08:45:36

It is very obvious. When you make in the third possibility the two collections (A1+B2) and (A2+B1) they will be "exactly" like the two collections formed in the second possibility. Why?!! Because their are just two rotten apples one less in weight then normal and the other more in weight then normal, so in the third possibility one of these two apples will be in a certain subgroup of group A and the other in a certain subgroup of group B, therefore when weighing the subgroups of the group which contains the less weight apple the subgroup of the normal apples will be heavier. On the other hand, when you weigh the subgroups of the group which contains the over weight apple the normal group will be lighter, so when you cross mix these subgroups of each group (the over weighted with the less weighted and vice veras) to form two new collections, one of them will contain the both rotten apples while the other will be compeletly normal!! So you will reach the same final situation of the second possibility.

Q.E.F

anna_gg · 2016-02-22 09:29:57

Right, I agree. The rotten apples will be either in A1+B2 or in A2+B1 and the scale will NOT balance. But how will we know which side contains the rotten apples and which side the good ones?

Grantingriver wrote:

It is very obvious. When you make in the third possibility the two collections (A1+B2) and (A2+B1) they will be "exactly" like the two collections formed in the second possibility. Why?!! Because their are just two rotten apples one less in weight then normal and the other more in weight then normal, so in the third possibility one of these two apples will be in a certain subgroup of group A and the other in a certain subgroup of group B, therefore when weighing the subgroups of the group which contains the less weight apple the subgroup of the normal apples will be heavier. On the other hand, when you weigh the subgroups of the group which contains the over weight apple the normal group will be lighter, so when you cross mix these subgroups of each group (the over weighted with the less weighted and vice veras) to form two new collections, one of them will contain the both rotten apples while the other will be compeletly normal!! So you will reach the same final situation of the second possibility.
Q.E.F

Grantingriver · 2016-02-22 10:36:52

You don't have to know which side of the scale (in both group) contains the rotten apple!!! Why? To explain this situation suppose "without loss of generality" that group A contains the heavier rotten apple in subgroup A1, then when you weigh the subgroups of this group (group A) the arm of the scale containing the subgroup which include the heavier rotten apple (it could be A1 or A2) will lift the other arm which containing the subgroup of the normal apples and when you do the same with group B which contains the lighter rotten apple the subgroup which contains that apple (it could be B1 or B2 no matter) will be lifted, so when you cross mix the subgroups of each group, the lifting side will match the lifted side and the rotten apples will be in the same collections.

Note: it is obvious that the thing which made you miss the point is confusing the heavier and the lighter sides with the notations of the subgrouops. I think you asked yourself the following question:
"If I donate a particular subgroup of group A "A1" and the other subgroup "A2" and do the same for group B then after weighing, the cross mix will be as follow :(A1+B2) and (A2+B1) but wait a minute!! How the hell could I know that B2 contains the lighter rotten apple or the heavier one and the same for B1?!!!" This is the mistake which you have made. We did not talk about the notations we talked about the heavier and the lighter sides!! So if group A contais the heavier side it could be A1 or A2 no matter and the same for the lighter side it could be B1 or B2 so focus on the heavier and the lighter side not on the notations. I think it is very clear now (please let me know in any case).

anna_gg · 2016-02-22 23:19:12

@Grantingriver: Thank you for your patience!!
I perfectly understand that by mixing the subgroups together, in the form of A2+B1 versus A1+B2, the two rotten apples are found together on one arm of the scale. This is clear.
For the sake of the example, let's suppose that A1>A2 and B1>B2. If A1 contains only good apples, then A2 contains the lighter rotten, so the other rotten must be the heavier, so it's in B1. So we have one rotten apple in A2 and one in B1. We put them together on one disk, while on the other we put A1 and B2, which we know that only contain good apples. Then of course we can deduce whether the two rotten apples are heavier or lighter, depending on which arm of the scale is lifted.
BUT: when we mix the subgroups, we DO NOT KNOW beforehand if A1 contains good or rotten apples. It might be that A1 contains the heavier rotten apple, in which case, B2 contains the lighter. So the two rotten apples are in A1+B2 (while in the first case, they were in A2+B1).
SO, how can you tell if they lighter or heavier, as compared to 2 good ones???

Grantingriver wrote:

You don't have to know which side of the scale (in both group) contains the rotten apple!!! Why? To explain this situation suppose "without loss of generality" that group A contains the heavier rotten apple in subgroup A1, then when you weigh the subgroups of this group (group A) the arm of the scale containing the subgroup which include the heavier rotten apple (it could be A1 or A2) will lift the other arm which containing the subgroup of the normal apples and when you do the same with group B which contains the lighter rotten apple the subgroup which contains that apple (it could be B1 or B2 no matter) will be lifted, so when you cross mix the subgroups of each group, the lifting side will match the lifted side and the rotten apples will be in the same collections.
Note: it is obvious that the thing which made you miss the point is confusing the heavier and the lighter sides with the notations of the subgrouops. I think you asked yourself the following question:
"If I donate a particular subgroup of group A "A1" and the other subgroup "A2" and do the same for group B then after weighing, the cross mix will be as follow :(A1+B2) and (A2+B1) but wait a minute!! How the hell could I know that B2 contains the lighter rotten apple or the heavier one and the same for B1?!!!" This is the mistake which you have made. We did not talk about the notations we talked about the heavier and the lighter sides!! So if group A contais the heavier side it could be A1 or A2 no matter and the same for the lighter side it could be B1 or B2 so focus on the heavier and the lighter side not on the notations. I think it is very clear now (please let me know in any case).

Grantingriver · 2016-02-23 02:24:18

You think that "because we don't know which subgroup will contain the good apples and which subgroup will contain the rotten apple" there is a possibility that the good apples will be mixed with a subgroup which contains a rotten apple, so we have to find a way to avoid this situation when cross mixing!! In fact, this is impossible to happen!! In cross mixing the subgroups to form two new collections (each consists of four apples) the rotten apples will "spontaneously" be both in one collection and the other collection will contain only good apples "in all cases" (this is the main point and the tricky one of the puzzle which you always miss).

anna_gg · 2016-02-23 03:04:10

No, I don't think there is such a possibility, I understand that ALL 2 rotten apples are in the same collection. They can either be at the left arm OR at the right arm, BUT in any case, together! ONLY THAT we don't know in which arm!! So, if we see, for example, that one side is raised and the other is lowered, we don't know which side has the good apples and which one has the rotten.

Grantingriver wrote:

You think that ... there is a possibility that the good apples will be mixed with a subgroup which contains a rotten apple.

Grantingriver · 2016-02-23 11:28:25

Ok...if the two arms of the scale balanced then the two rotten apple will be equal in weight to two normal apples, but if they unbalanced then this test even if it produces a new result in the third possibility (either the two rotten apples are more or less in weight to two good apples), it will be inconclusive. However, why we reject this kind of results, since it provides some useful information?!

Note: If we remove one of the restrictions (either more in weight or less in weight) then this test will be conclusive with only three weighing attempts. On the other hand, if we add fouth weighing attempt then the test will be conclusive with its restrictions.

phrontister · 2016-02-23 16:21:22

Hi;

You nearly solved it, Grantingriver!

I agree with possibilities 1 & 2 from post #9.

Grantingriver wrote:

...if they unbalanced then this test even if it produces a new result in the third possibility (either the two rotten apples are more or less in weight to two good apples), it will be inconclusive.

Yes, that's how I see it too.

The illustrations below confirm that, and are from my Excel spreadsheet, which is based on Grantingriver's post #9 method.

Spreadsheet explanation:-

I gave each good apple a weight of 20, the lighter rotten apple 18, and the heavier rotten apple either 21, 22 or 23 - by random choice, which gives an even chance of the 2 rotten apples together weighing (a) more than, (b) less than, or (c) equal to the combined weight of 2 good ones. Excel shuffles the order of the apples.

The boxes:-
Row 1: Each box shows the weight of one of the 8 apples.
Row 2: The boxes show the sum of the weights of the two row 1 apples directly above them.
Row 3: The results of the first two weighings, being either H or L (heavy or light).
Row 4: Cross-weighings (as per post #9), with the left side comprising group A's L + group B's H (the highlighted cells), and the right side comprising group A's H + group B's L (the non-highlighted cells).
Row 5: The result of the third weighing (either "2R = 2G", "Left < Right", or "Left > Right")...in the illustrations I've only used "Left < Right" examples.

The result of 2 of my simulations, with the first image saying that the 2 rotten apples are heavier, and the second that 2 good apples are heavier:

Same result as above but with the weights omitted, representing an actual weighing session:

Still can't see how to do this in 3 weighings...

Last edited by phrontister (2017-02-26 21:58:36)

anna_gg · 2016-02-23 19:19:12

Dear Grantingriver and Phrontister,
Many thanks for your efforts
It appears we must keep trying because the problem does have a solution, with the current restrictions and number of weightings

This is pretty much how far I could get myself...
So any further thoughts or ideas are most welcome!!

Grantingriver · 2016-02-24 03:32:26

It turns out that the third possibilty could be solved just in three weighing attempts, so the test ,in its three attempts and restrictions, is in fact conclusive!!!

In the third possibility, when we weigh the subgruops of group A and the subgroups of group B (using the first two attempts). We will have, in group A, one of the scale's arm be lifted and the same for group B, but we are using a scale!!, so we can know the following results:

the difference between the lighter and the heavier arms in in the first weighing attempt (weighing the subgroups of A), call it "d1", is equal to or more or less then the same difference in the second attempt (when weighing group B), call it "d2".

Now if they are equal then the two rotten apples are equal in weight to two good apples. If they are not, we cann't know which side of the balance contains the rotten apple in both weighing attempts of each group, so we remove the lighter subgroup of B from the arm (in the second attempt) and put on it the lighter side of the subgroups of A (using the third and last attempt), If the two arms balanced then we know ,for sure, that group A contains the heavier rotten apple and group B have the lighter one, So we have:

1) d1>d2 (in this case the two rotten apples weigh more then two normal ones)
2) d1<d2 ( in this case the two rotten apples weigh less then two normal)

Finally, if in the theird attempt, the scale which contains the heavier subgroup of group B remains unlifted, then we know, for sure, that group B contains the heavier rotten apple, and we have the results:

1) d2>d1 (the two rotten apples are more in weight then two normal ones)
2) d2<d1 (the two rotten are less in weight then two normal).

Note: the third possible case (the heavier subgroup of group B in the third attempt is lifted) can not happen, since the heavier rotten apple must be either in group "A" or in group "B".

Q.E.F

anna_gg · 2016-02-25 23:38:47

It seems that, as per the puzzle specs, we cannot rely upon differences in d1 and d2. Only information we can get from each weighting is whether the left side is heavier, or the right side is heavier or balance.

Grantingriver wrote:

It turns out that the third possibilty could be solved just in three weighing attempts, so the test ,in its three attempts and restrictions, is in fact conclusive!!!
In the third possibility, when we weigh the subgruops of group A and the subgroups of group B (using the first two attempts). We will have, in group A, one of the scale's arm be lifted and the same for group B, but we are using a scale!!, so we can know the following results:
the difference between the lighter and the heavier arms in in the first weighing attempt (weighing the subgroups of A), call it "d1", is equal to or more or less then the same difference in the second attempt (when weighing group B), call it "d2".
Now if they are equal then the two rotten apples are equal in weight to two good apples. If they are not, we cann't know which side of the balance contains the rotten apple in both weighing attempts of each group, so we remove the lighter subgroup of B from the arm (in the second attempt) and put on it the lighter side of the subgroups of A (using the third and last attempt), If the two arms balanced then we know ,for sure, that group A contains the heavier rotten apple and group B have the lighter one, So we have:
1) d1>d2 (in this case the two rotten apples weigh more then two normal ones)
2) d1<d2 ( in this case the two rotten apples weigh less then two normal)
Finally, if in the theird attempt, the scale which contains the heavier subgroup of group B remains unlifted, then we know, for sure, that group B contains the heavier rotten apple, and we have the results:
1) d2>d1 (the two rotten apples are more in weight then two normal ones)
2) d2<d1 (the two rotten are less in weight then two normal).
Note: the third possible case (the heavier subgroup of group B in the third attempt is lifted) can not happen, since the heavier rotten apple must be either in group "A" or in group "B".
Q.E.F

Grantingriver · 2016-02-26 02:12:17

We donn't have to find the differences "precisley" we just have to know "by eye inspection" which difference is more (in group A or in group B). This puzzle is very tricky!! Notice carefully: If we could not tell which difference is more "by eye inspection" then we can not tell the difference between the two rotten apples and two good apples "by eye inspection" even if we know them and weigh them directly, so they will be equal in this case. However, if the two rotten apples are heavier or lighter than two nomal apples "by eye inspection" then we can tell the difference between group A and group B "by eye inspection" also. It is know very clear if you get the point. (Please let me know in any case)

Q.E.F

anna_gg · 2016-03-02 07:54:07

Any ideas??

bobbym wrote:

Are we given that the sum of the two bad apples ≠ sum of two good apples?

Math Is Fun Forum

#1 2016-02-11 01:19:32

Rotten apples

#2 2016-02-18 07:12:02

Re: Rotten apples

#3 2016-02-18 18:59:50

Re: Rotten apples

#4 2016-02-19 06:16:57

Re: Rotten apples

#5 2016-02-19 13:28:55

Re: Rotten apples

#6 2016-02-19 19:32:44

Re: Rotten apples

#7 2016-02-20 02:20:28

Re: Rotten apples

#8 2016-02-20 03:05:19

Re: Rotten apples

#9 2016-02-22 00:43:36

Re: Rotten apples

#10 2016-02-22 07:50:09

Re: Rotten apples

#11 2016-02-22 08:45:36

Re: Rotten apples

#12 2016-02-22 09:29:57

Re: Rotten apples

#13 2016-02-22 10:36:52

Re: Rotten apples

#14 2016-02-22 23:19:12

Re: Rotten apples

#15 2016-02-23 02:24:18

Re: Rotten apples

#16 2016-02-23 03:04:10

Re: Rotten apples

#17 2016-02-23 11:28:25

Re: Rotten apples

#18 2016-02-23 16:21:22

Re: Rotten apples

#19 2016-02-23 19:19:12

Re: Rotten apples

#20 2016-02-24 03:32:26

Re: Rotten apples

#21 2016-02-25 23:38:47

Re: Rotten apples

#22 2016-02-26 02:12:17

Re: Rotten apples

#23 2016-03-02 07:54:07

Re: Rotten apples

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