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You have a bucket of 12lbs of a mixture of dry chemicals including both powders and coarse agglomerates. You actually need a 4.5lb sample. The rest will be desposed of.

You pour the sample into your other bucket and notice a serious issue: when such samples are poured from one container to another, they segrate by particle size to some random extent. In this case, the largest chunks poured first so the 4.5lb sample is far coarser than that remaining in the original bucket.

You find a Riffle-type splitter which is designed to split any material poured into it into two identical size samples with the same particle size distribution. This splitter, however, only can deposit the material into two tins which can only hold 1 lb each. If you let the sample pour onto the floor (no tin) the sample will be contaminated. Note: this means that you can only pour 2lbs at a time into the splitter but both 1lb tins it fills will have the sample distribution.

What is the minimum number of tins you need in order to get your sample? What is the minimum number of splits to get your sample with this many tins (to reduce splitting error)?

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Answer (number of tins):

You need three 1 lb tins.

Assumptions:

I've assumed that the splitter starts right away and you can't pour stuff into the splitter and then use the same Tin you've just poured from to catch stuff.

Terminology:

To make things simpler, I'll use the following terms:
tin 1/2/3= a (numbered) 1 pound tin
Bucket = final, 4.5 pound bucket

Big picture:

Essentially, you need 3/8 out of every pound, while preserving the overall distribution. (3/8 * 12 = 4.5). So out of every two pounds we need 3/4 of a pound.

First pour:

Use the splitter to split the first two pounds between the Tin 1 and Tin 2. Empty Tin 2.

Second pour:

Use the splitter to split the pound in Tin 1 between Tin 2 and Tin 3. Pour Tin 2 (1/2 pound) into the bucket. You now have half a pound in the bucket and half a pound in Tin 3, both have the same distribution as the original 2 pounds poured out of the 12 pound bag.

Third pour:

Use the splitter to split the material in Tin 3 between Tin 1 and Tin 2. Pour one out, pour the other into the bucket. The bucket now has 3/4 pound of material that has the same distribution as the first two pounds of material.

To finalize:

Repeat with the other 10 pounds of raw material, two pounds at a time.

If the splitter could pour into the bucket:

Then you'd only need two tins. Each of pours two and three could split into the bucket and the empty tin.

I haven't confirmed this is the minimum number of splits, but this algorithm does it in:

18 Splits, 3 for each 2 pounds of the material.

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  • $\begingroup$ Correct on the number of tins. "This splitter, however, only can deposit the material into two tins" so no it can't split directly into a bucket. $\endgroup$ – kaine Feb 9 '16 at 18:29
  • $\begingroup$ I thought so. I don't think I had it deposit directly into a bucket at any time. I just added that a side comment :) I'll edit it to make it more clear. $\endgroup$ – Duncan Feb 9 '16 at 19:08
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Assumptions:

  • I have 2 buckets (items capable of holding more than 1 lb of material). Based on the OP wording "your other bucket"
  • I can't measure exactly. I can either measure 2 pounds or "everything left".

I know that I need to measure out everything in my bucket to ensure that I have it split evenly.

Step one:

1 bucket with 12 pounds of material
1 empty bucket

First set of 6 pours (all 12 pounds from original bucket):
All poured using 1 tin and losing half my material. Tin is emptied into my other bucket.

This gives:

1 bucket with 6 pounds of material
1 empty bucket
6 pours
1 tin used

Step two:

Next set of 3 pours. All material needs to be saved.

Pour 7 uses 2 tins. One is emptied into the empty bucket, one is set aside.
Pour 8 uses 2 tins (3 total tins now). One is emptied into the bucket, one is set aside.
Pour 9 uses 2 tins (4 total tins now). One is emptied into the bucket. Now the 3 filled tins are emptied into the newly empty bucket.

This gives:

1 bucket with 3 pounds of material
1 bucket also with 3 pounds of material
9 pours
4 tins

Step three:

Pick either bucket to start emptying. Other bucket will be poured into. Only need to save half the material, so only 1 tin used.

Pour 10 gets you 1 pound of material. Pour into bucket (now has 4 pounds in it).
Pour 11 gets you 1/2 pound of material. Pour into bucket.

This gives:

1 bucket with 4 1/2 pounds of material
11 pours
4 tins used
A bit of a messy floor

TL/DR:

4 tins, 11 pours.

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  • $\begingroup$ I'm pretty sure you splits are right but you need 1 less tin...if you purchase directly from the bucket once. $\endgroup$ – kaine Feb 9 '16 at 23:27
  • $\begingroup$ @kaine Can the splitter hold an amount of material (say 2 pounds) without starting to split it? i.e. on pour 9 can I empty the bucket into the splitter, empty 2 tins into the bucket, then place the tins and start the splitter? That would get me down to 3 tins. $\endgroup$ – Joel Rondeau Feb 10 '16 at 16:47

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