Half-balls

[TimeSpaceLightForce]

A bearing factory cuts four steel balls into halves with a gram or 1% difference in weight due to laser error.
The eight pieces were all mixed up in a collecting bin and shall be paired into 4 similar spheres..
Using the balance scale and/or the digital scale..How can it be done quickly?

[hhh3]

  Reveal hidden contents

So if the machine makes one half heavier and the other one lighter by a gram, i will have to weigh 3 times using a balance scale to get the correct pairs.

This is unless am missing something big here  

[TimeSpaceLightForce]

  On 8/18/2015 at 2:43 PM, hhh3 said:

Hidden Content

Right but don't miss the solution for using the digital weighing machine..

[harey]

@hhh3: Can you tell how?

Edited August 20, 2015 by harey

[hhh3]

  Reveal hidden contents

well Harey, if you have a balance scale, with the first try - if the scale is balance that means, we can make one sphere with those 2 halves.

If the scale is not balanced, keep them aside and take another pair. If the scale is still unbalanced, that means you got 2 halves which are heavier and 2 which are lighter. Likewise if you continue, u can get all the pairs.

[k-man]

Assuming that all heavy halves are identical (and therefore all light halves are identical too), so we can pair any heavy half with any light half to create a sphere.

  Reveal hidden contents

Using the balance scale it can be done with 5 tries. You need to distinguish from 8C4=70 possible distributions (HHHHLLLL, HLHLHLHL, etc.), but we don't always have to identify every H and L. As soon as we found a HL pair we can make a sphere, so it may be possible to do it in 4 tries, but I seriously doubt it can be done in 3.

Using the digital scale, it can be done in 3 tries. Here is how...

  Reveal hidden contents

Take any 6 halves and weigh them on the digital scale. Based on the result you can distinguish from 3 possible heavy to light (H:L) ratios - 4:2, 3:3,  and 2:4. Note that 4:2 and 2:4 cases are symmetrical, so I will only cover 4:2 case as the other can be resolved the same way.

4:2 case 

The remaining 2 halves that we didn't weigh are light, so put them aside. Take any 4 of the initial 6 and weigh them on the digital scale. Possible results are 4:0, 3:1, and 2:2. 

-- 4:0 case is trivial as we've identified all heavy and all light halves and can put them together.
-- In 3:1 case we know that the 2 halves we didn't weigh the second time make up a sphere, so put them together. Weigh any 2 from 3:1 group to complete the remaining 3 spheres.
-- In 2:2 case we've already identified 2 heavy and 2 light halves, so we can put together 2 spheres. Weigh any 2 of the remaining 4 to complete the remaining 2 spheres.

3:3 case

The remaining 2 halves that we didn't weigh make up a sphere, so put them together. Take any 4 of the 6 and weigh them on the digital scale. Possible results are 3:1, 2:2 and 1:3. Again, 3:1 and 1:3 are symmetrical, so we don't need to review both of them.

-- In 3:1 case we've identified 2 light halves, so one more weighing of 2 will identify the third. 
-- In 2:2 case we can construct another sphere, so it's the same as the 2:2 case above.

Solved in 3 weighings.

Edited August 21, 2015 by k-man
formatting

[TimeSpaceLightForce]

All cleared . Good job !

What's the better weighing machine?

  Reveal hidden contents

[jasen]

I think just sort all the half sphere, then merge the lightest with the heavyest as 1 pair, Keep the procedure 4 times.