Order five balls

[bonanova]

Five balls, A, B, C, D and E weigh, in some order, 1, 2, 3, 4, and 5 pounds.

How many times must a balance scale be used to determine their weights?

If you did the weighing, you wouldn't need a scale, you could tell by lifting them.
So you use an assistant, who does the weighing in another room.
You say which ball(s) to place on the left and which ball(s) to place on the right.
Your assistant tells you Left > Right; Left = Right; or Left < Right.
That counts as one weighing.

Edited November 24, 2013 by bonanova
Re-label the spoiler as a clarification. No clues contined therein

[TimeSpaceLightForce]

nice weighing puzzle..we can use 2balls vs 1 or 2 only?

1+2=3
1+3=4
1+4=5
2+3=5
2+3=1+4
2+4=1+5
2+5=3+4

[bonanova]

No restrictions. The cases you list are all permitted.

You may read the spoiler for clarification. It gives no clues.

[wolfgang]

If we took 2 balls in each side, and we are told they are equal, so we should expect to have one out of 16 possibilities:

A+B= C+D
A+B= D+C
B+A= C+D
B+A= D+C
and so on...
for second weigh..
A against B....we should expect to have one of 8 results as( >) , or one of 8 results as( < ), and the 3rd weigh should be..
C against D...
now we can determine the two heavy balls and the two light ones
with 4th weigh, between the two heavies... and 5th between the two others, we can arrange them.

[bonanova]

Wolfgang, there are five balls. Do you include E here?

[TimeSpaceLightForce]

Weigh any 2balls (1 vs 1)

weigh another 2balls

weigh 2 heavier balls

weigh 2 lighter balls

After 4th weighing of 1 vs 1 we have
1 2345
2 1345
3 1245
4 1235
5 1234
then 5th weighing: 2mids vs 2ends
34=25
15<34
15=24
15>23
14=23
then 6th weighing: 1st two vs 3rd ball
23>4
12<4
12=3

Edited November 25, 2013 by TimeSpaceLightForce

[TimeSpaceLightForce]

missed the heavier light vs lighter heavy (1 vs 1) making it 7 weighing ..

[bonanova]

missed the heavier light vs lighter heavy (1 vs 1) making it 7 weighing ..

You can do it in fewer.

[plasmid]

There are 5! = 120 ways of arranging weights 1-5 to balls A-E.

We seek a solution where a set of N weighing outcomes leads to an unambiguous assignment to one of those 120 possible arrangements.

If possible weighing outcomes are "left side heavier", "right side heavier", or "both equal" then after N weighings there are 3^N possible sets of outcomes.

3⁵ = 243 while 3⁴ = 81, so 5 is the minimum theoretically possible that can distinguish among the 120 possible arrangements.

Weigh A+B versus C+D, then weigh A+C versus B+D, then weigh A+D vs B+C. There are then two possible scenarios to worry about: either 1) none of those weighings led to balance, or 2) one of those weighings led to balance.

If none of the first three weighings are balanced, then E must be either 2 or 4.

If any of the first three weighings are balanced, then E must be 1, 3, or 5.

The reason for this is that, if E is either 1, 3, or 5, then there exists some combination of pairs of A, B, C, and D such that the two pairs are equal weight, and the first three weighings tested all possible pairings. If E is 2 or 4, then A, B, C, and D must contain three odd numbers and one even number so they will never be balanced.

We know that E must be either 2 or 4. Suppose E is weight 4: then the mass with weight 5 will always be on the heaviest side of the first three weighings and can be identified, but the mass with weight 1 will not always be on the lightest side (1+5 > 2+3). Suppose E is weight 2: then the mass with weight 1 will always be on the lightest side but the mass with weight 5 will not always be on the heaviest side. So after the first three weighings, you will be able to identify the weight of E based on whether there was a mass that was always on the lighter side or a mass that was always on the heavier side, and you will know either the mass with weight 1 (if E is weight 2) or the mass with weight 5 (if E is weight 4). Label the remaining unknown masses X, Y, and Z.

If you know weights 1 and 2, you can figure out the rest by weighing 1+X versus Y followed by one more weighing.

Possible values for (X,Y) after the fourth weighing

1+X > Y: (4,3), (5,3), (5,4)... weigh Z versus Y+1 and you'll be able to distinguish these scenarios

1+X = Y: (3,4), (4,5)... weigh Z versus X and you'll be able to distinguish these scenarios

1+X < Y: (3,5)... you're done

If you know weights 4 and 5, you can figure out the rest by weighing X+Y versus 4 followed by X versus Y.

Possible values for (X,Y) after the fourth weighing

X+Y < 4: (1,2), (2,1)...

X+Y = 4: (1,3), (3,1)...

X+Y > 4: (2,3), (3,2)... in any of those cases you can weigh X versus Y to figure out the rest of the weights.

Relabel the masses A, B, C, and D with labels M, N, P, and Q such that M+N = P+Q is the weighing that was balanced. The other two weighings would be M+P versus N+Q and M+Q versus N+P. Without loss of generality, have label M go to the mass that was heaviest in both of those subsequent weighings (one mass must have been on the heaviest side in both of them), and you'll know that M is the heaviest out of the four and N must therefore be the lightest.

For the fourth weighing, weigh P versus Q. Label the heavier of them R and the lighter S. You now know that the order of the four weights is M > R > S > N, and all you have to do is find out whether E is 1, 3, or 5 and you'll have the problem solved.

Weigh E versus S+N. If E=1, then E < S+N. If E=3, then E = S+N. If E=5, then E > S+N.

[bonanova]

Kudos!