# Counting how many items can be weight on a scale if we pick the weights optimally [duplicate]

Suppose you have a balance and are allowed to choose the weights for its functionality. The objective is to pick the weights in a way that maximizes the number of items we can check by the scale. Given $$n$$ weights all of which are different how many different ways are there to arrange them on the scales of the balance? Assuming that the two sides of the balance are indistinguishable what conclusion can we make about the max range of items that we can use the scale to weigh?

The solution logic goes as follows:

Each weight can be placed on the left scale, the right scale or on the table. These are 3 options so for $$n$$ weights we have $$3^n$$ configurations of the balance. One of these is symmetric between left and right scale (when no weights are on the scales). The rest are asymmetric between the left and right scale. Which means the same configuration is counted twice.

So far it is clear. So we would have for e.g. $${1gram, 2gram}$$:

Left, Right, Table
-    -      (1,2) (--)
1    -      (2)
-    1      (2)  (*)
2    -      (1)
-    2      (1)  (*)
1    2      (-)
2    1      (-)  (*)
1,2  -      (-)
-    1,2    (-)  (*)


So we would have all the marked with $$*$$ as duplicates and the $$--$$ i.e. all on the table as the symmetric one we exclude since no weights are on the scales.

But then it states:

The total number of different configurations is $$\frac{3^n - 1}2 + 1$$

Where is the $$+1$$ coming from?

Going over an example for e.g. $$n = 2$$ for example $$1gram, 3gram$$
We can measure items of $${1gram, 2gram, 3gram, 4gram}$$ which is a total of $$\frac{3^2 - 1}2 =4$$ and not $$5$$ as the solution claims.

Am I misunderstanding something here?

• @RobPratt: No because I know how to derive the type. The $+1$ was the confusion on
– Jim
Jan 12 at 8:24
• please consider formatting your table using the newly available GitHub-flavored markdown syntax: meta.stackexchange.com/questions/356997/… Jan 13 at 11:05
– Jim
Jan 13 at 17:53
• @XanderHenderson: Thank you for checking my post. The other question is about proving that a similar formula to the one in my post solves the problem. My question is about why the specific formula in my post has an $+1$ and it was because of what the answer says i.e. include weight =0 case
– Jim
Jan 16 at 12:57
• @Jim You are making a different argument, now. Initially, you seemed to be arguing that the post should be reopened because it had an accepted answer. Now you are arguing that it should be reopened because it is not a duplicate. In response to that argument, I must admit that I am on the side of the close voters: the questions are very closely related, and the answers to the older question, if read carefully, address this newer question. Jan 16 at 13:09

• So I should be counting that I can measure ${0grams, 1grams, 2grams, 3grams, 4grams}$ which is 5 for $n = 2$ I gave as example? Seems odd that we measure $0grams$
• @Jim: It’s not so odd if you imagine that you always put the thing to be weighed on the left side and allow that thing to have negative weight (e.g., by attaching a helium-filled balloon to the pan): then you actually get all $3^n$ possible integral amounts from $-\frac{3^n-1}2$ to $\frac{3^n-1}2$, inclusive. Think of it as a physical model of balanced ternary. Jan 11 at 19:48
• @BrianM.Scott: I didn't understand the by attaching a helium-filled balloon to the pan. What do you mean by that?
• @Jim: Of course. The point is that in this model it’s quite clear that it makes sense to weigh something with zero weight, since in fact we are weighing everything from $-m$ to $m$ for some $m\in\Bbb Z^+$. If you make the two sides of the balance indistinguishable, you lose the possibility of weighing negative weights, but you can still weigh zero ounces. Jan 11 at 21:50