Deriving a Formula for IMP4 POW 6

[sequoia22]

In this problem you need to verify weights using a balance scale. To verify a range of weights varying from 0-infinity, you need to purchase weights to balance the scale.

There are two methods:

The first method allows you to only put weights on one side of the scale. You can of course add weights together, but you cannot place weights with the object you are trying to verify (therefore, i am limited to addition)

I have found that the weights needed fall neatly into 2^X. (So if i have a 4 pound weight, i can weight up to seven pounds (2x - 1)
I have a whole table of values, but it wont post here. I cant seem to get a correct formula, i can find any value, but i cant express it.

Weight to Verify---Combo of weights --- Number of weights
1---------------- - ----------- 1 --------------- 1
2 ----------------- ------------- 2 ---------------- 2
3 ---------------- ----------- 2 + 1 --------------- 2
4 ---------------- ----------- 4 ---------------- 3
5 ---------------- ---------- 4 + 1 ----------------- 3
6 -------------- ----------- 4 + 2 ---------------- -- 3
7 --------------- ---------- 4 + 2 + 1 ---------------- 3
8---------------- ---------- 8 --- ------------------ 4
9------------------ ------------ 8 + 1 ------------------ 4
10--------------- ---------- 8 + 2 ------------------ 4

This pattern continues. What i need to find, is for what weight i need to verify, how many weights i will need. (Column 1 and 3)


Method 2:
You can put weights on both sides, allowing subtraction. I have found this fits 3^x.

Can someone please help me arrive at a formula? If i find the first one, I'm sure i can arrive at one for method 2 similarly .

Thank you for all the help.

 

[sequoia22]

Do i need to clarify more?

The tables bad, i know, and sorry about that.

But what i need to find really is at X for pounds how many weights will i need.

Here is the data (weight on left) weights needed on right.

1-1
2-2
3-2
4-3
5-3
6-3
7-3
8-4
9-4
10-4
11-4
12-4
13-4
14-4
15-4
16-5


And it continues like this. I tried using regression, but it did not really work well (or i did not know how to use it.) I will even take a recursive function, at least from there i can quantify it.

 

[soroban]

Hello, sequoia22!

Method 1: You can place weights on one side only.
Method 2: You can place weights on both sides.


You are correct!

Method 1 uses binary weights: \(\displaystyle 1,2,4,8,16,\hdots\)
Method 2 uses ternary weights: \(\displaystyle 1,3,9,27,81, \hdots\)

In both cases, we have two pans:
. . the right pan, with the commodity to be weighed
. . the left pan, initially empty.



Binary weights: Convert the desired weight to base-2.

For example: 87 pounds.

. . We have: .\(\displaystyle 87 \:=\:1010111_2 \;=\;\:64 + 16 + 4 + 2 + 1\)

which tells us exactly which weights to place in the left pan.



Ternary weights: Convert the desired weight to base-3.

The resulting ternary number will have 0's, 1's and maybe some 2's.
We will "convert" the 2's.
. . Starting from the right, replace each "2" with "-1".
. . Add 1 to the digit to its immediate left.
. . If that digit becomes "2", repeat the procecure.


Example: \(\displaystyle 88 \;=\;10021_3\)

. . . \(\displaystyle \text{Then }1\;0\;0\;2\;1 \text{ becomes }1\;0\;1\;\text{-}1\;0\)

. . \(\displaystyle \text{That is: }\:88 \;=\;81 + 9 - 3 + 1\)

The "positive" weights are placed in the left pan.
The "negative" weights are placed in the right pan with the commodity.

\(\displaystyle \text{Hence, we have: }\;\boxed{81}\;\boxed{9}\;\boxed{1} \;\;\begin{array}{c}_{=} \\ ^{\wedge} \end{array}\;\;\boxed{3}\;\boxed{C}\)


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[sequoia22]

Thanks for the help, but i haven't really dealt with base 2 math, or converting it for that matter. Is there a way to get it without converting it to another base? Im not asking for the actual formula, but perhaps what type of formula it would be.