Best way to divide multiple equal groups into known unequal groups. Christmas Lights

Question

I'm not a math guru, but I do like Christmas lights.

I am helping set up a decent sized lighting display for my work, I have measured all outlines and props to figure out how many lights we are going to need for each area.

The lights come with 100 lights/string however, we would like to cut the strings into the appropriate number of bulbs per outline/prop and put connectors on the ends so we don't have a bunch of extra bulbs at the end of props (some only need 16 bulbs on a prop and we don't want to be limited in having to daisy chain everything using lit up bulbs in the strings. We want to splice in plain wire between props.)

I realize there are other factors such as voltage drop, amperage, etc. we need to think about (I'm getting help from an electrical engineer friend for that.)

But I am trying to figure out the best way to divide the 100 bulb strings without having more cuts than I need.

I was first thinking of making the largest cuts first (say I need 56) I now have 44 bulbs on that string, but I don't think it's the most efficient to then say I need a sting of 42 so then cut 2 off when I may have needed 2 sets of 22 that would have equaled 44 and used up the remaining bulbs on that string.

Without including the full strings of 100 that I need, here are the quantities and number of bulbs per string I need... what is the best way for me to start trying to divide these up while avoiding unnecessary cuts and splices?

$$\begin{array}{rl} \text{Quantity} & \text{Bulbs Per String} \\ 8 & 10\\ 1 & 16\\ 4 & 18\\ 1 & 20\\ 4 & 22\\ 1 & 23\\ 1 & 30\\ 4 & 34\\ 34 & 36\\ 1 & 38\\ 3 & 40\\ 2 & 42\\ 1 & 44\\ 1 & 47\\ 1 & 50\\ 2 & 51\\ 3 & 52\\ 4 & 54\\ 2 & 56\\ 1 & 58\\ 1 & 60\\ 1 & 76\\ 4 & 77\\ 1 & 78\\ 2 & 79\\ 3 & 80\\ 4 & 81\\ 2 & 96\\ 8 & 99\\ \end{array}$$

Answer 1

So you need a total of $4,944$ lamps, which means at least $50$ strings.

It would be a hard task also for a computer to proceed "brute-force" and check all possible combinations.

So I can suggest you just to proceed by "wise judgement" in a sort of a "greedy" algorithm. You are not going to achieve a fully optimal result, but the bulk discount you will get should compensate for that.

a) Clearly the 96 and 99 pieces will go by themselves: that makes 10 strings.

b) Next, the 36 lamps sections being the largest majority (34 out of 105) are those "leading the play". I would try and accomodate them first.

The complement of 36 is 64. Two pieces of 36 leave a remainder of 28.
There are not many ways to reuse such remainders.
Therefore proceed and accomodate these two measures first of all, then pass to the remaining ones.

-- Addendum --

I have been pondering upon the fact that the problem would be optimally solved by an "analogic gravitational computer": a prismatic box of width $100$ filled with rods (e.g. pencils) of the given lengths.

Once well shaken a number of times it will reach a (near) optimal filling.
And if the apex angle is progressively reduced to get a parallelepiped $100 \times 1 \times H$, and the material of the box is transparent you can read the solution.
I deem that such analogic calculator will perform the greedy algorithm said above (more or less).