# Is there a name for the "binary" sizing scheme?

Not quite sure how to express what I’m thinking, but here goes:

An integer expressed in base-2 notation looks like this:

10110

This requires five independent digits, each capable of two different states. To express the same quantity in base-10 (as 22) requires two independent digits, each capable of ten different states.

There’s an “efficiency” of sorts in base-2; I’m not sure how to express it, but it becomes more apparent in other venues.

Suppose I want to light a room with an occupant-selectable integer number of watts of light, ranging from zero to 22 watts.

Vendor A offers lamps with powers of 1 watt and 10 watts. If I select him, I’ll need to buy a pair of 10-watt lamps and nine 1-watt lamps, a total of eleven lamps to achieve my goal.

But vendor B offers lamps with 1, 2, 4, 8, and 16 watts. If I select him, I need only buy five lamps (16, 8, 4, 2 , and 1 watts). Clearly this is the more economical distribution of sizes, at least as concerns the customer.

You can imagine this arrangement in a lot of different situations where items are used simultaneously: resistors, pumps, lamps, shims, etc.

Is there a name for that sort of sizing scheme?

Not quite the same thing, but you may be interested in ISO 3 (Renard Numbers). These series are made specifically for the use case of what to offer from a selection products with a parameter in a range to efficiently cover the parameter space.

I think they are called “preferred numbers”. There are different sets of them for different purposes, but they generally more or less follow a geometric sequence pattern.

Cost-efficient?

Note that base-3 is optimal when subtractions are allowed:
What four weights allow you to measure any integer weight from 1 to 40 grams on an ordinary balance scale?

I don’t know of a name (besides “binary distribution”), but here’s a little puzzle along these lines:

You have a pan balance and a bunch of objects that weigh from 1 to 40 grams in 1-gram increments. What is the minimum number of reference weights that you need to balance anything in this range?

Damnit, septimus :)!

Can one of the weights be lighter than air and tied to the scale?

Can we use the objects as reference masses once we establish their mass?

Missed the edit window … that came out completely wrong … sorry …

Do you mean just four individual weights … or can we have more than one of each … like four 1 gm chunks?

Another common scheme is to have increments of 1, 2, 5, 10, 20, 50, etc. Many nations’ currency follows this pattern, and it’s also often used for sets of laboratory weights. This gives you most of the efficiency of binary (though you still need two 2s to make 4), but also fits nicely into a base-10 number system.

Just four individual weights, and you can’t use the objects themselves as references for other weighings. You can have more than one of the same number if you want, but note that it implies an inefficiency: you’ll have more than one way of weighing the same object if there’s any duplication.

No need for lighter-than-air weights or the like. The answer is clever, but not “tricky”.

Although not particularly on topic, I have to mention one of my favorite puzzles.

You have 12 coins which look identical. They all weigh the same, except for one which is a counterfeit, and is either heavier or lighter than the other 11, but you don’t know if it’s heavier or lighter. How can you find the counterfeit, and determine whether it is heavier or lighter than the others, by doing only 3 weighings on a two-pan balance scale? A balance scale, of course, simply lets you weigh some set of the 12 coins against another set and see which set is heavier.

I once stayed up most of a night working on this one (and was quite proud of myself when I came up with the solution, which is quite non-trivial).

–Mark

This is simply a tautology.

You decided that each lamp must be either on or off. So you pre-defined the problem to be binary. Given that, of course the powers-of-two bulbs would be best. Because nothing else maps more easily to binary than binary.

Instead, how about a single 22-watt bulb and a control system that can vary its output in 22 steps? By redefining the problem that way, one bulb beats the heck out of the massively less efficient binary-based idea of using five bulbs.

Also notice that your 5 binary bulbs can actually achieve 0 to 31 watts in 32 steps. You ought to subtract efficiency points for the overkill this represents. The two-10s-plus-9-ones solution is in some sense less inefficient because can only overkill up to 29, not 31.

The other posters have talked about various other aspects of meeting certain pre-defined combinations. That’s the more interesting area for exploration.

This can also be a three dimensional problem. Imagine that you are ‘stuffing’ boxes into a standard 40’ shipping container and want to leave the minimum amount of empty space.

Your boxes may be all the same size, which makes things fairly easy, but not trivial; or they may be random sizes.

You can buy a computer program that will do it for you these days like this

Just in case you were serious. You get the same effect by moving the weight to the other scale.

It’s true that I deliberately defined this problem so that a base-2 distribution was the best answer, even though a single bulb with a dimmer would have been possible. But some situations don’t lend themselves to an analog solution. This whole thing came to mind as I was researching critical flow nozzles; this company offers nozzles with flow sizes in a base-2 distribution, which allows you to buy the smallest number of nozzles able to span a given flow range with equal increments. This picture shows the installation of multiple nozzles in a manifold plate to achieve a particular flow rate (in this picture most of the nozzles appear to have been capped). The nozzles are inherently a binary thing: either they’re flowing or they aren’t, you can’t throttle them.

The efficiency is more apparent in the guess a number game, where it is most efficient and consistent way to pick a number is halfway in the range (binary), find out if it’s higher or lower, you now cut the range in half.

Doing it as a ‘base 10’ type guess most of the time you have your number in the 90% remaining range.

However, and a aside, if you are able to do that guess in a quantum form base 10 becomes more efficient as your guess is every number division at the same time, so you are effectively taking 9 guesses at once with base 10 and only 1 guess with base 2 (binary). So the efficiency also depends on the method.

Suppose that, to save space, you must mount the lightbulbs in a close row, ordered by size: e.g., … 4+8+16+32+… Unfortunately if two adjacent lightbulbs are both turned on their closeness means the combined heat will cause both of them to disintegrate after a short while. For example, if the 8 bulb is turned on, both the 4 and 16 bulbs must be off.

The binary scheme no longer allows you to produce all totals. What series of lightbulb sizes would you recommend now?

Holy almost kinda Gray code, Batman!

I’m surprised we didn’t see the usual 5-minute turnaround on this puzzle. (Perhaps I didn’t frame it clearly.) Don’t make me spoil my own puzzle — it has a lovely answer.