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

[CurtC]

@septimus

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.

It looks like 1, 3, 9, and 27 will do it.

[CurtC]

@markn_1

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?

I struggled with this one for a lot longer than one night! For a while I was thinking “wait - there’s not enough info. You have 24 possible situations to test for, and you only can eliminate half with each with each weighing, so how can you narrow it down to one situation in three weighings?” Then: “oh.”

[septimus]

@CurtC

[spoiler]

That’s the answer to #4. I’m waiting for the answer to #18.

[Dr.Strangelove]

@septimus

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.

[spoiler]I started as I usually do, by just enumerating the first N configurations and seeing what happens. I notice that with 1 bulb there are 2 valid combinations; with 2 bulbs there are 3; with 3, 5; with 4, 8… that’s a Fibonacci sequence if I ever saw one.

That’s not the question which was asked, but symmetry always seems to play a key role in these kinds of problems, so let’s see what happens with a Fibonacci bulb sequence. I can immediately find that I can make any combination work with bulbs of (1, 2, 3, 5, 8, …).

Although I have a hard time proving that this is optimal, there’s one bit of evidence that suggests to me that it is: a maximally lit sequence (all odd-numbered or all even-numbered bulbs lit), it is always one less than the next higher number in the sequence. So for instance, 1+3+8+21=33, and the next number is 34. Or 2+5+13+34=54, and the next number is 55. This suggests that the sequence is packed tightly with no dupes, and is thus the most efficient possible.[/spoiler]

[Dr.Strangelove]

[spoiler]Incidentally, I hadn’t realized the “sum of alternating elements sums to one less than the next number” property before, but it’s reasonably straightforward to prove with induction.

There are two cases to handle, even and odd. Even-index numbers are equal to the sum of all odd-index numbers (plus 1) below them, while odd-index numbers are equal to the sum of all even-index numbers (plus 1) below them.

The first odd index is 1 with a value of 1, and there are no numbers below it, and so our sum gives 0+1=1, which means our base case works. The next odd index is 3, and our Fibonacci recurrence relation is f(1) + f(2) = f(3). f(1) we just worked out, and gives the expected result. f(2) is a new even-indexed element that wasn’t in the original set, so adding it in still gives us “the sum of all even-index elements plus one”. This obviously extends to all f(n) with odd n (since f(n-1) must be even), stepping forward two elements at a time.

We then do the same thing for even n, solving the base case of f(2) and against stepping forward two at a time. We’ve then covered all n.[/spoiler]

[EdwardLost]

I was in a meeting with a realtor who described to me the process of buying foreclosed houses in his state: you attend an auction at the courthouse and, if you have the highest bid, you must pay for the house on the spot with cashier’s checks. He said buyers show up to the auction with stacks of checks of varying amounts.

From that point I struggled to stay focused on the meeting because I was busy trying to figure out which sized checks I should bring to an auction. Eventually I realized the answer was, just like the light bulbs: one check each in powers of two (assuming I wanted to buy only one house).

[ftg]

@EdwardLost

I was in a meeting with a realtor who described to me the process of buying foreclosed houses in his state: you attend an auction at the courthouse and, if you have the highest bid, you must pay for the house on the spot with cashier’s checks. He said buyers show up to the auction with stacks of checks of varying amounts.

From that point I struggled to stay focused on the meeting because I was busy trying to figure out which sized checks I should bring to an auction. Eventually I realized the answer was, just like the light bulbs: one check each in powers of two (assuming I wanted to buy only one house).

But you can control your bid. I.e., bid only in multiples of $500. So you only need powers of two times that values for you checks. Far fewer to deal with.

Of course, like on eBay, you might want to avoid round numbers and do multiples of $501 or something.

[Isilder]

@Ignotus

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.

In maths, an additive series is arithmetic series, and a multiplication step is geometric.

Perhaps The OP is asking, whats the efficency of base 2 ? whats the efficiency of base 10 ? base 100 ?

(b-1)* Log b (N ) where b is the base, N is the number of items.
base 2… suppose you had up to supply up to 250 (watts for example)… Well you need at most 8 … as efficiency is log 2 ( N) …

base 10… Suppose you had to supply up to 999 , you’d need up to (10-1) * log 10 (999)… 9 * 3 … 27…

base 1000… supply 999 , you need 999. Trivial
supply 999,999,999 … you need 999 * 3… You can see a pattern with the number required.
whats the efficiency of coins ?

1,2,5 … well basically 3 coins per decimal digit, so 3 * log 10 N.
Look at the “preferred values” of the standard available resisters, capacitors, inductors… if you are interested in other schemes of preferred numbers

[igor_frankensteen]

@Machine_Elf

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?

The answer to this, the original question, as described and framed, will be found in Business Science.

It has to do with taking a Modular approach to design, rather than an Integral approach.