**Title: Radix Economy for Alternating Bases**

__Subtitle: Efficiency of Bases for Specifying Information__

**Introduction:**

As a way of quantifying the cost of increasing the size of a base, radix economy is often mentioned. I will derive a definition for a version of radix economy from the number of options from which to select required to specify a piece of information or datum by a numerical base. To specify a piece of information, a sequence of questions must be answered with the number of options available to choose at each question being equal to the base of numeration. The possible pathways of answers to the sequence of questions may be depicted in a tree of nodes at each question branching into the number of answer options at each question. A tier of such a tree is a level of a stage in the sequence consisting of the nodes and branches at that stage. The total number, say \(R_{B}\), of options encountered in the sequence of questions for any one pathway in a tree would be the number of tiers, levels, or stages \(l\) multiplied by the number of options or available answers \(B\) at each stage.

\[R_{B} = B*l = B\log_{B}{N}\]

To illustrate this concept, consider a number \(N\) of data or pieces of information that is approximately a power of each of two different bases \(B\). For example, where the number \(N\) of data values is the sixth power of two, \(2^{6}\), and if the base \(B\) is four then in order to specify one piece of information out of the \(2^{6}\) pieces of information, three questions with four possible answers each time would have to be answered in sequence. The total number \(R_{B}\) of options in the pathway of answers to the piece of information in that tree would be four times three, which is twelve. If the base \(B\) were rather eight than four, then the number of options \(B\) as answers for each question at stages in the sequence would be eight, and only two questions would have to be answered in sequence to specify one datum out of \(2^{6}\) pieces of information. In this case of base eight, the total number of encountered options along a pathway route to a datum would be eight times two, which is more than the twelve from the base four. Thus, base four would produce fewer options to be considered as answers to questions in the route to specification or distinction of a single piece of information. If the base \(B\) were to be the number two, then the total number \(R_{B}\) of options along a pathway would also be twice six, because there would be six tiers or levels of questions with binary or bifurcating answers required to distinguish \(2^{6}\) data points.

To compare the base two versus the base three for radix economy in a way that can be demonstrated indisputably, a number \(N\) of data values may be chosen that is approximately a power of both bases two and three separately. Let \(N \approx 2\)

^{①⑦}\(\approx 3\)

^{①⓪}\(\approx\) ②①⑤⑤⑧⓪. Using this value of \(N\) in the formula for \(R_{B}\) produces about ③② total options in base two, which is worse than the approximately three dozen total options per route resulting from base three. Thus, base three is more economical than base two is.

**Radix Economy Extended to Alternating Bases**

If the base \(B\) were purely decimal without subdivisions, then the total number \(R_B\) of options would be about ④⑨, indicating that this form of a decimal base for classification of information is much less economical that the smaller bases three or two. However, in practice it is unlikely that as many as ten options would be allowed for answers at question stages. Instead, each decimal tier or level would be split into two tiers, one tier of two possible answers and the other of five possible answers. This produces combined decimal stages as mixed or alternating stages of bases two and five in sequence. In the calculation of the total number \(R_{B}\) of options resulting from such an alternating mixed decimal base, the number \(l\) of tiers would now be twice as many as from a pure decimal base because each combined decimal tier is split into two tiers, and the number of options for answers at a subbase tier can be supposed on average to be the number two half the time and five the other half of times, leading to an average of half the sum of two and five at each tier. The total number \(R_B\) of options in a pathway thus becomes \(R_{B} = 2 * \frac{2+5}{2} * \log_{10}{N}\), which for \(N\) as ②①⑤⑤⑧⓪ gives about ③④ total options per pathway. This indicates that even as an alternating base, decimal would be less economical than a binary power as base for classification of information.

In general, the total number \(R_{B}\) of options per pathway route for an alternating base \(B\) with \(n\) subbases \(b_{i}\) would be the multiple of the average \(A = \frac{ \sum_{i=1}^{i=n} b_{i}}{n}\) of the subbases by the number \(n\) of subbases and by the number \(\log_{B}{N}\) of pure base tiers:

\[ R_{B} = \frac{ \sum_{i=1}^{i=n} b_{i}}{n} * n * \log_{B}{N} \]

which simplifies to

\[ R_{B} = \left( \sum_{i=1}^{i=n} b_{i} \right)* \log_{B}{N} \]

Since 2 + 2 = 4 and 2

^{2}= 4, the economy of two halving subdivisions is the same as the economy of one quartering subdivision. The economy for base twelve with quartering and trisecting subdivisions for the same number ②①⑤⑤⑧⓪ as \(N\) would be better than for decimal and even for any binary power base because the total number \(R_{B}\) of options per pathway would be about ③①, fewer than for decimal or binary power bases. This means that base twelve would be a better way to classify information than by either decimal or any binary power base.

To form an efficiency score \(E_{B}\) ranging from zero to a maximum of one for the most economical base from the radix economy, the value of \(N\) and a scaling factor are chosen such as to make the most economical base \(e\) have an efficiency score of unity and the reciprocal of the total number \(R_{B}\) of options per route is used. The definition of this efficiency score for pure exponential bases \(B\) is the same as that defined in the topic on Potency in this Base Dozen Forum as

\[E_{B} = \frac{e\ln{B}}{B}\]

For alternating bases, this efficiency score is generalised to

\[E_{B} = \frac{e\ln{B}}{\sum_{i=1}^{i=n} b_{i}}\]

Based on this efficiency score from radix economy, a selection of different bases may be ranked in a table:

**Table of Bases Ranked by Radix Economy Efficiency**

Base | E_{B} | |

Decimally | Dozenally | |

e | 1 | |

3 | ③ | ⓪⁏⑪⑪④ |

54 | ④⑥ | ⓪⁏⑪⑨⑪ |

18 | ①⑥ | ⓪⁏⑪⑨⑤ |

6 | ⑥ | ⓪⁏⑪⑧③ |

72 | ⑥⓪ | ⓪⁏⑪⑦⑥ |

12 | ①⓪ | ⓪⁏⑪⑥⑪ |

24 | ②⓪ | ⓪⁏⑪⑥③ |

48 | ④⓪ | ⓪⁏⑪⑤⑨ |

270 | ①⑩⑥ | ⓪⁏⑪⑤⓪ |

2^{n} | ②^{n} | ⓪⁏⑪③⑧ |

720 | ⑤⓪⓪ | ⓪⁏⑪③⑥ |

120 | ⑩⓪ | ⓪⁏⑪①⑩ |

60 | ⑤⓪ | ⓪⁏⑪①⑦ |

30 | ②⑥ | ⓪⁏⑪①② |

15 | ①③ | ⓪⁏⑪⓪⑥ |

20 | ①⑧ | ⓪⁏⑩⑩④ |

10 | ⑩ | ⓪⁏⑩⑧⑨ |

14 | ①② | ⓪⁏⑨⑥⑨ |

I tested many other large bases. It can be seen that there are not many realistic bases more efficient for classification of information according to radix economy than base twelve as an alternating base. Base six has a better radix economy, but many believe that this base has disadvantages for storage of information because of its small size and the larger number of digits required. Base twice nine can be more efficient still, but many consider this base to be too large. Base three is often considered to be impractical because it is not divisible by the prime number two.

**References/See also:**

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