brutuscat

Chose one and tell the model to write into a file you’ll keep around while the implantation is on going so you won’t pollute the context and can start over each chunk of work in a clean prompt. Name the file refactor-<name >-plan.md tell it to write the plan step by step and dump a todo list having into account dependencies for tracking progress.

Review the plans, make fixes if needed. You need to have some sort of table reassembling a todo so it can track and make progress along.

Open a new prompt tell it analyze the plan file, to go to the todo list section and proceed with the next task. Verify it done, and update the plan.

Repeat until done.

Comment · 8 months ago

Look at Serena. Does something on this line.

> it can use code-centred tools like find_symbol, find_referencing_symbols and insert_after_symbol.

https://oraios.github.io/serena/01-about/000_intro.html

Comment · 1 year ago

Gemini thinks in a similar fashion:

https://gemini.google.com/share/1e59f734b43c

This is a fantastic observation, and yes, this pattern not only continues for larger bases, but the approximation to an integer becomes dramatically better.

The general pattern you've found is that for a number base $b$, the ratio of the number formed by digits $(b-1)...321$ to the number formed by digits $123...(b-1)$ is extremely close to $b-2$.

### The General Formula

Let's call your ascending number $N_{asc}(b)$ and your descending number $N_{desc}(b)$.

The exact ratio $R(b) = N_{desc}(b) / N_{asc}(b)$ can be shown to be:

$$R(b) = (b-2) + \frac{(b-1)^3}{b^b - b^2 + b - 1}$$

The "error" or the fractional part is that second term. As you can see, the numerator $(b-1)^3$ is roughly $b^3$, while the denominator $b^b$ grows much faster.

### Testing Your Examples

Let's check your two examples with this formula:

* *Base 10 (b=10):* * $R(10) = (10-2) + \frac{(10-1)^3}{10^{10} - 10^2 + 10 - 1}$ * $R(10) = 8 + \frac{9^3}{10,000,000,000 - 91}$ * $R(10) = 8 + \frac{729}{9,999,999,909} \approx 8 + 0.0000000729...$ * This matches your $8.0000000729$ perfectly.

* *Base 6 (b=6):* * $R(6) = (6-2) + \frac{(6-1)^3}{6^6 - 6^2 + 6 - 1}$ * $R(6) = 4 + \frac{5^3}{46,656 - 31}$ * $R(6) = 4 + \frac{125}{46,625} \approx 4 + 0.00268...$ * This also matches your $4.00268$.

### Answering Your Question: A Larger Base

So, what about a larger base? Let's use *Base 16* (hexadecimal), which uses digits $1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F$. Here, $b=16$.

* $N_{asc} = 123456789ABCDEF_{16}$ * $N_{desc} = FEDCBA987654321_{16}$

According to the pattern, the ratio should be extremely close to $b-2 = 16-2 = 14$.

Let's use the formula to see how close:

* $R(16) = (16-2) + \frac{(16-1)^3}{16^{16} - 16^2 + 16 - 1}$ * $R(16) = 14 + \frac{15^3}{16^{16} - 241}$ * $R(16) = 14 + \frac{3,375}{1.844... \times 10^{19} - 241}$ * $R(16) \approx 14 + (1.829... \times 10^{-16})$