Space Pledged Intuition

Research
1

Introduction

Every second the chain draws a random winning number (challenge).

Each 1 MB piece 8B tag wins if it happens to land close enough to challenge. “Close enough” is measured by a difficulty knob called solution range. The chain adjusts that knob so that, across all tickets in the world, only about one ticket in six seconds wins. The beauty is that a single division of the knob into a fixed constant tells you how many tickets—and therefore how many megabytes—must be online to make that six-second rhythm work.

1 — The tiny probability p that one ticket wins

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Every ticket is assigned a 64-bit number somewhere on the ruler from 0 to (2^{64}-1).

Each second the chain draws a fresh 64-bit challenge number.

A ticket wins if its number is within solution_range ⁄ 2 on either side of the challenge (wrapping around the ends of the ruler if necessary).

Because the ruler holds exactly (2^{64}) possible numbers and the highlighted window is

solution_range numbers wide, the chance for one ticket is

p = \frac{\text{solution\_range}}{2^{64}}

ref: subspace/crates/subspace-verification/src/lib.rs at fd11d4bd31bda1c04ca8b698c0832ec03f83e1d7 · autonomys/subspace · GitHub

2 — How many tickets n must exist to get one winner every six seconds

With n independent tickets checked every second, the chance that someone wins
in that second is

[1-(1-p)^{n}\;\;\approx\;\;n\,p \qquad\text{(because \(p\) is tiny)}]

The chain wants that probability to equal \frac{1}{6} so blocks average one every

six seconds:

[ n_{\text{tickets}}\;p \;=\;\frac16 \;\;\Longrightarrow\;\; n_{\text{tickets}} =\frac{1}{6*\,p} =\frac{2^{64}}{6 * \,\text{solution\_range}} ]

ref:

3 — Turn tickets into storage

level relation comment
ticket (8-byte tag) — the unit that wins
piece (1 MB record) 1 piece ⇒ ½ ticket (erasure coding) half the time the probed s-bucket is empty
sector (≈ 1 GB file) 1 000 pieces a fully-plotted sector holds ≈ 500 tickets

So the total pledged storage in megabytes is

[ \begin{aligned} n_{\text{pieces}} &= 2 \times n_{\text{tickets}} = \frac{2^{64}}{6*\,\text{solution\_range}} * 2, \\[6pt] n_{\text{sectors}} &= \frac{n_{\text{pieces}}}{1\,000} = \frac{2^{64}}{6*\,1000*\,\text{solution\_range}} * 2 \end{aligned} ]

ref:

4 — Concrete example

solution_range = 12 417 612

step formula value
tickets (n_{\text{tickets}} = 2^{64}/(6R)) 247 588 453 047
pieces (MB) (n_{\text{pieces}} = 2\times n_{\text{tickets}}) 495 176 906 094 MB
GB (÷ 1024) 483 571 197 GB
TB (÷ 1024 again) 472 237 TB
PB (÷ 1024 again) ≈ 461 PB

So a solution_range of 12 417 612 implies the network holds roughly

4.95 × 10¹¹ pieces (≈ 461 petabytes) of fully-plotted history.

(Using decimal units instead of binary would give ≈ 495 PB.)

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