One Time Pad Length Calculator

Work out one time pad length instantly with clear inputs, formula shown and shareable results.

About One Time Pad Length Calculator

A one-time pad is the only cipher with a proof of perfect secrecy, but the proof requires a truly random key at least as long as the message, used exactly once and then destroyed. That is why it is impractical: a year of modest traffic needs 75 GB of pre-shared random key, and the key distribution problem is strictly harder than the encryption problem it solves.

Formula

Pad requirement

key bytes = message bytes; total = message bytes x messages per day x days

Worked example

With the default values below, One Time Pad Length Calculator returns Key material required of 71.29 MiB. Change any input above and every figure here is recalculated in your browser.

Inputs used

Message lengthminimum 1 bytes: 4,096 bytes
Messages per dayminimum 1: 50
Days of operationminimum 1: 365

Results

Key material required: 71.29 MiB
Key bytes per message: 4.00 KiB
Key bytes per day: 200.00 KiB
Operational constraint: Every byte must be delivered in advance over a secure channel and destroyed after a single use - reuse breaks the perfect secrecy proof completely

Step by step

1Values used
2Pad requirement
3Key material required= 74,752,000
4Key bytes per message= 4,096
5Key bytes per day= 204,800
6Operational constraint= Every byte must be delivered in advance over a secure channel and destroyed after a single use - reuse breaks the perfect secrecy proof completely

Assumptions and limits

The result is derived only from the 3 inputs listed above. Anything not among them is not part of the calculation.
In particular, this calculator does not take inflation, tax, fees or charges or compounding frequency into account.

How these calculators are built and tested is described on the Methodology page. If a figure looks wrong, please report it.

Frequently Asked Questions

What happens if a pad is reused?

XORing two ciphertexts made with the same pad cancels the key and leaves the XOR of the two plaintexts, which is readily solvable. This is exactly how the VENONA decryptions succeeded.

Why not use a stream cipher instead?

That is precisely what a stream cipher is — a pseudorandom pad generated from a short key. It gives computational rather than perfect secrecy, but the key is 32 bytes instead of gigabytes.