[GUEST ACCESS MODE: Data is scrambled or limited to provide examples. Make requests using your API key to unlock full data. Check https://lunarcrush.ai/auth for authentication information.]  vitalik.eth [@VitalikButerin](/creator/twitter/VitalikButerin) on x 5.8M followers Created: 2025-05-21 18:04:50 UTC A fun math aside, on the idea of splitting a large zk proving workload between multiple provers. Suppose you have N provers, and you have a proving workload that you split into N parts (so, one part per prover). You require provers to pre-register, but registration is open-access. Suppose you have a constant fault rate (eg. 1/5 of registered provers fail). Provers expect to complete in one round (eg. 3s). If one prover fails, other provers have to come in and re-prove that load. How many rounds does it take for the entire workload to get proven? Answer: log*(N) (yes, that's the iterated-log function) Why: In the first round, you go from N unproven workloads to N/5 unproven workloads In the second round, each remaining workload gets assigned X provers, so per-workload failure rate becomes X in 5^5. So you go to N / X / 5^5 unproven workloads In the third round, each remaining workload gets assigned ~5^5 provers, so failure rate is X in 5^(5^5). So you go to N / X / 5^5 / 5^(5^5) unproven workloads XXXXXXX engagements  **Related Topics** [zk](/topic/zk) [Post Link](https://x.com/VitalikButerin/status/1925251453805027380)
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vitalik.eth @VitalikButerin on x 5.8M followers
Created: 2025-05-21 18:04:50 UTC
A fun math aside, on the idea of splitting a large zk proving workload between multiple provers.
Suppose you have N provers, and you have a proving workload that you split into N parts (so, one part per prover). You require provers to pre-register, but registration is open-access.
Suppose you have a constant fault rate (eg. 1/5 of registered provers fail). Provers expect to complete in one round (eg. 3s). If one prover fails, other provers have to come in and re-prove that load. How many rounds does it take for the entire workload to get proven?
Answer: log*(N)
(yes, that's the iterated-log function)
Why:
In the first round, you go from N unproven workloads to N/5 unproven workloads
In the second round, each remaining workload gets assigned X provers, so per-workload failure rate becomes X in 5^5. So you go to N / X / 5^5 unproven workloads In the third round, each remaining workload gets assigned ~5^5 provers, so failure rate is X in 5^(5^5). So you go to N / X / 5^5 / 5^(5^5) unproven workloads
XXXXXXX engagements
Related Topics zk