$$$$

Synth-PoRep Spec

Notation & Constants

$\mathbf{\text{const }}\mathsf{\text{NODE\_SIZE}}=32\text{ Bytes}$
Size in bytes of one node.

$\mathbf{\text{const }}\mathsf{\text{SECTOR\_SIZE}}\in \{32\text{GiB},64\text{GiB}\}$
The size in bytes of the sector being sealed.

$\mathbf{\text{const }}\mathsf{\text{SECTOR\_NODES}}=\mathsf{\text{SECTOR\_SIZE}}/\mathsf{\text{NODE\_SIZE}}$
The size in 32-byte nodes of the sector being sealed.

$\mathbf{\text{type }}\text{CommR}:\mathbb{F}=\mathtt{\text{poseidon}}(\text{CommC}\Vert \text{RootR})$
A commitment to sealed data.

$\text{CommC}$ is the root of

$\text{TreeC}$ and

$\text{RootR}$ is the root of

$\text{TreeR}$.

$\mathbf{\text{type }}\text{ReplicaID}:\mathbb{F}$
A unique identifier associated with a replica

$\text{R}$ of sector data

$\text{D}$.

Challenge Counts:

$\mathbf{\text{const }}{\text{N}}^{\text{syn}}=2^{14}=\mathrm{16,384}$
The number of synthetic challenges.

$\mathbf{\text{const }}\text{N}=176$
The number of PoRep challenges.

Challenge Sets:

$\mathbf{\text{type }}{\text{C}}^{\text{syn}}:{\mathtt{\text{u32}}}^{[{\text{N}}^{\text{syn}}]}=(c_i^{\text{syn}}{)}_{i\in [0,{\text{N}}^{\text{syn}})}$
The synthetic challenge set.

$\mathbf{\text{type }}\text{C}:{\mathtt{\text{u32}}}^{[\text{N}]}=(c_i{)}_{i\in [0,\text{N})}$
The PoRep challenge set.

Challenge Generation

SHA256

$\stackrel{―}{\underset{―}{\text{Function:}\mathtt{\text{gen\_synth\_challenges}}(\text{ReplicaID})\to {\text{C}}^{\text{syn}}}}$

$\mathbf{\text{for }}i:\mathtt{\text{u32}}\in [0,{\text{N}}^{\text{syn}})\mathbf{\text{:}}$

$\text{digest}:{\mathtt{\text{u32}}}^{[8]}=\mathtt{\text{sha256}}(\mathtt{\text{le\_bytes}}(\text{ReplicaID})\Vert \mathtt{\text{le\_bytes}}(i))$

$\text{bigint}=\mathtt{\text{u256::from\_le\_u32s}}(\text{digest})$

$c_i^{\text{syn}}:\mathtt{\text{u32}}=(\text{bigint}\mathrm{\%}(\mathsf{\text{SECTOR\_NODES}}-1))+1$

$\stackrel{―}{\underset{―}{\text{Function:}\mathtt{\text{gen\_porep\_challenges}}(\text{ReplicaID},{\text{C}}^{\text{syn}},\text{rand}:{\mathtt{\text{u8}}}^{[32]})\to \text{C}}}$

$\mathbf{\text{for }}i:\mathtt{\text{u32}}\in [0,\text{N}/8)\mathbf{\text{:}}$

$\text{digest}:{\mathtt{\text{u32}}}^{[8]}=\mathtt{\text{sha256}}(\mathtt{\text{le\_bytes}}(\text{ReplicaID})\Vert \text{rand}\Vert \mathtt{\text{le\_bytes}}(i))$

$\mathbf{\text{for }}j\in [0,8)\mathbf{\text{:}}$

$\text{idx}=\text{digest}[j]\mathrm{\%}{\text{N}}^{\text{syn}}$

$c_{8i+j}=c_{\text{idx}}^{\text{syn}}$

ChaCha20

$\mathbf{\text{const }}\mathsf{\text{NONCE}}:{\mathtt{\text{u8}}}^{[12]}=0^{[12]}$

$\mathbf{\text{const }}\mathsf{\text{EMPTY\_BLOCK}}:{\mathtt{\text{u8}}}^{[64]}=0^{[64]}$

$\stackrel{―}{\underset{―}{\text{Function:}\mathtt{\text{gen\_synth\_challenges}}(\text{ReplicaID})\to {\text{C}}^{\text{syn}}}}$

$\text{key}:{\mathtt{\text{u8}}}^{[32]}=\mathtt{\text{le\_bytes}}(\text{ReplicaID})$

$\text{stream}=\mathtt{\text{ChaCha20::new}}(\text{key},\mathsf{\text{NONCE}})$

$\mathbf{\text{for }}i\in [0,{\text{N}}^{\text{syn}}/2)\mathbf{\text{:}}$

$\text{block}:{\mathtt{\text{u8}}}^{[64]}=\text{stream}.\mathtt{\text{encrypt}}(\mathsf{\text{EMPTY\_BLOCK}})$

${\text{bigint}}_1=\mathtt{\text{u256::from\_le\_bytes}}(\text{block}[\text{..}32])$

${\text{bigint}}_2=\mathtt{\text{u256::from\_le\_bytes}}(\text{block}[32\text{..}])$

$c_{2i}^{\text{syn}}:\mathtt{\text{u32}}=({\text{bigint}}_1\mathrm{\%}(\mathsf{\text{SECTOR\_NODES}}-1))+1$

$c_{2i+1}^{\text{syn}}:\mathtt{\text{u32}}=({\text{bigint}}_2\mathrm{\%}(\mathsf{\text{SECTOR\_NODES}}-1))+1$

$\stackrel{―}{\underset{―}{\text{Function:}\mathtt{\text{gen\_porep\_challenges}}(\text{ReplicaID},{\text{C}}^{\text{syn}},\text{rand}:{\mathtt{\text{u8}}}^{[32]})\to \text{C}}}$

$\text{key}:{\mathtt{\text{u8}}}^{[32]}=\mathtt{\text{sha256}}(\mathtt{\text{le\_bytes}}(\text{ReplicaID})\Vert \text{rand})$

$\text{stream}=\mathtt{\text{ChaCha20::new}}(\text{key},\mathsf{\text{NONCE}})$

$\mathbf{\text{for }}i\in [0,\text{N}/16)\mathbf{\text{:}}$

$\text{block}:{\mathtt{\text{u8}}}^{[64]}=\text{stream}.\mathtt{\text{encrypt}}(\mathsf{\text{EMPTY\_BLOCK}})$

$\mathbf{\text{for }}j\in [0,16)\mathbf{\text{:}}$

$\text{idx}=\mathtt{\text{u32::from\_le\_bytes}}(\text{block}[4j..4j+4])\mathrm{\%}{\text{N}}^{\text{syn}}$

$c_{16i+j}=c_{\text{idx}}^{\text{syn}}$