make byte plural consistent
This commit is contained in:
Markus Teich
parent
ec0e5f0189
commit
2eb3cbaa3a
8
brandt.h
8
brandt.h
@ -79,7 +79,7 @@ typedef uint16_t
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* @param[in] auction_closure Closure pointer representing the respective
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* auction. This is the Pointer given to BRANDT_join() / BRANDT_new().
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* @param[in] msg The message to be delivered
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* @param[in] msg_len The length of the message @a msg in bytes.
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* @param[in] msg_len The length of the message @a msg in byte.
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* @return 0 on success, -1 on failure.
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*/
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typedef int
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@ -116,7 +116,7 @@ BRANDT_init (struct GNUNET_CRYPTO_EccDlogContext *dlogctx);
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* for an explanation of the different auction description fields.
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*
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* @param[in] auction_desc The auction description blob published by the seller.
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* @param[in] auction_desc_len Length of @a auction_desc in bytes.
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* @param[in] auction_desc_len Length of @a auction_desc in byte.
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* @param[out] time_start Starting time of the auction. May be NULL.
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* @param[out] time_round Maximum round time of the auction. May be NULL.
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* @param[out] num_prices Amount of possible prices. May be NULL.
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@ -176,7 +176,7 @@ BRANDT_join (BRANDT_CbResult result,
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* structure. It is generated by this function and should be distributed to
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* all possibly interested bidders. The seller MUST sign this data block before
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* publishing it!
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* @param[out] auction_desc_len The length in bytes of the @a auction_desc
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* @param[out] auction_desc_len The length in byte of the @a auction_desc
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* structure. Will be filled by BRANDT_new().
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* @param[in] time_start The time when the auction will start. Bidders have
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* until then to register.
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@ -243,7 +243,7 @@ BRANDT_destroy (struct BRANDT_Auction *auction);
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* which message @a msg was received.
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* @param[in] sender The id of the sender.
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* @param[in] msg The message that was received.
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* @param[in] msg_len The length in bytes of @a msg.
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* @param[in] msg_len The length in byte of @a msg.
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*/
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void
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BRANDT_got_message (struct BRANDT_Auction *auction,
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4
crypto.c
4
crypto.c
@ -636,7 +636,7 @@ smc_prep_keyshare (struct BRANDT_Auction *ad)
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* public key share
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*
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* @param[in] ad Pointer to the BRANDT_Auction struct to operate on
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* @param[out] buflen Size of the returned message buffer in bytes
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* @param[out] buflen Size of the returned message buffer in byte
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* @return A buffer containing the public key share
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* which needs to be broadcast
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*/
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@ -730,7 +730,7 @@ smc_prep_bid (struct BRANDT_Auction *ad)
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* into a message together with proofs of correctnes.
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*
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* @param[in] ad Pointer to the BRANDT_Auction struct to operate on
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* @param[out] buflen Size of the returned message buffer in bytes
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* @param[out] buflen Size of the returned message buffer in byte
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* @return A buffer containing the encrypted bid
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* which needs to be broadcast
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*/
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@ -149,7 +149,7 @@ fp_priv_prep_outcome (struct BRANDT_Auction *ad)
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* a message buffer together with proofs of correctnes.
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*
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* @param[in] ad Pointer to the BRANDT_Auction struct to operate on
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* @param[out] buflen Size of the returned message buffer in bytes
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* @param[out] buflen Size of the returned message buffer in byte
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* @return A buffer containing the encrypted outcome vectors
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* which needs to be broadcast
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*/
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@ -396,7 +396,7 @@ fp_priv_decrypt_outcome_bidder (struct BRANDT_Auction *ad, size_t *buflen)
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* buffer from all received decryption shares to broadcast back to all bidders.
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*
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* @param[in] ad Pointer to the BRANDT_Auction struct to operate on
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* @param[out] buflen Size of the returned message buffer in bytes
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* @param[out] buflen Size of the returned message buffer in byte
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* @return A buffer containing the share of the decrypted outcome
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* which needs to be broadcast
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*/
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4
fp_pub.c
4
fp_pub.c
@ -132,7 +132,7 @@ fp_pub_prep_outcome (struct BRANDT_Auction *ad)
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* correctnes.
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*
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* @param[in] ad Pointer to the BRANDT_Auction struct to operate on
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* @param[out] buflen Size of the returned message buffer in bytes
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* @param[out] buflen Size of the returned message buffer in byte
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* @return A buffer containing the encrypted outcome vectors
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* which needs to be broadcast
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*/
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@ -290,7 +290,7 @@ fp_pub_prep_decryption (struct BRANDT_Auction *ad)
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* message buffer together with proofs of correctnes.
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*
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* @param[in] ad Pointer to the BRANDT_Auction struct to operate on
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* @param[out] buflen Size of the returned message buffer in bytes
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* @param[out] buflen Size of the returned message buffer in byte
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* @return A buffer containing the own share of the decrypted outcome
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* which needs to be broadcast
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*/
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4
mp_pub.c
4
mp_pub.c
@ -182,7 +182,7 @@ mp_pub_prep_outcome (struct BRANDT_Auction *ad)
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* a message buffer together with proofs of correctnes.
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*
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* @param[in] ad Pointer to the BRANDT_Auction struct to operate on
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* @param[out] buflen Size of the returned message buffer in bytes
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* @param[out] buflen Size of the returned message buffer in byte
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* @return A buffer containing the encrypted outcome vectors
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* which needs to be broadcast
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*/
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@ -347,7 +347,7 @@ mp_pub_prep_decryption (struct BRANDT_Auction *ad)
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* message buffer together with proofs of correctnes.
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*
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* @param[in] ad Pointer to the BRANDT_Auction struct to operate on
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* @param[out] buflen Size of the returned message buffer in bytes
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* @param[out] buflen Size of the returned message buffer in byte
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* @return A buffer containing the own share of the decrypted outcome
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* which needs to be broadcast
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*/
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@ -112,7 +112,7 @@ the price $p_{b_a}$ bidder $a$ is willing to pay. $\forall j: p_j < p_{j+1}$.
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The message has $k$ parts, each consisting of $10$ Points plus an additional $3$
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Points for the last proof. Therefore the message is $10k*32 + 3*32 = 320k + 96$
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bytes large.
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byte large.
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\begin{enumerate}
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\item $\forall j:$ Set $B_{aj}=\begin{cases}G & \mathrm{if}\quad j=b_a\\0 & \mathrm{else}\end{cases}$ and publish $\alpha_{aj}=B_{aj}+r_{aj}Y$ and $\beta_{aj}=r_{aj}G$.
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@ -125,7 +125,7 @@ bytes large.
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\subsubsection{Round 2: Compute outcome}
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The message has $nk$ parts, each consisting of $5$ Points. Therefore the message
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is $5nk*32 = 160nk$ bytes large.
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is $5nk*32 = 160nk$ byte large.
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$\forall i,j:$ Compute and publish \\[2.0ex]
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$\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k\alpha_{hd}\right)+\left(\sum_{d=1}^{j-1}\alpha_{id}\right)+\left(\sum_{h=1}^{i-1}\alpha_{hj}\right)\right)$ and \\[2.0ex]
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@ -157,7 +157,7 @@ $\delta_j^{\times a} = m_j^{+a}\displaystyle\left(\sum_{h=1}^n\sum_{d=j+1}^k \be
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with a corresponding Proof 2 for $\displaystyle ECDL\left(m_j^{+a}\left(\sum_{h=1}^n\sum_{d=j+1}^k\alpha_{hd}\right)\right) = ECDL\left(m_j^{+a}\left(\sum_{h=1}^n\sum_{d=j+1}^k \beta_{hd}\right)\right)$. \\[2.0ex]
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The message has $k$ parts, each consisting of $5$ Points. Therefore the message
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is $5k*32 = 160k$ bytes large. Note that compared to auctions with private
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is $5k*32 = 160k$ byte large. Note that compared to auctions with private
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outcome the message size is reduced by a factor of $n$ because we don't need to
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compute different outcome functions for each bidder when the outcome should be
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public. Therefore we don't need $nk$ blinding factors $m_{ij}^{+a}$ in this
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@ -170,7 +170,7 @@ x_{+a}\left(\sum_{h=1}^n\delta_j^{\times h}\right)$ with a Proof 2 showing
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$ECDL(\varphi_j^{\times a}) = ECDL(Y_{\times a})$ \\[2.0ex]
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This message has $k$ parts, each consisting of $4$ Points. Therefore the message
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is $4k*32 = 128k$ bytes large.
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is $4k*32 = 128k$ byte large.
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\subsubsection{Epilogue: Outcome determination}
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@ -219,12 +219,12 @@ M+1st highest bid.
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The bidders also have to use Proof 2 to show that $\displaystyle ECDL_Y\left(\left(\sum_{j=1}^{k/n}\alpha_{a,jn+a}\right) - G\right) = ECDL_G\left(\sum_{j=1}^{k/n}\beta_{a,jn+a}\right)$. \\[2.0ex]
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This is to ensure bidders have only chosen valid bids for their bid index, since
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in M+1st price auctions the amount of possible prices is multiplied by $n$ to
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prevent ties. This increases the message size by $96$ bytes.
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prevent ties. This increases the message size by $96$ byte.
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\subsubsection{Round 2: Compute outcome}
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The message has $nk$ parts, each consisting of $5$ Points. Therefore the message
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is $5nk*32 = 160nk$ bytes large.
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is $5nk*32 = 160nk$ byte large.
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$\forall i,j:$ Compute and publish \\[2.0ex]
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$\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\sum_{h=1}^n\left(\sum_{d=j}^k\alpha_{hd}+\sum_{d=j+1}^k\alpha_{hd}\right)+\left(2M+2\right)\sum_{d=1}^{j}\alpha_{id} - \left(2M+1\right)G \right)$ and \\[2.0ex]
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