#
Minimum Round Card-Based Cryptographic Protocols Using Private Operations^{ †}

^{*}

^{†}

## Abstract

**:**

## 1. Introduction

#### 1.1. Motivation

#### 1.2. Related Works

## 2. Preliminaries

#### 2.1. Basic Notations

#### 2.2. Private Operations

**Primitive**

**1.**

**Primitive**

**2.**

**Primitive**

**3.**

#### 2.3. Definition of Round

**Protocol**

**1.**

- 1.
- Alice executes a private random bisection cut on $commit\left(x\right)$. Let the output be $commit\left({x}^{\prime}\right)$. Alice sends $commit\left({x}^{\prime}\right)$ and $commit\left(y\right)$ to Bob.
- 2.
- Bob executes a private reveal on $commit\left({x}^{\prime}\right)$. Bob privately sets the following:$${S}_{2}=\left(\right)open="\{"\; close>\begin{array}{cc}commit\left(y\right)\left|\right|commit\left(0\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime}=1\hfill \\ commit\left(0\right)\left|\right|commit\left(y\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime}=0\hfill \end{array}$$
- 3.
- Alice executes a private reverse selection on ${S}_{2}$ using the bit b generated in the private random bisection cut. Let the obtained sequence be ${S}_{3}$. Alice outputs ${S}_{3}$.

#### 2.4. Our Results

## 3. XOR, AND, and Copy with the Minimum Number of Rounds

#### 3.1. XOR Protocol

**Protocol**

**2.**

- 1.
- Alice executes a private random bisection cut on input ${S}_{0}=commit\left(x\right)$ and ${S}_{0}^{\prime}=commit\left(y\right)$, using the same random bit b. Let the output be ${S}_{1}=commit\left({x}^{\prime}\right)$ and ${S}_{1}^{\prime}=commit\left({y}^{\prime}\right)$, respectively. Note that ${x}^{\prime}=x\oplus b$ and ${y}^{\prime}=y\oplus b$. Alice sends ${S}_{1}$ and ${S}_{1}^{\prime}$ to Bob.
- 2.
- Bob executes a private reveal on ${S}_{1}=commit\left({x}^{\prime}\right)$. Bob executes a private reverse cut on ${S}_{1}^{\prime}$ using ${x}^{\prime}$. Let the result be ${S}_{2}$. Bob outputs ${S}_{2}$.

**Theorem**

**1.**

**Proof.**

#### 3.2. AND Protocol

**Protocol**

**3.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$ and ${S}_{0}^{\prime}=commit\left(0\right)\left|\right|$$commit\left(y\right)$ using the same random bit b. Two new cards are used to set $commit\left(0\right)$. Let the output be ${S}_{1}=commit\left({x}^{\prime}\right)$ and ${S}_{1}^{\prime}$, respectively. Note the following:$${S}_{1}^{\prime}=\left(\right)open="\{"\; close>\begin{array}{cc}commit\left(y\right)\left|\right|commit\left(0\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}b=1\hfill \\ commit\left(0\right)\left|\right|commit\left(y\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}b=0\hfill \end{array}$$
- 2.
- Bob executes a private reveal on ${S}_{1}$. Bob executes a private reverse selection on ${S}_{1}^{\prime}$ using ${x}^{\prime}$. Let the selected cards be ${S}_{2}$. Bob outputs ${S}_{2}$ as the result.

**Theorem**

**2.**

**Proof.**

#### 3.3. Copy Protocol

**Protocol**

**4.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$. Let the output be ${S}_{1}=commit\left({x}^{\prime}\right)$. Alice sets ${S}_{1}^{\prime}$ as m copies of $commit\left(b\right)$, where b is the bit selected in the random bisection cut. Note that ${x}^{\prime}=x\oplus b$. Alice sends ${S}_{1}$ and ${S}_{1}^{\prime}$ to Bob.
- 2.
- Bob executes a private reveal on ${S}_{1}$ and obtains ${x}^{\prime}$. Bob executes a private reverse cut on each pair of ${S}_{1}^{\prime}$ using ${x}^{\prime}$. Let the result be ${S}_{2}$. Bob outputs ${S}_{2}$.

**Theorem**

**3.**

**Proof.**

#### 3.4. Any Two-Variable Boolean Functions

**Theorem**

**4.**

**Proof.**

#### 3.5. n-Variable Boolean Functions

**Protocol**

**5.**

- 1.
- Alice executes a private random bisection cut on $commit\left({x}_{i}\right)$ $(i=1,2,\dots ,n)$. Let the results be $commit\left({x}_{i}^{\prime}\right)(i=1,2,\dots ,n)$. ${x}_{i}^{\prime}={x}_{i}\oplus {b}_{i}(i=1,2,\dots ,n)$. Note that one random bit ${b}_{i}$ is selected for each ${x}_{i}(i=1,2,\dots ,n)$. Alice generates ${2}^{n}$ commitment ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}$ $({a}_{i}\in \{0,1\},i=1,2,\dots ,n)$ as ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}=$ $commit\left(f({a}_{1}\oplus {b}_{1},{a}_{2}\oplus {b}_{2},\dots ,{a}_{n}\oplus {b}_{n})\right)$.Alice sends $commit\left({x}_{i}^{\prime}\right)(i=1,2,\dots ,n)$ and ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}$ $({a}_{i}\in \{0,1\},i=1,2,\dots ,n)$ to Bob.
- 2.
- Bob executes a private reveal on $commit\left({x}_{i}^{\prime}\right)$ $(i=1,2,\dots ,n)$. Bob outputs ${S}_{{x}_{1}^{\prime},{x}_{2}^{\prime},\dots ,{x}_{n}^{\prime}}$.

**Theorem**

**5.**

**Proof.**

## 4. Protocols without Private Reveals

#### 4.1. XOR Protocol without Private Reveals

**Protocol**

**6.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$ and ${S}_{0}^{\prime}=commit\left(y\right)$ using the same random bit b. Let the output be ${S}_{1}=commit\left({x}^{\prime}\right)$ and ${S}_{1}^{\prime}=commit\left({y}^{\prime}\right)$, respectively. Note that ${x}^{\prime}=x\oplus b$ and ${y}^{\prime}=y\oplus b$. Alice sends ${S}_{1}$ and ${S}_{1}^{\prime}$ to Bob.
- 2.
- Bob executes a private random bisection cut on ${S}_{1}$ and ${S}_{1}^{\prime}$ using a private bit ${b}^{\prime}$. Let the output be ${S}_{2}=commit\left({x}^{\prime \prime}\right)$ and ${S}_{2}^{\prime}=commit\left({y}^{\prime \prime}\right)$, respectively. ${x}^{\prime \prime}=x\oplus b\oplus {b}^{\prime}$ and ${y}^{\prime \prime}=y\oplus b\oplus {b}^{\prime}$ hold. Bob publicly opens ${S}_{2}$ and obtains ${x}^{\prime \prime}$. Alice can see ${x}^{\prime \prime}$. Bob publicly sets the following:$${S}_{3}=\left(\right)open="\{"\; close>\begin{array}{cc}commit\left(\overline{{y}^{\prime \prime}}\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime \prime}=1\hfill \\ commit\left({y}^{\prime \prime}\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime \prime}=0\hfill \end{array}$$

**Theorem**

**6.**

**Proof.**

#### 4.2. AND Protocol without Private Reveals

**Protocol**

**7.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$. Let the result be ${S}_{1}=commit\left({x}^{\prime}\right)$. Alice sends ${S}_{1}$ and ${S}_{0}^{\prime}=commit\left(y\right)$ to Bob.
- 2.
- Bob executes a private random bisection cut on ${S}_{1}$, using a private bit ${b}^{\prime}$. Let the result be ${S}_{1}^{\prime}=commit\left({x}^{\prime \prime}\right)$. ${x}^{\prime \prime}=x\oplus b\oplus {b}^{\prime}$ holds. Bob publicly opens ${S}_{1}^{\prime}$ and obtains value ${x}^{\prime \prime}$. Alice can see ${x}^{\prime \prime}$. Bob publicly sets the following:$${S}_{2}=\left(\right)open="\{"\; close>\begin{array}{cc}commit\left(y\right)\left|\right|commit\left(0\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime \prime}=1\hfill \\ commit\left(0\right)\left|\right|commit\left(y\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime \prime}=0\hfill \end{array}$$
- 3.
- Alice executes a private reverse selection on ${S}_{3}$ using the bit b generated in the private random bisection cut. Let the result be ${S}_{4}$. Alice outputs ${S}_{4}$.

**Theorem**

**7.**

**Proof.**

**Protocol**

**8.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$ and ${S}_{0}^{\prime}=commit\left(0\right)\left|\right|$ $commit\left(y\right)$ using the same random bit b. Two new cards are used to set $commit\left(0\right)$. Let the output be ${S}_{1}=commit\left({x}^{\prime}\right)$ and ${S}_{1}^{\prime}$, respectively. Note the following:$${S}_{1}^{\prime}=\left(\right)open="\{"\; close>\begin{array}{cc}commit\left(y\right)\left|\right|commit\left(0\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}b=1\hfill \\ commit\left(0\right)\left|\right|commit\left(y\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}b=0\hfill \end{array}$$
- 2.
- Bob executes a private random bisection cut on ${S}_{1}$ and ${S}_{1}^{\prime}$ using the same random bit ${b}^{\prime}$. Let the result be ${S}_{2}$ and ${S}_{2}^{\prime}$, respectively. Note that ${S}_{2}=commit(x\oplus b\oplus {b}^{\prime})$. Bob publicly opens cards of ${S}_{2}$ and obtains ${x}^{\prime \prime}=x\oplus b\oplus {b}^{\prime}$. Alice can see ${x}^{\prime \prime}$. Bob publicly selects the left pair of ${S}_{2}^{\prime}$ if ${x}^{\prime \prime}=0$, otherwise selects the right pair of ${S}_{2}^{\prime}$. Bob outputs the pair as the result.

**Theorem**

**8.**

**Proof.**

#### 4.3. Copy Protocol without Private Reveals

**Protocol**

**9.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$. Let the result be ${S}_{1}=commit\left({x}^{\prime}\right)$. Note that ${x}^{\prime}=x\oplus b$. Alice sends ${S}_{1}$ to Bob.
- 2.
- Bob executes a private random bisection cut on ${S}_{1}$ using a private random bit ${b}^{\prime}$. Let the result be ${S}_{2}=commit\left({x}^{\prime \prime}\right)$. Note that ${x}^{\prime \prime}=x\oplus b\oplus {b}^{\prime}$.Bob publicly opens ${S}_{2}$ and obtains ${x}^{\prime \prime}$. Alice can see ${x}^{\prime \prime}$. Bob publicly sets m pairs of cards of ${x}^{\prime \prime}$. Bob faces down the cards. Let the cards be ${S}_{3}$. Bob executes a private reverse cut on each pair of ${S}_{3}$ using ${b}^{\prime}$. Let the result be ${S}_{3}^{\prime}$. Bob sends ${S}_{3}^{\prime}$ to Alice.
- 3.
- Alice executes a private reverse cut on each pair of ${S}_{3}^{\prime}$ using b. Alice outputs the pairs.

**Theorem**

**9.**

**Proof.**

**Protocol**

**10.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$. Let the result be ${S}_{1}=commit\left({x}^{\prime}\right)$. Note that ${x}^{\prime}=x\oplus b$. Alice privately sets ${S}_{1}^{\prime}$ as m copies of $commit\left(b\right)$. Alice sends ${S}_{1}$ and ${S}_{1}^{\prime}$ to Bob.
- 2.
- Bob executes a private random bisection cut on ${S}_{1}$ and each pair of ${S}_{1}^{\prime}$ using a private random bit ${b}^{\prime}$. Let the output be ${S}_{2}=commit\left({x}^{\prime \prime}\right)$ and ${S}_{2}^{\prime}$, respectively.Bob publicly opens ${S}_{2}$ and obtains ${x}^{\prime \prime}$. Alice can see ${x}^{\prime \prime}$. Bob publicly swaps each pair of ${S}_{2}^{\prime}$ if ${x}^{\prime \prime}=1$. Otherwise, Bob does nothing. Let the result be ${S}_{3}$. Bob outputs ${S}_{3}$.

**Theorem**

**10.**

**Proof.**

#### 4.4. n-Variable Boolean Functions without Private Reveals

**Protocol**

**11.**

- 1.
- Alice executes a private random bisection cut on $commit\left({x}_{i}\right)$ $(i=1,2,\dots ,n)$. Let the results be $commit\left({x}_{i}^{\prime}\right)(i=1,2,\dots ,n)$. ${x}_{i}^{\prime}={x}_{i}\oplus {b}_{i}(i=1,2,\dots ,n)$. Note that one random bit ${b}_{i}$ is selected for each ${x}_{i}(i=1,2,\dots ,n)$. Alice generates ${2}^{n}$ commitment ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}$ $({a}_{i}\in \{0,1\},i=1,2,\dots ,n)$ as ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}=$ $commit\left(f({a}_{1}\oplus {b}_{1},{a}_{2}\oplus {b}_{2},\dots ,{a}_{n}\oplus {b}_{n})\right)$.Alice sends $commit\left({x}_{i}^{\prime}\right)(i=1,2,\dots ,n)$ and ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}$ $({a}_{i}\in \{0,1\},i=1,2,\dots ,n)$ to Bob.
- 2.
- Bob executes a private random bisection cut on $commit\left({x}_{i}^{\prime}\right)(i=1,2,\dots ,n)$. Note that one random bit ${b}_{i}^{\prime}$ is selected for each ${x}_{i}^{\prime}(i=1,2,\dots ,n)$. Let $commit\left({x}_{i}^{\prime \prime}\right)$ $(i=1,2,\dots ,n)$ be the obtained value. ${x}_{i}^{\prime \prime}={x}_{i}\oplus {b}_{i}\oplus {b}_{i}^{\prime}(i=1,2,\dots ,n)$ is satisfied. Bob privately relocates ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}({a}_{i}\in \{0,1\},i=1,2,\dots ,n)$ so that ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}^{\prime}={S}_{{a}_{1}\oplus {b}_{1}^{\prime},{a}_{2}\oplus {b}_{2}^{\prime},\dots ,{a}_{n}\oplus {b}_{n}^{\prime}}({a}_{i}\in \{0,1\},$ $i=1,2,\dots ,n)$. The cards satisfy ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}^{\prime}=$ $commit\left(f({a}_{1}\oplus {b}_{1}\oplus {b}_{1}^{\prime},{a}_{2}\oplus {b}_{2}\oplus {b}_{2}^{\prime},\dots ,{a}_{n}\oplus {b}_{n}\oplus {b}_{n}^{\prime})\right)$.Bob publicly reveals $commit\left({x}_{i}^{\prime \prime}\right)$ and obtains ${x}_{i}^{\prime \prime}(i=1,2,\dots ,n)$. Alice can see ${x}_{i}^{\prime \prime}(i=1,2,\dots ,n)$. Bob publicly selects ${S}_{{x}_{1}^{\prime \prime},{x}_{2}^{\prime \prime},\dots ,{x}_{n}^{\prime \prime}}^{\prime}$.

**Theorem**

**11.**

**Proof.**

## 5. Protocols That Preserve an Input

**Protocol**

**12.**

- 1.
- Alice executes a private random bisection cut on input ${S}_{0}=commit\left(x\right)$ and ${S}_{0}^{\prime}=commit\left(y\right)$ using the same random bit b. Let the output be ${S}_{1}=commit\left({x}^{\prime}\right)$ and ${S}_{1}^{\prime}=commit\left({y}^{\prime}\right)$, respectively. Note that ${x}^{\prime}=x\oplus b$ and ${y}^{\prime}=y\oplus b$. Alice sends ${S}_{1}$ and ${S}_{1}^{\prime}$ to Bob.
- 2.
- Bob executes a private reveal on ${S}_{1}=commit\left({x}^{\prime}\right)$. Bob executes a private reverse cut on ${S}_{1}^{\prime}$, using ${x}^{\prime}$. Let the result be ${S}_{2}$. Bob outputs ${S}_{2}$. Bob sends back ${S}_{1}=commit\left({x}^{\prime}\right)$ to Alice.
- 3.
- Alice executes a private reverse cut on ${S}_{1}$ using b and obtains $commit\left(x\right)$.

**Theorem**

**12.**

**Proof.**

**Protocol**

**13.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$ and ${S}_{0}^{\prime}=commit\left(y\right)$ using the same random bit b. Let the output be ${S}_{1}=commit\left({x}^{\prime}\right)$ and ${S}_{1}^{\prime}=commit\left({y}^{\prime}\right)$, respectively. Note that ${x}^{\prime}=x\oplus b$ and ${y}^{\prime}=y\oplus b$. Alice sends ${S}_{1}$ and ${S}_{1}^{\prime}$ to Bob.
- 2.
- Bob executes a private random bisection cut on ${S}_{1}$ and ${S}_{1}^{\prime}$ using a private bit ${b}^{\prime}$. Let the output be ${S}_{2}=commit\left({x}^{\prime \prime}\right)$ and ${S}_{2}^{\prime}=commit\left({y}^{\prime \prime}\right)$, respectively. ${x}^{\prime \prime}=x\oplus b\oplus {b}^{\prime}$ and ${y}^{\prime \prime}=y\oplus b\oplus {b}^{\prime}$ hold. Bob publicly opens ${S}_{2}$ and obtains ${x}^{\prime \prime}$. Alice can see ${x}^{\prime \prime}$. Bob publicly sets the following:$${S}_{3}=\left(\right)open="\{"\; close>\begin{array}{cc}commit\left(\overline{{y}^{\prime \prime}}\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime \prime}=1\hfill \\ commit\left({y}^{\prime \prime}\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime \prime}=0\hfill \end{array}$$
- 3.
- Alice executes a private reverse cut on ${S}_{1}$ using b and obtains $commit\left(x\right)$.

**Theorem**

**13.**

**Proof.**

**Protocol**

**14.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$ and ${S}_{0}^{\prime}=commit\left(0\right)\left|\right|$$commit\left(y\right)$ using the same random bit b. Two new cards are used to set $commit\left(0\right)$. Let the output be ${S}_{1}=commit\left({x}^{\prime}\right)$ and ${S}_{1}^{\prime}$, respectively. Note the following:$${S}_{1}^{\prime}=\left(\right)open="\{"\; close>\begin{array}{cc}commit\left(y\right)\left|\right|commit\left(0\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}b=1\hfill \\ commit\left(0\right)\left|\right|commit\left(y\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}b=0\hfill \end{array}$$
- 2.
- Bob executes a private reveal on ${S}_{1}$. Bob executes a private reverse selection on ${S}_{1}^{\prime}$ using ${x}^{\prime}$. Let the selected cards be ${S}_{2}$. Bob outputs ${S}_{2}$ as the result. Bob sends back ${S}_{1}=commit\left({x}^{\prime}\right)$ to Alice.
- 3.
- Alice executes a private reverse cut on ${S}_{1}$ using b and obtains $commit\left(x\right)$.

**Theorem**

**14.**

**Proof.**

**Protocol**

**15.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$. Let the result be ${S}_{1}=commit\left({x}^{\prime}\right)$. Alice sends ${S}_{1}$ and ${S}_{0}^{\prime}=commit\left(y\right)$ to Bob.
- 2.
- Bob executes a private random bisection cut on ${S}_{1}$ using a private bit ${b}^{\prime}$. Let the result be ${S}_{1}^{\prime}=commit\left({x}^{\prime \prime}\right)$. ${x}^{\prime \prime}=x\oplus b\oplus {b}^{\prime}$ holds. Bob publicly opens ${S}_{1}^{\prime}$ and obtains value ${x}^{\prime \prime}$. Alice can see ${x}^{\prime \prime}$. Bob publicly sets the following:$${S}_{2}=\left(\right)open="\{"\; close>\begin{array}{cc}commit\left(y\right)\left|\right|commit\left(0\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime \prime}=1\hfill \\ commit\left(0\right)\left|\right|commit\left(y\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime \prime}=0\hfill \end{array}$$
- 3.
- Alice executes a private reverse selection on ${S}_{3}$ using the bit b generated in the private random bisection cut. Let the result be ${S}_{4}$. Alice outputs ${S}_{4}$. Alice executes a private reverse cut on ${S}_{1}$ using b and obtains $commit\left(x\right)$.

**Theorem**

**15.**

**Proof.**

**Protocol**

**16.**

- 1.
- Alice executes a private random bisection cut on ${S}_{0}=commit\left(x\right)$. Let the result be ${S}_{1}=commit\left({x}^{\prime}\right)$. Alice sends ${S}_{1}$ and ${S}_{0}^{\prime}=commit\left(y\right)$ to Bob.
- 2.
- Bob executes a private random bisection cut on ${S}_{1}$, using a private bit ${b}^{\prime}$. Let the result be ${S}_{1}^{\prime}=commit\left({x}^{\prime \prime}\right)$. ${x}^{\prime \prime}=x\oplus b\oplus {b}^{\prime}$ holds. Bob publicly opens ${S}_{1}^{\prime}$ and obtains value ${x}^{\prime \prime}$. Alice can see ${x}^{\prime \prime}$. Bob publicly sets the following:$${S}_{2}=\left(\right)open="\{"\; close>\begin{array}{cc}commit\left(f\right(1,y\left)\right)\left|\right|commit\left(f\right(0,y\left)\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime \prime}=1\hfill \\ commit\left(f\right(0,y\left)\right)\left|\right|commit\left(f\right(1,y\left)\right)\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}{x}^{\prime \prime}=0\hfill \end{array}$$
- 3.
- Alice executes a private reverse selection on ${S}_{3}$ using the bit b generated in the private random bisection cut. Let the result be ${S}_{4}$. Alice outputs ${S}_{4}$. Let ${S}_{4}^{\prime}$ be the cards that are not selected.
- 4.
- Alice and Bob execute the XOR protocol that preserves an input without private reveals (Protocol 12) for ${S}_{4}$ and ${S}_{4}^{\prime}$. Let the preserved input, ${S}_{4}$, be the result. We obtain $commit\left(y\right)$ from the XOR result.

**Theorem**

**16.**

**Proof.**

**Protocol**

**17.**

- 1.
- Alice executes a private random bisection cut on $commit\left({x}_{i}\right)$ $(i=1,2,\dots ,n)$. Let the results be $commit\left({x}_{i}^{\prime}\right)(i=1,2,\dots ,n)$. ${x}_{i}^{\prime}={x}_{i}\oplus {b}_{i}(i=1,2,\dots ,n)$. Note that one random bit ${b}_{i}$ is selected for each ${x}_{i}(i=1,2,\dots ,n)$. Alice generates ${2}^{n}$ commitment ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}$ $({a}_{i}\in \{0,1\},i=1,2,\dots ,n)$ as ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}=$ $commit\left(f({a}_{1}\oplus {b}_{1},{a}_{2}\oplus {b}_{2},\dots ,{a}_{n}\oplus {b}_{n})\right)$.Alice sends $commit\left({x}_{i}^{\prime}\right)(i=1,2,\dots ,n)$ and ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}$ $({a}_{i}\in \{0,1\},i=1,2,\dots ,n)$ to Bob.
- 2.
- Bob executes a private reveal on $commit\left({x}_{i}^{\prime}\right)$ $(i=1,2,\dots ,n)$. Bob outputs ${S}_{{x}_{1}^{\prime},{x}_{2}^{\prime},\dots ,{x}_{n}^{\prime}}$. Bob sends back $commit\left({x}_{i}^{\prime}\right)(i=1,2,\dots ,n)$ to Alice.
- 3.
- Alice executes a private reverse cut on $commit\left({x}_{i}^{\prime}\right)$ using ${b}_{i}(i=1,2,\dots ,n)$. Alice obtains $commit\left({x}_{i}\right)(i=1,2,\dots ,n)$.

**Theorem**

**17.**

**Proof.**

**Protocol**

**18.**

- 1.
- 2.
- Bob executes a private random bisection cut on $commit\left({x}_{i}^{\prime}\right)(i=1,2,\dots ,n)$. Note that one random bit ${b}_{i}^{\prime}$ is selected for each ${x}_{i}^{\prime}(i=1,2,\dots ,n)$. Let $commit\left({x}_{i}^{\prime \prime}\right)$ $(i=1,2,\dots ,n)$ be the obtained value. ${x}_{i}^{\prime \prime}={x}_{i}\oplus {b}_{i}\oplus {b}_{i}^{\prime}(i=1,2,\dots ,n)$ is satisfied. Bob privately relocates ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}({a}_{i}\in \{0,1\},i=1,2,\dots ,n)$ so that ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}^{\prime}={S}_{{a}_{1}\oplus {b}_{1}^{\prime},{a}_{2}\oplus {b}_{2}^{\prime},\dots ,{a}_{n}\oplus {b}_{n}^{\prime}}({a}_{i}\in \{0,1\},$ $i=1,2,\dots ,n)$. The cards satisfy that ${S}_{{a}_{1},{a}_{2},\dots ,{a}_{n}}^{\prime}=$ $commit\left(f({a}_{1}\oplus {b}_{1}\oplus {b}_{1}^{\prime},{a}_{2}\oplus {b}_{2}\oplus {b}_{2}^{\prime},\dots ,{a}_{n}\oplus {b}_{n}\oplus {b}_{n}^{\prime})\right)$.Bob publicly reveals $commit\left({x}_{i}^{\prime \prime}\right)$ and obtains ${x}_{i}^{\prime \prime}(i=1,2,\dots ,n)$. Alice can see ${x}_{i}^{\prime \prime}(i=1,2,\dots ,n)$. Bob publicly selects ${S}_{{x}_{1}^{\prime \prime},{x}_{2}^{\prime \prime},\dots ,{x}_{n}^{\prime \prime}}^{\prime}$. Bob privately sets $commit({x}_{i}^{\prime \prime}\oplus {b}_{i}^{\prime})(i=1,2,\dots ,n)$. Note that ${x}_{i}^{\prime \prime}\oplus {b}_{i}^{\prime}={x}_{i}\oplus {b}_{i}(i=1,2,\dots ,n)$. Bob sends these pairs to Alice.
- 3.
- Alice privately executes a private reverse cut on $commit({x}_{i}\oplus {b}_{i})$ using ${b}_{i}$ for each $i(i=1,2,\dots ,n)$. Alice obtains $commit\left({x}_{i}\right)(i=1,2,\dots ,n)$.

**Theorem**

**18.**

**Proof.**

## 6. Parallel Computations

## 7. Asymmetric Card Protocols

**Protocol**

**19.**

- 1.
- Alice randomly selects bit b. If b=1, Alice turns $commit\left(x\right)$ and $commit\left(y\right)$ upside down, otherwise does nothing. Let the result be ${S}_{1}$ and ${S}_{1}^{\prime}$, respectively. Note that ${S}_{1}=commit\left({x}^{\prime}\right)$ and ${S}_{1}^{\prime}=commit\left({y}^{\prime}\right)$, where ${x}^{\prime}=x\oplus b$ and ${y}^{\prime}=y\oplus b$. Alice sends ${S}_{1}$ and ${S}_{1}^{\prime}$ to Bob.
- 2.
- Bob executes a private reveal on ${S}_{1}$ and obtains ${x}^{\prime}$. If ${x}^{\prime}=1$, Bob privately turns ${S}_{1}^{\prime}$ upside down and otherwise does nothing. Let the result be ${S}_{2}$. Bob outputs ${S}_{2}$.

**Protocol**

**20.**

- 1.
- Alice executes a private random bisection cut on $commit\left(0\right)\left|\right|commit\left(y\right)$ using a random bit b. Let the result be ${S}_{1}^{\prime}$. If b=1, Alice turns $commit\left(x\right)$ upside down, and otherwise does nothing. Let the output be ${S}_{1}=commit\left({x}^{\prime}\right)$. Note that ${x}^{\prime}=x\oplus b$. Alice sends ${S}_{1}$ and ${S}_{1}^{\prime}$ to Bob.
- 2.
- Bob executes a private reveal on ${S}_{1}$. Bob executes a private reverse selection on ${S}_{1}^{\prime}$, using ${x}^{\prime}$. Let the selected card be ${S}_{2}$. Bob outputs ${S}_{2}$ as the result.

**Protocol**

**21.**

- 1.
- Alice randomly selects bit b. If b=1, Alice privately turns $commit\left(x\right)$ upside down, otherwise does nothing. Let the result be ${S}_{1}=commit\left({x}^{\prime}\right)$. Note that ${x}^{\prime}=x\oplus b$. Alice privately sets ${S}_{1}^{\prime}$ as m copies of $commit\left(b\right)$. Alice sends ${S}_{1}$ and ${S}_{1}^{\prime}$ to Bob.
- 2.
- Bob executes a private reveal on ${S}_{1}$ and obtains ${x}^{\prime}$. Bob privately upside down all cards of ${S}_{1}^{\prime}$ if ${x}^{\prime}=1$, otherwise does nothing. Let the results be ${S}_{2}$. Bob outputs ${S}_{2}$.

**Theorem**

**19.**

**Proof.**

## 8. Conclusions

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Conflicts of Interest

## References

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Article | # of Rounds | # of Cards | Preserving an Input | Private Reveal |
---|---|---|---|---|

[9] | 3 | 4 | No | Use |

[9] | 3 | 4 | Yes | Use |

[7] | 2 | 4 | No | Does not use |

Protocol 2 | 2 | 4 | No | Use |

Protocol 6 | 2 | 4 | No | Does not use |

Protocol 12 | 3 | 4 | Yes | Use |

Protocol 13 | 3 | 4 | Yes | Does not use |

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Ono, H.; Manabe, Y.
Minimum Round Card-Based Cryptographic Protocols Using Private Operations. *Cryptography* **2021**, *5*, 17.
https://doi.org/10.3390/cryptography5030017

**AMA Style**

Ono H, Manabe Y.
Minimum Round Card-Based Cryptographic Protocols Using Private Operations. *Cryptography*. 2021; 5(3):17.
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**Chicago/Turabian Style**

Ono, Hibiki, and Yoshifumi Manabe.
2021. "Minimum Round Card-Based Cryptographic Protocols Using Private Operations" *Cryptography* 5, no. 3: 17.
https://doi.org/10.3390/cryptography5030017