# Nonasymptotic Upper Bounds on Binary Single Deletion Codes via Mixed Integer Linear Programming

## Abstract

**:**

## 1. Introduction

#### Notation

## 2. Preliminaries

#### 2.1. Single Deletion Code

**Definition**

**1.**

**Lemma**

**1.**

#### 2.2. Varshamov–Tenengolts Codes

#### 2.3. Maximum Independent Set Approach

#### 2.4. LP Relaxation

$\mathit{n}$ | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

$|V{T}_{0}\left(n\right)|$ | 10 | 16 | 30 | 52 | 94 | 172 | 316 |

Lovász | 10.00 | 16.84 | 30.00 | 53.03 | 95.98 | 174.73 | - |

LP | 10.25 | 17.15 | 30.32 | 53.56 | 96.52 | 175.19 | 321.27 |

## 3. Mixed Integer Linear Programming

#### 3.1. Main Results

Algorithm 1 Sequential MILP. |

Input: target threshold $\tau $ |

Output: new bound T |

procedure S_{EQ}MILP($\tau $) |

Set $\mathcal{S}=\varnothing $, and $T=inf$ |

do |

Solve $MIL{P}_{n}\left(\mathcal{S}\right)$ and let T be the objective function value and X be the solution |

${i}_{0}=arg{max}_{i\notin \mathcal{S}}d\left(i\right)\times {X}_{i}$ |

$\mathcal{S}\leftarrow \mathcal{S}\cup \left\{{i}_{0}\right\}$ |

while $\left|\mathcal{S}\right|<N$ and $T\ge \tau $ |

return T |

end procedure |

#### 3.2. Experiments

**Corollary**

**1.**

#### 3.3. Connection to Metaheuristics

## 4. Equivalent Conjecture

#### 4.1. VT-Sum Based Partition

**Lemma**

**2.**

- 1.
- $\left|{S}_{n,i}\right|=\left|{S}_{n,\frac{n(n+1)}{2}-i}\right|$.
- 2.
- If $|i-j|\ge n+1$, then ${\mathcal{B}}_{D}\left({S}_{n,i}\right)\cap {\mathcal{B}}_{D}\left({S}_{n,j}\right)=\varnothing $.
- 3.
- If $|i-j|\ge n+1$, then ${\mathcal{B}}_{I}\left({S}_{n,i}\right)\cap {\mathcal{B}}_{I}\left({S}_{n,j}\right)=\varnothing $.

**Proof.**

- There exists a one-to-one correspondence between ${S}_{n,i}$ and ${S}_{n,\frac{n(n+1)}{2}-i}$ which is an element-wise binary complement.
- For any element ${z}^{n-1}\in {\mathcal{B}}_{D}\left({S}_{n,i}\right)$, it is easy to show that $i-n\le {v}_{n-1}\left({z}^{n-1}\right)\le i$. If $|i-j|\ge n+1$, then $\{i,i-1,\dots ,i-n\}\cap \{j,j-1,\dots ,j-n\}=\varnothing $, and therefore ${\mathcal{B}}_{D}\left({S}_{n,i}\right)\cap {\mathcal{B}}_{D}\left({S}_{n,j}\right)=\varnothing $.
- For any ${x}^{n},{\stackrel{~}{x}}^{n}\in {\mathcal{X}}^{n}$, it is clear that ${\mathcal{B}}_{D}\left({x}^{n}\right)\cap {\mathcal{B}}_{D}\left({\stackrel{~}{x}}^{n}\right)=\varnothing $ if and only if ${\mathcal{B}}_{I}\left({x}^{n}\right)\cap {\mathcal{B}}_{I}\left({\stackrel{~}{x}}^{n}\right)=\varnothing $. Thus, the third property is a direct consequence of the second property.

**Remark**

**1.**

- There are no internal edges in ${S}_{n,i}$ for all i.
- There are no edges between ${S}_{n,i}$ and ${S}_{n,j}$ if $|i-j|\ge n+1$.

#### 4.2. Equivalent Conjecture

**Lemma**

**3.**

**Proof.**

**Lemma**

**4.**

**Proof.**

**Conjecture**

**1.**

- 1.
- For even n, any single deletion code$\mathcal{C}\subset {\mathcal{X}}^{n}$satisfies$$\begin{array}{c}\hfill |{\mathcal{C}}_{0}|+|{\mathcal{C}}_{1}|+\cdots +|{\mathcal{C}}_{(n+1)k}|\le |{S}_{n,0}|+|{S}_{n,n+1}|+\cdots +|{S}_{n,(n+1)k}|\end{array}$$
- 2.
- For odd n, any single deletion code$\mathcal{C}\subset {\mathcal{X}}^{n}$satisfies$$\begin{array}{c}\hfill |{\mathcal{C}}_{0}|+|{\mathcal{C}}_{1}|+\cdots +|{\mathcal{C}}_{(n+1)k+(n+1)/2}|\le |{S}_{n,(n+1)/2}|+|{S}_{n,n+1+(n+1)/2}|+\cdots +|{S}_{n,(n+1)k+(n+1)/2}|\end{array}$$

**Theorem**

**1.**

**Proof.**

#### 4.3. Special Case of $k=1$

**Lemma**

**5.**

**Proof.**

**Lemma**

**6.**

- 1.
- For all ${x}^{n}\in {\mathcal{X}}^{n}$, we have $f\left({x}^{n}\right)\in V{T}_{0}\left(n\right)$.
- 2.
- If ${x}^{n}\in V{T}_{0}\left(n\right)$, then $f\left({x}^{n}\right)={x}^{n}$.
- 3.
- For all ${x}^{n}\in {\mathcal{X}}^{n}$, we have $|{v}_{n}\left(f\left({x}^{n}\right)\right)-{v}_{n}\left({x}^{n}\right)|\le n$.

**Proof.**

- Let ${y}^{n+1}=g\left({x}^{n}\right)$, then we have$$\begin{array}{c}\hfill {v}_{n}\left(f\left({x}^{n}\right)\right)\equiv {v}_{n}\left(h\left({y}^{n+1}\right)\right)\equiv {v}_{n+1}\left({y}^{n+1}\right)\equiv 0(\mathrm{mod}n+1).\end{array}$$
- If ${v}_{n}\left({x}^{n}\right)\equiv 0\mathrm{mod}(n+1)$, then we have $g\left({x}^{n}\right)={x}^{n}0$, and therefore $f\left({x}^{n}\right)={x}^{n}$.
- If we let ${y}^{n+1}=g\left({x}^{n}\right)$, then we have$$\begin{array}{cc}\hfill {v}_{n}\left(f\left({x}^{n}\right)\right)-{v}_{n}\left({x}^{n}\right)=& {v}_{n}\left({y}^{n}\right)-{v}_{n}\left({x}^{n}\right)\hfill \\ \hfill =& {v}_{n+1}\left({y}^{n+1}\right)-(n+1){y}_{n+1}-{v}_{n}\left({x}^{n}\right)\hfill \\ \hfill =& {v}_{n+1}\left(g\left({x}^{n}\right)\right)-(n+1){y}_{n+1}-{v}_{n}\left({x}^{n}\right).\hfill \end{array}$$Recall that we have ${v}_{n}\left({x}^{n}\right)\le {v}_{n+1}\left({y}^{n+1}\right)\le {v}_{n}\left({x}^{n}\right)+n+1$. If ${v}_{n}\left({x}^{n}\right)\in V{T}_{0}\left(n\right)$, then $f\left({x}^{n}\right)={x}^{n}$, which implies ${v}_{n}\left(f\left({x}^{n}\right)\right)-{f}_{n}\left({x}^{n}\right)=0$. On the other hand, if ${v}_{n}\left({x}^{n}\right)\notin V{T}_{0}\left(n\right)$, we have ${v}_{n}\left({x}^{n}\right)+1\le {v}_{n+1}\left(g\left({x}^{n}\right)\right)\le {v}_{n}\left({x}^{n}\right)+n$. In such case, ${v}_{n}\left(f\left({x}^{n}\right)\right)-{v}_{n}\left({x}^{n}\right)$ achieves the maximum value when ${y}_{n+1}=0$ and ${v}_{n+1}\left(g\left({x}^{n}\right)\right)={v}_{n}\left({x}^{n}\right)+n$. In other words,$$\begin{array}{c}\hfill {v}_{n+1}\left(g\left({x}^{n}\right)\right)-{v}_{n}\left({x}^{n}\right)-(n+1){y}_{n+1}\le {v}_{n}\left({x}^{n}\right)+n-{v}_{n}\left({x}^{n}\right)=n.\end{array}$$On the other hand, it achieves the minimum value when ${y}_{n+1}=1$ and ${v}_{n+1}\left(g\left({x}^{n}\right)\right)={v}_{n}\left({x}^{n}\right)+1$. In other words,$$\begin{array}{c}\hfill {v}_{n+1}\left(g\left({x}^{n}\right)\right)-{v}_{n}\left({x}^{n}\right)-(n+1){y}_{n+1}\ge {v}_{n}\left({x}^{n}\right)+1-{v}_{n}\left({x}^{n}\right)-(n+1)=-n.\end{array}$$Thus, we have$$\begin{array}{cc}\hfill |{v}_{n}\left(f\left({x}^{n}\right)\right)-{v}_{n}\left({x}^{n}\right)|\le & n.\hfill \end{array}$$This concludes the proof.

**Theorem**

**2.**

**Proof.**

#### 4.4. Integer Programming

$\mathit{n}$ | 11 | 12 | 13 | 14 | 15 | 16 |

$k=1$ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ |

$k=2$ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ |

$k=3$ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ |

$k=4$ | ✓ | ✓ | ✓ | ✓ |

$\mathit{n}$ | 11 | 13 | 15 |

$k=1$ | ✓ | ✓ | ✓ |

$k=2$ | ✓ | ✓ | ✓ |

$k=3$ | ✓ | ✓ | ✓ |

## 5. Conclusions

## Funding

## Conflicts of Interest

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No, A. Nonasymptotic Upper Bounds on Binary Single Deletion Codes via Mixed Integer Linear Programming. *Entropy* **2019**, *21*, 1202.
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