XOR8 HASH FUNCTION
B04902073 洪崇凱
B02208025 蔡誌烜
B02705025 賴冠廷
Complexity
md5, sha1
djb2, sdbm
sum8, xor8
B02208025
B02407068
B03701137
B03208028
B04705045
︙
studentID
djb2
xor8
'B02208025'
hash
read
\times33\,+
×33+
djb2:
xor8:
5381
5381
42
42
0
0
=
=
177615
177615
\oplus
⊕
=
=
42
42
48
48
50
50
177615
177615
42
42
5861343
5861343
193424369
193424369
122
122
5861343
5861343
72
72
122
122
hash'
42
42
48
48
50
50
| B | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 2 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
| 2 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
| 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 8 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 2 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
| 5 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |
\oplus
⊕
Since XOR is bit-independent, symmetric and reflexive...
A=\begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 1 & 0 & 1 & 0 & 1
\end{bmatrix}
A=000000000100000000011111111011111111000001000000000001101100010000000001
H=\text{xor8}(A)\rightarrow f(\;[1\,1\,1\,1\,1\,1\,1\,1\,1]\cdot A\;)
H=xor8(A)→f([111111111]⋅A)
=[0\,1\,0\,0\,1\,1\,0\,1]
=[01001101]
f([x,y\cdots])=[x\,\text{mod}\,2,y\,\text{mod}\,2\cdots]
f([x,y⋯])=[xmod2,ymod2⋯]
每項有 n bits 做 XOR ⇒ n 維向量做運算 ⇒
\text{hash}\sim U[0,2^n]
hash∼U[0,2n]
Conclusion 1A
k\mid2^n\Rightarrow
k∣2n⇒
k\nmid2^n\Rightarrow
k∤2n⇒
最優（分散最平均）
個桶子有
2^n\,\text{mod}\,k
2nmodk
\frac{\lceil\frac{2^n}{k}\rceil}{\lfloor\frac{2^n}{k}\rfloor}
⌊k2n⌋⌈k2n⌉
倍碰撞
個桶子沒用
k-2^n
k−2n
Optimization 1
k=127
k=127
\rightarrow128
→128
\text{Data}\sim U[0,255]
Data∼U[0,255]
\sigma=43.7
σ=43.7
\rightarrow25.2
→25.2
Optimization 1
k=60
k=60
\rightarrow64
→64
\text{Data}\sim\text{studentID}
Data∼studentID
\sigma=1815
σ=1815
\rightarrow1782
→1782
P_A=\begin{bmatrix}
a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\
a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\
a_{3,1} & a_{3,2} & \cdots & a_{3,n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{m,1} & a_{m,2} & \cdots & a_{m,n}
\end{bmatrix}
PA=a1,1a2,1a3,1⋮am,1a1,2a2,2a3,2⋮am,2⋯⋯⋯⋱⋯a1,na2,na3,n⋮am,n
P_H=\begin{bmatrix}h_{1} & h_{2} & \cdots & h_{n}\end{bmatrix}
PH=[h1h2⋯hn]
P_A=\begin{bmatrix}x\\y\\z\end{bmatrix}
PA=xyz
\rightarrow\Pr(\text{hash}=1)=x(1-y)(1-z)+(1-x)y(1-z)
→Pr(hash=1)=x(1−y)(1−z)+(1−x)y(1−z)
+(1-x)(1-y)z+xyz
+(1−x)(1−y)z+xyz
=4(x-\frac{1}{2})(y-\frac{1}{2})(z-\frac{1}{2})+\frac{1}{2}
=4(x−21)(y−21)(z−21)+21
Conclusion 2A
h=\frac{1}{2}\Leftrightarrow x = \frac{1}{2}\vee y = \frac{1}{2}\vee z = \frac{1}{2}
h=21⇔x=21∨y=21∨z=21
\rightarrow h_n = \frac{1}{2}\;\Leftrightarrow\;\exists m\ni a_{m,n} = \frac{1}{2}
→hn=21⇔∃m∋am,n=21
P_A=\begin{bmatrix}x\\y\\z\end{bmatrix}
PA=xyz
P_H=\,\;[h]
PH=[h]
\rightarrow
→
當 xor 中有一項遵守 Uniform Distribution
最後結果同樣遵守 Uniform Distribution
A\sim U\Rightarrow (A\oplus B\oplus C)\sim U
A∼U⇒(A⊕B⊕C)∼U
By Induction
Conclusion 2B
x=y=z=t
x=y=z=t
當各項有相同性質，即
t
t
h
h
Optimization 2
'B02 208 025'
2828\oplus520\oplus25
2828⊕520⊕25
DEC
HEX
HEX
Optimization 2
\text{Data}\sim\text{studentID}
Data∼studentID
\text{(Re-grouped)}
(Re-grouped)
\sigma=1782
σ=1782
\rightarrow147
→147
'B02 208 025'
2828\oplus520\oplus25\,\oplus
2828⊕520⊕25⊕
- Random
- Hash
- Histogram Equilization
x
x
Histogram Equilization
h(\text{``B02208''})=60
h(“B02208”)=60
'B02 208 025'
2828\oplus520\oplus25\,\oplus
2828⊕520⊕25⊕
60
60
From Histogram Equilization:
Optimization 3
\sigma=147
σ=147
\rightarrow48.6
→48.6
\text{Data}\sim\text{studentID}
Data∼studentID
\text{(Histogram Equilization)}
(Histogram Equilization)
- 使用適當的桶子數
- 找到具 Uniform Distribution 的資料特徵
- 對已知分佈作 Histogram Equilization
2^i, i\leq n
2i,i≤n
xor8
djb2
xor8 (optimized)
\sigma=1815
σ=1815
\sigma=40.4
σ=40.4
\sigma=48.6
σ=48.6
6\text{x}
6x
but
(156 ms ⇒ 25 ms)
Faster!
Reference
XOR8 HASH FUNCTION B04902073 洪崇凱 B02208025 蔡誌烜 B02705025 賴冠廷