2023.11.13 闲话

图书馆

歌：酩酊パラノイア - Celery feat. 初音ミク .

怎么都放这个感觉的，那我也放个：運命線上のアリア (Vo. あよ) .

Five Nights at Freddy's - The Living Tombstone

We're waiting every night to finally roam and invite
Newcomers to play with us
For many years we've been all alone
We're forced to be still
And play the same songs we've known since that day
An imposter took our life away
Now we're stuck here to decay

Please let us get in
Don't lock us away
We're not like what you're thinking
We are poor little souls
Who have lost all control
And we're forced here to take that role

We've been all alone
Stuck in our little zone since 1987
Join us, be our friend
Or just be stuck and defend
After all you only got

Five Nights at Freddy's
Is this where you wanna be?
I just don't get it
Why do you want to stay?
Five Nights at Freddy's
Is this where you wanna be?
I just don't get it
Why do you want to stay?
Five Nights at Freddy's

We're really quite surprised
That we get to see you another night
You should have looked for another job
You should have said to this place goodbye

It's like there's so much more
Maybe you've been in this place before
We remember a face like yours
You seemed acquainted with those doors

Please let us get in
Don't lock us away
We're not like what you're thinking
We are poor little souls
Who have lost all control
And we're forced here to take that role

We've been all alone
Stuck in our little zone since 1987
Join us, be our friend
Or just be stuck and defend
After all you only got

Five Nights at Freddy's
Is this where you wanna be?
I just don't get it
Why do you want to stay?
Five Nights at Freddy's
Is this where you wanna be?
I just don't get it
Why do you want to stay?
Five Nights at Freddy's

---

问题 1：求长为 \(n\) 的环的 \(k\) 染色的方案数（相邻不能同色）.

固定 \(k\)，考察 \(1\) 和 \(n-1\) 是否同色：

• 同色：那么 \(n\) 有 \(k-1\) 种选择，再删了 \(n-1\) 剩下 \(n-2\) 个位置独立 .
• 不同色：\(n\) 有 \(k-2\) 种选择，剩下 \(n-1\) 个位置独立 .

那么可以导出递推：\(a_n=(k-2)a_{n-1}+(k-1)a_{n-2}\) .

所以就知道 \(a_n=(k-1)^n+(k-1)(-1)^n\) 了 .

问题 2：求 \(n\) 棱柱的 \(k\) 染色的方案数（相邻不能同色）.

也是一样做，分为五种情况（左边是 \(1\)，右边是 \(n-1\)）：

• Case 1：2 色同色相对

  A ... A
  B ... B
  

• Case 2：2 色异色相对

  A ... B
  B ... A
  

• Case 3：3 色同色相对

  A ... A     A ... C
  B ... C     B ... B
  

• Case 4：3 色异色相对

  A ... C     A ... B
  B ... A     B ... C
  

• Case 5：4 色

  A ... C
  B ... D
  

那么仔细地操作即可得出递推，为了方便设长为 5 的向量作为状态，那么转移矩阵为：

\[\small\begin{bmatrix}0&1&0&1&1\\1&0&1&0&1\\0&2(k-2)&(k-2)&(k-2)+(k-3)&2(k-3)\\2(k-2)&0&(k-2)+(k-3)&(k-2)&2(k-3)\\(k-2)(k-3)&(k-2)(k-3)&(k-3)^2&(k-3)^2&(k-4)^2+(k-3)\end{bmatrix} \]

因为我不会解五次方程所以通项就摆了 .

注意到色多项式和 Tutte 多项式的关系：

\[P_G(c)=(-1)^{|V|-k(E)}\cdot c^{k(E)}\cdot T_G(1-k,0) \]

其中 Tutte 多项式

\[T_G(x,y)=\sum_{A\subseteq E}(x-1)^{k(A)-k(E)}(y-1)^{k(A)+|A|-|V|} \]

其中 \(k(E)\) 是 \((V,E)\) 的连通分量数 .

通过这样的内容可以减少讨论的细节 .

问题 3：求 \(n\) 个点的图的 \(c\) 染色数，图的形态为一个长为 \(k\) 的环上连出一条链 .

直接代入色多项式：

\[\begin{aligned}P_G(c)&=(-1)^{n-1}c\sum_{A\subseteq E}(-c)^{k(A)-1}(-1)^{k(A)+|A|-n}\\&=\sum_{A\subseteq E}c^{k(A)}(-1)^{|A|}\\\end{aligned} \]

枚举选多少条边，分选不选环讨论：

\[\begin{aligned}P_G(c)&=\sum_{i=0}^n(-1)^i\left(\dbinom nic^{n-i}+\dbinom{n-k}{i-k}c^{n-i+1}-\dbinom{n-k}{i-k}c^{n-i}\right)\\&=(c-1)^n+\sum_{i=0}^n\dbinom{n-k}{i-k}(-1)^i(c^{n-i+1}-c^{n-i})\\&=(c-1)^n+(-1)^k(c-1)^{n-k+1}\end{aligned} \]

结束了！大家再见~