Site updated: 05/28/2024 23:47:07

source/_posts/data_dividing.md

@@ -11,7 +11,7 @@ tags:
 - Python
 title: Discuss the mathematics of apportionment when splitting the
   machine learning dataset into several parts by proportions
-updated: "2024-05-27 14:16:41"
+updated: "2024-05-28 23:45:58"
 ---
 
 This article discusses an operation that originated in machine learning,
@@ -1319,7 +1319,8 @@ and analysis step by step, until the conclusion is reached.
       - $h\in \mathbb{Z}^+$
       - If $h \ge 2$,
         - $\exists s,t \in \{1,2,\ldots,h\}, s < t, i_s \in G_s, i_t\in G_t, x_{i_s} < x_{i_t}$,
-        - $\forall b \in B_3$, \$ b ^n,~\_{i=1}^n b_i = m\$, and
+        - $\forall b \in B_3$,
+          $b \in \mathbb{Z}^n,~\sum_{i=1}^n b_i = m$, and
           $b_{i_s} < b_{i_t}$.
       - If $h=1$, $B_3 = \emptyset$.
     - If $h \ge 2$.
@@ -1412,7 +1413,7 @@ and analysis step by step, until the conclusion is reached.
           $D^{*'}_b = \complement_{D^{'}_b}B_1 \cap \complement_{D^{'}_b}B_2 \cap \complement_{D^{'}_b}B_3$
       - If $m > 0$, there will be
         $\exists h^{'} \in \{1,2,\ldots,h\} \sum_{i=1}^{h^{'}-1}g_i \le m < \sum_{i=1}^{h^{'}}g_i$.
-        - If \$\_{i=1}<sup>{h</sup>{’}-1}g_i = m \$,
+        - If $\sum_{i=1}^{h^{'}-1}g_i = m$,
           - then
             $\forall i \in G_1 \cup G_2 \cup \ldots \cup G_{h^{'}-1}, b_{i}=1$.
           - and,
@@ -1588,7 +1589,7 @@ and analysis step by step, until the conclusion is reached.
         - So, $D^{*}_b=\{\theta\}$, which is a single element set.
       - If $m > 0$, there will be
         $\exists h^{'} \in \{1,2,\ldots,h\} \sum_{i=1}^{h^{'}-1}g_i \le m < \sum_{i=1}^{h^{'}}g_i$.
-        - If \$\_{i=1}<sup>{h</sup>{’}-1}g_i = m \$,
+        - If $\sum_{i=1}^{h^{'}-1}g_i = m$,
           - then
             $\forall i \in G_1 \cup G_2 \cup \ldots \cup G_{h^{'}-1}, b_{i}=1$.
           - and,

source/_posts/【算法分析】round函数的值域分析.md

@@ -8,7 +8,7 @@ tags:
 - Mathematics
 - Algorithm
 title: Analyses of round function
-updated: "2024-05-23 01:02:52"
+updated: "2024-05-28 23:44:04"
 ---
 
 This article does a mathematical abstraction of the
@@ -169,8 +169,8 @@ which is base on $round(\cdot) \mathbb{R}\rightarrow \mathbb{Z}$.
 - Define
   $f(r,n,N):\{\mathbb{R}^n,\mathbb{Z}^+,\mathbb{Z}^+\}\rightarrow\{\mathbb{Z}\}$
   as:
-  - $\forall (r,n,N) \in D_{r,n,N}$, there is \$ f(r,n,N)=\[*{i=1}^n
-    round(r*{i}N)\]-N\$
+  - $\forall (r,n,N) \in D_{r,n,N}$, there is
+    $f(r,n,N)=[\sum_{i=1}^n round(r_{i}N)]-N$
 - $\forall n \in \mathbb{Z}^+$, define
   $D_{r,N}=\{(r,N)|N \in \mathbb{Z}^+, N\ge n,r = [r_1,r_2,\ldots,r_n]\in \mathbb{R}_+^n,\|r\|_{1}=1\}$.
 - $\forall n \in \mathbb{Z}^+$, define the function $f(r,n,N)$’s value
@@ -230,8 +230,8 @@ the $0$ is also necessary.
   - $D_{r,n,N}=\{(r,n,N)|n \in\mathbb{Z}^+,N \in\mathbb{Z}^+, n \le N,r = [r_1,r_2,\ldots,r_n]\in \mathbb{R}_+^n,\|r\|_{1}=1\}$
   - $f(r,n,N):\{\mathbb{R}^n,\mathbb{Z}^+,\mathbb{Z}^+\}\rightarrow\{\mathbb{Z}\}$
     as:
-    - $\forall (r,n,N) \in D_{r,n,N}$, there is \$ f(r,n,N)=\[*{i=1}^n
-      round(r*{i}N)\]-N\$.
+    - $\forall (r,n,N) \in D_{r,n,N}$, there is
+      $f(r,n,N)=[\sum_{i=1}^n round(r_{i}N)]-N$.
   - $\forall n \in \mathbb{Z}^+$,
     $D_{r,N}=\{(r,N)|N \in \mathbb{Z}^+, N\ge n,r = [r_1,r_2,\ldots,r_n]\in \mathbb{R}_+^n,\|r\|_{1}=1\}$.
 - Question:
@@ -372,8 +372,8 @@ Take the following steps:
   - $D_{r,n,N}=\{(r,n,N)|n \in\mathbb{Z}^+,N \in\mathbb{Z}^+, n \le N,r = [r_1,r_2,\ldots,r_n]\in \mathbb{R}_+^n,\|r\|_{1}=1\}$
   - $f(r,n,N):\{\mathbb{R}^n,\mathbb{Z}^+,\mathbb{Z}^+\}\rightarrow\{\mathbb{Z}\}$
     as:
-    - $\forall (r,n,N) \in D_{r,n,N}$, there is \$ f(r,n,N)=\[*{i=1}^n
-      round(r*{i}N)\]-N\$.
+    - $\forall (r,n,N) \in D_{r,n,N}$, there is
+      $f(r,n,N)=[\sum_{i=1}^n round(r_{i}N)]-N$.
   - $\forall n \in \mathbb{Z}^+$,
     $D_{r,N}=\{(r,N)|N \in \mathbb{Z}^+, N\ge n,r = [r_1,r_2,\ldots,r_n]\in \mathbb{R}_+^n,\|r\|_{1}=1\}$.
 - Question:
@@ -575,9 +575,8 @@ Take the following steps:
   - $D_{r,n,N}=\{(r,n,N)|n \in\mathbb{Z}^+,N \in\mathbb{Z}^+, n \le N,r = [r_1,r_2,\ldots,r_n]\in \mathbb{R}_+^n,\|r\|_{1}=1\}$
   - $f(r,n,N):\{\mathbb{R}^n,\mathbb{Z}^+,\mathbb{Z}^+\}\rightarrow\{\mathbb{Z}\}$
     as:
-    - $\forall (r,n,N) \in D_{r,n,N}$, there is \$ f(r,n,N)=\[*{i=1}^n
-      round(r*{i}N)\]-N=\[*{i=1}^n truncate(r*{i}N)\]-N=\[*{i=1}^n
-      floor(r*{i}N)\]-N\$.
+    - $\forall (r,n,N) \in D_{r,n,N}$, there is
+      $f(r,n,N)=[\sum_{i=1}^n round(r_{i}N)]-N=[\sum_{i=1}^n truncate(r_{i}N)]-N=[\sum_{i=1}^n floor(r_{i}N)]-N$.
   - $\forall n \in \mathbb{Z}^+$,
     $D_{r,N}=\{(r,N)|N \in \mathbb{Z}^+, N\ge n,r = [r_1,r_2,\ldots,r_n]\in \mathbb{R}_+^n,\|r\|_{1}=1\}$.
 - Question:
@@ -667,17 +666,17 @@ Take the following steps:
 
 ## A.4
 
-- Given $round(x)$, where \$x ^+{0},~round(x)=ceil(x)=x\$. Its domain of
-  definition is restricted to $\mathbb{R}^+\cup\{0\}$ instead of
-  $\mathbb{R}$, and in this domain, it is equivalent to
+- Given $round(x)$, where
+  $\forall x \in \mathbb{R}^+\cup\{0\},~round(x)=ceil(x)=\lceil x\rceil$.
+  Its domain of definition is restricted to $\mathbb{R}^+\cup\{0\}$
+  instead of $\mathbb{R}$, and in this domain, it is equivalent to
   $\eqref{round_4}$.
 - Define:
   - $D_{r,n,N}=\{(r,n,N)|n \in\mathbb{Z}^+,N \in\mathbb{Z}^+, n \le N,r = [r_1,r_2,\ldots,r_n]\in \mathbb{R}_+^n,\|r\|_{1}=1\}$
   - $f(r,n,N):\{\mathbb{R}^n,\mathbb{Z}^+,\mathbb{Z}^+\}\rightarrow\{\mathbb{Z}\}$
     as:
-    - $\forall (r,n,N) \in D_{r,n,N}$, there is \$ f(r,n,N)=\[*{i=1}^n
-      round(r*{i}N)\]-N=\[*{i=1}^n truncate(r*{i}N)\]-N=\[*{i=1}^n
-      floor(r*{i}N)\]-N\$.
+    - $\forall (r,n,N) \in D_{r,n,N}$, there is
+      $f(r,n,N)=[\sum_{i=1}^n round(r_{i}N)]-N=[\sum_{i=1}^n truncate(r_{i}N)]-N=[\sum_{i=1}^n floor(r_{i}N)]-N$.
   - $\forall n \in \mathbb{Z}^+$,
     $D_{r,N}=\{(r,N)|N \in \mathbb{Z}^+, N\ge n,r = [r_1,r_2,\ldots,r_n]\in \mathbb{R}_+^n,\|r\|_{1}=1\}$.
 - Question:

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@@ -19,7 +19,7 @@
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