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ai: finish hw7
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@tiankaima
tiankaima committed May 12, 2024
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266 changes: 257 additions & 9 deletions ea2724-ai_hw/hw7.typ
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Expand Up @@ -22,8 +22,13 @@ Due 2024.05.12

#table(
columns: (auto, auto, auto),
fill: (x, y) => if x == 0 or y == 0 {
blue.lighten(80%)
fill: (x, y) => {
if (x == 0 or y == 0) and not (x == 0 and y == 0) {
blue.lighten(80%)
}
if (x == 0 and y == 0) {
blue.lighten(60%)
}
},
stroke: blue,
align: center,
Expand Down Expand Up @@ -102,8 +107,13 @@ b. 如果你知道雅典的出租车 $10$ 辆中有 $9$ 辆是绿色的呢？

#table(
columns: (auto, auto, auto),
fill: (x, y) => if x == 0 or y == 0 {
blue.lighten(80%)
fill: (x, y) => {
if (x == 0 or y == 0) and not (x == 0 and y == 0) {
blue.lighten(80%)
}
if (x == 0 and y == 0) {
blue.lighten(60%)
}
},
stroke: blue,
align: center,
Expand Down Expand Up @@ -186,18 +196,26 @@ c. 题目中的条件独立性假设合理吗?请讨论。
[
#raw-render(```dot
digraph {
layout=neato;
rankdir=TD;
node [shape=circle];
F_1, F_2, M_1, M_2, N;
F_1 [pos="0,0!"];
F_2 [pos="1.5,0!"];
M_1 [pos="0,2!"];
M_2 [pos="1.5,2!"];
N [pos="0.5,1"];
M_1 -> F_1; M_1 -> M_2; M_1 -> N;
N -> F_1; N -> F_2; M_2 -> N; M_2 -> F_2;
}
```)
],[
],
[
(i)
],[
],
[
(ii)
],[
],
[
(iii)
],
)
Expand All @@ -206,17 +224,247 @@ c. 题目中的条件独立性假设合理吗?请讨论。
a. 这三种网络结构哪些是对上述信息的正确(但不一定高效)表示？

#ans[
(ii), (iii) 正确;

考虑到 $ P(N mid(|)M_1)!=P(N|M_1,F_1) $ 所以 $F_1$ 与 $M$ 应该同时连接到 $N$, (i) 不正确.
]

b.哪一种网络结构是最好的？请解释。

#ans[
(ii), 关系少, 更紧致.
]

c.当 $N in {1,2,3}, quad M_1 in {0,1,2,3,4}$时，请写出 $P(M_1 mid(|) N)$ 的条件概率表。概率分布表里的每个条目都应该表达为参数$e$和/或$f$的一个函数。

#ans[

$
P(M_1 mid(|) N) &= P(M_1 mid(|) N, F_1) P(F_1 mid(|) N) + P(M_1 mid(N), not F_1) P(not F_1 mid(|) N)\
&= P(M_1 mid(|) N, F_1) P(F_1) + P(M_1 mid(N), not F_1) P(not F_1)\
$

#table(
columns: (auto, auto, auto, auto, auto, auto),
align: center,
fill: (x, y) => {
if (x == 0 or y == 0) and not (x == 0 and y == 0) {
blue.lighten(80%)
}
if (x == 0 and y == 0) {
blue.lighten(90%)
}
},
stroke: blue,
[
$P(M_1 mid(|) N)$
],
[
$M_1 = 0$
],
[
$M_1 = 1$
],
[
$M_1 = 2$
],
[
$M_1 = 3$
],
[
$M_1 = 4$
],
[
$N = 1$
],
[
$f+e(1-f)$
],
[
$(1-2e)(1-f)$
],
[
$e(1-f)$
],
[
#set text(fill: red)
$0$
],
[
$0$
],
[
$N = 2$
],
[
$f$
],
[
#set text(fill: green)
$e(1-f)$
],
[
$(1-2e)(1-f)$
],
[
#set text(fill: green)
$e(1-f)$
],
[
$0$
],
[
$N = 3$
],
[
$f$
],
[
#set text(fill: red)
$0$
],
[
$e(1-f)$
],
[
$(1-2e)(1-f)$
],
[
$e(1-f)$
],
)
]

d.假设 $M_1 = 1, quad M_2 = 3$。如果我们假设 $N$ 取值上没有先验概率约束,可能的恒星数目是多少？

#ans[

#table(
columns: (auto, auto, auto, auto, auto, auto, auto),
align: center,
fill: (x, y) => {
if (x == 0 or y == 0) and not (x == 0 and y == 0) {
blue.lighten(80%)
}
if (x == 0 and y == 0) {
blue.lighten(90%)
}
},
stroke: blue,
[
$P(M_1 mid(|) N)$
],
[
$M_1 = 0$
],
[
$M_1 = 1$
],
[
$M_1 = 2$
],
[
$M_1 = 3$
],
[
$M_1 = 4$
],
[
$dots.c$
],
[
$N = 4$
],
[
$f$
],
[
#set text(fill: green)
$f$
],
[
$0$
],
[
#set text(fill: green)
$e(1-f)$
],
[
$(1-2e)(1-f)$
],
[
$dots.c$
],
[
$N=5$
],
[
$f$
],
[
$f$
],
[
$f$
],
[
#set text(fill: red)
$0$
],
[
$e(1-f)$
],
[
$dots.c$
],
)
#let color_r(x) = text(fill: red, $#x$)
$
P(M = 3 mid(|) N = 0) = 0 &quad& => &quad& N!=0\
P(M = 3 mid(|) N = 1) = 0 &quad& => &quad& N!=1\
P(M = 1 mid(|) N = 3) = 0 &quad& => &quad& N!=3\
P(M = 1 mid(|) N = 5) = 0 &quad& => &quad& N!=5\
$

$P(M = i mid(|) N = n) > 0 quad forall i = 1,3; n = 2,4$ 已经在表中标记出来了. (考虑到 $f approx 0$ 时=, 我们近似地认为每行加起来为 $1$. )

考虑 $n >= 6$ 时, $P(M = 1 mid(|) N = n) = P(M = 3 mid(|) N = n) = f > 0$.

因此可能的 $n$ 的取值为 $2,4$ 或 $n>=6$
]

e.在这些观测结果下，最可能的恒星数目是多少？解释如何计算这个数目，或者，如果不可能计算，请解释还需要什么附加信息以及它将如何影响结果。

#ans[
缺少 $N$ 的先验概率分布, 无法计算最可能的恒星数目.

考虑我们提供一个分布: $P(N=n) = p_n quad forall n = 2,4,6,7, dots$

$
&P(N=2, M_1 = 1, M_2 = 3) &=& p_2 e^2(1-f)^2\
&P(N=4, M_1 = 1, M_2 = 3) &=& p_4 e f(1-f)^2\
(n>=6) quad &P(N=n, M_1 = 1, M_2 = 3) &=& p_n f^2\
$

计算出并比较大小即可, 取 $n="argmax"_n P(N=n, M_1 = 1, M_2 = 3)$.
]

=== Question 14.13

考虑 图14.22(ii) 的网络,假设两个望远镜完全相同。$N in {1,2,3}$，$M_1, M_2 in {0,1,2,3,4}$，CPT表和习题14.12所描述的一样。使用枚举算法(图14.9)计算概率分布。
考虑 图14.22(ii) 的网络,假设两个望远镜完全相同。$N in {1,2,3}$，$M_1, M_2 in {0,1,2,3,4}$，CPT表和习题14.12所描述的一样。使用枚举算法(图14.9)计算概率分布 $P(N mid(|) M_1=1,M_2 = 2)$。

#ans[
$
cal(P)(N mid(|) M_1=2, M_2=2) &= alpha sum_(f_1,f_2) cal(P)(f_1, f_2, N, M_1=2,M_2=2)\
&=alpha sum_(f_1,f_2) P(f_1) P(f_2) cal(P)(N) P(M_1=2,M_2=2)
$

考虑到 $M_1=M_2=2$, 只有 $F_1=F_2="false"$ 时才能满足, 因此:

$
cal(P)(N mid(|) M_1=2, M_2=2) &= alpha (1-f)^2 angle.l p_1, p_2, p_3 angle.r angle.l e, (1-2e), e angle.r angle.l e, (
1-2e
), e angle.r\
&= alpha^' angle.l p_1 e^2, p_2(1-2e)^2, p_3 e^2 angle.r
$
]

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