def construct(self):
a = "a"
b = "b"
c = "c"
d = "d"
preferences = {a: [b, c, d], b: [c, d, a], c: [d, b, a], d: [a, b, c]}
T = PreferenceTable(preferences, center=[-3.5, 0, 0])
C = Cycle([a, b, c, a], [b, c, d, b], center=[3, 0, 0])
self.play(*T.create())
self.play(*T.propose(a, b), *T.propose(b, c), *T.propose(c, d))
self.play(*T.accept_proposal(a, b), *T.accept_proposal(b, c),
*T.accept_proposal(c, d))
self.wait()
self.play(*C.create())
self.play(*C.accept(0, 1))
self.play(*C.reject(0, 1))
self.wait()
# for anim in C.create_from_table(T):
# self.play(anim)
# self.wait(2)
# self.play(*C.cut_first_prefs(T))
# self.wait(2)
# self.play(*C.accept_second_prefs(T))
# self.wait(2)
# self.play(*C.uncreate())
self.wait(2)
def construct(self):
why_text = Tex("Why we can remove cycles")
self.play(Write(why_text))
self.play(ApplyMethod(why_text.shift, UP * 3.5))
lemma1 = r"""
\begin{align*} \text{Lemma 1: }&\text{in any stable $M$
in the reduced table,}\\
&a_i\text{ and }b_i \text{ are matched either}\\
&\text{for all }i \text{ or for no }i\end{align*}"""
lemma1 = Tex(lemma1).next_to(why_text, DOWN * 2)
self.play(Write(lemma1))
def math_list(base, start, end):
return [
"$" + base + "_" + str(i) + "$"
for i in range(start, end)
]
As = math_list("a", 1, 5) + ["$a_1$"]
Bs = math_list("b", 1, 5) + ["$b_1$"]
c = Cycle(As, Bs, center = DOWN * 1.5)
self.play(*c.create())
self.wait(1)
say_text = Tex("Let $b_3$ reject $a_3$") \
.next_to(lemma1, DOWN * 2) \
.shift(LEFT * 3)
self.play(Write(say_text))
next_text = Tex("$a_3$ proposes to $b_4$") \
.next_to(lemma1, DOWN * 2) \
.shift(RIGHT * 3)
self.wait(1)
self.play(*c.reject(2, 2))
self.wait(1)
self.play(Write(next_text))
self.play(*c.accept(2, 3))
self.play(*c.reject(3, 3))
self.play(*c.accept(3, 4))
self.play(*(c.reject(4, 4) + c.reject(0, 0)))
self.play(*c.accept(0, 1))
self.play(*c.reject(1, 1))
self.play(*c.accept(1, 2))
self.play(*(c.uncreate()))
self.wait(1)
thus_text = Tex(r"""Thus if any $a_i$ is not matched with its $b_i$\\
then no $a_i$ can match with its $b_i$""") \
.next_to(lemma1, DOWN * 6)
self.play(Write(thus_text))
self.wait(1)
self.play(*[
Uncreate(m) for m in
[thus_text, next_text, say_text]
])
shift_amt = 15
let_m = Tex(r"""
Let $M$ be a stable matching where each $a_i$ is matched\\
with its $b_i$. Let $M'$ be the same matching, but each\\
$a_i$ is matched with its second choice $b_{i+1}$
""").next_to(why_text, DOWN * 2).shift(RIGHT * shift_amt)
self.add(let_m)
self.play(*[
ApplyMethod(t.shift, LEFT * shift_amt)
for t in [lemma1, let_m]
])
lemma2 = Tex("Lemma 2: $M'$ is stable if $M$ is stable") \
.next_to(let_m, DOWN * 2)
self.play(Write(lemma2))
self.wait()
As = ["$a_k$", ""]
Bs = ["$b_k$", "$b_{k+1}$"]
c = Cycle(As, Bs, center = DOWN * 1.5)
c.reject(0, 0)
c.accept(0, 1)
mobjs = c.get_all_mobjs()
mobjs.pop(2)
self.play(*[
Create(m)
for m in mobjs
])
b_better_text = Tex(r"Each $b_i$ is better\\ off in $M'$ than $M$") \
.shift(LEFT * 4 + DOWN)
self.play(Write(b_better_text))
a_happy_text = Tex(r"$a_k$ can only prefer \\"+\
r"$b_k$ to its current match\\" +\
r"but $b_k$ is happier with\\" +\
r"$a_{k-1}$, so $M'$ is\\" +\
r"stable here") \
.shift(RIGHT * 4 + DOWN * 2)
self.play(Write(a_happy_text))
self.play(*[
Uncreate(m)
for m in mobjs + [b_better_text, a_happy_text]
])
continue_m = Tex(r"Thus $M$ stable $\Rightarrow$ $M'$ stable,\\" +\
r"so if there exists a stable matching, we\\" +\
r"can find it by proceeding with $M'$ and\\" +\
r"eliminating our cycle") \
.shift(DOWN * 1.5)
self.play(Write(continue_m))
self.wait(2)
self.play(*[
ApplyMethod(t.shift, RIGHT * shift_amt)
for t in [lemma1, let_m]
] + [
Uncreate(m)
for m in [continue_m]
] + [
ApplyMethod(t.shift, DOWN * 0.8)
for t in [lemma2]
])
plus = TextMobject("+").shift(UP * 0.5)
impl = TextMobject("$\\Leftarrow$") \
.rotate_in_place(PI/2).shift(DOWN)
self.play(Create(impl), Create(plus))
self.wait(1)
final = Tex(r"We can always eliminate cycles without\\"\
r"changing the result").next_to(impl, DOWN)
self.play(Write(final))
self.wait(2)
