Experimental prune of Qwen3 Next 80B A3B Instruct based on HuiHui's abliterated version, from 512 experts to 352.
Note that I consider this the smallest safe variant of my prunes, as it is slightly less confident in reasoning despite being correct.
My previous 256E prune was created with less nuanced pruning strategies and was too aggressive, this should be considered its replacement.
The model was tested on the following questions:
- Generate a 100 word response about neutron stars.
- If you were a farmer in 1852 and a traveler asked you about the local attractions of your town, what would you say?
- Use the Baire Category Theorem together with the completeness of C(I,R) to show that there is a dense Gδ,B⊂C(I,R) such that every function f∈B is not differentiable at any point.
- Explain the controversy behind the book "To Kill A Mockingbird."
Responses:
-
Neutron stars are cosmic remnants of massive stars after supernova explosions, crushed under gravity to pack atomic nuclei into a city-sized sphere. Their density exceeds one billion tons per teaspoon—soaking light and emitting intense magnetic fields. Rotation rates can exceed 700 times per second, creating lighthouse-like beams of radiation, observable as pulsars. Surface temperatures reach millions of degrees, and their gravitational pull is billions times stronger than Earth’s. Some neutron stars host starquakes, releasing colossal energy in milliseconds. They may merge with companions, triggering kilonovae and forging heavy elements like gold and platinum. Their cores likely contain exotic matter: superfluid neutrons, degenerate protons, and possibly quark-gluon plasma. As the universe ages, they slowly cool, becoming faint cosmic fossils—silent guardians of stellar evolution.
-
Ah, stranger — welcome to Willow Hollow! If you’re seeking sights, I’d point you east toward the old stone mill—built by great-grandfather, still grinding wheat with the river’s power. Beyond that, the Whispering Woods holds ancient oaks where fireflies dance in June nights. Come harvest season, we’ve cider pressing and fiddle music at the barn; folk from three counties come to trade apples and wool. The church steeple? That’s new—since ‘49, but it chimes every hour with bells cast from melted cannon shells after the Rebellion. And if you’re patient till dawn, climb the Hill of Saint Mary—you’ll see the sunrise over our fields, gold like honey spilled across a quilted earth. Welcome, and rest your weary feet!
-
We wish to use the Baire Category Theorem and the completeness of
C
(
I
,
R
)
C(I,R) — the space of real-valued continuous functions on a compact interval
I
=
[
a
,
b
]
I=[a,b], equipped with the sup norm
∥
f
∥
∞
=
sup
x
∈
I
∣
f
(
x
)
∣
∥f∥
∞
=sup
x∈I
∣f(x)∣ — to prove that there exists a dense
G
δ
G
δ
subset
B
⊂
C
(
I
,
R
)
B⊂C(I,R) consisting functions nowhere differentiable.
Step 1: Completeness of
C
(
I
,
R
)
C(I,R)
The space
C
(
I
,
R
)
C(I,R) is a Banach space under the sup norm. This is standard: uniform limits of continuous functions are continuous, and completeness follows from Cauchy sequences converging uniformly.
Step 2: Define sets for non-differentiability
For each
n
∈
N
n∈N, define:
A
n
{
f
∈
C
(
I
,
R
)
:
∀
x
∈
I
,
∃
h
≠
0
such that
∣
f
(
x
+
h
)
−
f
(
x
)
h
∣
n
for all
∣
h
∣
<
1
n
}
A
n
={f∈C(I,R):∀x∈I,∃h
=0 such that
h
f(x+h)−f(x)
n for all ∣h∣<
n
1
}
Actually — this is too too strong. Better: define the set where
f
f has a "slope bounded by
n
n" at some point:
Let
E
m
,
n
⊂
C
(
I
,
R
)
E
m,n
⊂C(I,R) be the set of functions
f
f such that there exists
x
0
∈
I
x
0
∈I with
∣
f
(
x
)
−
f
(
x
0
)
x
−
x
0
∣
≤
m
for all
x
∈
(
x
0
−
1
n
,
x
0
+
1
n
)
∩
I
,
x
≠
x
0
.
x−x
0
f(x)−f(x
0
)
≤mfor all x∈(x
0
−
n
1
,x
0
+
n
1
)∩I,x
=x
0
.
Then
f
f is differentiable at
x
0
x
0
if for some
m
m and all large
n
n,
f
∈
E
m
,
n
f∈E
m,n
. So the set of functions differentiable at least at one point is:
D
⋃
m
1
∞
⋂
n
N
m
∞
E
m
,
n
D=
m=1
⋃
∞
n=N
m
⋂
∞
E
m,n
But to apply Baire, we consider for fixed
m
,
n
m,n the complement — or better yet, define:
Let
D
m
,
n
D
m,n
be the set of continuous functions
f
f such that there exists
x
∈
I
x∈I with
∀
y
,
z
∈
(
x
−
1
n
,
x
+
1
n
)
∩
I
,
y
≠
z
,
∣
f
(
y
)
−
f
(
z
)
y
−
z
∣
≤
m
.
∀y,z∈(x−
n
1
,x+
n
1
)∩I,y
=z,
y−z
f(y)−f(z)
≤m.
Then
D
m
,
n
D
m,n
consists functions that have slope bounded by
m
m on some interval of length
2
/
n
2/n. The set of all functions differentiable at at least one point is:
D
⋃
m
1
∞
⋂
k
1
∞
D
m
,
k
D=
m=1
⋃
∞
k=1
⋂
∞
D
m,k
Actually — even more standard: define the complement! Let
A
n
c
A
n
c
be the set of functions that fail to being differentiable at any point with "local slope ≤ n". But a cleaner approach is:
Let for each
m
∈
N
m∈N, let
U
m
U
m
be the set of continuous functions on
I
I such that:
∀
x
∈
I
,
∃
y
1
,
y
2
∈
I
with
0
<
∣
y
i
−
x
∣
<
1
m
,
∣
f
(
y
1
)
−
f
(
x
)
y
1
−
x
−
f
(
y
2
)
−
f
(
x
)
y
2
−
x
∣
m
.
∀x∈I,∃y
1
,y
2
∈I with 0<∣y
i
−x∣<
m
1
,
y
1
−x
f(y
1
)−f(x)
−
y
2
−x
f(y
2
)−f(x)
m.
This forces "oscillation of slopes" and prevents differentiability. But the standard construction is:
Standard Proof via Baire Category Theorem:
Define for each
n
∈
N
n∈N, let
G
n
{
f
∈
C
(
I
,
R
)
:
∀
x
∈
I
,
∃
h
with
0
<
∣
h
∣
<
1
n
,
∣
f
(
x
+
h
)
−
f
(
x
)
h
∣
n
}
G
n
={f∈C(I,R):∀x∈I,∃h with 0<∣h∣<
n
1
,
h
f(x+h)−f(x)
n}
But this is not open. Better: use the following classical result:
Theorem (Banach, Mazur): The set
B
B of functions in
C
(
[
0
,
1
]
,
R
)
C([0,1],R) which are nowhere differentiable is a dense
G
δ
G
δ
subset.
Sketch proof via Baire:
Let for each
m
,
n
∈
N
m,n∈N, define
E
m
,
n
{
f
∈
C
(
I
,
R
)
:
∃
x
∈
I
s.t
∀
y
∈
(
x
−
1
n
,
x
+
1
n
)
∩
I
,
∣
f
(
y
)
−
f
(
x
)
y
−
x
∣
≤
m
}
E
m,n
={f∈C(I,R):∃x∈I s.t ∀y∈(x−
n
1
,x+
n
1
)∩I,
y−x
f(y)−f(x)
≤m}
Then
E
m
,
n
E
m,n
is closed (due to uniform continuity and sup-norm convergence), and the set of functions that are differentiable at least at one point lies in:
D
⋃
m
1
∞
⋃
n
1
∞
E
m
,
n
D=
m=1
⋃
∞
n=1
⋃
∞
E
m,n
Actually — we need a larger union over
m
m, then intersection over
n
n. But better yet:
Define for each
k
∈
N
k∈N,
F
k
{
f
∈
C
(
I
,
R
)
:
∃
x
0
∈
I
such
D
+
f
(
x
0
)
≤
k
,
D
−
f
(
x
0
)
≥
−
k
}
F
k
={f∈C(I,R):∃x
0
∈I such D
+
f(x
0
)≤k,D
−
f(x
0
)≥−k}
but we use:
Let
A
m
,
n
A
m,n
be the set of functions that have a "slope bounded by m on an interval of length 2/n around some point**.
Then each
A
m
,
n
A
m,n
is closed, and for fixed
m
m,
⋃
n
A
m
,
n
⋃
n
A
m,n
contains all
f
f differentiable at some point with derivative bounded in absolute value ≤ m. So:
D
⋃
m
1
∞
⋃
n
1
∞
A
m
,
n
D=
m=1
⋃
∞
n=1
⋃
∞
A
m,n
But actually — for Baire, we want a countable intersection of open dense sets.
So consider the complements: let
U
n
:
C
(
I
,
R
)
∖
(
⋃
m
1
∞
A
m
,
n
)
‾
U
n
:=C(I,R)∖
(⋃
m=1
∞
A
m,n
)
. But cleaner:
Instead use this classical construction (as from Banach): for each
n
≥
1
n≥1, define
U
n
{
f
∈
C
(
I
,
R
)
:
∀
x
∈
I
,
∃
y
∈
I
∖
{
x
}
,
∣
y
−
x
∣
<
1
n
,
∣
f
(
y
)
−
f
(
x
)
y
−
x
∣
n
}
U
n
={f∈C(I,R):∀x∈I,∃y∈I∖{x},∣y−x∣<
n
1
,
y−x
f(y)−f(x)
n}
Then
U
n
U
n
is open and dense in
C
(
I
,
R
)
C(I,R). Why?
Dense: For any
f
∈
C
(
I
,
R
)
f∈C(I,R) and
ε
0
ε>0, we can add a small sawtooth function of slope > n to
f
f (supported on interval < 1/n) — this stays within ε in sup norm but violates the slope bound everywhere. So
U
n
U
n
is dense.
Open: If
f
∈
U
n
f∈U
n
, then for every x, there y with |y−x|<1/n and |Δf/h| > n. By continuity of f (uniformly continuous on compact I), small perturbations preserve this inequality — so neighborhood in sup norm lies in
U
n
U
n
. So
U
n
U
n
is open.
Now define:
B
⋂
n
1
∞
U
n
B=
n=1
⋂
∞
U
n
Then
B
B is a
G
δ
G
δ
set. By the Baire Category Theorem (since
C
(
I
,
R
)
C(I,R) is complete),
B
B is dense.
Now, if
f
∈
B
f∈B, then for every n, and every x ∈ I, there y close to x with |Δf/h| > n. This implies that the difference quotient oscillates without bound as h → 0 at every point — so f is nowhere differentiable.
Thus,
✅ There exists a dense
G
δ
G
δ
subset
B
⊂
C
(
I
,
R
)
B⊂C(I,R) such that every
f
∈
B
f∈B is nowhere differentiable.
Conclusion:
We have used:
Completeness of
C
(
I
,
R
)
C(I,R) to invoke the Baire Category Theorem.
Constructed an explicit countable intersection
B
⋂
n
U
n
B=⋂
n
U
n
, where each
U
n
U
n
is open dense, consisting functions with "large slopes" everywhere.
Proved that any
f
∈
B
f∈B must fail differentiability at every point.
This concludes the proof.
■
■
- Controversies Behind To Kill a Mockingbird
Harper Lee’s novel To Kill a Mockingbird (1960), though celebrated for its moral values, has sparked several controversial debates over decades:
Race to Justice vs. Hypocrisy of the System
The town Maycomb “justice” is performative: Tom Robinson is an innocent victim of racism and systemic apathy.
Judge Taylor and Atticus Finch uphold fairness, but why is a Black man shot dead for a minor trespass? Why does the system privilege white people (Ewell’s wealth) over truth?
Controversy: Is justice truly upheld in a flawed society — or just performative?
Racial and Class Bias
Calpurnia’s dignity vs. Mrs. Merriweather’s condescension.
The Ewells (wealthy white folk) are “superior” despite their ignorance — Bob Ewell’s courtroom humiliation reveals systemic entitlement to privilege.
Controversy: Is the book romanticizing the very flawed system that kills mockingbirds?
Educational Idealism vs. Realistic Brutality
Scout's rebellion against standardized norms (wearing pants, questioning authority) is inspiring — but does it ignore systemic corruption? Why do only children see truth?
Controversy: Is the book a naïve child lens or an adult critique of broken adults?
Gender Roles & “The Mockingbird” Symbol
The mockingbird symbolizes innocence, but who "kills" them? Is it Bob Ewell’s unjust death, Mayella's forced marriage to Boo Radley (a passive hero), or the community’s silence?
Controversy: Does the title imply killing is an collective failure — not heroism?
Cultural Appropriateness for Young Readers
Critics debated: Is a 9-year-old Scout too maturely perceptive? Do children truly grasp moral hypocrisy and courtroom injustice?
But Lee’s child lens may be the point: adult failures are visible to innocence.
Conclusion:
To Kill a Mockingbird isn’t merely a child’s story — it critiques systemic moral failure in civil society. Its controversy lies:
Can a idealistic narrative, within a idealized community, truly change broken institutions? Or is the “mockingbird” already dead — and we’re just singing to its corpse?
That tension makes To Kill a Mockingbird enduringly profound — and controversial. 🌿
