Answer: —
Theorem 1 — class: algorithmic
PDF text-layer excerpt — equations and symbol order may be approximateThere is an nO(log ℓ)-time algorithm for computing a NE.
statement extracted from the paper’s text layer; consult the paper for the exact wording
Given candidates \(P\) partitioned among \(\ell=2^h\) parties, a fixed seeding, and a supported set \(T\) of complete voter rankings with rational masses \(\mu_t\) summing to \(1\), define each pairwise match by strict majority, restricting to societies without pairwise ties or fixing a tie-breaking rule. Compute a pure Nash equilibrium of the one-shot winner-take-all knockout game, or report that none exists.
Let \(P\) be a set of \(m\) candidates partitioned into nonempty parties \(C_1,\ldots,C_\ell\), where \(\ell=2^h\) and \(h\geq1\). A fixed, publicly known seeding bijectively assigns parties to the leaves of a perfect binary knockout tree.
A society is given by its support: a list \(T\) of \(\tau\) distinct strict complete rankings of \(P\), each with positive binary rational mass \(\mu_t\), with \(\sum_{t\in T}\mu_t=1\). Here \(\tau\) is the support size, not the number of possible rankings; \(L\) is the total input bit length.
For distinct candidates in different parties, \(a\) beats \(b\) exactly when \(\sum_{t:a\succ_t b}\mu_t>1/2\). Promise that every such comparison is tie-free.
Each party simultaneously chooses one nominee \(p_i\in C_i\) before play and retains that nominee throughout. Match winners advance up the tree; party \(i\) has utility \(1\) if its nominee wins the tournament and \(0\) otherwise. Only parties make strategic choices; the society is fixed.
Compute a pure Nash equilibrium profile \((p_1,\ldots,p_\ell)\), or report that none exists. Unilateral deviations may replace a party's nominee by any other member of that party.
Is this equilibrium computation problem polynomial-time solvable in \(m,\tau,L,\ell\), or is it NP-hard? State running times in \(m,\tau,L\) and \(h\) (equivalently, \(\ell\)).
For explicitly supplied deterministic pairwise outcomes, equilibrium computation takes \(m^{O(\log\ell)}\) time, and equilibrium recognition is polynomial-time solvable. [Verified results](https://ifmas.csc.liv.ac.uk/Proceedings/aamas2022/pdfs/p807.pdf)
No NP-hardness or W[1]-hardness result for this discrete problem, or exact complexity classification for the supported rational-mass formulation, was located in the literature search.
I do not resolve polynomial-time solvability versus NP-hardness for unrestricted $h$.
The new result proved below is a parameterized algorithm. Extend the cross-party comparison table to a tournament on all candidates by orienting within-party pairs arbitrarily. If deleting $k$ candidates makes this tournament transitive, equilibrium computation is fixed-parameter tractable in $k$. In particular, it is polynomial-time solvable when $k=O(\log m)$, even for unrestricted $h$.
The proof also simplifies the general dynamic program: its states need not explicitly identify the party that will win the entire tournament.
Write $a\triangleright b$ when $a$ beats $b$. Initially, this relation is defined only between different parties.
Suppose the rational masses are supplied as $\mu_t=a_t/b_t$. Set
$$
Q=\prod_{t\in T}b_t,
\qquad
A_t=a_t(Q/b_t).
$$
Normalization gives $\sum_t A_t=Q$. Hence
$$
a\triangleright b
\quad\Longleftrightarrow\quad
2\sum_{t:\,a\succ_t b}A_t>Q.
$$
The integers involved have $O(L)$ bits: the bit length of $Q$ is bounded by the sum of the denominator bit lengths, and $A_t=\mu_tQ\leq Q$.
After constructing candidate positions in each ranking, all cross-party comparisons can therefore be computed exactly in the conservative bound
$$
B=O(m^2\tau L^2)
$$
bit operations, using schoolbook arithmetic. The tie-free promise ensures that every required pair has exactly one winner.
Conversely, every deterministic cross-party table admits a polynomial-size supported representation. Here is a proof, because this equivalence explains why the supported formulation alone does not settle the classification.
Extend the table to all pairs arbitrarily. For each intended orientation $a\triangleright b$, list the remaining candidates as $r_1,\ldots,r_{m-2}$ and introduce the rankings
$$
a\succ b\succ r_1\succ\cdots\succ r_{m-2},
$$
$$
r_{m-2}\succ\cdots\succ r_1\succ a\succ b.
$$
These two rankings give margin $2$ to $a$ against $b$. They give margin zero to every other pair: their orders disagree on pairs of remaining candidates and on pairs involving one remaining candidate and either $a$ or $b$.
Applying this construction to all $\binom m2$ pairs produces
$$
N_0=m(m-1)
$$
ranking occurrences, with margin $2$ in every intended direction. Let $N$ be the least power of two with $N\geq N_0$. Both numbers are even. Pad with $(N-N_0)/2$ ranking-and-reverse pairs, each of which has zero margins.
Giving every occurrence mass $1/N$ yields winning support
$$
\frac{N+2}{2N}=\frac12+\frac1N.
$$
Merge identical rankings by adding their masses. The resulting support consists of distinct rankings with positive dyadic masses. Since $N<2m(m-1)$, it has fewer than $2m(m-1)$ rankings and encoding length $O(m^3\log m)$.
The construction preserves parties, seeding, and all cross-party outcomes. Consequently, it preserves every profile’s winner and every unilateral deviation’s winner, and hence preserves equilibrium profiles exactly.
Thus the supported and explicit-table formulations are polynomially equivalent.
For a fixed nomination profile, consider party $i$’s leaf-to-root path. At each of its $h$ rounds, let $o_{i,r}$ be the winner of the sibling subtree.
These sibling subtrees exclude party $i$, so their winners do not change when only party $i$ changes its nominee. Therefore a replacement $q\in C_i$ wins the tournament exactly when
$$
q\triangleright o_{i,r}
\qquad\text{for every }r\in\{1,\ldots,h\}.
$$
Indeed, conditional on surviving the preceding rounds, $q$ faces precisely $o_{i,r}$ at round $r$.
If $g$ is the winning party, its utility is already $1$. Every other party has a profitable deviation precisely when some replacement beats every opponent on its path. Hence the profile is an equilibrium exactly when
$$
\forall i\ne g\;\forall q\in C_i\;\exists r:
\quad o_{i,r}\triangleright q.
$$
Compute all subtree winners once, then check this condition. Recognition takes
$$
O(\ell+mh)
$$
table operations after preprocessing. Equilibrium existence therefore belongs to $\mathrm{NP}$: a profile is a polynomial-length certificate.
The algorithm below uses the following observation. For a collection $S$ of outside opponents, define its blocked set
$$
D(S)=\bigcup_{s\in S}N^+(s),
\qquad
N^+(s)=\{q\in P:s\triangleright q\}.
$$
Here and subsequently, orient within-party pairs arbitrarily, producing a tournament on $P$; set $q\notin N^+(q)$.
Whenever $S$ consists of actual outside opponents of a subtree, every candidate in that subtree belongs to a different party from every member of $S$. Thus the added within-party orientations do not affect the interpretation:
$$
q\in D(S)
\quad\Longleftrightarrow\quad
q\text{ loses to at least one opponent in }S.
$$
Only the blocked set matters, rather than the identities or order of all outside opponents separately.
For a tree node $v$, let $P_v$ be the candidates belonging to parties below $v$. For $w\in P_v$, let $i(w)$ denote its party.
For any set $D\subseteq P$, define two predicates.
These are mathematical definitions for arbitrary $D$. They do not require $D$ to arise from actual outside opponents.
At the root $r$, equilibrium existence is equivalent to
$$
\exists w\in P:\ F_1(r,w,\varnothing).
$$
To justify this, $q\notin\varnothing$ imposes no restriction. Thus $F_1$ forbids every winning deviation by every party other than the actual winning party. The latter already has utility $1$.
At the leaf of party $i$, every $q\in C_i$ wins the one-leaf subtree when nominated. Therefore, for every $w\in C_i$,
$$
F_0(v,w,D)\quad\Longleftrightarrow\quad C_i\subseteq D,
$$
whereas
$$
F_1(v,w,D)\quad\Longleftrightarrow\quad \text{true}.
$$
Let $u,z$ be the children of $v$. For prospective child champions $a\in P_u$ and $b\in P_z$, put
$$
D_a=D\cup N^+(a),
\qquad
D_b=D\cup N^+(b).
$$
If $a\triangleright b$, this pair witnesses $F_0(v,a,D)$ exactly when
$$
F_0(u,a,D_b)\ \land\ F_0(z,b,D_a),
$$
and it witnesses $F_1(v,a,D)$ exactly when
$$
F_1(u,a,D_b)\ \land\ F_0(z,b,D_a).
$$
If $b\triangleright a$, use the symmetric conditions:
$$
F_0(u,a,D_b)\ \land\ F_0(z,b,D_a)
$$
for $F_0(v,b,D)$, and
$$
F_0(u,a,D_b)\ \land\ F_1(z,b,D_a)
$$
for $F_1(v,b,D)$.
Each parent entry is true if at least one child-champion pair witnesses it.
Proof. Fix a party below $u$ and a replacement candidate $q$. The other child’s champion remains $b$. Thus $q$ wins subtree $v$ and lies outside $D$ exactly when it wins subtree $u$ and satisfies
$$
q\notin D
\quad\text{and}\quad
q\triangleright b.
$$
The comparison is between different parties. Because it is tie-free, these conditions are equivalent to
$$
q\notin D\cup N^+(b)=D_b.
$$
The analogous equivalence holds for parties below $z$.
For $F_0$, every party must be checked, giving two $F_0$ child states. For $F_1$ when $a$ wins the parent match, only party $i(a)$ is exempt. It belongs to $u$, giving an $F_1$ state in $u$ and an $F_0$ state in $z$.
Restricting a parent witness proves necessity. Conversely, combining witnessing child profiles proves sufficiency: their party sets are disjoint, their champions are $a,b$, and the equivalences above establish every required deviation condition. The symmetric case follows identically. $\square$
Storing a witnessing pair for each true internal entry permits reconstruction of a complete profile. If every eligible root entry is false, the recurrence proves nonexistence.
The recurrence recovers the unrestricted quasipolynomial algorithm without a global winning-party index.
At a node of depth $d$, represent its context by a set
$$
S\subseteq P\setminus P_v,
\qquad |S|\leq d,
$$
and interpret the state through $D(S)$. Moving to a child adds the sibling champion to $S$. The resulting context remains outside that child and has size at most $d+1$.
There are at most
$$
\sum_{j=0}^d\binom mj\leq(m+1)^d\leq(m+1)^h
$$
contexts per node.
Two counting identities control the work:
$$
\sum_v |P_v|=m(h+1),
$$
because each candidate occurs in one node per depth, and
$$
\sum_{\substack{v\text{ internal}\\u,z\text{ children of }v}}
|P_u|\,|P_z|
\leq \binom m2.
$$
For the second identity, a pair of candidates in different leaves is counted exactly at the lowest common ancestor of their leaves.
For each context and child-champion pair, the recurrence updates both exemption states and the uniquely determined parent winner. Context copying and leaf membership checks cost $O(h+1)$. Including initialization, this gives the conservative bound
$$
O\!\left((h+1)m^2(m+1)^h\right)
$$
table operations.
With exact-majority preprocessing, the overall bit-time bound is
$$
O(m^2\tau L^2)+m^{O(h)}\operatorname{poly}(\log m).
$$
Since nonempty parties imply $\ell\leq m$ and $h=\log_2\ell$, this is polynomial for fixed $h$ and quasipolynomial in general. The corresponding unrestricted upper bound is also Theorem 1 of [Lisowski, Ramanujan, and Turrini](https://ifmas.csc.liv.ac.uk/Proceedings/aamas2022/pdfs/p807.pdf).
Theorem. Fix any extension of the cross-party outcomes to a tournament on $P$. Suppose $X\subseteq P$, with $|X|=k$, is supplied such that the tournament on
$$
G=P\setminus X
$$
is transitive. Then equilibrium computation, including a correct nonexistence report, takes
$$
O\!\left(
4^k(m+1)\bigl[m^2(k+1)+m(h+1)\bigr]
\right)
$$
table operations after preprocessing.
Consequently, its bit complexity is bounded by
$$
O(m^2\tau L^2)
+
4^{O(k)}\operatorname{poly}(m,h,\log m).
$$
Transitivity means that $G$ has a strict total order $\prec$ satisfying
$$
a\triangleright b
\quad\Longleftrightarrow\quad
a\prec b
\qquad(a,b\in G,\ a\ne b).
$$
Earlier candidates are stronger.
For any opponent set $S\subseteq P$, record:
For $q\in G$, transitivity gives
$$
q\in D(S)
\quad\Longleftrightarrow\quad
\bigl(s\ne\bot\text{ and }s\prec q\bigr)
\ \lor\
\exists x\in U:\ x\triangleright q.
$$
Indeed, among the good opponents $S\cap G$, their earliest member defeats every candidate defeated by any later member. For $q\in X$, membership in $D(S)$ is recorded directly by $V$.
Thus the triple $(s,U,V)$ determines the entire blocked set.
For all triples, including ones not realized by an opponent set, define
$$
\mathcal D(s,U,V)
=
V
\ \cup\
\{q\in G:s\ne\bot,\ s\prec q\}
\ \cup\
\{q\in G:\exists x\in U,\ x\triangleright q\}.
$$
There are at most
$$
R=(|G|+1)4^k\leq(m+1)4^k
$$
such signatures. Distinct signatures may describe the same set; this causes no difficulty.
Given a signature $(s,U,V)$ and candidate $b$, define the updated signature as follows.
$$ V\cup\bigl(N^+(b)\cap X\bigr). $$
This update satisfies
$$
\mathcal D(\text{updated signature})
=
\mathcal D(s,U,V)\cup N^+(b).
$$
For candidates in $X$, this follows from the update to $V$. For candidates in $G$, adding a good candidate adds a suffix of the total order, while adding an exceptional candidate adds exactly its defeated good candidates.
Importantly, the equality holds for every signature, not merely realizable ones. Hence the signature family is closed under every context update required by the recurrence.
The empty blocked set is represented by
$$
(\bot,\varnothing,\varnothing).
$$
Run the two-state dynamic program using signatures instead of explicit blocked sets. Their exact set semantics and closure identity make every recurrence from Section 3 applicable. Induction on subtree height therefore proves that every computed entry has precisely its stated meaning.
Membership tests and signature updates take $O(k+1)$ elementary operations using $k$-bit vectors and the order on $G$. Updates can also be precomputed for every signature-candidate pair.
There are $R$ signatures. The counting identities from Section 4 give:
Substituting $R\leq(m+1)4^k$ gives the theorem’s stated table-operation bound. State indices and signatures have polynomial bit length in $k$, $h$, and $\log m$, yielding the displayed fixed-parameter bit bound. Backtracking takes polynomial additional time. $\square$
The set $X$ need not be supplied. For the fixed tournament extension, one can find a suitable set of size at most $k$, or prove that none exists, in
$$
O(3^k m^3)
$$
table operations.
First, a tournament is transitive exactly when it has no directed triangle.
To prove the nontrivial direction, suppose a tournament has a directed cycle and choose one of shortest length. If its length exceeds three, write its first three vertices as $v_1,v_2,v_3$. If $v_3\triangleright v_1$, they form a directed triangle. Otherwise $v_1\triangleright v_3$, and bypassing $v_2$ produces a shorter directed cycle. Thus every cyclic tournament has a directed triangle. An acyclic tournament has a topological order, which is a transitive order because every pair is adjacent.
The deletion algorithm finds a directed triangle. Any deletion set producing a transitive tournament must delete at least one of its three vertices. Branch on these three choices and decrease the remaining budget by one.
Each node searches for a triangle in $O(m^3)$ time. The search tree has $O(3^k)$ nodes. A triangle-free residual tournament supplies the required transitive order. Trying budgets successively up to the minimum successful value retains the same asymptotic bound.
Combining this search with Section 5 gives a fixed-parameter algorithm where $k$ is the minimum deletion-set size for the chosen extension.
In particular, if $k\leq c\log_2m$ for a fixed constant $c$, then
$$
4^k\leq m^{2c},
$$
so the resulting algorithm is polynomial in $m,\tau,L$, even when $h$ varies.
The unrestricted hypotheses impose no bound on this deletion parameter. Therefore the new algorithm does not establish unrestricted polynomial-time solvability.
The following conclusions are proved:
Neither an unrestricted polynomial algorithm nor an NP-hardness reduction is established here. A polynomial many-one NP-hardness proof for existence, combined with the general quasipolynomial algorithm, would imply deterministic quasipolynomial algorithms for every language in $\mathrm{NP}$; that consequence is not a contradiction and does not rule out hardness.
⟨structured verdict parsed — rendered above⟩
Overall assessment: partial, with correct substantive results. The output explicitly declines to settle the unrestricted polynomial-time versus NP-hardness question. Its representation equivalence, recognition criterion, dynamic program, and parameterized algorithm are sound as written. I found no false central step or mathematical gap in those partial results.
The blocking issue is scope: the parameterized tractability result does not classify the unrestricted problem requested in USER_TASK.
The common-denominator construction is valid:
$$
Q=\prod_{t\in T}b_t,
\qquad
A_t=a_t(Q/b_t).
$$
Each quotient is integral, and normalization gives
$$
\sum_{t\in T}A_t=Q.
$$
Consequently, the stated strict-majority test
$$
2\sum_{t:a\succ_t b}A_t>Q
$$
is exactly equivalent to the original comparison.
The bit-length argument is correct. The sum of denominator bit lengths is bounded by the input length, so $Q$ has $O(L)$ bits. Positivity and normalization imply $\mu_t\leq1$, hence $A_t\leq Q$. All comparison sums also have $O(L)$ bits.
The conservative preprocessing bound
$$
O(m^2\tau L^2)
$$
covers common-denominator arithmetic, ranking-position construction, and pairwise comparisons using schoolbook arithmetic. The tie-free promise supplies a unique orientation for every cross-party pair.
The two-ranking construction has the claimed margins:
Thus every gadget affects only its designated pair. Applying one gadget per unordered pair produces margin $2$ in each intended direction.
The padding argument is also correct. Because $h\geq1$ and parties are nonempty, $m\geq2$. Therefore $N_0=m(m-1)$ and its power-of-two ceiling $N$ are both even. Ranking-and-reverse padding preserves all margins, and the winning mass is exactly
$$
\frac{N+2}{2N}=\frac12+\frac1N.
$$
Merging identical rankings preserves these masses and produces distinct support entries with positive dyadic masses. The support-size and encoding-length bounds follow from
$$
N<2m(m-1).
$$
The edge case $m=2$ is valid: the two gadget rankings coincide, and merging them yields a single ranking of mass $1$.
Preserving every cross-party outcome preserves the winner under every profile and every unilateral deviation. Accordingly, the claimed polynomial equivalence applies to equilibrium computation, including nonexistence reports.
Assessment: correct.
A party’s sibling subtrees contain no member of that party. Their winners therefore remain unchanged under its unilateral nomination change.
A replacement $q$ wins the tournament exactly when it beats all $h$ sibling-subtree champions on its path. This criterion applies even when the original nominee was eliminated early: the sibling-subtree winners remain defined independently of that elimination.
The winning party cannot improve its utility beyond $1$. For every losing party, a replacement winning the tournament is precisely a profitable deviation. Thus the displayed recognition condition is necessary and sufficient.
The bound
$$
O(\ell+mh)
$$
table operations is valid: subtree winners take $O(\ell)$ operations, and checking each candidate against its party’s path takes at most $h$ comparisons.
The polynomial certificate claim is correct for equilibrium existence. Recognition alone does not supply a certificate for nonexistence; the output does not claim otherwise.
Assessment: correct.
The predicates are defined for arbitrary $D\subseteq P$, rather than only contexts realized by outside opponents. This stronger formulation supports the later use of unrealizable signatures.
These auxiliary predicates forbid specified winning nominations. They need not themselves describe profitable deviations within the subtree. At the root, however, exempting the actual winning party makes $F_1$ exactly the equilibrium condition.
There is no assumption that an equilibrium always exists: the root existential test can fail.
At a one-party leaf, every member of that party wins when nominated. Consequently:
Both base cases are correct, including singleton parties.
For a party below $u$, its unilateral change leaves the other child’s champion $b$ fixed. A candidate $q$ wins the parent subtree outside $D$ exactly when it wins $u$, lies outside $D$, and beats $b$.
Because $q$ and $b$ belong to different parties, the tie-free promise gives
$$
q\triangleright b
\quad\Longleftrightarrow\quad
q\notin N^+(b).
$$
Hence the updated blocked set $D_b$ is correct. The analogous statement holds for the other child.
The exemption is passed correctly. When $a$ wins the parent match, the exempt party is $i(a)$, which lies below $u$. Therefore the winning child receives an $F_1$ condition and the losing child an $F_0$ condition.
The necessity and sufficiency arguments are valid:
Witness storage supports reconstruction, and exhaustive failure of the eligible root states justifies a nonexistence report.
Assessment: correct.
At depth $d$, a context containing at most $d$ outside candidates is sufficient. Passing to a child adds its sibling champion, which is outside that child. The context family is therefore closed under the required updates.
The context bound
$$
\sum_{j=0}^d\binom mj\leq(m+1)^d
$$
is valid, including $d=0$.
Both work-counting statements are correct:
$$
\sum_v|P_v|=m(h+1),
$$
and
$$
\sum_{\substack{v\text{ internal}\\u,z\text{ children of }v}}
|P_u|\,|P_z|\leq\binom m2.
$$
For the second bound, only candidates belonging to different leaves are counted, and each such pair is counted at its unique lowest common ancestor.
These bounds support the conservative table-operation estimate
$$
O\!\left((h+1)m^2(m+1)^h\right).
$$
Since $\ell\leq m$ and $h=\log_2\ell$, this is polynomial for fixed $h$ and quasipolynomial in general. The corresponding cited result is indeed Theorem 1 of the referenced paper; its setting uses fixed seeding, one-shot nominations, and win-or-lose utilities. [Lisowski, Ramanujan, and Turrini](https://ifmas.csc.liv.ac.uk/Proceedings/aamas2022/pdfs/p807.pdf)
Assessment: correct. This upper bound does not settle the requested complexity dichotomy.
Within the transitive residual tournament, the out-neighborhood of an earlier candidate contains the out-neighborhood of every later candidate. Accordingly, the earliest member of $S\cap G$ determines all blocking contributed by good opponents.
The exceptional opponents $U$ determine their contribution on $G$, while $V$ records blocked exceptional candidates directly. Thus $(s,U,V)$ determines the entire blocked set.
The signature count
$$
R=(|G|+1)4^k
$$
is correct. The two exceptional subsets may be counted independently because the algorithm intentionally includes unrealizable signatures.
The update satisfies the claimed identity for every signature:
This remains valid when the signature is unrealizable. That point is essential: it permits the recurrence to operate on the whole signature family without introducing spurious set semantics.
The empty signature represents the empty blocked set, including when $G$ is empty.
Because signature updates implement exact union with $N^+(b)$, the previously verified recurrence applies unchanged. Induction on subtree height therefore establishes the stated semantics.
Membership testing by examining at most $k$ exceptional candidates and updating $k$-bit subsets fit the conservative $O(k+1)$ operation bound. The initialization, leaf-testing, recurrence, and update-precomputation counts support
$$
O\!\left(
4^k(m+1)\bigl[m^2(k+1)+m(h+1)\bigr]
\right).
$$
The stated bit-complexity form follows with polynomial overhead for indexing and signature manipulation.
Assessment: correct, conditional on a deletion set for the chosen extension. No proof of priority for the description “new result” is supplied; the mathematical result itself is verified.
The directed-triangle characterization is proved correctly. A shortest directed cycle longer than three either contains the displayed triangle or admits the stated shortcut. An acyclic tournament’s topological order orients every pair consistently.
Every deletion set making the tournament transitive must intersect each surviving directed triangle. Branching on its three vertices is therefore exhaustive.
The search-tree and triangle-search bounds yield
$$
O(3^k m^3).
$$
Successively trying budgets up to the first successful one retains this bound by a geometric sum. The argument includes $k=0$ through the triangle-free stopping condition.
Combining this search with the signature algorithm establishes fixed-parameter tractability in the minimum deletion-set size for the fixed, chosen extension. The output does not establish optimization over within-party extensions.
For a fixed constant $c$, the calculation
$$
k\leq c\log_2m
\quad\Longrightarrow\quad
4^k\leq m^{2c}
$$
is correct. Thus the asserted restricted polynomial-time result follows.
Assessment: correct.
Major scope issue — the unrestricted classification remains unresolved. Nothing in the task bounds the deletion parameter by $O(\log m)$, and the general algorithm remains quasipolynomial. The output supplies neither an unrestricted polynomial-time algorithm nor an NP-hardness proof.
This is an acknowledged limitation, not a contradiction within the partial proof. No clarification of the existing parameterized theorem suffices to complete USER_TASK; an additional classification result is required.
The final observation about the consequence of a polynomial many-one NP-hardness proof is correct: composing such a reduction with the established quasipolynomial decision algorithm would yield deterministic quasipolynomial algorithms for every language in $\mathrm{NP}$.
The direction is promising because the supplied compression and recurrence are valid. They nevertheless do not resolve the assigned unrestricted question.
⟨structured verdict parsed — rendered above⟩
Tier: machine-proved, adversarially reviewed by an independent model. No human has checked this proof. Generated from pipeline artifacts by tools/atlas_build.py.