Problem mit einfacher Entscheidung, Problem mit harter Suche

36

Es ist einfach zu entscheiden, ob ein Nash-Gleichgewicht besteht (es ist immer so); Es wird jedoch angenommen, dass es schwierig ist, eine zu finden (PPAD-Complete).

Was sind andere Beispiele für Probleme, bei denen die Entscheidungsversion einfach, die Suchversion jedoch relativ schwierig ist (im Vergleich zur Entscheidungsversion)?

Ich würde mich besonders für Probleme interessieren, bei denen die Entscheidungsversion nicht trival ist (im Gegensatz zum Nash-Gleichgewicht).

cc.complexity-theory big-list

— Travis Service
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Sollte wahrscheinlich Community-Wiki sein: meta.cstheory.stackexchange.com/questions/225/…

— Dave Clarke

2

@supercooldave: Ich würde mich in diesem Fall nicht mit CW beeilen. Es kann sich herausstellen, dass es nur sehr wenige natürliche Probleme mit einer nicht trivialen, aber einfachen Entscheidungsversion und einer Version mit harter Suche gibt. Dies ist nicht unbedingt eine "große Liste".

— Jukka Suomela

1

Ich habe mich für die Heuristik that big list = community wiki entschieden.

— Dave Clarke

5

Dies wirft also die Frage auf, was das natürliche Entscheidungsproblem ist, das mit einem Suchproblem verbunden ist. Ich denke, die Existenz von NE ist nicht das natürliche Entscheidungsproblem, das mit NE verbunden ist.

— Kaveh

1

@Kaveh: Sie können dieses Entscheidungsproblem für Nash definieren (wenn Sie eine Kodierung für eine Lösung für Nash angeben), aber das Problem ist, ob es die gleiche Komplexität wie Nash hat oder nicht, oder formal, ob dieses Entscheidungsproblem auf Nash reduzierbar ist . Ich bezweifle es, weil es oft NP-schwierig ist, ein Nash-Gleichgewicht zu finden, das eine zusätzliche Einschränkung erfüllt.

— Tsuyoshi Ito

37

Hat eine gegebene ganze Zahl einen nicht-trivialen Faktor? -> Nicht trivial in P.

Finden Sie bei einer gegebenen Ganzzahl einen nicht-trivialen Faktor, falls es einen gibt -> Nicht bekannt, dass er in FP enthalten ist.

— Slimton
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Oder könnte man fragen, hat es einen Primfaktor? Dann brauchen Sie nicht die PRIMES ist in P Papier

— Bjørn Kjos-Hanssen

28

Hier ist ein weiteres Beispiel: Wenn ein kubischer Graph G und ein Hamilton-Zyklus H in G gegeben sind, finden Sie in G einen anderen Hamilton-Zyklus. Ein solcher Zyklus existiert (nach dem Satz von Smith), aber meines Wissens ist offen, ob es möglich ist berechnet in polynomialer Zeit.

— Marek Chrobak
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20

Wenn Sie dem Folgenden den gleichen "Spielraum" geben, den Sie für Nash-Gleichgewichte tun, dann:

Ganzzahlfaktorisierung, wobei das Entscheidungsproblem lautet: "Gibt es eine faktorisierte Darstellung dieser Ganzzahl?" (trivial ja), und das Suchproblem besteht darin, es auszugeben

Es ist denkbar, dass hier eine Reihe von Gitterproblemen mit der gleichen großzügigen Zulage zur Definition des Entscheidungsproblems zusammenpassen:

Shortest Vector Problem (SVP) - Entscheide, ob es einen kürzesten Vektor gibt oder nicht
Closest Vector Problem (CVP) - Entscheiden Sie, ob es einen nächstgelegenen Vektor gibt oder nicht

Dies sind natürlich alle Fälle, in denen die von mir erwähnte Entscheidungsversion nicht sehr interessant ist (weil es trivial ist). Ein nicht ganz so einfaches Problem :

Planarer Graph -Färbbarkeit für $k$ $k \ge 4$

Das Entscheidungsproblem der planaren Graph-4-Färbbarkeit liegt in P. Die lexikografisch erste Lösung dieser Art ist jedoch NP-hart ( Khuller / Vazirani ).

Beachten Sie, dass die Eigenschaft, an der Sie wirklich interessiert sind, die Selbstreduzierbarkeit (oder besser gesagt die Nicht-Selbstreduzierbarkeit) ist. Bei dem Problem der planaren Graphfärbung besteht das wesentliche Problem darin, dass die Methode der Selbstreduzierung des allgemeinen Falls der Färbbarkeit die Planarität in einem Graphen zerstört. $k$

— Daniel Apon
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18

Lassen , der Zufall Graph auf , wobei jeder Rand mit einer Wahrscheinlichkeit unabhängig vorhanden ist . Wählen Eckpunkten gleichmäßig zufällig und fügen alle Kanten zwischen ihnen; rufen Sie das resultierende Graph . Dann hat eine Clique der Größe . $G=G(n,1/2)$ $1,\ldots,n$ $1/2$ $n^{1/3}$ $G$ $H$ $H$ $n^{1/3}$

Suchproblem: Finden Sie eine Clique mit einer Größe von mindestens . $10\log n$

— user1624
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Very neat! Is there a relevant paper about this?

— András Salamon

1

@András: To give a bit more background, this is called the "hidden clique problem". If the hidden clique that's planted is on Omega(sqrt(n log n)) vertices, one can easily see that the vertices of the clique are those with the highest degree, almost surely. [Alon-Krivelevic-Sudakov] (tau.ac.il/~nogaa/PDFS/clique3.pdf) improve this to Omega(sqrt(n)) using spectral techniques. For hidden cliques of smaller size, such as O(log n), nothing non-trivial is known.

— arnab

Another related intriguing problem, posed by Karp, is to find a clique of size (1+c)log(n) in G(n,1/2), for any constant 0<c<1. It's known that there exists a clique of size 2log(n) almost surely in G(n,1/2). The only polynomial time algorithms known (such as the greedy one) find cliques of size (1+o(1))log(n).

— arnab

@arnab: Feige and Ron recently simplified the AKS result (see reference at my question cstheory.stackexchange.com/questions/1406/… ). My question to @Louigi was really about the

10 \log n

$10\log n$ question: what motivates the particular constant, and has this question been asked in a paper one can cite?

— András Salamon

15

One more example; The Subset-sums equality: Given $a_1,a_2,a_3,...,,a_n$ natural numbers with $\sum_1^n{a_i} \lt 2^n -1$ . The pigeon-hole principle guarantees the existence of two subsets $I, J$ in ${1,2,..., n}$ such that $\sum_{i\in I} a_i=\sum_{j \in J} a_j$ (since the are more subsets than possible sums). The existence of polynomial time algorithm for finding sets $I$ and $J$ is a famous open problem.

Subset-sums equality (pigeonhole version)

— Mohammad Al-Turkistany
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13

Another number theory example, similar to the ones above. It's known by Bertrand's postulate that for every positive integer $n$ , there's a prime between $n$ and $2n$ . But we have no polynomial time algorithm currently to find such a prime, given $n$ . (The desired algorithm must run in polylog( $n$ ) time.) One can easily come up with polynomial time randomized algorithms because of the prime number theorem, and one can derandomize them by assuming some standard number theoretic conjectures (such as Cramer's conjecture), but no unconditionally polynomial time deterministic algorithm is known. Related work was recently done in the Polymath4 project; Tao's blog post on the project is a good summary of it.

— arnab
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1

Even without Bertrand's postulate, you have a deterministic algorithm with expected polynomial runtime due to the Prime Number Theorem and the AKS primality test.

— Joe Fitzsimons

@JoeFitzsimons, I am not sure what you mean by "deterministic algorithm with expected polynomial runtime".

— Chandra Chekuri

@ChandraChekuri, "deterministic" probably is meant to say that it always gets the correct answer.

— usul

@ChandraChekuri: Sorry, my choice of wording was poor. I meant that you can find a prime number with absolute certainty in expected polynomial time, rather than simply with bounded error. At least, I think that's what I meant. It was 3 years ago.

— Joe Fitzsimons

11

At the risk of being slightly off-topic, let me give a simple and natural example of a theory C answer: Eulerian cycles and distributed algorithms.

The decision problem is not completely trivial, in the sense that there are are both Eulerian and non-Eulerian graphs.

There is, however, a fast and simple distributed algorithm that solves the decision problem (in the sense that for yes-instances all nodes output "1" and for no-instances at least one node outputs "0"): each node just checks the parity of its own degree and outputs 0 or 1 accordingly.

But if you would like to find a Eulerian cycle (in the sense that each node outputs the structure of the cycle in its own neighbourhood), then we need essentially global information on the graph. It shouldn't be hard to come up with a pair of examples that shows that the problem requires $\Omega(n)$ communication rounds; on the other hand, $O(n)$ rounds is enough to solve any problem (assuming unique IDs).

In summary: $O(1)$ -time decision problem, $\Theta(n)$ -time search problem, and this is the worst possible gap.

Edit: This implicitly assumes that the graph is connected (or, equivalently, that we want to find an Eulerian cycle in each connected component).

— Jukka Suomela
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This might be a stupid question (because I know almost nothing about distributed computing), but is there a promise that the graph is connected, or is the connectedness easy to check efficiently in a distributed way?

— Tsuyoshi Ito

Thanks, not a stupid question at all. I clarified my answer, I had forgotten to add the assumption that we deal with connected graphs here. (Usually there is little point in studying disconnected graphs from the perspective of distributed algorithms, as by definition there is no way to transmit information between the connected components, but of course this should be made explicit.)

— Jukka Suomela

Thanks! After reading your answer, I think that it should have been obvious that the graph (= network topology) was assumed to be connected. :)

— Tsuyoshi Ito

10

Finding Tverberg partitions is of unknown complexity:

Theorem: Let $x_1,x_2,\dots, x_m$ be points in $R^d$ , $m \ge (r-1)(d+1)+1$ . Then there is a partition $S_1,S_2,\dots, S_r$ of ${1,2,\dots,m}$ such that $\cap _{j=1}^r \text{conv} (x_i: i \in S_j) \ne \emptyset$ .

Like with Nash equilibria, the partition is guaranteed by the theorem, but it's not known if a polytime algorithm exists to find one.

Gil Kalai wrote a wonderful series of posts on this topic: One, Two and Three.

— Suresh Venkat
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2

Actually, any problem that falls into TFNP would be a good candidate I think. When an answer is guaranteed to exist by a theorem -- then, define some apparently-harder-than-P search problem over the possible solutions to accompany it.

— Daniel Apon

7

In all the examples above the decision problem is in P and the search problem is not known to be in P but not known to be NP-hard either. I want to point out that it is possible to have an NP-hard search problem whose decision version is easy.

Consider the generalized satisfiability problem for given relations $R_1,\ldots,R_k$ over Boolean domain $\{0,1\}$ . An instance is an expression of the form

R_{i_{1}} (t_{11}, \dots, t_{1 r_{1}}) \land \dots \land R_{i_{m}} (t_{m 1}, \dots, t_{m r_{m}})

$R_{i_1}(t_{11},\ldots,t_{1r_1}) \wedge \cdots \wedge R_{i_m}(t_{m1},\ldots,t_{mr_m})$ where the

t_{i j}

$t_{ij}$ 's are either variables or constants in

0, 1

${0,1}$ , and

r_{1}, \dots, r_{m}

$r_1,\ldots,r_m$ are the arities of

R_{1}, \dots, R_{k}

$R_1,\ldots,R_k$ (this is the same framework as in Schaeffer's dichotomy theorem with constants, in case you know what it is). The search problem is: given such an expression, find a lexicographically minimal solution, if there is one.

It was shown by Reith and Vollmer here that there exists a choice of relations $R_1,\ldots,R_k$ that make this problem NP-hard (actually OptP-complete) but keep the satisfiability problem easy (quite trivial actually). An example given in the paper is $R = \{(1,0,0),(0,1,0),(1,1,1)\}$ (here $k = 1$ ). Once the satisfiability problem is solvable in polynomial-time, the question whether there exists a lexicographically minimal satisfying assignment is trivial.

See Corollary 13 and the example following it in the paper above (at least in this on-line version).

— slimton
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6

Decision version is (highly) non-trival in P: $k$ -colourability ( $k$ fixed) on graphs without induced path with five vertices; due to this paper.
Search version is NP-hard: Finding the chromatic number of graphs without induced path with five vertices; due to this paper.

— Peng O
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You perhaps meant to say that for fixed $k$ , the decision version is in P.

— András Salamon

4

Take a "pairing-friendly" elliptic curve. That is, a curve that has a one bilinear map $e$ associated with it - with $e (a + b, c + d) = e (a c) e (a d) e (b c) e (b d)$ such that $e$ is difficult to invert).

Such pairings are used widely in cryptography, partially since given $e$ , it is trivial to solve Decisional Diffie-Hellman (given $(g, h, g^a, h^b)$ , decide if $a = b$ : just verify whether $e (g, h^b) = e (h, g^a)$ ). However, it is still conjectured that the search/computational Diffie-Hellman problem is difficult.

Such groups are also generalized to "gap groups".

— cryptocat
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2

I guess Planar Perfect Matching got missed out from this list.

The decision version is in NC (even the counting version is in $\mathsf{NC}$ ) by a parallel version (see Mahajan-Subramanya-Vinay) of Kastelyn's algorithm
The search version remains unparallelised to date i.e there is no known deterministic $\mathsf{NC}$ algorithm for this problem (though if we drop either of the parallel or deterministic restrictions there are well known algorithms - Edmonds and Mulmuley-Vazirani-Vazirani/Karp-Upfal-Wigderson, respectively.

— SamiD
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2

Let's notch up the complexity a bit.

Many decision problems about vector addition systems (VAS) are EXPSPACE-complete, but may require much larger witnesses. For instance, deciding whether the language of a VAS is regular is EXPSPACE-complete (e.g. Blockelet & Schmitz, 2011), but the smallest equivalent finite-state automaton might be of Ackermannian size (Valk & Vidal-Naquet, 1981). The explanation behind this huge gap is that there exist much smaller witnesses of non-regularity.

— Sylvain
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