Bir poset üzerinden monotonik bir yüklem öğrenmek için en kötü soru sayısı

n öğe $(X, \leq)$üzerinde sonlu bir postayı düşünün ve P , X üzerinde bilinmeyen bir monotonik yüklem (yani herhangi bir x , y ∈ X , eğer P ( x ) ve x ≤ y sonra P ( y ) ise ). P'yi bir x x X düğümü sağlayarak ve P ( x ) 'in tutup tutmadığını bularak değerlendirebilirim . Amacım x düğümü setini tam olarak belirlemek$n$$P$$X$$x$$y \in X$$P(x)$$x \leq y$$P(y)$$P$$x \in X$$P(x)$$x \in X$ , öyle ki$P(x)$ birkaç değerlendirmeler olarak kullanılarak, tutan$P$ mümkün. (Sorgularımı önceki tüm sorguların cevabına bağlı olarak seçebilirim, tüm sorguları önceden planlamam gerekli değildir.)

Bir strateji $S$ üzerinde $(X, \leq)$ herhangi yüklem üzerinde şimdiye kadar koştum sorgular ve bunların cevapları, düğüm sorguya ve teminat altın bir fonksiyonu olarak, bana söyler bir fonksiyondur $P$ , strateji izleyerek, Tüm düğümlerde değerini bildiğim bir duruma ulaşacağım $P$. İşletim süresi $r(S, P)$ ve $S$ mesnet üzerinde $P$ sorgu sayısı değerini bilmek için gerekli olan $P$ tüm düğümlerde. En kötü çalışma süresi $S$ olduğu $wr(S) = \max_P r(S, P)$ . Optimal bir strateji$S'$ ,$wr(S') = \min_S wr(S)$ .

Sorum şu: giriş girişi olarak verildiğinde $(X, \leq)$, en uygun stratejilerin en kötü çalışma süresini nasıl belirleyebilirim?

[Boş bir poşet için açıktır $n$ sorgular ihtiyaç olacaktır (biz tek tek her düğüm hakkında sormak gerekir) ve bir toplam sipariş etrafında söz konusu $\lceil \log_2 n \rceil$ sorguları ihtiyaç olacaktır (sınır bulmak için ikili arama yapıyor ). Daha genel bir sonuç aşağıdaki bilgiler-teorik alt sınırı: yüklem için olası seçim sayısı $P$ sayısı $N_X$ ait antichains arasında $(X, \leq)$ monoton önermeler arasında bire-bir eşleme vardır, çünkü ( antichains maksimal olarak elementler şeklinde yorumlanabilir $P$Her sorgu bize bir bit bilgiyi verir çünkü), bu yüzden, biz en azından gerekir önceki iki vaka subsuming, sorgular. Bu sıkı mı, yoksa yapısı asimptotik olarak antikasit sayısından daha fazla sorgu gerektirebilecek şekildedir mi?]$\lceil \log_2 N_X \rceil$

Bu tam bir cevap değil, ama yorum yapmak için çok uzun.

I bağlı olan bir örnek bulduk sıkı değildir.$\lceil \log_2 N_X \rceil$

Aşağıdaki postayı düşünün. Zemin dizi ve bir ı daha küçük olan B j tüm i , j ∈ { 1 , 2 } . Diğer çiftler kıyaslanamaz. (Hasse diyagramı 4- devirdir).$X=\{a_1, a_2, b_1, b_2\}$$a_i$$b_j$$i,j\in\{1,2\}$$4$

Monoton özellikleri, postanın rahatsızlıklarıyla tanımlayayım. Bu postanın yedi üzümü vardır: , { b 1 } , { b 2 } , { b 1 , b 2 } , { a 1 , b 1 , b 2 } , { a 2 , b 1 , b 2 } , { a 1 , bir 2 , b 1$\emptyset$$\{b_1\}$$\{b_2\}$$\{b_1,b_2\}$$\{a_1,b_1,b_2\}$$\{a_2,b_1,b_2\}$$\{a_1,a_2,b_1,b_2\}$, and this poset has seven antichains since the antichains are in one-to-one correspondence with the upsets. So, $\lceil \log_2 N_X \rceil=\lceil \log_2 7 \rceil = 3$ for this poset.

Now, by adversary argument I'll show that any strategy needs at least four queries (so needs to query all elements). Let's fix an arbitrary strategy.

Strateji ilk sorgularsa , daha sonra düşman cevaplar " P ( a 1 ) tutmaz." Sonra beş olasılıkımız kalıyor : ∅ , { b 1 } , { b 2 } , { b 1 , b 2 } , { a 2 , b 1 , b 2 } . Böylece, olduğu belirlemek için biz, en azından mi ⌈ log 2 5 ⌉ = $a_1$$P(a_1)$$\emptyset$$\{b_1\}$$\{b_2\}$$\{b_1,b_2\}$$\{a_2,b_1,b_2\}$$\lceil \log_2 5\rceil = 3$daha fazla sorgu. Toplamda dört sorguya ihtiyacımız var. İlk sorgu ise aynı argüman geçerli .$a_2$

If the strategy first queries $b_1$, then the adversary answers "$P(b_1)$ holds." Then, we are left with five possibilities: $\{b_1\}$, $\{b_1,b_2\}$, $\{a_1,b_1,b_2\}$, $\{a_2,b_1,b_2\}$, $\{a_1,a_2,b_1,b_2\}$. Therefore, we need at least three more queries as before. In total, we need four queries. The same argument applies when the first query is $b_2$.

If we take $k$ parallel copies of this poset, then it has $7^k$ antichains, and thus the proposed bound is $\lceil \log_2 7^k \rceil = 3k$. But, since each of the copies needs four queries, we need at least $4k$ queries.

Probably, there is a larger poset with larger gap. But this argument can only improve the coefficient.

Here, the problem looks to be a situation where no query partitions the search space evenly. In such a case, the adversary can force the larger half to remain.

In their paper Every Poset Has a Central Element, Linial and Saks show (Theorem 1) that the number of queries required to solve the ideal identification problem in a poset $X$ is at most $K_0 \log_2 i(X)$, where $K_0 = 1/(2 - \log_2(1 + \log_2 5))$ and $i(X)$ is the number of ideals of $X$. What they call an "ideal" is actually a lower set and there is an obvious one to one correspondance between monotonic predicates and the lower set of the points at which they don't hold, besides their "identification problem" is to identify by querying nodes just like in my setting, so I think they are dealing with the problem I'm interested in and that $i(X) = N_X$.

$N_X$$\log_2 N_X$$K_0 \log_2 N_X$.

Linial and Saks quote a personal communication by Shearer to say that there are known orders for which we can prove a lower bound of $K_1 \log_2 N_X$ for some $K_1$ which is just slightly less than $K_0$ (this is in the spirit of Yoshio Okamoto's answer who tried this approach for a smaller value of $K_1$).

This does not fully answer my question of computing the number of questions required from $X$, however, since computing $N_X$ from $X$ is #P-complete, I have a feeling that there is little hope. (Comments about this point are welcome.) Still, this result by Linial and Saks is enlightening.

For the Boolean n-cube $(\{0, 1\}^n, \leq)$ (or, equivalently, for the poset $(2^S, \subseteq)$ of all subsets of an n-element set), the answer is given by Korobkov and Hansel's theorems (from 1963 and 1966, respectively). Hansel's theorem [1] states that an unknown monotone Boolean function (i.e., an unknown monotone predicate on this poset) can be learned by a deterministic algorithm making at most $\phi(n) = \binom{n}{\lfloor n/2 \rfloor} + \binom{n}{\lfloor n/2 \rfloor + 1}$ queries (that is, asking $\phi(n)$ questions in the worst case). This algorithm matches the lower bound of Korobkov's theorem [2], which says that $\phi(n) - 1$ queries do not suffice. (So Hansel's algorithm is optimal in the worst-case setting.) An algorithm in both statements is understood as a deterministic decision tree.

The logarithm of the number of antichains in $(\{0, 1\}^n, \le)$ is asymptotically equal to $\binom{n}{\lfloor n/2 \rfloor} \sim 2^n / \sqrt{\pi n / 2}$, so there is a constant-factor gap between $\log N_X$ and the optimal algorithm performance $\phi(n) \sim 2 \binom{n}{\lfloor n/2 \rfloor}$ for this poset.

Unfortunately, I have not been able to find a good treatment of Hansel's algorithm in English available on the web. It is based on a lemma that partitions the n-cube into $\phi(n)$ chains with special properties. Some description can be found in [3]. For the lower bound, I don't know any reference to a description in English.

Since I am familiar with these results, I can post a description on arXiv, if the treatment in Kovalerchuk's paper does not suffice.

If am not much mistaken, there have been attempts to generalize Hansel's approach, at least to the poset $(E_k^n, \le)$, where $(E_k, \le)$ is a chain $0 < 1 < \ldots < k - 1$, although I cannot give any reference straight away. For the Boolean case, people have also investigated notions of complexity other than worst-case for this problem.

[1] G. Hansel, Sur le nombre des fonctions booléennes monotones de n variables. C. R. Acad. Sci. Paris, 262(20), 1088-1090 (1966)

[2] V. K. Korobkov. Estimation of the number of monotonic functions of the algebra of logic and of the complexity of the algorithm for finding the resolvent set for an arbitrary monotonic function of the algebra of logic. Soviet Math. Doklady 4, 753-756 (1963) (translation from Russian)

[3] B. Kovalerchuk, E. Triantaphyllou, A. S. Deshpande, E. Vityaev. Interactive learning of monotone Boolean functions. Information Sciences 94(1), 87-118 (1996) (link)

[NOTE: The following argument doesn't seem to work, but I'm leaving it here so others don't make the same mistake / in case someone can fix it. The issue is that an exponential lower bound on learning/identifying a monotone function, as below, does not necessarily contradict an incrementally polynomial algorithm for the problem. And it is the latter which is equivalent to checking the mutual duality of two monotone functions in poly time.]

I believe your conjecture on $\log N_X$ is false in general. If it is indeed the case that $\log N_X$ queries are needed, that implies quite a strong lower bound on learning monotone functions using membership queries. In particular, let the poset $X$ be the Boolean cube with the usual ordering (if you like, $X$ is the powerset of $\{1,...,n\}$ with $\subseteq$ as its partial order). The number $M$ of maximal antichains in $X$ satisfies $\log M = (1 + o(1))\binom{n-1}{\lfloor n/2 \rfloor}$ [1]. If your idea on $\log N_X$ is correct, then there is some monotone predicate on $X$ that requires essentially $\binom{n-1}{n/2} \approx 2^n$ queries. In particular, this implies a lower bound of essentially $2^n$ for the complexity of any algorithm solving this problem.

However, if I've understood correctly [which I now know I hadn't], your problem is equivalent to checking the mutual duality of two monotone functions, which can be done in quasi-polynomial time (see the intro of this paper by Bioch and Ibaraki, which cites Fredman and Khachiyan), contradicting anything close to a $2^n$ lower bound.

[1] Liviu Ilinca and Jeff Kahn. Counting maximal antichains and independent sets. arXiv:1202.4427