如何并发是Prolog？ [英] How Concurrent is Prolog?

问题描述

我在线上找不到任何资讯...我也是Prolog的新用户...

我Prolog可能是高度并发，也许尝试许多可能性时，尝试匹配一个规则。现代Prolog编译器/解释器本质上是否并发？哪一个？默认情况下是否启用并发？我需要以某种方式启用它吗？

It seems to me that Prolog could be highly concurrent, perhaps trying many possibilities at once when trying to match a rule. Are modern Prolog compilers/interpreters inherently* concurrent? Which ones? Is concurrency on by default? Do I need to enable it somehow?

*我对多线程感兴趣，只是固有的并发。

* I am not interested in multi-threading, just inherent concurrency.

推荐答案

在理论上看起来很有吸引力，但是有很多问题使得这样的实现看起来不明智。

In theory that seems attractive, but there are various problems that make such an implementation seem unwise.

• 即使在Prolog中编程，人们习惯于将程序视为执行从左到右和自上而下。谓词和子句中的术语的子句的顺序在标准Prolog中在语义上是有意义的。并行化它们会改变过多的代码变得流行的行为。

• for better or worse, people are used to thinking of their programs as executing left-to-right and top-down, even when programming in Prolog. Both the order of clauses for a predicate and of terms within a clause is semantically meaningful in standard Prolog. Parallelizing them would change the behaviour of far too much exising code to become popular.

非关系语言元素，例如cut操作符只能被有意义地使用除非发明了非常复杂的依赖性跟踪。

non-relational language elements such as the cut operator can only be meaningfully be used when you can rely on such execution orders, i.e. they would become unusable in a parallel interpreter unless very complicated dependency tracking were invented.

所有现有的并行解决方案都会导致

all existing parallelization solutions incur at least some performance overhead for inter-thread communication.

Prolog通常用于高级深度递归问题，如图遍历，定理证明等。在现代机器上（理想地）对于一些常数 n 可以实现 n 的加速，递归解法成为可行的方法，因为这将需要指数加速。相比之下，Fortran和C程序员通常解决的数值问题通常具有高但是相当有限的计算成本;非常值得并行化的努力将10小时的工作变成1小时的工作。相比之下，将一个可以将6个移动的程序转换为可以（平均）看起来6.5个移动的程序并不令人信服。

Prolog is typically used for high-level, deeply recursive problems such as graph traversal, theorem proving etc. Parallelization on a modern machines can (ideally) achieve a speedup of n for some constant n, but it cannot turn an unviable recursive solution method into a viable one, because that would require an exponential speedup. In contrast, the numerical problems that Fortran and C programmers usually solve typically have a high but quite finite cost of computation; it is well worth the effort of parallelization to turn a 10-hour job into a 1-hour job. In contrast, turning a program that can look about 6 moves ahead into one that can (on average) look 6.5 moves ahead just isn't as compelling.

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