Question

我正在尝试为L =字符串集写一个无上下文语法，其中1的数字比0的数字多一个。因此，1011010,0001111等在L中，但001101,000011等字符串不在L中。

到目前为止，我有一个无上下文语法，L =一组字符串，其中1的数字大于0：

S→TS | 1T | 1S

T→TT | 0T1 | 1T0 |ε

如何更改此项以使1的数量仅比0的数量多1？

Answer 1

根据templatetypedef的答案，您可以在https://math.stackexchange.com/questions/2207708/context-free-grammar-for-language-a-b-where-the-number-of-as-the-number上找到这样的语法

所以最终的语法基本上是：

group = "com.github.imflog"
version = "0.10.0-SNAPSHOT"

plugins {
    kotlin("jvm").version("1.3.71")
    id("java-gradle-plugin")
    id("com.gradle.plugin-publish") version "0.11.0"
    id("maven-publish")
    id("com.github.ben-manes.versions") version "0.28.0"
}

repositories {
    jcenter()
    mavenCentral()
    maven("http://packages.confluent.io/maven/")
}

java {
    withSourcesJar()
}

// Dependencies versions
val confluentVersion = "5.4.1"
val avroVersion = "1.8.2"
dependencies {
    implementation(gradleApi())
    implementation(kotlin("stdlib"))
    implementation("io.confluent", "kafka-schema-registry", confluentVersion) {
        exclude("org.slf4j", "slf4j-log4j12")
    }
}

// Test versions
val junitVersion = "5.6.1"
val mockkVersion = "1.9.3"
val wiremockVersion = "2.26.3"
val assertJVersion = "3.15.0"
dependencies {
    testImplementation(gradleTestKit())
    testImplementation("org.junit.jupiter", "junit-jupiter-api", junitVersion)
    testImplementation("org.junit.jupiter", "junit-jupiter-engine", junitVersion)
    testImplementation("org.assertj", "assertj-core", assertJVersion)
    testImplementation("io.mockk", "mockk", mockkVersion)
    testImplementation("com.github.tomakehurst", "wiremock-jre8", wiremockVersion)
}

tasks.withType<Test> {
    useJUnitPlatform()
}

publishing {
    repositories {
        maven {
            name = "GitHubPackages"
            url = uri("https://maven.pkg.github.com/ImFlog/schema-registry-plugin")
            credentials {
                username = System.getenv("GITHUB_USERNAME")
                password = System.getenv("GITHUB_TOKEN")
            }
        }
    }
}

val registryPluginName = "com.github.imflog.kafka-schema-registry-gradle-plugin"
gradlePlugin {
    plugins {
        create("schema-registry") {
            id = registryPluginName
            description = "A plugin to download, register and test schemas from a Kafka Schema Registry"
            displayName = "Kafka schema registry gradle plugin"
            version = version

            implementationClass = "com.github.imflog.schema.registry.SchemaRegistryPlugin"
        }
    }
}

pluginBundle {
    website = "https://github.com/ImFlog/schema-registry-plugin"
    vcsUrl = "https://github.com/ImFlog/schema-registry-plugin.git"
    tags = listOf("schema", "registry", "schema-registry", "kafka")
}

Answer 2

作为提示，如果你的字符串只有一个1而不是0，那么有一些方法可以将它写成w1x，其中w和x具有完全相同的0和1。如果您可以找到一个CFG，它生成具有相同数字的0和1且带有起始符号X的字符串，那么您可以添加一个起始符号S和生产S→X1X，您应该很高兴。

Answer 3

考虑：

S → T1T

T → 10T | 01T | 0T1 | 1T0 | ε

然后，您从一个1 (S → T1T)开始，并在0's的每个应用程序中始终添加相同数量的1's和T（以任何可能的方式）。

Context Free Grammar - 一组字符串，其中1的数字比0的数字多一个

3 个答案: