UVa 134 - Loglan-A Logical Language

contents

1. 1. Problem
2. 2. Input and Output
3. 3. Sample input
4. 4. Sample output
5. 5. Solution

Problem

Loglan is a synthetic speakable language designed to test some of the fundamental problems of linguistics, such as the Sapir Whorf hypothesis. It is syntactically unambiguous, culturally neutral and metaphysically parsimonious. What follows is a gross over-simplification of an already very small grammar of some 200 rules.

Loglan sentences consist of a series of words and names, separated by spaces, and are terminated by a period (.). Loglan words all end with a vowel; names, which are derived extra-linguistically, end with a consonant. Loglan words are divided into two classes–little words which specify the structure of a sentence, and predicates which have the form CCVCV or CVCCV where C represents a consonant and V represents a vowel (see examples later).

The subset of Loglan that we are considering uses the following grammar:

A             →  a | e | i | o | u
MOD           →  ga | ge | gi | go | gu
BA            →  ba | be | bi | bo | bu
DA            →  da | de | di | do | du
LA            →  la | le | li | lo | lu
NAM           →  {all names}
PREDA         →  {all predicates}
<sentence>    →  <statement> | <predclaim>
<predclaim>   →  <predname> BA <preds> | DA <preds>
<preds>       →  <predstring> | <preds> A <predstring>
<predname>    →  LA <predstring> | NAM
<predstring>  →  PREDA | <predstring> PREDA
<statement>   →  <predname> <verbpred> <predname> | <predname> <verbpred>
<verbpred>    →  MOD <predstring>

Write a program that will read a succession of strings and determine whether or not they are correctly formed Loglan sentences.

Input and Output

Each Loglan sentence will start on a new line and will be terminated by a period (.). The sentence may occupy more than one line and words may be separated by more than one whitespace character. The input will be terminated by a line containing a single `#’. You can assume that all words will be correctly formed.

Output will consist of one line for each sentence containing either Good' orBad!’.

Sample input

la mutce bunbo mrenu bi ditca.
la fumna bi le mrenu.
djan ga vedma le negro ketpi.
#

Sample output

Good
Bad!
Good

Solution

其實這一題有比較好的解法，但是在正在學編譯器，因此把 LL(1) parser 運用在此題
，所以代碼會稍微長一點。

<sentence>    ->  <common_pre>
<sentence>    ->  DA <preds>
<common_pre>  ->  <predname> <suffix>
<suffix>      ->  <predclaim>
<suffix>      ->  <statement>
<predclaim>   ->  BA <preds>
<preds>       ->  <preds_head> <preds_tail>
<preds_head>  ->  <predstring>
<preds_tail>  ->  A <predstring> <preds_tail>
<preds_tail>  ->  l
<predname>    ->  LA <predstring>
<predname>    ->  NAM
<predstring>  ->  PREDA <predstr_tail>
<predstr_tail> -> PREDA <predstr_tail>
<predstr_tail> -> l
<statement>   ->  <verbpred> <statement_s>
<statement_s> ->  <predname>
<statement_s> ->  l
<verbpred>    ->  MOD <predstring>

為了要符合 LL(1) 的形式，要想辦法將 Grammar 符合規格，也就是每一條 production rule 的 predict set 不會有模稜兩個的情況，也就是說當我看到某個 non-terminal 在 stack 上，在去看下一個 token，然後可以對於 (non-terminal, token) 只找到一條規則去匹配。

predict set 是根據每一條 production rule 所產生的，也就是說這個推論可以產生的第一個 token 是什麼，假使有相同的前綴時，則必須拉出做調整。// 因為這將會造成模稜兩可的情況。

遇到

A -> BCx
A -> BCy

應調整成

A -> BCW
W -> x
W -> y

在 parsing 問題上，應避免替換的無限迴圈。
遇到

A -> AC
A -> X
A -> Y

應調整成

A -> WD
D -> CD
D -> lambda
W -> X
W -> Y

發現對照課本的 parsing function 撰寫起來才發現有些不完善的地方，對於 lldriver() 會沒有辦法應對輸入結束，要吐出 lambda 的問題。

#include <stdio.h> 
#include <vector>
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <ctype.h>
#include <assert.h>
using namespace std;

class Production {
    public:
        string 			LHS;
        vector<string> 	RHS;
        int 			label;
        
        Production(string L = "", vector<string> R = vector<string>()) {
            LHS = L;
            RHS = R;
        }
        
        void print() {
            printf("%s -> ", LHS.c_str());
            for(size_t i = 0; i < RHS.size(); i++)
                printf("%s", RHS[i].c_str());
        }
};
class Grammar {
    public:
        static const string 		lambda;
        string 						start_symbol;
        vector<Production> 			rules;
        map<string, set<string> > 	first_set;
        map<string, set<string> > 	follow_set;
        map<string, bool> 			derives_lambda;
        map<string, map<string, Production> >	lltable;
        
        map<string, bool> mark_lambda();
        void fill_first_set();
        void fill_follow_set();
        bool isNonterminal(string token);
        set<string> compute_first(vector<string> rhs);
        set<string> get_predict_set(Production p);
        void fill_lltable();
        void lldriver(queue<string> tokens);
};
const string Grammar::lambda("l");

set<string> Grammar::compute_first(vector<string> rhs) {
    set<string> result;
    size_t i;
    
    if(rhs.size() == 0 || rhs[0] == Grammar::lambda) {
        result.insert(Grammar::lambda);
    } else {
        result = first_set[rhs[0]];
        for(i = 1; i < rhs.size() && 
            first_set[rhs[i-1]].find(Grammar::lambda) != first_set[rhs[i-1]].end(); i++) {
            set<string> f = first_set[rhs[i]];
            f.erase(Grammar::lambda);
            result.insert(f.begin(), f.end());	
        }
        if(i == rhs.size() 
            && first_set[rhs[i-1]].find(Grammar::lambda) != first_set[rhs[i-1]].end()) {
            result.insert(Grammar::lambda);
        } else {
            result.erase(Grammar::lambda);
        }
    }
    
    return result;
} 
/*
 * please call get_predict_set() after fill_follow_set() and fill_first_set()
 */
set<string> Grammar::get_predict_set(Production p) {
    set<string> result;
    set<string> rfirst;
    
    rfirst = compute_first(p.RHS);
    if(rfirst.find(Grammar::lambda) != rfirst.end()) {
        rfirst.erase(Grammar::lambda);
        result.insert(rfirst.begin(), rfirst.end());
        rfirst = follow_set[p.LHS];
        result.insert(rfirst.begin(), rfirst.end());
    } else {
        result.insert(rfirst.begin(), rfirst.end());
    }
    
    return result;
}
/*
 * 
 */
void Grammar::fill_lltable() {
    string		A; // nonterminal
    Production 	p;
    
    for(size_t i = 0; i < rules.size(); i++) {
        p = rules[i];
        A = p.LHS;
        set<string> predict_set = get_predict_set(p);
        
        for(set<string>::iterator it = predict_set.begin();
            it != predict_set.end(); it++) {
            assert(lltable[A].find(*it) == lltable[A].end());
            lltable[A][*it] = p;
        }
    }
    
}
bool Grammar::isNonterminal(string token) {
    
    if(token[0] == '<' && token[token.length() - 1] == '>')
        return true;
    return false;

}
map<string, bool> Grammar::mark_lambda() {
    bool 				rhs_derives_lambda;
    bool 				changes;
    Production 			p;
    
    derives_lambda.clear();
    
    /* initially, nothing is marked. */
    for(size_t i = 0; i < rules.size(); i++) {
        derives_lambda[rules[i].LHS] = false;
    }
    do {
        changes = false;
        for(size_t i = 0; i < rules.size(); i++) {
            p = rules[i];
            if(!derives_lambda[p.LHS]) {
                if(p.RHS.size() == 0 || p.RHS[0] == Grammar::lambda) {
                    changes = derives_lambda[p.LHS] = true;
                    continue;
                }
                /* does each part of RHS derive lambda ? */
                rhs_derives_lambda = derives_lambda[string(p.RHS[0])];
                for(size_t j = 1; j < p.RHS.size(); j++) {
                    rhs_derives_lambda &= derives_lambda[p.RHS[j]];
                }
                if(rhs_derives_lambda) {
                    changes = true;
                    derives_lambda[p.LHS] = true;
                }
            }
        }
    } while(changes);
    return derives_lambda;
}
void Grammar::fill_first_set() {
    
    string 		A; // nonterminal
    string 		a; // terminal
    Production 	p;
    bool 		changes;
        
    mark_lambda();
    first_set.clear();
    
    for(size_t i = 0; i < rules.size(); i++) {
        A = rules[i].LHS;
        if(derives_lambda[A])
            first_set[A].insert(Grammar::lambda);
    }
    
    for(size_t i = 0; i < rules.size(); i++) {
        for(size_t j = 0; j < rules[i].RHS.size(); j++) {
            a = rules[i].RHS[j];
            if(!isNonterminal(a)) {
                if(a != Grammar::lambda)
                    first_set[a].insert(a);
                if(j == 0) { // A -> aXX
                    first_set[rules[i].LHS].insert(a);
                }
            }
        }
    }
    
    do {
        changes = false;
        for(size_t i = 0; i < rules.size(); i++) {
            p = rules[i];
            set<string> rfirst = compute_first(p.RHS);
            size_t oldsize = first_set[p.LHS].size();
            first_set[p.LHS].insert(rfirst.begin(), rfirst.end());
            size_t newsize = first_set[p.LHS].size();
            if(oldsize != newsize)
                changes = true;
        }
    } while(changes);
} 

void Grammar::fill_follow_set() {
    string 	A, B; // nonterminal
    Production 	p;
    bool 	changes;
    
    for(size_t i = 0; i < rules.size(); i++) {
        A = rules[i].LHS;
        follow_set[A].clear();
    }
    
    follow_set[start_symbol].insert(Grammar::lambda);
        
    do {
        changes = false;
        for(size_t i = 0; i < rules.size(); i++) {
            /* A -> a B b
			 * I.e. for each production and each occurrence
			 * of a nonterminal in its right-hand side. 
			 */
            p = rules[i];
            A = p.LHS;
            for(size_t j = 0; j < p.RHS.size(); j++) {
                B = p.RHS[j];
                if(isNonterminal(B)) {
                    vector<string> beta(p.RHS.begin() + j + 1, p.RHS.end());
                    set<string> rfirst = compute_first(beta);
                    size_t oldsize = follow_set[B].size();
                    
                    if(rfirst.find(Grammar::lambda) == rfirst.end()) {
                        follow_set[B].insert(rfirst.begin(), rfirst.end());	
                    } else {
                        rfirst.erase(Grammar::lambda);
                        follow_set[B].insert(rfirst.begin(), rfirst.end());	
                        rfirst = follow_set[A];
                        follow_set[B].insert(rfirst.begin(), rfirst.end());
                    }
                    size_t newsize = follow_set[B].size();			
                    if(oldsize != newsize)
                        changes = true;
                }
            }
        }
    } while(changes);
}

void Grammar::lldriver(queue<string> tokens) {
    stack<string> 	stk;
    string 			X;
    string			a;
    Production		p;
    /* Push the Start Symbol onto an empty stack */
    stk.push(start_symbol);
    
    while(!stk.empty() && !tokens.empty()) {
        X = stk.top();
        a = tokens.front();
        // cout << X << " " << a << endl;
        if(isNonterminal(X) && lltable[X].find(a) != lltable[X].end()) {
            p = lltable[X][a];
            stk.pop();
            for(int i = p.RHS.size() - 1; i >= 0; i--) {
                if(p.RHS[i] == Grammar::lambda)
                    continue;
                stk.push(p.RHS[i]);
            }
        } else if(X == a) {
            stk.pop();
            tokens.pop();
        } else if(lltable[X].find(Grammar::lambda) != lltable[X].end()) {
            stk.pop();
        } else {
            /* Process syntax error. */
            puts("Bad!");
            return;
        }
    }
    while(!stk.empty()) {
        X = stk.top();
        if(lltable[X].find(Grammar::lambda) != lltable[X].end())
            stk.pop();
        else
            break;
    }
    if(tokens.size() == 0 && stk.size() == 0)
        puts("Good");
    else
        puts("Bad!");
    return;
}
int aVowel(char c) {
    switch(c) {
        case 'a':case 'e':case 'i':case 'o':case 'u':
            return 1;
    }
    return 0;
}
string token2symbol(const string &str) {
    int n = str.length();
    if (!islower(str[n - 1]) || !aVowel(str[n - 1])) {
        return "NAM";
    }
    switch (n) {
    	case 1: return "A";
    	case 5:
        n = aVowel(str[4]);
        n |= ((aVowel(str[0]) << 4) | (aVowel(str[1]) << 3));
        n |= ((aVowel(str[2]) << 2) | (aVowel(str[3]) << 1));
        return (n == 5 || n == 9) ? "PREDA" : "UN";
    	case 2:
        	switch (str[0]) {
        		case 'g': return "MOD";
        		case 'b': return "BA";
        		case 'd': return "DA";
        		case 'l': return "LA";
        	}
    }
    return "UN";
}
void parsingProduction(string r, Grammar &g) {
    static int 		production_label = 0;
    stringstream 	sin(r);
    string 			lhs, foo;
    vector<string> 	tokens;
    sin >> lhs >> foo;
    while(sin >> foo)
        tokens.push_back(foo);
    Production p(lhs, tokens);
    p.label = ++production_label;
    g.rules.push_back(p);
} 

int main() {
    Grammar g;
    parsingProduction("<sentence>    ->  <common_pre>", g);
    parsingProduction("<sentence>    ->  DA <preds>", g);
    parsingProduction("<common_pre>  ->  <predname> <suffix>", g);
    parsingProduction("<suffix>      ->  <predclaim>", g);
    parsingProduction("<suffix>	  	 ->  <statement>", g);
    parsingProduction("<predclaim>   ->  BA <preds>", g);
    parsingProduction("<predclaim>   ->  DA <preds>", g);
    parsingProduction("<preds>       ->  <preds_head> <preds_tail>", g);
    parsingProduction("<preds_head>  ->  <predstring>", g);
    parsingProduction("<preds_tail>  ->  A <predstring> <preds_tail>", g);
    parsingProduction("<preds_tail>  ->  l", g);
    parsingProduction("<predname>    ->  LA <predstring>", g);
    parsingProduction("<predname>    ->  NAM", g);
    parsingProduction("<predstring>  ->  PREDA <predstr_tail>", g);
    parsingProduction("<predstr_tail> -> PREDA <predstr_tail>", g);
    parsingProduction("<predstr_tail> -> l", g);
    parsingProduction("<statement>   ->  <verbpred> <statement_s>", g);
    parsingProduction("<statement_s> ->  <predname>", g);
    parsingProduction("<statement_s> ->  l", g);
    parsingProduction("<verbpred>    ->  MOD <predstring>", g);
    
    g.start_symbol = "<sentence>";
        
    g.fill_first_set();
    g.fill_follow_set();
    g.fill_lltable();
    
    queue<string> SYMBOL;
    for(string token; cin >> token && token != "#";) {
        size_t pos = token.find('.');
        if(pos == string::npos) {
            SYMBOL.push(token2symbol(token));
            //cout << token2symbol(token) << " ";
            continue;
        }
        token.erase(token.length() - 1);
        //cout << token2symbol(token) << endl;
        if(token.length() > 0)
            SYMBOL.push(token2symbol(token));
        g.lldriver(SYMBOL);
        
        while(!SYMBOL.empty())
            SYMBOL.pop();
    }
    return 0;
}