How to resolve the algorithm Brace expansion step by step in the C programming language

Published on 7 June 2024 03:52 AM
#C

How to resolve the algorithm Brace expansion step by step in the C programming language

Table of Contents

Problem Statement

Brace expansion is a type of parameter expansion made popular by Unix shells, where it allows users to specify multiple similar string parameters without having to type them all out. E.g. the parameter enable_{audio,video} would be interpreted as if both enable_audio and enable_video had been specified.

Write a function that can perform brace expansion on any input string, according to the following specification. Demonstrate how it would be used, and that it passes the four test cases given below. In the input string, balanced pairs of braces containing comma-separated substrings (details below) represent alternations that specify multiple alternatives which are to appear at that position in the output. In general, one can imagine the information conveyed by the input string as a tree of nested alternations interspersed with literal substrings, as shown in the middle part of the following diagram: This tree can in turn be transformed into the intended list of output strings by, colloquially speaking, determining all the possible ways to walk through it from left to right while only descending into one branch of each alternation one comes across (see the right part of the diagram). When implementing it, one can of course combine the parsing and expansion into a single algorithm, but this specification discusses them separately for the sake of clarity. Expansion of alternations can be more rigorously described by these rules: Parsing the input string involves some additional complexity to deal with escaped characters and "incomplete" brace pairs: For every possible input string, your implementation should produce exactly the output which this specification mandates. Please comply with this even when it's inconvenient, to ensure that all implementations are comparable. However, none of the above should be interpreted as instructions (or even recommendations) for how to implement it. Try to come up with a solution that is idiomatic in your programming language. (See #Perl for a reference implementation.)

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Brace expansion step by step in the C programming language

This C code defines a program that uses a tree-like data structure to parse and expand text patterns. Let's break down the code step by step:

  1. Data Structure:

    • The program defines a tree-like data structure using the node struct.
    • Each node has a tag indicating its type (leaf, tree, or sequence) and a data union that stores either a string (for leaf nodes) or a pointer to another node (for tree and sequence nodes).
    • Nodes can be linked together to form a tree-like hierarchy.

  2. Node Manipulation Functions:

    • Several functions are defined for creating, modifying, and deallocating nodes:
      • allocate_node: Allocates a new node and initializes its tag and next pointer.
      • make_leaf: Creates a leaf node with a given string.
      • make_tree: Creates a tree node with an initially empty root.
      • make_seq: Creates a sequence node with an initially empty root.
      • deallocate_node: Recursively deallocates a node and its descendants.

  3. Tree Manipulation Functions:

    • append: Appends a node to the end of a tree or sequence node.
    • count: Recursively counts the number of nodes in a given tree or sequence.
    • expand: Recursively expands a tree or sequence node by visiting and printing its descendants.

  4. String Manipulation Functions:

    • allocate_string: Allocates and copies a new string from a source string.

  5. Parsing Functions:

    • parse_seq: Parses a sequence pattern and creates the corresponding tree structure.
    • parse_tree: Parses a tree pattern and creates the corresponding tree structure.

  6. Test Function:

    • test: Takes a string pattern as input, parses it using parse_seq, and then generates and prints all possible expansions of the pattern.

  7. Main Function:

    • The main function calls the test function on three different input patterns, demonstrating the program's parsing and expansion capabilities.

Here's an example of how the program works:

If you give the input pattern "~/{Downloads,Pictures}/*.{jpg,gif,png}" to the program, it will parse the pattern and create a tree structure that represents the possible expansions of the pattern. The expanded patterns could be:

  • ~/{Downloads,Pictures}/a.jpg
  • ~/{Downloads,Pictures}/b.gif
  • ~/{Downloads,Pictures}/c.png
  • ~/{Downloads}/a.jpg
  • ~/{Downloads}/b.gif
  • ~/{Downloads}/c.png
  • ~/{Pictures}/a.jpg
  • ~/{Pictures}/b.gif
  • ~/{Pictures}/c.png

Source code in the c programming language

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define BUFFER_SIZE 128

typedef unsigned char character;
typedef character *string;

typedef struct node_t node;
struct node_t {
    enum tag_t {
        NODE_LEAF,
        NODE_TREE,
        NODE_SEQ,
    } tag;

    union {
        string str;
        node *root;
    } data;

    node *next;
};

node *allocate_node(enum tag_t tag) {
    node *n = malloc(sizeof(node));
    if (n == NULL) {
        fprintf(stderr, "Failed to allocate node for tag: %d\n", tag);
        exit(1);
    }
    n->tag = tag;
    n->next = NULL;
    return n;
}

node *make_leaf(string str) {
    node *n = allocate_node(NODE_LEAF);
    n->data.str = str;
    return n;
}

node *make_tree() {
    node *n = allocate_node(NODE_TREE);
    n->data.root = NULL;
    return n;
}

node *make_seq() {
    node *n = allocate_node(NODE_SEQ);
    n->data.root = NULL;
    return n;
}

void deallocate_node(node *n) {
    if (n == NULL) {
        return;
    }

    deallocate_node(n->next);
    n->next = NULL;

    if (n->tag == NODE_LEAF) {
        free(n->data.str);
        n->data.str = NULL;
    } else if (n->tag == NODE_TREE || n->tag == NODE_SEQ) {
        deallocate_node(n->data.root);
        n->data.root = NULL;
    } else {
        fprintf(stderr, "Cannot deallocate node with tag: %d\n", n->tag);
        exit(1);
    }

    free(n);
}

void append(node *root, node *elem) {
    if (root == NULL) {
        fprintf(stderr, "Cannot append to uninitialized node.");
        exit(1);
    }
    if (elem == NULL) {
        return;
    }

    if (root->tag == NODE_SEQ || root->tag == NODE_TREE) {
        if (root->data.root == NULL) {
            root->data.root = elem;
        } else {
            node *it = root->data.root;
            while (it->next != NULL) {
                it = it->next;
            }
            it->next = elem;
        }
    } else {
        fprintf(stderr, "Cannot append to node with tag: %d\n", root->tag);
        exit(1);
    }
}

size_t count(node *n) {
    if (n == NULL) {
        return 0;
    }

    if (n->tag == NODE_LEAF) {
        return 1;
    }
    if (n->tag == NODE_TREE) {
        size_t sum = 0;
        node *it = n->data.root;
        while (it != NULL) {
            sum += count(it);
            it = it->next;
        }
        return sum;
    }
    if (n->tag == NODE_SEQ) {
        size_t prod = 1;
        node *it = n->data.root;
        while (it != NULL) {
            prod *= count(it);
            it = it->next;
        }
        return prod;
    }

    fprintf(stderr, "Cannot count node with tag: %d\n", n->tag);
    exit(1);
}

void expand(node *n, size_t pos) {
    if (n == NULL) {
        return;
    }

    if (n->tag == NODE_LEAF) {
        printf(n->data.str);
    } else if (n->tag == NODE_TREE) {
        node *it = n->data.root;
        while (true) {
            size_t cnt = count(it);
            if (pos < cnt) {
                expand(it, pos);
                break;
            }
            pos -= cnt;
            it = it->next;
        }
    } else if (n->tag == NODE_SEQ) {
        size_t prod = pos;
        node *it = n->data.root;
        while (it != NULL) {
            size_t cnt = count(it);

            size_t rem = prod % cnt;
            expand(it, rem);

            it = it->next;
        }
    } else {
        fprintf(stderr, "Cannot expand node with tag: %d\n", n->tag);
        exit(1);
    }
}

string allocate_string(string src) {
    size_t len = strlen(src);
    string out = calloc(len + 1, sizeof(character));
    if (out == NULL) {
        fprintf(stderr, "Failed to allocate a copy of the string.");
        exit(1);
    }
    strcpy(out, src);
    return out;
}

node *parse_seq(string input, size_t *pos);

node *parse_tree(string input, size_t *pos) {
    node *root = make_tree();

    character buffer[BUFFER_SIZE] = { 0 };
    size_t bufpos = 0;
    size_t depth = 0;
    bool asSeq = false;
    bool allow = false;

    while (input[*pos] != 0) {
        character c = input[(*pos)++];
        if (c == '\\') {
            c = input[(*pos)++];
            if (c == 0) {
                break;
            }
            buffer[bufpos++] = '\\';
            buffer[bufpos++] = c;
            buffer[bufpos] = 0;
        } else if (c == '{') {
            buffer[bufpos++] = c;
            buffer[bufpos] = 0;
            asSeq = true;
            depth++;
        } else if (c == '}') {
            if (depth-- > 0) {
                buffer[bufpos++] = c;
                buffer[bufpos] = 0;
            } else {
                if (asSeq) {
                    size_t new_pos = 0;
                    node *seq = parse_seq(buffer, &new_pos);
                    append(root, seq);
                } else {
                    append(root, make_leaf(allocate_string(buffer)));
                }
                break;
            }
        } else if (c == ',') {
            if (depth == 0) {
                if (asSeq) {
                    size_t new_pos = 0;
                    node *seq = parse_seq(buffer, &new_pos);
                    append(root, seq);
                    bufpos = 0;
                    buffer[bufpos] = 0;
                    asSeq = false;
                } else {
                    append(root, make_leaf(allocate_string(buffer)));
                    bufpos = 0;
                    buffer[bufpos] = 0;
                }
            } else {
                buffer[bufpos++] = c;
                buffer[bufpos] = 0;
            }
        } else {
            buffer[bufpos++] = c;
            buffer[bufpos] = 0;
        }
    }

    return root;
}

node *parse_seq(string input, size_t *pos) {
    node *root = make_seq();

    character buffer[BUFFER_SIZE] = { 0 };
    size_t bufpos = 0;

    while (input[*pos] != 0) {
        character c = input[(*pos)++];
        if (c == '\\') {
            c = input[(*pos)++];
            if (c == 0) {
                break;
            }
            buffer[bufpos++] = c;
            buffer[bufpos] = 0;
        } else if (c == '{') {
            node *tree = parse_tree(input, pos);
            if (bufpos > 0) {
                append(root, make_leaf(allocate_string(buffer)));
                bufpos = 0;
                buffer[bufpos] = 0;
            }
            append(root, tree);
        } else {
            buffer[bufpos++] = c;
            buffer[bufpos] = 0;
        }
    }

    if (bufpos > 0) {
        append(root, make_leaf(allocate_string(buffer)));
        bufpos = 0;
        buffer[bufpos] = 0;
    }

    return root;
}

void test(string input) {
    size_t pos = 0;
    node *n = parse_seq(input, &pos);
    size_t cnt = count(n);
    size_t i;

    printf("Pattern: %s\n", input);

    for (i = 0; i < cnt; i++) {
        expand(n, i);
        printf("\n");
    }
    printf("\n");

    deallocate_node(n);
}

int main() {
    test("~/{Downloads,Pictures}/*.{jpg,gif,png}");
    test("It{{em,alic}iz,erat}e{d,}, please.");
    test("{,{,gotta have{ ,\\, again\\, }}more }cowbell!");

    //not sure how to parse this one
    //test("{}} some }{,{\\{ edge, edge} \,}{ cases, {here} \\\\\}");

    return 0;
}

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