Exercise 10.2.8

$\star$ Explain how to implement doubly linked lists using only one pointer value x.np per item instead of the usual two (next and prev). Assume that all pointer values can be interpreted as $k$-bit integers, and define x.np = x.next XOR x.prev, the $k$-bit "exclusive-or" of x.next and x.prev. (The value NIL is represented by 0). Be sure to describe what information you need to access the head of the list. Show how to implement the SEARCH, INSERT and DELETE operations on such a list. Also show how to reverse such a list in $\O(1)$ time.

We can find the pointer to the next item by XOR-ing np with the pointer to the previous item and vice-versa. If the previous pointer of the head of the list is to NIL and the next pointer of the tail is to NIL, then we only need a pointer to either end of the list to access it. Reversing the list is just swapping the head and the tail.

C code

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

typedef struct node_t {
    int key;
    struct node_t *np;
} node_t;

typedef struct {
    struct node_t *head;
    struct node_t *tail;
} list_t;

node_t *xor(node_t *left, node_t *right) {
    return (node_t *) (((unsigned long) left) ^ ((unsigned long) right));
}

void init_list(list_t *list) {
    list->head = NULL;
    list->tail = NULL;
}

void destroy_list(list_t *list) {
    node_t *prev = NULL;
    node_t *node = list->head;
    node_t *next;

    while (node) {
        next = xor(node->np, prev);
        free(node);
        prev = node;
        node = next;
    }
}

void insert(list_t *list, int key) {
    node_t *new = (node_t *) malloc(sizeof(node_t));
    new->key = key;
    new->np = xor(NULL, list->tail);

    if (list->tail) {
        list->tail->np = xor(new, xor(NULL, list->tail->np));
    }

    if (!list->head) {
        list->head = new;
    }

    list->tail = new;
}

int get(list_t *list, int index) {
    node_t *node = list->head;
    node_t *prev = NULL;
    node_t *next;

    while (index--) {
        if (!node) {
            fprintf(stderr, "Index out of bounds\n");
            exit(1);
        }
        next = xor(node->np, prev);
        prev = node;
        node = next;
    }

    return node->key;
}

node_t *search(list_t *list, int key) {
    node_t *node = list->head;
    node_t *prev = NULL;
    node_t *next;

    while (node) {
        if (node->key == key) {
            return node;
        }

        next = xor(node->np, prev);
        prev = node;
        node = next;
    }

    return NULL;
}

void delete(list_t *list, int key) {
    node_t *node = list->head;
    node_t *prev = NULL;
    node_t *next;

    while (node) {
        if (node->key == key) {
            next = xor(node->np, prev);

            if (prev) {
                prev->np = xor(xor(prev->np, node), next);
            } else {
                list->head = next;
            }

            if (next) {
                next->np = xor(xor(next->np, node), prev);
            } else {
                list->tail = prev;
            }

            node = next;
        } else {
            next = xor(node->np, prev);
            prev = node;
            node = next;
        }
    }
}

void reverse(list_t *list) {
    node_t *tmp;
    tmp = list->head;
    list->head = list->tail;
    list->tail = tmp;
}