# 2024q1 Homework1 (lab0)
contributed by < `YangYeh-PD` >
## Development Environment
```shell
$ gcc --version
gcc (Ubuntu 13.2.0-4ubuntu3) 13.2.0
Copyright (C) 2023 Free Software Foundation, Inc.
Architecture: x86_64
CPU op-mode(s): 32-bit, 64-bit
Address sizes: 39 bits physical, 48 bits virtual
Byte Order: Little Endian
CPU(s): 8
On-line CPU(s) list: 0-7
Vendor ID: GenuineIntel
Model name: Intel(R) Core(TM) i5-8250U CPU @ 1.60GHz
CPU family: 6
Model: 142
Thread(s) per core: 2
Core(s) per socket: 4
Socket(s): 1
Stepping: 10
CPU(s) scaling MHz: 81%
CPU max MHz: 3400.0000
CPU min MHz: 400.0000
BogoMIPS: 3600.00
```
## Queue implementations
:::danger
The task is not simply to build APIs for queue operations, but rather to execute the fundamental operations adhering to the specified queue interface. Pay attention to how you express this.
:::
To construct APIs for queue, we first need to know the definitions and how it should be constructed. Consider the definition of the element `element_t` in a queue
```c
typedef struct {
char *value;
struct list_head list;
} element_t;
```
where we can store a string by the character pointer `value`, and `list_head` is defined in header file [<list.h>](https://github.com/YangYeh-PD/lab0-c/blob/master/list.h)
```c
struct list_head {
struct list_head *prev;
struct list_head *next;
};
```
```graphviz
digraph "Doubly Linked List" {
rankdir = "LR";
node1[shape = record, label = "{<prev> prev |<next> next}"];
node2[shape = record, label = "{<prev> prev |<next> next}"];
node3[shape = record, label = "{<prev> prev |<next> next}"];
node1:next:c -> node2;
node2:prev:c -> node1;
node2:next:c -> node3;
node3:prev:c -> node2;
}
```
Thus, the implementations of a node are
1. Declare a pointer of charater to store a string.
2. Embedding the whole `list_head` into the structure.
> I came up with the wrong queue picture, here is the correct one.[name=ChenYang Yeh] [time=Sat, Feb 24, 2024 19:41 PM]
```graphviz
digraph "queue"
{
rankdir = "LR"
subgraph "cluster queue 1"
{
node0_2[shape = record, label = "{<prev> prev |<next> next}"];
node0_1[shape = record, label = "head"];
node0_3[shape = record, label = "size"];
node0_4[shape = record, label = "id"];
}
subgraph "cluster node1"
{
node1_1[shape = record, label = "value"];
node1_2[shape = record, label = "{<prev> prev |<next> next}"];
}
subgraph "cluster node2"
{
node2_1[shape = record, label = "value"];
node2_2[shape = record, label = "{<prev> prev |<next> next}"];
}
subgraph "cluster queue 2"
{
node3_2[shape = record, label = "{<prev> prev |<next> next}"];
node3_1[shape = record, label = "head"];
node3_3[shape = record, label = "size"];
node3_4[shape = record, label = "id"];
}
subgraph "cluster queue 3"
{
node4_2[shape = record, label = "{<prev> prev |<next> next}"];
node4_1[shape = record, label = "head"];
node4_3[shape = record, label = "size"];
node4_4[shape = record, label = "id"];
}
node0[shape = record, label = "{<prev> prev |<next> next}"]
node0_2:next:c -> node3_2;
node3_2:prev:c -> node0_2;
node3_2:next:c -> node4_2;
node4_2:prev:c -> node3_2;
node0_1:head:c -> node0;
node0:next:c -> node1_2;
node1_2:prev:c -> node0;
node0:prev:c -> node2_2;
node2_2:next:c -> node0;
}
```
where the head of the queue is defined as
```c
typedef struct {
struct list_head *q;
struct list_head chain;
int size;
int id;
} queue_contex_t;
```
* q : the pointer of the linked list defined in [<lish.h>](https://github.com/YangYeh-PD/lab0-c/blob/master/list.h) that is embedded in `queue_contex_t`.
* chain : used to chain different queues.
* size : a integer that stores the size of the queue.
* id : the identification number.
Here is the list of implementations ([`queue.c`](https://github.com/YangYeh-PD/lab0-c/blob/master/queue.c)) of queue that support both FIFO and LIFO.
:::warning
Instead of focusing on the LIFO and FIFO properties, you should explain why the `queue_context_t` was designed with its specific members in a C structure.
> Thus, `queue_contex_t` not only stores queue basic information, but also connects each independent queues into a chain, which helps us manipulate data.[name=ChenYang Yeh] [time=Sun, Feb 25, 2024 03:16 AM]
:::
### `q_new()`
:::danger
Address your motivations and considerations here.
:::
Since we import [<lish.h>](https://github.com/YangYeh-PD/lab0-c/blob/master/list.h), we can use the function in the library rather than make the new one, or operate by myself.
> Inspired by <`csotaku0926`>
>
```diff
struct list_head *q_new()
{
+ struct list_head *head = malloc(sizeof(struct list_head));
- queue_contex_t *queue = malloc(sizeof(queue_contex_t));
+ if (!head) {
return NULL;
}
- queue->q->next = queue->q;
- queue->q->prev = queue->q;
+ INIT_LIST_HEAD(head);
return head;
}
```
### `q_free()`
:::danger
Address your motivations and considerations here.
:::
Because we need to ==free== all the memories of the queue, we need to know the pointer of each node, not only the pointer to the structure member `list_head`.
The macro `container_of()` is defined in [<lish.h>](https://github.com/YangYeh-PD/lab0-c/blob/master/list.h) which calculates address of structure that contains address pointer. Using macro `list_for_each_entry_safe()` to make code less complicated.
~~Since the member of `struct list_head *p` is a pointer itself, we need to pass the ==indirect pointer== `&head` which represent the address of the pointer `head`.~~
```diff
struct list_head *head = l;
+ l = l->next;
- l = l->next;
- while (l->next != head) {
- free(container_of(l->next, element_t, list));
- l = l->next;
+ q_release_element(container_of(l->prev, element_t, list));
- q_release_element(container_of(l->prev, element_t, list));
- }
- free(container_of(head, queue_contex_t, chain);
+ free(container_of(&head, queue_contex_t, q));
- free(container_of(&head, queue_contex_t, q));
+ l = l->next;
- l = l->next;
+ q_release_element(container_of(l->prev, element_t, list));
- q_release_element(container_of(l->prev, element_t, list));
+ if (!l) {
+ return;
+ }
+ element_t *curr, *next;
+ list_for_each_entry_safe (curr, next, l, lsit) {
+ list_del(&curr->list);
+ q_release_element(curr);
+ }
+ free(l);
+ return;
```
### `q_insert_head()`
:::danger
Address your motivations and considerations here.
:::
~~We can incorporate whether `head == NULL` or ==the failure memory allocation== in the same `if` statement. And again, we use the API `list_add()` instead of adding by myself.~~
> Well, actually we cannot, since it will cause ==memory leak== when first allocating memory to `node` and when `head == NULL`[name=ChenYang Yeh] [time=Sat, Mar 2, 2024 12:50 PM]
We need to allocated memory and copy the string into `s` and cannot directly assign the pointer to `value`. Note that we store `strlen(s)` into `len` to reduce the function call `strlen()`.
If `node` is not added into the list successfully, we need to free the space allocated by `node`, or it will again cause memory leak.
```diff
bool q_insert_head(struct list_head *head, char *s)
{
+ if (!head) {
+ return false;
+ }
element_t *node = malloc(sizeof(element_t));
- if (!head || !node) {
+ if (!node)
return false;
}
- (node->list).next = head->next;
- (node->list).prev = head;
- head->next = &(node->list);
- ((node->list).next)->prev = &(node->list);
- node->value = s;
+ size_t len = strlen(s);
+ node->value = malloc((len + 1) * sizeof(char));
+ if(!(node->value)) {
- q_release_element(node);
+ free(node);
+ return false;
+ }
- memcpy(node->value, s, strlen(s));
- *(node->value + strlen(s)) = '\0';
+ list_add(&node->list, head);
+ if (!node->list.next || !node->list.prev) {
+ q_release_element(node);
+ return false;
+ }
+ strncpy(node->value, s, len + 1);
return true;
}
```
> Inspired by <`marvin0102`>
<s>
Since the member `list` in `element_t` is ==not a pointer==, we have to use **dot (.) operator** to select the member.
* line 49 and 50 add the node into the queue by modifying its `prev` and `next`
* line 51 and 52 modify `next` of the `head` and `prev` of the next node to the inserted node.
</s>
:::warning
Given that the preceding is merely a brief segment of code, it's not necessary to indicate specific line numbers for reference. You may simply emphasize the code of interest by incorporating it directly into the ongoing discussion for clarity.
Always write meaningful and informative technical reports with great fluency.
:::
:::danger
To ensure this report is both informative and flows well, it's important to start by outlining the motivation, considerations, and examples for the suggested functions before compiling a list of code snippets.
$\to$ [Guide to Technical Report Writing](https://www.sussex.ac.uk/ei/internal/forstudents/engineeringdesign/studyguides/techreportwriting)
> Thanks![name=ChenYang Yeh] [time=Sat, Feb 24, 2024 08:10 AM]
:::
### `q_remove_head()`
According to the definition in [ISO/IEC 9899 (p)7.21.2.1](https://www.open-std.org/jtc1/sc22/wg14/www/docs/n1124.pdf#page=337), `memcpy()` ==only copies `n` characters==, which means we still need to add `\0` at the end of the string.
> The **memcpy** function copies **n** characters from the object pointed to by **s2** into the object pointed to **s1**. If copying takes place between objects that overlap, the behavior is undefined.
```diff
element_t *q_remove_head(struct list_head *head, char *sp, size_t bufsize)
{
- if(!head || !(head->next)) {
+ if(!head || list_empty(head)) {
return NULL;
}
- element_t *del_node = container_of(head->next, element_t, list);
+ element_t *del_node = list_entry(head->next, element_t, list);
+ char *del_str = del_node->value;
list_del_init(head->next);
if(sp) {
+ size_t del_len = strlen(del_str);
+ size_t len = (del_len > (bufsize - 1)) ? bufsize - 1 : del_len;
- memcpy(sp, del_str, len);
+ strncpy(sp, del_str, len);
*(sp + len) = '\0';
}
return del_node;
}
```
But why don't we use `strcpy()` instead of `memcpy()`?
In [ISO/IEC 9899 (P)7.21.2.4](https://www.open-std.org/jtc1/sc22/wg14/www/docs/n1124.pdf#page=338), `strcpy()` copies the string `s2` into `s1` ==without checking whether the memory allocated in `s1` is enough==, which may cause undefined behavior.
> The **strcpy** function copies the string pointed to by `s2` (including the terminating null character) into the array pointed to by `s1`.
In [(P)6.5.10](https://www.open-std.org/jtc1/sc22/wg14/www/docs/n1124.pdf#page=99) footnote 92)
> If prior invalid pointer operations (such as accesses outside array bounds), produced undefined behavior.
> However, it still cannot pass `make test` normally.
>> ERROR: Removed value dolphinU��� != expected value dolphin
>> ERROR: Removed value bearU��� != expected value bear
>>
> I don't know why I should add `-5` behind of the `strlen()`..., but it works.
[name=ChenYang Yeh] [time=Sun, Feb 25, 2024 00:48 AM]
:::warning
Remove the hacky code and utilize the tools to resolve memory issues.
> The problem is solved. It raised from `memcpy()` in `insert_head()` and `insert_tail()`, because I forgot to add `'\0'` at the end of the string.
[name=ChenYang Yeh] [time=Tue, Feb 27, 2024 06:19 AM]
:::
### `q_delete_mid()`
I figured out three different ways for implementation.
1. We defined fast and slow pointers. The fast pointer moving 2 nodes at once, while the slow one moving 1 node only.
```graphviz
digraph "Circular Linked List" {
rankdir = "LR";
node1[shape = circle, label = "1"];
node2[shape = circle, label = "2"];
node3[shape = circle, label = "3"];
node4[shape = circle, label = "4"];
node5[shape = circle, label = "5"];
node0__[shape = none, label = "fast", fontcolor = "blue"];
node1__[shape = circle, label = "1", style = invis];
node2__[shape = circle, label = "2", style = invis];
node0_[shape = none, label = "slow", fontcolor = "red"];
node1_[shape = circle, label = "1", style = invis];
node0_ -> node1_;
node0__ -> node1__ -> node2__;
node1 -> node2;
node2 -> node1;
node2 -> node3;
node3 -> node2;
node3 -> node4;
node4 -> node3;
node4 -> node5;
node5 -> node4;
}
```
Eventually, once the fast pointer come back to the `head`, the slow pointer would point to the middle point. But what is the time complexity?
Since the fast pointer traverse 2 nodes and the slow one traverse a node each time, the time complexity would be
$$
O(n) + O\left(\frac{n}{2}\right) = O\left(\frac{3}{2}n\right).
$$
2. We also can use `q_size()` to get the total size of queue, then divided it by 2 and traverse to the destination point. Since `q_size()` takes $O(n)$ time, and traversing takes $\displaystyle O\left(\frac{n}{2}\right)$, the time complexity would also be
$$
O(n) + O\left(\frac{n}{2}\right) = O\left(\frac{3}{2}n\right).
$$
3. Last but not least, we can make best use of the property of **circular doubly linked list**. We define left and right pointers, then traverse from head in different ways until they meet, then ==the meeting point will definitely be the midpoint of the queue.==
(Since both left and right pointers have traversed half of the queue.)
```graphviz
digraph {
rankdir = "LR";
subgraph "cluster node4"
{
node3_1[shape = record, label = "value"];
node3_2[shape = record, label = "{<prev> prev |<next> next}"];
}
subgraph "cluster node3"
{
node3_1[shape = record, label = "value"];
node3_2[shape = record, label = "{<prev> prev |<next> next}"];
}
node0[shape = record, label = "{<prev> prev |<next> next}"]
subgraph "cluster node2"
{
node2_1[shape = record, label = "value"];
node2_2[shape = record, label = "{<prev> prev |<next> next}"];
}
subgraph "cluster node1"
{
node1_1[shape = record, label = "value"];
node1_2[shape = record, label = "{<prev> prev |<next> next}"];
}
node0__[shape = none, label = "right", fontcolor = "blue"];
node1__[shape = circle, label = "1", style = invis];
node0_[shape = none, label = "left", fontcolor = "red"];
node1_[shape = circle, label = "1", style = invis];
node1_ -> node0_;
node0_ -> node1_[style = invis];
node0__ -> node1__;
node3_2:next:c -> node0;
node0:prev:c -> node3_2;
node0:next:c -> node2_2;
node2_2:prev:c -> node0;
node2_2:next:c -> node1_2;
node1_2:prev:c -> node2_2;
}
```
```c
bool q_delete_mid(struct list_head *head)
{
// https://leetcode.com/problems/delete-the-middle-node-of-a-linked-list/
if (!head || list_empty(head)) {
return false;
}
struct list_head *right = head->next;
struct list_head *left = head->prev;
while (!((left == right) || ((left->prev) == right))) {
right = right->next;
left = left->prev;
}
list_del(left);
q_release_element(container_of(left, element_t, list));
return true;
}
```
:::warning
Enforce the consistent style when you are going to list the code snip.
:::
Since both left and right pointers take only
$$
\displaystyle O\left(\frac{n}{2}\right)
$$
the time complexity would be
$$
O\left(\frac{n}{2}\right) + O\left(\frac{n}{2}\right) = O(n).
$$
:::warning
Check [this note](https://hackmd.io/@sysprog/ry8NwAMvT) and identify the performance concerns by measurements.
> Done![name=ChenYang Yeh] [time=Thu, Feb 29, 2024 07:32 AM]
:::
#### The operational mechanism behind the processor
In programs with good **temporal locality**, a memory location referenced once may be referred to ==multiple times shortly thereafter==. In programs with good **spatial locality**, if a memory location is referenced, the program is likely to reference ==nearby memory locations== in subsequent operations.
From the perspective of spatial locality, the storage locations of nodes are not regular in the linked list, meaning ==each node is scattered in memory==, resulting in poor spatial locality. Therefore, it can be inferred that the spatial locality performance of all algorithms above is the same.
From the view of temporal locality, in the "fast and slow pointer" approach, before the loop concludes and after the fast pointer has completed its traversal of the first half of the linked list, the slow pointer has accessed the same node. Thus time interval between two accesses of the same node is relatively shorter. However, in "single pointer" approach, the second access to the same node occurs after the completion of the first traversal. The time interval between two accesses of the same node is relatively longer.
| Method | Temporal Locality | Spatial Locality |
|:-----------------------:|:-----------------:|:----------------:|
| Single pointer | shorter | same |
| Fast and slow pointer | longer | same |
Thus, "Fast and slow pointer" is more cache friendly than single pointer, since it does not frequently trigger cache eviction by changing the contents in memory and reduces the occurrence of **cache misses**.
:::warning
TODO: Use `perf` to show preliminary measurements.
:::
#### Test the cache performance of both algorithms by `perf`
Consider a singly linked list defined as
```c
typedef struct list {
int value;
unsigned long buffer1, buffer2, buffer3;
struct list *next;
} list
```
:::warning
TODO: Increase the number of members in the structure above to render cache misses more expensive and enable straightforward comparisons between different approaches.
:::
After we construct the list, we can find the midpoint by both **single** or **fast-slow pointer** as we have mentioned above. The followings are c implementations respectively.
**Single porinter**
```c
list *list_mid(list *head)
{
int size = 0;
list *traverse = head;
while (traverse) {
size++;
traverse = traverse->next;
}
traverse = head;
for (int i = 0; i < size/2; i++) {
traverse = traverse->next;
}
return traverse;
}
```
**Fast-slow pointer**
```c
list *list_mid(list *head)
{
list *slow = head;
list *fast = slow->next;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
return slow->next;
}
```
Here are the results of cache reference and cache miss obtained through perf testing with different input sizes.
![Cache Reference](https://hackmd.io/_uploads/HJBVgpSAT.png)
![Cache Miss](https://hackmd.io/_uploads/HJiElaSCT.png)
In most cases, the version employing the fast-slow pointer tends to have fewer cache references and cache misses compared to the single pointer version.
### `q_delete_dup()`
We first define two pointers to `element_t`, `reference` and `traverse`, where `reference = head->next` and `traverse = reference->next` at the beginning, which means `traverse` is two steps ahead of the `head`.
```graphviz
digraph "Circular Linked List" {
rankdir = "LR";
head_[shape = record, label = "{<prev> prev |<next> next}"];
node1[shape = circle, label = "1"];
node2[shape = circle, label = "1"];
node3[shape = circle, label = "1"];
node4[shape = circle, label = "1"];
node5[shape = circle, label = "2"];
head[shape = none, label = "head"];
reference[shape = none, label = "reference", fontcolor = "red"];
traverse[shape = none, label = "traverse", fontcolor = "blue"];
head -> head_;
reference -> node1;
traverse -> node2;
head_:next:c -> node1;
node1 -> head_
node1 -> node2;
node2 -> node1;
node2 -> node3;
node3 -> node2;
node3 -> node4;
node4 -> node3;
node4 -> node5;
node5 -> node4;
}
```
If `traverse == reference`, then `traverse` will keep traversing until it reach the node where `traverse != reference`.
```graphviz
digraph "Circular Linked List" {
rankdir = "LR";
head_[shape = record, label = "{<prev> prev |<next> next}"];
node1[shape = circle, label = "1"];
node2[shape = circle, label = "1"];
node3[shape = circle, label = "1"];
node4[shape = circle, label = "1"];
node5[shape = circle, label = "2"];
head[shape = none, label = "head"];
reference[shape = none, label = "reference", fontcolor = "red"];
traverse[shape = none, label = "traverse", fontcolor = "blue"];
head -> head_;
reference -> node1;
traverse -> node5;
head_:next:c -> node1;
node1 -> head_
node1 -> node2;
node2 -> node1;
node2 -> node3;
node3 -> node2;
node3 -> node4;
node4 -> node3;
node4 -> node5;
node5 -> node4;
}
```
Then, `reference` will move ahead and delete ==the previous node== until it reach the `traverse` node.
```graphviz
digraph "Circular Linked List" {
rankdir = "LR";
head_[shape = record, label = "{<prev> prev |<next> next}"];
node1[shape = circle, label = "2"];
head[shape = none, label = "head"];
reference[shape = none, label = "reference", fontcolor = "red"];
traverse[shape = none, label = "traverse", fontcolor = "blue"];
head -> head_;
reference -> node1;
traverse -> node1;
head_:next:c -> node1;
node1 -> head_
}
```
```diff
bool q_delete_dup(struct list_head *head)
{
if (!head || list_empty(head)) {
return false;
}
+ struct list_head *reference = head->next, *traverse, *delete;
- struct list_head *traverse;
while ((reference->next != head) && (reference != head)) {
traverse = reference->next;
char *str_ref = container_of(reference, element_t, list)->value;
char *str_tra = container_of(traverse, element_t, list)->value;
if (!strcmp(str_ref, str_tra)) {
while (!(traverse == head) && !strcmp(str_ref, str_tra)) {
traverse = traverse->next;
+ str_tra = container_of(traverse, element_t, list)->value;
}
while (reference != traverse) {
+ delete = reference;
reference = reference->next;
- list_del(reference->prev);
+ list_del(delete);
- q_release_element(container_of(reference->prev, element_t, list));
+ q_release_element(container_of(delete, element_t, list));
}
} else {
reference = traverse;
}
}
```
> But there is a error message in test trace 06.
>> ERROR: Attempted to free unallocated block. Address = 0x5555555555555555
>>
> This may result from deleting the head node by `q_release_element`. But I have checked my code and ensured that such the behaviors is avoided. So I still cannot figure it out.
> The question has been resolved. There have two problems in the original code.
> 1. I forget to renew `str_tra` when traverse to the next node.
> 2. Because I delete the previous node of the reference node, if the deleted node is coincidentally the first node of the queue, then we call the function
>`q_release_element(container_of(reference->prev, element_t, list))`
>is equivalent to
>`q_release_element(container_of(head, element_t, list))`
> then the segmentation fault occured. Thus, we need the pointer `delete` to store the deleted node, then free it.
>[name=ChenYang Yeh] [time=Thu, Feb 29, 2024 07:02 AM]
### `q_reverseK()`
This took me very long time to deal with it. I devided this question into three different parts.
1. Check if the number of node is $N \leq k$.
It is reletively simple, we use `HEAD` to record the starting point, and `TAIL` to traverse the queue to ensure that the queue is long enough for reverse. To avoid the special case of the first node, we declare `TAIL` as the indirect pointer to the `next`;
```c
struct list_head **TAIL = &(head->next);
```
```graphviz
digraph "Circular Linked List" {
rankdir = "LR";
node0[shape = record, label = "{<prev> prev |<next> next}"]
subgraph "cluster node1"
{
node1_1[shape = record, label = "3"];
node1_2[shape = record, label = "{<prev> prev |<next> next}"];
}
subgraph "cluster node2"
{
node2_1[shape = record, label = "5"];
node2_2[shape = record, label = "{<prev> prev |<next> next}"];
}
subgraph "cluster node3"
{
node3_1[shape = record, label = "8"];
node3_2[shape = record, label = "{<prev> prev |<next> next}"];
}
head[shape = none, label = "head"];
HEAD[shape = none, label = "HEAD", fontcolor = "red"];
TAIL[shape = none, label = "TAIL", fontcolor = "blue"];
head -> node0;
HEAD -> node1_1;
TAIL -> node2_2:next;
node0:next:c -> node1_2;
node1_2:prev:c -> node0;
node1_2:next:c -> node2_2;
node2_2:prev:c -> node1_2;
node2_2:next:c -> node3_2;
node3_2:prev:c -> node2_2;
}
```
```c
// Check if the number of nodes is < k
struct list_head *HEAD = *TAIL;
for (int count = 0; count < k; count++) {
if (*TAIL == head) {
return;
}
TAIL = &(*TAIL)->next;
}
```
2. Delete the node from list and store it to stack.
We use `HEAD` declare before to delete each node and add it into the stack we just declared.
```graphviz
digraph "Circular Linked List" {
rankdir = "LR";
node0[shape = record, label = "{<prev> prev |<next> next}"]
subgraph "cluster node1"
{
node1_1[shape = record, label = "5"];
node1_2[shape = record, label = "{<prev> prev |<next> next}"];
}
subgraph "cluster node2"
{
node2_1[shape = record, label = "3"];
node2_2[shape = record, label = "{<prev> prev |<next> next}"];
}
stack[shape = none, label = stack];
stack -> node0;
node0:next:c -> node1_2;
node1_2:prev:c -> node0;
node1_2:next:c -> node2_2;
node2_2:prev:c -> node1_2;
}
```
```graphviz
digraph "Circular Linked List" {
rankdir = "LR";
node0[shape = record, label = "{<prev> prev |<next> next}"]
subgraph "cluster node3"
{
node3_1[shape = record, label = "8"];
node3_2[shape = record, label = "{<prev> prev |<next> next}"];
}
head[shape = none, label = "head"];
HEAD[shape = none, label = "HEAD", fontcolor = "red"];
head -> node0;
node0:next:c -> node3_2;
node3_2:prev:c -> node0;
HEAD -> node3_1;
}
```
```c
// Delete the nodes from the list
struct list_head stack;
INIT_LIST_HEAD(&stack);
for (int count = 0; count < k; count++) {
struct list_head *delete = HEAD;
HEAD = HEAD->next;
list_del(delete);
list_add(delete, &stack);
}
```
3. Last but not least, we use `list_splice()` to concatenate the list together.
(It will be automatically in reverse order because of the LIFO property of stack.)
```graphviz
digraph "Circular Linked List" {
rankdir = "LR";
node0[shape = record, label = "{<prev> prev |<next> next}"]
subgraph "cluster node1"
{
node1_1[shape = record, label = "5"];
node1_2[shape = record, label = "{<prev> prev |<next> next}"];
}
subgraph "cluster node2"
{
node2_1[shape = record, label = "3"];
node2_2[shape = record, label = "{<prev> prev |<next> next}"];
}
subgraph "cluster node3"
{
node3_1[shape = record, label = "8"];
node3_2[shape = record, label = "{<prev> prev |<next> next}"];
}
head[shape = none, label = "head"];
HEAD[shape = none, label = "HEAD", fontcolor = "red"];
TAIL[shape = none, label = "TAIL", fontcolor = "blue"];
head -> node0;
HEAD -> node3_1;
TAIL -> node2_2:next;
node0:next:c -> node1_2;
node1_2:prev:c -> node0;
node1_2:next:c -> node2_2;
node2_2:prev:c -> node1_2;
node2_2:next:c -> node3_2;
node3_2:prev:c -> node2_2;
}
```
```c
// Insert the list in reverse order
list_splice(&stack, HEAD->prev);
TAIL = &HEAD->prev->next;
```
:::warning
Can you shorten your code by facilitating List APIs?
> I can't. Since I don't want to iterate the whole queue, so I cannot use such a macro of `list_for_each_entry(_safe)`. I need to write the loop by myself.[name=ChenYang Yeh] [time=Sat, Mar 2, 2024 13:36 PM]
:::
### `q_sort()`
I use classic [merge sort](https://en.wikipedia.org/wiki/Merge_sort) as implementation.
1. Find the middle point of the list
We can use the algorithm in `q_delete_mid()` to find the midpoint.
```c
struct list_head *right = head->next;
struct list_head *left = head->prev;
while (!((left == right) || ((left->prev) == right))) {
right = right->next;
left = left->prev;
}
```
2. Break into 2 lists (head and stack)
After we find the midpoint of the list, we need to divided the list into 2 parts, which is called `head` and `stack` in my code.
```c
struct list_head stack;
INIT_LIST_HEAD(&stack);
for (struct list_head *traverse = head->next, *safe = traverse->next;
traverse != left; traverse = safe, safe = safe->next) {
list_del(traverse);
list_add_tail(traverse, &stack);
}
q_sort(head, descend);
q_sort(&stack, descend);
```
3. Merge two lists
Last but not least, we merge two lists into `head` in ascending or decending order.
```c
struct list_head *head_tra = head->next, *stack_del;
while (head_tra != head && !list_empty(&stack)) {
char *str_head = list_entry(head_tra, element_t, list)->value;
char *str_stack = list_entry(stack.next, element_t, list)->value;
if (descend ? (strcmp(str_head, str_stack) > 0)
: (strcmp(str_head, str_stack) < 0)) {
head_tra = head_tra->next;
} else {
stack_del = stack.next;
list_del(stack_del);
list_add(stack_del, head_tra->prev);
}
}
while (!list_empty(&stack)) {
stack_del = stack.next;
list_del(stack_del);
list_add(stack_del, head_tra->prev);
}
```
### `q_merge()`
I merge all of queue by the merging technique above. However, since I merge every queue into the first one **in order**, so the size of queues at the end of the algorithm may be quite ==unbalanced==.
```graphviz
digraph "queue"
{
rankdir = "LR"
subgraph "cluster queue 1"
{
node1_2[shape = record, label = "{<prev> prev |<next> next}"];
node1_1[shape = record, label = "head"];
node1_3[shape = record, label = "size"];
node1_4[shape = record, label = "queue 1"];
}
subgraph "cluster queue 2"
{
node2_2[shape = record, label = "{<prev> prev |<next> next}"];
node2_1[shape = record, label = "head"];
node2_3[shape = record, label = "size"];
node2_4[shape = record, label = "queue 2"];
}
subgraph "cluster queue 3"
{
node3_2[shape = record, label = "{<prev> prev |<next> next}"];
node3_1[shape = record, label = "head"];
node3_3[shape = record, label = "size"];
node3_4[shape = record, label = "queue 3"];
}
subgraph "cluster queue 4"
{
node4_2[shape = record, label = "{<prev> prev |<next> next}"];
node4_1[shape = record, label = "head"];
node4_3[shape = record, label = "size"];
node4_4[shape = record, label = "queue 1"];
}
subgraph "cluster queue 5"
{
node5_2[shape = record, label = "{<prev> prev |<next> next}"];
node5_1[shape = record, label = "head"];
node5_3[shape = record, label = "size"];
node5_4[shape = record, label = "queue 3"];
}
subgraph "cluster queue 6"
{
node6_2[shape = record, label = "{<prev> prev |<next> next}"];
node6_1[shape = record, label = "head"];
node6_3[shape = record, label = "size"];
node6_4[shape = record, label = "queue 1"];
}
node1_2:next:c -> node2_2;
node2_2:prev:c -> node1_2;
node2_2:next:c -> node3_2;
node3_2:prev:c -> node2_2;
node1_1 -> node4_1
node4_2:next:c -> node5_2;
node5_2:prev:c -> node4_2;
node4_1 -> node6_1;
}
```
```c
queue_contex_t *first = list_entry(head->next, queue_contex_t, chain);
if (head->next->next == head) {
return q_size(first->q);
}
queue_contex_t *traverse =
list_entry((first->chain).next, queue_contex_t, chain);
while (&traverse->chain != head) {
struct list_head *first_tra = first->q->next;
while (!list_empty(traverse->q) && first_tra != first->q) {
char *first_str = list_entry(first_tra, element_t, list)->value;
char *traverse_str =
list_entry(traverse->q->next, element_t, list)->value;
if (descend ? (strcmp(first_str, traverse_str) > 0)
: (strcmp(first_str, traverse_str) < 0)) {
first_tra = first_tra->next;
} else {
struct list_head *delete = traverse->q->next;
list_del(delete);
list_add(delete, first_tra->prev);
}
}
while (!list_empty(traverse->q)) {
struct list_head *delete = traverse->q->next;
list_del(delete);
list_add(delete, first_tra->prev);
}
traverse = list_entry((traverse->chain).next, queue_contex_t, chain);
}
return q_size(first->q);
```
---
## Fisher-Yates Shuffle
The [Fisher-Yates Shuffle](https://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle), also known as the Knuth Shuffle, is an algorithm used for randomly shuffling the elements of an array or a list. It was first proposed by Ronald Fisher and Frank Yates in 1938 and later popularized by Donald Knuth in his book "[The Art of Computer Programming.](https://www.amazon.com/Computer-Programming-Volumes-1-4A-Boxed/dp/0321751043)"
The basic idea of the Fisher-Yates Shuffle is to iterate through the array from the first element to the last, moving each element with a randomly selected element to the tail. This process continues until the entire array is shuffled.
```graphviz
digraph shuffle {
rankdir = "LR"
node [shape=box, style=filled]; // Setting default node properties
// Define nodes with different colors
node6_1 [label="1", fillcolor="white", color="black"];
node6_2 [label="2", fillcolor="white", color="black"];
node6_3 [label="3", fillcolor="white", color="black"];
node6_4 [label="4", fillcolor="white", color="black"];
node6_5 [label="5", fillcolor="white", color="black"];
tail5 [shape=none, label="tail", fillcolor=white];
node5_1 [label="1", fillcolor="white", color="black"];
node5_2 [label="2", fillcolor="#ffbd19", color="black"];
node5_3 [label="3", fillcolor="white", color="black"];
node5_4 [label="4", fillcolor="white", color="black"];
node5_5 [label="5", fillcolor="white", color="black"];
tail4 [shape=none, label="tail", fillcolor=white];
node4_1 [label="1", fillcolor="white", color="black"];
node4_2 [label="2", fillcolor="white", color="black"];
node4_3 [label="3", fillcolor="white", color="black"];
node4_4 [label="4", fillcolor="white", color="black"];
node4_5 [label="5", fillcolor="#ffbd19", color="black"];
tail3 [shape=none, label="tail", fillcolor=white];
node3_1 [label="1", fillcolor="white", color="black"];
node3_2 [label="2", fillcolor="white", color="black"];
node3_3 [label="3", fillcolor="lightgoldenrodyellow", color="black"];
node3_4 [label="4", fillcolor="white", color="black"];
node3_5 [label="5", fillcolor="grey", color="black"];
tail2 [shape=none, label="tail", fillcolor=white];
node2_1 [label="1", fillcolor="lightgoldenrodyellow", color="black"];
node2_2 [label="2", fillcolor="white", color="black"];
node2_3 [label="3", fillcolor="white", color="black"];
node2_4 [label="4", fillcolor="white", color="black"];
node2_5 [label="5", fillcolor="grey", color="black"];
tail1 [shape=none, label="tail", fillcolor=white];
node1_1 [label="1", fillcolor="white", color="black"];
node1_2 [label="2", fillcolor="white", color="black"];
node1_3 [label="3", fillcolor="white", color="black"];
node1_4 [label="4", fillcolor="lightgoldenrodyellow", color="black"];
node1_5 [label="5", fillcolor="grey", color="black"];
// Connect the nodes
node1_1 -> node1_2;
node1_2 -> node1_3;
node1_3 -> node1_4;
node1_4 -> node1_5;
tail1 -> node1_5;
node2_1 -> node2_2;
node2_2 -> node2_3;
node2_3 -> node2_5;
node2_5 -> node2_4;
tail2 -> node2_5;
node3_2 -> node3_3;
node3_3 -> node3_5;
node3_5 -> node3_4;
node3_4 -> node3_1;
tail3 -> node3_5;
node4_2 -> node4_5;
node4_5 -> node4_4;
node4_4 -> node4_1;
node4_1 -> node4_3;
tail4 -> node4_5;
node5_2 -> node5_4;
node5_4 -> node5_1;
node5_1 -> node5_3;
node5_3 -> node5_5;
tail5 -> node5_2;
node6_4 -> node6_1;
node6_1 -> node6_3;
node6_3 -> node6_5;
node6_5 -> node6_2;
}
```
### `q_shuffle()`
I added the `shuffle` command in `qtest.c` and implemented `q_shuffle()` using the Fisher-Yates method in `queue.c`.
```c
bool q_shuffle(struct list_head *head)
{
if (!head || list_empty(head)) {
return false;
}
/* Set seed */
srand(time(NULL));
int len = q_size(head);
while (len > 0) {
/* Select a random entry */
int number = rand() % len;
struct list_head *traverse = head;
for (; number >= 0; number--) {
traverse = traverse->next;
}
/* Move it to the tail */
list_del(traverse);
list_add_tail(traverse, head);
len--;
}
return true;
}
```
and here is the testing result in `qtest`.
```
cmd> sort
l = [alpha bravo charlie delta echo foxtrot golf hotel]
cmd> shuffle
l = [charlie alpha hotel golf bravo echo foxtrot delta]
```
### Testing the Randomness of Shuffling
To test the randomness of shuffling, I designed an experiment for the Fisher-Yates Shuffle.
Using the Fisher-Yates Shuffle, I changed the order of the string `"abc"`.
```c
/* The number of {abc, acb, bac, cab, bca, cba} */
int str_count[] = {0, 0, 0, 0, 0, 0};
```
After shuffling, I calculated the occurrences of the 6 possible different permutations. The experiment was repeated **90,000 times**.
```c
int main(int argc, char *argv[])
{
/* Initialize str to "abc" */
char str[3 + 1];
strcpy(str, "abc");
str[4 - 1] = '\0';
/* Set Seed */
srand(time(NULL));
/* Test 100000 times */
for (int i = 0; i < 90000; i++) {
str_shuffle(str);
/* Add it to the counter */
if (!strcmp(str, "abc")) {
str_count[0]++;
} else if (!strcmp(str, "acb")) {
str_count[1]++;
} else if (!strcmp(str, "bac")) {
str_count[2]++;
} else if (!strcmp(str, "cab")) {
str_count[3]++;
} else if (!strcmp(str, "bca")) {
str_count[4]++;
} else {
str_count[5]++;
}
/* reset the string */
strcpy(str, "abc");
str[4 - 1] = '\0';
}
/* Print the counter */
char *str_const[] = {"abc", "acb", "bac", "cab", "bca", "cba"};
for (int i = 0; i < 6; i++) {
printf("# of %s : %d\n", str_const[i], str_count[i]);
}
return 0;
}
```
The implementation of `shuffle()` is
```c
bool str_shuffle(char *str) {
if (!str) {
return false;
}
int len = strlen(str);
while (len > 0) {
/* Select a random character */
int random = rand() % len;
char final = str[random];
/* Move the character to the tail */
for (char traverse = str[random + 1]; traverse != '\0';
++random, traverse = str[random + 1]) {
str[random] = traverse;
}
str[random] = final;
len--;
}
return true;
}
```
and here is the result
```
# of abc : 14962
# of acb : 14999
# of bac : 14958
# of cab : 15090
# of bca : 15019
# of cba : 14972
```
![shuffle](https://hackmd.io/_uploads/SJ3sXQZCT.png)
### Chi-Square Test
In statistics, after shuffling, we hope that the probability of each permutation appearing is equal.
That is, if we denote by $X_i$ a random variable representing any possible permutation in $i$th experiment, then
$$
X_i \overset{\mathrm{iid}}{\sim} U(\alpha,\,\beta)\,.
$$
where $U(\alpha,\,\beta)\,$ is [uniform distribution](https://en.wikipedia.org/wiki/Continuous_uniform_distribution). ($\alpha$ = "abc", $\beta$ = "cba")
We use [hypothesis testing](https://en.wikipedia.org/wiki/Statistical_hypothesis_test) to verify whether if $X_i$ is uniformly distributed or not. Assume that
$$
\begin{cases}
H_0: P(X_i) = \frac{1}{6}, i = 1, 2, 3, 4, 5, 6 \\
H_a: \textrm{at least one } P(X_i) \textrm{ unequal}. \\
\end{cases}
$$
If the hypothesis $H_0$ is true (cannot be rejected), then the random variable
$$
Q_{k-1} = \sum_{1}^{k} \frac{X_i-np_{i0}}{np_{i0}}
$$
has an approximate [chi-square distribution](https://en.wikipedia.org/wiki/Chi-squared_test) with $k-1$ degrees of freedom.
> Reference: Introduction to Mathematical Statistics by Hogg, McKean and Craig 8/e P.286
Setting the significance value $\alpha = 0.05$, and $np_{i0} = nP(A_i)= 90000 \cdot \frac{1}{6} = 15000$, then
$$
\begin{split}
Q_{6-1} = Q_5 & = \frac{(14962-15000)^2}{15000} + \frac{(14999-15000)^2}{15000} + \frac{(14958-15000)^2}{15000}\\
& + \frac{(15090-15000)^2}{15000} + \frac{(15019-15000)^2}{15000} + \frac{(14972-15000)^2}{15000} \\
& \approx 0.830 \\
\end{split}
$$
![Chi-square Distribution](https://hackmd.io/_uploads/HJkdBEW06.png)
Since $\chi(X \leq 0.95\, , r = 5) \approx 11.070$, and $Q_5 = 0.830 < 11.070 = \chi(0.95, 5)$, **we cannot reject $H_0$** and hence pass chi-square test.
### Shannon Entropy
We expect that a sorted array or a list, after shuffling, will exhibit no discernible pattern in its arrangement, indicating a sufficiently high level of randomness (or ==maximum entropy==).
We first show that if $X_i \overset{\mathrm{iid}}{\sim} U(\alpha,\,\beta)\,$, then the [Shannon Entropy](https://en.wikipedia.org/wiki/Entropy_(information_theory)) reaches at maximum. Then we check whether the entropy of Fisher-Yates Shuffle is match the one of uniform distribution.
The Shannon Entropy is defined as
$$
S = - \sum_{i}^{k} p_i\log{p_i} = - \int P(x_i) \log{P(x_i)} \,dx.
$$
> Here, $\log{p_i} = \ln{p_i} = \log_{e} p_i.$
>
We first let $\phi(x) = x \log{x},$ then $S(p) = - \displaystyle\sum_{i}^{k} \phi(p_i), i = 1, 2, ... k$. Since
$$
\frac{d^2\phi(x)}{dx^2} = \frac{d}{dx}\left[\log{x}+ 1\right] = \frac{1}{x} > 0,
$$
by [Jensen's Inequality](https://en.wikipedia.org/wiki/Jensen%27s_inequality), for each convex function,
$$
\begin{split}
& \phi\left(\frac{\sum p_i}{k}\right) \leq \frac{\sum \phi(p_i)}{k} \\
\Rightarrow & \phi\left(\frac{\sum p_i}{k}\right) \leq -\frac{S(p)}{k}\\
\Rightarrow & S(p) \leq -k \cdot\phi\left(\frac{\sum p_i}{k}\right) = -k \left(\frac{\sum p_i}{k}\right) \log{\left(\frac{\sum p_i}{k}\right)}. \\
\end{split}
$$
Since $\sum p_i = 1$, it follows that
$$
S(p) \leq -\log{\left(\frac{1}{k}\right)} = \log{k}, \textrm{ where } k \in \textrm{constant}.
$$
And note that in uniform distribution,
$$
S = \sum_{i}^{k} p_i\log{p_i} = - \frac{1}{k} \sum_{i}^{k} \log{\left(\frac{1}{k}
\right)} = - \log{\left(\frac{1}{k}\right)} = \log{k},
$$
**Therefore, under uniform distribution, the entropy is at its maximum value.**
> Reference:
[Proving that Shannon Entropy is maximal for the uniform distribution using convexity.
](https://math.stackexchange.com/questions/2748388/proving-that-shannon-entropy-is-maximal-for-the-uniform-distribution-using-conve)
In our experiment, since there are six possible string permutations, the probability of each string appearing under a uniform distribution is $P(X_i) = \displaystyle\frac{1}{6}$, so the ideal entropy is
$$
S_i = - \sum_{i = 1}^{6} \frac{1}{6} \log{\frac{1}{6}} = \log{6} \approx 1.791759
$$
The experimental entropy is
$$
\begin{split}
S_e & = - \left[\frac{14962}{90000} \log{\left(\frac{14962}{90000}\right)} + \frac{14999}{90000} \log{\left(\frac{14999}{90000}\right)}+ \frac{14958}{90000} \log{\left(\frac{14958}{90000}\right)}\right] \\
& - \left[\frac{15090}{90000} \log{\left(\frac{15090}{90000}\right)} + \frac{15019}{90000} \log{\left(\frac{15019}{90000}\right)}+ \frac{14972}{90000} \log{\left(\frac{14972}{90000}\right)}\right]\\
& = 0.895166 + 896590 \approx 1.791756 \\
\end{split}
$$
Which is very close to $S_i$.
---
## Tic-tac-toe Incorporation
### Commands
To implement `ttt` command into `qtest`, I have done some updates and modifications in the branch of project and [qtest.c](https://github.com/YangYeh-PD/lab0-c/blob/master/qtest.c) respectively, including
Add `ADD_COMMAND()` in `console_init()`.
```diff
static void console_init()
{
...
+ ADD_COMMAND(ttt, "Activate tic-tac-toe", "");
...
}
```
`ADD_COMMAND()` is a macro which defined in [console.h](https://github.com/YangYeh-PD/lab0-c/blob/master/console.h).
```c
void add_cmd(char *name, cmd_func_t operation, char *summary, char *parameter);
#define ADD_COMMAND(cmd, msg, param) add_cmd(#cmd, do_##cmd, msg, param)
```
Obviously, `add_cmd()` helps us add commands into `qtest`.
Create `do_ttt()` function.
```c
static bool do_ttt();
```
Just copy the [main](https://github.com/jserv/ttt/blob/main/main.c) function into `do_ttt`, then add functions definition of
* `record_move()`
* `print_moves()`
* `get_input()`
above `do_ttt()`, or it will cause ==implicit function definition==.
Add all of relevant files
Here is the structure of my current project.
```
$ tree
.
├── console.c
├── console.h
├── CONTRIBUTING.md
├── dudect
│ ├── constant.c
│ ...
├── game.c
├── game.h
...
├── Makefile
├── mcts.c
├── mcts.h
├── mt19937-64.c
├── mt19937-64.h
├── negamax.c
├── negamax.h
├── qtest
├── qtest.c
...
├── scripts
│ ├── aspell-pws
│ ...
├── shannon_entropy.c
├── traces
│ ├── trace-01-ops.cmd
│ ...
├── util.h
├── web.c
├── web.h
├── zobrist.c
├── zobrist.h
```
> I tried to include `mcts.[ch]`, `negamax.[ch]` and `util.h` into my own created directory `agents`, however, it didn't work with makefile, as now I save them in the same hierarchy.[name=ChenYang Yeh] [time=Fri, Mar 8, 2024 3:17 AM]
Modify `Makefile`
Since all of files dependency and compile message has been constructed in original `Makefile`, all we need to do is add `.o` file in `OBJS`.
```diff
OBJS := qtest.o report.o console.o harness.o queue.o \
...
+ zobrist.o negamax.o mcts.o mt19937-64.o game.o \
```

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