# 2024q1 Homework1 (lab0) 執行人: popo8712 專題解說錄影 contributed by < `cheezad` > ### Reviewed by `ChenFuhuangKye` 1. 可以在開發過程中注記 `git commmit` 資訊。 ## 開發環境 ```shell $ gcc --version gcc (Ubuntu 11.4.0-1ubuntu1~22.04) 11.4.0 $ lscpu 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): 12 On-line CPU(s) list: 0-11 Vendor ID: GenuineIntel Model name: Intel(R) Core(TM) i7-9750H CPU @ 2.60GHz CPU family: 6 Model: 158 Thread(s) per core: 2 Core(s) per socket: 6 Socket(s): 1 Stepping: 10 CPU max MHz: 4500,0000 CPU min MHz: 800,0000 BogoMIPS: 5199.98 ``` ## 開發過程 ### lab0-c #### q_new() 原本的程式碼在 `malloc(sizeof(struct list_head))` 之後並沒有檢查是否 `malloc` 成功,在老師提醒之後有加上檢查。 ```diff struct list_head *q = malloc(sizeof(struct list_head)); + if(!q) + return NULL; ``` #### 了解 lab0 裡面的資料結構 <s>第一次</s> 接觸這類型的程式,所以做了一點整理。 :::danger 不用強調「第一次」,在資訊科技領域,每天都少不了「第一次」,若沒有遇到這樣的事,你可能不在這領域。 ::: list_head : 雙向鏈結串列 `prev` : 指向上一個節點 `next` : 指向下一個節點 如果是空的鏈結,則head指向自己 ```c struct list_head { struct list_head *prev; struct list_head *next; }; ``` :::danger 避免非必要的項目縮排 (即 `* ` 和 `- `),以清晰、明確,且流暢的漢語書寫。 ::: `element_t` : 取得該位置的值 `value` : 取得該節點的值 `list` : 雙向鏈結的節點 ``` typedef struct { char *value; struct list_head list; } element_t; ``` `queue_contex_t` `q` : 指向序列的head `chain` : 串起不同鏈結的head `size` : 序列的長度 `id` : 序列的編號 ```c typedef struct { struct list_head *q; struct list_head chain; int size; int id; } queue_contex_t; ``` #### [Linux Kernel API](https://www.kernel.org/doc/html/latest/core-api/kernel-api.html) 透過這些已經寫好的程式碼讓自己寫的程式更精簡。 :::danger 避免非必要的項目縮排 (即 `* ` 和 `- `),以清晰、明確,且流暢的漢語書寫。 ::: 範例:`list_splice` - 把`list` 接到 `head`前面 - 結果就是 `head` 後面先接 `list` ,再接 `head` 自己的節點 :::danger 改進你的漢語表達。 ::: ```c static inline void list_splice(struct list_head *list, struct list_head *head) { struct list_head *head_first = head->next; struct list_head *list_first = list->next; struct list_head *list_last = list->prev; if (list_empty(list)) return; head->next = list_first; list_first->prev = head; list_last->next = head_first; head_first->prev = list_last; } ``` --- ## 整合網頁伺服器 執行 `qtest` 裡面的 `web` 之後,在瀏覽器打 `http://<myip>:9999/new` ,<s>在未修改任何網頁相關程式碼之前</s> 更動網頁相關的程式碼之前,執行結果如下 ``` $ ./qtest cmd> web listen on port 9999, fd is 3 cmd> l = [] cmd> Unknown command 'favicon.ico' ``` 當在還沒執行 web 之前,按下 `tab` 會自動補齊剩下的字,但在開啟 web 之後,這個功能便會消失。 --- ## `lib/list_sort.c` Study - [lib/list_sort.c](https://github.com/torvalds/linux/blob/master/lib/list_sort.c) code ### Research Paper #### Golin, M., Sedgewick, R. (1993) Queue-mergesort In a list of queue Q 1. Pick out the first two list 2. Merge them together 3. Put the result at the end (tail) of the queue 4. Repeat until there is only one list left in the queue Queue Q ```graphviz digraph G { node[shape=plaintext]; Head; Tail; node[shape=box]; ele1[label=1]; ele2[label=1]; ele3[label=1]; ele4[label=1]; ele5[label=1]; { rank="same" ele1->ele2->ele3->ele4->ele5; } Head->ele1; Tail->ele5; node[shape=box]; ele1n[label=D]; ele2n[label=A]; ele3n[label=B]; ele4n[label=C]; ele5n[label=F]; { ele1->ele1n; ele2->ele2n; ele3->ele3n; ele4->ele4n; ele5->ele5n; } } ``` Pick out the first two list and merge them together ```graphviz digraph G1 { node[shape=plaintext]; Head; Tail; node[shape=box]; ele1[label=2]; ele2[label=1]; ele3[label=1]; ele4[label=1]; { rank="same" ele1->ele2->ele3->ele4; } Head->ele1; Tail->ele4; node[shape=box]; ele1n[label=D]; ele2n[label=A]; ele3n[label=B]; ele4n[label=C]; ele5n[label=F]; { ele1->ele2n->ele1n; ele2->ele3n; ele3->ele4n; ele4->ele5n; } } ``` Move it to the end of the list ```graphviz digraph G1 { node[shape=plaintext]; Head; Tail; node[shape=box]; ele1[label=2]; ele2[label=1]; ele3[label=1]; ele4[label=1]; { rank="same" ele2->ele3->ele4->ele1; } Head->ele2; Tail->ele1; node[shape=box]; ele1n[label=D]; ele2n[label=A]; ele3n[label=B]; ele4n[label=C]; ele5n[label=F]; { ele1->ele2n->ele1n; ele2->ele3n; ele3->ele4n; ele4->ele5n; } } ``` Repeat until there is only one list left in the queue ```graphviz digraph G1 { node[shape=plaintext]; Head; Tail; node[shape=box]; ele1[label=5]; Head->ele1; Tail->ele1; node[shape=box]; ele1n[label=D]; ele2n[label=A]; ele3n[label=B]; ele4n[label=C]; ele5n[label=F]; { ele1->ele2n->ele3n->ele4n->ele1n->ele5n; } } ``` Prove that queue-mergesort is optimal mergesort by checking number of the comparisons performed in worse case is no more then that performed by any mergesort. Define the weight of a binary tree T to be w(T) where the value is the sum of all internal nodes' weight of the tree. Merge tree T for queue-mergesort run on 11 elements ```graphviz digraph B{ node[shape=circle]; ele1[label=11]; ele2[label=7]; ele3[label=4]; ele4[label=4]; ele5[label=3]; ele6[label=2]; ele7[label=2]; ele8[label=2]; ele9[label=2]; ele10[label=2]; ele11[label=1]; ele12[label=1]; ele13[label=1]; ele14[label=1]; ele15[label=1]; ele16[label=1]; ele17[label=1]; ele18[label=1]; ele19[label=1]; ele20[label=1]; ele21[label=1]; { ele1->ele2; ele1->ele3; ele2->ele4; ele2->ele5; ele3->ele6; ele3->ele7; ele4->ele8; ele4->ele9; ele5->ele10; ele5->ele11; ele6->ele12; ele6->ele13; ele7->ele14; ele7->ele15; ele8->ele16; ele8->ele17; ele9->ele18; ele9->ele19; ele10->ele20; ele10->ele21; } } ``` After adding up the weight of merge tree T's internal nodes, w(T) = 39. Queue-mergesort is optimal mergesort -> w(T) <= w(T’) for all merge trees T’ with n leaves. Queue-mergesort and Top-down mergesort are both optimal mergesort, but bottom-up mergesort isn't. However, with a simple modification, we can make bottom-up mergesort optimal. At the end of a pass, in case there are an odd number of lists, we do not leave the last list by itself but merge it with the new list that was just created. By doing so, the merge-tree created becomes optimal. #### Bottom-up mergesort -- A detailed analysis Bottom-up Mergesort like the top-down variant follows the divide-and conquer paradigm. However, bottom-up uses "power of two" splitting strategy instead of "half and half" strategy in top-down variant, making it possible to be written in recursive. - Top-down mergesort and bottom-up mergesort has the exact same number of comparisons in base case. - Top-down has the best worst-case behavior among all possible splitting strategies. --- ## dudect Paper and Random Number - 〈[Dude, is my code constant time?](https://eprint.iacr.org/2016/1123.pdf)〉 - [dudect GitHub](https://github.com/oreparaz/dudect) ### 〈Dude, is my code constant time?〉 Time attack is a broad class of side-channel attacks that measure execution time of cryptographic implementation in order to infer secrets. People use tools to collect time attack information. Traditionally these tools usually rely on details of the hardware when CPU manufactures publish little information on, and may be subject to change on an update. Dedect is a tool that collects time attack information and checks whether a piece of code runs in constant-time or not. They use statistical analysis of execution timing based on leakage detection tests to determine whether the code runs in constant-time. Steps of their experiment are as following: 1. Collect data using cycle counters or timers. 2. Post-processing is applied, like cropping out the measurements that are interrupted by the OS. 3. A statistical test is performed to try to disprove the null hypothesis. #### Student's t-test / Welch's t-test t-test is a statistical test used to determine whether the difference between two groups is statistically significant or not. Normally used under the testing of null hypothesis. t-test requires two independent and random samples that have normal distributions. The difference between Student's t-test and Welch's t-test: - Student's t-test requires equal population variances and equal sample sizes - Welch's t-test can be used when two groups don't have equal sample variance or equal sample sizes Welch's t-test >General Form > $TestStat = {Estimate- ValueWe Hypothesize\over StandardError}$ $$ t= {(\bar x_A - \bar x_B)-(\mu_A - \mu_B)\over \sqrt{{s^2_A\over n_A} + {s^2_B\over n_B}}} $$ $\bar x$ : mean of group $\mu$ : difference we hypothesize for true population means (normally the result of subtraction is zero) $S$ : standard deviation $n$ : sample number Source: [Welch's t-test](https://sites.nicholas.duke.edu/statsreview/means/welch/) ### Processing data #### Setting threshold for cropping measurements in dudect.c ``` c static void prepare_percentiles(dudect_ctx_t *ctx) { qsort(ctx->exec_times, ctx->config->number_measurements, sizeof(int64_t), (int (*)(const void *, const void *))cmp); for (size_t i = 0; i < DUDECT_NUMBER_PERCENTILES; i++) { ctx->percentiles[i] = percentile( ctx->exec_times, 1 - (pow(0.5, 10 * (double)(i + 1) / DUDECT_NUMBER_PERCENTILES)), ctx->config->number_measurements); } } ``` This part of the code uses the function `percentile` ```c static int64_t percentile(int64_t *a_sorted, double which, size_t size) { size_t array_position = (size_t)((double)size * (double)which); assert(array_position < size); return a_sorted[array_position]; } ``` :::info Couldn't understand why `1 - (pow(0.5, 10 * (double)(i + 1) / DUDECT_NUMBER_PERCENTILES))` is the second parameter of `percentile` ::: ~~#### 〈Dude, is my code constant time?〉論文整理 dudect 運用定時攻擊的概念以及評估程式碼在特定平台上是否以常數時間運行的需求。現有工具檢測可變時間程式碼有缺點,因此作者利用洩漏檢測技術將dudect呈現為一種簡潔、易於使用和易於維護的工具。此外,它強調了與先前解決方案相比的方法差異,特別是不需要建立硬體行為模型。~~ :::danger 你忽略過多關鍵訊息,例如[常數時間的判定和硬體指令有關](https://www.facebook.com/groups/system.software2024/posts/1568358687275357/)。 ::: #### 程式碼改進 經過與 GitHub 上原始程式碼與 lab0-c 的程式碼比對之後可以發現程式之所以不能通過 `trace-17-complexity` 是因為他少了 `int64_t *percentiles` ,`percentiles` 在這裡的作用是把例外的數值去除。 :::danger 無論標題和內文中,中文和英文字元之間要有空白字元 (對排版和文字搜尋有利) ::: ### Random number #### Fisher-Yates Shuffle ![Durstenfeld_shuffle](https://hackmd.io/_uploads/HysuRgiAT.svg) Fisher-Yates Shuffle is an algorithm that randomizes a given sequence. Steps of this algorithm: 1. Get the number of cells in the sequence `n` 2. Loop from `n - 1` to `1` in a for loop with `i` 3. In each for loop, randomly pick a number from `0` to `i` , and swap it with the `i` th cell. 4. Repeat until `i` equals `1` > Question: How do we know that this sequence is uniformly distributed? How do we know it is random? > Answer: If an event is uniformly distributed, the frequency of every occasion happening is equal, and independent, we know that it is random. As for verifying uniform distribution, we can use Pearson's chi-squared test to verify null hypothesis. ##### Pearson's chi-sqared test Formula for Chi-Square $$ X_c^2 = \sum_{i=1}^n{(O_i - E_i)^2 \over E_i} $$ X~c~ : chi-square O~i~ : Observed value E~c~ : Expected value c : Degree of freedom Chi-square statistic is a test that measures how a model compares to actual observed data. It compares the size of any discrepancies between the expected results and the actual results. In this case, it compares the number of times a sequence appears to the uniform distributed number. ##### Null Hypothesis Null hypothesis is used to check if there is any relationship between two sets of data being analyzed. If null hypothesis returns true, it means that the datas are effected by chance alone. ##### Degrees of Freedom Degrees of freedom are the number of independent variables that can tell you how many items can be randomly selected before constraints must be put in place. In a data set, some initial numbers can be chosen at random, but when the data must meet a specific sum or mean, the data after the degrees of freedom need to be constrained to meet the set requirement. When it comes to verifying whether the data set accepts or rejects the null hypothesis, we set a significance level (⍺). If the null hypothesis is true, the error of the data would be less than the ⍺. Commonly, ⍺ is set to 0.05. We can check the matching chi-squared value matched by the degree of freedom and probability. With that value, we can evaluate whether X~c~^2^ accepts or rejects the null hypothesis. #### Randomness ##### Understand Entropy Entropy is the measurement of uncertainty or disorder of system. Shannon Entropy measures the uncertainty of a probability distribution. Shannon's entropy formula: $$ H(⍺):= \sum_{i=1}^{n}{p_i}log {1\over p_i} $$ H(⍺) can be understood as the "the amount of information" in a random message ⍺. We use the ideal entropy to compare to the entropy from our data, then calculate its chi-square value and check if it fits the null hypothesis. When it rejects the null hypothesis, it doesn't mean "H~0~ is true", it means "there is no evidence showing that H~1~ is false". (H~0~ here means it is uniformly distributed, H~1~ means it isn't uniformly distributed) Resource: [Shannon's Entropy](https://www.lirmm.fr/~romashchen/cours/crypto2022/crypto2022-lecture3.pdf), [Understanding random generators](https://www.redhat.com/en/blog/understanding-random-number-generators-and-their-limitations-linux) ## ttt ### Monte Carlo tree search (MCTS) Monte Carlo tree search is a heuristic search algorithm used on AI to play games. Heuristic means that it can proceed to a solution by tiral and error or by rules that are only loosely defined. MCTS can be stopped at any step and it can also find the optimal solution based on a partially constructed game tree. During the game, MCTS keeps track of the winning count and the total game count of each step, then applies Upper Confidence Bound (UCB) algorithm to the node. UCB is a function that is used to determine which step should be picked next. At first, the value of the UCB would be infinite, but as time progresses, the value would drop, leading to the node to pick other possible options. This way, it balances the exploration/exploitation of the algorithm. The formula of UCB is as below $$ {w_i\over n_i} + c{\sqrt{ln\ t \over n_i}} $$ w : number of the wins simulated over the i-th move at that node n : number of simulation after the i-th move at that node c : exploration parameter, normally using $\sqrt{2}$ t : total number of simulation done after the i-th move. In MCTS there are four basic steps: 1. selection : picking out the largest UCB node until get to root 2. expansion : add extra nodes to the tree 3. simulation : do random simulation to find who wins 4. back propogate : take the value from simulation and update the tree above In simulation, random moves are made. Meaning that it needs a random generator that is random enough, or else, it would lead to uneven exploration/exploitation or undiscovered path. Here I will provide an example 1. Selection: As it is the start of the game, it would select the parent node which is not making any move. ![Screenshot 2024-07-05 at 12.27.17 AM](https://hackmd.io/_uploads/H17u6rEPC.png) 2. Expansion: It expands a node. ![Screenshot 2024-07-05 at 12.29.06 AM](https://hackmd.io/_uploads/Bye10H4DR.png) 3. Simulation: MCTS simulates until there is a winner. ![Screenshot 2024-07-05 at 12.30.39 AM](https://hackmd.io/_uploads/BJyrRrVwC.png) 4. Back Propogate: It updates the values of its parent nodes - updates the parent node ![Screenshot 2024-07-05 at 12.32.27 AM](https://hackmd.io/_uploads/ryFsCB4w0.png) - updates the root node ![Screenshot 2024-07-05 at 12.32.58 AM](https://hackmd.io/_uploads/HycpRB4PA.png) 5. Repeat until the designated simulation iteration is completed. >The pictures above are screenshots from [this demo page](https://vgarciasc.github.io/mcts-viz/), for more iterations, please visit this page. ### Zobrist Hashing Zobrist Hashing is a hash function that represents a board game as hash values. It prevents the same board position to be analyzed more than once. For example, when MCTS is simulating, it checks if the position is already been played using a [transposition table](https://en.wikipedia.org/wiki/Transposition_table) without re-analyzing the board position each time. How it works: Zobrist hashing first generate keys, then use the keys to hash. For every available board cell and every possible game character, there is a random number generated as the key. Note that it is possible to have two positions with the same value, but it is very unlikely, thus it out weights the disadvantages. For example in a tic-tac-toe, there would be two players and 9 positions, meaning that there would be $2*9$ keys in total. The hash is calculated with `xor` operation. The previously generated number of the character at a certain position is combined through the `xor` operation. To undo the previous move, simply `xor` the value again. [Youtube: Transposition Tables and Zobrist Keys](https://www.youtube.com/watch?v=QYNRvMolN20) ### Implementing Timer that refreshes real time #### Ideas To use signal/setjmp/longjmp to keep calling the function while ttt is still in play. - Signal use `SIGALRM` to create preemptive scheduling. - setjmp / longjmp doesn't seem ideal in this situation. Wrote a program using signal that prints out the time every second ```c #include <stdio.h> #include <signal.h> #include <unistd.h> #include <time.h> time_t rawtime; struct tm * timeinfo; void display_message(int s){ time( &rawtime ); timeinfo = localtime( &rawtime ); printf( "Current local time and date: %s", asctime (timeinfo) ); signal(SIGALRM, display_message); alarm(1); } int main(void){ signal(SIGALRM, display_message); alarm(1); while(1){} return 0; } ``` #### Links related to this implementation: [Get the current time](https://stackoverflow.com/questions/5141960/get-the-current-time-in-c) [Refreshing program real time (Last answer)](https://stackoverflow.com/questions/18458064/how-to-refresh-the-screen-continuously-and-update-it-in-real-time) [SIGALRM to print time](https://stackoverflow.com/questions/21542077/c-sigalrm-alarm-to-display-message-every-second) [Signal Geeks](https://www.geeksforgeeks.org/signals-c-language/) [Jserv concurrency](https://hackmd.io/@sysprog/concurrency-sched) [preempt_sched GitHub](https://github.com/sysprog21/concurrent-programs/blob/master/preempt_sched/task_sched.c) ### Problems I encountered When I added `agents/mcts.c` to `lab0-c`, I only added `agents/mcts.o` under the Makefile `OBJS`. Not specifying `@mkdir -p .$(AGENTS_DIR)` under `%.o: %.c` prevents the makefile from finding `agents/mcts.c`. Leading to this error: ``` cheezad@cheezad-ubuntu:~/linux2024/lab0-c$ make CC agents/mcts.o agents/mcts.c:150:1: fatal error: opening dependency file .agents/mcts.o.d: No such file or directory 150 | } | ^ compilation terminated. make: *** [Makefile:58: agents/mcts.o] Error 1 ``` <Solution> This is caused by not knowing Makefile well enough. I will later dive deeper into how makefile works on my own. --- In fixed point numbers, how many bits should be fractional bits and how many should be integral bits? I am currently working on ttt to change `shannon_entropy.c` and `log2_lshift16.h` to fixed point, but I am not sure whether the bits of fraction and integer should be self defined or predefined. <Solution> Check how percise `shannon_entropy.c` needs to be and decide the bits. --- After two weeks of trying to understand `shannon_entropy.c` and `log2_lshift16.h`, still don't really understand how the mathematical part work.