為一描述所有可計算的函式的數學模型，也就是此模型可計算所有演算法上可計算的問題。 提出此模型當初是用來證明沒有有效的通用解法來解決 Entscheidungsproblem，也就是 decision problem。 On Computable Numbers, with an Application to the Entscheidungsproblem Deterministic Turing Machine (DTM) DTM 由一個 7 tuple 所定義 $$(Q, \Sigma, \Gamma, \delta, p_0, b, F)$$其中這 7 tuples 的定義為

古代希臘哲學 宇宙論時期 人事論時期 柏拉圖 亞里斯多德 希臘化時期

機率模型 什麼是機率模型? 結果不確定的實驗、現象、情況的量化 建立機率模型的兩步驟 描述可能的實驗結果 --- 樣本空間 (sample space) 樣本空間 (sample space): 可能的結果的集合，用 $\Omega$ 表示。 $\Omega$ 必須符合:Mutually exclusive: 一次只會出現一個結果 Collectively exhaustive: 所有結果的機率總合為 1 (i.e. 不會出現不再 $\Omega$ 的結果) At the right granularity: 用相關的事件當 $\Omega$

希臘自然哲學發展至西元前 5 世紀中葉，已逐漸喪失獨創性，大多重複先前已經定型的自然哲學理論。哲學思維轉向語言、文法、修辭、知識、法律、歷史、習俗、倫理、宗教、音樂等人存在本身的問題。 詭智學派 (辯士學派) (Sophist) 世襲制度結束，民主制度對於法律、演講等的需要，使得詭智學派的興起。其主要在文法修辭討論，為教師教人如何在法庭勝訴，在政壇上發揮雄辯才華。 詭智學派一方面想反對威權式的規範性概念，另一方面還要接受道德的底線，所以衍伸出道德的直覺。他們在打破傳統，並著重在主觀權力之中，但抹煞客觀的權力。 詭智學派是古希臘地一個懷疑主義，懷疑主義是哲學討論的重要核心。 宇宙論時期哲學家在探討知識的價值時有各式不同的觀點，有的是唯物主義、有的是變動等等。這衍伸出一個問題: 同樣的知識同樣的真理卻有不同的觀點，因此有人開始探討最核心的問題不是知識到底是什麼，而是知識本身到底有沒有固定的價值。詭智學派開始問說到底有沒有知識，如果有那為什麼答案一大堆，如果沒有那我們說什麼知識就是什麼 -- 【知識的本身是空的，一切都是說出來的。】

LeetCode 116 -- Populating Next Right Pointers in Each Node 題目描述 You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition: struct Node { int val; Node *left; Node *right; Node *next; }

在古埃及時代就有機率的遊戲，但到了 17 世紀才有人敢挑戰亞里斯多德的觀點 --- 『There can be no demonstrative knowledge of chance.』 :::info Demonstrative knowledge is when we perceive the agreement or disagreement indirectly through a series of intermediate ideas. For example, I know that A is greater than B and B is greater than C, thus I know demonstratively that A is greater than C. ::: 為什麼可靠的機率理論直到最近才出現? 機率通常是如此的違反我們的直覺，舉個例子:

Linked List 24. Swap Nodes in Pairs 61. Rotate List 82. Remove Duplicates from Sorted List II 86. Partition List 138. Copy List with Random Pointer 141. Linked List Cycle 206. Reverse Linked List

LeetCode 112 -- Path Sum 題目描述 Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum. A leaf is a node with no children. Example 1. Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true

LeetCode 129 -- Sum Root to Leaf Numbers 題目描述 You are given the root of a binary tree containing digits from 0 to 9 only. Each root-to-leaf path in the tree represents a number. For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123. Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.

這篇我們要講解用 RTL 實作一個轉換器將 one-hot code 轉成 binary number。 Definition 首先先定義這個轉換器的 interface，主要就是輸入一個 ONEHOT_WIDTH 長度的 one-hot code onehot_i (onehot_i 一定會有一個 bit 是 1) 然後輸出轉換後的 binary number binary_o，它的 bit 數為 BIN_WIDTH。 module onehot2bin_v1 #( parameter ONEHOT_WIDTH = 16, parameter BIN_WIDTH = ONEHOT_WIDTH == 1 ? 1 : $clog2(ONEHOT_WIDTH) )( input [ONEHOT_WIDTH-1 : 0] onehot_i,

AXI Architecture AXI 是 burst-based 的 protocol 其中有五個 channels Write Address Channel Write Data Channel Write Response Channel Read Address Channel Read Data Channel Read and Write Transactions

LeetCode 114 -- Flatten Binary Tree to Linked List 題目描述 Given the root of a binary tree, flatten the tree into a "linked list": The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null. The "linked list" should be in the same order as a pre-order traversal of the binary tree. Example. Input: root = [1,2,5,3,4,null,6]

LeetCode 133 -- Clone Graph 題目描述 Given a reference of a node in a connected undirected graph. Return a deep copy (clone) of the graph. Each node in the graph contains a value (int) and a list (List[Node]) of its neighbors. class Node { public int val;

LeetCode 99 -- Recover Binary Search Tree 題目描述 You are given the root of a binary search tree (BST), where exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure. Example 1. Input: root = [1,3,null,null,2] Output: [3,1,null,null,2] Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.

LeetCode 131 -- Palindrome Partitioning 題目描述 Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. A palindrome string is a string that reads the same backward as forward. Example 1: Input: s = "aab" Output: [["a","a","b"],["aa","b"]]

條件機率問題 Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers? 從題目描述，我們列出幾個事件: $A$: 至少骰出 1 個 6 $B$: 兩個骰子的值不一樣 $C$: 兩個骰子都骰到 6 $D$: 只有一個骰子是 6 並且我們可以知道題目要問的是當 $B$ 發生時 $A$ 發生的機率，也就是 $P(A|B)$。我們先得出 $A$ 的機率為$$P(A) = P(C) + P(D) = \frac{1}{36} + \frac{10}{36} = \frac{11}{36}$$ $B$ 的機率為$$P(B)=\frac{6\cdot 5}{6\cdot 6}=\frac{5}{6}$$而同時發生 $A$ 跟 $B$ 的機率為 $$P(A\cap B)=P(A)-P(C)=\frac{10}{36}$$所以 $$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{1}{3}$$

LeetCode 199 -- Binary Tree Right Side View 題目描述 Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. Example 1. Input: root = [1,2,3,null,5,null,4] Output: [1,3,4] 本題的目的是要找出 tree 的每一層最右邊的 node，但這並不代表就是從 root 開始的每一個 right child，因為並不一定向上面範例的 tree 從 root 開始都有 right child 存在。

LeetCode 108 - Convert Sorted Array to Binary Search Tree 題目描述 Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree. A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one. 本題的目的是要將一個由小到大的數列轉換成 Binary Search Tree (BST)。BST 的定義是 tree 中的每一個 node 其左 subtree 的所有 nodes 都要比它小，其右 subtree 的所有 node 都要比它大，同時它的左右 subtree 也要是 BST。 另外，其實只要把 BST 壓扁來看也就是一個由小到大的數列。

LeetCode 98 -- Validate Binary Search Tree 題目描述 Given the root of a binary tree, determine if it is a valid binary search tree (BST). A valid BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys greater than the node's key. Both the left and right subtrees must also be binary search trees.

LeetCode 100 -- Same Tree 題目描述 Given the roots of two binary trees p and q, write a function to check if they are the same or not. Two binary trees are considered the same if they are structurally identical, and the nodes have the same value. Example 1. Input: p = [1,2,3], q = [1,2,3] Output: true