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2025 on Tumblr: Trends That Defined the Year

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noise dept.

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@compute-info

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Typical job.
Class Envy
Android vs. iPhone
Why you shouldn’t interrupt a programmer.

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Growth of Functions
- Θ notation -
f(n) = Θ(g(n))
There exist positive constants c1, c2, and n0 such that         Â
0 ≤ c1g(n) ≤ f(n) ≤ c2g(n) for all n ≥ n0.
We say g(n) is an asymptotically tight bound for f(n) if f(n) = Θ(g(n)). Informally we can say, as far as time complexity or order of growth, these functions are equal.
- O notation -
f(n) = O(g(n)) There exist positive constants c and n0 such that
0 ≤ f(n) ≤ cg(n) for all n ≥ n0.
We say g(n) is an asymptotic upper bound for f(n) if f(n) = O(g(n)).
- Ω notation -
f(n) = Ω(g(n)) There exist positive constants c and n0 such that          Â
0 ≤ cg(n) ≤ f(n) for all n ≥ n0.
We say g(n) is an asymptotic lower bound for f(n) if f(n) = Ω(g(n)).
- o notation -
f(n) = o(g(n)) For any positive constant c > 0, there exists a constant          Â
n0 > 0 such that 0 ≤ f(n) < cg(n) for all n ≥ n0.
We say f(n) is asymptotically smaller than g(n) if f(n) = o(g(n)). We can also express this as the following limit
limn→∞(f(n)/g(n)) = 0
- ω notation -
f(n) = ω(g(n)) For any positive constant c > 0, there exists a constant          Â
n0 > 0 such that 0 ≤ cg(n) < f(n) for all n ≥ n0.
We say f(n) is asymptotically bigger than g(n) if f(n) = ω(g(n)). We can also express this as the following limit.
limn→∞(f(n)/g(n)) = ∞
- Properties -
Transitivity:
  f(n) = Θ(g(n)) and g(n) = Θ(h(n)) ⇒ f(n) = Θ(h(n))   f(n) = O(g(n)) and g(n) = O(h(n)) ⇒ f(n) = O(h(n))   f(n) = Ω(g(n)) and g(n) = Ω(h(n)) ⇒ f(n) = Ω(h(n))   f(n) = o(g(n)) and g(n) = o(h(n)) ⇒ f(n) = o(h(n))   f(n) = ω(g(n)) and g(n) = ω(h(n)) ⇒ f(n) = ω(h(n))
Reflexivity:
 f(n) = Θ(f(n))   f(n) = O(f(n))   f(n) = Ω(f(n))
Symmetry:
  f(n) = Θ(g(n)) iff g(n) = Θ(f(n))
Transpose symmetry:
 f(n) = O(g(n)) iff g(n) = Ω(f(n))   f(n) = o(g(n)) iff g(n) = ω(f(n))
- Informal Comparison -
  f(n) = O(g(n)) ≈ a ≤ b   f(n) = Ω(g(n)) ≈ a ≥ b   f(n) = Θ(g(n)) ≈ a = b   f(n) = o(g(n)) ≈ a < b   f(n) = ω(g(n)) ≈ a > b
- Note -
From the definitions of Θ, O, and Ω notations we can conclude the following:   * For any two functions f(n) and g(n), we have f(n) = Θ(g(n)) iff f(n) = O(g(n)) and f(n) = Ω(g(n)).   * If f(n) = O(g(n)) then g(n) = Ω(f(n)) and vice versa.
Powershell TCP Listener
tcplistener.ps1:
function tcplisten ($port) {    $endpoint = new-object System.Net.IPEndPoint ([ipaddress]::any, $port)    $listener = new-object System.Net.Sockets.TcpListener $endpoint    $listener.start()    $listener.AcceptTcpClient()    $listener.stop() } tcplisten($args[0])
Extract Hostname of Machine in KD
Extract hostname of the machine being debugged either in live KD or from a kernel dump.
1: kd> x srv!SrvComputerName fffff800`19851f98 srv!SrvComputerName = struct _UNICODE_STRING "TEST1"
Windows Settings
Make a folder with the following name to access all available settings on Windows 8.x.
Settings.{ED7BA470-8E54-465E-825C-99712043E01C}

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The Corporate Ladder
Test Remote Port
Powershell one-liners to see if a remote port is open. The target port is open if the command returns with no errors.
#Check TCP port (New-Object Net.Sockets.TcpClient).Connect("remote_machine", port) #Check UDP port (New-Object Net.Sockets.UdpClient).Connect("remote_machine", port)

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Binary Tree to Array Conversion
Given a binary tree, convert it to an array and vice versa.
int *BinaryTreeToArray(Node *root) {     int *arr = new int[8];     // Skipping index 0 for simplicity     BinaryTreeToArrayHelper(root, arr, 8, 1);     return arr; } void BinaryTreeToArrayHelper(Node *n, int *arr, int size, int index) {     if (!n || index > size - 1)         return;     arr[index] = n->value;     BinaryTreeToArrayHelper(n->left, arr, size, index * 2);     BinaryTreeToArrayHelper(n->right, arr, size, index * 2 + 1); }
Node *ArrayToBinaryTree(int *arr, int size) {     if (!arr || 1 > size)         return nullptr;     Node *root = new Node(arr[1]), *curr = nullptr;     queue<Node*> q;     q.push(root);     int index = 1;     while (!q.empty())     {         curr = q.front();         if (index < size / 2)         {             curr->left = new Node(arr[index * 2]);             q.push(curr->left);             curr->right = new Node(arr[index * 2 + 1]);             q.push(curr->right);         }         q.pop();         index++;     }     return root; }