I'd rather be in outer space 🛸

Cosmic Funnies
Cosimo Galluzzi

JBB: An Artblog!

titsay
Acquired Stardust
todays bird
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⁂
"I'm Dorothy Gale from Kansas"
Not today Justin

Product Placement
RMH

pixel skylines
cherry valley forever
Jules of Nature
$LAYYYTER
styofa doing anything
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seen from United States
@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
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
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
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
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; }