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isAnagram = (a,b)-> | |
if a.length != b.length | |
return false | |
[a,b] = [a,b].map (x)->x.toLowerCase().split('') | |
length = a.length | |
if length > 12852 | |
throw new Error "This algorithm can only reliably handle strings smaller than 12,852 characters" | |
anagramHash = (hash, letter)-> | |
charCode = letter.charCodeAt(0)-96 | |
if !(1 <= charCode <= 26) |
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prefixString = "* + 2 3 4" | |
evaluatePrefix = (prefix)-> | |
operator = prefix.shift() | |
leftValue = if isNaN parseFloat prefix[0] then evaluatePrefix prefix else prefix.shift() | |
rightValue = if isNaN parseFloat prefix[0] then evaluatePrefix prefix else prefix.shift() | |
eval leftValue+operator+rightValue | |
prefix = prefixString.split ' ' | |
console.log evaluatePrefix prefix |
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l=Math.log;r=((n)->z=[0..l(n)/l 2].reduce(((a,b)->`1<<b&n?3+a:4+a`),0);`z==n?n:r(z)`);c={};[1..1000000].map((n)->c[r n]?=0;c[r n]++);c |
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# set l to Math.log because you use it twice | |
l = Math.log | |
# create a recursive function, r, to iterate on a chain until it reaches a fixed point | |
# n is the input for the current iteration | |
r = (n)-> | |
### | |
This next line is a mouthful. "l(n)/l 2" computes the log base 2 of n. This is the index of the highest set bit in n | |
The reduce function is basically a sum across the bits of the number. | |
It adds 3 to the sum if the bit is set ("one".length) and 4 if the bit is not set ("zero".length) | |
So, in a nutshell, z is set to the number of letters in the binary phrase for the number n |
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L?-b=dsm?.&.<1db3 4hlbydb@mXdH1mydSQ0 | |
L create lambda function named y with param b | |
? ternary if | |
- minus used for expressing "if d != b" | |
b lambda param | |
= assign | |
d d = | |
s sum | |
m map |
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y = function(b){ | |
nums = [] | |
for(i=0; i<=Math.floor(Math.log(b)/Math.log(2)); i++) | |
nums.push(i); | |
d = nums.reduce(function(n, a){1<<n&b?3+a:4+a},0) | |
if(b == d) | |
return b | |
else | |
return y(b) | |
} |
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numBars = 73 | |
bars = [0..numBars-1].map -> 0 | |
getDisplayBars = (audioData)-> | |
deltaT = audioData.length/numBars | |
bars[0] = audioData[0] | |
for barNum in [1..numBars-1] | |
timeIndex = barNum*deltaT | |
leftIndex = Math.floor timeIndex |
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A.Q@c6csm?-i+OG1+1OG1Z1HH2 | |
Two commands: | |
A .Q - Parse the input (.Q), assign 1st line (max dice number) to G and 2nd line (number of trials) to H | |
@c6csm?-i+OG1+1OG1Z1HH2 - The output | |
The output: | |
@c6csm?-i+OG1+1OG1Z1HH2 | |
@ c6csm?-i+OG1+1OG1Z1HH 2 - Take the square root of the middle chunk | |
c 6 csm?-i+OG1+1OG1Z1HH - Divide 6 by the 3rd chunk |
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const crypto = require('crypto') | |
describe('Predictable Randomness', () => { | |
describe('Math.random', () => { | |
const firstResult = 0.37454011430963874 | |
const secondResult = 0.7965429842006415 | |
beforeEach(() => Math.__clearChances__()) | |
it('should produce these two numbers first by default', () => { | |
expect(Math.random()).toEqual(firstResult) | |
expect(Math.random()).toEqual(secondResult) |
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[ | |
"L2 F2 R2 D' L2 F2 U' L2 D' L2 U B D B2 D R U' F' L B' F R", | |
"L2 D2 B2 U R2 F2 L2 D2 B2 D B2 F U' L U2 B2 D' L F2 L U2", | |
"D F2 U' F2 R2 B2 L2 U' B2 R2 U' L2 R D' B U' F U' B' L R U", | |
"D2 B2 U' B2 U' R2 D F2 U2 L2 F2 D' R' U' R2 D' B2 F' U L2 D2 F'", | |
"B2 U F2 D F2 D2 F2 D B2 L2 B2 R D2 F U' L' F' D' L F R F2", | |
"D' B2 R2 F2 U' R2 D' B2 U2 F2 D' U2 F' U B' R' F L' B2 R2 U' F", | |
"L2 B2 U F2 U' B2 F2 U2 B2 L2 U B2 F' U' L' B' F' L' U' B2 L2 U2", | |
"B2 D2 L2 F2 D2 L2 U B2 F2 U2 B' R' B' U F' R B' F2 U F2 D2 R", | |
"R2 D2 L2 U B2 D' B2 D' L2 U L2 U' B R' D' R' D' R2 D2 U2 B R'", |
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