I hereby claim:
- I am kayladnls on github.
- I am kayladnls (https://keybase.io/kayladnls) on keybase.
- I have a public key whose fingerprint is FF80 F5CF AE4E 35F9 5AF3 2585 844E 177D 1843 F73C
To claim this, I am signing this object:
| git clone https://github.com/ProgrammingAreHard/ResourceBundle.git | |
| git checkout -b "rewriteHistory" | |
| git log --oneline --graph | |
| git checkout ac97baa | |
| git merge origin/master --squash | |
| git commit | 
| // Use Gists to store code you would like to remember later on | |
| console.log(window); // log the "window" object to the console | 
I hereby claim:
To claim this, I am signing this object:
| Array | |
| ( | |
| [query] => select *, sum((ifnull(`debit`,0) - ifnull(`credit`,0))) AS `amount`, `acct_journal_details`.*, `acct_journal`.`employee_id` from `acct_journal_details` inner join `acct_journal` on `acct_journal`.`acct_journal_id` = `acct_journal_details`.`acct_journal_id` where `is_per_diem` <> ? and `acct_journal`.`employee_id` in (?, ?, ?, ?) group by `acct_journal`.`employee_id`, `acct_journal_details`.`ref_date` having `amount` <> ? | |
| [bindings] => Array | |
| ( | |
| [0] => FirstWhere | |
| [1] => Having | |
| [2] => Relationship | |
| [3] => Relationship | |
| [4] => Relationship | 
| <?php | |
| function computeErrorForLineGivenPoints($b, $m, array $points) | |
| { | |
| $total_error = 0; | |
| foreach ($points as $point) | |
| { | |
| $total_error += pow(($point['y'] - ($m * $point['x'] + $b)), 2); | |
| } | 
Ingredients
| <?php | |
| $lines = array(); | |
| $first_line = array_fill(0, 51, '*'); | |
| $first_line[24] = '/'; | |
| $lines[] = $first_line; | |
| $rules = array( | |
| '///'=> '*', | |
| '//*' => '*', | 
| <?php | |
| $lines = array(); | |
| $first_line = array_fill(0, 51, '*'); | |
| $first_line[24] = '/'; | |
| $lines[] = $first_line; | |
| $rules = array( | |
| '///'=> '*', | |
| '//*' => '/', | 
B3(z) = Original notation
T(z) == name he's giving for function
T(z) = 1 + zT(z)3
What this says: Either Tree is empty (1) or Tree has one node (z) + 3 more ternary trees (T(z)) Because there are 3 of them, we cube (T(z))
| <?php | |
| /** | |
| * Created by PhpStorm. | |
| * User: kayladnls | |
| * Date: 12/13/14 | |
| * Time: 7:24 PM | |
| */ | |
| // Bt(z)^r == sigma(k) * (tk + r, k) * r / tk + r * z^k | |
| // Problem: Compute the number of 4th order ternary trees; |