The website of the guild for MapleStory oddjobbers.
https://oddjobs.codeberg.page/
htmlcsswebsitehtml5maplestoryagplv3css3maplelegendsoddjoboddjobsagplv3pagesagplwebpagejoblessmsjoblessguild
You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('') and can be up to 35 characters long.
203 lines
6.8 KiB
203 lines
6.8 KiB
/*


* @licstart The following is the entire license notice for the JavaScript


* code in this page.


*


* This file is part of oddjobsdmgcalc.


*


* oddjobsdmgcalc is free software: you can redistribute it and/or modify it


* under the terms of the GNU Affero General Public License as published by the


* Free Software Foundation, either version 3 of the License, or (at your


* option) any later version.


*


* oddjobsdmgcalc is distributed in the hope that it will be useful, but


* WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY


* or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Affero General Public


* License for more details.


*


* You should have received a copy of the GNU Affero General Public License


* along with oddjobsdmgcalc. If not, see <https://www.gnu.org/licenses/>.


*


* @licend The above is the entire license notice for the JavaScript code in


* this page.


*/




/**


* Returns the sum of the set {`n`, `n`+1, `n`+2, ..., `m`3, `m`2, `m`1}.


* Note that the upper bound is exclusive. Another way of saying this is: the


* sum of all of the integers in the interval [`n`, `m`).


*


* This function assumes that `m >= n`, and that both `m` and `n` are integers,


* but usable results are still returned even if these assumptions are


* violated. Swapping `n` and `m` just flips the sign of the result, and


* nonintegral values for `n` and/or `m` will yield intermediate values, as


* expected.


*


* This function is so called because it computes the difference between a


* larger triangular number and a smaller triangular number.


*/


export function trapezoid(n: number, m: number): number {


return (m * (m  1)  n * (n  1)) / 2;


}




/**


* Returns the sum of the set {`n`^2, (`n`+1)^2, (`n`+2)^2, ..., (`m`3)^2,


* (`m`2)^2, (`m`1)^2}. Note that the upper bound is exclusive. Another way


* of saying this is: the sum of the squares of all of the integers in the


* interval [`n`, `m`).


*


* This function assumes that `m >= n`, and that both `m` and `n` are integers,


* but usable results are still returned even if these assumptions are


* violated. Swapping `n` and `m` just flips the sign of the result, and


* nonintegral values for `n` and/or `m` will yield intermediate values, as


* expected.


*


* This function is so called because it computes the difference between a


* larger square pyramidal number and a smaller square pyramidal number. It


* should probably be called `squareFrustum`, but `frustum` is cuter.


*/


export function frustum(n: number, m: number): number {


return (m * (m  1) * (2 * m  1)  n * (n  1) * (2 * n  1)) / 6;


}




/**


* Gets the expected value for a uniform distribution over the interval


* [`min`, `max`] that is **_actually not uniform_**, because the outcomes are


* clamped to a minimum of 1, **and** the outcomes' fractional parts are


* truncated.


*/


export function truncClampedExpectation(a: number, b: number): number {


if (a === b) {


return Math.max(Math.trunc(a), 1);


}




if (b >= 1) {


const [aInt, bInt] = [Math.trunc(a), Math.trunc(b)];




return (


((a >= 1 ? aInt * (1  (a % 1)) : 2  a) +


trapezoid(Math.max(aInt + 1, 2), bInt) +


bInt * (b % 1)) /


(b  a)


);


}




return 1;


}




/**


* Gets the expected value for a uniform distribution over the interval


* [`min`, `max`] that is **_actually not uniform_**, because the outcomes are


* clamped to a minimum of 1.


*/


/*


function clampedExpectation(min: number, max: number): number {


if (min >= 1) {


return (min + max) / 2;


}




// The logic below is there because it's possible that the lower end of the


// damage range (and possibly the higher end as well) is strictly less than


// 1, in which case we no longer have a uniform distribution! This means


// no simple `(minValue + maxValue) / 2` will calculate the expectation for


// us. Instead, we have to split the distribution out into two parts: the


// clamped bit (everything at or below 1), which is always clamped to an


// outcome of 1, and the uniform bit (everything above 1). These are then


// weighted and summed. Note that it's possible for the uniform bit to


// have a measure/norm of zero (particularly, in the case that


// `maxValue <= 1`).




if (min >= max) {


return 1;


}




const rawRangeNorm = max  min;


const rawClampedNorm = Math.min(1  min, rawRangeNorm);




const uniformExpected = (1 + max) / 2;




const clampedWeight = Math.min(rawClampedNorm / rawRangeNorm, 1);


const uniformWeight = 1  clampedWeight;




return clampedWeight + uniformWeight * uniformExpected;


}


*/




/**


* Gets the variance for a uniform distribution over the interval [`a`, `b`]


* that is **_actually not uniform_**, because the outcomes are clamped to a


* minimum of 1, **and** the outcomes' fractional parts are truncated. The


* `mu` parameter is the expectation of the distribution, which can be obtained


* from the `truncClampedExpectation` function.


*


* This function assumes that `b >= a`.


*


* This function computes the variance based on its definition. In LaTeX:


*


* ```latex


* \operatorname{Var}(X) = \operatorname{E}[\left(X  \mu\right)^2]


* ```


*/


export function truncClampedVariance(


a: number,


b: number,


mu: number,


): number {


if (a === b  b <= 1) {


return 0;


}




const [aInt, bInt] = [Math.trunc(a), Math.trunc(b)];


const oneMinusMu = 1  mu;


const bIntMinusMu = bInt  mu;




return (


((a >= 1


? (aInt  mu) ** 2 * (1  (a % 1))


: (2  a) * oneMinusMu ** 2) +


frustum(Math.max(aInt + oneMinusMu, 2  mu), bIntMinusMu) +


bIntMinusMu ** 2 * (b % 1)) /


(b  a)


);


}




/**


* Gets the variance for a uniform distribution over the interval [`a`, `b`]


* that is **_actually not uniform_**, because the outcomes are clamped to a


* minimum of 1. The `mu` parameter is the expectation of the distribution,


* which can be obtained from the `clampedExpectation` function.


*


* This function assumes that `b >= a`, so if `b < a && b > 1`, you will get


* `undefined`.


*


* In LaTeX:


*


* ```latex


* \sigma^2 = \begin{cases}


* 0 & \text{when } a = b \lor b \leq 1 \\


* \frac{(b  \mu)^3  (\text{max}\left\{a, 1\right\}  \mu)^3}{3(b  a)} +


* (1  \mu)^2\,\text{max}\!\left\{\frac{1  a}{b  a}, 0\right\} &


* \text{when } b > \text{max}\!\left\{a, 1\right\}


* \end{cases}


* ```


*/


/*


function clampedVariance(


a: number,


b: number,


mu: number,


): number  undefined {


if (a === b  b <= 1) {


return 0;


}


if (a > b) {


return;


}




const bMinusA = b  a;




return (


((b  mu) ** 3  (Math.max(a, 1)  mu) ** 3) / (3 * bMinusA) +


(1  mu) ** 2 * Math.max((1  a) / bMinusA, 0)


);


}


*/


