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 /*   * @licstart The following is the entire license notice for the JavaScript   * code in this page.   *   * This file is part of oddjobs-dmg-calc.   *   * oddjobs-dmg-calc 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.   *   * oddjobs-dmg-calc 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 oddjobs-dmg-calc. If not, see .   *   * @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   * non-integral 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   * non-integral 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)   );  }  */