如何评价简单代数运算的难度

Question

我正在编写一个简单的应用程序来帮助我女儿练习基础代数（一年级）。我想根据操作的难度为每个响应分配一个分数，例如，2+2 的价值小于 12-7。你知道我可以查找并适应我的需求的任何现有算法吗？

编辑

如果您想查看该应用程序 https://algebro.herokuapp.com
如果你想看代码https://gitlab.com/etozzato/algebro

**编辑←**

试图使问题更具体 ;)

给定两个整数和一个基本的代数运算

算法输入：int a,b和string operation

算法输出：float difficulty

→ 哪些因素可以帮助推断一个困难的系数？

我肯定会看看输入的数字，它们的距离可以是在确定操作的复杂性方面具有重要意义。 10 + 1 明显比 7 + 5 容易，因为（当不记和立即反应时）它需要更长的计算时间；
作为对先前声明的修正，common/simple 参数应降低操作的复杂性：0 或 10 是一个很好的例子；

Answer 1

我不知道有什么算法可以找到方程的 "difficulty"，但如果我有更多时间，我可能会尝试使用 this 之类的东西......即使仅供阅读，概念可能适用于算术。

无论如何，这里有一个超级愚蠢的 post-midnight crack，可以通过一些调整来实现极其 basic 的算法。您可以调整 factors/weights 但这可能会让您入门。祝你好运！

function get_difficulty (eq) {
    var difficulty = 0;
    var settings = {
        terms_factor : 3, //Multiply by the number of terms in an equation
        digits_factor : 2, //Multiply by the number of digits in each term
        negative_weight : 2, //Add this if subtracting two numbers in the equation yields a negative number

        operations : {
            "+" : 1,
            "-" : 2,
            "*" : 4,
            "/" : 6,

            "=" : 0
        }
    };

    eq += "=";

    var ptr = 0;
    var terms = 0;
    var prev_term = null;
    var len = eq.length;
    var stack = [ ];

    var is_numeric = function (n) {
        return /\d+/.test (n); //Not a brilliant way but works for basic arithmetic
    };

    while (ptr < len) {
        var tok = eq [ptr];

        if (tok !== " " && tok !== "(" && tok !== ")") {
            if (is_numeric (tok)) {
                stack.push (tok);   
            } else if (tok in settings.operations) {
                var curr_term = parseInt (stack.join (""));
                if (prev_term !== null && curr_term > prev_term && ["-", "="].indexOf (tok) !== -1) {
                    difficulty += settings.negative_weight;
                }

                difficulty += stack.length * settings.digits_factor;
                prev_term = curr_term;
                stack = [ ];
                terms++;

                difficulty += settings.operations [tok];
            } else {
                console.log ("Unknown token: " + tok);   
            }
        }

        ptr++;
    }

    difficulty += terms * settings.terms_factor;

    return difficulty;
}


console.log (get_difficulty (" 2 + 2 ")); //11
console.log (get_difficulty (" 12 - 7 ")); //14
console.log (get_difficulty (" 7 - 12 ")); //16
console.log (get_difficulty (" 5 - 5 ")); //12 
console.log (get_difficulty (" 5 - 1205 ")); //20
console.log (get_difficulty (" 5 - 1205 * 35 ")); //29
console.log (get_difficulty (" 5 * 40 ")); //18
console.log (get_difficulty (" 102 - 5 / 13 + 32 ")); //39
console.log (get_difficulty (" 100 - 100 ")); //20
console.log (get_difficulty (" 32 - 12 ")); //16
console.log (get_difficulty (" 12 - 32 ")); //18

如何评价简单代数运算的难度

How to rate to difficulty of simple algebra operations

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