JS求解三元一次方程组值的方法

本文实例讲述了JS求解三元一次方程组值的方法。分享给大家供大家参考，具体如下：

// 求用js 码一段代码求 三元一次方程组的值！
// a1 = y * b1 + z * c1 - d1 * x; (1)
// a2 = y * b2 + z * c2 - d2 * x; (2)
// a3 = y * b3 + z * c3 - d3 * x; (3)
// (1) - (2)
// a1 - a2 = y * (c1 - c2) + z * (d1 - d2)  (4)
// (1) - (3)
// a1 - a3 = y * (c1 - c3) + z * (d1 - d3)  (5)
// (4) * (c1 - c3)/(c1 - c2)
// (a1 - a2) * ((c1 - c3)/(c1 - c2)) = y * (c1 - c3) + z * (d1 - d2) * ((c1 - c3)/(c1 - c2));  (6)
// (6) - (5)
// (a1 - a2) * ((c1 - c3)/(c1 - c2)) - (a1 - a3) = z * ((d1 - d2) * ((c1 - c3)/(c1 - c2)) - (d1 - d3));  (7)
//
//==================================================================================================
// (1) * (d2/d1)
// ( a1 * (d2/d1) ) = y * b1 * (d2/d1) + z * c1 * (d2/d1) - d1 * x * (d2/d1)  (4)
// ( a1 * (d2/d1) ) = y * b1 * (d2/d1) + z * c1 * (d2/d1) - d2 * x       (4.1)
// (4.1) - (2)
// ( ( a1 * (d2/d1) ) - a2 ) = y * ( (b1 * (d2/d1)) - b2 ) + z * ( (c1 * (d2/d1)) - c2 );  (5)
// (1) * (d3/d1)
// ( a1 * (d3/d1) ) = y * ( b1 * (d3/d1) ) + z * ( c1 * (d3/d1) ) - d1 * x * (d3/d1)  (6)
// ( a1 * (d3/d1) ) = y * ( b1 * (d3/d1) ) + z * ( c1 * (d3/d1) ) - d3 * x       (6.1)
// (6.1) - (3)
// ( ( a1 * (d3/d1) ) - a3 ) = y * ( ( b1 * (d3/d1) ) - b3 ) + z * ( ( c1 * (d3/d1) ) - c3 )  (7)
// (5) * ( ( (b1 * (d3/d1)) - b3 ) / ( ( b1 * (d2/d1) ) - b2 ) )
// ( ( ( a1 * (d2/d1) ) - a2 ) * ( ( (b1 * (d3/d1)) - b3 ) / ( ( b1 * (d2/d1) ) - b2 ) ) ) = y * ( (b1 * (d3/d1)) - b3 ) + z * ( ( (c1 * (d2/d1)) - c2 ) * ( ( (b1 * (d3/d1)) - b3 ) / ( ( b1 * (d2/d1) ) - b2 ) ) )  (8)
// (8) - (7)
// ( ( ( ( a1 * (d2/d1) ) - a2 ) * ( ( (b1 * (d3/d1)) - b3 ) / ( ( b1 * (d2/d1) ) - b2 ) ) ) - ( ( a1 * (d3/d1) ) - a3 ) )
// =
// z * ( ( ( (c1 * (d2/d1)) - c2 ) * ( ( (b1 * (d3/d1)) - b3 ) / ( ( b1 * (d2/d1) ) - b2 ) ) ) - ( ( c1 * (d3/d1) ) - c3 ) )
//var a1 = 41778.58102, a2 = 60925.87223, a3 = 88229.03849,
//b1 = 72931.77839, b2 = 61296.01212, b3 = 67138.30916,
//c1 = 120.1318412, c2 = 120.1308286, c3 = 120.1573771,
//d1 = 30.27345376, d2 = 30.30174637, d3 = 30.32151101;
//var a1 = 1, a2 = -9, a3 = -3,
//b1 = -1, b2 = 3, b3 = -3,
//c1 = 1, c2 = -2, c3 = -5,
//d1 = -2, d2 = -1, d3 = -4;
var a1 = 11, a2 = 0, a3 = -2,
b1 = 3, b2 = 1, b3 = -1,
c1 = 1, c2 = 1, c3 = -1,
d1 = -2, d2 = -1, d3 = -3;
var equations = [
  "a1 = y * b1 + z * c1 - d1 * x",
  "a2 = y * b2 + z * c2 - d2 * x",
  "a3 = y * b3 + z * c3 - d3 * x"
];
for (var i = 0; i < equations.length; i++) {
  equations[i] = equations[i].replace("a1", "(" + a1 + ")").replace("b1", "(" + b1 + ")").replace("c1", "(" + c1 + ")").replace("d1", "(" + d1 + ")")
  .replace("a2", "(" + a2 + ")").replace("b2", "(" + b2 + ")").replace("c2", "(" + c2 + ")").replace("d2", "(" + d2 + ")")
  .replace("a3", "(" + a3 + ")").replace("b3", "(" + b3 + ")").replace("c3", "(" + c3 + ")").replace("d3", "(" + d3 + ")");
}
console.log(equations);
// 计算三元一次方程
// 确保常量和未知数分开在2边,逻辑安装这样处理的
var LEquat = {
  formatFloat: function(f, digit) {
    var m = Math.pow(10, digit);
    return parseInt(f * m, 10) / m;
  },
  // 是否常量一边, 返回0常量的一边索引,1变量一边的索引，返回一个数组
  separateSide: function(equations) {
    var arr = [], equa = equations.split("=");
    if (equa[0].search(/x|y|z/) == -1) {
      arr = equa.slice(0);
    } else {
      arr[0] = equa[1];
      arr[1] = equa[0];
    }
    return arr;
  },
  // 分隔变量的一边，把x，y，z分隔
  // x 是索引0，y是索引1，z是索引2
  separateVariableSide: function(variableSide) {
    var variables = variableSide.split(/\s+(\+|\-)\s+/);
    var arr = [];
    for (var i = 0; i < variables.length; i++) {
      if (variables[i].search(/x/) != -1) {
        arr[0] = variables[i];
      } else if (variables[i].search(/y/) != -1) {
        arr[1] = variables[i];
      } else if (variables[i].search(/z/) != -1) {
        arr[2] = variables[i];
      }
    }
    return arr;
  },
  // 取出方程里面变量一边对应x，y，z的常量系数
  // 0: x的系数，1：y的系数，2：z的系数
  getCoefficient: function(variableSideArray) {
    var me = LEquat;
    var arr = [];
    for (var i = 0; i < variableSideArray.length; i++) {
      var coeffs = variableSideArray[i].split("*");
      if (coeffs[0].search(/x|y|z/) == -1) {
        arr[i] = coeffs[0];
      } else {
        arr[i] = coeffs[1];
      }
    }
    return arr;
  },
  step: function() {
    var me = LEquat;
    // 分隔第一个方程
    var equation1 = me.separateSide(equations[0]);
    // 分隔第二个方程
    var equation2 = me.separateSide(equations[1]);
    // 分隔第三个方程
    var equation3 = me.separateSide(equations[2]);
    // 第一个方程的常量一边
    var a1 = equation1[0];
    // 第一个方程x系数
    var equa1 = me.getCoefficient(me.separateVariableSide(equation1[1]));
    // y的系数     z的系数     x的系数
    var b1 = equa1[1], c1 = equa1[2], d1 = equa1[0];
    // 第二个方程的常量一边
    var a2 = equation2[0];
    // 第二个方程变量一边系数
    var equa2 = me.getCoefficient(me.separateVariableSide(equation2[1]));
    var b2 = equa2[1], c2 = equa2[2], d2 = equa2[0];
    // 第三个方程的常量一边
    var a3 = equation3[0];
    // 第三个方程变量一边系数
    var equa3 = me.getCoefficient(me.separateVariableSide(equation3[1]));
    var b3 = equa3[1], c3 = equa3[2], d3 = equa3[0];
    // ( a1 * (d2/d1) )
    var left4_1 = " ( " + a1 + " * ( " + d2 + " / " + d1 + " ) ) ";
    // (4.1) 右边
    var right4_1 = " y " + " * " + b1 + " * ( " + d2 + " / " + d1 + " ) + z * " + c1 + " * ( " + d2 + " / " + d1 + " ) - " + d2 + " * x ";
    // 消x (4.1) - (2)
    // ( ( a1 * (d2/d1) ) - a2 ) = y * ( (b1 * (d2/d1)) - b2 ) + z * ( (c1 * (d2/d1)) - c2 );  (5)
    // (5) 中常量的一边
    var left5_1 = " ( " + left4_1 + " - " + a2 + " ) ";
    // (5) 中变量一边
    var right5_1 = " y * ( ( " + b1 + " * ( " + d2 + " / " + d1 + " ) ) - " + b2 + " ) + z * ( ( " + c1 + " * ( " + d2 + " / " + d1 + " ) ) - " + c2 + " ) ";
    // (1) * (d3/d1)
    // ( a1 * (d3/d1) ) = y * ( b1 * (d3/d1) ) + z * ( c1 * (d3/d1) ) - d1 * x * (d3/d1)  (6)
    // ( a1 * (d3/d1) ) = y * ( b1 * (d3/d1) ) + z * ( c1 * (d3/d1) ) - d3 * x       (6.1)
    var left6_1 = " ( " + a1 + " * ( " + d3 + " / " + d1 + " ) ) ";
    var right6_1 = " y * ( " + b1 + " * ( " + d3 + " / " + d1 + " ) ) + z * ( " + c1 + " * ( " + d3 + " / " + d1 + " ) ) - " + d3 + " * x ";
    // (6.1) - (3)
    // ( ( a1 * (d3/d1) ) - a3 ) = y * ( ( b1 * (d3/d1) ) - b3 ) + z * ( ( c1 * (d3/d1) ) - c3 )  (7)
    var left7_1 = " ( " + left6_1 + " - " + a3 + " ) ";
    var right7_1 = " y * ( ( " + b1 + " * ( " + d3 + " / " + d1 + " ) ) - " + b3 + " ) + z * ( ( " + c1 + " * ( " + d3 + " / " + d1 + " ) ) - " + c3 + " ) ";
    // (5) * ( ( (b1 * (d3/d1)) - b3 ) / ( ( b1 * (d2/d1) ) - b2 ) )
    // ( ( ( a1 * (d2/d1) ) - a2 ) * ( ( (b1 * (d3/d1)) - b3 ) / ( ( b1 * (d2/d1) ) - b2 ) ) )
    // = y * ( (b1 * (d3/d1)) - b3 ) + z * ( ( (c1 * (d2/d1)) - c2 ) * ( ( (b1 * (d3/d1)) - b3 ) / ( ( b1 * (d2/d1) ) - b2 ) ) )  (8)
    var left8_1 = " ( " + left5_1 + " * ( ( ( " + b1 + " * ( " + d3 + " / " + d1 + " ) ) - " + b3 + " ) / ( ( " + b1 + " * ( " + d2 + " / " + d1 + " ) ) - " + b2 + " ) ) ) ";
    var right8_1 = " y * ( ( " + b1 + " * ( " + d3 + " / " + d1 + " ) ) - " + b3 + " ) + z * ( ( ( " + c1 + " * ( " + d2 + " / " + d1 + " ) ) - " + c2 + " ) * ( ( ( " + b1 + " * ( " + d3 + " / " + d1 + " ) ) - " + b3 + " ) / ( ( " + b1 + " * ( " + d2 + " / " + d1 + " ) ) - " + b2 + " ) ) ) ";
    // (8) - (7)
    // ( ( ( ( a1 * (d2/d1) ) - a2 ) * ( ( (b1 * (d3/d1)) - b3 ) / ( ( b1 * (d2/d1) ) - b2 ) ) ) - ( ( a1 * (d3/d1) ) - a3 ) )
    // =
    // z * ( ( ( (c1 * (d2/d1)) - c2 ) * ( ( (b1 * (d3/d1)) - b3 ) / ( ( b1 * (d2/d1) ) - b2 ) ) ) - ( ( c1 * (d3/d1) ) - c3 ) )  (9)
    var left9_1 = " ( " + left8_1 + " - " + left7_1 + " ) ";
    var right9_1 = " z * ( ( ( ( " + c1 + " * ( " + d2 + " / " + d1 + " ) ) - " + c2 + " ) * ( ( ( " + b1 + " * ( " + d3 + " / " + d1 + " ) ) - " + b3 + " ) / ( ( " + b1 + " * ( " + d2 + " / " + d1 + " ) ) - " + b2 + " ) ) ) - ( ( " + c1 + " * ( " + d3 + " / " + d1 + " ) ) - " + c3 + " ) ) ";
    var z = left9_1 + " / " + " ( ( ( ( " + c1 + " * ( " + d2 + " / " + d1 + " ) ) - " + c2 + " ) * ( ( ( " + b1 + " * ( " + d3 + " / " + d1 + " ) ) - " + b3 + " ) / ( ( " + b1 + " * ( " + d2 + " / " + d1 + " ) ) - " + b2 + " ) ) ) - ( ( " + c1 + " * ( " + d3 + " / " + d1 + " ) ) - " + c3 + " ) ) ";
    // ( ( a1 * (d2/d1) ) - a2 ) = y * ( (b1 * (d2/d1)) - b2 ) + z * ( (c1 * (d2/d1)) - c2 );
    var y = " ( " + left5_1 + " - " + " z * ( ( " + c1 + " * ( " + d2 + " / " + d1 + " ) ) - " + c2 + " ) ) " + " / " + " ( ( " + b1 + " * ( " + d2 + " / " + d1 + " ) ) - " + b2 + " ) ";
    // 把z代入到(5)中
    y = y.replace("z", z);
    // 把y，z代入方程1
    // a1 = y * b1 + z * c1 - d1 * x
    var x = " ( y * " + b1 + " + z * " + c1 + " - " + a1 + " ) / " + d1;
    x = x.replace("z", z).replace("y", y);
    x = me.formatFloat(eval(x), 2);
    y = me.formatFloat(eval(y), 2);
    z = me.formatFloat(eval(z), 2);
    return {x: x,y: y,z: z};
  }
};
console.log("-----------计算结果--------------");
console.log(LEquat.step());

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希望本文所述对大家JavaScript程序设计有所帮助。