Javascript 该函数中使用的方法背后的数学推理是什么，从而产生期望的结果？

我有一个函数，可以生成给定字符串的所有组合；不包括排列。我相信我完全理解了给定字符串如何通过函数获得所需结果的逻辑。我想了解的是函数将字符串元素处理成包含所有可能组合的数组的主要方法背后的数学推理

下面是注释较多的代码，显示了我目前对函数逻辑处理的理解

function generateAllCombinationsOfString (str1) { //let str1 = "dog"
    // convert str1 into an array
    var array1 = []; // create var in order to capture elements of str1 into an array
        for (var x = 0, y = 1; x < str1.length; x++,y++) {
            array1[x] = str1.substring(x, y); // we will add each character to array1, in order, starting with the first then ending the loop once the last character has been included in array1
            // for each iteration: x = 0, y = 1     x < str1.length     x++, y++        array1[x] = str1.substring(x, y)
            // iteration 1:      x = 0, y = 1         yes                1    2         array1[0] = str1.substring(0, 1) = "d"
            // iteration 2:      x = 1, y = 2       yes                2    3         array1[1] = str1.substring(1, 2) = "o"
            // iteration 3:      x = 2, y = 3       yes                3    4         array1[2] = str1.substring(2, 3) = "g"
            // iteration 4:      x = 3, y = 4       no
            // end of loop
        }
    // create array containing all combinations of str1
    var combi = []; // create var to capture each possible combination within an array
    var temp = "";  // create cache to temporarily hold elements each possible combination before adding to an array
    var slent = Math.pow(2, array1.length);
        for (var i = 0; i < slent; i++){
            temp = "";
            // for each iteration:      i = 0       i < slent       i++     var combi = []; temp = ""; var slent = Math.pow(2, array1.length)
            // iteration 1:             i = 0         yes            1      var combi = []; temp = ""; var slent = Math.pow(2, 3) = 8
            // iteration 2:             i = 1         yes            2      var combi = [d]; temp = ""; var slent = Math.pow(2, 3) = 8
            // iteration 3:             i = 2         yes            3      var combi = [d, o]; temp = ""; var slent = Math.pow(2, 3) = 8
            // iteration 4:             i = 3         yes            4      var combi = [d, o, do]; temp = ""; var slent = Math.pow(2, 3) = 8
            // iteration 5:             i = 4         yes            5      var combi = [d, o, do, g]; temp = ""; var slent = Math.pow(2, 3) = 8
            // iteration 6:             i = 5         yes            6      var combi = [d, o, do, g, dg]; temp = ""; var slent = Math.pow(2, 3) = 8
            // iteration 7:             i = 6         yes            7      var combi = [d, o, do, g, dg, og]; temp = ""; var slent = Math.pow(2, 3) = 8
            // iteration 8:             i = 7         yes            8      var combi = [d, o, do, g, dg, og, dog]; temp = ""; var slent = Math.pow(2, 3) = 8
            // iteration 9:             i = 8         no
            // end of loop
            for (var j = 0; j < array1.length; j++) {
                if ((i & Math.pow(2, j))) {
                    temp += array1[j];
                    // for each iteration:      j = 0       j < array1.length       j++     if((i & Math.pow(2, j)))?                             {temp += array1[j]}
                    // iteration 1-1:           j = 0         yes                    1      if((0 & Math.pow(2, 0)))? => if((0 & 1))? // false
                    // iteration 1-2:           j = 1         yes                    2      if((0 & Math.pow(2, 1)))? => if((0 & 2))? // false
                    // iteration 1-3:           j = 2         yes                    3      if((0 & Math.pow(2, 2)))? => if((0 & 4))? // false
                    // iteration 1-4:           j = 3         no
                    // end of loop
                    // iteration 2-1:           j = 0         yes                    1      if((1 & Math.pow(2, 0)))? => if((1 & 1))? // true  // {temp += array1[0]} => temp = "d"}
                    // iteration 2-2:           j = 1         yes                    2      if((1 & Math.pow(2, 1)))? => if((1 & 2))? // false
                    // iteration 2-3:           j = 2         yes                    3      if((1 & Math.pow(2, 2)))? => if((1 & 4))? // false
                    // iteration 2-4:           j = 3         no
                    // end of loop
                    // iteration 3-1:           j = 0         yes                    1      if((2 & Math.pow(2, 0)))? => if((2 & 1))? // false
                    // iteration 3-2:           j = 1         yes                    2      if((2 & Math.pow(2, 1)))? => if((2 & 2))? // true  // {temp += array1[1] => temp = "o"}
                    // iteration 3-3:           j = 2         yes                    3      if((2 & Math.pow(2, 2)))? => if((2 & 4))? // false
                    // iteration 3-4:           j = 3         no
                    // end of loop
                    // iteration 4-1:           j = 0         yes                    1      if((3 & Math.pow(2, 0)))? => if((3 & 1))? // true  // {temp += array1[0] => temp = "d"}
                    // iteration 4-2:           j = 1         yes                    2      if((3 & Math.pow(2, 1)))? => if((3 & 2))? // true  // {temp += array1[1] => temp = "do"}
                    // iteration 4-3:           j = 2         yes                    3      if((3 & Math.pow(2, 2)))? => if((3 & 4))? // false //
                    // iteration 4-4:           j = 3         no
                    // end of loop
                    // iteration 5-1:           j = 0         yes                    1      if((4 & Math.pow(2, 0)))? => if((4 & 1))? // false //
                    // iteration 5-2:           j = 1         yes                    2      if((4 & Math.pow(2, 1)))? => if((4 & 2))? // false //
                    // iteration 5-3:           j = 2         yes                    3      if((4 & Math.pow(2, 2)))? => if((4 & 4))? // true  // {temp += array1[2] => temp = "g"}
                    // iteration 5-4:           j = 3         no
                    // end of loop
                    // iteration 6-1:           j = 0         yes                    1      if((5 & Math.pow(2, 0)))? => if((5 & 1))? // true  // {temp += array1[0] => temp = "d"}
                    // iteration 6-2:           j = 1         yes                    2      if((5 & Math.pow(2, 1)))? => if((5 & 2))? // false //
                    // iteration 6-3:           j = 2         yes                    3      if((5 & Math.pow(2, 2)))? => if((5 & 4))? // true  // {temp += array1[2] => temp = "dg"}
                    // iteration 6-4:           j = 3         no
                    // end of loop
                    // iteration 7-1:           j = 0         yes                    1      if((6 & Math.pow(2, 0)))? => if((6 & 1))? // false // 
                    // iteration 7-2:           j = 1         yes                    2      if((6 & Math.pow(2, 1)))? => if((6 & 2))? // true  // {temp += array1[1] => temp = "o"}
                    // iteration 7-3:           j = 2         yes                    3      if((6 & Math.pow(2, 2)))? => if((6 & 4))? // true  // {temp += array1[2] => temp = "og"}
                    // iteration 7-4:           j = 3         no         
                    // end of loop
                    // iteration 8-1:           j = 0         yes                    1      if((7 & Math.pow(2, 0)))? => if((7 & 1))? // true  // temp += array1[0] => temp = "d"}
                    // iteration 8-2:           j = 1         yes                    2      if((7 & Math.pow(2, 1)))? => if((7 & 2))? // true  // temp += array1[1] => temp = "do"}
                    // iteration 8-3:           j = 2         yes                    3      if((7 & Math.pow(2, 2)))? => if((7 & 4))? // true  // temp += array1[2] => temp = "dog"}
                    // iteration 8-4:           j = 3         no 
                    // end of loop
                }
            }
            if (temp !== "") { // if var temp is not an empty string then we add elements of var temp to end of the array var combi at the end of each loop cycle
                combi.push(temp); 
                // for each iteration       if(temp !== "")?
                // iteration 1:             if(temp !== "")? // false //
                // iteration 2:             if(temp !== "")? // true // combi.push(temp) => combi.push("d") 
                // iteration 3:             if(temp !== "")? // true // combi.push(temp) => combi.push("o") 
                // iteration 4:             if(temp !== "")? // true // combi.push(temp) => combi.push("do")
                // iteration 5:             if(temp !== "")? // true // combi.push(temp) => combi.push("g")
                // iteration 6:             if(temp !== "")? // true // combi.push(temp) => combi.push("dg")
                // iteration 7:             if(temp !== "")? // true // combi.push(temp) => combi.push("og")
                // iteration 8:             if(temp !== "")? // true // combi.push(temp) => combi.push("dog")
            }
        }
        // join array of combinations while seperating each by a new line
        console.log(combi.join("\n"))
        /*  expected output if str1 = "dog"

            d
            o
            do
            g
            dg
            og
            dog

        */
}

函数generateAllCombinationsOfString（str1）{//let str1=“dog”
//将str1转换为数组
var array1=[]；//创建var以将str1的元素捕获到数组中
对于（变量x=0，y=1；xif（（0&1））？//错误
//迭代1-2:j=1是2如果（（0&Math.pow（2，1））？=>if（（0&2））？//错误
//迭代1-3:j=2是3如果（（0&Math.pow（2，2））？=>if（（0&4））？//错误
//迭代1-4:j=3个
//循环结束
//迭代2-1:j=0是1如果（（1&Math.pow（2，0））？=>if（（1&1））？///true/{temp+=array1[0]}=>temp=“d”}
//迭代2-2:j=1是2如果（（1&Math.pow（2，1））？=>if（（1&2））？//错误
//迭代2-3:j=2是3如果（（1&Math.pow（2，2））？=>if（（1&4））？//错误
//迭代2-4:j=3个
//循环结束
//迭代3-1:j=0是1如果（（2&Math.pow（2，0））？=>if（（2&1））？//错误
//迭代3-2:j=1是2如果（（2&Math.pow（2，1））？=>if（（2&2））？//true/{temp+=array1[1]=>temp=“o”}
//迭代3-3:j=2是3如果（（2&Math.pow（2，2））？=>if（（2&4））？//错误
//迭代3-4:j=3个
//循环结束
//迭代4-1:j=0是1如果（（3&Math.pow（2，0））？=>if（（3&1））？//true/{temp+=array1[0]=>temp=“d”}
//迭代4-2:j=1是2如果（（3&Math.pow（2，1））？=>if（（3&2））？//true/{temp+=array1[1]=>temp=“do”}
//迭代4-3:j=2是3如果（（3&Math.pow（2，2））？=>if（（3&4））？//错误//
//迭代4-4:j=3个
    var combi = [];
    var temp = "";  
    var slent = Math.pow(2, array1.length);
        for (var i = 0; i < slent; i++){
            temp = "";

            for (var j = 0; j < array1.length; j++) {
                if ((i & Math.pow(2, j))) {
                    temp += array1[j];
                }
            }
            if (temp !== "") { 
                combi.push(temp); 
            }
        }