java - Finding square roots using long division method -

so class assignment. can't seem grasp complexity of long division. quite frankly, i'm not sure how begin solve algorithm make need.

for second part, must take in number user inputs , find square root specific number of decimal places user inputs. algorithm need replicate find square root can found here.

the following have far (clearly not progress made far):

import java.util.*; import java.math.*; import terminalio.*;  public class squarerootprogram{      public static float squarerootdiv(string number, int decimals){          string[] test = new string[2];         float answer = 0;         string groups = "";          test = number.split("\\.");         test = number.split("\\.");         system.out.println(test[0]);         system.out.println(test[1]);          for(int = 0; groups != test[i].tostring(); i++){             test[i].length();         }         return answer;     }  public static void main(string[]args){      keyboardreader reader = new keyboardreader();     string response = " ";     string initnumber = " ";     float number = 0;     int decimals = 0;      system.out.println("welcome square root program.");     do{         addline();         system.out.print("please enter number of want find square root:  ");         initnumber = reader.readline();         number = float.valueof(initnumber);         system.out.println(number);         system.out.print("please enter number of decimal places in want answer:  ");         decimals = reader.readint();          system.out.println("the answer provided division algorithm " + squarerootdiv(initnumber,decimals));          addline();         system.out.print("would try number?: ");         response = reader.readline();     }while(response.compareto("no")!=0);  } } 

an example of required output is:

welcome square root program.  please enter number of want find square root: 3.14159 please enter number of decimal places in want answer: 3  answer provided division algorithm 1.772  try number?: no  thank using program. 

obviously needs work number input , amount of decimal places user have square root returned to.

so if can me either explaining me how long division in java, or leaving code or without comments, appreciated!

i have created working version of program. instructions provided give bunch of steps used in code.

as procedure indicates, first locate integer part , try find "closest" square. in given example, "closest" square 40 36. let's go through in steps.

public static double squarerootdiv (string number, int decimals) {      if (decimals <= 0) return -1; //sanity check      int ipart = integer.parseint(number.split("\\.")[0]); 

that code give integer part can locate closest square, running loop (you improve binary search) so:

int closestsquare = 0; (closestsquare = 0; closestsquare <= ipart; closestsquare++) {      if (closestsquare * closestsquare > ipart) {           closestsquare -= 1;           break;      } } 

what check every single number integer part found. first number greater integer part indicates have overstepped. step backwards once , closest square. make sure make variable remembers decimal place should go:

int decimalix = ans.length(); 

step 2: find decimals.

the procedure asks keep track of numbers have found far, remainder of closest square subtracted integer part. in case of finding 40, remainder 40 - 36 = 4. our current answer far 6.

the procedure asks double current answer , multiply 10, multiplying current answer 20. if that, 120. let's call simplicity's sake. then, must find digit add such (a + digit) * digit < remainder * 100. in concrete terms, have find digit 1 through 9 such (120 + (some digit)) * digit < 4*100.

we can use decimals parameter bound our loop, so:

for (int = 0; < decimals; i++) {      remainder *= 100;      int base = integer.parseint(ans) * 20;      //now check digits      (int j = 9; j >= 0; j--) {           int trial = (base + j) * j; //use digit           if (trial < remainder) {                //we have found first digit less remainder!                remainder -= trial;                ans += j;                break;           }      } } 

now remains send answer:

return double.parsedouble(ans.substring(0, decimalix) + "." + ans.substring(decimalix)); 

here full code.

public static double squarerootdiv(string number, int decimals){         if (decimals <= 0) return -1;          string ans = "";          int ipart = integer.parseint(number.split("\\.")[0]);          int first = getnext(ipart);         int remainder = ipart - first*first;         ans += first;         int decimalix = ans.length();         int numdecimalsneeded = decimals - ans.length();          (int = 0; <= numdecimalsneeded; i++) {              remainder *= 100;             int base = integer.parseint(ans) * 20;              (int j = 9; j >= 0; j--) {                  int trial = (base + j) * j;                  if (trial < remainder) {                     remainder -= trial;                     ans += j;                     break;                 }             }         }            return double.parsedouble(ans.substring(0, decimalix) + "." + ans.substring(decimalix));     }      public static int getnext (int ipart) {         (int = 0; <= ipart; i++) {             if (i*i > ipart) {                 return - 1;             }         }         return -1;     }