BigInt problem

[kafka]

Hello guys. This is the problem I have to solve for class until next week:

Given: n
Info about n: natural number with at least 20 digits and maximum 1000 digits
Requested: Find the integer part of its square root.


I`ve searched for any answers or any useful tips for the last week but I haven`t found any. Maybe here I will be more lucky because I have no idea how could I solve this :). I hope you understand what I have to do and sorry for any grammatical mistakes I`ve made.

Edit: I cannot use any libraries.

[Thomas1965]

What you could use is a "big integer library"
https://github.com/connormanning/little-big-int

[kafka]

I forgot to mention I can not use any libraries.

[tpb]

Have yo written any big integer stuff before? Do you know how to multiply two 500 digit numbers?

[kafka]

Not even once. That s why I have such big problems. I don't know how i could even start.

[Thomas1965]

Who gave you such an difficult assignment without giving any hints ? Try to get a PhD ?
Maybe just forget it.
OR
Maybe you study the code from the link above and try to see how they implemented it.
Then you know the general principle can can apply to your project.

[kafka]

I m a student, first year at Computer Science. That s the homework I have to complete so I can participate at finals. Class : Data structures and algorithms.

[tpb]

If you're not worried about performance you could store the bignum in decimal as a string and only implement addition and subtraction for use in this algorithm: https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Decimal_(base_10)

* multiplying by a single decimal digit can be reasonably implemented with addition.
E.g., 123 * 5 = 123+123=246 246+246=492 492+123=615

* subtraction is used to calculate the remainder.

The digits are calculated one at a time from the most significant downwards, so it's easy to stop at the decimal point giving the integer square root.

[kafka]

Yea..I have to worry about performance. It has to have memory and execution time performance. I`m confused :\

[tpb]

Whatever. Go ask your teacher. Talk to other students. Read a book. Find out more. You don't know enough about it.

[lastchance]

@kafka

I don't see any easy way of avoiding having some sort of big integer class.

I wrote one some time ago - but I'm afraid it would not benefit your learning to simply post it here. Also, it contains many operations not necessary for your problem, making it confusing.

FWIW, I stored the digits of a large number in a vector<int> and defined member functions to do things like adding or multiplying big integers ... using exactly the same techniques as you would use for long addition and long multiplication in the primary-school days before you were allowed calculators!

If YOU write such a class then the integer square root can be found relatively easily (if not necessarily terribly efficiently) by simply taking the digits of the original number in pairs (see @tpb's link) and finding successive digits of the square root by trial.

So, here's a tantalising starter for you ... but YOU have to write the BigInt class. I've no idea how efficient it would be for a 1000-digit number (here, entered as a string).

For some completely illogical reason I find this particular challenge entirely suited to your username!

#include <iostream>
#include <vector>
#include <string>

#include "BigInt.h"          // YOU have to write this class

// YOU have to write enough BigInt members in your class to do the following:
//        Constructor (from a normal int)
//        Copy assignment from a BigInt
//        Add an int to a BigInt (or vice versa)
//        Multiply an int by a BigInt (or vice versa)
//        Multiply a BigInt by a BigInt
//        Compare a BigInt with a BigInt using <=
//        Output a BigInt with <<

using namespace std;

int main()
{
   string Nstring = "9876543210123456789"
                    "9876543210123456789"
                    "9876543210123456789"
                    "9876543210123456789"
                    "9876543210123456789"
                    "9876543210123456789"
                    "9876543210123456789";

   string number = Nstring;
   if ( number.size() % 2 ) number = "0" + number;         // Use digit pairs, so make sure that length is even

   BigInt C = 0, T = 0;                                    // Start with nothing
   while ( number != "" )                                  // While there are still digit pairs to consider
   {
      T = 100 * T + stoi( number.substr( 0, 2 ) );         // Append the next digit pair to the number to be square-rooted
      number.erase( 0, 2 );                                // Remove this digit pair from further consideration

      BigInt tenC = 10 * C;                                // Find largest x such that (10C+x)^2 <= T
      int x = 9;
      for ( ; x; x-- )
      {
         BigInt test = tenC + x;
         if ( test * test <= T ) break;
      }
      C = tenC + x;                                         // Update C with the next digit
   }

   cout << "ISQRT(" << Nstring << ")\n= " << C << '\n';
}

ISQRT(9876543210123456789987654321012345678998765432101234567899876543210123456789987654321012345678998765432101234567899876543210123456789)
= 3142696805312828262936709352546076540242174622210824369830188050565

[tpb]

lastchance wrote:

I don't see any easy way of avoiding having some sort of big integer class.

I gave an "easy" way in the sense that you can store the bigint as a string and only implement addition and subtraction. I implemented it in 20 minutes and it works just fine. But he's not interested. *sigh*

[kafka]

@tpb

Is not about me being uninterested, is about the fact I`m very frustrated because every solution you gave me sounds good, works, but it`s not what I need. I asked the teacher to give me a feedback about those ideas and she told me are not good enough. I`m sorry for wasting your time, tbh. Have a nice day.

@lastchance

Thanks for your help too, but apparently that`s not good enough for my teacher. Have a nice day.

[lastchance]

kafka wrote:

apparently that`s not good enough for my teacher

I'm mortified!

Any suggestions from your teacher how to improve?

[helios]

So, I'm confused. What exactly are the parameters of this assignment? It seems so far that no solution is acceptable.

[icy1]

https://stackoverflow.com/questions/23854975/whats-the-efficient-algorithm-to-find-the-integer-square-root-of-a-very-large-n

and apparently a bigint library from one of the comments http://codepad.org/T2AcSJzN

Since it appears he has multiplication working, you could prob square big numbers until their "string length" is greater than or equal to the target's length. Then it'd be a matter of refining the guess...

In general this doesn't sound like fun, and from some googling I saw this was some sort of 'School Olympiad' thing?

[helios]

Since the string length of a number n is always approximately k*log(n), and since log(x^y) = y*log(x), the string length of an integer square root is approximately k/2*log(n). That is, in any base, the integer square root of an integer always has half as many digits as the number, plus or minus half a digit.

[kafka]

Guys, that is a lot of informations. I have to read and think about everything you wrote here. I will come in the morning with an update about my situation.

@lastchance

I mailed her again today and I m waiting for a response.

[lastchance]

This do you @kafka?

#include <iostream>
#include <vector>
#include <string>

using namespace std;


//======================================================================


string operator * ( unsigned n, const string &text )       // Multiples of strings
{
   string result;
   while ( n-- ) result += text;
   return result;
}


//======================================================================


void multiplyByTens( vector<int> &V, int tens )            // Multiply by powers of ten
{
   V.resize( V.size() + tens );
   for ( int j = V.size() - 1; j >= tens; j-- ) V[j] = V[j-tens];
   for ( int j = 0; j < tens; j++ ) V[j] = 0;
}


//======================================================================


vector<int> operator * ( int n, const vector<int> &V )     // Multiply int * BigInt
{
   if ( n == 0 ) return vector<int>( 1, 0 );

   vector<int> result;

   int i = 0;
   int carry = 0;
   while ( i < V.size() || carry > 0 )
   {
      if ( i < V.size() ) carry += n * V[i];
      int digit = carry % 10;
      carry /= 10;
      result.push_back( digit );
      i++;
   }
   return result;
}


//======================================================================


vector<int> operator - ( const vector<int> &A, const vector<int> &B )       // Subtraction
{
   vector<int> result = A;

   int i = 0;
   int borrow = 0;
   while ( i < B.size() || borrow > 0 )
   {
      int digit = result[i] - borrow;
      if ( i < B.size() ) digit -= B[i];
      borrow = 0;
      while ( digit < 0 )
      {
         borrow++;
         digit += 10;
      }
      result[i] = digit;
      i++;
   }

   // Remove leading zeroes (unless number itself is zero)
   i = result.size() - 1;
   while( i > 0 && result[i] == 0 )
   {
      result.pop_back();
      i--;
   }

   return result;
}


//======================================================================


bool operator <= ( const vector<int> &A, const vector<int> &B )      // Compare BigInts
{
   if ( A.size() < B.size() ) return true;
   if ( A.size() > B.size() ) return false;

   for ( int i = A.size() - 1; i >= 0; i-- )
   {
      if ( A[i] < B[i] ) return true;
      if ( A[i] > B[i] ) return false;
   }

   return true;
}


//======================================================================


ostream &operator << ( ostream &strm, const vector<int> &V )
{
   for ( int i = V.size() - 1; i >= 0; i-- ) strm << V[i];
   return strm;
}


//======================================================================


vector<int> isqrt( const string &Nstring )
{
   string number = Nstring;
   if ( number.size() % 2 ) number = "0" + number;         // Use digit pairs, so make sure that length is even

   vector<int> C, RHS;                                     // C is growing root; for RHS see below
   while ( number != "" )                                  // While there are still digit pairs to consider
   {
      // Multiply RHS ( = T - 100 * C * C )  by 100 and grab next two digits
      multiplyByTens( RHS, 2 );
      RHS[1] = number[0] - '0';   RHS[0] = number[1] - '0';
      number.erase( 0, 2 );                                

      // Multiply C by 10 and try for last digit
      multiplyByTens( C, 1 );
      vector<int> twentyC = 2 * C;                         // Actually 20C
      int x = 9;
      for ( ; x; x-- )
      {
         vector<int> test = twentyC;
         test[0] = x;                                      // Actually 20C + x
         test = x * test;                                  // Actually x(20C + x)
         if ( test <= RHS )                                // Find largest x such that (10C+x)^2 <= T
         {                                                 //                     i.e.  x(20C+x) <= T-(10C)^2 = RHS
            C[0] = x;
            RHS = RHS - test;
            break;
         }
      }
   }

   return C;
}


//======================================================================


int main()
{
   string Nstring = "9876543210";
// Nstring = 100 * Nstring;                                // 1000-digit string (when uncommented)
   Nstring = 5   * Nstring;                                //   50-digit string (when uncommented)

   cout << "isqrt(" << Nstring << ") \n = " << isqrt( Nstring ) << '\n';
}


//====================================================================== 

isqrt(98765432109876543210987654321098765432109876543210) 
 = 9938079900558082311789231