Esempio n. 1
0
 /**
  * Utility method to split the list into two
  *
  * @param left : Starting index for split
  * @param right : Ending index for the split
  * @return : Returns an object of the BigArithmetic class with the list split
  */
 public G05_BigArithmetic split(int left, int right) {
   G05_BigArithmetic temp = new G05_BigArithmetic();
   for (int i = left; i < right; i++) {
     temp.number.add(this.number.get(i));
   }
   temp.effectiveSize = temp.number.size();
   return temp;
 }
Esempio n. 2
0
 /**
  * DAC Algorithm to multiply two long numbers
  *
  * @param a : Input object of BigArithmetic class
  * @param b : Input object of BigArithmetic class
  * @return : Returns an object of the BigArithmetic class which is a product of the two inputs
  */
 public static G05_BigArithmetic product(G05_BigArithmetic a, G05_BigArithmetic b) {
   // If any of the two lists is null return null
   if (a == null || b == null) {
     return null;
   }
   // We remove the MS zeros to get the effective size of the two numbers.
   G05_BigArithmetic x = a.removeMSZeros();
   G05_BigArithmetic y = b.removeMSZeros();
   /*If the size is zero, then the list is 0, so return zero - Multiply by zero */
   if (x.effectiveSize == 0 || y.effectiveSize == 0) {
     return new G05_BigArithmetic(0L);
   }
   // Call the productBA for DAC
   G05_BigArithmetic c = productBA(x, y);
   return c;
 }
Esempio n. 3
0
 /**
  * Method to multiply the two Big Arithmetic objects
  *
  * @param a : Input - object of class BigArithmetic
  * @param b : Input - object of class BigArithmetic
  * @return : Returns the product of the two inputs.
  */
 public static G05_BigArithmetic productBA(G05_BigArithmetic a, G05_BigArithmetic b) {
   int maxSize = Math.max(a.effectiveSize, b.effectiveSize); // Get the max size of the two lists
   // If max size is one - normal single digit multiplication
   if (maxSize <= 1) {
     return new G05_BigArithmetic(a.number.get(0) * b.number.get(0));
   }
   // If any of the lists is small - pad zeros to make the size equal
   if (a.number.size() < maxSize) {
     a.addZero(maxSize - a.number.size());
   } else if (b.number.size() < maxSize) {
     b.addZero(maxSize - b.number.size());
   }
   // Get the middle of the two lists
   int mid = (maxSize + 1) / 2;
   // Find a1, a2, b1 and b2
   G05_BigArithmetic a1 = a.split(0, mid);
   G05_BigArithmetic a2 = a.split(mid, maxSize);
   G05_BigArithmetic b1 = b.split(0, mid);
   G05_BigArithmetic b2 = b.split(mid, maxSize);
   G05_BigArithmetic temp1 = productBA(a1, b1); // a1*b1
   G05_BigArithmetic temp2 = productBA(a2, b2); // a2*b2
   G05_BigArithmetic temp12 =
       productBA(G05_BigArithmetic.add(a1, a2), G05_BigArithmetic.add(b1, b2)); // (a1+a2)*(b1+b2)
   // [(a1+a2)*(b1+b2) - {(a1*b1) + (a2*b2)}]
   G05_BigArithmetic temp =
       G05_BigArithmetic.subtract(temp12, G05_BigArithmetic.add(temp1, temp2));
   temp12 = temp;
   // Pad zeroes - similar to raising to base
   temp12 = temp12.addLeastZero(mid);
   temp2 = temp2.addLeastZero(mid * 2);
   // Return the product while removing MS Zeros
   return G05_BigArithmetic.add(G05_BigArithmetic.add(temp2, temp12), temp1).removeMSZeros();
 }
Esempio n. 4
0
 /**
  * Method to subtract two large numbers
  *
  * @param a : Input a - an object of the BigArithmetic Class.
  * @param b : Input b - Second object of the BigArithmetic Class.
  * @return : An object of the type BigArithmetic, that contains the result of the subtraction.
  */
 public static G05_BigArithmetic subtract(G05_BigArithmetic a, G05_BigArithmetic b) {
   boolean sign = false, bo = false; // Two boolean flags to represent sign and borrow
   /*
    * If second number is bigger than first, swap the two and set sign to true
    * We will return zero if sign is still set to true - Level 1 - No negative nos.
    */
   if (a.number.size() < b.number.size()) {
     G05_BigArithmetic t = b;
     b = a;
     a = t;
     sign = true;
   }
   /*
    * If the size of the two numbers are equal, then iterate to see if second is bigger than first.
    * If yes, swap and set sign as true like the previous case.
    * Else, we are good to continue with the subtraction, so break out of the loop.
    */
   else if (a.number.size() == b.number.size()) {
     for (int i = a.number.size() - 1; i >= 0; i--) {
       if (a.number.get(i) < b.number.get(i)) {
         sign = true;
         G05_BigArithmetic t = b;
         b = a;
         a = t;
         continue;
       } else break;
     }
   }
   // Iterators for looping through the linked list.
   Iterator<Long> ita = a.number.iterator();
   Iterator<Long> itb = b.number.iterator();
   G05_BigArithmetic result = new G05_BigArithmetic();
   Long diff, firstElement = nextElement(ita), secondElement = nextElement(itb);
   while (secondElement != null) {
     if (bo) {
       if (firstElement == 0) {
         firstElement = 9L;
         bo = true;
       } else {
         firstElement = firstElement - 1;
         bo = false;
       }
     }
     if (firstElement < secondElement) {
       firstElement += 10;
       bo = true;
     }
     diff = firstElement - secondElement;
     result.number.add(diff % B);
     firstElement = nextElement(ita);
     secondElement = nextElement(itb);
   }
   while (firstElement != null) {
     if (bo) {
       if (firstElement == 0) {
         firstElement = 9L;
         bo = true;
       } else {
         firstElement = firstElement - 1;
         bo = false;
       }
     }
     result.number.add(firstElement);
     firstElement = nextElement(ita);
   }
   // If sign, result is a negative number - so return zero
   if (sign) {
     return new G05_BigArithmetic(0L);
   }
   // Remove the most significant zeros from the list.
   result.removeMSZeros();
   // If after removing the zeros the size is zero, we return the answer as zero
   if (result.effectiveSize == 0) {
     return new G05_BigArithmetic(0L);
   }
   return result;
 }