src/java.base/share/classes/java/math/BigInteger.java

8247782: typos in java.math
Reviewed-by: rriggs, lancea, darcy

*** 2749,2759 ****
  
              // Calculate new base from m1
              BigInteger base2 = (this.signum < 0 || this.compareTo(m1) >= 0
                                  ? this.mod(m1) : this);
  
!             // Caculate (base ** exponent) mod m1.
              BigInteger a1 = (m1.equals(ONE) ? ZERO :
                               base2.oddModPow(exponent, m1));
  
              // Calculate (this ** exponent) mod m2
              BigInteger a2 = base.modPow2(exponent, p);
--- 2749,2759 ----
  
              // Calculate new base from m1
              BigInteger base2 = (this.signum < 0 || this.compareTo(m1) >= 0
                                  ? this.mod(m1) : this);
  
!             // Calculate (base ** exponent) mod m1.
              BigInteger a1 = (m1.equals(ONE) ? ZERO :
                               base2.oddModPow(exponent, m1));
  
              // Calculate (this ** exponent) mod m2
              BigInteger a2 = base.modPow2(exponent, p);
*** 2903,2913 ****
       * the buffer and save a squaring.
       *
       * This means that if you have a k-bit window, to compute n^z,
       * where z is the high k bits of the exponent, 1/2 of the time
       * it requires no squarings.  1/4 of the time, it requires 1
!      * squaring, ... 1/2^(k-1) of the time, it reqires k-2 squarings.
       * And the remaining 1/2^(k-1) of the time, the top k bits are a
       * 1 followed by k-1 0 bits, so it again only requires k-2
       * squarings, not k-1.  The average of these is 1.  Add that
       * to the one squaring we have to do to compute the table,
       * and you'll see that a k-bit window saves k-2 squarings
--- 2903,2913 ----
       * the buffer and save a squaring.
       *
       * This means that if you have a k-bit window, to compute n^z,
       * where z is the high k bits of the exponent, 1/2 of the time
       * it requires no squarings.  1/4 of the time, it requires 1
!      * squaring, ... 1/2^(k-1) of the time, it requires k-2 squarings.
       * And the remaining 1/2^(k-1) of the time, the top k bits are a
       * 1 followed by k-1 0 bits, so it again only requires k-2
       * squarings, not k-1.  The average of these is 1.  Add that
       * to the one squaring we have to do to compute the table,
       * and you'll see that a k-bit window saves k-2 squarings