3

At the bottom of Page 264 of CLRS, the authors say after obtaining r0 = 17612864, the 14 most significant bits of r0 yield the hash value h(k) = 67. I do not understand why it gives 67 since 67 in binary is 1000011 which is 7 bits.

EDIT In the textbook: As an example, suppose we have k = 123456, p = 14, m = 2^14 = 16384, and w = 32. Adapting Knuth's suggestion, we choose A to be the fraction of the form s/2^32 that is closest to (\sqrt(5) - 1) / 2, so that A = 2654435769/2^32. Then k*s = 327706022297664 = (76300 * 2^32) + 17612864, and so r1 = 76300 and r0 = 17612864. The 14 most significant bits of r0 yield the value h(k)=67.

squancy
  • 565
  • 1
  • 7
  • 25
zxia31
  • 43
  • 5
  • 2
    I do not know how possible it would be, but including here a bit more of the context instead of just the page number on the book might help to find someone that can answer your question. – David Bejar Nov 02 '12 at 02:35
  • Look at the figure 11.4 (at the top of the page 264), `w` is 32, you extract `p` (14) bits from the left (most significant) of a 32-bit number. – Esailija Nov 02 '12 at 02:51
  • Thank you! Now I can understand this figure. – zxia31 Nov 02 '12 at 03:00

2 Answers2

9

17612864 = 0x010CC040 = 

0000 0001 0000 1100 1100 0000 0100 0000

Most significant 14 bits of that is

0000 0001 0000 11

Which is 0x43, which is 67

Also:

int32 input = 17612864;
int32 output = input >> (32-14); //67
Esailija
  • 138,174
  • 23
  • 272
  • 326
  • I did not take the machine word size into account. Indeed, the authors mention before that suppose the word size of the machine is w bits and that k fits into a single word. I did not fully understand its meaning. – zxia31 Nov 02 '12 at 02:54
3

In a 32 bit world

17612864 = 00000001 00001100 11000000 01000000 (binary)

top fourteen bits = 00000001 000011 = 67

Thilo
  • 257,207
  • 101
  • 511
  • 656