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2

$begingroup$

In "Introduction to Algorithms" by C. E. Leiserson, R. L. Rivest and C. Stein (ISBN: 978-0262033848), p. 264 they state this:

I get everything but the last part stating $h(k) = 67$

>>> r = 17612864
>>> bin(r) # r's binary representation
'0b1000011001100000001000000'
>>> int(bin(r)[: 14 + 2], 2) # extract 14 most significant bits and convert to int
8600

hash python

share|cite|improve this question

edited Apr 8 at 2:34

user02814

1031

asked Apr 7 at 17:57

tedted

1134

New contributor

ted is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

$begingroup$

In "Introduction to Algorithms" by C. E. Leiserson, R. L. Rivest and C. Stein (ISBN: 978-0262033848), p. 264 they state this:

I get everything but the last part stating $h(k) = 67$

>>> r = 17612864
>>> bin(r) # r's binary representation
'0b1000011001100000001000000'
>>> int(bin(r)[: 14 + 2], 2) # extract 14 most significant bits and convert to int
8600

hash python

share|cite|improve this question

edited Apr 8 at 2:34

user02814

1031

asked Apr 7 at 17:57

tedted

1134

New contributor

ted is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

In "Introduction to Algorithms" by C. E. Leiserson, R. L. Rivest and C. Stein (ISBN: 978-0262033848), p. 264 they state this:

I get everything but the last part stating $h(k) = 67$

>>> r = 17612864
>>> bin(r) # r's binary representation
'0b1000011001100000001000000'
>>> int(bin(r)[: 14 + 2], 2) # extract 14 most significant bits and convert to int
8600

hash python

share|cite|improve this question

edited Apr 8 at 2:34

user02814

1031

asked Apr 7 at 17:57

tedted

1134

New contributor

ted is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

>>> r = 17612864
>>> bin(r) # r's binary representation
'0b1000011001100000001000000'
>>> int(bin(r)[: 14 + 2], 2) # extract 14 most significant bits and convert to int
8600

share|cite|improve this question

edited Apr 8 at 2:34

user02814

1031

asked Apr 7 at 17:57

tedted

1134

New contributor

ted is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Apr 8 at 2:34

user02814

1031

asked Apr 7 at 17:57

tedted

1134

New contributor

ted is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Apr 8 at 2:34

user02814

1031

edited Apr 8 at 2:34

user02814

1031

asked Apr 7 at 17:57

tedted

1134

New contributor

ted is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 7 at 17:57

tedted

1134

1 Answer
1

active

oldest

votes

3

$begingroup$

You haven't extracted the 14 most significant bits. First, you have to write $r$ as a $w$-bit number:
$$
00000001000011001100000001000000
$$
Now you extract the 14 most significant bits:
$$
00000001000011
$$
Converting to decimal, this is 67.

share|cite|improve this answer

answered Apr 7 at 20:32

Yuval FilmusYuval Filmus

197k15185349

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Makes sense, I had forgotten about this step thanks
  $endgroup$
  – ted
  Apr 7 at 20:41
  
  

  
  
  
  

add a comment | 

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3

$begingroup$

You haven't extracted the 14 most significant bits. First, you have to write $r$ as a $w$-bit number:
$$
00000001000011001100000001000000
$$
Now you extract the 14 most significant bits:
$$
00000001000011
$$
Converting to decimal, this is 67.

share|cite|improve this answer

answered Apr 7 at 20:32

Yuval FilmusYuval Filmus

197k15185349

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Makes sense, I had forgotten about this step thanks
  $endgroup$
  – ted
  Apr 7 at 20:41
  
  

  
  
  
  

add a comment | 

3

$begingroup$

You haven't extracted the 14 most significant bits. First, you have to write $r$ as a $w$-bit number:
$$
00000001000011001100000001000000
$$
Now you extract the 14 most significant bits:
$$
00000001000011
$$
Converting to decimal, this is 67.

share|cite|improve this answer

answered Apr 7 at 20:32

Yuval FilmusYuval Filmus

197k15185349

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Makes sense, I had forgotten about this step thanks
  $endgroup$
  – ted
  Apr 7 at 20:41
  
  

  
  
  
  

add a comment | 

$begingroup$

You haven't extracted the 14 most significant bits. First, you have to write $r$ as a $w$-bit number:
$$
00000001000011001100000001000000
$$
Now you extract the 14 most significant bits:
$$
00000001000011
$$
Converting to decimal, this is 67.

share|cite|improve this answer

answered Apr 7 at 20:32

Yuval FilmusYuval Filmus

197k15185349

$endgroup$

share|cite|improve this answer

answered Apr 7 at 20:32

Yuval FilmusYuval Filmus

197k15185349

answered Apr 7 at 20:32

Yuval FilmusYuval Filmus

197k15185349

answered Apr 7 at 20:32

Yuval FilmusYuval Filmus

197k15185349