Compute hash value according to multiplication method












2












$begingroup$


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



enter image description here



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





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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:



    enter image description here



    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









    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.







    $endgroup$















      2












      2








      2





      $begingroup$


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



      enter image description here



      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









      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.







      $endgroup$




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



      enter image description here



      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




      hash python






      share|cite|improve this question









      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









      New contributor




      ted is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question








      edited 1 hour ago









      user02814

      1031




      1031






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      asked 9 hours ago









      tedted

      1134




      1134




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          1 Answer
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          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Makes sense, I had forgotten about this step thanks
            $endgroup$
            – ted
            7 hours ago














          Your Answer





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          1 Answer
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          1 Answer
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          active

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          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Makes sense, I had forgotten about this step thanks
            $endgroup$
            – ted
            7 hours ago


















          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









          $endgroup$













          • $begingroup$
            Makes sense, I had forgotten about this step thanks
            $endgroup$
            – ted
            7 hours ago
















          3












          3








          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









          $endgroup$



          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












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          Yuval FilmusYuval Filmus

          196k15184349




          196k15184349












          • $begingroup$
            Makes sense, I had forgotten about this step thanks
            $endgroup$
            – ted
            7 hours ago




















          • $begingroup$
            Makes sense, I had forgotten about this step thanks
            $endgroup$
            – ted
            7 hours ago


















          $begingroup$
          Makes sense, I had forgotten about this step thanks
          $endgroup$
          – ted
          7 hours ago






          $begingroup$
          Makes sense, I had forgotten about this step thanks
          $endgroup$
          – ted
          7 hours ago












          ted is a new contributor. Be nice, and check out our Code of Conduct.










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