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Compute hash value according to multiplication method


Collision resistant hash functionMin/max of hash function (Whirlpool)Hashing by doing modulo $m$ for $m=p^2$ for a prime $p$ instead of using a prime $m$ - is it that bad?Why having a simple multiplication loop and very good avalanche isn't enough to produce well-distributed hash values?Building static hash table with particular collisionsUnderstanding of hash tablesUniversal family of hash functionsTruth value of a propositionRolling Hash calculation with Horner's methodIs the capacity of a hash table a constant value?













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









share|cite|improve this question











$endgroup$
















    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









    share|cite|improve this question











    $endgroup$














      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









      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 8 at 2:34









      user02814

      1031




      1031










      asked Apr 7 at 17:57









      tedted

      1134




      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









          $endgroup$












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












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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          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









          $endgroup$












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
















          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
            Apr 7 at 20:41














          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 Apr 7 at 20:32









          Yuval FilmusYuval Filmus

          197k15187350




          197k15187350











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

















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
















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





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


















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