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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?
$begingroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
hash python
New contributor
New contributor
edited 1 hour ago
user02814
1031
1031
New contributor
asked 10 hours ago
tedted
1134
1134
New contributor
New contributor
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
7 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
7 hours ago
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.
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
7 hours ago
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.
$endgroup$
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.
answered 7 hours ago
Yuval FilmusYuval Filmus
196k15184349
196k15184349
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
7 hours ago
add a comment |
$begingroup$
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
add a comment |
ted is a new contributor. Be nice, and check out our Code of Conduct.
ted is a new contributor. Be nice, and check out our Code of Conduct.
ted is a new contributor. Be nice, and check out our Code of Conduct.
ted is a new contributor. Be nice, and check out our Code of Conduct.
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