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Compute hash value according to multiplication method
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?How would you implement truly random hash functions in practice?Why does this particular hashCode function help decrease collisions?Constraint on Universal set of hash functionsChoosing a non-cryptographic hash function for language with no unsigned integersHash size: do prime numbers “near” powers of two are bad?Function to generate longer bit-sequence from shorter sequence with certain propertiesRolling Hash calculation with Horner's methodWhat is a minimal, pseudo-random hash function?
$begingroup$
In Introduction to Algorithms, CLR, p264 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
asked 3 hours ago
tedted
1083
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, CLR, p264 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
asked 3 hours ago
tedted
1083
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, CLR, p264 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
asked 3 hours ago
tedted
1083
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, CLR, p264 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
asked 3 hours ago
tedted
1083
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 3 hours ago
tedted
1083
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 3 hours ago
tedted
1083
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 3 hours ago
tedted
1083
asked 3 hours ago
tedted
1083
1083
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.
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.
ted is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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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.
answered 47 mins ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
38 mins ago
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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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.
answered 47 mins ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
38 mins 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.
answered 47 mins ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
38 mins 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.
answered 47 mins ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
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 47 mins ago
Yuval FilmusYuval Filmus
196k15184349
answered 47 mins ago
Yuval FilmusYuval Filmus
196k15184349
answered 47 mins ago
Yuval FilmusYuval Filmus
196k15184349
answered 47 mins ago
Yuval FilmusYuval Filmus
196k15184349
196k15184349
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
38 mins ago
add a comment |
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
38 mins ago
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
38 mins ago
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
38 mins 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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