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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 8 hours ago
tedted
1133
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 8 hours ago
tedted
1133
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 8 hours ago
tedted
1133
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 8 hours ago
tedted
1133
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 8 hours ago
tedted
1133
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 8 hours ago
tedted
1133
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 8 hours ago
tedted
1133
asked 8 hours ago
tedted
1133
1133
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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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 5 hours ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
5 hours 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
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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 5 hours ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
5 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.
answered 5 hours ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
5 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.
answered 5 hours 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 5 hours ago
Yuval FilmusYuval Filmus
196k15184349
answered 5 hours ago
Yuval FilmusYuval Filmus
196k15184349
answered 5 hours ago
Yuval FilmusYuval Filmus
196k15184349
answered 5 hours ago
Yuval FilmusYuval Filmus
196k15184349
196k15184349
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
5 hours ago
add a comment |
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
5 hours ago
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
5 hours ago
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
5 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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