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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
edited yesterday
user02814
1031
asked yesterday
tedted
1134
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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
edited yesterday
user02814
1031
asked yesterday
tedted
1134
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" 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
edited yesterday
user02814
1031
asked yesterday
tedted
1134
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:
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
edited yesterday
user02814
1031
asked yesterday
tedted
1134
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.
edited yesterday
user02814
1031
asked yesterday
tedted
1134
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.
edited yesterday
user02814
1031
edited yesterday
user02814
1031
edited yesterday
user02814
1031
1031
asked yesterday
tedted
1134
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 yesterday
tedted
1134
asked yesterday
tedted
1134
1134
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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$begingroup$
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 yesterday
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
yesterday
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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.
answered yesterday
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
yesterday
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 yesterday
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
yesterday
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 yesterday
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 yesterday
Yuval FilmusYuval Filmus
196k15184349
answered yesterday
Yuval FilmusYuval Filmus
196k15184349
answered yesterday
Yuval FilmusYuval Filmus
196k15184349
answered yesterday
Yuval FilmusYuval Filmus
196k15184349
196k15184349
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
yesterday
add a comment |
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
yesterday
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
yesterday
$begingroup$
Makes sense, I had forgotten about this step thanks
$endgroup$
– ted
yesterday
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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