Why was the term “discrete” used in discrete logarithm? Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Trying to better understand the failure of the Index Calculus for ECDLPWhat is so special about elliptic curves?Why is the discrete logarithm problem assumed to be hard?What is the difference between discrete logarithm and logarithm?Calculating the discrete logarithmWhy is NON DISCRETE logarithm problem not hard as the DISCRETE logarithm problem (so computationally hard)?How to construct a hash function into a cyclic group such that its discrete log is intractable?Discrete logarithm key sizes for very short term usageDiscrete Logarithm NotationDescribing Discrete Logarithm Assumption
Withdrew £2800, but only £2000 shows as withdrawn on online banking; what are my obligations?
Storing hydrofluoric acid before the invention of plastics
When -s is used with third person singular. What's its use in this context?
Can a non-EU citizen traveling with me come with me through the EU passport line?
What are the motives behind Cersei's orders given to Bronn?
Bonus calculation: Am I making a mountain out of a molehill?
What do you call a phrase that's not an idiom yet?
Does surprise arrest existing movement?
What is the longest distance a 13th-level monk can jump while attacking on the same turn?
What causes the vertical darker bands in my photo?
ListPlot join points by nearest neighbor rather than order
Letter Boxed validator
Gastric acid as a weapon
Check which numbers satisfy the condition [A*B*C = A! + B! + C!]
How to deal with a team lead who never gives me credit?
What makes black pepper strong or mild?
Models of set theory where not every set can be linearly ordered
How can I make names more distinctive without making them longer?
How widely used is the term Treppenwitz? Is it something that most Germans know?
Why is "Captain Marvel" translated as male in Portugal?
What does the "x" in "x86" represent?
Why one of virtual NICs called bond0?
Is 1 ppb equal to 1 μg/kg?
Does accepting a pardon have any bearing on trying that person for the same crime in a sovereign jurisdiction?
Why was the term “discrete” used in discrete logarithm?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Trying to better understand the failure of the Index Calculus for ECDLPWhat is so special about elliptic curves?Why is the discrete logarithm problem assumed to be hard?What is the difference between discrete logarithm and logarithm?Calculating the discrete logarithmWhy is NON DISCRETE logarithm problem not hard as the DISCRETE logarithm problem (so computationally hard)?How to construct a hash function into a cyclic group such that its discrete log is intractable?Discrete logarithm key sizes for very short term usageDiscrete Logarithm NotationDescribing Discrete Logarithm Assumption
$begingroup$
Is there anything especially "discrete" about a discrete logarithm? This is not a question of what is a discrete logarithm or why the discrete logarithm problem is an "intractable problem" given certain circumstances. I'm just trying to determine if there's some additional meaning to the term "discrete" as it's used in name discrete logarithm?
The definition of "discrete" is "individually separate and distinct". Could it be that the term "discrete" is a reference to the least non-negative residues of a modulus or the order of points for a particular cyclic group on an elliptic curve?
discrete-logarithm terminology
$endgroup$
add a comment |
$begingroup$
Is there anything especially "discrete" about a discrete logarithm? This is not a question of what is a discrete logarithm or why the discrete logarithm problem is an "intractable problem" given certain circumstances. I'm just trying to determine if there's some additional meaning to the term "discrete" as it's used in name discrete logarithm?
The definition of "discrete" is "individually separate and distinct". Could it be that the term "discrete" is a reference to the least non-negative residues of a modulus or the order of points for a particular cyclic group on an elliptic curve?
discrete-logarithm terminology
$endgroup$
4
$begingroup$
Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
$endgroup$
– Mikero
7 hours ago
2
$begingroup$
See also discrete mathematics
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
add a comment |
$begingroup$
Is there anything especially "discrete" about a discrete logarithm? This is not a question of what is a discrete logarithm or why the discrete logarithm problem is an "intractable problem" given certain circumstances. I'm just trying to determine if there's some additional meaning to the term "discrete" as it's used in name discrete logarithm?
The definition of "discrete" is "individually separate and distinct". Could it be that the term "discrete" is a reference to the least non-negative residues of a modulus or the order of points for a particular cyclic group on an elliptic curve?
discrete-logarithm terminology
$endgroup$
Is there anything especially "discrete" about a discrete logarithm? This is not a question of what is a discrete logarithm or why the discrete logarithm problem is an "intractable problem" given certain circumstances. I'm just trying to determine if there's some additional meaning to the term "discrete" as it's used in name discrete logarithm?
The definition of "discrete" is "individually separate and distinct". Could it be that the term "discrete" is a reference to the least non-negative residues of a modulus or the order of points for a particular cyclic group on an elliptic curve?
discrete-logarithm terminology
discrete-logarithm terminology
asked 7 hours ago
JohnGaltJohnGalt
24816
24816
4
$begingroup$
Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
$endgroup$
– Mikero
7 hours ago
2
$begingroup$
See also discrete mathematics
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
add a comment |
4
$begingroup$
Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
$endgroup$
– Mikero
7 hours ago
2
$begingroup$
See also discrete mathematics
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
4
4
$begingroup$
Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
$endgroup$
– Mikero
7 hours ago
$begingroup$
Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
$endgroup$
– Mikero
7 hours ago
2
2
$begingroup$
See also discrete mathematics
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
$begingroup$
See also discrete mathematics
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The word discrete is used as an antonym of 'continuous', that is, it is the normal logarithmic problem, just over a discrete group.
The standard logarithmic problem is over the infinite group $mathbbR^*$, this group is called 'continuous', because for any element $x$, there are other elements that are arbitrarily close to it.
The discrete logarithmic problem is over a finite group (for example, $mathbbZ_p^*$); in contrast to $mathbbR^*$, we don't have group elements arbitrarily close together; we call this type of group 'discrete'.
$endgroup$
1
$begingroup$
yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
$endgroup$
– Geoffroy Couteau
7 hours ago
3
$begingroup$
@JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
$endgroup$
– Mark
7 hours ago
1
$begingroup$
my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
$endgroup$
– Geoffroy Couteau
7 hours ago
1
$begingroup$
@GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
$endgroup$
– Mark
7 hours ago
2
$begingroup$
@Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
$endgroup$
– Geoffroy Couteau
7 hours ago
|
show 5 more comments
$begingroup$
While I agree completely with poncho's answer, this other viewpoint might be useful.
Specifically, I think a better comparison isn't between $mathbbZ_p^*$ and $mathbbR^*$, but with $mathbbZ_p^*$ and $S^1$. We can view $S^1 cong zinmathbbC mid $. It's not hard to show that any $zin S^1$ is able to be written as $z = exp(2pi i t)$ for $tinmathbbR$ (we don't strictly need the factor $2pi$ here, but it's traditional). Due to $exp(x)$ being periodic, it's in fact enough to have $tin[0,1)$.
This has an obvious group structure, in that:
$$exp(2pi i t_0)exp(2pi i t_1) = exp(2pi i (t_0+t_1))$$
If we're making the restriction that $t_iin[0,1)$, then we have to take $t_0+t_1mod 1$, but this is fairly standard.
More than just having an obvious group structure, we actually have that any $mathbbZ_p^*$ injects into it.
Specifically, we always have:
$$
phi_p:mathbbZ_p^*to S^1,quad phi_p(x) = exp(2pi i x/(p-1))
$$
Here, $p-1$ in the denominator is because $|mathbbZ_p^*| = p-1$.
We can define the discrete logarithm problem for both of these groups in the standard way (here, it's important to restrict $t_iin[0, 1)$ if we want a unique answer).
Then, we can relate these problems to each via the aforementioned injection.
Through this image, we see that $S^1$ is "continuous" in the sense that it takes up the full circle, but the image of $mathbbZ_p^*$ in $S^1$ will always be "discrete" --- there will always be "some space" between points (they can't get arbitrarily close).
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "281"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcrypto.stackexchange.com%2fquestions%2f68801%2fwhy-was-the-term-discrete-used-in-discrete-logarithm%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The word discrete is used as an antonym of 'continuous', that is, it is the normal logarithmic problem, just over a discrete group.
The standard logarithmic problem is over the infinite group $mathbbR^*$, this group is called 'continuous', because for any element $x$, there are other elements that are arbitrarily close to it.
The discrete logarithmic problem is over a finite group (for example, $mathbbZ_p^*$); in contrast to $mathbbR^*$, we don't have group elements arbitrarily close together; we call this type of group 'discrete'.
$endgroup$
1
$begingroup$
yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
$endgroup$
– Geoffroy Couteau
7 hours ago
3
$begingroup$
@JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
$endgroup$
– Mark
7 hours ago
1
$begingroup$
my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
$endgroup$
– Geoffroy Couteau
7 hours ago
1
$begingroup$
@GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
$endgroup$
– Mark
7 hours ago
2
$begingroup$
@Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
$endgroup$
– Geoffroy Couteau
7 hours ago
|
show 5 more comments
$begingroup$
The word discrete is used as an antonym of 'continuous', that is, it is the normal logarithmic problem, just over a discrete group.
The standard logarithmic problem is over the infinite group $mathbbR^*$, this group is called 'continuous', because for any element $x$, there are other elements that are arbitrarily close to it.
The discrete logarithmic problem is over a finite group (for example, $mathbbZ_p^*$); in contrast to $mathbbR^*$, we don't have group elements arbitrarily close together; we call this type of group 'discrete'.
$endgroup$
1
$begingroup$
yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
$endgroup$
– Geoffroy Couteau
7 hours ago
3
$begingroup$
@JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
$endgroup$
– Mark
7 hours ago
1
$begingroup$
my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
$endgroup$
– Geoffroy Couteau
7 hours ago
1
$begingroup$
@GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
$endgroup$
– Mark
7 hours ago
2
$begingroup$
@Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
$endgroup$
– Geoffroy Couteau
7 hours ago
|
show 5 more comments
$begingroup$
The word discrete is used as an antonym of 'continuous', that is, it is the normal logarithmic problem, just over a discrete group.
The standard logarithmic problem is over the infinite group $mathbbR^*$, this group is called 'continuous', because for any element $x$, there are other elements that are arbitrarily close to it.
The discrete logarithmic problem is over a finite group (for example, $mathbbZ_p^*$); in contrast to $mathbbR^*$, we don't have group elements arbitrarily close together; we call this type of group 'discrete'.
$endgroup$
The word discrete is used as an antonym of 'continuous', that is, it is the normal logarithmic problem, just over a discrete group.
The standard logarithmic problem is over the infinite group $mathbbR^*$, this group is called 'continuous', because for any element $x$, there are other elements that are arbitrarily close to it.
The discrete logarithmic problem is over a finite group (for example, $mathbbZ_p^*$); in contrast to $mathbbR^*$, we don't have group elements arbitrarily close together; we call this type of group 'discrete'.
answered 7 hours ago
ponchoponcho
94.2k2148247
94.2k2148247
1
$begingroup$
yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
$endgroup$
– Geoffroy Couteau
7 hours ago
3
$begingroup$
@JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
$endgroup$
– Mark
7 hours ago
1
$begingroup$
my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
$endgroup$
– Geoffroy Couteau
7 hours ago
1
$begingroup$
@GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
$endgroup$
– Mark
7 hours ago
2
$begingroup$
@Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
$endgroup$
– Geoffroy Couteau
7 hours ago
|
show 5 more comments
1
$begingroup$
yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
$endgroup$
– Geoffroy Couteau
7 hours ago
3
$begingroup$
@JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
$endgroup$
– Mark
7 hours ago
1
$begingroup$
my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
$endgroup$
– Geoffroy Couteau
7 hours ago
1
$begingroup$
@GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
$endgroup$
– Mark
7 hours ago
2
$begingroup$
@Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
$endgroup$
– Geoffroy Couteau
7 hours ago
1
1
$begingroup$
yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
$endgroup$
– Geoffroy Couteau
7 hours ago
$begingroup$
yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
$endgroup$
– Geoffroy Couteau
7 hours ago
3
3
$begingroup$
@JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
$endgroup$
– Mark
7 hours ago
$begingroup$
@JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
$endgroup$
– Mark
7 hours ago
1
1
$begingroup$
my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
$endgroup$
– Geoffroy Couteau
7 hours ago
$begingroup$
my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
$endgroup$
– Geoffroy Couteau
7 hours ago
1
1
$begingroup$
@GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
$endgroup$
– Mark
7 hours ago
$begingroup$
@GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
$endgroup$
– Mark
7 hours ago
2
2
$begingroup$
@Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
$endgroup$
– Geoffroy Couteau
7 hours ago
$begingroup$
@Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
$endgroup$
– Geoffroy Couteau
7 hours ago
|
show 5 more comments
$begingroup$
While I agree completely with poncho's answer, this other viewpoint might be useful.
Specifically, I think a better comparison isn't between $mathbbZ_p^*$ and $mathbbR^*$, but with $mathbbZ_p^*$ and $S^1$. We can view $S^1 cong zinmathbbC mid $. It's not hard to show that any $zin S^1$ is able to be written as $z = exp(2pi i t)$ for $tinmathbbR$ (we don't strictly need the factor $2pi$ here, but it's traditional). Due to $exp(x)$ being periodic, it's in fact enough to have $tin[0,1)$.
This has an obvious group structure, in that:
$$exp(2pi i t_0)exp(2pi i t_1) = exp(2pi i (t_0+t_1))$$
If we're making the restriction that $t_iin[0,1)$, then we have to take $t_0+t_1mod 1$, but this is fairly standard.
More than just having an obvious group structure, we actually have that any $mathbbZ_p^*$ injects into it.
Specifically, we always have:
$$
phi_p:mathbbZ_p^*to S^1,quad phi_p(x) = exp(2pi i x/(p-1))
$$
Here, $p-1$ in the denominator is because $|mathbbZ_p^*| = p-1$.
We can define the discrete logarithm problem for both of these groups in the standard way (here, it's important to restrict $t_iin[0, 1)$ if we want a unique answer).
Then, we can relate these problems to each via the aforementioned injection.
Through this image, we see that $S^1$ is "continuous" in the sense that it takes up the full circle, but the image of $mathbbZ_p^*$ in $S^1$ will always be "discrete" --- there will always be "some space" between points (they can't get arbitrarily close).
$endgroup$
add a comment |
$begingroup$
While I agree completely with poncho's answer, this other viewpoint might be useful.
Specifically, I think a better comparison isn't between $mathbbZ_p^*$ and $mathbbR^*$, but with $mathbbZ_p^*$ and $S^1$. We can view $S^1 cong zinmathbbC mid $. It's not hard to show that any $zin S^1$ is able to be written as $z = exp(2pi i t)$ for $tinmathbbR$ (we don't strictly need the factor $2pi$ here, but it's traditional). Due to $exp(x)$ being periodic, it's in fact enough to have $tin[0,1)$.
This has an obvious group structure, in that:
$$exp(2pi i t_0)exp(2pi i t_1) = exp(2pi i (t_0+t_1))$$
If we're making the restriction that $t_iin[0,1)$, then we have to take $t_0+t_1mod 1$, but this is fairly standard.
More than just having an obvious group structure, we actually have that any $mathbbZ_p^*$ injects into it.
Specifically, we always have:
$$
phi_p:mathbbZ_p^*to S^1,quad phi_p(x) = exp(2pi i x/(p-1))
$$
Here, $p-1$ in the denominator is because $|mathbbZ_p^*| = p-1$.
We can define the discrete logarithm problem for both of these groups in the standard way (here, it's important to restrict $t_iin[0, 1)$ if we want a unique answer).
Then, we can relate these problems to each via the aforementioned injection.
Through this image, we see that $S^1$ is "continuous" in the sense that it takes up the full circle, but the image of $mathbbZ_p^*$ in $S^1$ will always be "discrete" --- there will always be "some space" between points (they can't get arbitrarily close).
$endgroup$
add a comment |
$begingroup$
While I agree completely with poncho's answer, this other viewpoint might be useful.
Specifically, I think a better comparison isn't between $mathbbZ_p^*$ and $mathbbR^*$, but with $mathbbZ_p^*$ and $S^1$. We can view $S^1 cong zinmathbbC mid $. It's not hard to show that any $zin S^1$ is able to be written as $z = exp(2pi i t)$ for $tinmathbbR$ (we don't strictly need the factor $2pi$ here, but it's traditional). Due to $exp(x)$ being periodic, it's in fact enough to have $tin[0,1)$.
This has an obvious group structure, in that:
$$exp(2pi i t_0)exp(2pi i t_1) = exp(2pi i (t_0+t_1))$$
If we're making the restriction that $t_iin[0,1)$, then we have to take $t_0+t_1mod 1$, but this is fairly standard.
More than just having an obvious group structure, we actually have that any $mathbbZ_p^*$ injects into it.
Specifically, we always have:
$$
phi_p:mathbbZ_p^*to S^1,quad phi_p(x) = exp(2pi i x/(p-1))
$$
Here, $p-1$ in the denominator is because $|mathbbZ_p^*| = p-1$.
We can define the discrete logarithm problem for both of these groups in the standard way (here, it's important to restrict $t_iin[0, 1)$ if we want a unique answer).
Then, we can relate these problems to each via the aforementioned injection.
Through this image, we see that $S^1$ is "continuous" in the sense that it takes up the full circle, but the image of $mathbbZ_p^*$ in $S^1$ will always be "discrete" --- there will always be "some space" between points (they can't get arbitrarily close).
$endgroup$
While I agree completely with poncho's answer, this other viewpoint might be useful.
Specifically, I think a better comparison isn't between $mathbbZ_p^*$ and $mathbbR^*$, but with $mathbbZ_p^*$ and $S^1$. We can view $S^1 cong zinmathbbC mid $. It's not hard to show that any $zin S^1$ is able to be written as $z = exp(2pi i t)$ for $tinmathbbR$ (we don't strictly need the factor $2pi$ here, but it's traditional). Due to $exp(x)$ being periodic, it's in fact enough to have $tin[0,1)$.
This has an obvious group structure, in that:
$$exp(2pi i t_0)exp(2pi i t_1) = exp(2pi i (t_0+t_1))$$
If we're making the restriction that $t_iin[0,1)$, then we have to take $t_0+t_1mod 1$, but this is fairly standard.
More than just having an obvious group structure, we actually have that any $mathbbZ_p^*$ injects into it.
Specifically, we always have:
$$
phi_p:mathbbZ_p^*to S^1,quad phi_p(x) = exp(2pi i x/(p-1))
$$
Here, $p-1$ in the denominator is because $|mathbbZ_p^*| = p-1$.
We can define the discrete logarithm problem for both of these groups in the standard way (here, it's important to restrict $t_iin[0, 1)$ if we want a unique answer).
Then, we can relate these problems to each via the aforementioned injection.
Through this image, we see that $S^1$ is "continuous" in the sense that it takes up the full circle, but the image of $mathbbZ_p^*$ in $S^1$ will always be "discrete" --- there will always be "some space" between points (they can't get arbitrarily close).
answered 7 hours ago
MarkMark
21115
21115
add a comment |
add a comment |
Thanks for contributing an answer to Cryptography Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcrypto.stackexchange.com%2fquestions%2f68801%2fwhy-was-the-term-discrete-used-in-discrete-logarithm%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
4
$begingroup$
Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
$endgroup$
– Mikero
7 hours ago
2
$begingroup$
See also discrete mathematics
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago