Comparing exact expressions vs real numbers
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Often I need to generate some data using some symmetry operations and I usually keep them as exact expressions (for example, consider the points on a triangular grid -(1/2), Sqrt[3]/2, -(1/2), -(Sqrt[3]/2), 1, 0, ...) and I need to compare different points, something like if a1+b==a2. I am trying to find an efficient way to do that.
Consider this
a=-2/Sqrt[3] + Sqrt[3]
b=Sqrt[1/4 + (2/Sqrt[3] - Sqrt[3]/2)^2]
N[a],N[b]
0.57735,0.57735
Now, a==b does not do anything.
N[a] == N[b]
N[a]-N[b] == 0
N[a - b] == 0
True
False
False
Because N[a-b]=-3.33067*10^-16. So the way out is
Chop@N[a - b] == 0
True
However,
RepeatedTiming[N[a] == N[b]]
RepeatedTiming[Chop@N[a - b] == 0]
5.4*10^-6, True
9.07*10^-6, True
On the other hand,
a1 = N[a]; b1 = N[b];
RepeatedTiming[a1 == b1]
2.7*10^-7, True
So my questions are
Is it better to use real numbers if I have to do such comparisons?
What would be the best (least time consuming when dealing with a large number of inputs) way to compare exact expressions if I have to use exact expressions?
numerical-value
asked Mar 8 at 10:19
SumitSumit
11.7k21956
$endgroup$
add a comment |
$begingroup$
Often I need to generate some data using some symmetry operations and I usually keep them as exact expressions (for example, consider the points on a triangular grid -(1/2), Sqrt[3]/2, -(1/2), -(Sqrt[3]/2), 1, 0, ...) and I need to compare different points, something like if a1+b==a2. I am trying to find an efficient way to do that.
Consider this
a=-2/Sqrt[3] + Sqrt[3]
b=Sqrt[1/4 + (2/Sqrt[3] - Sqrt[3]/2)^2]
N[a],N[b]
0.57735,0.57735
Now, a==b does not do anything.
N[a] == N[b]
N[a]-N[b] == 0
N[a - b] == 0
True
False
False
Because N[a-b]=-3.33067*10^-16. So the way out is
Chop@N[a - b] == 0
True
However,
RepeatedTiming[N[a] == N[b]]
RepeatedTiming[Chop@N[a - b] == 0]
5.4*10^-6, True
9.07*10^-6, True
On the other hand,
a1 = N[a]; b1 = N[b];
RepeatedTiming[a1 == b1]
2.7*10^-7, True
So my questions are
Is it better to use real numbers if I have to do such comparisons?
What would be the best (least time consuming when dealing with a large number of inputs) way to compare exact expressions if I have to use exact expressions?
numerical-value
asked Mar 8 at 10:19
SumitSumit
11.7k21956
$endgroup$
2
$begingroup$
You'll want to be careful if you find cases like this.N[Sin[2017 2^(1/5)]] - N[-1]
$endgroup$
– J. M. is away♦
Mar 8 at 10:29
1
$begingroup$
PossibleZeroQcould help.
$endgroup$
– Roman
Mar 8 at 10:48
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Rebel-Scum
Mar 8 at 16:00
add a comment |
$begingroup$
Often I need to generate some data using some symmetry operations and I usually keep them as exact expressions (for example, consider the points on a triangular grid -(1/2), Sqrt[3]/2, -(1/2), -(Sqrt[3]/2), 1, 0, ...) and I need to compare different points, something like if a1+b==a2. I am trying to find an efficient way to do that.
Consider this
a=-2/Sqrt[3] + Sqrt[3]
b=Sqrt[1/4 + (2/Sqrt[3] - Sqrt[3]/2)^2]
N[a],N[b]
0.57735,0.57735
Now, a==b does not do anything.
N[a] == N[b]
N[a]-N[b] == 0
N[a - b] == 0
True
False
False
Because N[a-b]=-3.33067*10^-16. So the way out is
Chop@N[a - b] == 0
True
However,
RepeatedTiming[N[a] == N[b]]
RepeatedTiming[Chop@N[a - b] == 0]
5.4*10^-6, True
9.07*10^-6, True
On the other hand,
a1 = N[a]; b1 = N[b];
RepeatedTiming[a1 == b1]
2.7*10^-7, True
So my questions are
Is it better to use real numbers if I have to do such comparisons?
What would be the best (least time consuming when dealing with a large number of inputs) way to compare exact expressions if I have to use exact expressions?
numerical-value
asked Mar 8 at 10:19
SumitSumit
11.7k21956
$endgroup$
Often I need to generate some data using some symmetry operations and I usually keep them as exact expressions (for example, consider the points on a triangular grid -(1/2), Sqrt[3]/2, -(1/2), -(Sqrt[3]/2), 1, 0, ...) and I need to compare different points, something like if a1+b==a2. I am trying to find an efficient way to do that.
Consider this
a=-2/Sqrt[3] + Sqrt[3]
b=Sqrt[1/4 + (2/Sqrt[3] - Sqrt[3]/2)^2]
N[a],N[b]
0.57735,0.57735
Now, a==b does not do anything.
N[a] == N[b]
N[a]-N[b] == 0
N[a - b] == 0
True
False
False
Because N[a-b]=-3.33067*10^-16. So the way out is
Chop@N[a - b] == 0
True
However,
RepeatedTiming[N[a] == N[b]]
RepeatedTiming[Chop@N[a - b] == 0]
5.4*10^-6, True
9.07*10^-6, True
On the other hand,
a1 = N[a]; b1 = N[b];
RepeatedTiming[a1 == b1]
2.7*10^-7, True
So my questions are
Is it better to use real numbers if I have to do such comparisons?
What would be the best (least time consuming when dealing with a large number of inputs) way to compare exact expressions if I have to use exact expressions?
numerical-value
numerical-value
asked Mar 8 at 10:19
SumitSumit
11.7k21956
asked Mar 8 at 10:19
SumitSumit
11.7k21956
asked Mar 8 at 10:19
SumitSumit
11.7k21956
asked Mar 8 at 10:19
SumitSumit
11.7k21956
asked Mar 8 at 10:19
SumitSumit
11.7k21956
11.7k21956
2
$begingroup$
You'll want to be careful if you find cases like this.N[Sin[2017 2^(1/5)]] - N[-1]
$endgroup$
– J. M. is away♦
Mar 8 at 10:29
1
$begingroup$
PossibleZeroQcould help.
$endgroup$
– Roman
Mar 8 at 10:48
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Rebel-Scum
Mar 8 at 16:00
add a comment |
2
$begingroup$
You'll want to be careful if you find cases like this.N[Sin[2017 2^(1/5)]] - N[-1]
$endgroup$
– J. M. is away♦
Mar 8 at 10:29
1
$begingroup$
PossibleZeroQcould help.
$endgroup$
– Roman
Mar 8 at 10:48
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Rebel-Scum
Mar 8 at 16:00
2
2
$begingroup$
You'll want to be careful if you find cases like this.
N[Sin[2017 2^(1/5)]] - N[-1]$endgroup$
– J. M. is away♦
Mar 8 at 10:29
$begingroup$
You'll want to be careful if you find cases like this.
N[Sin[2017 2^(1/5)]] - N[-1]$endgroup$
– J. M. is away♦
Mar 8 at 10:29
1
1
$begingroup$
PossibleZeroQ could help.$endgroup$
– Roman
Mar 8 at 10:48
$begingroup$
PossibleZeroQ could help.$endgroup$
– Roman
Mar 8 at 10:48
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Rebel-Scum
Mar 8 at 16:00
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Rebel-Scum
Mar 8 at 16:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
PossibleZeroQ is rather fast and does precisely what you're looking for:
RepeatedTiming[PossibleZeroQ[a - b]]
3.2*10^-6, True
@JM's difficult case is handled correctly:
PossibleZeroQ[Sin[2017 2^(1/5)] - (-1)]
False
The limits of PossibleZeroQ can be fine-tuned with $MaxExtraPrecision.
answered Mar 8 at 13:00
RomanRoman
4,4801127
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
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
,
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%2fmathematica.stackexchange.com%2fquestions%2f192869%2fcomparing-exact-expressions-vs-real-numbers%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
PossibleZeroQ is rather fast and does precisely what you're looking for:
RepeatedTiming[PossibleZeroQ[a - b]]
3.2*10^-6, True
@JM's difficult case is handled correctly:
PossibleZeroQ[Sin[2017 2^(1/5)] - (-1)]
False
The limits of PossibleZeroQ can be fine-tuned with $MaxExtraPrecision.
answered Mar 8 at 13:00
RomanRoman
4,4801127
$endgroup$
add a comment |
$begingroup$
PossibleZeroQ is rather fast and does precisely what you're looking for:
RepeatedTiming[PossibleZeroQ[a - b]]
3.2*10^-6, True
@JM's difficult case is handled correctly:
PossibleZeroQ[Sin[2017 2^(1/5)] - (-1)]
False
The limits of PossibleZeroQ can be fine-tuned with $MaxExtraPrecision.
answered Mar 8 at 13:00
RomanRoman
4,4801127
$endgroup$
add a comment |
$begingroup$
PossibleZeroQ is rather fast and does precisely what you're looking for:
RepeatedTiming[PossibleZeroQ[a - b]]
3.2*10^-6, True
@JM's difficult case is handled correctly:
PossibleZeroQ[Sin[2017 2^(1/5)] - (-1)]
False
The limits of PossibleZeroQ can be fine-tuned with $MaxExtraPrecision.
answered Mar 8 at 13:00
RomanRoman
4,4801127
$endgroup$
PossibleZeroQ is rather fast and does precisely what you're looking for:
RepeatedTiming[PossibleZeroQ[a - b]]
3.2*10^-6, True
@JM's difficult case is handled correctly:
PossibleZeroQ[Sin[2017 2^(1/5)] - (-1)]
False
The limits of PossibleZeroQ can be fine-tuned with $MaxExtraPrecision.
answered Mar 8 at 13:00
RomanRoman
4,4801127
edited Mar 9 at 2:29
answered Mar 8 at 13:00
RomanRoman
4,4801127
answered Mar 8 at 13:00
RomanRoman
4,4801127
answered Mar 8 at 13:00
RomanRoman
4,4801127
4,4801127
add a comment |
add a comment |
Thanks for contributing an answer to Mathematica 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%2fmathematica.stackexchange.com%2fquestions%2f192869%2fcomparing-exact-expressions-vs-real-numbers%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
2
$begingroup$
You'll want to be careful if you find cases like this.
N[Sin[2017 2^(1/5)]] - N[-1]$endgroup$
– J. M. is away♦
Mar 8 at 10:29
1
$begingroup$
PossibleZeroQcould help.$endgroup$
– Roman
Mar 8 at 10:48
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Rebel-Scum
Mar 8 at 16:00