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Comparing exact expressions vs real numbers
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$begingroup$
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 16 hours ago
SumitSumit
11.7k21955
$endgroup$
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 16 hours ago
SumitSumit
11.7k21955
$endgroup$
1
$begingroup$
You'll want to be careful if you find cases like this.N[Sin[2017 2^(1/5)]] - N[-1]
$endgroup$
– J. M. is computer-less♦
16 hours ago
1
$begingroup$
PossibleZeroQcould help.
$endgroup$
– Roman
16 hours ago
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Loki
10 hours ago
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 16 hours ago
SumitSumit
11.7k21955
$endgroup$
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 16 hours ago
SumitSumit
11.7k21955
asked 16 hours ago
SumitSumit
11.7k21955
asked 16 hours ago
SumitSumit
11.7k21955
asked 16 hours ago
SumitSumit
11.7k21955
asked 16 hours ago
SumitSumit
11.7k21955
11.7k21955
1
$begingroup$
You'll want to be careful if you find cases like this.N[Sin[2017 2^(1/5)]] - N[-1]
$endgroup$
– J. M. is computer-less♦
16 hours ago
1
$begingroup$
PossibleZeroQcould help.
$endgroup$
– Roman
16 hours ago
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Loki
10 hours ago
add a comment |
1
$begingroup$
You'll want to be careful if you find cases like this.N[Sin[2017 2^(1/5)]] - N[-1]
$endgroup$
– J. M. is computer-less♦
16 hours ago
1
$begingroup$
PossibleZeroQcould help.
$endgroup$
– Roman
16 hours ago
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Loki
10 hours ago
1
1
$begingroup$
You'll want to be careful if you find cases like this.
N[Sin[2017 2^(1/5)]] - N[-1]$endgroup$
– J. M. is computer-less♦
16 hours ago
$begingroup$
You'll want to be careful if you find cases like this.
N[Sin[2017 2^(1/5)]] - N[-1]$endgroup$
– J. M. is computer-less♦
16 hours ago
1
1
$begingroup$
PossibleZeroQ could help.$endgroup$
– Roman
16 hours ago
$begingroup$
PossibleZeroQ could help.$endgroup$
– Roman
16 hours ago
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Loki
10 hours ago
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Loki
10 hours ago
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 13 hours ago
RomanRoman
2,735717
$endgroup$
add a comment |
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
votes
active
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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 13 hours ago
RomanRoman
2,735717
$endgroup$
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 13 hours ago
RomanRoman
2,735717
$endgroup$
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 13 hours ago
RomanRoman
2,735717
$endgroup$
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 13 hours ago
RomanRoman
2,735717
edited 20 mins ago
answered 13 hours ago
RomanRoman
2,735717
answered 13 hours ago
RomanRoman
2,735717
answered 13 hours ago
RomanRoman
2,735717
2,735717
add a comment |
add a comment |
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1
$begingroup$
You'll want to be careful if you find cases like this.
N[Sin[2017 2^(1/5)]] - N[-1]$endgroup$
– J. M. is computer-less♦
16 hours ago
1
$begingroup$
PossibleZeroQcould help.$endgroup$
– Roman
16 hours ago
$begingroup$
@J.M.iscomputer-less Damn, this almost integer stuff is fascinating. Thanks!
$endgroup$
– Loki
10 hours ago