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Comparing exact expressions vs real numbers
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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 5 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 5 hours ago
SumitSumit
11.7k21955
$endgroup$
$begingroup$
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♦
5 hours ago
$begingroup$
PossibleZeroQcould help.
$endgroup$
– Roman
5 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 5 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 5 hours ago
SumitSumit
11.7k21955
asked 5 hours ago
SumitSumit
11.7k21955
asked 5 hours ago
SumitSumit
11.7k21955
asked 5 hours ago
SumitSumit
11.7k21955
asked 5 hours ago
SumitSumit
11.7k21955
11.7k21955
$begingroup$
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♦
5 hours ago
$begingroup$
PossibleZeroQcould help.
$endgroup$
– Roman
5 hours ago
add a comment |
$begingroup$
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♦
5 hours ago
$begingroup$
PossibleZeroQcould help.
$endgroup$
– Roman
5 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♦
5 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♦
5 hours ago
$begingroup$
PossibleZeroQ could help.$endgroup$
– Roman
5 hours ago
$begingroup$
PossibleZeroQ could help.$endgroup$
– Roman
5 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
answered 2 hours ago
RomanRoman
2,714717
$endgroup$
add a comment |
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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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
answered 2 hours ago
RomanRoman
2,714717
$endgroup$
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
answered 2 hours ago
RomanRoman
2,714717
$endgroup$
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
answered 2 hours ago
RomanRoman
2,714717
$endgroup$
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
answered 2 hours ago
RomanRoman
2,714717
edited 1 hour ago
answered 2 hours ago
RomanRoman
2,714717
answered 2 hours ago
RomanRoman
2,714717
answered 2 hours ago
RomanRoman
2,714717
2,714717
add a comment |
add a comment |
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$begingroup$
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♦
5 hours ago
$begingroup$
PossibleZeroQcould help.$endgroup$
– Roman
5 hours ago