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Comparing exact expressions vs real numbers

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1

$begingroup$

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

1. Is it better to use real numbers if I have to do such comparisons?




2. 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

share|improve this question

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
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  PossibleZeroQ could help.
  $endgroup$
  – Roman
  5 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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

1. Is it better to use real numbers if I have to do such comparisons?




2. 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

share|improve this question

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
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  PossibleZeroQ could 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

1. Is it better to use real numbers if I have to do such comparisons?




2. 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

share|improve this question

asked 5 hours ago

SumitSumit

11.7k21955

$endgroup$

a=-2/Sqrt[3] + Sqrt[3]

b=Sqrt[1/4 + (2/Sqrt[3] - Sqrt[3]/2)^2]



{N[a],N[b]}

N[a] == N[b]

N[a]-N[b] == 0

N[a - b] == 0

Chop@N[a - b] == 0

RepeatedTiming[N[a] == N[b]]

RepeatedTiming[Chop@N[a - b] == 0]

a1 = N[a]; b1 = N[b];

RepeatedTiming[a1 == b1]

share|improve this question

asked 5 hours ago

SumitSumit

11.7k21955

share|improve this question

asked 5 hours ago

SumitSumit

11.7k21955

asked 5 hours ago

SumitSumit

11.7k21955

asked 5 hours ago

SumitSumit

11.7k21955

1 Answer
1

active

oldest

votes

4

$begingroup$

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

share|improve this answer

edited 1 hour ago

answered 2 hours ago

RomanRoman

2,714717

$endgroup$

add a comment | 

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4

$begingroup$

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

share|improve this answer

edited 1 hour ago

answered 2 hours ago

RomanRoman

2,714717

$endgroup$

add a comment | 

4

$begingroup$

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

share|improve this answer

edited 1 hour ago

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

share|improve this answer

edited 1 hour ago

answered 2 hours ago

RomanRoman

2,714717

$endgroup$

RepeatedTiming[PossibleZeroQ[a - b]]

PossibleZeroQ[Sin[2017 2^(1/5)] - (-1)]

share|improve this answer

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