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How to set a tolerance level for equality constraints

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2

$begingroup$

Given two equality constraints: x+y==250 and z+p==65 where x=190, y=50, z=45, p=15, I want to specify an error-tolerance level of e=0.05 below which the first equality should be TRUE and above which the second equality be FALSE.

{x=190, y=50, z=45, p=15};

eq1= x+y==250;   (* e=0.0416 is the percentage change from 250*)

eq2= z+p==65;    (* e=0.0833 is the percentage change from 65*)



(*Mathematica output*)

(*FALSE, FALSE*)



(*I like to receive for given e=0.05*)

(*TRUE, FALSE*)

How can I set the error-tolerance level of e=0.05 for the two constraints?

Any idea?

programming

share|improve this question

asked 9 hours ago

Tugrul TemelTugrul Temel

1,11833 silver badges1313 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Search the site for $EqualTolerance
  $endgroup$
  – Michael E2
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelE2: I can give a specific tolerance level for the system of equations, but my question is about an error-tolerance calculated as percentage change in the equality concerned. As far as I know you can not set the tolerance level to percentage changes because every equation has a different level of percentage change.
  $endgroup$
  – Tugrul Temel
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A proper objection to $EqualTolerance would be that you do not want to use machine precision floats, but would like something that works on, say, exact input such as shown (except for e=0.05). Otherwise, it does what you desire: Block[{Internal`$EqualTolerance = MachinePrecision + Log10[0.05]}, {eq1 = x + y == 250., eq2 = z + p == 65.} ]
  $endgroup$
  – Michael E2
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichelE2: Thank you very much for this answer. Here, what does MachinePrecision+Log10[0.05] do? I know it generates the output I want, but I did not understand what your code does.
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

Given two equality constraints: x+y==250 and z+p==65 where x=190, y=50, z=45, p=15, I want to specify an error-tolerance level of e=0.05 below which the first equality should be TRUE and above which the second equality be FALSE.

{x=190, y=50, z=45, p=15};

eq1= x+y==250;   (* e=0.0416 is the percentage change from 250*)

eq2= z+p==65;    (* e=0.0833 is the percentage change from 65*)



(*Mathematica output*)

(*FALSE, FALSE*)



(*I like to receive for given e=0.05*)

(*TRUE, FALSE*)

How can I set the error-tolerance level of e=0.05 for the two constraints?

Any idea?

programming

share|improve this question

asked 9 hours ago

Tugrul TemelTugrul Temel

1,11833 silver badges1313 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Search the site for $EqualTolerance
  $endgroup$
  – Michael E2
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelE2: I can give a specific tolerance level for the system of equations, but my question is about an error-tolerance calculated as percentage change in the equality concerned. As far as I know you can not set the tolerance level to percentage changes because every equation has a different level of percentage change.
  $endgroup$
  – Tugrul Temel
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A proper objection to $EqualTolerance would be that you do not want to use machine precision floats, but would like something that works on, say, exact input such as shown (except for e=0.05). Otherwise, it does what you desire: Block[{Internal`$EqualTolerance = MachinePrecision + Log10[0.05]}, {eq1 = x + y == 250., eq2 = z + p == 65.} ]
  $endgroup$
  – Michael E2
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichelE2: Thank you very much for this answer. Here, what does MachinePrecision+Log10[0.05] do? I know it generates the output I want, but I did not understand what your code does.
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Given two equality constraints: x+y==250 and z+p==65 where x=190, y=50, z=45, p=15, I want to specify an error-tolerance level of e=0.05 below which the first equality should be TRUE and above which the second equality be FALSE.

{x=190, y=50, z=45, p=15};

eq1= x+y==250;   (* e=0.0416 is the percentage change from 250*)

eq2= z+p==65;    (* e=0.0833 is the percentage change from 65*)



(*Mathematica output*)

(*FALSE, FALSE*)



(*I like to receive for given e=0.05*)

(*TRUE, FALSE*)

How can I set the error-tolerance level of e=0.05 for the two constraints?

Any idea?

programming

share|improve this question

asked 9 hours ago

Tugrul TemelTugrul Temel

1,11833 silver badges1313 bronze badges

$endgroup$

{x=190, y=50, z=45, p=15};

eq1= x+y==250;   (* e=0.0416 is the percentage change from 250*)

eq2= z+p==65;    (* e=0.0833 is the percentage change from 65*)



(*Mathematica output*)

(*FALSE, FALSE*)



(*I like to receive for given e=0.05*)

(*TRUE, FALSE*)

share|improve this question

asked 9 hours ago

Tugrul TemelTugrul Temel

1,11833 silver badges1313 bronze badges

share|improve this question

asked 9 hours ago

Tugrul TemelTugrul Temel

1,11833 silver badges1313 bronze badges

asked 9 hours ago

Tugrul TemelTugrul Temel

1,11833 silver badges1313 bronze badges

asked 9 hours ago

Tugrul TemelTugrul Temel

1,11833 silver badges1313 bronze badges

3 Answers
3

active

oldest

votes

2

$begingroup$

ClearAll[choppedEqual]

SetAttributes[choppedEqual, {HoldFirst, Listable}]

choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;

Examples:

choppedEqual[x + y == 250, .0416]

True

choppedEqual[x + p == 65, .0833]

False

choppedEqual[{x + y == 250, x + p == 65}, {0.0416, 0.0833}]

{True, False}

share|improve this answer

edited 2 hours ago

answered 7 hours ago

kglrkglr

217k1010 gold badges247247 silver badges497497 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you solved my problem. Thank you so much. Regards..
  $endgroup$
  – Tugrul Temel
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel, my pleasure. Thank you for the accept.
  $endgroup$
  – kglr
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In real case, I have many equations to be tested. It will be time consuming to enter each equation in the format you use. Is there any other way to format choppedEqual[x + y, 250, .0416] automatically, separating the left (x+y) and right 250 sides of each equation?
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Tugrul, please see the new version.
  $endgroup$
  – kglr
  2 hours ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

Some more ways, with the relative error e = 0.05:

Block[{Internal`$EqualTolerance = MachinePrecision + Log10[e]},

 {x + y == 250., z + p == 65.} (* advantage: equations written in terms of == *)

 ]

(*  {True, False}  *)



{SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]], 

 SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]}

(*  {True, False}  *)



svn = NDSolve`ScaledVectorNorm[Infinity, {e, 0}];

{svn[{x + y - 250}, {250}] < 1, svn[{z + p - 65}, {65}] < 1}

(*  {True, False}  *)

share|improve this answer

answered 5 hours ago

Michael E2Michael E2

159k1313 gold badges217217 silver badges516516 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The first seems to be the obviously correct way to do this, to my eyes at least.
  $endgroup$
  – b3m2a1
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelE2: The first option will be my choice because it uses the equations in terms of ==. What I do not understand is what MachinePrecision+Log10[e] imposes as the tolerance level? What kind of number is that?
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel See mathematica.stackexchange.com/questions/86815/… and stackoverflow.com/a/6626748 for some discussion. Setting $EqualTolerance to a causes two numbers to be considered equal if they agree except for the lower-order a digits (base 10), which means that machine-precision numbers would be considered equal if the relative error is less than 10^(MachinePrecision - a). Hence if it is set to MachinePrecision - a, they would be equal if the relative error is less than 10^a, where a = Log10[e].
  $endgroup$
  – Michael E2
  14 mins ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

You can use Congruent.

ClearAll[Congruent]

TOL = 0.05;

Congruent[a_, b_] := 

  If[Abs[a - b] > 0, Abs[a - b] / Norm[{a, b}, Infinity] <= TOL, True];

Addendum

your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?

Please note that lines 37 and 39 on the attached screenshot use Congruent.

It can be seen in the linked function page that the infix symbol of Congruent ("≡") can be entered as "Esc === Esc".

x + y ≡ 250    

(* True *)



z + p ≡ 65  

(* False *)

Using Block we can also make bulk evaluations of many equalities.

Block[{Equal = Congruent},

 {x + y == 250, z + p == 65}

]



(* {True, False} *)

share|improve this answer

edited 1 hour ago

answered 8 hours ago

Anton AntonovAnton Antonov

25.7k11 gold badge6868 silver badges122122 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You seemed to answer the question, but I did not understand what you propose. I have two specific equations and your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?
  $endgroup$
  – Tugrul Temel
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel See the update of my answer.
  $endgroup$
  – Anton Antonov
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now I understand the code, though it is still not so clear how I enter the constraints to be tested.
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel You can use Block[{Equal = Congruent},...] -- see my answer update.
  $endgroup$
  – Anton Antonov
  1 hour ago
  

  
  
  
  

add a comment
 | 

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2

$begingroup$

ClearAll[choppedEqual]

SetAttributes[choppedEqual, {HoldFirst, Listable}]

choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;

Examples:

choppedEqual[x + y == 250, .0416]

True

choppedEqual[x + p == 65, .0833]

False

choppedEqual[{x + y == 250, x + p == 65}, {0.0416, 0.0833}]

{True, False}

share|improve this answer

edited 2 hours ago

answered 7 hours ago

kglrkglr

217k1010 gold badges247247 silver badges497497 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you solved my problem. Thank you so much. Regards..
  $endgroup$
  – Tugrul Temel
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel, my pleasure. Thank you for the accept.
  $endgroup$
  – kglr
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In real case, I have many equations to be tested. It will be time consuming to enter each equation in the format you use. Is there any other way to format choppedEqual[x + y, 250, .0416] automatically, separating the left (x+y) and right 250 sides of each equation?
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Tugrul, please see the new version.
  $endgroup$
  – kglr
  2 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

ClearAll[choppedEqual]

SetAttributes[choppedEqual, {HoldFirst, Listable}]

choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;

Examples:

choppedEqual[x + y == 250, .0416]

True

choppedEqual[x + p == 65, .0833]

False

choppedEqual[{x + y == 250, x + p == 65}, {0.0416, 0.0833}]

{True, False}

share|improve this answer

edited 2 hours ago

answered 7 hours ago

kglrkglr

217k1010 gold badges247247 silver badges497497 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you solved my problem. Thank you so much. Regards..
  $endgroup$
  – Tugrul Temel
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel, my pleasure. Thank you for the accept.
  $endgroup$
  – kglr
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In real case, I have many equations to be tested. It will be time consuming to enter each equation in the format you use. Is there any other way to format choppedEqual[x + y, 250, .0416] automatically, separating the left (x+y) and right 250 sides of each equation?
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Tugrul, please see the new version.
  $endgroup$
  – kglr
  2 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

ClearAll[choppedEqual]

SetAttributes[choppedEqual, {HoldFirst, Listable}]

choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;

Examples:

choppedEqual[x + y == 250, .0416]

True

choppedEqual[x + p == 65, .0833]

False

choppedEqual[{x + y == 250, x + p == 65}, {0.0416, 0.0833}]

{True, False}

share|improve this answer

edited 2 hours ago

answered 7 hours ago

kglrkglr

217k1010 gold badges247247 silver badges497497 bronze badges

$endgroup$

ClearAll[choppedEqual]

SetAttributes[choppedEqual, {HoldFirst, Listable}]

choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;

choppedEqual[x + y == 250, .0416]

choppedEqual[x + p == 65, .0833]

choppedEqual[{x + y == 250, x + p == 65}, {0.0416, 0.0833}]

share|improve this answer

edited 2 hours ago

answered 7 hours ago

kglrkglr

217k1010 gold badges247247 silver badges497497 bronze badges

answered 7 hours ago

kglrkglr

217k1010 gold badges247247 silver badges497497 bronze badges

answered 7 hours ago

kglrkglr

217k1010 gold badges247247 silver badges497497 bronze badges

3

$begingroup$

Some more ways, with the relative error e = 0.05:

Block[{Internal`$EqualTolerance = MachinePrecision + Log10[e]},

 {x + y == 250., z + p == 65.} (* advantage: equations written in terms of == *)

 ]

(*  {True, False}  *)



{SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]], 

 SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]}

(*  {True, False}  *)



svn = NDSolve`ScaledVectorNorm[Infinity, {e, 0}];

{svn[{x + y - 250}, {250}] < 1, svn[{z + p - 65}, {65}] < 1}

(*  {True, False}  *)

share|improve this answer

answered 5 hours ago

Michael E2Michael E2

159k1313 gold badges217217 silver badges516516 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The first seems to be the obviously correct way to do this, to my eyes at least.
  $endgroup$
  – b3m2a1
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelE2: The first option will be my choice because it uses the equations in terms of ==. What I do not understand is what MachinePrecision+Log10[e] imposes as the tolerance level? What kind of number is that?
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel See mathematica.stackexchange.com/questions/86815/… and stackoverflow.com/a/6626748 for some discussion. Setting $EqualTolerance to a causes two numbers to be considered equal if they agree except for the lower-order a digits (base 10), which means that machine-precision numbers would be considered equal if the relative error is less than 10^(MachinePrecision - a). Hence if it is set to MachinePrecision - a, they would be equal if the relative error is less than 10^a, where a = Log10[e].
  $endgroup$
  – Michael E2
  14 mins ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

Some more ways, with the relative error e = 0.05:

Block[{Internal`$EqualTolerance = MachinePrecision + Log10[e]},

 {x + y == 250., z + p == 65.} (* advantage: equations written in terms of == *)

 ]

(*  {True, False}  *)



{SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]], 

 SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]}

(*  {True, False}  *)



svn = NDSolve`ScaledVectorNorm[Infinity, {e, 0}];

{svn[{x + y - 250}, {250}] < 1, svn[{z + p - 65}, {65}] < 1}

(*  {True, False}  *)

share|improve this answer

answered 5 hours ago

Michael E2Michael E2

159k1313 gold badges217217 silver badges516516 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The first seems to be the obviously correct way to do this, to my eyes at least.
  $endgroup$
  – b3m2a1
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelE2: The first option will be my choice because it uses the equations in terms of ==. What I do not understand is what MachinePrecision+Log10[e] imposes as the tolerance level? What kind of number is that?
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel See mathematica.stackexchange.com/questions/86815/… and stackoverflow.com/a/6626748 for some discussion. Setting $EqualTolerance to a causes two numbers to be considered equal if they agree except for the lower-order a digits (base 10), which means that machine-precision numbers would be considered equal if the relative error is less than 10^(MachinePrecision - a). Hence if it is set to MachinePrecision - a, they would be equal if the relative error is less than 10^a, where a = Log10[e].
  $endgroup$
  – Michael E2
  14 mins ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Some more ways, with the relative error e = 0.05:

Block[{Internal`$EqualTolerance = MachinePrecision + Log10[e]},

 {x + y == 250., z + p == 65.} (* advantage: equations written in terms of == *)

 ]

(*  {True, False}  *)



{SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]], 

 SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]}

(*  {True, False}  *)



svn = NDSolve`ScaledVectorNorm[Infinity, {e, 0}];

{svn[{x + y - 250}, {250}] < 1, svn[{z + p - 65}, {65}] < 1}

(*  {True, False}  *)

share|improve this answer

answered 5 hours ago

Michael E2Michael E2

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$endgroup$

Block[{Internal`$EqualTolerance = MachinePrecision + Log10[e]},

 {x + y == 250., z + p == 65.} (* advantage: equations written in terms of == *)

 ]

(*  {True, False}  *)



{SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]], 

 SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]}

(*  {True, False}  *)



svn = NDSolve`ScaledVectorNorm[Infinity, {e, 0}];

{svn[{x + y - 250}, {250}] < 1, svn[{z + p - 65}, {65}] < 1}

(*  {True, False}  *)

share|improve this answer

answered 5 hours ago

Michael E2Michael E2

159k1313 gold badges217217 silver badges516516 bronze badges

answered 5 hours ago

Michael E2Michael E2

159k1313 gold badges217217 silver badges516516 bronze badges

answered 5 hours ago

Michael E2Michael E2

159k1313 gold badges217217 silver badges516516 bronze badges

3

$begingroup$

You can use Congruent.

ClearAll[Congruent]

TOL = 0.05;

Congruent[a_, b_] := 

  If[Abs[a - b] > 0, Abs[a - b] / Norm[{a, b}, Infinity] <= TOL, True];

Addendum

your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?

Please note that lines 37 and 39 on the attached screenshot use Congruent.

It can be seen in the linked function page that the infix symbol of Congruent ("≡") can be entered as "Esc === Esc".

x + y ≡ 250    

(* True *)



z + p ≡ 65  

(* False *)

Using Block we can also make bulk evaluations of many equalities.

Block[{Equal = Congruent},

 {x + y == 250, z + p == 65}

]



(* {True, False} *)

share|improve this answer

edited 1 hour ago

answered 8 hours ago

Anton AntonovAnton Antonov

25.7k11 gold badge6868 silver badges122122 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You seemed to answer the question, but I did not understand what you propose. I have two specific equations and your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?
  $endgroup$
  – Tugrul Temel
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel See the update of my answer.
  $endgroup$
  – Anton Antonov
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now I understand the code, though it is still not so clear how I enter the constraints to be tested.
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel You can use Block[{Equal = Congruent},...] -- see my answer update.
  $endgroup$
  – Anton Antonov
  1 hour ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

You can use Congruent.

ClearAll[Congruent]

TOL = 0.05;

Congruent[a_, b_] := 

  If[Abs[a - b] > 0, Abs[a - b] / Norm[{a, b}, Infinity] <= TOL, True];

Addendum

your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?

Please note that lines 37 and 39 on the attached screenshot use Congruent.

It can be seen in the linked function page that the infix symbol of Congruent ("≡") can be entered as "Esc === Esc".

x + y ≡ 250    

(* True *)



z + p ≡ 65  

(* False *)

Using Block we can also make bulk evaluations of many equalities.

Block[{Equal = Congruent},

 {x + y == 250, z + p == 65}

]



(* {True, False} *)

share|improve this answer

edited 1 hour ago

answered 8 hours ago

Anton AntonovAnton Antonov

25.7k11 gold badge6868 silver badges122122 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You seemed to answer the question, but I did not understand what you propose. I have two specific equations and your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?
  $endgroup$
  – Tugrul Temel
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel See the update of my answer.
  $endgroup$
  – Anton Antonov
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now I understand the code, though it is still not so clear how I enter the constraints to be tested.
  $endgroup$
  – Tugrul Temel
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TugrulTemel You can use Block[{Equal = Congruent},...] -- see my answer update.
  $endgroup$
  – Anton Antonov
  1 hour ago
  

  
  
  
  

add a comment
 | 

$begingroup$

You can use Congruent.

ClearAll[Congruent]

TOL = 0.05;

Congruent[a_, b_] := 

  If[Abs[a - b] > 0, Abs[a - b] / Norm[{a, b}, Infinity] <= TOL, True];

Addendum

your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?

Please note that lines 37 and 39 on the attached screenshot use Congruent.

It can be seen in the linked function page that the infix symbol of Congruent ("≡") can be entered as "Esc === Esc".

x + y ≡ 250    

(* True *)



z + p ≡ 65  

(* False *)

Using Block we can also make bulk evaluations of many equalities.

Block[{Equal = Congruent},

 {x + y == 250, z + p == 65}

]



(* {True, False} *)

share|improve this answer

edited 1 hour ago

answered 8 hours ago

Anton AntonovAnton Antonov

25.7k11 gold badge6868 silver badges122122 bronze badges

$endgroup$

ClearAll[Congruent]

TOL = 0.05;

Congruent[a_, b_] := 

  If[Abs[a - b] > 0, Abs[a - b] / Norm[{a, b}, Infinity] <= TOL, True];

x + y ≡ 250    

(* True *)



z + p ≡ 65  

(* False *)

Block[{Equal = Congruent},

 {x + y == 250, z + p == 65}

]



(* {True, False} *)

share|improve this answer

edited 1 hour ago

answered 8 hours ago

Anton AntonovAnton Antonov

25.7k11 gold badge6868 silver badges122122 bronze badges

answered 8 hours ago

Anton AntonovAnton Antonov

25.7k11 gold badge6868 silver badges122122 bronze badges

answered 8 hours ago

Anton AntonovAnton Antonov

25.7k11 gold badge6868 silver badges122122 bronze badges