Fast way to compare angles w/o their measures












1














I need an efficient way to compare the measures of two angles in $mathbb{R^2}$ that ideally relies on the smallest number of arithmetic operations and no trigonometric operations (no $arccos$) or square roots. Given two angles $angle{p_1p_0p_2}$ and $angle{p_3p_0p_4}$ whose measures are in the range $big(0,{pi}big)$ radians with vertex $p_0$ at the origin, what is the fastest way to compare the measures of the angles (i.e. find which angle has the largest measure/which one has the smallest measure)?



I understand that the $z$ component of the cross product of two or more pairs of vectors can be used to compare angles where $theta in big(0,frac{pi}{2}big)$ radians. As stated above, however, I need to compare angles with $theta in big(0,{pi}big)$.



Any ideas?






linear-algebra rotations angle







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  • 1




    What does it mean to "compare"?
    – AlexanderJ93
    Nov 27 at 3:08










  • So the points are given as $(x, y)$ pairs, and you wish to compare what exactly about the angles?
    – Joppy
    Nov 27 at 3:09










  • Compare as in find which angle has the largest measure.
    – rbjacob
    Nov 27 at 3:10










  • As in, when the angles lie in the range $[0, pi]$, which is larger?
    – Joppy
    Nov 27 at 3:16










  • Yes, but for range $big(0,pibig)$, not$big[0,pibig]$.
    – rbjacob
    Nov 27 at 3:20


















1














I need an efficient way to compare the measures of two angles in $mathbb{R^2}$ that ideally relies on the smallest number of arithmetic operations and no trigonometric operations (no $arccos$) or square roots. Given two angles $angle{p_1p_0p_2}$ and $angle{p_3p_0p_4}$ whose measures are in the range $big(0,{pi}big)$ radians with vertex $p_0$ at the origin, what is the fastest way to compare the measures of the angles (i.e. find which angle has the largest measure/which one has the smallest measure)?



I understand that the $z$ component of the cross product of two or more pairs of vectors can be used to compare angles where $theta in big(0,frac{pi}{2}big)$ radians. As stated above, however, I need to compare angles with $theta in big(0,{pi}big)$.



Any ideas?






linear-algebra rotations angle







share|cite|improve this question




















  • 1




    What does it mean to "compare"?
    – AlexanderJ93
    Nov 27 at 3:08










  • So the points are given as $(x, y)$ pairs, and you wish to compare what exactly about the angles?
    – Joppy
    Nov 27 at 3:09










  • Compare as in find which angle has the largest measure.
    – rbjacob
    Nov 27 at 3:10










  • As in, when the angles lie in the range $[0, pi]$, which is larger?
    – Joppy
    Nov 27 at 3:16










  • Yes, but for range $big(0,pibig)$, not$big[0,pibig]$.
    – rbjacob
    Nov 27 at 3:20
















1












1








1







I need an efficient way to compare the measures of two angles in $mathbb{R^2}$ that ideally relies on the smallest number of arithmetic operations and no trigonometric operations (no $arccos$) or square roots. Given two angles $angle{p_1p_0p_2}$ and $angle{p_3p_0p_4}$ whose measures are in the range $big(0,{pi}big)$ radians with vertex $p_0$ at the origin, what is the fastest way to compare the measures of the angles (i.e. find which angle has the largest measure/which one has the smallest measure)?



I understand that the $z$ component of the cross product of two or more pairs of vectors can be used to compare angles where $theta in big(0,frac{pi}{2}big)$ radians. As stated above, however, I need to compare angles with $theta in big(0,{pi}big)$.



Any ideas?






linear-algebra rotations angle







share|cite|improve this question















I need an efficient way to compare the measures of two angles in $mathbb{R^2}$ that ideally relies on the smallest number of arithmetic operations and no trigonometric operations (no $arccos$) or square roots. Given two angles $angle{p_1p_0p_2}$ and $angle{p_3p_0p_4}$ whose measures are in the range $big(0,{pi}big)$ radians with vertex $p_0$ at the origin, what is the fastest way to compare the measures of the angles (i.e. find which angle has the largest measure/which one has the smallest measure)?



I understand that the $z$ component of the cross product of two or more pairs of vectors can be used to compare angles where $theta in big(0,frac{pi}{2}big)$ radians. As stated above, however, I need to compare angles with $theta in big(0,{pi}big)$.



Any ideas?






linear-algebra rotations angle




linear-algebra rotations angle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 3:11

























asked Nov 27 at 3:01









rbjacob

223




223








  • 1




    What does it mean to "compare"?
    – AlexanderJ93
    Nov 27 at 3:08










  • So the points are given as $(x, y)$ pairs, and you wish to compare what exactly about the angles?
    – Joppy
    Nov 27 at 3:09










  • Compare as in find which angle has the largest measure.
    – rbjacob
    Nov 27 at 3:10










  • As in, when the angles lie in the range $[0, pi]$, which is larger?
    – Joppy
    Nov 27 at 3:16










  • Yes, but for range $big(0,pibig)$, not$big[0,pibig]$.
    – rbjacob
    Nov 27 at 3:20
















  • 1




    What does it mean to "compare"?
    – AlexanderJ93
    Nov 27 at 3:08










  • So the points are given as $(x, y)$ pairs, and you wish to compare what exactly about the angles?
    – Joppy
    Nov 27 at 3:09










  • Compare as in find which angle has the largest measure.
    – rbjacob
    Nov 27 at 3:10










  • As in, when the angles lie in the range $[0, pi]$, which is larger?
    – Joppy
    Nov 27 at 3:16










  • Yes, but for range $big(0,pibig)$, not$big[0,pibig]$.
    – rbjacob
    Nov 27 at 3:20










1




1




What does it mean to "compare"?
– AlexanderJ93
Nov 27 at 3:08




What does it mean to "compare"?
– AlexanderJ93
Nov 27 at 3:08












So the points are given as $(x, y)$ pairs, and you wish to compare what exactly about the angles?
– Joppy
Nov 27 at 3:09




So the points are given as $(x, y)$ pairs, and you wish to compare what exactly about the angles?
– Joppy
Nov 27 at 3:09












Compare as in find which angle has the largest measure.
– rbjacob
Nov 27 at 3:10




Compare as in find which angle has the largest measure.
– rbjacob
Nov 27 at 3:10












As in, when the angles lie in the range $[0, pi]$, which is larger?
– Joppy
Nov 27 at 3:16




As in, when the angles lie in the range $[0, pi]$, which is larger?
– Joppy
Nov 27 at 3:16












Yes, but for range $big(0,pibig)$, not$big[0,pibig]$.
– rbjacob
Nov 27 at 3:20






Yes, but for range $big(0,pibig)$, not$big[0,pibig]$.
– rbjacob
Nov 27 at 3:20












2 Answers
2






active

oldest

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1














Let $theta$ be the angle between $p_1$ and $p_2$, and $phi$ the one between $p_3$ and $p_4$. Since $p_1 cdot p_2 = |p_1| |p_2| cos theta$, the sign (positive or negative) of $p_1 cdot p_2$ tells you whether $theta in [0, pi/2)$ or $(pi/2, pi)$. Similarly for $phi$, and so you may already be able to tell which is larger.



If both dot products have the same sign, then define
$$ r_theta := (p_1 cdot p_2)^2 |p_3|^2 |p_4|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 cos^2 theta$$
and
$$ r_phi := (p_3 cdot p_4)^2 |p_1|^2 |p_2|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 cos^2 phi$$
and so comparing $r_theta$ and $r_phi$ tells you which has the largest cosine squared, from which you can figure out what is the larger angle. At worst, this takes 6 dot products, 4 multiplications, and 4 comparisons.






share|cite|improve this answer





















  • Thanks, this is clever and does the job perfectly!
    – rbjacob
    Nov 27 at 4:16



















-1














Try the dot product. You will like the results.



Because, dot product of two vectors will be equal to the cosine of the angle between the vectors, times the lengths of each of the vectors.



You do the rest.






share|cite|improve this answer























  • The dot product is equal to the product of the magnitudes times the cosine of the angle. Since the magnitude involves taking square roots, I believe this violates the requirements in the question.
    – AlexanderJ93
    Nov 27 at 3:14










  • @AlexanderJ93 The rest was done for the student here. No roots taken.
    – John L Winters
    Nov 27 at 4:17













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Let $theta$ be the angle between $p_1$ and $p_2$, and $phi$ the one between $p_3$ and $p_4$. Since $p_1 cdot p_2 = |p_1| |p_2| cos theta$, the sign (positive or negative) of $p_1 cdot p_2$ tells you whether $theta in [0, pi/2)$ or $(pi/2, pi)$. Similarly for $phi$, and so you may already be able to tell which is larger.



If both dot products have the same sign, then define
$$ r_theta := (p_1 cdot p_2)^2 |p_3|^2 |p_4|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 cos^2 theta$$
and
$$ r_phi := (p_3 cdot p_4)^2 |p_1|^2 |p_2|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 cos^2 phi$$
and so comparing $r_theta$ and $r_phi$ tells you which has the largest cosine squared, from which you can figure out what is the larger angle. At worst, this takes 6 dot products, 4 multiplications, and 4 comparisons.






share|cite|improve this answer





















  • Thanks, this is clever and does the job perfectly!
    – rbjacob
    Nov 27 at 4:16
















1














Let $theta$ be the angle between $p_1$ and $p_2$, and $phi$ the one between $p_3$ and $p_4$. Since $p_1 cdot p_2 = |p_1| |p_2| cos theta$, the sign (positive or negative) of $p_1 cdot p_2$ tells you whether $theta in [0, pi/2)$ or $(pi/2, pi)$. Similarly for $phi$, and so you may already be able to tell which is larger.



If both dot products have the same sign, then define
$$ r_theta := (p_1 cdot p_2)^2 |p_3|^2 |p_4|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 cos^2 theta$$
and
$$ r_phi := (p_3 cdot p_4)^2 |p_1|^2 |p_2|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 cos^2 phi$$
and so comparing $r_theta$ and $r_phi$ tells you which has the largest cosine squared, from which you can figure out what is the larger angle. At worst, this takes 6 dot products, 4 multiplications, and 4 comparisons.






share|cite|improve this answer





















  • Thanks, this is clever and does the job perfectly!
    – rbjacob
    Nov 27 at 4:16














1












1








1






Let $theta$ be the angle between $p_1$ and $p_2$, and $phi$ the one between $p_3$ and $p_4$. Since $p_1 cdot p_2 = |p_1| |p_2| cos theta$, the sign (positive or negative) of $p_1 cdot p_2$ tells you whether $theta in [0, pi/2)$ or $(pi/2, pi)$. Similarly for $phi$, and so you may already be able to tell which is larger.



If both dot products have the same sign, then define
$$ r_theta := (p_1 cdot p_2)^2 |p_3|^2 |p_4|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 cos^2 theta$$
and
$$ r_phi := (p_3 cdot p_4)^2 |p_1|^2 |p_2|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 cos^2 phi$$
and so comparing $r_theta$ and $r_phi$ tells you which has the largest cosine squared, from which you can figure out what is the larger angle. At worst, this takes 6 dot products, 4 multiplications, and 4 comparisons.






share|cite|improve this answer












Let $theta$ be the angle between $p_1$ and $p_2$, and $phi$ the one between $p_3$ and $p_4$. Since $p_1 cdot p_2 = |p_1| |p_2| cos theta$, the sign (positive or negative) of $p_1 cdot p_2$ tells you whether $theta in [0, pi/2)$ or $(pi/2, pi)$. Similarly for $phi$, and so you may already be able to tell which is larger.



If both dot products have the same sign, then define
$$ r_theta := (p_1 cdot p_2)^2 |p_3|^2 |p_4|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 cos^2 theta$$
and
$$ r_phi := (p_3 cdot p_4)^2 |p_1|^2 |p_2|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 cos^2 phi$$
and so comparing $r_theta$ and $r_phi$ tells you which has the largest cosine squared, from which you can figure out what is the larger angle. At worst, this takes 6 dot products, 4 multiplications, and 4 comparisons.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 at 3:30









Joppy

5,573420




5,573420












  • Thanks, this is clever and does the job perfectly!
    – rbjacob
    Nov 27 at 4:16


















  • Thanks, this is clever and does the job perfectly!
    – rbjacob
    Nov 27 at 4:16
















Thanks, this is clever and does the job perfectly!
– rbjacob
Nov 27 at 4:16




Thanks, this is clever and does the job perfectly!
– rbjacob
Nov 27 at 4:16











-1














Try the dot product. You will like the results.



Because, dot product of two vectors will be equal to the cosine of the angle between the vectors, times the lengths of each of the vectors.



You do the rest.






share|cite|improve this answer























  • The dot product is equal to the product of the magnitudes times the cosine of the angle. Since the magnitude involves taking square roots, I believe this violates the requirements in the question.
    – AlexanderJ93
    Nov 27 at 3:14










  • @AlexanderJ93 The rest was done for the student here. No roots taken.
    – John L Winters
    Nov 27 at 4:17


















-1














Try the dot product. You will like the results.



Because, dot product of two vectors will be equal to the cosine of the angle between the vectors, times the lengths of each of the vectors.



You do the rest.






share|cite|improve this answer























  • The dot product is equal to the product of the magnitudes times the cosine of the angle. Since the magnitude involves taking square roots, I believe this violates the requirements in the question.
    – AlexanderJ93
    Nov 27 at 3:14










  • @AlexanderJ93 The rest was done for the student here. No roots taken.
    – John L Winters
    Nov 27 at 4:17
















-1












-1








-1






Try the dot product. You will like the results.



Because, dot product of two vectors will be equal to the cosine of the angle between the vectors, times the lengths of each of the vectors.



You do the rest.






share|cite|improve this answer














Try the dot product. You will like the results.



Because, dot product of two vectors will be equal to the cosine of the angle between the vectors, times the lengths of each of the vectors.



You do the rest.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 at 3:12

























answered Nov 27 at 3:06









John L Winters

829




829












  • The dot product is equal to the product of the magnitudes times the cosine of the angle. Since the magnitude involves taking square roots, I believe this violates the requirements in the question.
    – AlexanderJ93
    Nov 27 at 3:14










  • @AlexanderJ93 The rest was done for the student here. No roots taken.
    – John L Winters
    Nov 27 at 4:17




















  • The dot product is equal to the product of the magnitudes times the cosine of the angle. Since the magnitude involves taking square roots, I believe this violates the requirements in the question.
    – AlexanderJ93
    Nov 27 at 3:14










  • @AlexanderJ93 The rest was done for the student here. No roots taken.
    – John L Winters
    Nov 27 at 4:17


















The dot product is equal to the product of the magnitudes times the cosine of the angle. Since the magnitude involves taking square roots, I believe this violates the requirements in the question.
– AlexanderJ93
Nov 27 at 3:14




The dot product is equal to the product of the magnitudes times the cosine of the angle. Since the magnitude involves taking square roots, I believe this violates the requirements in the question.
– AlexanderJ93
Nov 27 at 3:14












@AlexanderJ93 The rest was done for the student here. No roots taken.
– John L Winters
Nov 27 at 4:17






@AlexanderJ93 The rest was done for the student here. No roots taken.
– John L Winters
Nov 27 at 4:17




















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