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Circle divided by lines between a blue dots
Difficult IQ test question: What is the box suggesting?What is the minimum number of straight lines to connect all the dots on this grid?Hikers Meeting in the MiddleYet another adventitious triangleMy roommate is back add it!Letters and dots and paperInner Triangles in the circleAsk for suggestion on a hard IQ questionMissing Number in a Seven Segment Circle
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
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What is the solution for this IQ test question?
Source: https://www.quora.com/What-are-some-extremely-difficult-genius-level-160-IQ-questions
mathematics visual geometry
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add a comment
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What is the solution for this IQ test question?
Source: https://www.quora.com/What-are-some-extremely-difficult-genius-level-160-IQ-questions
mathematics visual geometry
New contributor
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Added source now
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– CuriousSuperhero
10 hours ago
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$begingroup$
What is the solution for this IQ test question?
Source: https://www.quora.com/What-are-some-extremely-difficult-genius-level-160-IQ-questions
mathematics visual geometry
New contributor
$endgroup$
What is the solution for this IQ test question?
Source: https://www.quora.com/What-are-some-extremely-difficult-genius-level-160-IQ-questions
mathematics visual geometry
mathematics visual geometry
New contributor
New contributor
edited 10 hours ago
Rand al'Thor
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76k15 gold badges249 silver badges499 bronze badges
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asked 10 hours ago
CuriousSuperheroCuriousSuperhero
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Added source now
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– CuriousSuperhero
10 hours ago
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Added source now
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– CuriousSuperhero
10 hours ago
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Added source now
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– CuriousSuperhero
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– CuriousSuperhero
10 hours ago
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4 Answers
4
active
oldest
votes
$begingroup$
The answer is
57.
This is a well-known problem called
Moser's circle problem. The sequence given by "maximal number of regions with $n$ blue dots" for increasing values of $n$ is $1,2,4,8,16,31,57,dots$. It's famously deceptive because the first few terms make it look like it's going to be simply the powers of 2, as another answer guessed, but it isn't.
$endgroup$
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
2 hours ago
add a comment
|
$begingroup$
An answer from @Randal'Thor was posted while I prepared this.
My (independent) answer is
57
Which I obtained by counting successive diagrams.
This is confirmed by the sequence
2,4,8,16,31,57
which is shown by OEIS to be A000127
Maximal number of regions obtained by joining n points around a circle by straight lines.
$endgroup$
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
10 hours ago
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
6 hours ago
add a comment
|
$begingroup$
My answer is reference
Regions of a Circle Cut by Chords to n Points
---------------------------------------------- n points are distributed round the circumference of a circle and each point is
joined to every other point by a chord of the circle. Assuming that
no three chords intersect at a point inside the circle we require the
number of regions into which the circle is divided.
With no lines the circle has just one region. Now consider any
collection of lines. If you draw a new line across the circle which
does not cross any existing lines, then the effect is to increase the
number of regions by 1. In addition, every time a new line crosses an
existing line inside the circle the number of regions is increased by
1 again.
So in any such arrangement
number of regions = 1 + number of lines + number of interior
intersections
= 1 + C(n,2) + C(n,4)
Note that the number of lines is the number of ways 2 points can be
chosen from n points. Also, the number of interior intersections is
the number of quadrilaterals that can be formed from n points, since
each quadrilateral produces just 1 intersection where the diagonals
of the quadrilateral intersect.
Examples:
n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8
n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16
n=6 " " = 1 + C(6,2) + C(6,4) = 31
n=7 " " = 1 + C(7,2) + C(7,4) = 57
$endgroup$
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
9 hours ago
add a comment
|
$begingroup$
64 - the number appears to be doubling with each additional point.
New contributor
$endgroup$
2
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
10 hours ago
add a comment
|
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is
57.
This is a well-known problem called
Moser's circle problem. The sequence given by "maximal number of regions with $n$ blue dots" for increasing values of $n$ is $1,2,4,8,16,31,57,dots$. It's famously deceptive because the first few terms make it look like it's going to be simply the powers of 2, as another answer guessed, but it isn't.
$endgroup$
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
2 hours ago
add a comment
|
$begingroup$
The answer is
57.
This is a well-known problem called
Moser's circle problem. The sequence given by "maximal number of regions with $n$ blue dots" for increasing values of $n$ is $1,2,4,8,16,31,57,dots$. It's famously deceptive because the first few terms make it look like it's going to be simply the powers of 2, as another answer guessed, but it isn't.
$endgroup$
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
2 hours ago
add a comment
|
$begingroup$
The answer is
57.
This is a well-known problem called
Moser's circle problem. The sequence given by "maximal number of regions with $n$ blue dots" for increasing values of $n$ is $1,2,4,8,16,31,57,dots$. It's famously deceptive because the first few terms make it look like it's going to be simply the powers of 2, as another answer guessed, but it isn't.
$endgroup$
The answer is
57.
This is a well-known problem called
Moser's circle problem. The sequence given by "maximal number of regions with $n$ blue dots" for increasing values of $n$ is $1,2,4,8,16,31,57,dots$. It's famously deceptive because the first few terms make it look like it's going to be simply the powers of 2, as another answer guessed, but it isn't.
answered 10 hours ago
Rand al'ThorRand al'Thor
76k15 gold badges249 silver badges499 bronze badges
76k15 gold badges249 silver badges499 bronze badges
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
2 hours ago
add a comment
|
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
2 hours ago
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
2 hours ago
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
2 hours ago
add a comment
|
$begingroup$
An answer from @Randal'Thor was posted while I prepared this.
My (independent) answer is
57
Which I obtained by counting successive diagrams.
This is confirmed by the sequence
2,4,8,16,31,57
which is shown by OEIS to be A000127
Maximal number of regions obtained by joining n points around a circle by straight lines.
$endgroup$
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
10 hours ago
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
6 hours ago
add a comment
|
$begingroup$
An answer from @Randal'Thor was posted while I prepared this.
My (independent) answer is
57
Which I obtained by counting successive diagrams.
This is confirmed by the sequence
2,4,8,16,31,57
which is shown by OEIS to be A000127
Maximal number of regions obtained by joining n points around a circle by straight lines.
$endgroup$
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
10 hours ago
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
6 hours ago
add a comment
|
$begingroup$
An answer from @Randal'Thor was posted while I prepared this.
My (independent) answer is
57
Which I obtained by counting successive diagrams.
This is confirmed by the sequence
2,4,8,16,31,57
which is shown by OEIS to be A000127
Maximal number of regions obtained by joining n points around a circle by straight lines.
$endgroup$
An answer from @Randal'Thor was posted while I prepared this.
My (independent) answer is
57
Which I obtained by counting successive diagrams.
This is confirmed by the sequence
2,4,8,16,31,57
which is shown by OEIS to be A000127
Maximal number of regions obtained by joining n points around a circle by straight lines.
answered 10 hours ago
Weather VaneWeather Vane
6,2661 gold badge4 silver badges26 bronze badges
6,2661 gold badge4 silver badges26 bronze badges
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
10 hours ago
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
6 hours ago
add a comment
|
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
10 hours ago
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
6 hours ago
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
10 hours ago
1
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
10 hours ago
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
6 hours ago
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
6 hours ago
add a comment
|
$begingroup$
My answer is reference
Regions of a Circle Cut by Chords to n Points
---------------------------------------------- n points are distributed round the circumference of a circle and each point is
joined to every other point by a chord of the circle. Assuming that
no three chords intersect at a point inside the circle we require the
number of regions into which the circle is divided.
With no lines the circle has just one region. Now consider any
collection of lines. If you draw a new line across the circle which
does not cross any existing lines, then the effect is to increase the
number of regions by 1. In addition, every time a new line crosses an
existing line inside the circle the number of regions is increased by
1 again.
So in any such arrangement
number of regions = 1 + number of lines + number of interior
intersections
= 1 + C(n,2) + C(n,4)
Note that the number of lines is the number of ways 2 points can be
chosen from n points. Also, the number of interior intersections is
the number of quadrilaterals that can be formed from n points, since
each quadrilateral produces just 1 intersection where the diagonals
of the quadrilateral intersect.
Examples:
n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8
n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16
n=6 " " = 1 + C(6,2) + C(6,4) = 31
n=7 " " = 1 + C(7,2) + C(7,4) = 57
$endgroup$
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
9 hours ago
add a comment
|
$begingroup$
My answer is reference
Regions of a Circle Cut by Chords to n Points
---------------------------------------------- n points are distributed round the circumference of a circle and each point is
joined to every other point by a chord of the circle. Assuming that
no three chords intersect at a point inside the circle we require the
number of regions into which the circle is divided.
With no lines the circle has just one region. Now consider any
collection of lines. If you draw a new line across the circle which
does not cross any existing lines, then the effect is to increase the
number of regions by 1. In addition, every time a new line crosses an
existing line inside the circle the number of regions is increased by
1 again.
So in any such arrangement
number of regions = 1 + number of lines + number of interior
intersections
= 1 + C(n,2) + C(n,4)
Note that the number of lines is the number of ways 2 points can be
chosen from n points. Also, the number of interior intersections is
the number of quadrilaterals that can be formed from n points, since
each quadrilateral produces just 1 intersection where the diagonals
of the quadrilateral intersect.
Examples:
n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8
n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16
n=6 " " = 1 + C(6,2) + C(6,4) = 31
n=7 " " = 1 + C(7,2) + C(7,4) = 57
$endgroup$
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
9 hours ago
add a comment
|
$begingroup$
My answer is reference
Regions of a Circle Cut by Chords to n Points
---------------------------------------------- n points are distributed round the circumference of a circle and each point is
joined to every other point by a chord of the circle. Assuming that
no three chords intersect at a point inside the circle we require the
number of regions into which the circle is divided.
With no lines the circle has just one region. Now consider any
collection of lines. If you draw a new line across the circle which
does not cross any existing lines, then the effect is to increase the
number of regions by 1. In addition, every time a new line crosses an
existing line inside the circle the number of regions is increased by
1 again.
So in any such arrangement
number of regions = 1 + number of lines + number of interior
intersections
= 1 + C(n,2) + C(n,4)
Note that the number of lines is the number of ways 2 points can be
chosen from n points. Also, the number of interior intersections is
the number of quadrilaterals that can be formed from n points, since
each quadrilateral produces just 1 intersection where the diagonals
of the quadrilateral intersect.
Examples:
n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8
n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16
n=6 " " = 1 + C(6,2) + C(6,4) = 31
n=7 " " = 1 + C(7,2) + C(7,4) = 57
$endgroup$
My answer is reference
Regions of a Circle Cut by Chords to n Points
---------------------------------------------- n points are distributed round the circumference of a circle and each point is
joined to every other point by a chord of the circle. Assuming that
no three chords intersect at a point inside the circle we require the
number of regions into which the circle is divided.
With no lines the circle has just one region. Now consider any
collection of lines. If you draw a new line across the circle which
does not cross any existing lines, then the effect is to increase the
number of regions by 1. In addition, every time a new line crosses an
existing line inside the circle the number of regions is increased by
1 again.
So in any such arrangement
number of regions = 1 + number of lines + number of interior
intersections
= 1 + C(n,2) + C(n,4)
Note that the number of lines is the number of ways 2 points can be
chosen from n points. Also, the number of interior intersections is
the number of quadrilaterals that can be formed from n points, since
each quadrilateral produces just 1 intersection where the diagonals
of the quadrilateral intersect.
Examples:
n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8
n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16
n=6 " " = 1 + C(6,2) + C(6,4) = 31
n=7 " " = 1 + C(7,2) + C(7,4) = 57
edited 6 hours ago
answered 10 hours ago
Sayed Mohd AliSayed Mohd Ali
54716 bronze badges
54716 bronze badges
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I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
10 hours ago
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It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
10 hours ago
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In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
9 hours ago
add a comment
|
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
9 hours ago
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
10 hours ago
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
10 hours ago
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
9 hours ago
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
9 hours ago
add a comment
|
$begingroup$
64 - the number appears to be doubling with each additional point.
New contributor
$endgroup$
2
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
10 hours ago
add a comment
|
$begingroup$
64 - the number appears to be doubling with each additional point.
New contributor
$endgroup$
2
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
10 hours ago
add a comment
|
$begingroup$
64 - the number appears to be doubling with each additional point.
New contributor
$endgroup$
64 - the number appears to be doubling with each additional point.
New contributor
New contributor
answered 10 hours ago
AndyJ97AndyJ97
1
1
New contributor
New contributor
2
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
10 hours ago
add a comment
|
2
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
10 hours ago
2
2
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
10 hours ago
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
10 hours ago
add a comment
|
CuriousSuperhero is a new contributor. Be nice, and check out our Code of Conduct.
CuriousSuperhero is a new contributor. Be nice, and check out our Code of Conduct.
CuriousSuperhero is a new contributor. Be nice, and check out our Code of Conduct.
CuriousSuperhero is a new contributor. Be nice, and check out our Code of Conduct.
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Added source now
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– CuriousSuperhero
10 hours ago