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A plane divides the 3D space in two regions. A good method to discern is to apply the plane equation to the point and see if it yields negative or positive result:

$$Ax + By + Cz + D = 0$$

is the equation of a plane in 3D. So if you put the point coordinates you'll get a positive, zero or negative number. (If zero, the point belongs to the plane and so, it's neither in either side of the plane.)

The question asks to count the points that belong to each of the delimited regions for each plane. Well, you have n planes each dividing the space in two halves (so you have $2^n$ regions, except in case you have paralell planes, but this procedure is applicable also in that case ---you'll get regions not accounted for as no point can lie in them), just consider each point and apply it to each plane. Then construct a number X, making bit $i$ (for plane $i$) of X a 1 in case applying plane to it gives a positive number and 0 in case applying it gives a negative one (in case it returns 0 you have to decide what side to count it ---what value you select for bit $i$--- for (if any) as this is a point that belongs to the plane) then with that number as a key for region X, increment counter of region X selected, so in one pass over all points you'll get all counters actualized with how many points belong to each region. Each key X, so constructed, identifies each of the possible regions you have divided the space into, so you must cope with $2^n$ regions (and counters) for $n$ planes.

A plane divides the 3D space in two regions. A good method to discern is to apply the plane equation to the point and see if it yields negative or positive result:

$$Ax + By + Cz + D = 0$$

is the equation of a plane in 3D. So if you put the point coordinates you'll get a positive, zero or negative number. (If zero, the point belongs to the plane and so, it's neither in either side of the plane.)

The question asks to count the points that belong to each of the delimited regions for each plane. Well, you have n planes each dividing the space in two halves (so you have $2^n$ regions at most, except in case you have paralell planes, but this procedure is applicable also in that case ---you'll get regions not accounted for as no point can lie in them), just consider each point and apply it to each plane. Then construct a number X, making bit $i$ (for plane $i$) of X a 1 in case applying plane to it gives a positive number and 0 in case applying it gives a negative one (in case it returns 0 you have to decide what side to count it ---what value you select for bit $i$--- for (if any) as this is a point that belongs to the plane) then with that number as a key for region X, increment counter of region X selected, so in one pass over all points you'll get all counters actualized with how many points belong to each region. Each key X, so constructed, identifies each of the possible regions you have divided the space into, so you must cope with $2^n$ regions (and counters) for $n$ planes.

A plane divides the 3D space in two regions. A good method to discern is to apply the plane equation to the point and see if it yields negative or positive result:

$$Ax + By + Cz + D = 0$$

is the equation of a plane in 3D. So if you put the point coordinates you'll get a positive, zero or negative number. (If zero, the point belongs to the plane and so, it's neither inside nor outside of the region.)

The question asks to count the points that belong to each of the delimited regions for each plane. Well, you have n planes each dividing the space in two halves (so you have $2^n$ regions, except in case you have paralell planes, but this procedure is applicable also in that case), just consider each point and apply it to each plane. Then construct a number, making bit $i$ (for plane $i$) a 1 in case applying plane to it gives a positive number and 0 in case applying it gives a negative one (in case it returns 0 you have to decide what side to count it for (if any) as it's a point that belongs to the plane) then with that number as a key for region X, increment counter of region selected, so in one pass over all points you'll get all counters actualized with how many points belong to each region. Each key X, so constructed, identifies each of the possible regions you have divided the space into, so you must cope with $2^n$ regions (and counters) for $n$ planes.

A plane divides the 3D space in two regions. A good method to discern is to apply the plane equation to the point and see if it yields negative or positive result:

$$Ax + By + Cz + D = 0$$

is the equation of a plane in 3D. So if you put the point coordinates you'll get a positive, zero or negative number. (If zero, the point belongs to the plane and so, it's neither in either side of the plane.)

The question asks to count the points that belong to each of the delimited regions for each plane. Well, you have n planes each dividing the space in two halves (so you have $2^n$ regions, except in case you have paralell planes, but this procedure is applicable also in that case ---you'll get regions not accounted for as no point can lie in them), just consider each point and apply it to each plane. Then construct a number X, making bit $i$ (for plane $i$) of X a 1 in case applying plane to it gives a positive number and 0 in case applying it gives a negative one (in case it returns 0 you have to decide what side to count it ---what value you select for bit $i$--- for (if any) as this is a point that belongs to the plane) then with that number as a key for region X, increment counter of region X selected, so in one pass over all points you'll get all counters actualized with how many points belong to each region. Each key X, so constructed, identifies each of the possible regions you have divided the space into, so you must cope with $2^n$ regions (and counters) for $n$ planes.

A plane divides the 3D space in two regions, and if you have several planes, in general you cannot say wich one of the halves is inside or outside. You have to select which side of the plane will be inside and them make a selection (iterating with each plane) and discarding the points that lie outside this new plane. A good method to discern is to apply the plane equation to the point and see if it yields negative or positive result:

Ax + By + Cz + D = 0

is the equation of a plane in 3D, so if you put the point coordinates you'll get a positive, zero or negative number (if zero, te point belongs to the plane and so, it's neither inside nor outside of the region)

for a tetraedron (just four planes) you can have only one bounded region (or even none, in case the four planes coincide in one point), but you have 15 more unbounded ones. In case of five or more planes, you can have more than one bounded region, so you cannot apply the bounded property to consider being inside.

I think the most practical thing (instead of searching all possibilities, as the number of regions grow exponentially with the number of planes) is to consider which side of the plane makes the points to be inside or outside of your region, and apply the plane function to them to see if they are inside or outside.

Another issue to consider is what you define the region inside, as I'm supposing you mean being at one side to be inside and being inside only if all planes agree (this deals you to a convex polyedron) but the definition can be modified to only some of them, to consider nonconvex, bounded regions or several other criterion...

I expect this explanation gives you some light on the subject.

As @FrankW points out, the question asks to count the points that belong to each of the delimited regions for each plane. Well, you have n planes each dividing the space in two halves (so you have 2^n regions, except in case you have paralell planes, but this procedure is applicable also in that case), just consider each point and apply it to each plane. Then construct a number, making bit i (for plane i) a 1 in case applying plane to it gives a positive number and 0 in case applying it gives a negative one (in case it returns 0 you have to decide if it is in what side as it's a point that belongs to the plane) then with that number as a key for region X, increment counter of region selected, so in one pass over all points you'll get all counters actualized with how many points belong to each region. Each key X, so constructed, identifies each of the possible regions you have divided the space into, so you must cope with 2^n regions (and counters) for n planes.

A plane divides the 3D space in two regions. A good method to discern is to apply the plane equation to the point and see if it yields negative or positive result:

$$Ax + By + Cz + D = 0$$

is the equation of a plane in 3D. So if you put the point coordinates you'll get a positive, zero or negative number. (If zero, the point belongs to the plane and so, it's neither inside nor outside of the region.)

The question asks to count the points that belong to each of the delimited regions for each plane. Well, you have n planes each dividing the space in two halves (so you have $2^n$ regions, except in case you have paralell planes, but this procedure is applicable also in that case), just consider each point and apply it to each plane. Then construct a number, making bit $i$ (for plane $i$) a 1 in case applying plane to it gives a positive number and 0 in case applying it gives a negative one (in case it returns 0 you have to decide what side to count it for (if any) as it's a point that belongs to the plane) then with that number as a key for region X, increment counter of region selected, so in one pass over all points you'll get all counters actualized with how many points belong to each region. Each key X, so constructed, identifies each of the possible regions you have divided the space into, so you must cope with $2^n$ regions (and counters) for $n$ planes.