
You should always read "Find the last two digits" as "calculate stuff mod 100".
The described method works, because the order of Z/100Z (as a ring) is phi(100). This means that the order of any element of Z/100Z is a factor of 40 and thus, g^40 = 1 for any member of Z/100Z.
You could also simply try to find the order of 3 (mod 100) by calculating 3 9 27 81=19 57=43 121=21 63 189=89=11 33 99=1
and thus derive that the order of 3 mod 100 is actually 10, and that you can thus remove any multiples of 10 from the exponent in question, once again leading you to 3^3 as your result. Takes a bit longer, but not that much longer, and takes less stuff you need to realize. This is probably how i would have done it.
Didn't we have a similar question a few pages back?

United States23507 Posts
On May 18 2019 02:19 Simberto wrote: You should always read "Find the last two digits" as "calculate stuff mod 100".
The described method works, because the order of Z/100Z (as a ring) is phi(100). This means that the order of any element of Z/100Z is a factor of 40 and thus, g^40 = 1 for any member of Z/100Z.
You could also simply try to find the order of 3 (mod 100) by calculating 3 9 27 81=19 57=43 121=21 63 189=89=11 33 99=1
and thus derive that the order of 3 mod 100 is actually 10, and that you can thus remove any multiples of 10 from the exponent in question, once again leading you to 3^3 as your result. Takes a bit longer, but not that much longer, and takes less stuff you need to realize. This is probably how i would have done it.
Didn't we have a similar question a few pages back?
I am sure we did. I appreciate everyone's patience with me in this thread, since it's mostly me asking questions here I realize I am just absorbing everyone's knowledge and don't really have an opportunity to contribute much. Maybe now that I am back on ADD medication I'll be able to learn this stuff a bit quicker.
ok so, This means that the order of any element of Z/100Z is a factor of 40 and thus, g^40 = 1 for any member of Z/100Z seems to be the crucial part. I am sure this will be on the exam, thank you.
I actually like your method too, though. I remember that being done in this thread, and it's pretty straightforward if I can remember to use the negative modulus

On May 18 2019 02:09 travis wrote:Here is another question. I don't mean to spam, I just think it's better in a separate post. We have a 3d cube, each point(vertex) is labeled 1...8 G is the group of rotational symmetries of this cube Is G cyclic? Give a brief justification I say: Show nested quote + No, it is not. Any of these permutations are their own inverse in the group and thus the group is not cyclic.
Is my answer accurate? Would i get full credit for that?
Your answer is correct, but if you would get full credit relies on how explicitly you have established some of those things in class.
I would go a bit more into detail here, and basically find two different objects of order 2 (180° Rotation on x and y axis, for example), and then explain why there can not be more than one object with an order of 2 in a cyclical group. Thus, you have an actual proof.
I actually like your method too, though. I remember that being done in this thread, and it's pretty straightforward if I can remember to use the negative modulus
The problem with my method is that it doesn't scale well at some point. If in another situation you have an object of order 800 or so, you would have to do 800 calculations to discover this. It works well for kinda small groups, but at some point it can be really annoying, and other ways to quickly discover the result can be helpful.

On May 18 2019 02:09 travis wrote:Here is another question. I don't mean to spam, I just think it's better in a separate post. We have a 3d cube, each point(vertex) is labeled 1...8 G is the group of rotational symmetries of this cube Is G cyclic? Give a brief justification I say: Show nested quote + No, it is not. Any of these permutations are their own inverse in the group and thus the group is not cyclic.
Is my answer accurate? Would i get full credit for that?
This is true of the cyclic group Z/2Z so no. Your argument would need at least one extra detail to suffice.
EDIT: Post above this covers it fully. The extra detail could simply be observing that G>2.

United States23507 Posts
I have two finite line segments on a cartesian plane. The line segments are defined by their endpoints ((x1,y1),(x2,y2)) and ((x3,y3),(x4,y4)).
The lines are not exactly the same, but they can intersect, touch, or be completely disconnected from each other.
Here is the question:
Imagine that we form a single polygon from the endpoints of our two lines. It could be a triangle, or a trapezoid  depending on if the an endpoint of one line lays on the other line or not . Given an arbitrary point (x5,y5) on my cartesian plane, I want to know if that point lies within the bounds of the polgyon.
Hooow do I do this?

You need to find a way to describe the polygon, for example by finding some sort of "paramertrization". I think this could be done by definining a mapping f: [0,1]^2 > R^2 where f([0,1]x{0}) is the first line segment and f([0,1]x{1}) is the second line segment. Then the question is equivalent to checking if the point lays in the image of this map.

How is your original polygon even constructed? Is it just p1> p2 > p3 > p4 > p1? That would result in basically 2 triangles with a common point in the case of the two original lines intersecting. (or in cases where the connection lines intersect)
Or are you magically reordering your input points so it always ends as a single polygon? Then how is this magic reordering defined? Because there are multiple possibilities for a given input of 4 points... And if you magically reorder, then what was the meaning of those 2 initial lines anyway? "Here are two lines, they might be opposing edges or maybe diagonals of your new polygon"

United States23507 Posts
The easiest way to put it is that the constructed polgyon is the one with the maximum size, and that the original lines can be included within the bounds of the polygon (for example if the line criss crossed such that they made a shape like a plus sign or a cross, then the polygon would be the diamond around that cross)

On July 07 2019 08:41 travis wrote: The easiest way to put it is that the constructed polgyon is the one with the maximum size, and that the original lines can be included within the bounds of the polygon (for example if the line criss crossed such that they made a shape like a plus sign or a cross, then the polygon would be the diamond around that cross)
That is not guaranteeing a unique solution.

I wouldve read the assignment as taking the convex hull of the endpoints (prob mostly because convex combinations are what I usually encounter...). Then it's easy to describe the points that lie inside the set. However, I absolutely agree that the task is not worded in a welldefined way from what I can tell.

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On July 07 2019 16:48 mahrgell wrote:Show nested quote +On July 07 2019 08:41 travis wrote: The easiest way to put it is that the constructed polgyon is the one with the maximum size, and that the original lines can be included within the bounds of the polygon (for example if the line criss crossed such that they made a shape like a plus sign or a cross, then the polygon would be the diamond around that cross) That is not guaranteeing a unique solution.
Can you give me an example of why not?

On July 08 2019 04:41 travis wrote:Show nested quote +On July 07 2019 16:48 mahrgell wrote:On July 07 2019 08:41 travis wrote: The easiest way to put it is that the constructed polgyon is the one with the maximum size, and that the original lines can be included within the bounds of the polygon (for example if the line criss crossed such that they made a shape like a plus sign or a cross, then the polygon would be the diamond around that cross) That is not guaranteeing a unique solution. Can you give me an example of why not?
Offhand, imagine the following scenario I whipped up in paint:
The two line segments are perpendicular, and C and D are equidistant from B. Therefore A is equidistant from C and D as well. Do you connect AC and BD, or AD and BC?

United States23507 Posts
Ah, maybe I miscommunicated that all endpoints needed to be along the boundary of the polygon. They don't. In that example, we would construct the triangle AC,CD,DA

Okay, so we are simply talking about the convex hull of the 4 input points?
And let's just never mention those two lines again, because they really also don't seem to matter here.
Okay.... so now that it seems like we finally figured out the task.... for your other problem in the most generic way and compatible with the x/y description you gave originally:
As your shape is convex, you can simply describe the inside of your polygon by 3 or 4 inequalities (one for each side), so you only have to check if your point of question is fulfilling those.
So if we are talking about the most brute forcish algorithm possible: (the final states are bolded) 1) You loop over all possible unordered pairs of points from your 4 points. 1a) There are no pairs left: You are done. The point is inside the polygon 1b) There are still pairs left: > 2) 2) You calculate the line through those points 3) You check on which side of your line the other two original points are. 3a) They are on different sides > go back to 1) 3b) They are on the same side > 4) 4) You check if your point of question is on the same side: 4a) yes > go back to 1) 4b) no > abort, the point is not inside your polygon
The cases where a point is on the line in 3) are trivial, in the case of the point being on the line in 4) it really depends on how you define "in the polygon"



