Section+4.2+Art+Gallery+Theorem

We are all going to use the wiki to discuss problems assigned. Look below and you will find the problem that you are responsible to post the first answer for based on your Aquinas e-mail address. It is NOT o.k. to just write "I don't get it". Start by typing the problem (click on edit to start) word for word from the book. You Post your answer as if it were a test question ( You would not leave those blank you would take a stab at it!), HOW you got it (list specific page references and show work) and WHY you did that way. The rest of the class will make comments, on at least 3 other problems, corrections, and post their thoughts below your original response.

Mindscapes for Chapter 4.2 please post original response by Friday 10/1(noon) and classmate responses by Sunday 10/3 (noon). FOR SECTIONS WHERE DRAWINGS ARE NECESSARY AND YOU DO NOT HAVE THE ABILITY TO SCAN IN A PICTURE...PLEASE SUBMIT A PAPER COPY TO SHARI LEWIS MAILBOX (LOWER LEVEL AB) BY THURSDAY AT NOON AND I WILL POST THEM FOR YOU!

4.2 (1) [nmg002] 3 Vertices/3=1 guard. The circle is the guard and the lines show that it can see everywhere SUPER JOB! and explanation. [Sharil]

4.2 (3) [fab001] (comment jeh003) How many guards do you need for a gallery with 12 vertices? with 13? with 11? Does anybody know if there's a simple answer for this or if I just have to draw out some shapes and figure it out that way? It seems like the answer would depend on what kind of shape it was completely, not necessarily just how many vertices there are. it should be a multiple of 3. That's what we did in class with the shapes. We connected the dots of each vertice and made many little triangles. Having a label for three different kinds of vertices (ex: triangle, square, circle.) I think 12 should be four. But I'm not somewhere where I can draw it out. I'll post back later! (fab001)I think the number of guards needed is simply v/3, rounded up. So with 12 vertices 12/3 will find the number of guards. 12/3=4 4 is the number of guards needed. The same method is used for 13 and 11 however you round up to the nearest integer so with 11 vertices 11/3=3.66 but 4 guards are needed. The same method tells us 13/3=4.33 and 5 guards are needed. I think that if we know the vertices we don't have to draw anything out

(CommentSMD001): I think you are right this is how I worked it out and I checked the examples in the book so this makes sense to me because you can't have .66 of a camera or a guard so you would have to round up.

Be careful - reread the theorem...for 11/3 = 3.6666... It tells you to take the largest integer LESS THAN the answer - we discussed in class that you would not concern yourself with any remainder so for 11/3 you need 3 cameras/guards not 4. So what would you need for 13? [ShariL] (Fab001) If you round down you would need 4 for 13 rather than 5

4.2 (4) [trc002] The little circles are the cameras. This was a pain due to I couldn't find the CD so I did it on paint and inserted it here. Yeah, that looks good! There are lots of different places we could place the cameras. On the arrow looking one, would we be able to put it on the point? Comment klk002: Those look good! I thought about my problem, 4.2, 14 and I was struggling to draw a 6 sided area with one camera.. But, I had to simplify my way of thinking ! Your post was helpful ! (CommentSMD001): I feel like you could just use the one dot that you have in the arrow shapes and if you wanted to there could be some that were placed on the side points of the arrow. Nice discussion and drawings [ShariL] 4.2 (7) [mrh003] (comment jeh003) I thought this problem was very straightforward, I just drew the shapes out and I came out with 3 cameras for the first one and 2 for each of the last two drawings respectively, its all just a matter of judging the angles correctly but overall I feel like this is one of the simplest problems from this section.

Will someone please post a response here...anyone! Make sure you put your aq address so that you get credit for posts and comments. mrh003 where are you??? [ShariL]

4.2 (9) [SMD001]

4.2 (12) [jtl001] Will someone please post a response here...anyone! Make sure you put your aq address so that you get credit for posts and comments. mrh003 where are you??? [ShariL]

This is Chris! csv001, I'll post a response to this.

Since this is last moment and I don't have access to a scanner, I'll just say what my findings were with the problem.

all of the shapes we had to fill in the vertices for were to be expected. HOwever, the first shape was very odd. There were 5 green verts and 5 red verts, but there were only 2 blue verts. This must be due to the strange nature of the shape. Take a look at it and say what you think. 4.2 (14) [klk002] Im sorry this is mine, my computer never works on wiki.. it takes me forever to load anything !

4.2 (15) [ljj001] Still trying to figure this one out lots of drawings but no results yet. and not sure how to draw it on here. Any help would be great! You can draw it on paper and scan it in as a .jpeg and it will insert right in this document. Anyone want to help? [ShariL]