More Law & Order Commentary

Watched awesome episode of Law & Order today. Jeffrey Tambor was an incompetent judge presiding over the case of a senator, played by somebody I should recognize but forgot. The defending lawyer was the guy who now plays scruffy cop on the current season. It was like a wormhole in the L & O timeline.

I have been thinking about it, and I think there are a few reasons I actually like Law & Order and can't stand other dramas.

1) Each episode manages to tell a whole story, so I don't have to wait forever for the conclusion of plotlines I don't care about.

2) The characters are not universally irritating. The motivations for the recurring cast are almost irrelevant because they are just doing their jobs, and the motivations for the suspects, etc., just make sense.

That I guess leads to

3) The writing is just better. Usually the second half of each episode centers on some sort of interesting (some readers might say "gimmicky") legal argument, so it's not wholly dependent on "what happens to so-and-so" type stories, which depend on you liking, or at least caring about, the character. The first half of the show is generally just pretty solid mystery-ing.

To expand on my points in a rambling and unstructured way, I'd like to mention that we do get to see more details about the main characters (that is, DAs and detectives) rolled out over the course of many episodes, but it's not generally essential to the plot, and I think the characters are actually more endearing because we are seeing them work and trying to figure out stuff along with them rather than just having their stories shoved at us. Maybe it is a Japanese way of thinking, but I feel a greater connection with the ADAs, whose lives we see very little of outside of the office than with people on other tv dramas who spend their time talking about their messed up childhoods or lost loves and the like.

Law & Order is clearly the best brand of the three that are still on (it is also better to Trial by Jury, I think, but I didn't see much of its lone season), and I think the reasons I talked about before show why. SVU focuses far too much on each of the detectives' overwrought backstories. For example, the episode I watched last night was just a story about Eliot's daughter and his mother who both have some sort of mental illness. It was the kind of story that if it happened to someone in real life would be tragic, but as it was, was just kind of a boring hour of poorly lighted emoting. That is another strike against SVU, which is sort of unrelated to my previous points. It's way too dark, not in subject matter, but in the sense that it looks like the whole show is shot using only a flashlight for lighting. This helps obscure Mariska Hargitay's face, though, so that is a plus (she is ugly and looks like a dude).

Criminal Intent is really a different kind of show. It's pretty much like Sherlock Holmes if Holmes were living in present day Manhattan and also basically a mental patient. The real selling point of the show is watching Gorin twist up his face and body, then own some suspect through psychology or pick out some bizarre clue. It's in no way realistic, but pretty great. Eames makes a great Watson, too. I've only seen one episode with Jeff Goldblum, but he seems pretty great thus far. The Chris Noth episodes were meh, but mostly due to none of his partners having any personality whatsoever. I don't know if that was mostly a writing thing or if they just look shabby compared to D'Onofrio, who is just plain awesome.

Happy Jack

I just watched an episode of Law & Order, original recipe, which ended with Jack McCoy meeting his daughter (?) for dinner, and then he smiled. That is the only time I think that has ever happened on that show. Amazing.

"Live" blogging

For some reason it is funny to me to liveblog a rerun of a show. That show is SVU, which USA likes to show in huge blocks, leading to me having seen pretty much all of them.

7:05 - They are looking for someone named Anika (sp?) using cell phones. Stabler got in a plug for some kind of 911 thing with cell phones. SVU loves putting public services anouncements in their dialogue as clunkily as possible.

7:08 - Oh noes! She was pregnant! I didn't see the beginning of this episode, so I have no idea what's up.

7:09 - Benson brushes off somebody's question like always. Amazing police work. By police work, I mean being a jerk and making a big show of being offended by crimes.

7:10 - Circle camera!

7:11 - This may be the legendary episode where the Asian tech guy gets to flip out. I've seen it, but I forget which one is which. I just remember thinking it hilarious that the writers decided this minor character should get his own spotlight episode. Usually we get nonsense about Stabler's marriage or Benson not being able to find a relationship because of her job [actually it is because she looks like a dude]. The fact that this guy who basically just runs audio programs on a laptop gets so attached to a case is just hilarious.

7:12 - Really I am just annoyed with Miller and Bud commercials now because they keep trying to convince us that they are good beer. They should really just say, "You will buy this because it is fairly cheap but doesn't taste like Steel Reserve." I would respect them a lot more if they were honest about it being pretty bad but still alright if you don't really care about how it tastes. Beats the Beast any day.

7:16 - I really hate Ben Stein now, too. He keep advertising for one of these "free" credit score sites that is no doubt a scam, though I haven't figured out how yet because I don't care. If you didn't know, he is a vocal creationist, which should tell you that he is a conman or an idiot or both, so I wouldn't suggest using whatever crap he is trying to shill.

7:20 - The budget at SVU must be really tight. They don't seem to be able to afford to turning on the lights EVER.

7:22 - Cragen is always under pressure from the brass. That must really ruin his otherwise cushy job of standing around in his office and telling the detectives that he "wants this guy."

7:25 - Looking inconspicuous in your long coats and black sunglasses, ENTIRE POLICE DEPARTMENT.

7:27 - So that guy is also dead. I'm thinking it is all an overly complicated conspiracy. Commercial break.

7:29 - How many times can they redesign this Nasonex bee without ever making him endearing or not creepy?

7:30 - Finn: "Prints on the dead guy came back!" Ha ha

7:37 - Sorry for the lack of updates. Firefox stopped cooperating for a few minutes. In the meantime, they've managed to arrest the guy who apparently had a guy kidnap his ex-girlfriend and then push a "kidnapper" into a moving car during a payoff. Paula Dean is telling me how easy it is to cook tenderloins in a bag.

7:40 - "I didn't kidnap Anika (?)" A likely story. They found BROCHURES in your HOUSE, dude. Lock him up.

7:42 - Benson: "Baby, baby, baby, baby..." I am paraphrasing. Stabler: "I am going to punch someone I am so angry."

7:44 - Stabler: "She's definitely going to put her money where her mouth is." A joke is too easy here because he is talking about a rather large woman.

7:47 - I'm tired of commercials bossing me around. No, TV, I won't "chill out with Coke products." Please phrase your shtick in the form of a question.

7:50 - Coming this fall, Stabler and Finn are Law & Order: Beach Patrol.

7:51 - I don't think I've seen a single episode of this show where Stabler has failed to mention that he has four kids. I can't help noticing he doesn't seem to own a single World's #1 Dad T-shirt, though. hmm...

7:52 - On TV they always make a big deal of how hard it is to deliver a baby, and then they end up doing it in about 30 seconds by the power of saying "push!"

7:55 - I take back my comments that Mariska Hargitay is a dude. I now believe she is some sort of alien sent to earth to "act with her eyes" by moving them back and forth as if there were a permanent fly in front of the camera.

7:58 - Another hour of police work wasted. I could have told you who did it at the beginning: stodgy, fat, rich lady.

Well, that was fun. Saturday, Saturday, Saturday night's alright!

Stripes

I am watching Stripes. It's a pretty good movie. I think the cherries in the kitchen are attracting little flies. Biteface is sleeping on the couch, getting his blanket all covered with hair. A fascinating day, indeed.

TV post

Again I was going to liveblog a rerun of Law & Order, but TNT decided to show golf instead this afternoon, so I didn't get to. Then I forgot to do it for one from last season. Since Dan was posting about TV, though, I will, too.

Law & Order is great and pretty much always has been. Tonight's episode was lame because the motivation was really far fetched and there's no way that people would remember minor incidents from twenty years ago, but the plot hinged on them doing just that. Also, the guy who took Jack's place continues to be a shallow imitation of Jack. Also also it is sexist that his assistant DA wasn't promoted ahead of him. But, whatever, fat black cop and scraggly white cop make a good detective pair. I don't always remember the characters' names on that show, but if you watch it, you can probably tell who I mean.

First Post in a While

I was going to "live" blog a rerun of Law & Order today, but blogger was not cooperating, so I couldn't. It is too bad. Because this was an episode with Alana De La Garza who is very cute even though she has kind of an alien face, and now you are all missing out on my insights. My heart belongs to this No! Drug girl, however. That poster was up in the BOE and I looked at it a lot over my last week or so because I had nothing to do and she is mesmerizing.

Maze Problem

Have you ever read a Games Magazine? I guess "read" is not the proper word. It's a magazine of puzzles, like crossword puzzles and whatnot, so just reading through it wouldn't be very enjoyable. If you are a member of my family, and since you are reading this blog, the odds are pretty good you are, you have at least tried some of the puzzles. Today I am going to talk about a puzzle that I liked doing because it's a math-y thing and I'm pleased with myself for having solved it. If Will Shores or whoever wants to complain, then he can email me and I'll take this down. I sort of doubt that will happen.

Anyway, on the cover of this latest issue, there is a puzzle entitled "Lost in the Pyramid." It is a 7 x 7 grid of squares in different colors. The object is to get from the center square to one of the edges, with the path touching one and only one of each color square along the way. The last square, then, has to be one of the edge squares, but not one of the four corner squares. I have translated the colors into letters so as to make this possible to blog. Hopefully the formatting works out, but if not, you should at least be able to follow along by making your own representation. Here's what the puzzle looks like (the letters-color correspondence is obviously arbitrary):

ADDLLLT
AEJMTTT
AFJNOSU
BBJVQSU
CGKKQRU
CHIPQRR
CHIPPPR

Obviously not quite square when I type it like that (I think J's are too thin), but hopefully you get the idea. You start at the V, and have to draw a path touching one of each letter and end at an edge. I am going to refer to each square with a redundant nomenclature of X(y, z) where X is the letter and y and z the row and column number, starting from the top left [A(1,1)] and proceeding to the bottom right [R(7,7)].

I'll get to the solution, but since you might want to figure it out yourself, I'll include a picture break here.


Mmm, Stag. I'm not sure why this beer brewed in Milwaukee is the local beer of choice, but it seems to be.

My method for solving this was to turn it into a graph theory problem. That is, each square is a vertex, and adjacent squares are connected by edges. We proceed by removing edges and vertices until only one path remains. So, initially, the graph should look like a big grid [I would be a poor explainer of graph theory to note that you can draw the graph in any shape you'd like as long as you preserve the relationship between vertices and edges], which is what it is. The obvious first move would be to remove all edges between two letters of the same kind, since no path could contain two of the same letters. There are too many to list here specifically and anyone should be able to do this step on their own. Another convenient preliminary step is to mark all squares which are the only ones with their corresponding letters. That may be confusing wording. If there is only one of a letter in the grid, mark it, as the path necessarily contains that square. That is E(2,2); F(3,2); M(2,4); N(3,4); O(3,5); G(5,2); V(4,4). I am using bold here to represent a vertex we know (or are showing) to be in the path. I will try to keep that notation going throughout.

Note that some of the squares we have marked as included in the path are adjacent. One might be tempted to suppose the path runs from one marked square to the next, but take heed that this is not necessarily true. Note that N(3,4) is adjacent to both M(2,4) and O(3,5). We can't distinguish which of these it will be, if it is even one of these at all. The path could even pass through J(3,3); we don't know if it is in the path or not. Now, onto the parts that require a bit more reasoning. I am going to label my steps so that I can refer to them like a real mathematician might.

1. If we look at A(1,1), we can see that there is only one edge for us to use [we already eliminated the A-A edge]. Then A(1,1) must be the final vertex on our path. However, it is a corner vertex, and, thus, cannot be final. So, we may eliminate A(1,1) and the corresponding edge. The same process can be used to remove T(1,7); C(7,1); and R(7,7), all our corner vertices.

2. Examine R(6,7). Only one edge remains, to U(5,7). From that vertex, only one other edge remains, to R(5,6). Then any path through R(6,7) contains two R's; we may eliminate R(6,7). I won't mention eliminating the edges from an eliminated vertex anymore, this should be obvious.

3. This is really a key step, so if you are paying attention, pay attention. U(4,7) and U(5,7) have only one edge and are possible final vertices. U(3,7) has only two, one of which leads to T(2,5), which similarly can only be final. Thus, if a U is contained in the path, either that U is final, or T(2,5) is final. Since a U is necessarily contained in the path, we have determined that our final vertex is one of these four vertices [a U or T(2,5)]. Then obviously, no other vertex can be final.

4. By applying (3), we may now remove any vertex along the edge of the grid (sorry for the edge/edge) confusion with only one edge. That is A(2,1); H(7,2); P(7,6); D(1,2); C(6,1); L(1,6).

5. Another application of (3) yields I(7,3); L(1,5); and P(7,5) out.

6. Another application yields P(7, 4) out.

7. R(6,6) now has only one remaining edge but can't be final, as it is not along an edge, so we may eliminate it, as well.

8. We have removed vertices, so we may mark some others as in the path as in our preliminary step. D(1,3); L(1,4); A(3,1); C(5,1); H(6,2); I(6,3); R(5,6); P(6,4).

9. Since A(3,1) has only two edges, B(4,1).

10. Then B(4,2) is out.

11. E(2,2) has only two edges, so J(2,3).

12. Then J(3,3) and J(4,3) are out.

13. As in (6), Q(6,5) is out.

14. We know C(5,1)-G(5,2) is in our path. Assume G(5,2)-K(5,3). Then K(5,3)-I(6,3). If we continue to P(6,4), then the path contains no H (it is impossible to get to). If, instead, we suppose I(6,3)-H(6,2), we have essentially cut ourselves off and must end at H, which is impossible. Thus G-K is impossible, so we must have G(5,2)-H(6,2)-I(6,3)-P(6,4).

15. Then K(5,4).

16. Since each vertex in the obvious path E(2,2)-...-P(6,4) has only two edges, we can confirm our suspicions that J(2,3)-E(2,2)-...-P(6,4)-K(5,4) is part of our path (a sub-path, I suppose). Note, we don't yet know the order, that is, which end connects to the V, and which to the edge.

17. From J(2,3), we have two options, D(1,3) or M(2,4). If we assume the latter, we have cut off our only D and L, so we must have J(2,3)-D(1,3)-L(1,4)-M(2,4).

At this point, the solution is close enough that one could probably guess one's way to the end, but I have my reasons for wanting to continue in this fashion, despite the next two steps being quite involved in terms of logic trees. Regardless of where we look, we are going to have to start using more complicated logic trees to find contradictions, etc., so let's just start at the beginning, that is V, and try each of the wrong first steps. By eliminating them, we leave ourselves with only the correct solution.

18. Assume V(4,4)-Q(4,5). Then either Q(4,5)-S(4,6) or Q(4,5)-O(3,5). The former yields two short (they leave out letters) paths, both to U's, so we must have Q(4,5)-O(3,5). Passing to S(3,6) again leads to a short path, so that possibility is out. Passing to T(2,5) means leaving out N(3,4), so that is out. Our final possibility here is V(4,4)-Q(4,5)-O(3,5)-N(3,4)-M(2,4)-...-K(5,4)-Q(5,5), but this path contains two Q's and is thus out. With all possibilities eliminated, we conclude V-Q is out.

19. Assume V-K. Then we must have V(4,4)-K(5,4)-...M(2,4). Passing to T(2,5) again leaves out N(3,4) as in (18), so that is out. Passing to N(3,4) leaves out T(2,5), so T(2,7) must be (the final step) in the path. Then working backwards we have O(3,5)-S(3,6)-U(3,7)-T(2,7). One more step back leads to either N(3,4) or Q(4,5). If we suppose N(3,4), then our path is set and leaves out R, so that possibility is out. Stepping back to Q(4,5) leads to another V-Q, which is out by our assumption V-K. Then our assumption [V-K] is out, so we must have V(4,4)-N(3,4).

That was an awful lot of work to show one edge between two vertices we already knew were in the path. It gets easier, though.

20. K(5,4) has only two edges, so K(5,4)-Q(5,5); then Q(4,5) is out.

21. Similarly, Q(5,5)-R(5,6).

22. Then either U(4,7) or U(5,7) must be final, so we may remove T(2,7) and U(3,7).

23. As in (6), we may remove S(3,6).

24. Then S(4,6) and R(5,6)-S(4,6)-U(4,7).

25. Then U(5, 7) is out.

26. T(2,5).

27. We now know every vertex in the path, and all but a few edges. We just need to determine how to get from N(3,4) to M(2,4). As in 17, we can easily verify N-O-T-M is the only path with all the letters.

That's it, then, we have the full path:
V(4,4)
N(3,4)
O(3,5)
T(2,5)
M(2,4)
L(1,4)
D(1-3)
J(2,3)
E(2,2)
F(3,2)
A(3,1)
B(4,1)
C(5,1)
G(5,2)
H(6,2)
I(6,3)
P(6,4)
K(5,4)
Q(5,5)
R(5,6)
S(4,6)
U(4,7)

It took a bit, but it's worth it, right?