tag:blogger.com,1999:blog-25469513700168837072014-10-04T18:38:32.853-07:00Celestial Adventuresnadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.comBlogger21125tag:blogger.com,1999:blog-2546951370016883707.post-2008517233136565502011-07-20T13:31:00.001-07:002011-07-20T13:32:02.433-07:00SOLVED 4 PROBLEMS!<div>Granted, they were homework problems from the beginning of the course, but I solved them! Myself! In only 2 hours!</div><div><br /></div><div>Today has been very productive. :)</div>nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-73036513923726144652011-07-17T16:55:00.000-07:002011-07-17T16:57:43.152-07:00Today: <div><br /></div><div>7:30-7:40 -- listened to Dan explain a solution that I had copied</div><div>7:40-7:50 -- found lost homework files</div><div><br /></div>nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-89538349430658949852011-07-12T10:09:00.001-07:002011-07-12T10:46:20.478-07:00Once upon a time, I did math research. I loved it. Part of why it was so enjoyable is inextricably connected to my adviser, Leah W. Berman, who gave me a healthy perspective on productivity. It's been two years since I've done any truly creative math, which really puts a damper on my perseverance...<div><br /></div><div>Nevertheless, there is hope! In the spirit of the work begun two years ago, I re-plunge myself into this blog for the sake of accountability and thus much-hoped-for productivity!</div><div><br /></div><div>10:30 -- woke up</div><div>12:30 -- left for campus</div><div>1-1:30-- course calender posted and algebra email list made</div><div>1:30-1:45--blogged.</div><div><br /></div><div>Things left to do:</div><div><br /></div><div>MAC and teaching 3-8:30. :(</div><div>Lesson plan??</div><div><br /></div><div>... which clearly leaves no time for prelim studying. Or prelim memorizing. You know. $%^#.</div><div><br /></div>nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-70103555203802671912010-04-09T02:44:00.000-07:002010-04-09T02:50:06.805-07:00BLOGGER IS WORKING IN UST!While I was sitting in the middle of a conference given in Russian, I was wondering what would happen if you merged two 2-cfgs together. I'm pretty sure it gets you a 4-cfg. The question is whether or not it gets you a set of superimposable ones... and I think I need more of my notes before I determine that. I forget the proof... that is what I'll have to look up. Who knows! Maybe I'll figure out what I couldn't finish!nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-66234325612318346712009-07-07T06:49:00.000-07:002009-07-07T07:51:25.890-07:00Wk 6, TuBrussel sprouts and lima beans! There are less than 3 weeks left!<br /><br />Progress, progress... I proved that given a configuration <span style="font-style: italic;">m</span>#(<span style="font-style: italic;">a,b,c,d,b,a,d,c</span>), it has complements <span style="font-style: italic;">m#(b,c,a,d,c,b,d,a)</span> and <span style="font-style: italic;">m#(c,a,b,d,a,c,d,b).</span> Now I'm trying to generalize this.<br /><br />After a questionable proof and a feeble attempt at coding, I discovered a pair of nontrivial configurations that trivially satisfied the conditions for being superimposable/having extra intersections!<br /><br />Furthermore, while my code yields non-realizable configurations (i.e., useless), it still (I think) shows where it's not possible, and I think I'm finding a correlation:<br /><br />1. According to my code (and my analysis of the results it yielded), there are NO superimposable configurations for <span style="font-style: italic;">m</span> = 8, 9, 10, 11, 13, 14, 16, 17, 19, or 20 (15 is questionable: it was too extensive <i> </i>to check according to one coding [this code was yielding pattern 2 cfgs, which was way bad], and the second coding of restrictions said there were none. so, it's very, very unlikely <by> that there are any superimposable cfgs for </by><i><by>m</by></i><by>=15).<br /><br />2. Conjecture: a configuration is superimposable if it is half-trivial, i.e., two elements of <span>S</span> are the same as two elements of <span>T</span>.<br />*I'm having a little bit of trouble determining the direction of this implication, so:<br /><br />3. According to Angela's data, there are no half-trivial configurations for</by><i><by> <span style="font-style: italic;">m</span> </by></i><by>= 8, 9, 10, 11, 13, 14, 15, 16, 17, or 19, (haven't checked 20 yet) but there IS one in 12, which just so happens to yield the two pairs of superimposables.</by>nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-23466813510950086322009-06-29T23:59:00.000-07:002009-06-30T06:41:23.149-07:00Wk 5, M8:30-11<br />2-3<br />10-11<br /><br />i have two pages of notes trying to figure out how to generalize my theorem! geeze!nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-67389232213757243532009-06-25T13:15:00.000-07:002009-06-25T13:17:23.484-07:00Wk 4, Th9-11:30 -- paper revisions; 6-celestial pre-work<br />1:30-3 -- made a (3,4) cfg<br />3-4 -- 6-celestial expsnadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-74365568351781057322009-06-24T07:18:00.000-07:002009-06-24T12:50:56.662-07:00Wk 4, W7:20am-9 -- typed paper<br />9-10 -- meeting<br />10-11:30 -- surfed and slept<br />11:30-12 -- configured<br />2-3 -- i think i was working on categorizing the properties?<br /><br />i was up at 6 this morning! how have i only worked 4 hours?!nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-15111049771859028982009-06-19T07:56:00.000-07:002009-06-19T07:58:10.112-07:00Wk 3, F8:30-9:30--"finished" proof, i.e., wrote up angle problem solution<br />9:30-10:30--meeting w/berman<br />10:30... added more pictures to papernadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-70752441988897900472009-06-18T16:58:00.000-07:002009-06-19T08:00:20.585-07:00Wk 3, Th9-12 -- wrote half a page; other half of page: picture<br />12-3:30 -- lunch, talk, tutored<br />3:30-5:30 -- papered<br />6-7-- papered futile-y<br /><br />i hate papers.nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-58022317863290687392009-06-16T11:51:00.000-07:002009-06-17T06:01:34.475-07:00Wk 3, Tu9:30-11 -- tried to fix "101 error(s)" in Latex in my tables (i was missing some dollar signs)<br />i was wondering about formatting. how do my tables look? (i was also thinking they could be equation lists, like to show the middle step, and forget about having columns.<br />1-2:30 -- worked on showing (and typing up) that the angles must be the same. got half way there.<br />i think this also leads to some interesting questions about configurations... i wonder if anyone's done any research on the placement of vertices? (especially with the pairing that occurs with trivial configurations). for instance, when are the symmetry classes of points collinear?<br />2:30-3:30 -- outlined proof, for organizational purposes; blogged.<br />11-12 -- reviewed proof for tomorrow<br /><br />today, like yesterday, was very much a writing day.nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-53097884525832593752009-06-12T11:07:00.000-07:002009-06-12T14:51:22.778-07:00Wk 2, F8:30-9 -- worked on paper<br />9 -- meeting<br />9-12:15 -- worked on writing down exactly what i was proving; chose better forms for the complements (b,c,a,d,... instead of a,d,c,b,...); made a table of spans for the three cfgs; hashed out branko's explanation of how to calculate where the next point goes. hopefully, this will be enlightening for why there <span style="font-style: italic;">must</span> be the third (fifth and sixth 2PI rings) set of intersections.<br /><br />2-2:30 -- blogged<br />2:30-5 -- conjectured that the innermost radius of the L_i intersections was determined by t_(i-1); further modified to hold only for pattern 1 (first look at pattern 2 yielded a counterexample). looked at more examples. figured out why the s_i weren't mapping to the predicted color: it's s_i <span style="font-style: italic;">and</span> t_(i-1) dependent, not just s_i dependent. Fun.nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-83320894765840826972009-06-11T07:05:00.000-07:002009-06-11T19:11:52.645-07:00Wk 2, Th9-10 -- meeting<br />10-11 -- made final decisions about the colors:<br />Lines: Blue, Red, (Dark)Green, (Dark)Magenta<br />Points: Cyan, Orange, Green, Magenta, Yellow, Purple<br />Additionally, mathematica is STILL like, NO, go figure out spans and colors and whatnot yourself. :P see notebook.<br />11:30-12:30 -- fiddled with<br />4:30-5:30 -- wasted. like duh. obviously you can't preserve the line labels: take your initial 2PIs, BdG & RdM. They become your new points. Your new 2PIs, however, are still BdG and RdM, which are supposed to be in locations that <span style="font-style: italic;">were</span> points in the initial cfg. But they couldn't be points in the initial cfg because they were 2PIs!nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-29959663387361779072009-06-08T10:54:00.000-07:002009-06-08T11:16:57.238-07:00Wk 2, MMore evidence of the BG, RM pairings:<br />Their orbits are always cyclically adjacent! It even applies when you include the missing 2PIs.nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-41268509126996810292009-06-05T13:38:00.000-07:002009-06-05T13:50:54.017-07:00Wk 1, F9-10 meeting<br />10-12:30 weekly update<br />2:30 wireless broke<br />3-4:30 pattern 2 experimentsnadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-61366169168743951292009-06-05T12:01:00.000-07:002009-06-05T09:21:53.164-07:00Weekly Reflections 1A General Documentation of our Conjectures:<br /><ol><li>Fixing <span style="font-style: italic;">a,b,c,d</span> as the spans of lines (i.e., <span style="font-style: italic;">m#(a, _; b, _; c, _; d, _)</span>) there are only two ways to fill in the blanks so as to have a coherent/realizable/valid configuration: <span style="font-style: italic;">m</span>#(<span style="font-style: italic;">a, c; b, d; c, a; d, b </span>) and <span style="font-style: italic;">m</span>#(<span style="font-style: italic;">a, d; b, a; c, b; d, c</span>). The former we will call <span style="font-style: italic;">Pattern 1</span>; the latter, <span style="font-style: italic;">Pattern 2</span>.</li><li>These two patterns have different properties:<br /></li><li>Pattern 1 configurations <span style="font-weight: bold;">seem to always</span> have two extra orbits of 4-fold intersections, (abbreviated here as 2-Pair-intersection, or 2PIs) where 2 of the lines are from one symmetry class of lines, and the other 2 are from another. Furthermore, these 2PIs are different from the ones constructed by the symbol, resulting in every pair existing: 4 symmetry classes of lines, 2 classes through each of these intersections; 4 choose 2 = 6 possible combinations.<br /></li><li>Pattern 2 configurations do not always have two extra 2PIs. In some cases, there is only 1 extra; other cases, there are none. There are also 3-fold intersections, usually of three different symmetry classes. There is a possibility that the properties have connections to the parity of <span style="font-style: italic;">m</span>; however, there are too many possibilities to make any conjectures at this point. Further analysis is needed.</li><li>You can always check algorithmically to see if a 2PI exists. There are only <span style="font-style: italic;">floor</span>[<span style="font-style: italic;">m</span>/2] distinct possibilities to check. Furthermore, if all the intersections of one color occur inside the innermost intersection of another color, you know certainly that a 2PI of those colors does not exist.</li><li>Pseudo Conjecture: Based on Branko Grunbaum's work, there are 6 (isomorphically) distinct configurations given <span style="font-style: italic;">a,b,c,d,m</span>, 3 of Pattern 1 and 3 of Pattern 2 (they are polars of each other).</li><li>Conjecture: The 3 configurations of Pattern 1 given <span style="font-style: italic;">a</span>,<span style="font-style: italic;">b</span>,<span style="font-style: italic;">c</span>,<span style="font-style: italic;">d</span>,<span style="font-style: italic;">m</span> all have the same line structure. What this means is that you could place the configurations right on top of each other and you would get a [6,4] configuration.</li><li>Conjecture: The 2PIs that are missing from each of the Pattern 1 configurations are always the same: the innermost (orbit 1) and outermost (orbit 6) symmetry classes, orbits 2 & 5, or orbits 3&4.</li></ol>nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-80127551735226533432009-06-04T10:55:00.000-07:002009-06-04T10:56:22.421-07:00What i Actually know so Farthere are 4 distinct configurations, but there are also 2 more. lamez. or great!nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-76526836274620627972009-06-04T05:43:00.000-07:002009-06-04T06:31:24.467-07:00What I know so Far<ul><li>You can always check algorithmically to see if a 4-folder exists. Furthermore, if all the points of intersection of one color are inside the innermost intersection of another color, there can be no 4-folder of those two colors.<br /></li><li>Given <span style="font-style: italic;">a,b,c,d,m</span>, there are 4 distinct configurations.</li><li>All the 9 configurations have projective intersections of 1 of each of the colors</li><li>For <span style="font-style: italic;">m</span>=9 (and maybe for all values of m??), Type 1 (abcdbadc) intersections give you the two remaining 4-folders; type 2 sometimes gives you 1, sometimes none. Further analysis needed.</li><li><span style="font-style: italic;"></span>12#(2,4,3,5,4,2,5,3) has a LOT of extra intersections. So does <span style="font-style: italic;"></span>24#(11,8,3,8,3,7,2,6), even when it's no longer trivial.<span style="font-style: italic;"></span><br /></li></ul>nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-81659876382746089172009-06-03T09:38:00.001-07:002009-06-03T11:34:46.244-07:00Wk 1, W9am -- meeting<br />10-1:15pm --worked on presentation<br />2:15 -- messed with cfgs and compositionsnadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-78458226043874631832009-06-02T14:41:00.000-07:002009-06-12T14:36:44.730-07:00Wk 1, Tu9am -- met with berman<br />10am -- cried to rachel<br />11am -- listed out possible configuration symbols and found their equivalences<br />noon -- lunch<br />noon:30 -- gave tech support my computer<br />2pm -- retrieved computer; mathematica: drew (all) 12 configurations for m = 10<br />3pm -- started messing with GS<br />5:30 -- started logging progress by means of email<br />6pm -- dinnernadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0tag:blogger.com,1999:blog-2546951370016883707.post-16759785792384553462009-05-31T19:20:00.000-07:002009-05-31T19:24:21.589-07:00IntroductionSummer Fellows starts tomorrow! <br /><br />This blog is for the purposes of recording my progress through the program.nadinehttp://www.blogger.com/profile/13768429794877693658noreply@blogger.com0