SSL Minutes

19 April 2001
Dominic Johann-Berkel

Nick will bring the beverage for next Tuesday's meeting.

We discussed what we would like to have on our shirts. Possible ideas:

  1. Hexagonal FPL
  2. ASMs with a superimposed Aztec Diamond
  3. Show everthing (TOAD, ASMs, DPFL, etc.) so people can see how they are related.

It was decided that Dan and Joel (since they were not present at the meeting) will be responsible for organizing, choosing, and producing the shirts. Ha. Seriously, if anybody has an idea for a shirt design, make a mock up design to present at Tuesday's meeting. Otherwise Jim and Mike will draft somebody to head the shirt project, or we might scrap the idea all together.

Hal announced that he will be giving a talk on Monday at 4:30 in room 2321 Sterling Hall (the Physics Club room) regarding undergraduate research. Coffee and cookies will be served.

Hal continued with his proof of the Baxter Conjecture. He outlined his proof for the second half of the conjecture. That is, Hal showed that, when viewed as a permutation matrix, a Baxter permutation in the larger ASM and the existence of a -1 in the smaller ASM are the same thing.

Dominic then presented a different method of approaching the Baxter Conjecture. While Hal's method works from the outside inward, Dominic used a method that worked from the inside outward. That is, he used Hal's method to determine all the zeros of the smaller ASM. Then he looked at all the remaining "undetermined elements." There are three cases:

  1. The undetermined element is flanked by two 0's and two 1's. 
    0   1
  x
1   0
    In this case, the undetermined element must be a 1. If it were anything other than a 1, there would be no possible way to tile the area with dominos; there must be a 1-by-1 square somewhere.

  2. The undetermined element is flanked by three 0's and one 1. 
    0   1
  x
0   0
    In this case, the undetermined element must be a 1. If it were a -1, then there would be no possible way to tile the area with dominos; there must be a 1-by1 square somewhere. The undetermined element cannot be a 0. To see this, we must examine three cases. In each case, a shape is forced to propagate to the edge of the Aztec Diamond. But this shape does not fit the edge shape of an Aztec Diamond. Thus, we have a contradiction, and the undetermined element cannot be 0.

  3. The undetermined element is flanked by four 0's. 
    0   0
  x
0   0
    In this case, the undetermined element cannot be a 0. To see this, we assume it is a 0 and derive a contradiction. We must examine four cases. In each case, we propagate a shape through the larger ASM until we hit a 1. Then the shape changes to a shape that continues propagating to the edge of the Aztec Diamond. But this shape does not match the edge of an Aztec Diamond. This is a contradiction, so the undetermined element cannot be a 0.

Thus, we can always determine all the 0's of a smaller ASM. We can then alternate between 1 and -1 for the undetermined elements in a given row or column, and proceed with the proof as Hal described.

Next, we went around the room to see what people had accomplished since Tuesday's meeting.

Abe: nothing
Dom: the above method of proving Baxter
Nick: nonBaxter stuff
Geir: not done much since Tuesday
Kristin: not done much since Tuesday
B'cho: not done much since Tuesday
Hal: Baxter stuff

We then retired to the computer lab where we each worked on our own stuff.

SSL Minutes (Part 2)

19 April 2001
Nick Pongratz

Here's the few notes I took dealing with the programming of reusable components for further research:

We want an ASM class. 

Variables: storage array (char)
           size of ASM

Constructors:
         fromMonotone
         fromHF
         fromPermutation
         fromFPL

Methods: find_bigger_compatible( *function(ASM))
         find_smaller_ "        "        "
         display
         elementary moves

That's all I have. I know we talked about more stuff, but my notes kind of drift off. What else do we need? What do we have implemented, Boytcho?