Based on a Nash-Williams conjecture, Daykin and H¨aggkvist conjectured that all 14-dense partial Latin squares are complete. In Chapter 2 we construct completions for completely dense partial Latin squares containing no more than δn2filled cells in total, given that < 121, δ <.
Summary of Results
If we interpret these rules as triples (r, c, s), then the rules above define a partial Latin square. This class of partial Latin squares was first introduced in a paper by Daykin and H¨aggkvist [10] where they made the following conjecture.
Proper and Improper Trades
As mentioned before, these methods work for 10−5-dense partial Latin squares when nis equals and is not less than 107. The following process outlines how we will construct a completion of such a -dense partial Latin squareP that contains no more endδn2filled cells.
The Proof
Naively, we can hope that for most pairs of cells in our row, we can find the following trade. Therefore, whenever we can make all these choices, we have built the trade we claimed was possible.
Future Directions
From there we will show that the algorithm of Theorem 2.4.4 works in polynomial time. We assume that the task of completing such squares is NP-complete for any > 14, and prove that it is NP-complete for any > 12. For example, if we were to assert that the answer to a particular example of the traveling salesman problem was yes , we could accompany this statement with a particular tour of the cities that requires less than B units of distance to cover.
In fact, if S proves that the problem instance given by I has “yes” as an answer, it checks this proof and returns “yes”. Otherwise, if S is not proof that the problem instance given by I has “yes” as an answer, it notes that S fails as proof of I, and returns “no”. It is worth noting that any problem in P is in NP, because we can simply use the polynomial time solver of the P problem to decide whether or not a given instance is true in polynomial time, and thus complete the whole 'checking process' to skip.
We say that the problem Π1 is polynomially reducible to the problem Π2 if there is a function f with the following properties. Is there an assignment of true and false to the variables of this formula so that the entire expression evaluates to true?
Completing -Dense Squares in Polynomial Time
Runtime: This takes at most O(n2) steps to complete; we need O(n2) steps to fill each mesh, and at most O(n) steps to build and use a transversal to add our mesh in the odd case. Determine if the symbols in cells (r1, c1) and (r1, c2) are both still in the correct squares, if exactly one is out of place, or both are out of place. Specifically, start from the first row in the opposite half from rowstr3, r4 defined by our data (r1, c1) and (r1, c2).
For each row, check whether it meets the criteria set in the lemma: there are a constant number of checks to be performed, for each possible row choice. Runtime: This takes O(n) many steps to perform: determining whether the symbols (r1, c1) and (r1, c2) are still in the correct quadrants takes a constant amount of time, and creating our choices for r2, s6. As discussed in the lemma, choose the cell (r4, c4) with 1 so that all conditions of the lemma are met.
Running time: Again, it takes at most O(n) steps to complete, since at each stage we either allocate no objects and perform a constant number of checks for each object, or simply use Lemma 2.4.2 four times, which we already know takes O(n) steps. This is because Lemma 2.4.3 takes O(n) steps to run in each case, we need to run Lemma 2.4.3 at most O(n2) times, and Lemma 2.4.1/update steps only O(n2) steps in total throughout the implementation of the theorem.
Colbourn’s Theorem on NP-Completeness and Completing Partial Latin Squares
Therefore, membership of the NP is instant; so it is sufficient to reduce the task of completing an arbitrary partial Latin square to another NP-complete problem. From here, Colbourn reduces the above problem to the problem of completing an arbitrary partial Latin square. As mentioned before, the resulting Latin frame is a partial Latin square of order 2n, such that any completion of this partial Latin square corresponds to a triangulation of the tripartite graph G.
Therefore, we have reduced the task of completing a uniform tripartite graph to that of completing a partial Latin square. The task of completing a partial Latin square of arbitrary density is NP-complete, for every >12. Now we seek to find a clever way to place most of the cells in our empty square, such that each termination of this new partial Latin square will correspond to a triangulation of G.
By the logic established above, any completion of the resulting partial Latin square P0 will still correspond to a triangulation of G. Therefore, completing a dense partial Latin square is an NP-complete task, for any > 12.
Future Directions
Therefore, the expected number of cells in rowr, which are both overlapping cells and row-dependent cells, is no more than n·=2n, and the expected total number of such cells in our entire Latin square is no more than 2n2. Therefore, the number of row-dependent cells in this set is always no greater than n, as is the number of column-dependent cells. Thus, the total expected number of such collisions over our entire Latin square is no more than 2n2.
After we change the symbols of P, what is the expected number of cells in our array that are overlapping cells and symbol-dependent cells, or cells that are symbol-dependent in two different ways. Therefore, the expected number of such cells in the entire array is no more than jn2. Assume for the moment that the number of symbol-dependent cells in our array is no more than ln, for some l > 1.
Accordingly, the expected number of both row- and symbol-dependent cells generated by these choices is no greater than ln−12n2. As a result, the expected number of cells in our set that depend on both row and symbol is at most.
Future Directions
Each triangle of the tripartite complementG0 of this new graph will, as before, correspond to a partial Latin square LG0. Furthermore, each termination of LG0 will correspond to a triangulation of G0, and therefore to itself, by "adding back" the triangles we deleted earlier. So to triangulate G, it suffices to show that we can find 2 many such trades, since there are at most 3n2 edges in G0.
Since w1, w2 are nonγ-overloaded, this eliminates at most 2·(3γn) many edges from situations where one of these two vertices was used as anxi in a previous transaction, because each xi in our transaction has at most three edges for a given part . Note that each vertex in a cycle in such a transaction uses at most one edge to another part. Furthermore, note that no vertex appears in more than n many cycles, and that for any given cycle no vertex is used more than twice (as described in our earlier discussion of how we find these transactions iteratively). Therefore, this eliminates at most 4 choices. .
The 3γ comes from the fact that a node loses at most 3 edges per completed trade and no node is ever used when overloaded with γ. The extra 7δn2 comes from the fact that we need to find at most δn2 many such trades, each using 21 edges.
Future Directions
In this chapter, we take a break from Latin squares and study the concept of quasirandom graphs, a concept first introduced by Chung, Graham, and Wilson [8] in 1989. Roughly speaking, a sequence of graphs is called quasirandom if it has a row properties possessed by the random graph, all of which (surprisingly) turn out to be equivalent. In this chapter we study possible extensions of these results to random k-edge colorings and create an analogue of Chung, Graham, and Wilson's result for such colorings.
Given this model, a natural set of questions to ask is the following: What properties does a random graph generated by the above process likely have, asn goes to infinity. In most introductory classes on the probabilistic method in combinatorics, a student will quickly calculate a number of such properties as the random graph. Similarly, NG(H) denotes the number of labeled occurrences of H as a subgraph of G (not necessarily induced).
For example, suppose that a set of graphs has asymptotically the same number of 4-cycles in its members as the arbitrary graph Gn,1/2. Since their result, a number of authors have extended the results of Chung, Graham, and Wilson to generalized random graph models (Lov´asz and S´os, [24]), graph sets with given degree sets (Chung and Graham, [7]), and series hypergraphs (notably by Chung [5], [6] and more recently by Lenz and Mubayi. The aim of this chapter is similar: we want to extend the results of Chung, Graham and Wilson to a notion of quasi-random k-edge dyes .
Examples of Quasirandom Graphs
Using this transformation, we can create quasi-random graphs from affine planes as illustrated by the following construction. Given any k, we can create a "random" coloring of the k edges of the complete graph Kn by rotating a cover by k for each edge and coloring each edge with the result. Because P3⇒P0, we know that all buto(n) of these vertices have degree (1 +o(1))nk; therefore, we know we can write.
Now, if we think about what this means for the eigenvalues of A, we can use P3 to show it. On the other hand, if we use the observation that s is orthogonal by construction of finger1, we can see that In the first case, we can prove our claim by expressing ei(T) as the average value of setsei(S0), over.
Note, on the other hand, that we can calculate ei(X+i, n(v0)) strictly relative to the sizes of other sets whose sizes we can check with P4. Because we sum over all possible values of, we can see that we actually just choose α, v, w like this. But if you now imagine that we choose v, first before we choose α, we can see that the choices of vertices for α must be exactly those where the colors color(v, αi) = color(w, αi): in other words, we pick from just the pool of vertices counted bys(v, w).
Hence, if we return to our desire to control the second moment of the fr(α, )'s, we can see that.
