Addressing and hashing Question: For a network of computers it is desirable to be able to send a message from A to B without B knowing that a message is on its way. Model the network as a graph G. We can assign an address for each vertex. The address is from {0,1} k. The distance of two vertices in the graph is equal to the Hamming distance of the addresses. This is equivalent to regarding G as an induced subgraph of the hypercube H k. This is impossible for K * * * 1 * 3 * * *

Introduce a new alphabet {0,1,*} k and form addresses by taking n-tuples from it. The distance between two addresses is defined to be the number of places where one has a 0 and the other a 1. For an addressing of a graph G, we require that the distance of any two vertices in G is equal to the distance of their addresses. Denote by N(G) the minimum value of n for which there exists an addressing of G with length n.

For a tree we can do without the stars. By induction, it is clear for a tree of two vertices, which is trivial addressing of length 1. Suppose we can address trees with k vertices. If x 0,..,x n are the vertices of the tree T and x 0 has degree 1, consider the addressing for T/{x 0 } and let a i be the address of x i. Suppose x 0 is connected to x 1. Then address x 0 as (1, a 1 ) and change the other addresses to (0, a i ) for i=1,..,k. Thus N(T)  |V(T)|-1. Address a tree

Address a complete graph K m Consider the identity matrix of size m-1. Replace the zeros above the diagonal with stars and add a new row of zeros. N(K m )  m-1. 1 * * * * * 0 1 * * * 1 * * 3 * * *

x * * x2 1 0 * 1 * x3 * x * * x From the first column we have (x 1 +x 2 )(x 4 +x 5 ). From the 2nd column we have (x 1 )(x 2 +x 3 +x 4 +x 5 ). From the 3rd column we have (x 1 +x 4 )(x 3 +x 5 ). From the 4th column we have (x 2 )(x 3 +x 5 ). From the 5th column we have (x 3 )(x 5 ). Sum them up and have:  d ij x i x j. What does this mean?

Thm 9.1. Let n+, respectively n-, be the number of positive, respectively negative, eigenvalues of the distance matrix (d ij ) of the graph G. Then N(G)  max {n+, n-}. Pf: From the previous example:(x 1 +x 2 )(x 4 +x 5 ) can be represented by xAx/2, where x=(x 1, x 2, x 3,x 4,x 5 ) and A is the following matrix of rank 2 and trace 0: Therefore it has one positive and one negative eigenvalue Since (d ij ) is the sum of the matrices corresponding to the quadratic forms, it can have at most n positive eigenvalues Thm 9.2. N(K m ) = m-1. Thm 9.3. If T is a tree on n vertices, then N(T)=n-1.

The conjecture that N(G)  |V(G)|-1 for all connected graphs G was proved by Peter Winkler in ($200) Pick a vertex x 0, then construct a spanning tree T by BFS, and number the vertices by DFS. Define P(i):={j: x j is on the path from x 0 to x i in T}. Define i  j :=max {P(i)  P(j)), i’:=max{P(i)\{i}}. x0x0 x1x1 x2x2 x3x3 x4x4 x5x5 x6x6 x7x7 x8x8 x9x9

Define i ~ j  P(i)  P(j) or P(j)  P(i). Define c(i,j) := d T (x i,x j ) - d G (x i,x j ). Lemma 9.4. (i) c(i,j) =c(j,i)  0; (ii) if i ~ j, then c(i,j) =0. (iii) if not (i ~ j), then c(i, j’)  c(i,j)  c(i,j’)+2. Pf: (i) is trivial. (ii) d G (x i,x j )  |d G (x i,x 0 ) - d G (x j,x 0 )|= d T (x i,x j ). (Why?) (iii) By |d G (x i,x j ) - d G (x i,x j’ )|  1 and d T (x i,x j ) =1+ d T (x i,x j’ ). d T (x i,x j ) - d G (x i,x j ) =1+ d T (x i,x j’ ) - d G (x i,x j ).  x0x0 xjxj xixi

c(i,j) = d T (i,j) - d G (i,j) Example: C(0,1)=0, C(0,2)=0, C(0,3)=0, C(0,4)=0, C(0,5)=0 C(1,2)=0, C(1,3)=1, C(1,4)=1, C(1,5)=1 C(2,3)=1, C(2,4)=3, C(2,5)=3 C(3,4)=0, C(3,5)=0 C(4,5)= a 0 =(0,0,0,0,0) a 1 =(1,0,0,0,0) a 2 =(1,1,0,*,0) a 3 =(*,0,1,0,0) a 4 =(*,*,1,1,0) a 5 =(*,*,1,1,1) a 0 =(0,0,0,0,0) a 1 =(1,0,0,0,0) a 2 =(1,1,0,0,0) a 3 =(0,0,1,0,0) a 4 =(0,0,1,1,0) a 5 =(0,0,1,1,1)

Let n=V(G)-1. Let a i  {0,1,*} n be the address of x i. a i =(a i (1),..., a i (n)), where j>i i 0 j’ j<i

Thm 9.5. d(a i, a k )= d G (x i, x k ). Pf: Assume i < k. (1) Suppose i~k, i.e. i is in P(k): For j  P(i), a k (j) =1 and a i (j) = 1. Thus for j  P(k)\P(i), a k (j) =1 and a i (j)  1. For these j, c(i,j)=0, hence a i (j) =0. Then d G (x i, x k )=|P(k)\P(i)|. x0x0 xkxk xixi xjxj

Thm 9.5. d(a i,a k )= d G (x i,x k ). (2) Suppose not (i~k), i.e. i is not in P(k): Exactly one of {a i (j), a k (j)} equals 1 when j is in P(i)-P(k) or P(k)-P(i). Need to prove that there are exactly c(i,k) *’s in these coordinates. So this will yield d(a i, a k ) = |P(k)\P(i)| + |P(i)\P(k)| - c(i,k) = d T (x i,x k ) – c(i,k) = d G (x i,x k ). Consider c(i,k)  c(i,k’)  c(i,k’’) ...  c(i, i  k) =0 c(i,k)  c(i’,k)  c(i’’,k) ...  c(i  k, k) =0. We will obtain one * in a i for each even m with 0< m  c(i,k) and one * in a k for each odd m with 0< m  c(i,k). For each even m with 0 c(i,j’). Note m may not be in the list but j does exist! Each j is distinct, because c changes at most 2 with each step. The DFS ordering guarantees i < j for each j in P(k) - P(i). Thus a i (j) = * for j in P(k) - P(i) corresponding to some even m. x0x0 xkxk xixi

Similarly, for each odd m with 0 c(j’, k). Each j is distinct because c changes at most 2 with each step. The DFS ordering guarantees j < k for each j in P(i) - P(k). Thus a k (j) = * for j in P(i) - P(k) corresponding to some odd m. Thus we’ve counted the *’s in P(i)-P(k) and P(k)-P(i). The number of * is exactly the number of even integers plus the number of odd integers between 1 and c(i,k), which together equals c(i,k).  Thm 9.6. N(G)  |V(G)| - 1. x0x0 xkxk xixi

Def: An ABD(k,w), associative block design, is a set of b:= 2 w elements of {0,1,*} k with the following properties: if the elements are the rows of a b × k matrix C, then (i) each row of C has k-w stars; (ii) each column of C has b(k-w)/k stars; (iii) any two distinct rows have distance at least 1. Ex: ABD(4, 3) * * * * * * * * This implies each k-bit 0-1 vector has distance 0 to exactly one row of C. å0    represents 2 4 = vectors. So an ABD(k,w) uses 2 w k-bit 0-1-  vectors to indicate 2 k k-bit 0-1 vectors.

Thm 9.7. If an ABD(k,w) exists, then: (1) it has exactly bw/(2k) 0’s and bw/(2k) 1’s in each column; (2) for each k-bit x it has exactly C(w, u) rows which agrees with x in u positions; (3) the parameters satisfy w 2  2k(1-1/b); (4) for any row, the number of rows with stars in the same position is even. Pf: Let C be the ABD(k,w).

(i) each row of C has k-w stars; (ii) each column of C has b(k-w)/k stars; (iii) any two distinct rows have distance at least 1. (1) It has exactly bw/(2k) 0’s and bw/(2k) 1’s in each column. Pf: Each row in C represents 2 (k-w) k-bit 0-1 vectors. Consider the j-th column. There are 2 (k-1) k-bit 0-1 vectors with the j-th bit as 0. How many rows of C are needed to cover these 2 (k-1) 0-1 vectors? Each row in C with the j-th bit as zero can cover 2 (k-w) 0-1 vectors and each row in C with the j-th bit as * can cover 2 (k-w-1) vectors. Thus (# of 0 in the j-th col) 2 (k-w) + (# of * in the j-th col) 2 (k- w-1) = 2 (k-1). (# of 0 in the j-th col) = 2 (w-1) - b(k-w)/2k = b/2 - b(k-w)/2k = bw/2k. Similarly, for 1.

(i) each row of C has k-w stars; (ii) each column of C has b(k-w)/k stars; (iii) any two distinct rows have distance at least 1. (2) For each k-bit x, it has exactly C(w, u) rows which agrees with x in u positions. Pf: Let x be any k-bit 0-1 vector. Denote n i the number of rows of C which agree with x in i positions. There are C(k, l) k-bit vectors which agree with x in exactly l positions. Therefore C(k,l) =  i n i C(k-w, l-i).  l C(k,l)z l =  l  i n i C(k-w, l-i) z l =  i n i z i  l C(k-w, l-i) z l-i. I.e., (1+z) k = (1+z) k-w  n i z i. This proves n i =C(w, i).

(i) each row of C has k-w stars; (ii) each column of C has b(k-w)/k stars; (iii) any two distinct rows have distance at least 1. (3) The parameters satisfy w 2  2k(1-1/b). Pf: The sum of the distances between pairs of rows of C is k(bw/2k) 2 by (1). Why? Since any two rows have distance at least 1, this sum is at least C(b,2). Thus k(bw/2k) 2  C(b,2)= b(b-1)/2. (bw) 2  (4k) b(b-1)/2. w 2  2k (1-1/b).

(i) each row of C has k-w stars; (ii) each column of C has b(k-w)/k stars; (iii) any two distinct rows have distance at least 1. (4) For any row, the number of rows with stars in the same position is even. Pf: Consider a row of C. Count k-bit vectors which have 0’s in the positions where the row has *’s. e.g. For (***1100), there are 2 4 (000_ _ _ _) vectors with 0’s in the *-positions. In general, 2 w. Let R = (# of rows with the same *-pattern). Then these R rows indicate R 0-1 vectors with 0’s in the *- positions. Each row with a different star pattern represents an even number of such vectors whereas a row with the same *-pattern represents exactly one such vector. Why? Thus R + (even number) = 2 w, which implies R is even. **0001 **1010 **0101 ** ** 0**001 11**01

(i) each row of C has k-w stars; (ii) each column of C has b(k-w)/k stars; (iii) any two distinct rows have distance at least 1. Thm 9.8. Let C be an ABD(k,w) with w >3. (1) If two rows of C agree in all but one position, then C(w, 2)  k. (2) Otherwise w 2 > 2k. Proof of (1): Let c 1 and c 2 be two rows of C differing only in position 1. Then all the other rows of C must differ from c 1 in some other position. Thus, b - 2 ≤ (w-1) (bw/2k). kb – 2k ≤ (w-1)bw/2. k ≤ C(w, 2) (b/(b-2))= C(w, 2) (1+ 2/(b-2)),. k ≤ C(w, 2) because 2C(w,2)/(b-2) <1. 0 * * * * 0 _ _ _ _ _ 1 _ _ _ _ _ * _ _ _ _ _

Thm 9.8. Let C be an ABD(k,w) with w >3. (1) If two rows of C agree in all but one position, then C(w, 2)  k, i.e., w(w-1)  2k. (2) Otherwise w 2 > 2k. Proof of (2) : Let c 1 and c 2 be two rows of C with the same star–pattern differing in more than one position. Count the sum of the distances from c 1 to the other rows, which is w(bw/2k), by Thm 9.7.1, and at least (b-2) + 2 = b. Thus, w(bw/2k)  b, i.e., w 2  2k. Need to show the equality does NOT hold. If the equality holds, it implies c 1 and c 2 have distance 2 and the other rows have distance 1 to c 1 and c 2. And this holds for any pair of the same star-pattern. 0 * * 1 1 * 1 * * 0 1 * 0 _ _ _ _ _ 1 _ _ _ _ _ * _ _ _ _ _

Assume c 1 = (* * *....  ) and c 2 = (* * *....  ) to be such a pair. The other bw/(2k) – 1 rows ending in a ‘1’ would end with ’01’. Why? Similarly, there were bw/(2k) – 1 rows ending in ’10’. Thus, there exist two rows with distance at least 2. But with w(bw/2k) = b, we must have bw/(2k) – 1 = 1, which implies bw = 4k, i.e., bw=2w 2. 2 w =b=2w, which is impossible for w > 3. Therefore w 2 > 2k.

Cor. An ABD(8,4) does not exist. Pf: By Thm 9.8. Open problem: The smallest open case is the question whether an ABD(12, 6) exists? Thm 9.9. If an ABD(k i, w i ) exists for i=1, 2, then an ABD(k 1 k 2, w 1 w 2 ) exists. * Pf: Assume w 2 > 0. Partition the rows of ABD(k 2, w 2 ) into two 0 * 1 0 classes R 0 and R 1 of equal size. 0 0 * 1 In ABD(k 1, w 1 ) replace each star by a row of k 2 stars, each * by a row from R 0 and each 1 by a row from R 1 in all possible * ways. Then check it is an ABD as claimed. (Exercise!) 1 * * * Cor: An ABD(4 t, 3 t ) exists.

Thm Let w >0. Suppose an ABD(k, w) exists, k = s 2 h, s odd. Then an ABD(k, w+i  s) exists for i=0,..., (k-w)/s. Pf: It suffices to consider i=1. Let C be the ABD(k, w) and define a matrix A of the same size by requiring a ij =1 if C ij =  and a ij =0 otherwise. Then A has (k-w) 1’s in each rows and b(k-w)/k 1’s in each column. By Thm 7.3, there is a matrix A’ from A with (k-w-s) 1’s in each rows and b(k-w-s)/k 1’s in each column. Note that bs/k=2 w-h is integral. Thus the matrix A’’=A-A’ has s 1’s in each rows and bs/k=2 w-h 1’s in each column. In a row of C, replace  by – if the  occurs in a position where A’’ has a ‘1’. This produces an ABD(k, w+s). Use k-tuple of 0,1,*, – to represent all possible 0, 1, * by replacing each – by 0 or 1.

Thm If an ABD(k, w) exists and R ≥1 is a number such that Rk and Rw are integers, then an ABD(Rk, Rw) exists. Pf: It suffices to show that ABD(k+l, w+m) exists for (k+l)/(w+m)=k/w and gcd(l,m)=1. Let k = s 2 h, s odd. By (ii) of the ABD’s definition, s|w. From (k+l)/(w+m)=k/w, we have w(k+l)=(w+m)k, i.e., wl=mk. Since gcd(l,m)=1, we have l is a power of 2. Consider the l  l circulant matrix with a row of l-m stars and m minus signs as the first row. Since l|b, adjoin a column of b/l copies of this circulant to the matrix C of ABD(k,w). The larger matrix is an ABD(k+l, w+m). 2w2w k l l b/l...