Coloring Paths in Directed Symmetric Trees with Applications to WDM Routing By L. Gargano, P. Hell, S. Perennes Presented By: Gal Ostfeld

A Tree Definition: Let G (V,E) be a graph. G is called a “tree” if it’s connected and circuit-free. Also: Any new edge added to G will form a circuit. Every 2 vertices have a unique path connecting them. If any edge is deleted the connectivity of G is interrupted. |E| = |V|-1.

A Symmetric Directed Tree Equivalent to an undirected tree with each edge viewed as two opposite arcs.

A Dipath The dipaths P(x,y) and P(y,x) are different and do not traverse any edge in the same direction Let T be a tree and x,y two vertices of T. The “dipath” P(x,y) in T is the undirected path joining x to y, in which each edge is considered traversed in the direction from x to y.

Dipaths P(x,y) & P(y,x) xy P(x,y) P(y,x)

Our Goal – The All to All Problem (AKA “Gossiping”) Coloring the set of dipaths P(x,y), for all ordered pairs x,y of vertices of T, in such a way that two dipaths using the same edge of T in the same direction obtain different colors with minimum number of colors (This way is called “proper” coloring).

So, what’s “proper”?…

Then, what’s not “proper”?…

We shall prove that… Given c(T) is the minimum number of colors in such a coloring of the dipaths in tree T. Given  (T) is the maximum number of dipaths P(x,y) which all pass through the same edge of T in the same direction. c(T) =  (T)

A weighted Tree A weighted Tree is a tree T with positive integer weights w(x) on the vertices x of T. (The intention is to have a w weight vertex represent w unweighted vertices) The total weight of a set X of vertices of T is w(X)= w(x) => w(T) is tree weight.

Load / Maximum Load Let e be an edge in a weighted tree T. The removal of e from T results in two weighted subtrees T 1 and T 2. The “load” of e is the product w(T 1 )w(T 2 ) The “forwarding index”,  (T), is the maximum load of any edge in T. (when all w(x) = 1,  (T) is the maximum number of dipaths P(x,y) which all pass through the same edge of T in the same direction)

Minimum Number of Colors In a weighted tree T, the “multiset of all dipaths” consists of the w(a)w(b) copies of the dipath from a to b for every ordered pair of vertices a,b. c(T) is the minimum number of colors in a proper coloring of the multiset of all dipaths. (when all w(x) = 1, c(T) is the minimum number of colors in a proper coloring of the set of all dipaths)

In(v) / Out (v) Let v be a vertex in a weighted tree T. In(v) consists of the dipaths from the multiset of all dipaths which end with v. Out(v) consists of the dipaths from the multiset of all dipaths which begin with v.

Theorem Any weighted tree T satisfies c(T) =  (T). And there exists an efficient algorithm which colors T with c(T) colors. Proof: Inductively…

Definitions: T – weighted tree w(T) = W x - leaf of T f - parent of x δ - integer, 0<δ<w(x) x f w(x)>δ T w(T) = W

Definition: AddLeaf AddLead δ (x) modifies T as follows: –w(x) = w(x) – δ –A new node y is added: w(y) = δ connected to x x f x f y w(x) w(x)–δ δ w(x)

Definition: SplitLeaf SplitLead δ (x) modifies T as follows: –w(x) = w(x) – δ –A new node y is added: w(y) = δ connected to f x f x f y w(x) w(x)–δ δ w(x)

Definition: Legal actions AddLeaf or SplitLeaf are “legal” if w(x)  W/2 (δ+w(x)  W).

Lemma 1 If AddLeaf is legal then in the new tree the load of [x,y] cannot be larger than the load of [x,f] in T. If SplitLeaf is legal then in the new tree the load of [x,f] and [y,f] cannot be larger than the load of [x,f] in T.

Lemma 1 Proof Max = W/2 n*(W-n) n Before Add/Split After Add/Split w(x)w(x)- δ δ Finding the max: (n*(W-n))’ = 0 W-2n = 0 n = W/2 x f y w(x)–δ δ x f y δ Add Split

Lemma 2 If AddLeaf or SplitLeaf on T is legal then the forwarding index of the new tree doesn’t exceed the forwarding index T. Proof: Trivial – based on the preceding lemma.

Definition: W/C Tree Let T be a weighted tree. Let w(T) = W. T is called a “W/C tree” if the two trees resulting from the removal of an edge of maximum load have weights C and W-C, with C≥W/2. In a W/C tree:  (T) = C(W-C)

Definition: W/C Star In a W/C star the maximum load on an edge leading to a leaf is  (T)=C(W-C). In case a W/C tree is a weighted star then T is called a “W/C star”.

What are we trying to achieve? Given a W/C Tree, T. We shall recursively reconstruct T from some initial W/C star S, with AddLeaf and SplitLeaf legal operations. In the W/C star S:  (S) = C(W-C) By Lemma 2, eventually we’ll get  (T) =  (S) = C(W-C)

Lemma 3 A W/C tree T can be generated from some W/C Star T* by repeated applications of legal operations of AddLeaf or SplitLeaf

Lemma 4 (Needed for lemma 3’s proof) Any W/C tree T contains a vertex u such that the maximum weight of a component of T\{u} is W-C. Proof: We know that exists an edge e 1 (between vertices u 1 and u 2 ) that divides T to T 1 and T 2 (respectively) that hold w(T 1 )=C, w(T 2 )=W-C. Let’s choose u 1 and prove its suitable for our “needs”.

Lemma 4 Proof u1u1 u2u2 T2T2 T1T1 W(T 2 )=W-C W(T 1 )=C TiTi Let’s assume w(T i )>W-C, for example: w(T i )=W-C+n=W-(C-n). TjTj TkTk u3u3 T is a W/(C-n) tree (via the e 2 edge). e1e1 e2e2 Contradicts T being a W/C tree (since (C-n)(W-(C-n))>C(W-C). All T i -s “inside” T 1 should hold w(T i )<=W-C. QED w(T 2 )=W-C, by definition.

Construction of W/C tree T’s W/C star T* Start with a W/C star T* consisting of the vertex u and all its neighbors in T. Each neighbor v, should have his w(v) set to the weight of the component of T\{u} that contains v.

W/C tree T’s W/C star T* u T1T1 W(T 2 )=W-C W(T 1 )=C TjTj TkTk T2T2 TiTi W(T i )=D W(T j )=E W(T k )=F u uiui ujuj ukuk u2u2 W(u i )= D W(u j )=E W(u k )=F W(u 2 )=W-C

From W/C star T* to W/C tree T Let t be the number of nodes in T which are not adjacent to u. If t=0 then T is a star and we’re done! u uiui ujuj ukuk u2u2 W(u i )=D W(u j )=E W(u k )=F W(u 2 )=W-C So, we have the induction base.

From W/C star T* to W/C tree T Suppose that the result holds for if t<=k. Let t=k+1. (Induction Step) Let z be a leaf of T of maximum distance from u, and let p be z’s parent. If degree(p)=2 then Let T’ be the weighted tree obtained from T by removing z and increasing w(p) by w(z). T is generated from T’ by AddLeaf w(z) (p).

From W/C star T* to W/C tree T (When degree(p)=2) p f p f z [w(p)+w(z)]–w(z) w(z) T’T w(p)+w(z)

From W/C star T* to W/C tree T If degree(p)>2 then Let x be a leaf neighbor of p other than z. Let T’ be the weighted tree obtained from T by removing z and increasing w(x) by w(z). T is generated from T’ by SplitLeaf w(z) (x).

From W/C star T* to W/C tree T (When degree(p)>2) x p x p z [w(x)+w(z)]–w(z) w(z) T’T w(x)+w(z) Lemma 3 QED

Lemma 5 The multiset of dipaths of any W/C star can be efficiently colored with C(W-C) colors.

Hall’s Marriage Theorem (Needed for lemma 5’s proof) Let’s have two sets of N elements denoted M, F. Each of the elements in M is willing to “marry” a number of elements from F. If for every K sets from M, 1<=K<=N, the number of elements from F is at least K, then we would have a perfect matching, i.e. we can choose N couples according to the “wish list”.

Hall’s Marriage Theorem - Example 5 x MF

Lemma 5 Proof In a star two dipaths conflict if and only if they have the same beginning or the same end. Two dipaths of the same color must belong to two different multisets In(v) and to two different multisets Out(v). For each vertex v |In(v)| = |Out(v)|. This size differs from vertex to vertex. Maximum In(v) per edge = C(W-C).

Lemma 5 Proof - continued Add to each In(v) and Out(v), at each vertex v, C(W-C)-|In(v)| artificial nodes (consisting only of v). Denote the set of all In(v)-s as Hall’s Theorem’s M. Denote the set of all Out(v)-s as Hall’s Theorem’s F. Denote “exists a dipath that begins in Out(v) and ends in In(v)” as Hall’s Theorem’s “willing to marry”.

Lemma 5 Proof – Step #1 Out(v1) Out(v2)Out(vn) … In(v1) In(v2)In(vn) … ……… C(W-C) ……… n x(By color #1)

Lemma 5 Proof - end Since all the Out(v)-s have C(W-C) out edges, and In(v)-s have C(W-C) in edges, every K Out(v)-s point to at least K different In(v)-s. After matching each In(v) with an Out(v) perfectly, we use color #1 to “marry” them. Then we can delete from consideration the edges we used for the matching. Now we have the same problem, but with one less edge from/to each node. We can go on, till all the edges are used. We used C(W-C) colors. QED

Going On... (Some definitions) Let x be a fixed leaf in tree T (on which we shall perform the legal AddLeaf, SplitLeaf). T is properly colored. In(x) (with the In set of colors) and Out(x) (with the Out set of colors) might have used different colors. |In|=|Out| Let “  ” be the fixed bijection between Out and In, such that for any c belonging to Out  In,  (c)=c.

And On... (Some more definitions) Let z be any vertex of T\{x}. All the w(z)*w(x) dipaths between x and z must obtain different colors, since there’s a unique edge out (in) of (to) x (x is a leaf). Arbitrarily fix (for each z) a partition of these w(z)w(x) colors into w(z) classes of size w(x) denoted O 1 z, O 2 z …, O w(z) z (I 1 z, I 2 z …, I w(z) z ). We shall say that two colors on dipaths ending (starting) in x are “I-equivalent” (“O-equivalent”) if they belong to the same class I i z (O i z ).

And On... (And more definitions) A “supercolor” is a set U of colors such that no colors from U are I-euivalent, and no colors from  (U) are O-euivalent. Let X be the set of w(x)(W-w(x)) colors used by the dipaths starting in x.

Lemma 6 The set X of colors can be partitioned into w(x) supercolors.

Lemma 6 - Proof Let E i (1  i  W-w(x)) be the I-equivalence classes of x on X. Let F i (1  i  W-w(x)) be the O-equivalence classes of x on X. Let E’ i =  (E i ) (1  i  W-w(x)). Denote the set of all E’ i -s as Hall’s Theorem’s M. Denote the set of all F i -s as Hall’s Theorem’s F. Denote “a color exists both in E’ i and in F i ” as Hall’s Theorem’s “willing to marry”. QED

Lemma 6 Proof - Illustration E’ 1 E’ 2 E’ W-w(x) … F1F1 F1F1 F W-w(x) … ……… w(x) ……… (W-w(x)) x(x w(x) colors)

Proposition 7 Let T be a W/C tree. With a proper coloring of its multiset of all dipaths. T’ is obtained from T by performing the legal operation AddLeaf or SplitLeaf. T’ is a W/C tree which also admits a proper coloring of its multiset of all dipaths.

Proposition 7 Proof Partition the colors into w(x) supercolors. Partition the supercolors into S 1, S 2 such that |S 1 |=w(x)-δ and |S 2 |=δ. Perform AddLeaf δ (x) (SplitLeaf δ (x)) on T, to create node y. Let T’\{x,y} be T ~. Use S 1 to color the paths from (to) x to (from) T ~. Use S 2 to color the paths from (to) y to (from) T ~. If we dealt with a SplitLeaf operation – we’re done!

Proposition 7 Proof If we dealt with an AddLeaf operation, we must check if we have enough supercolors to color the new dipaths between x and y (both directions). We used δ supercolors [=w(T ~ )*δ colors] to color the dipaths between y and T ~ (both directions). The w(x)- δ supercolors [=w(T ~ )*(w(x)-δ) colors] to color the dipaths between x and T ~ (both directions), can be reused for the dipaths between x and y (different edge of x).

Proposition 7 Proof There are w(x)-δ supercolors [=w(T ~ )*(w(x)-δ) colors] left to color the dipaths between x and y (both directions). We need δ(w(x)-δ) colors to color the δ(w(x)-δ) dipaths between x and y (both directions). δ < w(T ~ ) (it’s a legal operation). δ(w(x)-δ) < w(T ~ )*(w(x)-δ). QED

Proposition 7 Proof - Illustration x f x f y w(x) w(x)–δ δ w(x) x f f y w(x)–δ δ w(x) x

c s (T) c s (T) = The minimum number of colors in a coloring of the paths from a subset S of all the paths on a directed symmetric tree T, such that conflicting paths obtain different colors. In the general situation of an arbitrary set S of dipaths, it’s known that the problem of optimally coloring the paths in the set is NP-hard. Only approximation algorithms are known. We shall consider cases in which c s (T) can be efficiently evaluated.

T Is A Path “Proper vertices coloring” = Coloring of the vertices, such that adjacent vertices obtain different colors. “Chromatic number” = The minimum number of colors in such a coloring of the vertices in tree T. If T is a path, c s (T) is equivalent to the fact that for an interval graph the chromatic number is equal to the maximum clique size.

Path Coloring Clique Number = 3 Clique Number =

T Is A Star “Proper edges coloring” = Coloring of the edges, such that adjacent edges obtain different colors. “Chromatic index” = The minimum number of colors in such a coloring of the edges in tree T. If T is a star, c s (T) is equivalent to the fact that for a bipartite graph the chromatic index is equal to the maximum degree.

Star Coloring Maximum Degree = 3

A Conflict Graph The “conflict graph” of set of paths S on a tree T is the undirected graph whose vertices are the the dipaths from S, and two dipaths are adjacent if and only if they conflict, i.e. use an edge of T in the same direction

A Generalized Star It’s a tree obtained from a star by replacing each edge with a path (the paths may have different lengths).

Facts About Generalized Stars A generalized star is a tree in which at most one vertex has a degree greater than two. Any tree in which at most one vertex has a degree greater than two is a generalized star. Stars and paths are generalized starts.

Odd Hole / Odd Antihole An “odd hole” of an undirected graph is an induced cycle without chords, of odd length greater than three. An “odd antihole” of an undirected graph is an induced complement of a cycle without chords, of odd length greater than three.

Lemma 8 The conflict graph of any set of paths on a generalized star cannot contain an odd hole or an odd antihole.

Lemma 8 Proof – Odd Holes Let “c” (for center) denote the vertex with degree more than two (if exists). Let “A type” be a path traversing via the c. Let “B type” be a dipath which is not an A Type. Let P 1,…,P m (m>3) be such that their conflict graph forms an odd hole, i.e. P i conflicts with P i-1 and P i+1 (cyclically) for any 1  i  m. P 1,…,P m cannot be all B Types, since the closest one and the farthest one from c cannot conflict without conflicting with all other dipaths.

All B Types - Illustration c The red dipaths conflict with only one dipath.

Lemma 8 Proof – Odd Holes A types can conflict either in both of their first part (before c) or in both of their last part (after c). In a hole there must be an even number of A types.

Even A Types - Illustration c The red dipath is needed to complete the hole.

Lemma 8 Proof – Odd Holes There can be no B types between any two A types, because either the B types conflict with only one of them, so that A type conflict with at least three dipaths (and that’s not a chordless hole), or the B types conflict with both of them, so both A types conflict with at least three dipaths (and that’s not a chordless hole). If there can be no B types and there can only be an even number of A types, any conflict graph’s hole must be even.

B Types Between A Types - Illustration c The red B types conflict with both A types conflicts The purple B types conflict one A type.

Lemma 8 Proof – Odd Antiholes An antihole can be viewed as some general case of a hole (when n=5): –Each vertex has the exact same number of neighbours (n-3 in an antihole, 2 in a hole). –It has a sense of direction. a cb d e = a ed c b

Lemma 8 Proof – Odd Antiholes We’re left with antiholes of size  7. Its complement is a hole of size  7. Its complement has a chordless path on  6 vertices. But a conflict graph’s complement cannot contain a chordless path on  6 vertices. QED

Corollary 9 For any set S of dipaths in a generalized star T we have c S (T)=  S (T).

Corollary 9 Proof A graph is perfect if (and only if) it contains no odd hole and no odd antihole (The perfect graph conjecture). Conflict graphs in generalized stars are perfect. c S (T)=  S (T). QED

Proposition 10 c S (T)=  S (T) for all sets S on T if and only if T is a generalized star.

Proposition 10 Proof T isn’t a generalized star. T contains at least two vertices of degree  3. Let’s build the following S: And since we’ve proved that for a generalized star c S (T)=  S (T). QED c S (T)=3  S (T)=2

Well Distributed Set of Paths Let T be a tree. Let S be a set of paths in T. S is “well distributed” in T if T does not contain an odd number of edges [v,a 1 ], [v,a 2 ],…, [v,a 2k+1 ] such that some dipath of S contains both edges [v,a i ] and [v,a i+1 ] (in some direction, addition on index is taken modulo 2k+1), for any index i= 1,…2k+1.

Proposition 11 If S is well distributed in T then c S (T)=  S (T) and c S (T) can be found in polynomial time.

Proposition 11 Proof Intuition Let’s look at what was the “problem” in the beneath example: Although all the non-dashed edges were all used only in one direction, the dashed edge was used in both, in a way that it “mixed” the constraints of paths from different edges (or from similar edges but from different directions, in the general case). c S (T)=3  S (T)=2

Proposition 11 Proof Intuition So, let’s try and see how we can keep apart constraints on different edges and on different directions on each edge. We shall first show that if S is well distributed in T, then T admits an orientation such that each dipath in S either uses all edges in the chosen direction, or all in the opposite direction. And since those two “groups” of paths cannot conflict, we can color them separately. Then we shall easily see that for each set c S (T)=  S (T).

Proposition 11 Proof Let’s say that S is well distributed in T. Either there’s a cycle as in the well distributed definition, but with even size (instead of odd). If both paths’ common edge is in the same direction we shall put both of them in the same “group”, else in two different “groups” – The even size, assures that the “last” path can agree (disagree) with both “neighbors” together. Red group – in the “chosen direction” Blue group – in the “opposite direction” Dashed line – spoils well distribution. Black Arrowheads – The edges’ “directions”.

Proposition 11 Proof Or there isn’t a cycle as in the well distributed definition. If both paths’ common edge is in the same direction we shall put both of them in the same “group”, else in two different “groups” – The lack of cycle, assures that the “last” path has to agree (disagree) with only one “neighbor” (which is no problem). Red group – in the “chosen direction” Blue group – in the “opposite direction” Dashed line – spoils well distribution. Black Arrowheads – The edges’ “directions”.

So, How Do We Decide On The “Chosen Direction” Loop: Choose a non-chosen path from S (mark as chosen). If it doesn’t conflict with any chosen path choose all of the path’s edges’ direction in T according to the path’s, else if this path agrees (disagrees) with the conflicting path’s direction, choose all of the other’s path’s edges direction according to (against) the conflicting path’s relative direction.

Deciding On The “Chosen Direction” - Example

Proposition 11 Proof There’s only one path between any two vertices in tree T. For any at least four vertices from which every exactly two “neighbors” conflict, there is at least one “most close to the beginning vertex” and at least one “most close to the end vertex” (“direction”-wise). These two can “complete” a “conflict cycle”, only if at least one of them conflicts with all others in the cycle. Each “group’s” conflict graph cannot contain cycles of size > 3 without chords in them.

Proposition 11 Proof Each “group’s” conflict graph contains no odd hole and no odd antihole. Each “group’s” conflict graph is perfect. c S (T)=  S (T). QED

Thanks to: Prof. Shmuel Zaks and Msc. Mordo Shalom for their help and guidance.

References L. Gargano, P. Hell, S. Perennes “Coloring Paths in Directed Symmetric Trees with Applications to WDM Routing”, 1997.