The minimal set of Ingleton inequalities

The minimal set of Ingleton inequali ties Laurent Guill ´ e ENST Paris T erence H. Cha n and Alex Grant Institute for T elecommunicatio ns Research University of South Australia Abstract — The In g leton-LP bound is an outer bound for the multicast capacity r egion, assuming the use of linear network codes. Computation of the bound is perfo rmed on a p o lyhedral cone obtained by taking the intersection of half-spaces induced by the basic (Shan non-type) inequalities and Ingleton in equalities. This paper simplifies the characterization of this cone, by obtaining the uniq ue minimal set of Ingleton inequaliti es. As a result, the effort required for computation of the Ingleton-LP bound can be grea tly reduced. I . I N T RO D U C T I O N The network coding ap proach introd uced in [1], [2] gen eral- izes routing by allowing intermediate nodes to forward packets that are coded combin ations of all received data packets. One fundam ental prob lem in network cod ing is to unde rstand the capacity region and the classes of cod es that a chieve capacity . In the single session multicast scen ario, the prob lem is well understoo d – the cap acity r egion is characterized b y max- flow/min-cut bou nds and linear n etwork codes are sufficient to achieve m aximal through put [2], [3]. Significant practical and the oretical comp lications arise in m ore g eneral multicast scenarios, in volving mo re th an on e session. There are o nly a few too ls in the literature for study of the cap acity region. One powerful theoretical tool bound s th e capacity region by the inter section of a set of hyperp lanes (specified b y the network topo logy and conn ection require - ment) and the set o f entropy fun ctions Γ ∗ (inner bound) , or its closure ¯ Γ ∗ (outer boun d) [4 ]–[6] . Th e set of entropy functions is fo rmally intro duced in Section II. In fact an exact expression for the capacity region has been obtained, again in terms of Γ ∗ [7]. Unfortu nately , the capacity region, or even the b ounds cannot be computed in practice, due to the lack of an explicit char acterization of the set of entropy function s fo r m ore than thr ee rando m variables. One way to resolve this difficulty is via relaxation of the bound, replacin g the set of entropy f unction s with the set of po lymatroid s Γ . This yields the linea r pr ogramm ing ( LP) bo und [6 ]. The LP bo und can howe ver be q uite loose. While th e set of polymatr oids Γ ∗ is po lyhedra l, the set of entropy function s Γ is not [8]. The addition of any finite number of linea r inequalities to the LP bo und can not tighten it to the capacity region. Fur thermor e, the LP bo und h olds for any choice of network co des (linear o r n on-line ar). Hence, the LP boun d can be even looser when r estricted to linear network c odes. T o address th e issue, a modified LP bou nd was prop osed in [9] . The id ea of the m odified bo und is quite simple. Giv en any network code, the sour ce messages an d transmitted lin k mes- sages are random v ariables. Restriction to linear codes requires that th e c orrespon ding en tropy fu nction satisfies the Ingleton inequality . As a result, we can tig hten the LP b ound for line ar network co des by replacin g Γ (the set o f p olymatr oids) with Γ In (a subset of Γ that satisfies all I ngleton in equalities). Efficiently compu tation of this Ingleto n-LP bou nd requires a compac t and explicit characterization of Γ In . As the set Γ In is the intersection o f many half-spac es, one can greatly simplify the characteriza tion by identifying which inequalities (or half- spaces) are redund ant, meaning that they are imp lied by other inequalities. The main objective of this p aper is to understan d the rela tionship betwe en th ese h alf-spaces, so as to simplify the characterization of Γ In . Our main result, Theor em 3 in Section III is the id entification of the uniqu e m inimal set o f Ingleton inequalities. For r easons of space, all simple pro ofs have been omitted, an d long er proof s are given in sketch fo rm. I I . E N T RO P Y F U N C T I O N S Let N , { 1 , 2 , · · · , n } induce a 2 n − 1 -dimension al real Euclidean spac e F n whose coordin ates are in dexed by the set of all no nempty subsets α ⊆ N . Each h ∈ F n is defined b y ( h ( α ) : α ⊆ N ) . Alth ough h ( α ) is n ot defined for the emp ty set ∅ , we will assume h ( ∅ ) = 0 . Points h ∈ F n can also b e considered as fu nctions h : 2 N 7→ R . Definition 1 ( Entr opic function): A fun ction h ∈ F n is entr o pic if there exists discrete ra ndom variables X 1 , . . . , X n such that the joint en tropy of { X i : i ∈ α } is h ( α ) fo r all ∅ 6 = α ⊆ N . Furthermo re, h is almost en tr o pic if it is the limit o f a seque nce of entropic func tions. Let Γ ∗ n be the set of all en tropic functions. Its closure ¯ Γ ∗ n (i.e., the set of all almost e ntropic functio ns) is well-known to be a closed, co n vex cone [ 10]. An impo rtant recen t re sult with significant implicatio ns fo r ¯ Γ ∗ n is a series of linear in formatio n inequalities obtained in [ 11]. Usin g this series, ¯ Γ ∗ n was p roved to be non- polyh edral for n ≥ 4 . This means tha t ¯ Γ ∗ n cannot be defined by an in tersection of any finite numb er o f linear informa tion inequalities. T o simplify n otation, set u nion will be denoted by con- catenation, an d singleton s and sets with one elements are no t distinguished . For any α, β ⊆ N define h ( α | β ) , h ( αβ ) − h ( β ) I h ( α ; β | δ ) , h ( αδ ) + h ( β δ ) − h ( δ ) − h ( αβ δ ) . Pr o position 1: Let h ∈ ¯ Γ ∗ n . Then for all α, β , δ ⊆ N , h ( α | β ) ≥ 0 (1) I h ( α ; β | δ ) ≥ 0 . (2) Proposition 1 gives the basic inequa lities , namely the non- negativity of (con ditional) entropy (1) an d of (conditional) mu- tual informatio n (2). This set of basic ineq ualities is redundant, and the uniq ue m inimal set of basic inequ alities th at yields all basic inequalities ( as linear combin ations) is as f ollows h ( i |N \ i ) ≥ 0 (3) I h ( i ; j | δ ) ≥ 0 (4) where i 6 = j ∈ N and δ ⊆ N \{ i, j } . In equalities (3) and ( 4) are called elemental basic inequalities. See [5] for discussion of b asic an d elem ental in equalities. For any linear expression P α ⊆N c α h ( α ) , d efine the p rojec- tion o nto a sub set β of N X α ⊆N c α h ( β ∩ α ) . (5) Similarly , d efine the p rojection a way fr om β X α ⊆N c α h ( α \ β ) . (6) Clearly , if two linear expressions P α ⊆N c α h ( α ) and P α ⊆N d α h ( α ) are the same (i.e , the two expressions map to the same real n umber f or all h ∈ F n ), then their p rojection s onto (or away fr om) any sub set β are identical. I I I . I N G L E T O N I N E Q U A L I T I E S : P R O P E RT I E S A N D T H E M I N I M A L S E T Definition 2 ( Ingleton Ine quality): An In gleton inequality J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 is a linea r ine quality over F n defined in terms of four subsets α 1 , α 2 , α 3 , α 4 of th e groun d set N where th e In gleton term J ( h ; α 1 , α 2 , α 3 , α 4 ) is defined as h ( α 1 α 2 ) + h ( α 1 α 3 ) + h ( α 1 α 4 ) + h ( α 2 α 3 ) + h ( α 2 α 4 ) − h ( α 1 ) − h ( α 2 ) − h ( α 3 α 4 ) − h ( α 1 α 2 α 3 ) − h ( α 1 α 2 α 4 ) . Remark: Originally , the In gleton inequality refers to the case when α 1 , . . . , α 4 are distinct subsets of sin gletons. W e extend its use to allow arbitrar y subsets α 1 , · · · , α 4 of N . W ithin the class of I ngleton inequ alities indexed by four subsets of N , some are trivial ineq ualities w hile some can be derived from basic inequ alities o r oth er Ingleton inequ alities. For example, if α 1 = α 2 = α 4 = α 4 , then J ( h, α 1 , . . . , α 4 ) = 0 for all h and the co rrespond ing Ingleto n in equality is tr ivial. W e now list several pro perties o f the Ingleton term, leading to the min imal set of non -redu ndant In gleton ineq ualities. Pr o perty 1 (S ymmetry): J ( h ; α 1 , α 2 , α 3 , α 4 ) = J ( h ; α 2 , α 1 , α 3 , α 4 ) = J ( h ; α 1 , α 2 , α 3 , α 4 ) = J ( h ; α 1 , α 2 , α 4 , α 3 ) . Thus exchangin g α 1 with α 2 or α 3 with α 4 does n ot change the value of the Ingleton term, and the number of distinct Ingleton inequalities is at most 2 4 n − 2 . Pr o perty 2 (Ex tending basic ineq ualities): J ( h ; α 1 , α 2 , ∅ , α 4 ) = I h ( α 1 ; α 2 | α 4 ) J ( h ; α 1 , α 1 , ∅ , α 2 ) = h ( α 1 | α 2 ) . Thus all basic inequalities are sp ecial cases o f Ing leton in- equalities via proper selection of α 1 , . . . , α 4 . Hence, Γ In n ⊂ Γ n . Pr o perty 3: Let α 1 , . . . , α 4 , β ⊆ N . If β ⊆ α 1 ∩ α 2 , then J ( h ; α 1 , α 2 , α 3 , α 4 ) = J ( h ; α 1 , α 2 , α 3 β , α 4 β ) + h ( β | α 3 , α 4 ) . If β ⊆ α 1 ∩ α 3 , then we have J ( h ; α 1 , α 2 , α 3 , α 4 ) = J ( h ; α 1 , α 2 β , α 3 , α 4 β ) + I h ( β ; α 4 | α 2 ) . On the oth er hand , if β ⊆ α 3 ∩ α 4 , then J ( h ; α 1 , α 2 , α 3 , α 4 ) = J ( h ; α 1 β , α 2 β , α 3 , α 4 ) + I h ( β ; α 2 | α 1 ) + h ( β | α 2 ) . T o sum marize, if an element ap pears in at least two subsets of an Ingleton term , then we add th at elem ent to the remainin g two subsets with out in creasing the value of th e Ing leton term. Pr o perty 4: Let a ⊆ α 2 , b ⊆ α 3 and c ⊆ α 4 , then J ( h ; abc, α 2 , α 3 , α 4 ) = I h ( α 3 ; α 4 | abc ) + I h ( α 3 ; c | α 2 a ) + I h ( α 4 ; b | α 2 ) + h ( a | α 3 α 4 ) . Similarly , if a ⊆ α 1 , b ⊆ α 2 and c ⊆ α 3 , then J ( h ; α 1 , α 2 , α 3 , abc ) = I h ( α 2 ; c | α 1 b ) + I h ( α 3 ; b | α 1 ) + I h ( α 3 ; a | α 2 c ) + I h ( α 1 ; α 2 | α 3 ab ) + h ( c | α 2 ) . Consequently , if one of the subsets in the Ingleton term J ( h ; α 1 , α 2 , α 3 , α 4 ) is contain ed in the u nion of the other three subsets, then the co rrespon ding Ingleton inequa lity is implied by the basic inequ alities. In fact, we will prove that the converse is also true in the f ollowing theorem. Theor em 1: An In gleton in equality J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 is imp lied by the basic in equalities if an d on ly if one of th e four subsets α 1 , α 2 , α 3 , α 4 is contained in the union of the other thre e subsets. Pr o of: The if -p art fo llows directly from Proper ty 4. A sketch pro of for the co n verse is given as follows. Suppose to the con trary that non e of the four subsets α 1 , α 2 , α 3 , α 4 are co ntained in the u nion o f the remain ing three subsets. T o pr ove the con verse, it suffices to show that there exists h ∗ ∈ F n satisfying every basic inequ ality but not the Ingleton ineq uality . T o this end, write α i = δ i ∪ β i , wh ere (1) | δ i | = 1 , (2) β i ∩ δ i = ∅ a nd (3) ∅ 6 = δ i ∩ α j = ∅ f or all j 6 = i . This ter m re-writin g is possible by assum ption. In [12] , an entro py f unction g in volving fou r e lements was constructed which satisfies the basic inequalities but no t the Ingleton inequ ality . Since J ( h ; δ 1 , δ 2 , δ 3 , δ 4 ) in volves only four elements, we ca n easily construct an h ∗ ∈ F n that satisfies all the basic inequa lities and J ( h ∗ ; α 1 , α 2 , α 3 , α 4 ) = J ( h ∗ ; δ 1 , δ 2 , δ 3 , δ 4 ) = J ( g ; δ 1 , δ 2 , δ 3 , δ 4 ) < 0 . Hence, J ( h ∗ ; δ 1 , δ 2 , δ 3 , δ 4 ) < 0 and th e theo rem follows. Define the set of f unctions with n on-negative Ingleto n term Γ In , { h ∈ F n : J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 , ∀ α i ⊆ N } . It is c lear th at Γ In is a closed an d co n vex cone. T o char acterize Γ In , we first defin e the following sets of in equalities. ∆ 1 = { J ( h ; i, j, ∅ , α ) ≥ 0 : i, j ∈ N , i 6 = j, N \{ i , j } } (7) ∆ 2 = { J ( h ; i, i, ∅ , N \{ i } ) ≥ 0 : i ∈ N } . (8) By Property 2, ∆ 1 and ∆ 2 together imply all the basic inequalities. Pr o position 2: Let α 1 , α 2 , α 3 , α 4 be any four subsets of N . Define δ 1 , δ 2 , δ 3 , δ 4 and β as follows. δ i , α i \ [ j 6 = i α j (9) β , [ i ( α i \ δ i ) . (10) Then β contains elem ents th at appea r in at lea st two subsets α i . Then, J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ J ( h ; δ 1 β , δ 2 β , δ 3 β , δ 4 β ) . Pr o of: From Pr operty 3, if an elemen t appea rs in more than two subsets, then the elemen t can b e “added” to the remaining subsets without increasing the value of the I ngleton term. T he re sult th en fo llows. Let ∆ 0 be the set o f Ingleton ineq ualities o f the fo rm J ( h ; δ 1 β , δ 2 β , δ 3 β , δ 4 β ) ≥ 0 wh ere δ 1 , δ 2 , δ 3 , δ 4 and β are disjoint subsets and δ 1 , δ 2 , δ 3 , δ 4 are nonemp ty . Furthermo re, denote J ( h ; δ 1 β , δ 2 β , δ 3 β , δ 4 β ) as J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β ) . Theor em 2: All Ingleton inequalities are implied by the subset of In gleton in equalities ∆ = ∆ 0 ∪ ∆ 1 ∪ ∆ 2 Pr o of: If one sub set α i is contained in the union of the other thr ee subsets, then by Theorem 1, the in- equality J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 is implied by inequ alities in ∆ 1 ∪ ∆ 2 . Otherwise, by Propo sition 2, the inequality J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 is imp lied by J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β ) ≥ 0 in ∆ 0 where β , δ 1 , · · · , δ 4 are defined as in (9) and (1 0). Hence, Γ In contains all h ∈ F n satisfying all inequ alities in ∆ . Ad ditionally , the set is full-dimen sional, as shown b elow . Pr o position 3: Ther e exists a function h ∗ ∈ F n such th at J ( h ∗ ; α 1 , α 2 , α 3 , α 4 ) > 0 for all non trivial ineq ualities (i. e., not always zero) J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 . Pr o of: Let h ∗ ( α ) , 2 n (1 − 2 −| α | ) . The proposition then follows by d irect verification . Cor o llary 1 : Su ppose c i ≥ 0 for all i ∈ I . The n X i ∈I c i J ( h ; α i 1 , α i 2 , α i 3 , α i 4 ) = 0 (11) for a ll h ∈ F n if and o nly if c i = 0 for all i ∈ I . Pr o of: I n the proof of Proposition 3, we construc ted h such th at J ( h ; α i 1 , α i 2 , α i 3 , α i 4 ) ≥ 1 for all i . Ther efore, X i c i J ( h ; α i 1 , α i 2 , α i 3 , α i 4 ) ≥ X i c i . (12) Hence, if P i c i J ( h ; α i 1 , α i 2 , α i 3 , α i 4 ) = 0 , the n P i c i = 0 or equiv alently , c i = 0 fo r all i . By Pr oposition 3, we also have the fo llowing lem ma. Lemma 1: Supp ose that δ 1 , δ 2 , δ 3 , δ 4 , β are disjoint. T hen 1) J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β ) = 0 if and o nly if (1) eithe r δ 1 or δ 2 are empty , a nd ( 2) either δ 3 or δ 4 are empty . 2) J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β ) = J ( h ; δ ′ 1 , δ ′ 2 , δ ′ 3 , δ ′ 4 | β ′ ) if and on ly if (1) β = β ′ , (2) { δ 1 , δ 2 } = { δ ′ 1 , δ ′ 2 } an d (3 ) { δ 3 , δ 4 } = { δ ′ 3 , δ ′ 4 } . So far, we have proved that the set of ineq ualities ∆ imp lies all Ingleto n inequalities and hence charac terizes Γ In . In the following, we will prove that ∆ is indeed the unique minim al set ch aracterizing Γ In . T o obtain the minimal set of In gleton inequalities, we need to overcom e an ob stacle – that two different choices o f { α i , i = 1 , · · · , 4 } migh t g iv e the same Ing leton inequality . Therefo re, the “rep eated” inequ alities must be removed. Example 1: Supp ose that α 1 = { 1 , 5 } , α 2 = { 2 , 5 } , α 3 = { 3 , 5 } and α 4 = { 4 , 5 } . If β i = α i for i = 1 , 2 , 3 and β 4 = { 4 } , the n J ( h ; α 1 , · · · , α 4 ) = J ( h ; β 1 , · · · , β 4 ) . Fortunately , b y our choice of ∆ 1 and ∆ 2 , no two inequalities are the same, an d by Lemm a 1 , two inequalities are the same if and on ly the sub sets in the Ing leton term are per mutations of each oth er as specified in Lemma 1. Hen ce, those r epeated inequalities can be easily rem oved. From now on , we assume that no inequalities in ∆ are the same b y removin g all these duplication s. Theor em 3: No In gleton inequality in ∆ can b e implied by other s in ∆ . Con sequently , the set of Ingleto n inequalities ∆ is th e uniq ue minimal set of Ingleton inequalities that characterizes Γ In . Pr o of: The pr oof for Theorem 3 when n ≤ 5 can be obtained b y br ute-for ce verification. When n > 5 , we will prove Th eorem 3 by co nsidering several cases. The proof is rather leng thy and will be given in the next section . By d irect co unting, the size of ∆ can be shown to be n +  n 2  2 n − 2 + 1 4 6 n − 5 n + 3 2 4 n − 3 n + 1 4 2 n . I V . P R O O F O F T H E O R E M 3 Suppose to the contra ry of Theorem 3 that an in equality J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 in ∆ is implied by oth er ineq ualities J ( h ; α i 1 , α i 2 , α i 3 , α i 4 ) ≥ 0 in ∆ . By Farkas’ Lemma ( [13 ], p.61), there exists no n-negative c onstants c i such th at J ( h ; α 1 , α 2 , α 3 , α 4 ) = X i ∈I c i J ( h ; α i 1 , α i 2 , α i 3 , α i 4 ) (13) = X i ∈I 0 c i J ( h ; δ i 1 , δ i 2 , δ i 3 , δ i 4 | β i ) + X i ∈I 1 c i J ( h ; k i , l i , ∅ , µ i ) + X i ∈I 2 c i J ( h ; m i , m i , ∅ , N \{ m i } ) . (14) T o p rove T heorem 3, it suffices to show that c i = 0 for all i ∈ I , I 0 ∪ I 1 ∪ I 2 and hence J ( h ; α 1 , α 2 , α 3 , α 4 ) = 0 contradicting the fact that ∆ does not contain th e tri vial inequality 0 ≥ 0 . W e will prove Theor em 3 by considering three exha ustiv e cases, Case A, Case B and Case C. Case A . J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 is in ∆ 1 In this ca se, the ine quality J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 is of the form J ( h ; i 0 , j 0 , ∅ , µ 0 ) ≥ 0 for so me i 0 6 = j 0 ∈ N and µ ⊆ N \{ i 0 , j 0 } . Now , consider ag ain the eq uality (1 4) and pr oject both sides of it away from i 0 . Then th e le ft hand side becomes zero by Lemma 1. Hen ce, we have 0 = X i ∈I 0 c i J ( h ; δ i 1 \ i 0 , δ i 2 \ i 0 , δ i 3 \ i 0 , δ i 4 \ i 0 | β i \ i 0 ) + X i ∈I 1 c i J ( h ; k i \ i 0 , l i \ i 0 , ∅ , µ i \ i 0 ) + X i ∈I 2 c i J ( h ; m i \ i 0 , m i \ i 0 , ∅ , N \{ m i , i 0 } ) . By Co rollary 1, J ( h ; δ i 1 \ i 0 , δ i 2 \ i 0 , δ i 3 \ i 0 , δ i 4 \ i 0 | β i \ i 0 ) = 0 for a ll i ∈ I 0 . Then, by Lemma 1, we have eith er δ i 1 or δ i 2 is empty , and either δ i 3 or δ i 4 is empty . Howe ver , this is impo ssible as δ i 1 , δ i 2 , δ i 3 , δ i 4 are disjoint and nonemp ty . Ther efore, c i = 0 for a ll i ∈ I 0 . Consequently , J ( h ; i 0 , j 0 , ∅ , µ ) = X i ∈ I 1 c i J ( h ; k i , l i , ∅ , µ i ) + X i ∈ I 2 c i J ( h ; m i , m i , ∅ , N \{ m i } ) . It is known th at the elemental basic inequalities ∆ 1 ∪ ∆ 2 are not r edund ant [5]. Th e theo rem is thu s proved in this case. Case B . J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 is in ∆ 2 In this case, we can write the inequality as J ( h ; i 0 , i 0 , ∅ , N \{ i 0 } ) ≥ 0 fo r som e i 0 ∈ N . Again , we can project bo th sides o f th e inequality away fr om i 0 . Using the same argument, we can conclud e that c i = 0 for i ∈ I 0 . Then the theor em again fo llows from that ∆ 1 ∪ ∆ 2 is not r edund ant. Case C. J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 is in ∆ 0 In this ca se, J ( h ; α 1 , α 2 , α 3 , α 4 ) ≥ 0 can be rewritten in the form J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β ) ≥ 0 for som e disjoint subsets δ 1 , δ 2 , δ 3 , δ 4 , β such that δ i are all nonemp ty . Again , assume that J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β ) can be wr itten as a linear c ombina- tion o f oth er In gleton ineq ualities as in (14). First, we will show that c i = 0 f or all i ∈ I 2 . Let h ∗ be the entropy function for rand om variables { X 1 , . . . , X n } such that h ∗ ( α ) = H ( X i : i ∈ α ) = | α | . Then , it is straightfor ward to prove tha t J ( h ∗ ; δ 1 , δ 2 , δ 3 , δ 4 | β ) = 0 , J ( h ∗ ; δ i 1 , δ i 2 , δ i 3 , δ i 4 | β i ) = 0 J ( h ∗ ; k i , l i , ∅ , µ i ) = 0 . Substituting ba ck into ( 14), we have 0 = X i ∈ I 2 c i . Consequently , c i = 0 fo r all i ∈ I 2 and h ence J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β ) = X i ∈I 0 c i J ( h ; δ i 1 , δ i 2 , δ i 3 , δ i 4 | β i ) + X i ∈I 1 c i J ( h ; k i , l i , ∅ , µ i ) . (15) Our secon d task is to show that c i = 0 for all i ∈ I 1 . Aga in, we will u se a similar p rojection trick. Consider any i 0 ∈ I 1 and th e cor respond ing in equality J ( h ; k i 0 , l i 0 , ∅ , µ i 0 ) ≥ 0 . W e can pro ject both sides of (15) o nto k i 0 and l i 0 . Clearly , the right han d side con tains the term c i 0 J ( h ; k i 0 , l i 0 , ∅ , ∅ ) . Thus, the lef t hand side after projection canno t be zero. As a result, (1) { k i 0 , l i 0 } ∩ ( δ 1 δ 2 ) and { k i 0 , l i 0 } ∩ ( δ 3 δ 4 ) are nonemp ty , and (2 ) k i 0 and l i 0 are not in th e same subset. Therefo re, we may assume witho ut loss of ge nerality that k i 0 ∈ δ 1 and that l i 0 ∈ δ 3 . Since δ 2 and δ 4 are no nempty , we can pick a ∈ δ 2 and b ∈ δ 4 . The n we can project (1 5) onto { k i 0 , l i 0 , a, b } . After p rojection, th e left han d side become s J ( h ; k i 0 , a, l i 0 , b ) and the righ t h and side is a sum mation of several In gle- ton inequ alities (inv olving at m ost four variables) inclu ding c i 0 J ( h ; k i 0 , l i 0 , ∅ , µ i ) . As Theo rem 3 hold s wh en n = 4 , we have c i 0 = 0 . Repeating the same argu ment for all i ∈ I 1 , we prove tha t c i = 0 for all i ∈ I 1 . Now (15) c an be rewritten as J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β ) = X i ∈I 0 c i J ( h ; δ i 1 β i , δ i 2 β i , δ i 3 β i , δ i 4 β i ) (16) Assume that c i > 0 fo r all i ∈ I 0 in (1 6). Now , to prove Theorem 3, it suffices to prove the fo llowing statem ent. Pr o position 4 (I nduction Hypo thesis H ( n ) ) : Let n be the number of set elemen ts inv olved in the left han d side of the expression in (16). Su ppose that the equality ( 16) hold s. Then J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β ) = J ( h ; δ i 1 , δ i 2 , δ i 3 , δ i 4 | β i ) . W e have verified cases u p to n ≤ 5 . The case when n ≥ 6 will be proved by ind uction. T o this end , we first p rove the claim that any elemen t appearin g in th e right hand side o f ( 16) must also appear in the left hand side . Suppose to the contr ary that ( 1) the eq uality (16) holds and (2) there exists an element a appearing only on the right h and side. Furth er supp ose that a ∈ δ i 0 j for som e i 0 ∈ I 0 . Th en it is easy to fin d ano ther e lement b such that after projectio n onto { a, b } , th e right hand side of (1 6) h as a term I h ( a, b ) . Howe ver , the left hand side of (16) can b e shown to be zer o, contradictin g to Cor ollary 1 On the other hand , if a ∈ β i 0 , th en we can pick elements b ∈ δ i 0 1 and c ∈ δ i 0 2 . Projecting b oth sides of (16) onto { a, b, c } , the left hand side is either zero or I h ( b ; c ) , while the rig ht hand side of ( 16) is nonzero as it con tains a ter m c i 0 I h ( b ; c | a ) . Contradiction occurs and h ence all eleme nts app earing in the right h and side will also appear in th e left han d side. Suppose that the ind uction h ypoth esis H ( n ) holds. W e n ow aim to prove tha t H ( n + 1) also holds. Suppose (16) in volv es at most n + 1 set elements wh ere n ≥ 5 . W e co nsider two sub-cases, C.1 an d C.2 . Case C.1. | β | ≥ 2 : Let a, b ∈ β and a 6 = b . Pro jecting both sides o f (1 6) away fr om a . Then (16) becomes J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β \ a ) = X i ∈I 0 c i J ( h ; δ i 1 \ a, δ i 2 \ a, δ i 3 \ a, δ i 4 \ a | β i \ a ) . (17) As (17) inv olves only n variables, ap plying the induction hypoth esis, we have β i = β \ a or β i = β . Similarly , we can prove that β i = β \ b or β i = β . Con sequently , β i = β for all i ∈ I 0 . Again , fro m (17), J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β \ a ) = J ( h ; δ i 1 \ a, δ i 2 \ a, δ i 3 \ a, δ i 4 \ a | β i ) . By Lemma 1 and the induction hypothe sis, we have { δ 1 , δ 2 } = { δ i 1 , δ i 2 } and { δ 3 , δ 4 } = { δ i 3 , δ i 4 } . The h ypothesis H ( n + 1) then holds. Case C .2. | β | ≤ 1 : Since n +1 ≥ 6 and | β | ≤ 1 , there e xists distinct a and b in a subset δ i for some i . Assume without lo ss of g enerality that i = 1 . Projecting (16) away from a , J ( h ; δ 1 \ a, δ 2 , δ 3 , δ 4 | β ) = X i ∈I 0 c i J ( h ; δ i 1 \ a, δ i 2 \ a, δ i 3 \ a, δ i 4 \ a | β i \ a ) . By Lemma 1 a nd the induc tion hypoth esis, β i = β or β i = β ∪ { a } . Similar ly , by p rojecting (1 6) away fr om b , we have β i = β or β i = β ∪ { b } . Consequen tly , β i = β for all i ∈ I 0 . Thus ( 17) beco mes J ( h ; δ 1 , δ 2 , δ 3 , δ 4 | β ) = X i ∈I 0 c i J ( h ; δ i 1 , δ i 2 , δ i 3 , δ i 4 | β ) . and h ence after p rojection away fr om a , J ( h ; δ 1 \ a, δ 2 , δ 3 , δ 4 | β ) = X i ∈I 0 c i J ( h ; δ i 1 \ a, δ i 2 \ a, δ i 3 \ a, δ i 4 \ a | β ) . (18) Similarly , p rojecting ( 16) away from b , we have J ( h ; δ 1 \ b, δ 2 , δ 3 , δ 4 | β ) = X i ∈I 0 c i J ( h ; δ i 1 \ b, δ i 2 \ b, δ i 3 \ b, δ i 4 \ b | β ) . (19) By ( 18) and (19), we can then prove that { δ 1 , δ 2 } = { δ i 1 , δ i 2 } and { δ 3 , δ 4 } = { δ i 3 , δ i 4 } . The hy pothesis H ( n + 1) th en ho lds. Combining the two cases, we can thus conc lude that the set of inequalities in ∆ is not red undan t. Since Γ In is full- dimensiona l, ∆ is indeed the uniqu e, no n-redu ndant set of Ingleton inequalities ch aracterizing Γ In , [13], p .64. V . C O N C L U S I O N W e have identified the uniqu e minimal ch aracterization for the set of polyma troids satisfy ing all Ingleton inequalities. Knowing th is set can greatly simplify computation of Ingleton- LP b ound s f or the multicast capacity of linear network co des. Compared to na¨ ıve enumera tion of all In gleton ineq ualities, approx imately of the order 16 n , the actu al number of necessary inequalities has size appro ximately of the orde r 6 n / 4 − 5 n . The complexity reductio n is sign ificant, in p articular fo r large n . A C K N O W L E D G E M E N T This work was sup ported by the Australian Government under ARC gran t DP0557310. This w ork was perf ormed while L. Guill ´ e was visiting the University of South Autralia. R E F E R E N C E S [1] R. Ahlswede, N. Cai, S.-Y . R. Li, and R. W . 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