A Mehod for Recovering Arbirary Graphs Nicolas Addingon January 4, 005 Absrac We develop a mehod for recovering arbirary graphs, a generalizaion of he sar k mehod ha avoids spurious parameers. They key is o recognize ha he inverse problem amouns o undoing he Schur complemen and o analyze he residue erm BC 1 B. Conens 1 Inroducion Graphs wih One Inerior Node.1 Basic Recovery Technique....................................... Example: The Kie.......................................... 3.3 Quadrilaerals: A Picorial Tool................................... 3.4 The Square Roo Trick........................................ 4 3 The Sar K Mehod 4 3.1 Wih Marices............................................ 4 3. Wih Quadrilaerals......................................... 5 3.3 Example: The Marshmallow..................................... 5 4 A General Recovery Mehod 6 4.1 Shorcomings of he Sar K Mehod................................ 6 4. Looking a Muliple Layers..................................... 6 4.3 Recovery wih Larger Submarices................................. 7 4.4 Tesing for Singulariy........................................ 8 4.5 Recovering an Arbirary Graph................................... 9 4.6 Example: Circles, 4 Rays..................................... 9 4.7 Connecions and Deerminans................................... 11 5 Direcions for Fuure Research 11 6 Acknowledgemens 13 1
1 Inroducion We assume ha he reader is familiar wih he basics of Curis and Morrow s inverse problem [1]: graphs, Kirchhoff marices, response marices, and he Schur complemen. The forward problem is o ake a Kirchhoff marix K and find is response marix Λ. This is easily done using he Schur complemen: if we wrie K in he block form ( ) A B B C where C represens he inerior-oinerior connecions, Λ = A BC 1 B = K/C, he Schur complemen of C in K. A imes i will be convenien, albei awkward-sounding, o say K Schur C raher han he Schur complemen of C in K. I will also be convenien o refer o BC 1 B as he residue erm. The inverse problem is o ake a graph and is response marix and find he Kirchhoff marix, ha is, knowing K Schur C and knowing which enries of K are zero, o recover he remaining enries of K. The inverse problem, hen, amouns o undoing he Schur complemen. This is difficul because he Schur complemen desroys a lo of informaion: if a graph has eigh boundary nodes and eigh inerior nodes, we are rying o recover a 16 16 marix from an 8 8 wih he help of a few zeros. The inverse problem has been solved for circular planar graphs [1] and annular graphs have been sudied some [3]; in his paper we are ineresed in he recoverabiliy of arbirary graphs. Our plan is o se up a sequence of inermediae marices beween he Kirchhoff marix and he response marix and pull he informaion in he response marix back hrough hese inermediae marices o he Kirchhoff marix. In we sudy he simples case, graphs wih a single inerior node. In 3 we describe he sar k mehod, which amouns o ieraing he one-node case. In 4 we develop a general recovery mehod, which is our main resul. In 5 we sugges direcions for fuure research. Graphs wih One Inerior Node.1 Basic Recovery Technique Consider a graph wih n boundary nodes and one inerior node. Fix an ordering of he nodes wih he boundary nodes firs. Le γ 1,...,γ n be he conducances of he edges joining he inerior node o he boundary nodes 1,...,n, le γ = (γ 1 γ n ) be he column vecor of hese conducances, and le σ = γ 1 + + γ n be heir sum. Then when we wrie K in block form, A γ K = γ σ so Λ = A γγ σ. We wish o view his in he following manner: Λ is jus he upper lef corner of K superimposed wih a residue R = γγ σ from he eliminaion of he inerior node. Le us inroduce he noaion K for he upper lef corner of K (he size of which will always be clear from conex), so Λ = K +R. If we imagine laying hese marices fla, we can wrie R + K Λ. (.1.1) Recall ha K and Λ are symmeric, and observe ha R mus be as well. We are given all of Λ and he posiion of he zeros in K, so wherever here is a zero in K, we can read he corresponding enry of R direcly from Λ. Now γ 1 γ 1 γ γ 1 γ 3... γ 1 γ n R = γγ σ = 1 γ 1 γ γ γ γ 3... γ γ n γ 1 γ 3 γ γ 3 γ 3... γ γ n σ....... γ 1 γ n γ γ n γ 3 γ n... γ n
so any submarix of R has deerminan 0. Thus if we know hree enries in some submarix of R, we can recover he fourh. If here are sufficienly many well-placed zeros in K, we may be able o recover all of R his way. Once we have R, we can ge all bu he las row and column of K since K = Λ R. Since K is a Kirchhoff marix, is rows and columns sum o zero, so we can recover he las row and column as well.. Example: The Kie Consider he kie graph (Figure 1). We are given Λ = (λ ij ) and we wish o recover K. We know ha k 13, k 3, k 4, and k 34 are all zero since here are no edges beween he corresponding pairs of nodes, so we know he following enries of R:?? λ 13? R =?? λ 3 λ 4 λ 13 λ 3? λ 34.? λ 4 λ 34? Since r 13 r 4 r 14 r 3 = 0, we recover r 14 = λ13λ4 λ 3. We can recover all of R similarly: R = λ 13 λ 4 λ 13λ 4 λ λ 3λ 34 λ 34 λ 13λ 4 13 λ 3 λ 13λ 4 λ 34 λ 3λ 4 λ 34 λ 3 λ 4 λ 13 λ 3 λ 3λ 34 λ 4 λ 34 λ 13λ 4 λ 3 λ 4 λ 34 λ 4λ 34 λ 3 Observe ha here are several ways o recover a given enry, for example r 11 = r13r14 r 34 = r1r13 r 3 = r14 r 44, bu hese all give he same resul. (Observe also ha he indices cancel in each expression.) I appears ha in general, if we begin wih a se of enries in R which is independen, ha is, if no enry in our se can be goen from he ohers, hen no maer how we recover he remaining enries we ge he same values. Now ha we have R, we can recover he upper lef corner of K: λ 11 λ13 λ 4 λ 3λ 34 λ 1 λ13λ4 λ 34 0 λ 14 λ13λ4 λ 3? λ 1 λ13λ4 λ 34 λ λ3λ4 λ 34 0 0? K = 0 0 λ 33 λ3λ34 λ 4 0? λ 14 λ13λ4 λ 3 0 0 λ 14 λ4λ34 λ 3?????? and since he rows and columns of K sum o 0, we can recover all of K. Observe ha k 1 = λ 1 λ 13λ 4 λ 1 λ 13 λ 4 λ 34 = λ 34 and ha he deerminan in he las expression corresponds o he -connecion from nodes 1 and 4 o and 3. Since here is an edge beween nodes 1 and, k 1 is negaive, so his deerminan is posiive, and he connecion exiss. Removing his edge would break his connecion, bu i would also make k 1 = 0 and hence make his deerminan 0. Thus we see he link beween -connecions and deerminans very clearly by sudying graphs wih one inerior node..3 Quadrilaerals: A Picorial Tool We can idenify a nework (i.e. a graph wih conducances) wih is Kirchhoff marix, and a response marix is he Kirchhoff marix of a complee graph, so we can undersand aking he Schur complemen of he Kirchhoff marix as replacing he inerior nodes and he edges inciden o hem wih a response-equivalen complee graph (Figure ). λ 34. 3
For graphs wih one inerior node, we can idenify he superimposed complee graph wih he off-diagonal enries of he residue R, since R = Λ excep on he diagonal if here are no boundary-o-boundary connecions. Thus our saemen Λ = K + R can be inerpreed picorially as in Figure 3. In his inerpreaion, a submarix of R ( rik ) r il r jk r jl corresponds o a quadrilaeral in he superimposed complee graph as in Figure 4, so if we know hree sides of his quadrilaeral we can recover a fourh. Quadrilaerals provide a quick way o check he recoverabiliy of a graph. To show ha he kie is recoverable, we begin by drawing he picure in Figure 5. Where here are doubled edges, knowing he response marix gives us heir sum, so we need o separae ha number ino he par ha came from he boundary-o-boundary connecion and he par ha came from he superimposed complee graph. For he edge 1, we know he oher hree sides of he quadrilaeral 1 3 4, and he quadrilaeral 1 3 4 gives us he edge 1 4. Now we know he boundary-o-boundary connecions and he superimposed complee graph, i.e. he off-diagonal enries of R, so we know we can recover he diagonal R, so he kie is recoverable. If insead we were rying o recover he bowie, we would draw he picure in Figure 6. Any quadrilaeral wih 1 4 as a side also has 3 as a side, so we canno ge hree sides of any quadrilaeral o recover 1 4 or 3, and he bowie is no recoverable..4 The Square Roo Trick In.1 we said ha since any submarix of R is singular, we can recover an enry in a submarix given he oher hree. Since R is symmeric, however, more is rue. If our submarix is principal and we know only he wo diagonal enries, we can recover he off-diagonal enries: 0 = r ii r ij = r iir jj rij, r ij r jj so r ij = r ii r jj ; he enries of R are necessarily negaive. Observe again ha he indices cancel. The Sar of David graph (Figure 7) can be recovered using his square roo rick bu no wihou i. 3 The Sar K Mehod 3.1 Wih Marices In his secion we shall assume ha our graph has 1 boundary nodes and 4 inerior nodes o avoid drowning he reader in a sea of index variables; wha is mean in general should be clear. Le K 16 = K be he Kirchhoff marix of he original nework. Le K 15 equal K 16 Schur is lower righ 1 1 corner. Le R 16 be he residue γγ σ from his Schur complemen, i.e. he residue from he eliminaion of node 16, so K 15 = K 16 + R 16. Similarly, le K 14 equal K 15 Schur is lower righ 1 1 corner and R 15 be he residue, so K 14 = K 15 + R 15. Coninue in his fashion down o K 1. Proposiion 1. K 1 equals K 16 Schur is lower righ 4 4 corner, ha is, K 1 = Λ. Proof. In [], Crabree and Haynsworh show ha if C is an inverible square submarix of M and D is an inverible square submarix of D hen M/D = (M/C)/(C/D). I is easy o check ha he lower righ 1 1 corner of K 15 equals he lower righ corner of K 16 Schur is lower righ 1 1 corner, so K 14 equals K 16 Schur is lower righ corner. The desired resul is obained by inducion. We are given he graph, or equivalenly, he zeros of K = K 16. From his we can deermine he zeros of R 16 and hence he zeros of K 15, and so on down o K 13. Given Λ = K 1 and he zeros of K 13 we may be able, using our one-inerior-node mehod, o recover R 14 and hence K 14. From his we may be able o recover K 15, and so on back up o K 16. If his process succeeds, we know ha he graph is recoverable. If a some poin during his process we ge suck, ha is, we know a few enries of some R n bu canno recover any more using submarices, we can paramerize one of he unknown enries and coninue. The 4
number of parameers gives a measure of he irrecoverabiliy of our graph. I may also happen ha a one sage we do no have enough informaion o coninue, so we inroduce a parameer, bu a a laer sage we have more informaion han we need and are able o use submarices o eliminae he parameer. An example will be given in 3.3. 3. Wih Quadrilaerals In.3, we replaced he inerior node and he edges inciden o i, or he sar around he inerior node, wih he response-equivalen complee graph, or k. Now ha we are dealing wih several inerior nodes, we make his sar k replacemen several imes and obain a sequence of inermediae graphs as in Figure 8. The K n of he previous secion are he Kirchhoff marices of hese inermediae graphs; specifically, K n is he inermediae graph wih n nodes. If each inermediae graph is recoverable, he original graph is recoverable. Here again, quadrilaerals provide a quick check of recoverabiliy. Consider he hexagons on a erahedron in Figure 9. Wih he quadrilaeral 1 5 4 we can separae he wo edges 1, and by symmery all he doubled edges, so we can recover all four inerior nodes. This example is ineresing because i is a flower: i has neiher a boundary-o-boundary connecion nor a boundary spike (a boundary node conneced o a single inerior node). Curis and Morrow have shown ha no circular planar flower is recoverable; his flower is, of course, no circular planar. Jeff Russell and Tracy Lovejoy s oy drum (Figure 10) is anoher example of a recoverable flower. 3.3 Example: The Marshmallow Consider he marshmallow graph (Figure 11), so called because he presenaion in which he auhor inroduced i involved a marshmallow model. For his graph,? 0 0 0 0? 0 0? 0 0 0??? 0 0 0 0? 0 0? 0 0?? 0????? λ 11 λ 1 λ 13 λ 14 λ 15 K 7 = 0 0 0? 0?? K 6 = 0????? λ 1 λ λ 3 λ 4 λ 5 0 0 0 0??? 0????? K 5 = λ 13 λ 3 λ 33 λ 34 λ 35?????? 0 0????? λ 14 λ 4 λ 34 λ 44 λ 45 λ?????? 15 λ 5 λ 35 λ 45 λ 55 0???? 0? so? λ 1 λ 13 λ 14 λ 15 λ 1???? R 6 = λ 13???? λ 14????. λ 15???? No furher enries of R 6 can be recovered, bu if we paramerize he 1,1 enry by < 0, we can ge everyhing else: λ 1 λ 13 λ 14 λ 15 λ λ 1 λ 1λ 13 λ 1λ 14 λ 1λ 15 1 λ R 6 = λ 1λ 13 λ 13 λ 13λ 14 λ 13λ 15 13. λ 13λ 14 λ 14 λ 14λ 15 λ λ 1λ 14 14 λ λ 1λ 15 15 λ 13λ 15 λ 14λ 15 From here we can ge K 6 and complee he recovery. Along he way, however, an ineresing hing happens. In R 5, rows and 3 by columns 4 and 5 are ( ) λ4 λ1λ14 λ 5 λ1λ15 = 1 λ 1 λ 14 λ 4 λ 1 λ 15 λ 5 λ 34 λ13λ14 λ 35 λ13λ15 λ 13 λ 14 λ 34 λ 13 λ 15 λ 35 λ 15 5
bu since his is a submarix of an R n, i is singular, so by he six-erm ideniy, 0 = 1 λ 1 λ 14 λ 4 λ 1 λ 15 λ 5 λ 13 λ 14 λ 34 λ = 1 λ 14 λ 15 λ 1 λ 4 λ 5 13 λ 13 λ 34 λ 35. λ 15 λ 35 If ( λ 4 λ 5 ) λ 34 λ 35 is inverible, his allows us o solve for, ha is, was a spurious parameer. Observe ha ( λ 4 λ 5 ) λ 34 λ 35 corresponds o he -connecion from nodes and 3 o 4 and 5. This connecion exiss, bu he corresponding deerminan may be zero because we can also connec and 3 o 5 and 4 (he opposie permuaion). For mos graphs in which spurious parameers arise, i is possible o decide wheher he necessary deerminan is inverible or no by examining connecions; he marshmallow is somehing of a pahalogical case in his respec. If ( λ 4 λ 5 ) ( λ 34 λ 35 is inverible, he marshmallow is recoverable. By symmery, if λ3 λ 5 ) ( λ 34 λ 45 or λ3 λ 4 ) λ 35 λ 45 is inverible, i is also recoverable. If all hree of hese are singular, i is possible o exhibi a one-parameer family of Kirchhoff marices wih he same response marix. The marshmallow, hen, is almos always recoverable. Tracy s J (Figure 1) has he same propery. I is ineresing o compare his sor of graph wih he -o-1 graphs sudied by Ernie Esser and Tracy Lovejoy [5], which are almos never recoverable he former are recoverable unless cerain submarices of Λ are singular, whereas he laer are recoverable only if cerain submarices of Λ are singular. 4 A General Recovery Mehod 4.1 Shorcomings of he Sar K Mehod The sar k mehod has wo main shorcomings. Firs, i only works unequivocally when each of he inermediae graphs is recoverable. This is ofen rue for small graphs, such as he opha (Figure 8), and graphs wih few connecions, such as hexagons on a erahedron, bu i fails for many denser graphs. There is no order of inerior nodes for which he sar k mehod can recover Ernie Esser s circles, 4 rays (Figure 13a) or even he well-conneced graph on 5 nodes shown in Figure 14a, which is circular planar. Boh hese graphs are known o be recoverable by oher means. Second, he sar k mehod is very sensiive o he order in which he inerior nodes are eliminaed. Figure 15a is recoverable using he sar k mehod, bu Figure 15b is no. In he previous secion, he recovery of he marshmallow required one parameer, bu if we swich he order of nodes 6 and 7, i requires four parameers (hree of which end up being spurious). Parameers which come and go provide a way around hese shorcomings, and i may be ha all recoverable graphs can be recovered using he sar k mehod and spurious parameers. This mehod is inelegan, however, and does no reflec wha is really going on. If a spurious parameer arises in he sar k recovery of a graph, i is because one of he inermediae graphs was no recoverable bu he original graph was; some informaion was presen in he chain of inermediae graphs ha could no be seen by looking only from one graph o he nex. In he previous secion we saw how when we look across several seps, 3 3 deerminans arise from deerminans of deerminans by way of he six-erm ideniy. In his secion we will use 3 3 and larger deerminans direcly, wihou he help of spurious parameers. 4. Looking a Muliple Layers In.1 we wroe Λ = K + R, or when we laid K and R fla, R + K Λ We were given cerain enries of he marices in his sum: he zeros of K and R and all he enries of Λ. We used verical informaion knowing k ij and λ ij we could find r ij, and knowing r ij and λ ij we could find. 6
k ij and horizonal informaion he singulariy of cerain submarices of R and he fac ha he rows and columns of K summed o zero o recover all he enries of everyhing, and paricular o recover K. In 3.1 we had or if we lay he marices fla, K 1 = K 13 + R 13 = K 14 + R 14 + R 13 = K 15 + R 15 + R 14 + R 13 = K 16 + R 16 + R 15 + R 14 + R 13 R 13 R 14 R 15 R 16 + K 16 K 1. (4..1) Here we did he same hing: we began wih he known zeros of K 16 (and hence of he R n ) and he known enries of Λ = K 1 and used verical and horizonal informaion o peel off he R n one a a ime and recover K 16. The following sor of hing happens ofen: he i,j enries of K 16, R 16, and R 15 are zero bu he i,j enries of R 13 and R 14 are no, so we can recover he sum of hese wo enries, bu we do no know wha porion of he number comes from R 13 and wha porion from R 14. To make use of his informaion, we inroduce a marix R14 13 = R 14 + R 13 ; we ypically know more abou his marix han we do abou R 13 and R 14 separaely. This is our plan: in addiion o considering he single layers R 13,...,R 16, we will also consider all sums of coniguous blocks of hem, for example R15 13 = R 15 + R 14 + R 13 and R16 15 = R 16 + R 15 and even K 14 = K 16 + R 16 + R 15. These muli-layer gadges add up nicely: R16 13 = R16 14 + R 13 = R16 15 + R14 13 = R 16 + R15. 13 Thus we undersand he verical informaion for he R m n. To recover a graph, we wish o use horizonal informaion abou he R m n o recover heir enries from a few given enries. For single layers, we used he fac ha submarices were singular. For muliple layers, we will use larger submarices. 4.3 Recovery wih Larger Submarices The following Lemma was used acily in 3.3 and will be used exensively in wha follows. Lemma. If an n n marix M is singular, we know all bu one enry m ij, and he cofacor M ij is inverible, we can recover he unknown enry. Proof. By cofacor expansion along he i h row, 0 = dem = ( 1) i+1 m i1 de M i1 + + ( 1) i+j m ij de M ij + + ( 1) i+n m in dem in, so since de M ij 0, we can solve for m ij. In.4, we used he symmery of R o do slighly beer for marices. I is unclear wheher i is worh doing his for larger marices. Suppose ha a square marix M is singular and symmeric, wih he block form a B x M = B C D x D e 7
where he corners a, x, and e are 1 1. We wish o solve for x. By he six-erm ideniy, 0 = M C = a B B C C = a B B C C = a B B C C D D D D e B x C D B C x D D e B x C D D ( e B 0 C D + ( 1)n x C ) so if C is inverible, x = ( 1)n+1 C ( B 0 C D ± a B B C C D D ) e. If he plus-or-minus erm is nonzero, we have wo possible values for x. If one is posiive and one negaive, in our applicaion we will know which one we wan, bu here is no obvious way of elling wheher he wo soluions are of he same or opposie sign. On he oher hand, if he plus-or-minus erm is zero, B C D x = 0, so we can find x from ha. I is no known wheher any graphs can be recovered using his generalized square roo rick bu no wihou i. We guess ha here are none, bu suspec ha his rick is a work in -o-1 graphs. 4.4 Tesing for Singulariy To make use of Lemma, we need o be able o es submarices of he Rn m for singulariy. We coninue o assume ha our graph has 16 nodes. We defined R 15 (for example) as he residue erm γγ σ from K 15 Schur is lower righ 1 1 corner. We wish o show ha R15 14 (for example), which we defined as R 15 +R 14, is he residue from K 15 Schur is lower righ corner. Le C15 14 be ha corner: rows 14 o 15 by columns 14 o 15 of K 15. Le K 15 be rows 1 o 13 by columns 1 o 13 and B be rows 1 o 13 by columns 14 o 15, so K 15 B K 15 =. B C15 14 Now K 13 = K 15 + R 15 + R 14 = K 15 + R15, 14 bu by Proposiion 1, K 13 = K 15 B (C15) 14 1 B as well, so R15 14 = B (C15) 14 1 B, i.e. he residue from he eliminaion of nodes 15 and 14. K 13 is he Schur complemen of C15 14 in K 15. R15 14 is he Schur complemen of C15 14 in a closely relaed marix: wih B as above, define 0 B Z15 14 =, B C15 14 ha is, K 15 wih he upper lef 13 13 corner replaced wih zeros. Now R 15 14 = 0 B (C 14 15) 1 B = Z 14 15/C 14 15. Proposiion 3. Le M be he submarix of R15 14 consising of rows r 1,...,r n by columns c 1,...,c n. Le N be he submarix of Z15 14 consising of rows r 1,...,r n,14,15 by columns c 1,...,c n,14,15. Then M is singular if and only if N is. Proof. M = N/C15, 14 so de M = (den)/(de C15). 14 C15 14 is no singular since i is a principal proper submarix of a Kirchhoff marix. 8
In oher words, o decide wheher a square submarix M of R15 14 is singular, we can ake he same submarix of Z15 14 (which is jus K 15 wih he upper lef corner suppressed), ack on he rows and columns ha would be chopped off in aking he Schur complemen of C15, 14 and decide if ha is singular. We know which enries of K 15 are zero, which are posiive (he diagonal enries), and which are negaive (he off-diagonal enries ha are no zero). We can ry o decide wheher a submarix N of Z15 14 is singular as follows. Expand he deerminan as a polynomial in he enries of N (hence of K 15 ). If he erms are all zero, N is singular. If he erms are all posiive or all negaive, N is inverible. If some of he erms are posiive and some negaive, he signs of he enries alone are no enough o decide. 4.5 Recovering an Arbirary Graph Now we are in a posiion o ouline a mehod for recovering he Kirchoff marix of an arbirary graph from is response marix. Suppose our graph has M boundary nodes and N nodes alogeher. 1. Wrie down he signs of all he enries of K N, which we know from he graph. From hese, deermine he signs of all he enries of K M,...,K N 1. We will use hese o es submarices for singulariy using Proposiion 3.. Make empy marices K M,...K N, R M+1,...,R N, and R m n, M < m < n N of he appropriae sizes: K n is n n, and R n and R m n are n 1 n 1. Fill in he zeros of all hese, which can be derived from he zeros of K N. Fill in he enries of K M, he response marix. 3. Whenever we know wo of hree enries from somehing of he form R16 13 = R16 15 + R 13 K 13 = K 15 + R15, 14 recover he hird. 4. Whenever we know all bu one enry of a submarix of any marix, if he submarix is singular and he cofacor of he unknown enry is inverible, recover he unknown enry using Lemma. Also use he basic square roo rick from.4. 5. Whenever we know all bu one enry in a row of a K n, recover i using he fac ha he rows of Kirchhoff marices sum o zero. 6. If a any poin no more enries can be recovered bu some are sill missing, paramerize an unknown enry. The firs single layer R n wih unknown enries (firs in he sense ha n is leas) seems o be he bes place o parameerize. Observe ha if we resric our recovery o he single layers R n and omi he muli-layers R m n, his is exacly he sar k mehod. This mehod, while powerful (here is no known recoverable graph ha i fails o recover), is impracical o work by hand for large graphs, as we shall see in he nex secion, and unnecessary for small graphs, where he sar k mehod usually works. I is ideally suied o a compuer, however. The auhor has implemened he mehod as a C++ program, which is available a hp://www.mah.washingon.edu/ reu/. 4.6 Example: Circles, 4 Rays Consider circles, 4 rays (Figure 13a), which has 8 boundary nodes and 8 inerior nodes. An ad hoc proof of his graph s recoverabiliy is given by Ernie Esser in [3]. Here we recover i using our general mehod. These are he signs of he enries in K 16, he Kirchhoff marix: K 16 = + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 + 9 14 or
We are given K 8, he response marix. Firs we wish o recover R 9, he residue from he eliminaion of node 9. This is an 8 8 marix, all of whose enries are negaive. Now R 10 16 = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 K 9 = K 16 + R 10 16 = + 0 + 0 + 0 + 0 0 0 0 0 + 0 0 0 0 0 + 0 + 0 + 0 + so we know he following enries of R 9 = K 8 K 9 :???? λ 15??????? λ 5??????? λ 35??? R 9 = K 8 K 9 =???? λ 45??? λ 15 λ 5 λ 35 λ 45? λ 56 λ 57 λ 58.???? λ 56??????? λ 57??????? λ 58??? To recover R 9, i suffices o recover he 5,5 enry, for hen we can he res using submarices as we did in.. We will recover he 5,5 enry by way of R16. 9 Consider he submarix M of R16 9 consising of rows 1,, 4, 5 by columns 3, 5, 7, 8. Since R16 9 = K 8 K 16 and he submarix 1,,4,5 3,5,7,8 of K 16 is zero excep a 5,5, we have λ 13 λ 15 λ 17 λ 18 M = λ 3 λ 5 λ 7 λ 8 λ 34 λ 45 λ 47 λ 48 λ 35? λ 57 λ 58 I suffices o show ha 1,,4,5 3,5,7,8 is singular bu 1,,4 3,7,8 is inverible, for hen we can recover he 5,5 enry by Lemma ; hen since R 9 = R16 9 R16, 10 and he 5,5 enry of R16 10 is 0, we can recover he 5,5 enry of K 9. From 4.4 we know ha he submarix 1,,4 3,7,8 of R16 9 is inverible if and only if he submarix 1,,4,9 16 3,7,8,9 16 of K 16 wih he upper lef 3 3 corner changed o zeros is inverible: 3 7 8 9 10 11 1 13 14 15 16 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 9 0 0 0 + 0 0 0 0 10 0 0 0 + 0 0 0 0. 11 0 0 0 + 0 0 0 1 0 0 0 + 0 0 0 13 0 0 0 0 0 0 + 0 14 0 0 0 0 0 0 + 0 15 0 0 0 0 0 0 + 16 0 0 0 0 0 0 0 + By cofacor expansion along he op row, his submarix is inverible if and only if he submarix,4,9 16 3,7,8,9 1,14 16 is inverible; ha is, since here is a single nonzero enry in he op row, we can cross ou row 1 by column 13 (he reader is encouraged o ge a pencil and do his). Similarly, we can cross ou 14, 4 16, 15 3 (going down he lef-hand column now), 11 7, 1 8, 13 9, and 10 11. This leaves us wih he submarix 9, 14, 16 10, 1, 15: 10 1 15 9 0 14 0 16 0 which is necessarily inverible, so he submarix 1,,4 3,7,8 of R 9 16 is inverible. For 1,,4,5 3,5,7,8, we play he same game, bu afer crossing ou a few hings we find a row of all zeros, so ha marix is singular. Thus R 9 is recoverable, so K 9 is recoverable. The recovery of K 10,...,K 16 is similar. Exercise. Show ha he well-conneced graph on 5 nodes (Figure 14a) is recoverable. (Hin: Recover he 1,1 enry of R 6 9 using he submarix 1,,3 1,4,5.) 10
4.7 Connecions and Deerminans In he previous example, we saw ha a cerain submarix of R 9 16 was inverible because a relaed submarix of Z 9 16 was. Because of he placemen of zeros in K 16, he submarix of R 9 16 was equal o a submarix of K 8 and he submarix of Z 9 16 was equal o a submarix of K 16. Bu we know ha he subdeerminan 1,,4 3,7,8 of K 8, he response marix, corresponds o a 3-connecion from nodes 1,, and 4 o 3, 7, and 8. The calculaion we jus made shows us how. The reader is encouraged o mark Figure 13a wih a pencil. Firs we crossed ou row 1 by column 13. A 3-connecion from 1,, 4 o 3, 7, 8, mus include he edge from node 1 o node 13; from node 1, we canno go anywhere else. Similarly, we mus go from o 14, 4 o 16, 15 o 3, 11 o 7, and 1 o 8. Now we could go from 13 o 14, 16, or 9, bu if we wen o 14 or 16 we would collide wih he connecions coming from and 4, so we mus in fac go o 9. Similarly, we mus ge o 11 from 10. Now here are only wo ways o complee he connecion: eiher 9 o 10, 14 o 15, and 16 o 1, or 9 o 1, 14 o 10, and 16 o 15; hese correspond o he wo non-zero erms in he deerminan of 10 1 15 9 0. 14 0 16 0 How subdeerminans of Rn m ranslae ino connecions like his has no been sudied carefully. We can say a few hings. The lower index n indicaes he inermediae graph hrough which he connecion is going; subdeerminans of R14, 9 for example, correspond o connecions no hrough he original graph bu hrough he inermediae graph wih 14 nodes (where nodes 15 and 16 have been sar k d ). The upper index m indicaes how many nodes are considered boundary nodes; in R16, 9 nodes 1 8 were boundary nodes, bu for R16, 11 nodes 1 10 would ac as boundary nodes. The fac ha any submarix of a single layer, say R 11, is singular reflecs he fac ha here are no -connecions hrough a sar. Similarly, for an Rn m comprising k = n m + 1 layers, any k + 1 k + 1 submarix is singular since here are no k + 1 connecions hrough a graph wih k inerior nodes. There are many quesions, however. Wha does i mean o replace he upper righ corner of a Kirchhoff marix wih zeros? If he zeros in K 16 had no been so foruiously placed, how would a subdeerminan of R16 9 differ from a subdeerminan of K 8? Wha does i mean when a subdeerminan inersecs he diagonal does i make sense o speak of he connecion from nodes 1,, and 3 o 1, 4, and 5, as we seem o have been doing in 3.3? A circular planar graph is recoverable if and only if i is criical, ha is, if deleing or conracing any edge (replacing is conduciviy wih 0 or ) would break some connecion. Wha is he analoguous resul for non-planar graphs? To recover a piece of informaion wih our mehod, we need some deerminan o be zero, bu become nonzero when we crossed off he unknown enry s row and column (i.e. consider he unknown enry s cofacor); his mus be closely relaed o criicaliy. 5 Direcions for Fuure Research Our sudy on undoing he Schur complemen raises many quesions. 1. If our mehod succeeds, he graph is recoverable. Is he converse rue? Are here any recoverable graphs for which his mehod fails?. If we have o inroduce a parameer, does i maer where we inroduce i? Suppose we can recover some bu no all enries of he K n, R n, and R m n. We paramerize an unknown enry α, which allows us o recover one or more addiional enries β. We would hope ha paramerizing any β would have allowed us o recover α; ha is, any enry in he se we gain is as good as any oher. This does no appear o be he case. If we paramerize enries of single layers a he op of he pile (4..1), say R 14 if R 13 is compleely recovered, his seems o give us more han if we paramerize near he boom, say in K 16. Things propogae down beer han hey propogae up. Thus we suspec ha we have overlooked some informaion available o us, jus as he sar k mehod overlooked he muli-layers R m n. 11
One soluion may be o look a hings like his. Using he noaion of.1, ) ( γiγ j σ γ iγ k σ is singular. Half of his marix comes from R and half from K; i is as hough siing in R in (.1.1), we have peered over he ledge, down ino K. Now γ j is he sum of he j h row of R, so more generally, we can consider sums of enries in R: he marix ( ) rik + r il r im γ j γ k r jk + r jl r jm is singular, for example, and as we have seen, we ofen know more abou he sum han abou he summands, and invesigaing his can be fruiful. Some hings can be said abou he row sums of he R m n as well. We do no wan o use hese facs piecemeal, however. Wha is he general phenomenon ha is a work here? 3. Is any of he informaion we are using redundan? Can our recovery mehod be simplified? In paricular, is i necessary o consider submarices of he inermediae K n or do he he R m n suffice? Are here any graphs ha can be recovered by looking a he K n bu no by looking a he R m n alone? 4. Is our recovery mehod independen of he order of inerior nodes? Tha is, if i succeeds for one order, does i succeed for all possible orders? This appears o be rue. Some orders are nicer han ohers, however. I is much easier o recover circles, 4 rays wih he order shown in Figure 13b (due o Jeff Russell) han wih he usual order (Figure 13a). Using he sar k mehod, he former requires fewer spurious parameers. Figure 14b is similar. Why is his? Wha makes one order nicer han anoher? How should one choose an ordering of he inerior nodes o make he recovery as fas as possible? 5. During he recovery process, mos of he work is done a he surface, in he single-layer R n s, bu we usually need a leas one deerminan buried in he hickes R m n s. In 4.6, his was he submarix 1,,4,5 3,5,7,8 of R 9 16, which is a residue marix eigh layers hick. Currenly, he only way o find hese is for a human o have some inuiion abou he graph or for he compuer o search exhausively. Is here some way o make his search smarer? How can we know in advance which deeply buried submarices will be helpful? Where should we look for hem? 6. If we have o inroduce parameers, i appears ha he graph is no recoverable. However, for -o-1 graphs, which are almos recoverable, he parameer is almos spurious presumably some deerminan somewhere forces i o ake only finiely many values. Where exacly does his -o-1 ness come from in he R m n? Wha role does he generalized square roo rick of 4.3 play? Are here any 3-o-1 graphs? 7. The quesions abou deerminans, connecions, and criical graphs from he end of he previous secion. 8. Aside from he square roo rick, we did no use he symmery of our marices. How easily can our mehod be applied o direced graphs? How can we undersand ordinary (undireced) graphs as direced graphs wih some addiional consrains, jus like layered neworks? Wha abou verex conduciviy neworks? 9. Michael Goff makes an exensive sudy of neworks wih negaive conduciviies in [4]. The Schur complemen sill plays a key role for such neworks, bu hey can be quie pahalogical principal proper submarices of Kirchhoff marices need no be inverible, for example. Our mehod relied heavily on he signs of he enries of Kirchhoff marices being nice. Can i be modified o work wih negaive conduciviies? 10. Wha does our mehod allow us o say abou special classes of graphs, such as hree-dimensional laices? 1
6 Acknowledgemens The work on he sar k mehod was done in collaboraion wih Jeff Russell, whose exposiion can be found in [6]. The idea of inroducing parameers is due o Jeff Giansiracusa. The auhor would also like o hank Sam Coskey and Jim Morrow for many helpful discussions and suggesions. References [1] E. B. Curis and J. A. Morrow, Inverse problems for elecrical neworks, World Scienific, (000). [] D. Crabree and E. Haynsworh, An ideniy for he Schur complemen of a marix, Proc. Amer. Mah Soc. (1969), 364-366. [3] Ernie Esser, On solving he inverse conduciviy problem for annular neworks, (000). [4] Michael Goff, Recovering neworks wih signed conduciviies, (003). [5] Tracy Lovejoy, Applicaions of he sar k ool, (003). [6] Jeff Russell, Sar and K solve he inverse problem, (003). 13