Numbering Boundary Nodes
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- Jonah Morgan
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1 Numring Bounry Nos Lh MBri Empori Stt Univrsity August 10, Introution Th purpos of this ppr is to xplor how numring ltril rsistor ntworks ffts thir rspons mtrix, Λ. Morovr, wht n lrn from Λ out th topology of th ntwork? This ppr onsirs spifi ntworks n thir Λ, n shows how th ounry nos wr pl in th originl ntwork. In othr wors, no mttr how th ounry nos r numr, n th informtion from Λ n out th ntwork still rovr? Most of th ntworks look t in this ppr r g onutivity ntworks whr urrnt t p, φ(p), is fin s suh: γ p,q (u p u q ) q p whr γ p,q is th onutivity of th g joining no p to q. Th Kirhhoff mtrix, K, for ntwork with n nos, is n n n symmtri mtrix whr K i,j = γ i,j whn i j K i,i = j i γ i,j It is usful to writ K in lok form whr th intrior nos fll in lok C n th ounry nos fll in A. [ ] A B K = B T C Thn th rspons mtrix, Λ is form y tking th Shur omplmnt of K in C. Tht is Λ = A BC 1 B T In orr to trmin sign onitions in Λ, slt two istint sts of ounry nos, P = (p 1, p 2,..., p k ) n Q = (q 1, q 2,..., q k ). Thn tλ(p ; Q) tk(i; I) = ( 1) { } k sgn(τ) γ() D α τ S k α E α 1 τ α=τ
2 shows wht th sign of t Λ(P ; Q) is. So if thr is only on istint piring twn th nos in P n th nos in Q, thn th sign of t Λ(P ; Q) is known. This kgroun informtion oms from [1]. This ppr looks t nnulr ntworks, mny of whih hv onutivitis onstnt on lyrs. Du to th symmtry of th ntworks, th orring of th ounry nos n foun in ths ss, minly us of th symmtris foun in Λ n sign onitions of Λ. 2 Annulr Ntwork with Thr Rys n Two Cirls Figur 1: G(3,2) First onsir G(3,2), tht is grph fin s suh: G(# of rys, # of irls), n ssum tht onutivity is onstnt on lyrs. Thn th rspons mtrix, Λ, is s follows: 2
3 whr α = Λ = Σ α α β γ γ α Σ α γ β γ α α Σ γ γ β β γ γ Σ δ δ γ β γ δ Σ δ γ γ β δ δ Σ 2 ( ) ( + + )( ) β = ( ) ( + + )( ) γ = ( ) ( + + )( ) n δ = ( ) 2 ( + + )( ) Σ = (Sum of th row ntris) Rgrlss of how th ounry nos r numr th rspons mtrix still ontins four istint ntris whih provi informtion out th strutur of th ntwork. Th informtion my not s ovious s for us th orring of th ntris in Λ is no longr th sm, ut th informtion is still thr. For instn, if λ x,y is β thn x n y r ounry nos on th sm ry. It is firly sy to intify th β trm in th rspons mtrix us it is th only trm tht shows up six tims n is in vry row n vry olumn. Now, whih nos ppr on th insi of th grph n whih r on th outsi? Thnilly, thr is no wy to tll twn th insi n outsi of this ntwork us it oul sily invrt. But, it is possil to tll whih thr nos r group togthr on th insi or outsi. Th trms in Λ tht show this r δ n α, h of whih r th two othr trms whih ppr six tims. So for vry x n y suh tht λ x,y is δ, thn thos x n y s r group togthr. Likwis for vry x n y suh tht λ x,y is α. Thr n mny iffrnt Λ mtris tht hv th sm strutur s th Λ for this ntwork. In orr for Λ to rspons mtrix for this ntwork, not only must th ov symmtris hol, ut ths sign onitions must lso hol, whr (x 1, x 2, x 3 ) r group togthr on th insi (or outsi) n (y 1, y 2, y 3 ) r th outsi, n (x 1, y 1 ), (x 2, y 2 ), n (x 3, y 3 ) r pir on th sm ry. 3
4 1. t Λ(x 1, x 2, x 3 ; y 1, y 2, y 3 ) < 0 2. t Λ(x 1, x 2, y 2 ; x 3, y 1, y 3 ) > 0 3. t Λ(x 1, x 2 ; y 1, y 2 ) > 0 4. t Λ(x 1, y 1 ; x 2, y 2 ) > 0 3 Annulr Ntwork with Thr Rys n Two Cirls (With Lss Symmtry) 1 f 10 g h j 12 3 k Figur 2: G(3,2) with lss symmtry 4
5 This is nothr G(3,2) grph, ut symmtri only with rspt to th yxis. Th rspons mtrix, Λ is s follows: Σ α α β δ δ α Σ ɛ σ ρ θ Λ = α ɛ Σ σ θ ρ β σ σ Σ γ γ δ ρ θ γ Σ ω δ θ ρ γ ω Σ whr α = (( g + 2g + 2g + 2g)h)/( g + 2g + 4g + 2g + 2g + 2g + 4g + 4g + 4g + + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + 2h + 2h + + 2h + 2h + 2h + 2gh + 2gh + 2gh + 2gh + 2gh + 2gh + + 2gh + 2gh) β = (f(2 + g + g + g + g + g + g + g + g)h)/( g + 2g + 4g + 2g + 2g + 2g + 4g + + 4g + 4g + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g h + 2h + 2h + 2h + 2h + 2gh + 2gh + 2gh + 2gh + + 2gh + 2gh + 2gh + 2gh) δ = ((2 + g + g + 2g + g)h)/( g + + 2g + 4g + 2g + 2g + 2g + 4g + 4g + 4g + + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + 2h + 2h + + 2h + 2h + 2h + 2gh + 2gh + 2gh + 2gh + 2gh + + 2gh + 2gh + 2gh) ɛ = ( 2 ( g + + 2g g g g g g g g g h j j + 8j j j gj gj + 4gj gj gj gj + + 8gj gj + 8gj + 8gj gj + 2 g 2 j + 2g 2 j + 2 g 2 j + + 2g 2 j + 2g 2 j + 2 g 2 j hj hj + 4hj hj hj ghj + 4ghj ghj + 4ghj + 4ghj ghj k k k gk + 4gk gk gk gk gk + 2 g 2 k hk ghk + 8jk + 8jk + 4gjk + 4gjk + 5
6 + 8gjk + 8gjk + 8gjk + 2g 2 jk + 2g 2 jk + 2g 2 jk + 4hjk + + 4hjk + 4ghjk + 4ghjk + 4ghjk)/( g + + 2g + 4g + 2g + 2g + 2g + 4g + 4g + 4g + + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + 2h + 2h + + 2h + 2h + 2h + 2gh + 2gh + 2gh + 2gh + 2gh + + 2gh + 2gh + 2gh) σ = (f(2 + g + g + g + g + h))/( g + + 2g + 4g + 2g + 2g + 2g + 4g + 4g + 4g + + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + 2h + 2h + + 2h + 2h + 2h + 2gh + 2gh + 2gh + 2gh + 2gh + + 2gh + 2gh + 2gh) ρ = (( g g + 3g g g g + 2 g g g g g + 2g + 2 g g + g g + 4g + 5g g + 2 g g 2 + g 2 + g g g 2 + g g h h + 2h + 2 h h + 2h h h gh gh + 2gh + 2gh gh + 2gh + + 2gh gh + 4j j j + 2gj gj gj + + 2gj gj + 4gj + 6gj gj + g 2 j + g 2 j + 2 g 2 j + + 2hj hj hj + 2ghj + 2ghj ghj k k + + 4k gk + 4gk gk gk gk + 2gk + 2gk + + 4gk + 2 g 2 k + g 2 k + g 2 k hk + 2hk + 2hk ghk + + 2ghk + 2ghk + 8jk + 4gjk + 4gjk + 8gjk + 2g 2 jk + 4hjk + + 4ghjk))/( g + 2g + 4g + 2g + + 2g + 2g + 4g + 4g + 4g + g 2 + g 2 + g 2 + g g 2 + g 2 + g 2 + g 2 + 2h + 2h + 2h + 2h + 2h + + 2gh + 2gh + 2gh + 2gh + 2gh + 2gh + 2gh + 2gh) θ = (( g + g g g + 2 g g g g g + 2 g + g + 2 g + + g h + 2 h + 2 h + 4j j j + 2gj gj gj + 2gj gj + 4gj + 6gj gj + g 2 j + g 2 j g 2 j + 2hj hj hj + 2ghj + 2ghj ghj k k + 4k gk + 4gk gk gk gk + 2gk + + 2gk + 4gk + 2 g 2 k + g 2 k + g 2 k hk + 2hk + 2hk + 6
7 + 2 2 ghk + 2ghk + 2ghk + 8jk + 4gjk + 4gjk + 8gjk + 2g 2 jk + + 4hjk + 4ghjk))/( g + 2g + 4g + 2g + + 2g + 2g + 4g + 4g + 4g + g 2 + g 2 + g 2 + g g 2 + g 2 + g 2 + g 2 + 2h + 2h + 2h + 2h + 2h + + 2gh + 2gh + 2gh + 2gh + 2gh + 2gh + 2gh + 2gh) γ = f( g + g + g + g + h + h + h))/( g + 2g + 4g + 2g + 2g + 2g + 4g + 4g + + 4g + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + 2h + + 2h + 2h + 2h + 2h + 2gh + 2gh + 2gh + 2gh + 2gh + + 2gh + 2gh + 2gh) ω = ( 2 ( g g + n g g g g g g + 2g g h h h h h h j j j gj gj gj + 4gj gj gj + 2 g 2 j hj hj hj hj ghj k k + 8k k k gk + 4gk gk gk gk + 4gk + + 4gk + 8gk gk gk gk + 2 g 2 k + 2g 2 k + 2 g 2 k + + 2g 2 k + 2g 2 k + 2 g 2 k hk + 4hk hk + 4hk + 4hk hk ghk + 4ghk ghk + 4ghk + 4ghk ghk + 8jk + + 8jk + 4gjk + 4gjk + 8gjk + 4gjk + 8gjk + 2g 2 jk + 2g 2 jk + + 2g 2 jk + 4hjk + 4hjk + 4hjk + 4ghjk + 4ghjk + 4ghjk)/( g + 2g + 4g + 2g + 2g + 2g + 4g + 4g + + 4g + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + g 2 + 2h + + 2h + 2h + 2h + 2h + 2gh + 2gh + 2gh + 2gh + 2gh + + 2gh + 2gh + 2gh) Σ = (Sum of th row ntris) Rgrlss of how th ounry nos r numr th rspons mtrix still ontins nin istint ntris whih provi informtion out th strutur of th ntwork. Th informtion my not s ovious s for us th orring of th ntris in Λ is no longr th sm, ut th informtion is still thr. Th first thing to noti in Λ is tht thr r thr ntris tht only ppr twi: β, ɛ, n ω. Upon furthr insption, β n istinguish from ɛ n ω us vry row n olumn tht ontins β lso ontins Σ n two pirs. Not this is only tru for β. So if λ x,y is β, thn ounry 7
8 nos x n y r ontin on th sm ry. Morovr x n y r on th non-symmtri ry. Consir th othr two ntris tht ppr only twi: ɛ n ω. If λ x,y is ɛ or ω thn x n y r group togthr on th insi or th outsi. Now on of th insi nos will pir with on of th outsi nos, ut whih on? In orr to trmin this, first slt on of th ounry nos not on th symmtri ry, ll it x. Nxt, slt th group of nos (y, z) tht os not ontin x. Tht is if x is on th insi, thn slt th group known to on th outsi. Thr r two hois, x n y n on th sm ry or x n z. If λ x,y > λ x,z, thn x n z r on th sm ry. Othrwis x n y r togthr. Th rmining two nos r on th lst ry. This is tru us θ > ρ u to t Λ(1, 2; 4, 5) > 0 n t Λ(2, 1, 4; 3, 5, 6) > 0. Now, whih nos r group togthr on th insi of th grph n whih r on th outsi? Thnilly, thr is no wy to tll twn th insi n outsi of this ntwork us it oul sily invrt. But, it is possil to tll whih thr nos r togthr on th insi or outsi, two r lry known u to ɛ n ω. To hlp simplify things ll th nos on th non-symmtri ry (x 1, x 2 ) n ll th two sts of nos tht r group togthr on th insi or outsi (y 1, y 2 ) n (z 1, z 2 ). Now thr r two options: (x 1, y 1, y 2 ) n (x 2, z 1, z 2 ) n group togthr or (x 1, z 1, z 2 ) n (x 2, y 1, y 2 ). If λ x1,y 1 λ x2,z 1 < λ x1,z 1 λ x2,y 1, thn (x 1, z 1, z 2 ) n (x 2, y 1, y 2 ) r th orrt grouping, othrwis (x 1, y 1, y 2 ) n (x 2, z 1, z 2 ) r th orrt grouping. This is tru us γα > σδ u to t Λ(1, 4; 2, 5) > 0. Not, in th inqulity y 2 n z 2 oul rpl y 1 n z 1 rsptivly us thy r th sm. Thr n mny iffrnt Λ mtris tht hv th sm strutur s th Λ for this ntwork. In orr for Λ to rspons mtrix for this ntwork, not only must th ov symmtris hol, ut ths sign onitions must lso hol, whr (x 1, x 2, x 3 ) r group togthr on th insi (or outsi) n (y 1, y 2, y 3 ) r on th outsi, n (x 1, y 1 ), (x 2, y 2 ), n (x 3, y 3 ) r pir on th sm ry, not (x 1, y 1 ) is th non-symmtri ry. 1. t Λ(x 1, x 2, x 3 ; y 1, y 2, y 3 ) < 0 2. t Λ(x 1, x 2, y 2 ; x 3, y 1, y 3 ) > 0 3. t Λ(x 2, x 1, y 1 ; x 3, y 2, y 3 ) > 0 4. t Λ(x 1, x 2 ; y 1, y 2 ) > 0 5. t Λ(x 2, x 3 ; y 2, y 3 ) > 0 6. t Λ(x 1, y 1 ; x 2, y 2 ) > 0 7. t Λ(x 2, y 2 ; x 3, y 3 ) > 0 4 Grph with Thr Hols 4.1 Compltly Symmtri Cs 8
9 Figur 3: Grph with Thr Hols Consir this grph with onutivitis on lyrs ompltly onstnt. Σ α β δ β γ γ β κ β δ κ α Σ γ β δ β β γ κ δ β κ β γ Σ α β δ κ γ β κ β δ δ β α Σ γ β κ β γ κ δ β β δ β γ Σ α β κ γ δ κ β Λ = γ β δ β α Σ γ κ β β κ δ γ β κ κ β γ Σ δ δ α β β β γ γ β κ κ δ Σ δ β α β κ κ β γ γ β δ δ Σ β β α β δ κ κ δ β α β β Σ γ γ δ β β δ κ κ β α β γ Σ γ κ κ δ β β δ β β α γ γ Σ whr α = ( ) ( + 3)( )( ) β = ( ) ( + 3)( )( ) 9
10 δ = ( ) ( + 3)( )( ) γ = ( ) ( + 3)( )( ) κ = 2 ( + 3)( )( ) Σ = (Sum of th row ntris) Agin, rgrlss of how th ounry nos r numr th rspons mtrix still ontins fiv istint ntris whih provi informtion out th strutur of th ntwork. Th informtion my not s ovious s for us th orring of th ntris in Λ is no longr th sm, ut th informtion is still thr. Th first thing to noti in Λ is tht α is th only ntry tht pprs just 12 tims in th mtrix. If λ x,y = α, thn x n y r pir on th sm ry. Nxt thr r thr trms, δ, γ, n κ tht ppr 24 tims h. If λ x,y = α, thn κ will in th sm olumn of th x th n y th rows. No othr lmnt will hv tht hrtristi. β is sy to fin us it pprs 48 tims in th mtrix. Now th qustion is how n γ istinguish from δ? Using two sign onitions of th ntwork, t Λ(1, 2; 5, 4) > 0 n t Λ(1, 2; 6, 3) > 0, it is known tht γ 2 > δ 2. If λ x,y = γ, thn x n y r group in th sm irl. Hn ll ounry nos in h irl r known. Thr n mny iffrnt Λ mtris tht hv th sm strutur s th Λ for this ntwork. In orr for Λ to rspons mtrix for this ntwork, not only must th ov symmtris hol, ut rtin sign onitions must lso hol. A list of som of ths sign onitions n foun in th ppnix. It is not omplt list u to th grt quntity of trminnts n omplxity of th ntwork. Th xtr symmtry in this ntwork my mn tht som of th trminnts in th ppnix r quivlnt. 4.2 Lss Symmtri Cs 10
11 f f f Figur 4: Grph with Thr Hols(with lss symmtry) Hr is nothr grph to onsir. Assum tht onutivity is onstnt on lyrs. Thn th rspons mtrix, Λ, is s follows: Σ α β γ β δ θ κ π ψ η µ α Σ δ β γ β κ θ π η ψ µ β δ Σ α β γ π θ κ µ ψ η γ β α Σ δ β π κ θ µ η ψ β γ β δ Σ α κ π θ η µ ψ Λ = δ β γ β α Σ θ π κ ψ µ η θ κ π π κ θ Σ σ σ ɛ τ τ κ θ θ κ π π σ Σ σ τ ɛ τ π π κ θ θ κ σ σ Σ τ τ ɛ ψ η µ µ η ψ ɛ τ τ Σ χ χ η ψ ψ η µ µ τ ɛ τ χ Σ χ µ µ η ψ ψ η τ τ ɛ χ χ Σ Rgrlss of how th ounry nos r numr th rspons mtrix still ontins fourtn istint ntris whih provi informtion out th strutur of th ntwork. Th informtion my not s ovious s for us th orring of th ntris in Λ is no longr th sm, ut th informtion is still thr. Th first thing to noti in Λ is tht in th uppr tringulr prt of th mtrix, β is th only trm tht pprs six tims in th uppr tringl n is only in four olumns. Hn β is istinguish from th othr ntris 11
12 of th mtrix. Now looking t th whol mtrix, τ n lso istinguish. In th mtrix, β pprs in six iffrnt rows n olumns; τ lso pprs in six iffrnt rows n olumns, non of whih ontin β. So, τ n β n intifi in Λ. Nxt look t th 6 6 sumtrix ontining only rows n olumns whih hv β. In this sumtrix thr r thr othr ntris: α, γ, n δ. γ n istinguish from th othr two y using som sign onitions of trminnts. γ is th only on of th thr, suh tht γ 2 < β 2. This is us t Λ(1, 2; 6, 3) > 0, t Λ(1, 4; 2, 3) > 0, n t Λ(1, 2; 5, 4) > 0. Now tht β n γ hv n intifi in Λ th rrngmnt of nos in th innr tringl n trmin. If λ x,y = β, thn x is two nos wy from y. If λ x,y = γ, thn x is thr nos wy from y. Using this informtion it is now known whih nos r pir on th intrior rys n two of th nos group togthr in h of th thr intrior irls. Anothr lmnt n intifi y tking th six y six sumtrix of vry row n olumn ontining τ. Within this mtrix, ɛ is th only lmnt ppring six tims n is in vry row n vry olumn. If λ x,y = ɛ thn x n y r on th sm outr ry. Upon furthr insption, it oms lr tht this sumtrix hs th sm strutur s G(3,2), thus hving four iffrnt lmnts. As in th G(3,2), σ n χ n not istinguish from h othr, ut thy r iffrnt ntris. Thus if λ x,y = σ, thn x n y r group togthr on th outsi irl or with th nos from th innr tringl. Similrly for λ x,y = χ. Furthrmor, it woul ni to know whih outr ry is group with h innr irl. For h of th innr irls, lt (x 1, y 1 ), (x 2, y 2 ), n (x 3, y 3 ) th pir of ounry nos tht r group togthr in on of th innr irls. Now is givn ry group with (x 1, y 1 ), (x 2, y 2 ), or (x 3, y 3 )? Slt on of th ounry nos on th outr ry, ll it z. If λ x1,z = λ y1,z, thn x 1, y 1, n z r group togthr in th sm innr irl. If λ x1,z λ y1,z, thn try th sm thing using (x 2, y 2 ) n (x 3, y 3 ); thy will only qul in on of ths ss. So h outr ry is group with on of th innr irls. In orr to trmin whih nos of th outr rys r in th outr irl n whih r in th innr irls trminnt must hk. Lt (x 1, x 2, x 3 ) n (y 1, y 2, y 3 ) th group togthr on th outr irl or in th innr irls, this is know from σ n χ. Now for th momnt, group x 1, x 2,n x 3 with y 1. Thn group y 2 with on of th lmnts in th innr irl group with y 1 ll it w, n with th two lmnts, (z 1, z 2 ), tht shr n innr ry tht onnts th two innr irls whih o not group with y 1. If t Λ(x 1, x 2, x 3, y 1 ; y 2, w, z 1, z 2 ) = 0, thn (x 1, x 2, x 3 ) r th nos on th outr irl n (y 1, y 2, y 3 ) r th nos in th innr irls. If it is not qul to zro, thn (x 1, x 2, x 3 ) r th nos in th innr irls n (y 1, y 2, y 3 ) r th nos on th outr irl. Thr n mny iffrnt Λ mtris tht hv th sm strutur s th Λ for this ntwork. In orr for Λ to rspons mtrix for this ntwork, not only must th ov symmtris hol, ut rtin sign onitions must lso hol. A list of som of ths sign onitions n foun in th ppnix. It is not omplt list u to th grt quntity of trminnts n omplxity of th ntwork. 12
13 5 Strutur of Entris in Λ Whil working with th rspons mtrix pttrn mrg. No mttr how lrg or smll th ntris wr, th off-igonl ntris wr ll rtios of th opposit of sums of monomils with positiv intgr offiints. n ntry in th r- Monomils with Positiv Coffiints 5.1 Lt λ i,j spons mtrix. It n writtn in th following form. λ i,j = mi,j, whn i j λ i,i = mi,i, whn i = j Whr m i,j,m i,i, n r sums of monomils with positiv intgr offiints. Proof: (y inution) In orr to go from th Kirhhoff mtrix, K, to th rspons mtrix, Λ, Gussin limintion n on on row t tim until Λ is th rsult. In othr wors tk th Shur omplmnt ut on lmnt t tim. In th first s this is oviously tru, tht is whn λ i,j is n lmnt of th Kirhhoff mtrix. This is us K i,j = γ i,j whn i j n K i,i = j i K i,j, n γ i,j is sum of positiv monomils with intgr offiints. Now ssum this is tru for λ i,j. So, λ i,j = mi,j whn i j n λ i,i = mi,i whn i = j. Whr m i,j, m i,i, n r sums of monomils with positiv intgr offiints. (This is th inutiv hypothsis.) Thn y Gussin limintion whn i j. So λ i,j = 1 λ i,j = m i,j m i,n m n,j m n,n ( mi,j m i,nm n,j ) m n,n m i,j + m i,n m n,j = m n,n m n,n y th inutiv hypothsis. Whn i = j, thn Thus th thorm hs n provn. λ i,i = 1 j i m i,j 6 G(3,2) with Vrtx Conutivity = m i,j 13
14 Figur 5: G(3,2) with vrtx onutivity Now onsir ntwork with vrtx onutivity inst of g onutivity. So th urrnt t no is fin s: γ q (u q u p ) q p whr γ q is th onutivity t no q. Using this finition of urrnt, thn th rspons mtrix, Λ, is s follows: Σ α α β γ γ α Σ α γ β γ Λ = α α Σ γ γ β δ κ κ Σ ɛ ɛ κ δ κ ɛ Σ ɛ κ κ δ ɛ ɛ Σ whr α = β = 2 ( ) ( + + )( ) ( ) ( + + )( ) 14
15 γ = δ = κ = ɛ = ( ) ( + + )( ) ( ) ( + + )( ) ( ) ( + + )( ) 2 ( )) ( + + )( ) Σ = (Sum of th row ntris) Rgrlss of how th ounry nos r numr th rspons mtrix still ontins six istint ntris whih provi informtion out th strutur of th ntwork. Th informtion my not s ovious s for us th orring of th ntris in Λ is no longr th sm, ut th informtion is still thr. Th first thing to rmmr is tht Λ is no longr symmtri. This is us Λ is for ntwork with vrtx onutivity. β n δ r th only two ntris tht ppr thr tims. If λ x,y = β or λ x,y = δ, thn x n y r pir on th sm ry. Now, whih nos ppr on th insi of th grph n whih r on th outsi? Thnilly, thr is no wy to tll twn th insi n outsi of this ntwork us it oul sily invrt. But, it is possil to tll whih thr nos r group togthr on th insi or outsi. Consir ll λ x,y = β, or ɛ. It os not mttr whih ntry is hosn, just tht on of th ntris tht only pprs thr tims. Thn ll x r group on th insi n ll y r group on th outsi. Thr n mny iffrnt Λ mtris tht hv th sm strutur s th Λ for this ntwork. In orr for Λ to rspons mtrix for this ntwork, not only must th ov symmtris hol, ut rtin sign onitions must lso hol. Rfrns [1] Curtis, E.B. n J.A. Morrow. Invrs Prolms for Eltril Ntworks, Worl Sintifi, Nw Jrsy,
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