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Finding your way in a graph Finding your way in a graph Finding - PowerPoint PPT Presentation

Nov 29, 2022 •466 likes •2.13k views

Finding your way in a graph Finding your way in a graph Finding your way in a graph Finding your way in a graph = station One loop = junction A system of interconnected loops L L What is the best way to go from L to ? L L L 1 ? L

  1. 110 111 000 a 001 010 011 100 b f e c 100 011 101 110 d 111 001 000 G

  2. A valid addressing of G. 110 111 000 e d a 001 010 011 c f 100 b f e c 100 011 101 110 d a b 111 001 000 G

  3. 110 111 000 e d a 001 010 011 c f 100 b f e c 100 011 101 110 d a b 111 001 000 G 000 a 001 010 b f An invalid addressing e c 011 110 d 111 G

  4. Trees e a d b f c g A tree T

  5. Trees e a a - 0 b - 1 d b f So far, so good! c g A tree T

  6. Trees e a a - 0 0 b - 1 0 d b f c - 1 1 c g A tree T

  7. Trees e a a - 0 0 0 b - 1 0 0 d b f c - 1 1 0 d - 1 0 1 c g A tree T

  8. Trees e a a - 0 0 0 0 b - 1 0 0 0 d b f c - 1 1 0 0 d - 1 0 1 0 c g e - 1 0 1 1 A tree T

  9. Trees e a a - 0 0 0 0 0 b - 1 0 0 0 0 d b f c - 1 1 0 0 0 d - 1 0 1 0 0 c g e - 1 0 1 1 0 f - 1 0 1 0 1 A tree T

  10. Trees e a a - 0 0 0 0 0 0 b - 1 0 0 0 0 0 d b f c - 1 1 0 0 0 0 d - 1 0 1 0 0 0 c g e - 1 0 1 1 0 0 f - 1 0 1 0 1 0 g - 1 0 1 0 1 1 A tree T

  11. Trees e a a - 0 0 0 0 0 0 b - 1 0 0 0 0 0 d b f c - 1 1 0 0 0 0 d - 1 0 1 0 0 0 c g e - 1 0 1 1 0 0 f - 1 0 1 0 1 0 g - 1 0 1 0 1 1 A tree T A valid addressing of T

  12. Now we can address a triangle What about a triangle ? ? Introduce a new symbol * , and define d (0,*) = d (1,*) = 0. H H For example, d (0 1 * 0 * 1 0 1, * 0 1 * * 0 1 1) = 3 H

  13. Now we can address a triangle What about a triangle ? ? Introduce a new symbol * , and define d (0,*) = d (1,*) = 0. H H For example, d (0 1 * 0 * 1 0 1, * 0 1 * * 0 1 1) = 3 H

  14. Now we can address a triangle What about a triangle ? ? Introduce a new symbol * , and define d (0,*) = d (1,*) = 0. H H For example, d (0 1 * 0 * 1 0 1, * 0 1 * * 0 1 1) = 3 H

  15. Now we can address a triangle What about a triangle ? ? Introduce a new symbol * , and define d (0,*) = d (1,*) = 0. H H For example, d (0 1 * 0 * 1 0 1, * 0 1 * * 0 1 1) = 3 H

  16. 0 0 Now we can address a triangle 0 1 What about a triangle ? ? Introduce a new symbol * , and define d (0,*) = d (1,*) = 0. H H For example, d (0 1 * 0 * 1 0 1, * 0 1 * * 0 1 1) = 3 H

  17. 0 0 Now we can address a triangle 1 * 0 1 What about a triangle ? ? Introduce a new symbol * , and define d (0,*) = d (1,*) = 0. H H For example, d (0 1 * 0 * 1 0 1, * 0 1 * * 0 1 1) = 3 H

  18. A valid extended addressing of G is an assignment A(v) to each vertex v in G an N-tuple of 0, 1, and *’s so that for all vertices u and v in G, d (u,v ) = d (A(u),A(v)) H G Theorem: Valid extended addresses exist for every graph G.

  19. A valid extended addressing of G is an assignment A(v) to each vertex v in G an N-tuple of 0, 1, and *’s so that for all vertices u and v in G, d (u,v ) = d (A(u),A(v)) H G Theorem: Valid extended addresses exist for every graph G.

  20. Proof: d (v ,v ) 1 2 G { A(v ) = 0......0 1 A(v ) = 1……1 2

  21. Proof: d (v ,v ) 1 2 d (v ,v ) 1 3 G G { { A(v ) = 0......0 0……0 1 A(v ) = 1……1 *……* 2 A(v ) = *……* 1……1 3

  22. Proof: d (v ,v ) 1 2 d (v ,v ) d (v ,v ) 1 3 G G i j G { { { A(v ) = 0......0 0……0……………………*……*……………… 1 A(v ) = 1……1 *……*…………………….*……*……………… 2 A(v ) = *……* 1……1…………………….*……*……………… 3 . . . . A(v ) = *……* *……*…………………….0……0……………… i . . . . . A(v ) = *……* *……*…………………….1……1……………… j . . . .

  23. Unfortunately, the length of the addresses maybe very long by using this method! Define N(G) to be the least N such that a valid (extended) addressing of G of length N exists. Conjecture: If G has n vertices then N(G) ! n – 1. Conjecture:

  24. Unfortunately, the length of the addresses maybe very long by using this method! Define N(G) to be the least N such that a valid (extended) addressing of G of length N exists. Conjecture: If G has n vertices then N(G) ! n – 1. Conjecture:

  25. Unfortunately, the length of the addresses maybe very long by using this method! Define N(G) to be the least N such that a valid (extended) addressing of G of length N exists. Conjecture: If G has n vertices then N(G) ! n – 1. Conjecture: Theorem (Peter Winkler) Theorem

  26. Unfortunately, the length of the addresses maybe very long by using this method! Define N(G) to be the least N such that a valid (extended) addressing of G of length N exists. Conjecture: If G has n vertices then N(G) ! n – 1. Conjecture: Theorem (Peter Winkler - $100) Theorem

  27. Unfortunately, the length of the addresses maybe very long by using this method! Define N(G) to be the least N such that a valid (extended) addressing of G of length N exists. Conjecture: If G has n vertices then N(G) ! n – 1. Conjecture: Theorem (Peter Winkler - $100) Theorem

  28. A A B C D E F . . . . . . . A 0 1 1 2 2 3 . . . . . . . B 1 0 1 1 2 2 C B . . . . . . . C 1 1 0 2 1 2 . . . . . . . D 2 1 2 0 2 1 D E . . . . . . . 2 2 1 2 0 1 E F . . . . . . . F 3 2 2 1 1 0 . . . . . . . Distance matrix D(G) = (d ) ij

  29. A vertex - address A B C D E F . . . . . . . A 0 1 1 2 2 3 A - 0 0 0 0 0 . . . . . . . B 1 0 1 1 2 2 C B B - 1 * 0 0 * . . . . . . . C - 0 1 0 0 * C 1 1 0 2 1 2 . . . . . . . D 2 1 2 0 2 1 D - 1 * 1 D E * 0 . . . . . . . 2 2 1 2 0 1 E E - 0 1 0 1 * F . . . . . . . F 3 2 2 1 1 0 F - * * 1 1 1 . . . . . . . Distance matrix D(G) = (d ) ij

  30. A vertex - address A B C D E F . . . . . . . A 0 1 1 2 2 3 A - 0 0 0 0 0 . . . . . . . B 1 0 1 1 2 2 C B B - 1 * 0 0 * . . . . . . . C - 0 1 0 0 * C 1 1 0 2 1 2 . . . . . . . D 2 1 2 0 2 1 D - 1 * 1 D E * 0 . . . . . . . 2 2 1 2 0 1 E E - 0 1 0 1 * F . . . . . . . F 3 2 2 1 1 0 F - * * 1 1 1 . . . . . . . Distance matrix D(G) = (d ) ij

  31. A vertex - address A B C D E F . . . . . . . A 1 1 A - 0 0 0 0 0 . . . . . . . B C B B - 1 * 0 0 * . . . . . . . C - 0 1 0 0 * C 1 1 . . . . . . . D D - 1 * 1 D E * 0 . . . . . . . E 1 1 E - 0 1 0 1 * F . . . . . . . F F - * * 1 1 1 . . . . . . .

  32. A vertex - address A B C D E F . . . . . . . A 1 1 A - 0 0 0 0 0 . . . . . . . B C B B - 1 * 0 0 * . . . . . . . C - 0 1 0 0 * C 1 1 . . . . . . . D D - 1 * 1 D E * 0 . . . . . . . E 1 1 E - 0 1 0 1 * F . . . . . . . F F - * * 1 1 1 . . . . . . . A C E x B D

  33. A vertex - address A B C D E F . . . . . . . A A - 0 0 0 0 0 . . . . . . . B C B 1 B - 1 * 0 0 * 1 1 . . . . . . . C - 0 1 0 0 * C . . . . . . . D 1 1 D - 1 * 1 1 D E * 0 . . . . . . . E E - 0 1 0 1 * F . . . . . . . F F - * * 1 1 1 . . . . . . . A C E x B D = B D x A C E

  34. A vertex - address A B C D E F . . . . . . . A 1 1 A - 0 0 0 0 0 . . . . . . . B C B B - 1 * 0 0 * . . . . . . . C - 0 1 0 0 * C . . . . . . . D D - 1 * 1 D E * 0 . . . . . . . E E - 0 1 0 1 * F . . . . . . . F F - * * 1 1 1 . . . . . . . A x C E

  35. A vertex - address A B C D E F . . . . . . . A 0 1 1 2 2 3 A - 0 0 0 0 0 . . . . . . . 1 0 1 1 2 2 B C B B - 1 * 0 0 * . . . . . . . C - 0 1 0 0 * C 1 1 0 2 1 2 . . . . . . . D 2 1 2 0 2 1 D - 1 * 1 D E * 0 . . . . . . . 2 2 1 2 0 1 E E - 0 1 0 1 * F . . . . . . . F 3 2 2 1 1 0 F - * * 1 1 1 . . . . . . . column contribution Distance matrix D(G) = (d ) ij 1 ACE x BD " Q( G ) d xx (x x x )(x x ) = = + + + 12 ij i j 1 3 5 2 4 1 i ,j n 2 A x CE ! ! x (x x ) + + 1 3 5 3 ABCE x DF (x x x x )(x x ) + + + + + 1 2 3 5 4 6 4 ABC x EF (x x x ) x x + + + ( ) + 1 2 3 5 AD x F 5 6 (x x ) x + + 1 4 6

  36. A valid extended addressing of G using N-tuples corresponds exactly to " Q(G) d xx j a decomposition of = into a sum of N terms of form 12 ij i , 1 i j n ! ! (x x ... x )(x x ... x ) . + + + + + + i i i j j j 1 2 r 1 2 s AB [(A B) 2 (A B) ] 2 = + ! ! However, since 1 4 then ! Q(G) (x x ... x )(x x ... x ) = + + + + + + j j j i i i r 1 2 s 1 2 N terms " [(x ... x x ... x ) 2 (x ... x x ... x ) ] 2 + + + + + ! + + ! ! ! 1 = j j j j 4 i i i i r s r s 1 1 1 1 N Thus, Q(G) is congruent to a quadratic form which has N positive squares and N negative squares.

  37. A valid extended addressing of G using N-tuples corresponds exactly to " Q(G) d xx j a decomposition of = into a sum of N terms of form 12 ij i , 1 i j n ! ! (x x ... x )(x x ... x ) . + + + + + + i i i j j j 1 2 r 1 2 s AB [(A B) 2 (A B) ] 2 = + ! ! However, since 1 4 then ! Q(G) (x x ... x )(x x ... x ) = + + + + + + j j j i i i r 1 2 s 1 2 N terms " [(x ... x x ... x ) 2 (x ... x x ... x ) ] 2 + + + + + ! + + ! ! ! 1 = j j j j 4 i i i i r s r s 1 1 1 1 N Thus, Q(G) is congruent to a quadratic form which has N positive squares and N negative squares.

  38. A valid extended addressing of G using N-tuples corresponds exactly to " Q(G) d xx j a decomposition of = into a sum of N terms of form 12 ij i , 1 i j n ! ! (x x ... x )(x x ... x ) . + + + + + + i i i j j j 1 2 r 1 2 s AB [(A B) 2 (A B) ] 2 = + ! ! However, since 1 4 then ! Q(G) (x x ... x )(x x ... x ) = + + + + + + j j j i i i r 1 2 s 1 2 N terms " [(x ... x x ... x ) 2 (x ... x x ... x ) ] 2 + + + + + ! + + ! ! ! 1 = j j j j 4 i i i i r s r s 1 1 1 1 N Thus, Q(G) is congruent to a quadratic form which has N positive squares and N negative squares.

  39. Hence, by Sylvester’s law of inertia, N n (G) ! = number of positive eigenvalues of D(G); + and N n (G) ! = number of negative eigenvalues of D(G); -

  40. Hence, by Sylvester’s law of inertia, N n (G) ! = number of positive eigenvalues of D(G); + and N n (G) ! = number of negative eigenvalues of D(G); - Theorem: (Graham, Pollak, Witsenhausen) Theorem: N(G) max{n (G),n (G)} ! + -

  41. Hence, by Sylvester’s law of inertia, N n (G) ! = number of positive eigenvalues of D(G); + and N n (G) ! = number of negative eigenvalues of D(G); - Theorem: (Graham, Pollak, Witsenhausen) Theorem: N(G) max{n (G),n (G)} ! + -

  42. Hence, by Sylvester’s law of inertia, N n (G) ! = number of positive eigenvalues of D(G); + and N n (G) ! = number of negative eigenvalues of D(G); - Theorem: (Graham, Pollak, Witsenhausen) Theorem: N(G) max{n (G),n (G)} ! + -

  43. Hence, by Sylvester’s law of inertia, N n (G) ! = number of positive eigenvalues of D(G); + and N n (G) ! = number of negative eigenvalues of D(G); - Theorem: (Graham, Pollak, Witsenhausen) Theorem: N(G) max{n (G),n (G)} ! + - Question: How close to the truth is this bound?

  44. T - a tree with n vertices n T ’’ T ’ T 5 5 5 _ _ _ _ _ _ 0 1 1 1 1 0 1 2 3 4 0 1 2 3 3 1 0 2 2 2 1 0 1 2 3 1 0 1 2 2 1 2 0 2 2 2 1 0 1 2 2 1 0 1 1 1 2 2 0 2 3 2 1 0 1 3 2 1 0 2 _ _ _ _ _ _ 1 2 2 2 0 4 3 2 1 0 3 2 1 2 0 T ’ T ’’ D( ) T 5 D( ) D( ) D( ) 5 5 n = 1 n = 1 n = 1 + + + n = 4 n = 4 n = 4 - - - T ’’ T ’ det = 32 D( ) det = 32 ! D( ) det = 32 !! D( ) D( ) T 5 5 5 A coincidence ? (or an example of the law of small numbers?)

  45. T - a tree with n vertices n T ’’ T ’ T 5 5 5 _ _ _ _ _ _ 0 1 1 1 1 0 1 2 3 4 0 1 2 3 3 1 0 2 2 2 1 0 1 2 3 1 0 1 2 2 1 2 0 2 2 2 1 0 1 2 2 1 0 1 1 1 2 2 0 2 3 2 1 0 1 3 2 1 0 2 _ _ _ _ _ _ 1 2 2 2 0 4 3 2 1 0 3 2 1 2 0 T ’ T ’’ D( ) T 5 D( ) D( ) D( ) 5 5 n = 1 n = 1 n = 1 + + + n = 4 n = 4 n = 4 - - - T ’’ T ’ det = 32 D( ) det = 32 ! D( ) det = 32 !! D( ) D( ) T 5 5 5 A coincidence ? (or an example of the law of small numbers?)

  46. T - a tree with n vertices n T ’’ T ’ T 5 5 5 _ _ _ _ _ _ 0 1 1 1 1 0 1 2 3 4 0 1 2 3 3 1 0 2 2 2 1 0 1 2 3 1 0 1 2 2 1 2 0 2 2 2 1 0 1 2 2 1 0 1 1 1 2 2 0 2 3 2 1 0 1 3 2 1 0 2 _ _ _ _ _ _ 1 2 2 2 0 4 3 2 1 0 3 2 1 2 0 T ’ T ’’ D( ) T 5 D( ) D( ) D( ) 5 5 n = 1 n = 1 n = 1 + + + n = 4 n = 4 n = 4 - - - T ’’ T ’ det = 32 D( ) det = 32 ! D( ) det = 32 !! D( ) D( ) T 5 5 5 A coincidence ? (or an example of the law of small numbers?)

  47. T - a tree with n vertices n T ’’ T ’ T 5 5 5 _ _ _ _ _ _ 0 1 1 1 1 0 1 2 3 4 0 1 2 3 3 1 0 2 2 2 1 0 1 2 3 1 0 1 2 2 1 2 0 2 2 2 1 0 1 2 2 1 0 1 1 1 2 2 0 2 3 2 1 0 1 3 2 1 0 2 _ _ _ _ _ _ 1 2 2 2 0 4 3 2 1 0 3 2 1 2 0 T ’ T ’’ D( ) T 5 D( ) D( ) D( ) 5 5 n = 1 n = 1 n = 1 + + + n = 4 n = 4 n = 4 - - - T ’’ T ’ det = 32 D( ) det = 32 ! D( ) det = 32 !! D( ) D( ) T 5 5 5 A coincidence ? (or an example of the law of small numbers?)

  48. T - a tree with n vertices n T ’’ T ’ T 5 5 5 _ _ _ _ _ _ 0 1 1 1 1 0 1 2 3 4 0 1 2 3 3 1 0 2 2 2 1 0 1 2 3 1 0 1 2 2 1 2 0 2 2 2 1 0 1 2 2 1 0 1 1 1 2 2 0 2 3 2 1 0 1 3 2 1 0 2 _ _ _ _ _ _ 1 2 2 2 0 4 3 2 1 0 3 2 1 2 0 T ’ T ’’ D( ) T 5 D( ) D( ) D( ) 5 5 n = 1 n = 1 n = 1 + + + n = 4 n = 4 n = 4 - - - T ’’ T ’ det = 32 D( ) det = 32 ! D( ) det = 32 !! D( ) D( ) T 5 5 5 A coincidence ? (or an example of the law of small numbers?)

  49. T - a tree with n vertices n T ’’ T ’ T 5 5 5 _ _ _ _ _ _ 0 1 1 1 1 0 1 2 3 4 0 1 2 3 3 1 0 2 2 2 1 0 1 2 3 1 0 1 2 2 1 2 0 2 2 2 1 0 1 2 2 1 0 1 1 1 2 2 0 2 3 2 1 0 1 3 2 1 0 2 _ _ _ _ _ _ 1 2 2 2 0 4 3 2 1 0 3 2 1 2 0 T ’ T ’’ D( ) T 5 D( ) D( ) D( ) 5 5 n = 1 n = 1 n = 1 + + + n = 4 n = 4 n = 4 - - - T ’’ T ’ det = 32 D( ) det = 32 ! D( ) det = 32 !! D( ) D( ) T 5 5 5 A coincidence ? (or an example of the law of small numbers?)

  50. T - a tree with n vertices n T ’’ T ’ T 5 5 5 _ _ _ _ _ _ 0 1 1 1 1 0 1 2 3 4 0 1 2 3 3 1 0 2 2 2 1 0 1 2 3 1 0 1 2 2 1 2 0 2 2 2 1 0 1 2 2 1 0 1 1 1 2 2 0 2 3 2 1 0 1 3 2 1 0 2 _ _ _ _ _ _ 1 2 2 2 0 4 3 2 1 0 3 2 1 2 0 T ’ T ’’ D( ) T 5 D( ) D( ) D( ) 5 5 n = 1 n = 1 n = 1 + + + n = 4 n = 4 n = 4 - - - T ’’ T ’ det = 32 D( ) det = 32 ! D( ) det = 32 !! D( ) D( ) T 5 5 5 A coincidence ? (or an example of the law of small numbers?)

  51. T - a tree with n vertices n T ’’ T ’ T 5 5 5 _ _ _ _ _ _ 0 1 1 1 1 0 1 2 3 4 0 1 2 3 3 1 0 2 2 2 1 0 1 2 3 1 0 1 2 2 1 2 0 2 2 2 1 0 1 2 2 1 0 1 1 1 2 2 0 2 3 2 1 0 1 3 2 1 0 2 _ _ _ _ _ _ 1 2 2 2 0 4 3 2 1 0 3 2 1 2 0 T ’ T ’’ D( ) T 5 D( ) D( ) D( ) 5 5 n = 1 n = 1 n = 1 + + + n = 4 n = 4 n = 4 - - - T ’’ T ’ det = 32 D( ) det = 32 ! D( ) det = 32 !! D( ) D( ) T 5 5 5 A coincidence ? (or an example of the law of small numbers?)

  52. If T is a tree with n vertices then n ) ( ) (n ) 2 -2 n-1 n detD(T = 1 ! 1 ! n independent of the structure of the tree. (T ) , n (T ) n n = 1 = ! 1 This implies n n + - and so, N(T ) n = ! 1 n for any tree T tree with n vertices. n

  53. If T is a tree with n vertices then n ) ( ) (n ) 2 -2 n-1 n detD(T = 1 ! 1 ! n independent of the structure of the tree. (T ) , n (T ) n n = 1 = ! 1 This implies n n + - and so, N(T ) n = ! 1 n for any tree T tree with n vertices. n

  54. Some questions Some questions N(G) max{n (G),n (G)}? = Is it true that + - No! Take G = K . 2,3 n (G) = 2, n (G) = 3, N(G) = 4 - + What is the value of N(K ) in general? s,t (It is between s+t-2 and s+t-1). Why is n (G) so small in general? +

  55. Some questions Some questions N(G) max{n (G),n (G)}? = Is it true that + - No! Take G = K . 2,3 n (G) = 2, n (G) = 3, N(G) = 4 - + What is the value of N(K ) in general? s,t (It is between s+t-2 and s+t-1). Why is n (G) so small in general? +

  56. Some questions Some questions N(G) max{n (G),n (G)}? = Is it true that + - No! Take G = K . 2,3 n (G) = 2, n (G) = 3, N(G) = 4 - + What is the value of N(K ) in general? s,t (It is between s+t-2 and s+t-1). Why is n (G) so small in general? +

  57. Some questions Some questions N(G) max{n (G),n (G)}? = Is it true that + - No! Take G = K . 2,3 n (G) = 2, n (G) = 3, N(G) = 4 - + What is the value of N(K ) in general? s,t (It is between s+t-2 and s+t-1). Why is n (G) so small in general? +

  58. What does det D(G) mean? ) ( ) n-1 (n ) 2 -2 n detD(T = 1 ! 1 For example, ! for any tree T . n n n - 1 is the number of edges in T . n In general, one could look at the characteristic polynomial of D(G), i.e., det (D(G) – xI) (where I denotes the n by n identity matrix). The constant term is just det D(G). What do the other coefficients of det (D(G) – xI) mean? For G = T , we understand them (Graham/Lovász). n For example, the coefficient of x is 4 #( ) + 2 #( ) + 4 #( ) - 4

  59. What does det D(G) mean? ) ( ) n-1 (n ) 2 -2 n detD(T = 1 ! 1 For example, ! for any tree T . n n n - 1 is the number of edges in T . n In general, one could look at the characteristic polynomial of D(G), i.e., det (D(G) – xI) (where I denotes the n by n identity matrix). The constant term is just det D(G). What do the other coefficients of det (D(G) – xI) mean? For G = T , we understand them (Graham/Lovász). n For example, the coefficient of x is 4 #( ) + 2 #( ) + 4 #( ) - 4

  60. What does det D(G) mean? ) ( ) n-1 (n ) 2 -2 n detD(T = 1 ! 1 For example, ! for any tree T . n n n - 1 is the number of edges in T . n In general, one could look at the characteristic polynomial of D(G), i.e., det (D(G) – xI) (where I denotes the n by n identity matrix). The constant term is just det D(G). What do the other coefficients of det (D(G) – xI) mean? For G = T , we understand them (Graham/Lovász). n For example, the coefficient of x is 4 #( ) + 2 #( ) + 4 #( ) - 4

  61. Which graphs have valid addressings which use only 0’s and 1’s (i.e., no *’s)? That is, which graphs can be isometrically embedded in an N-cube? Theorem (Djokovi č ) Theorem G can be isometrically embedded into an N-cube if and only if for every edge {u,v} of G, the set of vertices S(u) which are closer to u than to v is closed under taking shortest paths, i.e., all shortest paths between any two vertices in S(u) stay within S(u). More generally, there is now a rather complete theory as to when graphs can be isometrically embedded in a cartesian product of smaller graphs (Graham/Winkler).

  62. Which graphs have valid addressings which use only 0’s and 1’s (i.e., no *’s)? That is, which graphs can be isometrically embedded in an N-cube? Theorem (Djokovi č ) Theorem G can be isometrically embedded into an N-cube if and only if for every edge {u,v} of G, the set of vertices S(u) which are closer to u than to v is closed under taking shortest paths, i.e., all shortest paths between any two vertices in S(u) stay within S(u). More generally, there is now a rather complete theory as to when graphs can be isometrically embedded in a cartesian product of smaller graphs (Graham/Winkler).

  69. Define N*(G) to be the least N for which a valid addressing of the directed graph G exists.

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