1. SL = L An Alternative Proof
Derandomized Squaring of Graphs
Eyal Rozenman and Salil Vadhan
Slides by Adam Segoli Schubert

2. Transition Matrix For a K-regular graph X :
Where AX is the adjacency matrix of X

3. λ(X)
Gives a measurement of the rate at which the random walk converges to uN.
X is undirected Equals to the second largest eigenvalue of MX in absolute value.
X is directed Equals to the square root of the second largest eigenvalue of MXtMX in absolute value.
Notice: MXtMX is an Hermitian operator - All eigenvalues are real.
מדד למהירות ההתלכדות עם ההתפלגות האחידה

4. Spectral Gap
We define the spectral gap of X as:

5. Graph Notation We denote a graph X as (N, D, λ)-graph when
X’s vertex set is of size N
X is D-regular graph
λ(X) ≤ λ

6. Labelling
For a K-outregular graph X. lab: E(X)  [K] For every vertex All outgoing edges have distinct labels.
v[a] – The neighbor of v via the edge labelled a  [K]
כל גרף רגולרי ניתן ללייבלינג
Undirected consistent labelling
לא טוב:
לא תמיד אפשרי (מס, אי זוגי של קודקודים)
לא נשמר תחת הפעולה שנציג

7. Consistent Labelling
Consistent Labelling A labeling where for every vertex all incoming edges have distinct labels. v 1 1 2 1 2 u w 2

8. Consistent Labelling
Notice: If a graph has consistent labelling it is K-inregular. Hence K-regular.
Undirected consistent labelling lab[(v, u)] = lab[(u, v)] in an undirected graph. v 1 1 2 2 2 u w 1

9. Graph Squaring X is an (N, K, λ)-graph.
The square X2 is an (N, K2, λ2)-graph.
When X is undirected Placing a clique on the K neighbors of each vertex.
Connecting all paths of length 2.
נשים לב שהדרגה גדלה בריבוע
בפועל נעבור עם גרפים מכוונים...

11. Derandomized Squaring
X is an (N, K, λ)-graph.
Using an auxiliary graph G on K vertices
Connecting only some paths of length 2.

12. Proposition 3.2
Let M be the transition matrix of an (N, D, λ)-graph then:
קליק עם סלף לופ
אינטואיציה
Operator Norm - Induced by Norm 2

13. Proof. Write .  . For :  It therefore suffice to show that Finally ■
For : 
It therefore suffice to show that
Finally ■
??????????????????????????
לצערנו C לאו דווקא מטריצת מעברים

14. Derandomized Squaring Definition
v[x] X:
y = x[a] x
X is a labelled (N, K, λ)
G is a labelled (K, D, μ)
X © G is on the N vertices of X.
Neighbors of v  [N] in X © G are:
v[x][y] such that y is a neighbor of x in G.
x, y  [K] labels in X and vertices in G.
Or Equivalently
v[x][x[a]] via the edge (x, a), where
x  [K] is an edge label in X and a vertex in G.
a  [D] is an edge label in G.
(x, a) v w G:
לוח
הוא לא בהכרח אין רגולרי אבל אם X consistently labelled
להוכיח בסוף????? a x y

15. Example u 1 1 1 3 3 3 t 2 w 2 2 2 1 2
(2, 1) 2 3 3 1 1 1 1 2 3 1 v 3 1 s
2[1]=3 1

16. Undirected Graphs u
Oy Vey! 2 2 2 2 2 1 1
(2,1) 1 3 1 1 2 4 2 1 4 v

17. Undirected Consistent Labelling2 w v 2 1 1 u
Oy Vey!! It is not Undirected
Consistently Labelled
(2, a) a 1 2 a

18. Derandomized Squaring Observations
X © G is KD-outregular
V[x][x[a]]
y = x[a]
(x, a) D v
v[x] K x

19. Derandomized Squaring Observations
X © G is KD-inregular
Is X © G consistently labelled?
(x, a) K w v D

20. Derandomized Squaring Observations
X © G is KD-inregular
Is X © G consistently labelled?
And if X is consistently labelled?
(y, b) y w
y[b] v y
(y, b)

21. Derandomized Squaring Observations
X © G is KD-inregular
Is X © G consistently labelled?
And if X is consistently labelled? y s
y[b]
 [K] w
y[b]
 [K] v y
Yes.

22. Theorem 4.4 X is an (N, K, λ) consistently labelled G is an (K, D, μ)
then X © G is an (N, KD, f(λ, μ)), where
f(λ, μ) = (1-μ)∙λ2 + μ∙1
f is monotone increasing in λ and μ and satisfies:
f(λ, μ) ≤ λ2 + μ
1 - f(1-γ, 1/100) ≥ (3/2) ∙ γ, when γ < ¼
נשים לב שאם 0  μ אזי λ2 f(λ, μ) כלומר X ® G טוב כמעט כמו שהיינו מצפים ש- 2 X יהיה.
אינטואיציה – כפי שראינו ניתן לפרק
מונוטונית עולה כי f(λ, μ) = 1-(1-λ2)∙(1-μ)
המטרה להגדיל גאפ
נראה (4.5) בהמשך כי החסם העליון ל-f הדוק.

23. 1 - f(1-γ, 1/100) ≥ (3/2) ∙ γ, when γ < ¼

24. Theorem 4.4 Proof.
Let M be the transition matrix of X © G we shall show that
Meaning that
נוכיח החסם לעה"ע השני של הפרודקט
להסביר על מטריצת מעברים

25. Theorem 4.4 Proof.
But how can we build it? We should first understand what is a random walk on X © G…

26. v1 K 1 2 4 v1 3 v1 4 K 2 v1 3 1 v1 2 1 v1 3 K 4 v2 K v2 1 4 2 3 v8 2 1 K 4 3 v8 vN vN 2 3 VN 1 K 4 1 3 4 K 2 VN

27. Step 1 Choose a  [K] uniformly at random, go to state (v, a).
Lifting the probability distribution on [N] to one on [N] x [K]
Choose a  [K] uniformly at random, go to state (v, a).
Go to state (v[a], a).
Go to state (v[a], b) where b is a random neighbor of a (in G).
Go to state (v[a][b], b).
Output v[a][b].

28. Step 2 Choose a  [K] uniformly at random, go to state (v, a).
A move on X
The NK x NK permutation matrix
Choose a  [K] uniformly at random, go to state (v, a).
Go to state (v[a], a)
Go to state (v[a], b) where b is a random neighbor of a (in G).
Go to state (v[a][b], b).
Output v[a][b].
עבור צומת v1 הלייבל a1 עובר רק לצומת אחר
פרמוטציה - בכל שורה ועמודה יהיה רק 1 יחיד כי X consistently labelled
A(v,a),(v’,a’) is 1 iff a’ = a and v’ = v[a]

29. Step 3 Choose a  [K] uniformly at random, go to state (v, a).
A move on G
Choose a  [K] uniformly at random, go to state (v, a).
Go to state (v[a], a).
Go to state (v[a], b) where b is a random neighbor of a (in G).
Go to state (v[a][b], b).
Output v[a][b].
טנזור
where B is the transition matrix of G

30. Step 4 Choose a  [K] uniformly at random, go to state (v, a).
Second move on X
The NK x NK permutation matrix.
Choose a  [K] uniformly at random, go to state (v, a).
Go to state (v[a], a).
Go to state (v[a], b) where b is a random neighbor of a (in G).
Go to state (v[a][b], b).
Output v[a][b].
?????

31. Step 5 Choose a  [K] uniformly at random, go to state (v, a).
Projection of the probability distributions on [N] x [K] to ones on [N]
Choose a  [K] uniformly at random, go to state (v, a).
Go to state (v[a], a).
Go to state (v[a], b) where b is a random neighbor of a (in G).
Go to state (v[a][b], b)
Output v[a][b].
This is inverse to Step 1 in the sense that

32. Proof. We can now conclude the transition matrix of X © G
By Proposition 3.2 we can decompose B:
Inducing:
Therefore:
A מטריצת המעברים של X
דפים

33. Proof.
We Have:
Therefore:

34. Proof.  Therefore for some matrix D: Implying that:
C טילדה היא טנזור ||A®B|| ≤ ||A||*||B|| ן-C קטן מ-1
||AB|| ≤ ||A||*||B|| מן המכפלה נקבל
דפים

35. Proposition 3.4
Let X be an (N, D, 1/(2N1.5))-graph. Then X contains an edge between any pair of vertices.
Indeed, for any pair of vertices v, w:
קליק כמעט מושלם

36. Proof. - Probability distribution of a random neighbor of v.
We shall show that every coordinate of ≥
,  It suffices to show that every coordinate of has absolute value ≥

37. As  
Let m be the minimal absolute value of coordinates: ■

38. A Log-Space Algorithm for Undirected Connectivity
USTCON
Decision problem: Given (X, s, t)
YES: There is a path between s and t in G.
NO: There is no such path.
Search problem: Given (X, s.t)
Output a s  t path if exists.
NO: If no such path exists.
X is an undirected graph.

39. USTCON Search problem Algorithm Overview
Given (X, s, t)
Construct a sequence of graphs Xm
Xm1 with degree poly(N). Contains a clique on each CC.
Enumerate all the neighbors of s, searching for t.
Finally we will show that neighbors in Xm1 are log-space computable.
שבוע שעבר אלגוריתם של עומר ריינגולד סדרה של גרפים עם מספר עולה של צמתים.
שלב ראשון - דרגה קבועה 100logN
שלב שני – דרגה פעמיים אספוננט loglogN+O(1)

40. Proposition 5.3 We may assume X is: 4-regular. consistently labelled.
contains a loop on each vertex.

41. Proof. Given an undirected graph X.
Define the 4-regular directed graph Xreg:
Its vertices are (v, i) for every v  V(X) and 0 ≤ i < deg(v).
The neighbors of (v, i) are:
(v, i)[1] = (v, i + 1 (mod deg(v))
(v, i)[2] = (v, i – 1 (mod deg(v))
(v, i)[3] = (v[i], location of v in the array of v[i])
(v, i)[4] = (v, i)
This operation can be done in logarithmic space.
דפים

42. Constructing a 4-regular graphv1 v v0 v2 s s0 s1 s2 w w1 w0

43. Constructing a 4-regular graph3 u0 v2 3 2 2 1 1 1 u1 4 v1 v0 4 2 3 3 s2 s1 w0 s0 w1

44. The Auxiliary Expanders
Lemma 5.1 For some constant Q = 4q, there exist a sequence of consistently labelled graphs:
Hm = (Qm, Q, 1/100)
Neighbors in Hm are computable in space O(m)
אנחנו הרי עושים פרודקט בין X למשהו...
כלומר בהנתן v є [Qm] ו-לייבל x є [Q] ניתן לחשב את v[x] ב- space O(m) and time poly(m)
למה צריך Log-space constructible??????????????????
לכתוב פרמטרים Hm

45. The Gm Sequence For a positive integer N, we set
We define a graph sequence Gm by:
לכתוב פרמטרים Gm
???????????????? הוכחה
Neighbors in Gm are computable in space

46. The Analysis Let X be a 4-regular graph with a loop on every vertex.
Assume X is connected.
Define inductively:
X1 = Xq is a
Xm+1 = Xm © Gm is a for m ≤ m0
for m > m0
אם הוא לא קשיר נבצע את התהליך על כל רכיב קשירות בנפרד זה גרף לא מכוון נקודה מעניינת
X1 כל המסלולים באורך לכל היותר q
הפעולה אכן מוגדרת כי Dm דרגת Xm
m ≤ m0 Dm = Qm
m > m0 Dm = Q^(m0+2m-m0-1)
ודרגת Gm
m ≤ m dm = Q
m > m0 dm = Q^(2m-m0)

47. Phase One Reducing the λ(Xm) to ¾
Lemma 3.1 Let X be a connected D-regular graph with a loop on every vertex then:
נגזר מהלמה שראינו שבוע שעבר אצל שרון
++להוכיח??

48. Phase One 1 - f(1-γ, 1/100) ≥ (3/2) ∙ γ, when γ < ¼
By lemma 3.3. we have
From Theorem 4.4 it follows that
As long as
Therefore for some m < 100logN we will get
Let then
הגאפ גדל אבל בסוף הגאפ של Xm כבר גדול מרבע וכבר לא ניתן להשתמש בגרפי עזר עם דרגה קבועה כי הע"ע השני של גרף העזר יהיה דומיננטי

49. Phase Two
Decreasing λ(Xm) to
Proposition 5.5 For m ≥ m0 we have

50. Proposition 5.5 Proof. Define . We shall show that for m ≥ m0.
For m = m0 we have
Suppose it holds for some m > m0.
Since : ■
קטן מ-0.75 כדרוש פשוט קטן ממיאית

51. Setting m1
Let
We have
The degree of Xm1 is

52. The Algorithm
Proposition 5.7 Neighborhoods in Xm1 are computable in space O(logN).

53. Proof. Edge labels in Xm are vectors
ym = (ym-1, am-1) = (y1, a1, …, am-1) where
y1 - edge label in X1
ai - edge label in Gi
ym corresponds to a path of length 2m in X.
Log-space algorithm:
given v, ym and an integer b  [1, 2m], returns the b-th edge label in X.
???????????????? 2m מה עם q

54. A Path in X2 v y1
y2=(y1, a1)
v[y1]
רמאות. u
y1[a1]
Remember X1 = Xq

55. A Path in Xm Two paths of length 2m-1 in Xm-1. uv
ym-1=(y1, a1, …, am-2)
ym=(y1, a1, …, am-1)
v[ym-1] u
ym-1[am-1]
Two paths of length 2m-1 in Xm-1.

56. Proof. The algorithm Given the problem encoded by (Xm, b, ym):
If b ≤ 2m-1 solve (Xm-1, b, ym-1)
If b > 2m-1 solve (Xm-1, b  b - 2m-1, ym-1[am-1]) v
v[ym-1]
ym=(y1, a1, …, am-1)
ym-1=(y1, a1, …, am-2)
ym-1[am-1]
דפים

57. Finally
When m = m1.
The input length is log|b|+log(ym) = log(2m) + log(Dm1) = O(logN).
A neighbor computation in Gi is preformed in space O(m + 2m-m0) = O(logN).
And we have a log-space algorithm.

58. Finally. Given (G, s, t) Enumerate all edges of s searching for t.
Every path in X is poly(N).
If t was found compute and output the path.