2 Setup for proof: vertex labels v 1 , v 2 , . . . , v n such that - PowerPoint PPT Presentation

I SOPERIMETRIC PROBLEMS S ATU E LISA S CHAEFFER elisa.schaeffer@tkk.fi T-79.7001, A PRIL 24 2006

T HE ISOPERIMETRIC PROBLEM Among all closed curves of length ℓ , which one encloses the maximum area ? For graphs : separator problems (vertex and edge cuts ) — relations between the cut sizes and the sizes of the separated parts

V OLUME AND BOUNDARY • Notation : graph G = ( V, E ( G )) , set S ⊂ V , | V | = n � • Volume : vol S = d v v ∈ S • Edge boundary : ∂S = {{ u, v } ∈ E ( G ) | u ∈ S, v / ∈ S } • Vertex boundary : δS = { v / ∈ S | { u, v } ∈ E ( G ) , u ∈ S }

R ELATED PROBLEMS Given a fixed integer m , find a subset S with m ≤ vol S ≤ vol ¯ S s.t. 1. the boundary ∂S = {{ u, v } ∈ E ( G ) | u ∈ S, v / ∈ S } contains as few edges as possible 2. the boundary δS = { v / ∈ S | { u, v } ∈ E ( G ) , u ∈ S } contains as few vertices as possible

C HEEGER CONSTANT | ∂S | h G = min vol S, vol ¯ � � min S S From the definition, we get for S s.t. vol S < vol ¯ S that | ∂S | ≥ h G · vol S . Also, G is connected iff h G > 0 .

V ERTEX EXPANSION | δS | | δS | g G = min S } , Regular graphs: g G ( S ) = min { vol S, vol ¯ min {| S | , | ¯ S |} S Definition : (volume replaced by unit measure) | δS | g G = min ¯ min {| S | , | ¯ S |} S

L EMMA : 2 h G ≥ λ 1 Setup for the proof: • C is a cut that achieves h G • C splits V into sets A and B 1  vol A, if v ∈ A,     • Definition: f ( v ) = 1   − vol B , if v ∈ B  

E XPRESSION FOR λ 1 ( f ( u ) − f ( v )) 2 � u ∼ v λ 1 = λ G = inf ( f ( v )) 2 d v � f ⊥ T 1 v

P ROOF OF 2 h G ≥ λ 1 , PART 1 2 Using the definition of λ 1 with definitions of vol S , C and f , we get the result. First we simply “partition” the expression using A and B : � ( f ( u ) − f ( v )) 2 u ∼ v λ 1 = inf ( f ( v )) 2 d v � f ⊥ T 1 v ( f ( u ) − f ( v )) 2 + ( f ( u ) − f ( v )) 2 + � � � ( f ( u ) − f ( v )) 2 u ∈ A, u ∈ A, u ∈ B, v ∈ B v ∈ A v ∈ B = � � ( f ( v )) 2 d v + ( f ( v )) 2 d v v ∈ A v ∈ B

P ROOF CONTINUES , PART 2 2 We use the definitions of f and vol : « 2 „ 1 1 X vol A + + 0 + 0 vol B u ∈ A,v ∈ B λ 1 = d v d v X X ( vol A ) 2 − ( vol B ) 2 v ∈ A v ∈ B « 2 „ 1 1 | C | vol A + vol B = 1 1 ( vol A ) 2 · vol A + ( vol B ) 2 · vol B „ « 1 1 = | C | vol A + vol B

T HEOREM : λ 1 > h 2 G 2 Setup for proof: • vertex labels v 1 , v 2 , . . . , v n such that f ( v i ) ≤ f ( v i +1 ) ( 1 ≤ i ≤ n − 1 ) � � • w.l.o.g. d v ≥ d v f ( v ) < 0 f ( v ) ≥ 0 • cuts C i = {{ v j , v k } ∈ E ( G ) | 2 ≤ j ≤ i < k ≤ n } , 1 ≤ i ≤ n

D EFINITIONS FOR THE PROOF | C i | • Definition : α = min � � � 1 ≤ i ≤ n � min d j , d j j ≤ i j>i • By definition α ≥ h G (divisors are the volumes of the parts) • V + = { v ∈ V | f ( v ) ≥ 0 } � • E + = { u, v } ∈ E ( G ) | u ∈ V + , v ∈ V }  f ( v ) , if v ∈ V + ,  • g ( v ) = 0 , otherwise 

H ARMONIC EIGFN f OF L WITH EIGVAL λ 1 For any v ∈ V , it holds for f that 1 � � � f ( v ) − f ( u ) = λ 1 f ( v ) d v u ∼ v 1 � � � ⇒ λ 1 = f ( v ) − f ( u ) ( † ) d v f ( v ) u ∼ v (a lemma from the previous chapter)

P ROOF OF THE THEOREM , PART 1 8 Substituting λ 1 = ( † ) and summing over V + 1 X ` ´ = f ( v ) − f ( u ) ( † ) λ 1 d v f ( v ) u ∼ v X X ` ´ f ( v ) f ( v ) − f ( u ) v ∈ V + { u,v }∈ E + = ( △ ) X ´ 2 d v ` f ( v ) v ∈ V + because for any subset S ⊆ V , we have X ` ´ λ 1 f ( v ) d v = f ( v ) − f ( u ) u ∼ v X λ 1 ( f ( v )) 2 d v ` ´ = f ( v ) f ( v ) − f ( u ) u ∼ v X X X ( f ( v )) 2 d v ` ´ λ 1 = f ( v ) f ( v ) − f ( u ) v ∈ S v ∈ S u ∼ v

P ROOF OF THE THEOREM , PART 2 8 From the defs of g , V + and E + (as ( f ( u )) 2 > 0 and g ( v ) ≥ f ( v ) ), X X ` ´ f ( v ) f ( v ) − f ( u ) v ∈ V + { u,v }∈ E + = ( △ ) λ 1 X ´ 2 d v ` f ( v ) v ∈ V + ( f ( v )) 2 − f ( v ) f ( u ) X ` ´ v ∈ V + { u,v }∈ E + = X ´ 2 d v ` f ( v ) v ∈ V + ´ 2 X ` g ( u ) − g ( v ) { u,v }∈ E + > ( ∗ ) X ´ 2 d v ` g ( v ) v ∈ V

P ROOF OF THE THEOREM , PART 3 8 Using the Cauchy-Schwarz inequality ( � x i y i ) 2 ≤ ( � x 2 i )( � y 2 i ) with x i = | g ( u ) − g ( v ) | and y i = g ( u ) + g ( v ) , we get � 2 � 2 � � � � g ( u ) + g ( v ) g ( u ) − g ( v ) { u,v }∈ E + { u,v }∈ E + λ 1 > � 2 · ( ∗ ) � 2 d v � � � � g ( u ) + g ( v ) g ( v ) v ∈ V { u,v }∈ E + � � �� � | g ( u ) − g ( v ) | g ( u ) + g ( v ) u ∼ v ≥ � 2 �� � 2 d v � 2 g ( v ) v

P ROOF OF THE THEOREM , PART 4 8 Now using ( a + b )( a − b ) = a 2 − b 2 , we get � � �� � | g ( u ) − g ( v ) | g ( u ) + g ( v ) u ∼ v λ 1 ≥ � 2 �� � 2 d v � 2 g ( v ) v � 2 � � 2 − � 2 | � � | g ( u ) g ( v ) ≥ � 2 �� � 2 d v � 2 g ( v ) v

P ROOF OF THE THEOREM , PART 5 8 Now from the definition of C i and “partitioning” the edges to “steps” over the cuts C i , we continue � 2 �� � 2 − � 2 | � � | g ( u ) g ( v ) u ∼ v λ 1 ≥ � 2 �� � 2 d v � 2 g ( v ) v � 2 �� | ( g ( v i )) 2 − ( g ( v i +1 )) 2 | · | C i | i = . � 2 �� ( g ( v )) 2 d v 2 v

P ROOF OF THE THEOREM , PART 6 8 Using the definition of α together with the fact that � � d v ≥ d v and the vertex ordering, we get f ( v ) < 0 f ( v ) ≥ 0 � 2 � � | ( g ( v i )) 2 − ( g ( v i +1 )) 2 | · | C i | i λ 1 ≥ � 2 � � ( g ( v )) 2 d v 2 v � 2 � � ( g ( v i )) 2 − ( g ( v i +1 )) 2 · α � d j i j>i ≥ . � 2 � � ( g ( v )) 2 d v 2 v

P ROOF OF THE THEOREM , PART 7 8 � 2 � � i ( g ( v i )) 2 − ( g ( v i +1 )) 2 � j>i d j � 2 d i +1 � n − 1 � g ( v i +1 ) i =0 = = 1 � 2 � 2 d v � n � � � g ( v ) v ( g ( v )) 2 d v v =1 as when we multiply the nominator “open”, all but one of the � 2 cancel out, appearing both positive and negative, except � g ( v i +1 ) for once for j = i + 1 , which leaves the same summation than we have in the denumerator.

P ROOF OF THE THEOREM , PART 8 8 Now we simply take out α 2 and use the previous observation and the definition of α to complete the proof: � 2 � � ( g ( v i )) 2 − ( g ( v i +1 )) 2 · α � d j = α 2 2 ≥ h 2 i j>i G λ 1 ≥ 2 . � 2 � � ( g ( v )) 2 d v 2 v

C HEEGER INEQUALITY Putting together the lemma and the theorem, we have 2 h G ≥ λ 1 > h 2 G 2 .

� 1 − h 2 I MPROVEMENT : λ 1 > 1 − G From the proof of the previous theorem we have λ 1 = ( △ ) and we define W = ( ∗ ) : � � � � f ( v ) f ( v ) − f ( u ) v ∈ V + { u,v }∈ E + λ 1 = � 2 d v � � f ( v ) v ∈ V + � 2 � � g ( u ) − g ( v ) { u,v }∈ E + > = W � 2 d v � � g ( v ) v ∈ V

P ROOF OF THE SECOND THEOREM Again we extend and use some already familiar tricks (plugging in the def. of W itself): � 2 � 2 � � � � g ( u ) + g ( v ) g ( u ) − g ( v ) { u,v }∈ E + { u,v }∈ E + W = � 2 · � 2 d v � � � � g ( u ) + g ( v ) g ( v ) v ∈ V { u,v }∈ E + � 2 � � | ( g ( u )) 2 − ( g ( v )) 2 | u ∼ v ≥ � � � � � � � ( g ( v )) 2 d v ( g ( v )) 2 d v − W ( g ( v )) 2 d v · 2 v v v

P ROOF CONTINUES Rewriting the nominator just as in the previous proof, simple factorization of the denominator gives � 2 � � | ( g ( v i )) 2 − ( g ( v i +1 )) 2 | · | C i | i W ≥ � 2 � � ( g ( v )) 2 (2 − W ) d v v � 2 � � | ( g ( v i )) 2 − ( g ( v i +1 )) 2 | · α � d j i j>i ≥ � 2 � � ( g ( v )) 2 (2 − W ) d v v α 2 = 2 − W

α 2 I NTERMEDIATE RESULT : W ≥ 2 − W ⇒ W 2 − 2 W + α 2 ≤ 0 . √ 1 − α 2 . Solving the zeroes gives W ≥ 1 − By definitions of W and α , we have λ 1 > W and α ≥ h G . Hence we � 1 − h 2 have proved the theorem λ 1 > 1 − G . Note that h 2 � G 1 − h 2 2 < 1 − G whenever h G > 0 (i.e., for any connected graph), meaning that this is always an improvement to the previous lower bound.

C ONSTRUCTIONAL “ COROLLARY ” In a graph G with eigfn f associated with λ 1 , define for each v ∈ V � � C v = { u, w } ∈ E ( G ) | f ( u ) ≤ f ( v ) < f ( w ) and � − 1 � � � α = min | C v | · min d u , d u . v u u f ( u ) ≤ f ( v ) f ( u ) >f ( v ) √ 1 − α 2 . Then λ 1 > 1 −

L OWER BOUND ON λ 1 2 For a connected simple graph G , h G ≥ vol G . From Cheegers inequality, 2 h G ≥ λ 1 > h 2 G 2 , we have � 2 � λ 1 > 1 2 . 2 vol G As vol G = 2 | E ( G ) | ≤ n ( n − 1) ≤ n 2 , we get a lower bound λ 1 ≥ 2 n 4 .