class 11 - stanford universityweb.stanford.edu/.../cs265/lectures/lecture11/inclass11.pdf · 2020....

Class 11Practice with LLL and Second Moment Method

Plan for today

• More practice with LLL• Application to k-SAT• (Closure on the example set up in the minilecture video!)

• More practice with the second-moment method• Isolated vertices in random graphs

𝑘-SATLLL for h - SAT Question Say each var appears in E X clauses

.

What's the biggest you can take X to be andstill conclude that y is satisfiable? Elin lams of k)

Hint LLL . Bad events are the event that y is NOTsat .

if= ( x, VXZVXJ V - - - VX

, ) A (XaV XIV - - -Vxao) A - - .

I• n variables

• k literals per clause X, ,Xz , . . . . Xn

• m clauses

The Say each clause has EXACTLY k literals,

and each variable appears in E 2k-4k Clauses.

-Then y is satisfiable .

this is our t

• For today, each clause has exactly 𝑘 distinct variables.• Goal: a statement of the form:

As long as each variable appears in no more than ______ clauses, then 𝜑 is satisfiable.

Group work! Use the LLL to do this.

Solutions to group work.

LLL for h - SAT Question Say each var appears in E X clauses.




, ) A (XaV XIV - - -Vxao) A - - .

I• n variables


• m clauses




this is our t

Setting up the LLLUse Lovasg Local Lemma .

Let Xi, Xz , . . ., Xn be TRUE or FALSE randomly (

"Ini form)

Ai = { ith clause NOT satisfied }

Pf Ai ] = 112k ← only one out of 2k waysto NOT satisfy a clause .

Use Lovasg Local Lemma .


"Ini form)





"Ini form)



What is the parameter “d”?



"Ini form)



Ai is mutually indeep . ofSi .- { ai : TI:b.bg" him:b::info }

(No matter how we set variables in clause j , won't affect Ai )

f#i sit . I mating" Inimitable:" ) #Igf . tk vars in

( each of themClause i

is in s tclauses

.





is in s tclauses

.





is in s tclauses

.





is in s tclauses

.

⇒ Each Ai is mutually indep . ofall but d : = ht clauses.

other events.

LLL with D= ht, p - 2-

k:

If d p e 14 , then Pf Ai Ai ] s ( t - 2pm > O- -

aka, k t - 2- KE 44 aka, y is satisfied

aka t = 2h44


other events.


k:



aka t = 2h44

Applying the LLL⇒ Each Ai is mutually indep . ofall but d : = ht clauses

.

other events.


k:



aka t = 2h44


other events.


k:



aka t = 2h44k

Conclusion

LLL for h - SAT Question Say each var appears in E X clauses.




, ) A (XaV XIV - - -Vxao) A - - .

I• n variables


• m clauses




this is our t

• For example, if 𝑘 = 10, then as long as each variable appears in at most !!

"# = 25.6 clauses (aka, in ≤ 25 clauses), then 𝜑 is ALWAYS satisfiable!!• No matter how many variables or how many clauses!

Next up:

Practice with the second moment method• Consider a random graph 𝐺!,#• 𝑛 vertices, and each of the

𝑛2 edges is present with probability 𝑝,

independently.

• If 𝑝 < $%& !'!

, say, then in expectation, there are isolated vertices:• Pr[ 𝑣 is isolated ] = 1 − 𝑝 $%" ≈ 𝑒%& $%" ≈ "

$• Thus 𝐄 number of isolated vertices ≈ 𝑛 ≫ 1.

• How likely is it that we actually have isolated vertices?• Maybe it’s the case that with tiny probability, tons of vertices are

isolated, but with some decent probability there are no isolated vertices?

v

The 𝑛 − 1 other vertices

None of these edges exist.

Group work!

SolutionsIt’s unlikely that there are no isolated vertices.

• Let G - Gn , p , p = ctongn for cc 1.

• Let X = # isolated vertices inG , so X = ¥ I{ v is isolated}• Notice that IP{v is isolated} = ( t -p)

"- t

a exp f- in -hp) -- n'" ""sync

• EX = [vP{ vis isolated ) - Mnc = n'"← since cc e , ni

-c ⇒ I

• Varlxl = LECK ] - CEXT



"- t



-c ⇒ I




"- t



-c ⇒ I




"- t



-c ⇒ I


Let’s quickly sanity-check that the expected number of isolated vertices is large:

Second moment methodWe need to compute the variance



"- t



-c ⇒ I

• Varlxl = LECK ] - CEXT• ElX' ] = E@If isolated})

'

= IE Er Er If isolated } I { isolated}

rPl ::: :::B

-

- Erm:*!iii.is +⇐Misiak:'T }=n . (n

- c

) t n in - i ) ( n-" )

• ElX' ] = E@If isolated})'


rPl ::: :::B

-


- c

) t n in - i ) ( n-" )



rPl ::: :::B

-


- c

) t n in - i ) ( n-" )



rPl ::: :::B

-


- c

) t n in - i ) ( n-" )

What is thisterm?

This term is ≈ 𝑛 ⋅ 𝑛!"

Pr[ both 𝑢 and 𝑣 are isolated ]for 𝑢 ≠ 𝑣

n-2 edges¥#-

•⇐-• U

v←n - 2

edges.

U and vboth isolated

⇐ noneof theseun - 2) t1 edgesexisting

n-2 edges¥#-

•⇐-• U

v←n - 2

edges.

U and vboth isolated

⇐ noneof theseun - 2) t1 edgesexisting

P {9%3%7} - H -plan -3 - exp f- Can -3) p )

= N- c ( 2 - 3h )

un n- 2C

SECOND MOMENT METHOD :

PIX -03 ⇐ TEETH - Tna .- no

P {9%3%7} - H -plan -3 - exp f- Can -3) p )

= N- c ( 2 - 3h )

un n- 2C



P {9%3%7} - H -plan -3 - exp f- Can -3) p )

= N- c ( 2 - 3h )

un n- 2C



P {9%3%7} - H -plan -3 - exp f- Can -3) p )

= N- c ( 2 - 3h )

un n- 2C



𝑝 =𝑐 ⋅ ln 𝑛𝑛

Returning to our computation



rPl ::: :::B

-


- c

) t n in - i ) ( n-" )



rPl ::: :::B

-


- c

) t n in - i ) ( n-" )



rPl ::: :::B

-


- c

) t n in - i ) ( n-" )



rPl ::: :::B

-


- c

) t n in - i ) ( n-" )



rPl ::: :::B

-


- c

) t n in - i ) ( n-" )

T Nt- c

t n211 - c )

• Var [XI -- Efx. ] - (EXTanti-o,Te Gi-ok. nai -d

un nl - C

To finish up…T Nt- c

t n211 - c )

• Var [XI -- Efx. ] - (EXTanti-o,Te Gi-ok. nai -d

un nl - C

P {9%3%7} - H -plan -3 - exp f- Can -3) p )

= N- c ( 2 - 3h )

un n- 2C


PIX -03 ⇐ TEETH - Tna .- noSince cat , this is of 1) AM

Recap

• More practice with the LLL and Second Moment Method!• You’ll get even more on your HW J

• We saw how the LLL applies to k-SAT – if you haven’t already watched minilectures for Wednesday on the algorithmic LLL, this will come upagain.

class 11 - stanford universityweb.stanford.edu/.../cs265/lectures/lecture11/inclass11.pdf · 2020....

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