Complexity Lower Bounds, P vs NP
& Gowers Blog
𝑷 ≠ 𝑵𝑷?
Try to prove they are different by using circuit complexity.
Define a function to be 𝑓: {0,1}𝑛 → {±1}
To built a circuit we define the following:
1) "Basic Functions" – 𝑓𝑖 (𝑥1 , … , 𝑥𝑛 ) = (-1)
2) "Basic Operations":
𝑥𝑖 +1
a. 𝑓 → -𝑓
b. 𝑓, 𝑔 → 𝑓 ∨ 𝑔
c. 𝑓, 𝑔 → 𝑓 ∧ 𝑔
3) Straight line composition: 𝑓1 , 𝑓2 , … , 𝑓𝑛 such that 𝑓𝑖 is either a basic function or
obtained from 𝑖1 , 𝑖2 < 𝑖 by basic operations.
The definition represents a DAG (Directional Acyclic Graph)
Definition: a function 𝐹 has (circuit) complexity 𝑚 if ∃ a circuit 𝑓1 , … , 𝑓𝑚 as above.
𝑃⁄
𝑃𝑜𝑙𝑦: All functions with a polynomial circuit complexity are our equivalent of 𝑃.
𝑓{{0,1}𝑛 → {±1}|𝑓 ℎ𝑎𝑠 𝑎 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑐𝑜𝑚𝑝𝑙𝑒𝑥𝑖𝑡𝑦 < 𝑛log log 𝑛 }
If we find ℎ ∈ 𝑁𝑃 𝑠. 𝑡. ℎ ∉ 𝑃⁄𝑃𝑜𝑙𝑦 then 𝑃 ≠ 𝑁𝑃.
For that purpose, lets define a complexity measuring function 𝜅.
Such 𝜅 must satisfy:
1) 𝜅(𝑓) = 1 if 𝑓 is basic
2) 𝜅(𝑓) = 𝜅(-𝑓)
3) If 𝜅(𝑓), 𝜅(𝑔) are small then 𝑓(𝑓 ∨ 𝑔) is small
4) 𝜅(𝑓) is large for some ℎ ∈ 𝑁𝑃
Attempts to find such 𝜿
Idea 1
Take the fourier representation of 𝐹 = ∑𝑆⊂[𝑛] 𝑓(𝑠)𝜒𝑠
𝑓̂𝑚𝑎𝑥 = max|𝑓(𝑠)|
𝑠
Define 𝜅(𝑓) = 𝑓
1
𝑚𝑎𝑥
𝑓: {0,1}𝑛 → {±1}
𝑛
𝑓 ∈ ℝ2
Instead of the standard base, we use the Fourier base:
𝑥𝑖
𝜒{𝑖} = (-1) .
∀𝑠 ⊆ {1, … , 𝑛}. 𝜒𝑠 = ∏ 𝜒(𝑖)
𝑖∈𝑠
Fourier base is: {𝜒𝑠 }𝑠
Problem
2
1 = 𝑇𝑂𝐷𝑂 = ‖𝑓‖22 = ∑|𝑓̂(𝑠)|
∀𝑓{0,1}2 → {𝐼1 } ∃𝑠 𝑠. 𝑡. |𝑓̂(𝑠)| >
And the function 𝑔(𝑥) = (-1)
𝑥1 𝑥2 +𝑥2 𝑥3 +𝑥3 𝑥4 ,…,𝑥𝑛 𝑥1
1
2𝑛
also has 𝑔̂𝑚𝑎𝑥 =
1
𝑛
22
Could there be a good measure 𝜿?
𝜅(𝑓 ∧ 𝑔) ≤ 𝜅 (𝑓) + 𝜅(𝑔)
} - a formal complexity measure
𝜅(𝑓 ∧ 𝑔) ≤ 𝜅(𝑓) + 𝜅(𝑔)
Related to the formula size of 𝑓 – 𝜆(𝑓)?
Formula trees with basic formulas on leaves and basic operations in internal nodes. Formula
size is the number of leaves.
Claim: For any formal complexity measure 𝜅 – 𝜅(𝑓) ≤ 𝜆(𝑓)
Proof: By induction
𝜅(𝑓 ∨ 𝑔) ≤ 𝜅(𝑓) + 𝜅(𝑔) ≤ 𝜆(𝑓) + 𝜆(𝑔)
Assume the smallest formula for f writes 𝑓 = 𝑔 ∨ ℎ, then
𝜆(𝑔) = 𝑟, 𝜆(ℎ) = 𝑠 → 𝜆(𝑓) = 𝑟 + 𝑠.
Note that the smallest formula might be even smaller!
𝜅(𝑓) = 𝜅(𝑔) + 𝜅(ℎ) ≤ 𝜆(𝑔) + 𝜆(ℎ) ≤ 𝜆(𝑓)
So far it seems as if there isn't a good 𝜅…
On one hand: 𝜅(𝑓) = 𝑓̂
1
𝑚𝑎𝑥
is bad for easy functions.
On the other hand, 𝜅(𝑓) = 𝜆(𝑓) is bad because tautological!
We haven't made any progress...
Natural Proofs – Razborov & Rudich
𝑛
{0,1}𝐴={0,1}
X
(0,1,1,0, … )
𝑥=⏟
|𝐴|
𝑋 ⊂ {0,1}𝐴
𝐹 ⊂ {{0,1}𝐴 → {±1}}
𝑋 is 𝜖 pseudo-random w.r.t. 𝐹 if
∀𝐹 ∈ ℱ |𝑃[𝐹(𝑥) = 1|𝑥 ∈ 𝑋] − 𝑃[𝐹(𝑥) = 1]| < 𝜖
Extreme opposite – 1𝑋 ∈ ℱ
𝑋 is pseudo random if every 𝐹 ∈ ℱ cannot distinguish 𝑥 ≈ 𝑋 from 𝑥 ≈ {0,1} 𝐴
Main point: Random functions of lowe complexity look like random functions w.r.t. a poly
time distinguisher.
Let 𝑋 = all points in {0,1}𝐴 with respect to polyline functions
𝛿 = {𝐹: {0,1}𝐴 → {±1}|𝐹 𝑖𝑠 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑡𝑖𝑚𝑒 (𝑖𝑛 𝑖𝑡𝑠 𝑖𝑛𝑝𝑢𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝐴)}
Statement: ∀𝜖 > 0. 𝑋 is 𝜖-pseudorandom w.r.t. ℱ.
A "natural proof" for 𝑃 ≠ 𝑁𝑃 would
- Devise a "simplicity" probability S of Boolean functions, so that 𝑠(𝑓) = 1 for all
simple (poly-time complexity) functions 𝐹 ∈ 𝑋.
- If 𝑆 itself is poly-time computable (in its input length - |𝐴| = 2𝑛 ) then since X is 𝜖pseudorandom for ℱ and 𝑠 ∈ ℱ, it follows that 𝑆(𝑓) = 1 for almost all 𝐹 ∈ {0,1}𝐴 .
This is bad because a random function 𝐹 ∈ {0,1}^𝐴 shouldn't be simple!
So either 𝑆 ∉ ℱ or 𝑆(𝑓) = 1 for almost all functions.
A proof is "natural" if it defines a simplicity property 𝑠 such that:
(1) All low complexity functions are simple
(2) A random function is not simple
(3) Whether or not a function is simple can be determined in poly-time
(4) Some NP-function is not simple
1,2,3 cannot hold together!
-------- end of lesson 1
Connection between P, NP and Circuits
𝐿 ⊂ {0,1}∗
𝐿𝑛 ≔ 𝐿 ∩ {0,1}𝑛
𝐿 = {𝐿𝑛 }∞
𝑛=1
The Clique language has a circuit complexity ≥ 𝑆(𝑛)
↔
∞
{𝑐
}
For any sequence of circuits 𝑛 𝑛=1 solves Clique
∃𝑛0 ∀𝑛 > 𝑛0 𝑠(𝐶𝑛 ) > 𝑠(𝑛)
The class 𝑃⁄𝑃𝑜𝑙𝑦 ≡ all languages computable by poly circuits.
The set of poly-time functions looks like the set of all functions
Looks like = To a simple observer (another polynomial time algorithm).
Exercise: Let 𝜅 be a formal complexity measure. Prove that if there exist ℎ: {0,1}𝑛 → {±1}
1
𝜅(ℎ) > 4 ∙ 𝑠, then 𝑃𝑟𝑜𝑏[𝜅(𝑓) > 𝑠] ≥ 4
𝑓 random, 𝑓: {0,1}𝑛 → {±1}
Things that are known

The discrete log function
Let ℤ𝑛 be the cyclic group with N elements
Let g be a generator
𝐺 = {𝑔1 , 𝑔2 , 𝑔3 , … } ℤ𝑝∗
𝑓𝑔 : 𝐺 → 𝐺 𝑓(𝑥) = 𝑔 𝑥
𝑓 -−1 ← discrete log function, is 1-1, believed hard to compute.
𝜖
CONJ – There exists some 𝜖 > 0 𝑠. 𝑡. the complexity of this problem is ≥ 2𝑛 .
Goldreich-Levin “Hard core bit”: Any one way permutation → gives rise to a pseudorandom generator.
A pseudo random generator is:
{0,1}𝑘 → {0,1}𝑘+1
Such that you cannot tell the difference between the half that emerged from the domain
and the half that didn’t (in the range).
𝑃𝑅𝐺 ( ⏟
𝑥 , ⏟
𝑟 ) = (𝑓(𝑥)
𝑟 , ∑ 𝑥𝑖 𝑟𝑖 𝑚𝑜𝑑 2)
⏟ ,⏟
⏟
𝑘
𝑘
𝑘
2
𝑏𝑖𝑡𝑠
2
𝑘
2
𝑏𝑖𝑡𝑠
{0,1}𝑘
𝑜𝑛𝑒 𝑚𝑜𝑟𝑒 𝑏𝑖𝑡
2𝑘
{0,1}
2
𝑃𝑅𝐺:
→
Now they constructed a pseudo random function generator
The took a seed (denoted 𝑦).
y
y(𝑔1 (𝑦))
y(𝑔0 (𝑦))
y(𝑔0 (𝑔0 (𝑦)))
y(𝑔1 (𝑔0 (𝑦)))
y(𝑔0 (𝑔1 ((𝑦)))
y(𝑔1 (𝑔1 ((𝑦)))
𝐹(𝑦, 𝑥) ≔ 𝑀𝑆𝐵 (𝑔𝑥𝑘 ∘ … ∘ 𝑔𝑥2 𝑔𝑥1 (𝑦))
Define 𝑓𝑦 : {0,1}𝑛 → {0,1} 𝑓𝑦 (𝑥) = 𝐹(𝑦, 𝑥)
Consider the distribution {𝑓𝑦 }𝑦∈{0,1}𝑘 , 𝑘 > 𝑛𝑐 𝑐 > 2

For each y 𝑓𝑦 is poly-time computable

This distribution is pseudo-random against polynomial time(in 𝟐𝒏 ) distinguishers
On the other hand…
𝜅, or any property being used in a lower bound proof, shouldn’t be too complex either!
𝜅(𝑓) = 1 iff 𝑓 has low circuit complexity is not good (trivial).
Note 𝜅 ∈ 𝑁𝑃!
Take the basic functions:
(- − 1)
⋮
100
𝑛
𝑥1
(- − 1)
⋮
𝑥2
(- − 1)
⋮
𝑥3
… (- − 1)
⋮
𝑥𝑛
A model for generating a random formula.
I have 𝑛 ∙ 𝑛100 ∙ 2 functions.
But this is false! Why? Because using AND or OR changes the distribution from
Gowers Norms (𝑼𝒌 𝑵𝒐𝒓𝒎𝒔)
Fix a finite set 𝐴 and consider ℝ𝐴 (the vector space of functions 𝑓: 𝐴 → ℝ.
A norm on this space is a function ‖𝑥‖: ℝ 𝐴 → ℝ+
s.t.
‖𝛼𝑓‖ = |𝛼|‖𝑓‖
𝑓 ≠ 0 → ‖𝑓‖ ≠ 0
‖𝑓 + 𝑔‖ ≤ ‖𝑓‖ + ‖𝑔‖
1
2
3
4
𝑓𝑜𝑟 0 to …
Example: ‖𝑓‖ = max|𝑓(𝑥)| (𝑜𝑛 𝑥 ∈ 𝐴)
Definition (dual norm):
1
‖𝑓‖∗ = max{⟨𝑓, 𝑔⟩|‖𝑔‖ ≤ 1} where ⟨𝑓, 𝑔⟩ = ∑𝑥∈𝐴 𝑓(𝑥)𝑔(𝑥) ← “Correlation”
|𝐴|
Example: For ‖𝑓‖∞, the dual is:
‖𝑓‖∗∞ = max{< 𝑓, 𝑔 > |‖𝑔‖∞ ≤ 1, ∀𝑥: |𝑔(𝑥)| ≤ 1}
1
1
max {∑ 𝑓(𝑥)𝑔(𝑥) ||𝑔(𝑥)| ≤ 1} =
∑ 𝑓(𝑥)𝑆𝑖𝑔𝑛(𝑓(𝑥)) =
∑|𝑓(𝑥)| = ‖𝑓‖1
|𝐴|
|𝐴|
In general: If ‖∙‖is in P, it doesn’t mean that ‖∙‖∗is in P.
Another Example: Let𝐴̂ be an abelian group
‖𝑓‖4 42 = 𝐸 [𝑓(𝑥)𝑓(𝑥 + 𝑎)𝑓(𝑥 + 𝑏)𝑓(𝑥 + 𝑎 + 𝑏)]
𝑥,𝑎,𝑏
Is it in P? YES.
1
Is ‖𝑓‖∗42 in P? Turns out that ‖𝑓‖∗42 = ‖𝑓̂‖4 = (∑𝑠⊆[𝑛](𝑓̂(𝑥)4 ))4
Exercise: Prove that the 𝑢2 norm is a norm. Hint: Cauchy’s norm.
However, can define 𝑢2 norm
TODO: Did not have enough time to copy the formula:
‖𝑓‖𝑢2 = ( 𝐸 (𝑓(𝑥)𝑓(𝑥 + 𝑎)𝑓(𝑥 + 𝑏)𝑓(𝑥 + 𝑎 + 𝑏)𝑓(𝑥 + 𝑐)𝑓(𝑥 + 𝑎 + 𝑐)𝑓()
𝑥,𝑎,𝑏,𝑐
Do not know a poly-time algorithm for ‖ ∙ ‖∗𝑢2
“Goal” for introducing these norms was to extend fourier analysis to “higher degree”.
𝑓: {0,1}𝑛 → {±1}. 𝑓̂(𝑠) = 𝑐𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒𝑎𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 (−1)∑𝑖∈𝑆 𝑋𝑖
𝑥𝑠 : {0,1}𝑛 → {±1} 𝑥2 (𝑥1 , … , 𝑥𝑛 ) = ∏𝑖∈𝑆(−1)𝑥𝑖 ←linear phase functions
Consider degree d phase functions (−1)^𝑔(𝑥̅ ) where deg 𝑔 ≤ 𝑑
Fix 𝑓: {0,1}𝑛 → {±1}
𝑓𝑦 :
{0,1}𝑛
𝑦 ∈ {0,1}𝑛
→ {±1} defined by 𝑓𝑦 (𝑥) = 𝑓(𝑥) ∙ 𝑓(𝑥 + 𝑦)
𝑓(𝑥) = (−1)(𝜙(𝑥))
Say, for instance:
𝑓(𝑥) = (−1)𝑥1 +𝑥2 +𝑥3 ∙ (−1)𝑥1 +𝑥2 +𝑥3 +𝑦1 +𝑦2 +𝑦3 = (−1)𝑦1 +𝑦+𝑦3 =constant! Doesn’t depend
on x.
If 𝑓(𝑥) = (−1)𝑞(𝑥) for deg 𝑞 ≤ 𝑑
Then 𝑓𝑦 (𝑥) = (−1)𝑞
′ (𝑥)
𝑤ℎ𝑒𝑟𝑒 deg 𝑞 ′ ≤ 𝑑 − 1
(−1)𝑞(𝑥) ∙ (−1)𝑞(𝑥+𝑦)
Define 𝑓𝑦,𝑧 ≡ (𝑓𝑦 )𝑧
Similarily: 𝑓𝑦1 …𝑦𝑑 = (… ((𝑓𝑦1 ) ) … )
𝑦2
---- end of lesson 2
𝑦𝑑
𝑓: {0,1}𝑛 → {±1}
𝑘
‖𝑓‖2𝑢𝑘 =
𝐸
𝑥1 …𝑥𝑘 ∈{0,1}𝑛
∏
𝑓(𝑥0 + 𝜖1 𝑥1 + ⋯ + 𝜖𝑘 𝑥𝑘 )
𝜖1 …𝜖𝑘 ∈{0,1}
𝑥0 = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑠𝑝𝑎𝑛 (𝑥1 … 𝑥𝑘 ) (dimension k affine subspace)
𝑥0 , 𝑥0 + 𝑥1 , 𝑥0 + 𝑥2 , 𝑥0 + 𝑥1 + 𝑥2
Example: Suppose 𝑓 is a linear function
∃𝑎1 , … , 𝑎𝑛 𝑠. 𝑡. 𝑓(𝑥) = ∑ 𝑎𝑖 𝑥𝑖 (𝑚𝑜𝑑2)
Then for any choice of 𝑥0 , 𝑥1 , 𝑥2
𝑓(𝑥0 ) + +𝑓(𝑥0 ) + 𝑓(𝑥1 ) + + ⋯ = 0
Hence for any choise of 𝑥0 , 𝑥1 , 𝑥2
+ + 𝑓(𝑥1 + 𝜖1 𝑥1 + 𝜖2 𝑥2 ) ≡ 0
𝜖1 ,𝜖2
Similarly the expectancy is 0 as well.
Linearity Testing
(proven by Blum-Luby-Rubinfeld)
𝑓: {0,1}𝑛 → {±1}
Question: Is 𝑓 a linear function?
Definition 1: ∃𝑎1 , … , 𝑎𝑛 𝑠. 𝑡. ∀𝑥. 𝑓(𝑥) = ∑ 𝑎𝑖 𝑥𝑖
Definition 2: ∀𝑥, 𝑦 𝑓(𝑥) + 𝑓(𝑦) = 𝑓(𝑥 + 𝑦)
Definition 2 implies definition 1 since:
We can define 𝑎𝑖 = 𝑓(𝑒𝑖 ), then 𝑓(𝑥) = 𝑓(∑ 𝑥𝑖 𝑒𝑖 )
𝑙𝑖𝑛𝑒𝑎𝑟𝑖𝑡𝑦
=
∑ 𝑥𝑖 ∙ 𝑓(𝑒𝑖 )…
Testing
Global object, e.g. 𝑓: {0,1}𝑛 → {0,1}
Want to test if 𝑓 ∈ 𝒫
In our example – all linear functions.
Only willing to invest limited resources, but willing to randomize.
Question: Can we deduce global property by considering local behavior?
If the answer is yes we say that this 𝒫 is testable.
Non testable property:
𝑥1 … 𝑥𝑛 - Boolean variables. ~𝑥1 … ~𝑥𝑛 - their negations.
Fom the list above, I select 𝑚 3-CNF clauses indices at andom.
Most of the time we will find clauses that don’t have any shared variables
If 𝑚 > 50𝑛 than with high probability 𝜑 is unsatisfiable!
PCP “theory” implies that every polynomialy varifiable property (e.g. formula satisfiability)
can be cast (“encoded”) in testable form.
Definition: ∀𝑓, 𝑔 {0,1}𝑛 → {0,1}
Distance(𝑓, 𝑔) = 𝑃𝑟𝑜𝑏𝑛 [𝑓(𝑥) ≠ 𝑓(𝑥)]
𝑥∈{0,1}
Distance(𝑓, 𝑆) = min 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒(𝑓, 𝑔)
f∈S
Theorem (BLR):
Let 𝑓: {0,1}𝑛 → {0,1} If distance (𝑓, 𝐿𝐼𝑁𝐸𝐴𝑅) ≥ 0
Then 𝑃𝑟𝑜𝑏(𝑓(𝑥) + 𝑓(𝑦) + 𝑓(𝑥 + 𝑦) ≠ 0) ≥ Ω(𝛿)
𝑥,𝑦
Theorem (AKKLR):
Let 𝑓: {0,1}𝑛 → {0,1} if 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒(𝑓, 𝐷𝐸𝐺𝑅𝐸𝐸(𝑘)) ≥ δ
Then 𝑃𝑟𝑜𝑏 (𝑋𝑂𝑅(𝑓(𝑥0 + 𝜖1 𝑥1 + ⋯ + 𝜖𝑘+1 𝑥𝑘+1 ))) ≥ Ω(𝛿 ∙ 2−𝑘 )
𝑥0 ,…,𝑥𝑘+1
So we select an affine space of 𝑘 + 1 points.
We look at all of these points and check that they are not zero.
𝑓 is a degree k function if
𝑓(𝑥1 , … , 𝑥𝑘 ) = ∑ 𝑎𝑆 ∏ 𝑥𝑖
|𝑆|⊆𝑘
𝑖∈𝑆
𝑥, 𝑦
𝑓(𝑥1 , … , 𝑥𝑘 ) = 𝑥1 𝑥2 … 𝑥𝑘
If 𝑓 has degree 𝑘 then ∀𝑥0 , … , 𝑥𝑘+1
𝑋𝑂𝑅(𝑓(𝑥0 + 𝜖1 𝑥1 + ⋯ + 𝜖𝑘+! 𝑥𝑘+1 ))
Proof: Let 𝑓𝑦 (𝑥) = 𝑓(𝑥 + 𝑦)
The function 𝑔 = 𝑓𝑥𝑘+1 ,𝑥𝑘 ,…,𝑥2 has degree 0 (it is constant)
The above expression equals 𝑔(𝑥0 )𝑋𝑂𝑅𝑔(𝑥0 + 𝑥1 ) ≡ 0
Claim: Let 𝑐𝑜𝑟𝑟(𝑓, 𝑘 − 1) be the correlation of 𝑓 with degree 𝑘 − 1 polynomials.
𝑐𝑜𝑟𝑟 = (1 − 𝛿) − 𝛿 = 1 − 2 ∙ 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒(𝑓, 𝑘 − 1 𝐷𝐸𝐺𝐸𝐸 𝐹𝑈𝑁𝐶𝑇𝐼𝑂𝑁𝑆)
𝑐𝑜𝑟𝑟(𝑓, 𝑘 − 1) ≤ ‖𝑓‖𝑢𝑘
Reed-Mulle (Low-Degee) Test
Let 𝑝 be the closest degree 𝑘 polynomial to 𝑓.
𝛿 ≔ 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒(𝑝, 𝑓)
1. 𝛿 is tiny – 𝛿 < 𝛽2−𝑘
If an affine 𝑘 + 1 space contains exactly one point 𝑥 𝑠. 𝑡. 𝑓(𝑥) ≠ 𝑝(𝑥) then the test
rejects.
We will prove that this happens with constant probability.
Assume 𝛿~2−𝑘
Choose a random affine subspace 𝐴 by choosing 𝑀𝑛×𝑙 a random full rank matrix
over 𝔽2 and a random 𝑏 ∈ 𝔽𝑛2
𝐴 = {𝑎𝑥 = 𝑀𝑥 + 𝑏|𝑥 ∈ 𝔽𝑙2 }
For each 𝑥 𝐸𝑥 − The event 𝑓(𝑥) ≠ 𝑝(𝑥)
𝐹𝑥 − 𝐸𝑥 and ∀ 𝑦 ≠ 𝑥 𝑓(𝑎𝑦 ) = 𝑝(𝑎𝑦 )
𝑎𝑥 is distributed uniformly in 𝔽𝑛2
𝑎𝑦 is distributed uniformly on 𝔽𝑛2 \{𝑎𝑥 }
∀𝑥 ≠ 𝑦𝑃𝑟𝑜𝑏[𝐸𝑥 ] = 𝛿
𝑃𝑟𝑜𝑏[𝐸𝑥 𝑎𝑛𝑑 𝐸𝑦 ] ≤ 𝛿 2
𝑀,𝑏
𝑃𝑟𝑜𝑏[𝐹𝑥 ] ≥ 𝑃𝑟𝑜𝑏[𝐸𝑥 ] − ∑ 𝑃𝑟𝑜𝑏[𝐸𝑦 ∧ 𝐸𝑥 ] ≥ 𝛿 − 2𝑙 ∙ 𝛿 2 ≈ 𝛿
𝑥≠𝑦
𝑃𝑟𝑜𝑏 (⋃ 𝐹𝑥 ) = ∑ 𝑃𝑟𝑜𝑏[𝐹𝑥 ] = 2𝑙 ∙ 𝛿 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑀,𝑏
2.
--- End of lesson 3
𝑥
𝑥
Last week we:
 Defined the 𝑢𝑘 norm ⇔Degree 𝑘 − 1 test (“low degree test”)
 Proved the following theorem: If ‖𝑓‖𝑢𝑘 > 1 − 𝛿 Then there ∃𝑝 of degree 𝑘 − 1
⟨𝑓, 𝑝⟩ = 𝐸 [𝑓(𝑥)𝑝(𝑥)] > 1 − 𝛿 𝑙
𝑥
Lemma 1: Let 𝑓: {0,1}𝑛 → {0,1}
𝑝 - some polynomial
Denote 𝑐𝑜𝑟𝑟(𝑓, deg 𝑘) = max{⟨𝑓, 𝑔⟩|𝑔 = (−1) 𝑓(𝑥) , 𝑝 − deg 𝑘}
So:
𝑐𝑜𝑟𝑟(𝑓, deg 𝑘) ≤ ‖𝑓‖𝑢𝑘+1
Lemma 2: For every ℎ: {0,1}𝑛 → {0,1}
‖ℎ‖𝑢𝑘 ≤ ‖ℎ‖𝑢𝑘+1
Proof of 2:
We shall use the fact that:
𝐸[𝑍 2 ] ≥ (𝐸[𝑍])2
𝑘+1
𝑘+1
‖ℎ‖2𝑢𝑘+1
=
𝐸
∏
𝑥
𝑦1 ,…,𝑦𝑘 𝜖 ,…,𝜖
𝑘+1
𝑦𝑘+1 1
ℎ (𝑥 + ∑ 𝜖𝑖 𝑦𝑖 )
𝑖=1
𝑘
=
𝐸
𝐸
𝑘
∏ ℎ (𝑥 + ∑ 𝜖𝑖 𝑦𝑖 ) ∏ ℎ (𝑥 + 𝑦𝑘+1 ∑ 𝜖𝑖 𝑦𝑖 )
𝑦1 ,…,𝑦𝑘 𝑥
𝑦𝑘+1 𝜖 ,…,𝜖
1
𝑘
𝑖=1
𝜖1 ,…,𝜖𝑘
𝑖=1
′
Now let’s fix: 𝑥 ← 𝑥
𝑦 ′ ← 𝑥 + 𝑦𝑘+1
𝑘
=
𝐸
𝐸
𝑘
′
′
∏ ℎ (𝑥 + ∑ 𝜖𝑖 𝑦𝑖 ) ∏ ℎ (𝑦 ∑ 𝜖𝑖 𝑦𝑖 )
𝑦1 ,…,𝑦𝑘 𝑥
𝑦𝑘+1 𝜖 ,…,𝜖
1
𝑘
𝑖=1
𝜖1 ,…,𝜖𝑘
𝑖=1
2
𝑘
=
𝐸
𝑦1 ,…,𝑦𝑘
(𝐸 ∏ ℎ (𝑥 ′ + ∑ 𝜖𝑖 𝑦𝑖 ))
𝑥
𝜖1 ,…,𝜖𝑘
𝑖=1
2
𝑘
≥( 𝐸
𝑥
𝑘+1
∏ ℎ (𝑥 ′ + ∑ 𝜖𝑖 𝑦𝑖 )) = ‖ℎ‖2𝑢𝑘
𝑦! ,…,𝑦𝑘 𝜖 ,…,𝜖
1
𝑘
∎
𝑖=1
Proof of 1:
For any ℎ: {0,1}𝑛 → {±1}
1
1) |𝐸 ℎ(𝑥)| = ‖ℎ‖𝑢′
𝑥
𝑝×3
(‖ℎ‖𝑢2
4 4
= (𝐸 |ℎ̂(𝑠)| ) )
𝑥
2) ∀𝑘 ‖ℎ‖𝑢𝑘 ≤ ‖ℎ‖𝑢𝑘+1
‖𝑓 ∙ 𝑝‖𝑢𝑘
3) ∀𝑓. ∀𝑝: {0,1}𝑛 → {±1} degree 𝑘 polynomial ⏟
𝑝𝑜𝑖𝑛𝑡𝑤𝑖𝑠𝑒
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛
= ‖𝑓‖𝑢𝑘+1
Proof for 1:
‖ℎ‖2𝑢1 = 𝐸
∏ ℎ(𝑥 + 𝜖𝑦1 ) = 𝐸
∏ ℎ(𝑥 + 𝜖𝑦1 )
𝑥
𝑥
𝑦1 𝜖=0,1
′
′
𝑦1 𝜖=0,1
Denote 𝑥 = 𝑥, 𝑦 = 𝑥 + 𝑦
= (𝐸 ℎ(𝑥)) (𝐸 ℎ(𝑦))
𝑥
𝑦
Proof for 3:
𝑘+1
‖𝑓 ∙
𝑘+1
𝑝‖2𝑢𝑘+1
=
𝐸
𝑥
𝑘+1
𝑘+1
∏
𝑦1 ,…,𝑦𝑘+1 𝜖 ,…,𝜖
1
𝑘+1
𝑓 (𝑥 + ∑ 𝜖𝑖 𝑦𝑖 ) 𝑝 (𝑥 + ∑ 𝜖𝑖 𝑦𝑖 ) = ‖𝑓‖2𝑢𝑘+1
𝑖=1
Because ∀𝑥, 𝑦1 , … , 𝑦𝑘+1 ∏𝜖1 ,…,𝜖𝑘 𝑥 +
𝑖=1
∑𝑘+1
𝑖=1 𝜖𝑖 𝑦𝑖
=1
Let 𝑝: {0,1}𝑛 → {±1} be the degree 𝑘 function closest to 𝑓 (i.e. attaining max correlation).
Define ℎ(𝑥) = 𝑓(𝑥) ∙ 𝑝(𝑥)
𝑐𝑜𝑟𝑟(𝑓, 𝑝) = |𝐸 𝑓(𝑥)𝑝(𝑥)|
𝑏𝑦 𝑠𝑡𝑒𝑝 1
𝑥
-
=
𝑏𝑦 𝑠𝑡𝑒𝑝 2
‖ℎ‖𝑢𝑘
≤
‖ℎ‖𝑢𝑘+1
𝑏𝑦 𝑠𝑡𝑒𝑝 3
=
‖𝑓‖𝑢𝑘+1 ∎
considered ‖𝑓‖𝑢3 as a “formal complexity measure”
Consider dual norms:
‖𝑓‖∗ = max{⟨𝑓, 𝑔⟩|‖𝑔‖ ≤ 1}
Motivation for dual:
1) “NP-ish” definition possibly circumvents RR
2) More robust
TODO: Draw world
Suppose we have two parts in our world. A and B
And we have two functions 𝑓, 𝑔 such that 𝑓 is random on A and is 1 on B and 𝑔 is the exact
opposite.
‖𝑓‖𝑢3 =constant.
‖𝑔‖𝑢3 =constant as well.
ℎ =𝑓∨𝑔
‖ℎ‖𝑢3 = 𝑧𝑒𝑟𝑜!
This is a problem! We just use an or and got such a dramatic difference
‖ℎ‖∗ ≥ ⟨ℎ, ℎ⟩ = 1
⟨ℎ, 𝛼ℎ⟩ = 𝛼
1
Can take 𝛼 = ‖ℎ‖ - very large!
‖𝑓‖∗𝑢3 =?
‖𝑓‖∗𝑢3
Need to find a “norming function”
Use ℎ!
‖
ℎ
‖=1
‖ℎ‖
1
ℎ
1
1
‖𝑓‖∗ ≥ ⟨𝑓, ⟩ =
𝐸 𝑓(𝑥)ℎ(𝑥) = 𝑃𝑟𝑜𝑏(𝐴) ⏞
𝐸𝑓(𝑥)𝑔(𝑥) + 𝑃𝑟𝑜𝑏[𝐴] ∙ 𝐸ℎ(𝑥) ∙ 1 = 2 ∙
‖ℎ‖
‖ℎ‖
𝑥
1
‖ℎ‖
=very large!
Note: ⟨𝑓, 𝑓⟩ ≤ ‖𝑓‖ ∙ ‖𝑓‖∗
Question 1: Given 𝑓: {0,1}𝑛 → {±1} Can we compute ‖𝑓‖∗𝑢𝑘 in polytime?
(an open question even for k=3)
Question 2: Suppose you know that ‖𝑓‖∗𝑢3 ≥ 𝜖 by [Samorodnitsky ‘07] ∃ deg 2 polynomial
2
that correlates with 𝑓. Can P be found in time poly(2𝑛 )? (search space is 2𝑛 )
Gappalan-Klivans-Zuckerman ’08: “list-decoding Reed-Muller Codes”.
If 𝑓 is 𝜖-correlated with some degree-2 function, then the following is true:
1) Number of deg 2 polynomials correlation with 𝑓 ≤ 2𝑂𝜖 (𝑛)
2) Can find list above in time 2𝑂𝜖 (𝑛)
Interpretation: if ‖𝑓‖ small then ‖𝑓‖∗ is large
Simplicity Property = {𝑓|‖𝑓‖∗ 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙} ⊆ {𝑓|‖𝑓‖ 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒}- This is a property in 𝑃!
If the first if 𝑃1 and the second is 𝑃2 , then the first one that contains all functions.
Next idea: 𝑢𝑘 norm for super-constant 𝑘.
Now the naïve algorithm for computing ‖𝑓‖𝑢𝑘
Takes (2𝑛 )𝑘 −time. If 𝑘 = 𝑘(𝑁) this is not polytime
𝑁 = 2𝑛 , 𝑁^𝐾
Still intent to use dual ‖ ‖∗𝑢𝑘 norm (want robustness)
Question 3: Is there an algorithm for ‖ ‖𝑢𝑘 running in time better than 𝑁 𝑘−1 ∙ log 𝑁.