Structure and (pseudo-)randomness in combinatorics FOCS 2007 tutorial October 20, 2007
Terence Tao (UCLA)
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Structure and (pseudo-)randomness in combinatorics FOCS 2007 - - PDF document
Structure and (pseudo-)randomness in combinatorics FOCS 2007 tutorial October 20, 2007 Terence Tao (UCLA) 1 Large data In combinatorics, one often deals with high-complexity objects, such as Functions f : F n 2 R on a Hamming cube;
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2 → R on a Hamming cube;
2 in that Hamming cube Fn 2; or
2| = 2n and N as being very large,
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2 → R with
2 f(x) :=
1 2n
2 |f(x)| large;
2 with |A|/2n large;
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2 → R form a Hilbert space H
2 f(x)g(x).
2 can be identified with its indicator
2 → {0, 1}, which lies in H.
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2 → R which exhibit linear (Fourier)
2 → R which exhibit low-degree
2 which only depend on a few of the
2 (dictators, juntas);
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2 → {−1, +1} is “100%-linear” if we
2;
2 → {−1, +1} is “99%-linear” if we
2;
2 → {−1, +1} is “1%-linear” if we
2 + ε of all
2.
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2 → R which are
2 → R which are
2 in which each coordinate has small
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2 → R whose pair correlations
2 f(x)f(x + h) are small for most h ∈ Fn
2;
2 → R whose k-point correlations
2 f(x + h1) . . . f(x + hk) are small for most
2;
2 → R whose Gowers norms
2 ) := (EL:Fd 2→Fn 2 Ex∈Fn 2
2 f(x + Lω))1/2d
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2 → R has many large
2 → R has large Gowers
2 ) if and only if it has large correlation
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H over all fstr ∈ V . The latter proof is more
H.
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i∈I civi is such that |I| ≤ 1/ε2 and
i∈I civi is such that |fpsd, vi| ≤ ε
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if, vivi and the Plancherel
H = i |f, vi|2. But it is instructive to see
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H of fstr increases by at least ε2 (by Pythagoras’
H decreases
H ≤ f2 H ≤ 1. So the algorithm must
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2 (Fourier characters).
2 (Reed-Muller codewords).
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i=1 civi for
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H by at least ε2. Now
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H is increasing. By the pigeonhole
H − fstr,i−12 H ≤ ε.
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2 → R, (X, X, µ) is the space
2 with uniform probability measure µ and the
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L2 = E(f|Y)2 L2 + f − E(f|Y)2 L2
L2 = E(f|Y)2 L2+E(f|Y∨Y′)−E(f|Y)2 L2 40
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2 ) := (EL:Fd 2→Fn 2 Ex
2
2 → R for d ≥ 1. The dth Gowers norm
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2 ) = |Ex∈Fn 2 f(x)f(x + h)|1/2
2 f(x)|
2 ) = |Ex,h,k∈Fn 2 f(x)f(x + h)f(x + k)f(x + h + k)|1/4
2 ) = |Ex,h,k,l∈Fn 2 f(x)f(x + h)f(x + k) . . . f(x + h + k + l)|1/8
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2→Fn 2 Ex
2
2
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Ud = 1 − ε, then we have the identity
2 with probability
U2 = 1 − ε, then
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2 → {−1, 1} and d ≥ 1. Then fUd = 1
2 → {−1, 1}, d ≥ 1, and ε > 0.
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2 → {−1, 1}, d ≥ 1, and ε > 0.
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2 ) exactly. Clearly fU1(Fn 2 ) requires O(2n)
2 ) in O(n2n) computations. But even
2 )
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