# ppt - IBM ```The Complexity of Linear
Dependence Problems in
Vector Spaces
David Woodruff
Joint work with Arnab Bhattacharyya, Piotr
Indyk, and Ning Xie from MIT
The 3-SUM Problem
• Given a set S containing r real numbers, are
there: a, b, c 2 S with a+b+c = 0?
• Solve in O(r2) time
– Interview question
• Conjectured to require ~(r2) time
• Useful for hardness results in P. Many problems
are “3-SUM Hard”
Generalizations
• We study generalizations of this problem:
– Replace 3 summands with k summands
– Replace real field R with a finite field
– Replace sum of field elements with sum of vectors
– Replace sum with a fixed linear combination
– Replace sum with any linear combination
– Require vectors be minimally linearly dependent
– Replace target 0 with an arbitrary vector
– and so on…
Applications
• Maximum Likelihood Decoding
- Given x1, …, xr in Fqn and z in Fqn, do there exist xi1, …, xik that
contain z in their span?
- xi are the columns of a parity-check matrix
- z is the syndrome
- there is a codeword corrupted in at most k positions with
syndrome z iff the k-span contains z
• Weight Distribution Problem
– Let A be an n x r matrix over F2
– Define the code C = {x | Ax = 0}
– C has a codeword of weight k iff k columns of A sum to 0
Formal Definitions
• In this talk, we focus on two problems:
• (k,r)-LinDependence: given r elements x1, …, xr in F2n
and z in F2n, do there exist
xi1, …, xik that span z?
• (k,r)-ZeroSum: given r elements x1, …, xr in F2n, do there
exist xi1, …, xik with
xi1 + xi2 + … + xik = 0?
• We allow k and r to be functions of n
• First problem at least as hard as second
Results
• Assume 3-SAT cannot be solved in time
less than 2cn for a constant c &gt; 0
• Then (k,r)-ZeroSum requires min(rk, 2n) time, up
to polynomial factors
– So, (k,r)-LinDependence requires min(rk, 2n) time
– Other variants also require this time
• Have matching upper bound:
– rk is trivial. Can get roughly rk/2
– Can get 2n with the FFT
Implications
• (k,r)-LinDependence reduces to Maximum Likelihood
Decoding, so min(rk, 2n) lower bound
• (k,r)-ZeroSum reduces to the Weight Distribution
Problem, so min(rk, 2n) lower bound
1/4
k
r
• Results improve previous best
lower bounds for these
coding theory problems [Downey, Fellows, Vardy, Whittle]
• Hold for r and k functions of n
Our starting point: [PW] showed an r(k) bound for k-SUM
over R assuming 3-SAT on n variables requires 2cn time:
3-SAT formula F with n variables and m clauses
[CIP]
- s = 2εn. Each &Aacute;i
has n variables
and O(n) clauses
This ensures bit complexity
of resulting numbers is
small
…
&Aacute;1
&Aacute;2
&Aacute;s
- &Aacute;i replaced with 1in-3-SAT formula &Atilde;i
&Atilde;1
&Atilde;s
&Atilde;2
…
- &Atilde;i converted to
k-SUM instance
Each k-SUM instance on a set of r = 2Θ(n/k) real numbers. If can
solve k-SUM in time ro(k), can solve 3-SAT in time ro(k) &cent;2εn
Reducing a 1-in-3-SAT formula &Atilde;i on n variables and O(n)
clauses to k-SUM on r = 2Θ(n/k) real numbers
•
Partition variables into k groups G1, …, Gk of n/k variables
…
Gk
1 true iff…
&Atilde;G
there areGki real numbers
i is
that sum to 1k + O(n)
vi,1
vi,2
vi,3
…
vi, 2n/k
• In each group Gi, create a real number vi,j for each possible
assignment to its n/k variables
vi,j:
Base-k
representation
•
•
•
k group indicator digits
O(n) clause digits
i-th indicator digit is 1 iff v 2 Gi
j-th clause digit is 1 iff A(v) sets exactly 1 literal of j-th clause to 1
All other digits are 0
Can we do the same for F2?
•
Partition variables into k groups G1, …, Gk of n/k variables
- A sum of k vectors over F2 can equal
k+O(n), but just means an odd number
1G
…
Gi
…
1
of literals in each clause are true
v
v
i,1
i,2
- Odd-SAT
is
easy
vi,3
…
Gk
vi, 2n/k
• In each group Gi, create a real
vector
number
vi,j for veach
each possible
assignment to its
i,j for possible
assignment
n/k
variablesto its n/k variables
vi,j:
k + O(n)
Base-k
representation
coordinates
•
•
•
k kgroup
groupindicator
coordinates
digits
O(n) clause
O(n) coordinates
clause digits
i-th indicator coordinate
digit is 1 iff is
v 21 G
iffi v 2 Gi
j-th clause coordinate
digit is 1 iff A(v)
is 1 iff
sets
A(v)exactly
sets exactly
1 literal1 of
literal
j-th clause
of j-th clause
to 1 to 1
All other digits
coordinates
are 0 are 0
Our Modifications
3-SAT formula
F withthis
n variables
and
- Before
was used
form clauses
bit complexity.
- s = 2εn. Each &Aacute;i
has n variables
and O(n) clauses
- &Aacute;i replaced with
NAE-SAT
Formula &Atilde;i
- &Atilde;i converted to
(k,r)-ZeroSum
- Now it determines the
of &Aacute;
dimensions…
&Aacute;number
2 formula &Atilde; is 1 if
-1We
A NAE-SAT
need interaction i
for
each clause,
at least one
between
groups
but not
all literals are
- With
1-in-3-SAT
overtrue
R,
&Atilde;1
&Atilde;2
…
variables in different groups
independently
update the
clause digit
[CIP]
&Aacute;s
&Atilde;s
Each (k,r)-ZeroSum instance on a set of r = 2Θ(n/k) vectors.
If can solve (k,r)-ZeroSum in time ro(k), solve 3-SAT in time ro(k) &cent;2εn
Interacting Variables
• We can replace duplicates of a variable with distinct variables and
introduce equality constraints
– preserve NAE-SAT and &middot; 3 literals per clause
– each variable occurs in a constant number of clauses
Gi still has O(n/k)
variables
┐
• For each clause (a &Ccedil; b &Ccedil; c), we introduce pairvairs
– 1 variable is [a, b], 1 variable is [b, c], and 1 variable is [c, a]
• Partition original n variables into k groups Gi of n/k variables
• For a pairvar [a,b],
– if original variables a and b occur in the same group Gi, place [a,b] in Gi
– else, if a 2 Gi and b 2 Gj, place [a,b] in Gmin(i, j)
New Reduction
• In each group Gi, create a vector vi,j for each assignment to its n/k
variables as well as variables in Gi’s pairvars
vi,j1:
k+O(n)
coordinates
pair of consistency coordinates for
each pairvar (a,b)
k group
coordinates
O(n) clause
coordinates
O(n) consistency
coordinates
• i-th group coordinate is 1, the others are 0
• clause coordinates more complicated
• depend on variables and pairvars assigned to the group
• consistency coordinates allow for assignments to the same
variable from different groups to be patched together
Clause Coordinates
• Clause coordinates are set so that for a consistent
assignment (i.e., group and consistency coordinates are
ok), then for clause with literals a, b, c
– v(a) + v(b) + v(c) – v(a) &cent; v(b) – v(b) &cent; v(c) – v(a) &cent; v(c)
– v(.) denotes the value assigned
• Case analysis
– Clause only equals 1 if exactly 1 or 2 literals are true
Upper Bounds
• Consider functions f: Z2n ! {0,1}
• Fourier transform F: Z2n ! R is F(x) = 2-n&cent;y f(y)&cent;(-1)&lt;x,y&gt;
• Fast Fourier Transform computes F from f in O(n&cent;2n) time
• Let f be indicator function of input set of r vectors. Then
sum v1+ v2 + … vk = 0 f(v1) &cent; f(v2)  f(vk) is what we want
• This is just 2n times the 0n-Fourier coefficient of fk
• So we can get O(n&cent;2n) time instead of the trivial 2nk
Conclusion
Assuming 3-SAT cannot be solved in time less than 2cn
for a constant c &gt; 0,
– (k,r)-LinDependence and (k,r)-ZeroSum require
min(rk, 2n) time (up to polynomial factors)
– Same bound holds for many similar problems
– Almost matching upper bounds
– New way to prove hardness in coding theory
– Optimal hardness of basic problems in coding theory
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