Shuffling by semi-random transpositions

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Shuffling by semi-random transpositions
Elchanan Mossel, U.C. Berkeley Joint work
with Yuval Peres and Alistair Sinclair
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Shuffling by random transpositions

• At each step choose two independent uniformly
  chosen cards and exchange them.

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Shuffling by random transpositions

• ThmDiaconis-Shahshahani-81 The mixing time of
  the random transpositions shuffle is (½ o(1)) n
  log n.
• One can prove an O(n log n) upper bound can
  using marking (more later).
• Proof of an ?(n log n) lower bound
• At each step touch 2 random cards.
• Until time (n log n)/4 there are ?(n1/2)
  untouched cards
• ) permutation is not random.

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The cyclic to random shuffle

• At step i exchange card at location (i mod n)
  with a uniformly chosen card.

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History of the cyclic to random shuffle

• Shuffle introduced by Thorp (65).
• Aldous and Diaconis (86) asked what is the mixing
  time?
• Mironov posed again and proved O(n log n) upper
  bound using marking.

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Why do we care?

• General question Is systematic scan faster than
  random update? (other examples Diaconis-Ram
  Benjamini-Berger-Hoffman-M for asymmetric
  exclusion Gaussian fields etc.).
• Would be nice to find a natural problem where
  the mixing time is strictly between ?(n) and ?(n
  log n)
• Mironov Cyclic to random may tell us a lot
  about a widely used crypto algorithm RC4.

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The RC4 algorithm
More than 106 hits in google

• Mironov Lets study algorithm assuming j is
  random.
• Slow mixing corresponds to weak crypto.

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Upper Bounds - Broders Marking

• Broders Marking argument
• Call the two pointers Lt and Rt.
• Start by marking the first card that is pointed
  by L1.
• At time t, mark card pointed by Lt if either
• The card at Rt is marked or
• Rt Lt.

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Broders Marking
L
R
RL
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Broders marking

• By induction Given the time and
• set of marked cards and
• their positions,
• the permutation on the marked cards is uniform.
• ) The time when all cards are marked is a
  strong uniform time (permutation is random given
  the time).
• In order to prove upper bound, need to bound the
  marking time.
• For random transpositions easy By coupon
  collector estimate this time is O(n log n).
• Mironov delicate analysis for cyclic to random.

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A general n log n upper bound

• Thm M-Peres-Sinclair An O(n log n) upper bound
  on the mixing time holds for any shuffle where
• At step t we exchange cards Lt and Rt where
• Rt are i.i.d. uniform in 0,,n-1.
• The sequence Lt is independent of Rt.
• Lt can be random, deterministic etc.
• Cyclic to random is given by Lt t mod n.
• Top to random is given by Lt 0 for all n.
• Random transpositions is given by Lt i.i.d
  uniform.
• Pf Careful analysis of the marking process.

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A general n log n upper bound

• Proof In more detail
• May assume that Lt is deterministic.
• Partition time into intervals of length 2n.
• In such an interval look at pairs of times s lt t
  such that Ls Lt (there are at least n such
  pairs).
• We can mark card x if
• at time s, x is chosen by Rs.
• Rr ? Lt for s lt r lt t.
• Rt is one of the marked cards.
• Letting mi (ui) be the (un)-marked card at
  interval i, gives
• Eui1 Fi ui (1 c mi) for c gt 0.
• Will skip the rest of the proof.

Rs
Ls
x
Lt
x
Rt
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Cyclic to random shuffle lower bound?

• Mironov proved c n lower bound for some c gt 1
  using parity as a test function
• Each shuffle changes the parity with probability
• (1 1/n).
• After t steps, resulting parity original parity
  with probability

• Q Is next to random faster than random
  transpositions?
• Note All cards are touched by time n.

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n log n lower bound for cyclic to random shuffle

• ThmM-Peres-Sinclair
• The cyclic to random shuffle has a mixing time
  ?(n log n).
• More precisely

• And here is how the proof goes

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Step 1 Homogenizing the chain

• Problem The chain is not time homogenous.
• Can be easily fixed Consider a chain where at
  time t
• ?(0) swaped with ?(U), where U is uniform.
• Rotate all cards to the left ?(k) ?(k1 mod
  n).
• Clearly chain is equivalent
• It is homogenous.
• From now on study homogenized chain.

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One card chain
Markov chain for a single card

• Eigenvalues satisfy ? (1 1/n) ? where
• (n-1)?n n ?n-1 1 0.
• Want to show slow mixing ) want ? close to 1.

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Asymptotics of eigen values and functions

• ? (1 1/n) ? where (n-1)?n n ?n-1 1 0.
• Let ?-1 1 z/n and get
• (1z/n)n n (1z/n) (n-1) 0 ! ez z 1
  0.
• Lemma 1 ez z 1 has non-zero complex roots.
• Lemma 2 If ? is a root, then M has an eigenvalue
  ? such that 1-? (1lt ?)/n O(1/n2).
• Lemma 3 The eigenvector f corresponding to ? is
  smooth f1 C f2. Will write f for
  either.
• Pfs Complex analysis
• Remark Numerically, the smallest non-zero root
  is
• ? 2.088 7.416 i

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The test function

• Take f to be an eigenfunction of M corresponding
  to the eigenvalue ? closest to 1.
• Define the test function F

• Easy Ef 0 ) EF 0.
• Easy EF(idt) ?t f2.
• A Longer calculation gives

E(F2) f4/n
E(F) 0
E(F(idt)) ?t f2
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The main Lemma
E(F2) f4/n

• Remains to bound EF(idt)2.
• Main Lemma

E(F) 0
E(F(idt)) ?t f2

• ) as long as ?2t (4t n)/n2 the idt and ?
  (where ? is uniform) have large total variation
  distance (2nd moment method).
• Since 1 - ? O(1/n)
• ) ?1 ?(n log n)

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Proof of main Lemma

• The main lemma can be proved using Wilsons
• method and the properties of ? and f.
• Or it can be done more directly using coupling
• Lemma

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Proof of main Lemma

• Pf idea Couple the following two processes
• Process 1 cards i and j move independently.
• Process 2 The location of cards i and j in the
  real process.
• In process 1

• Remains to bound the difference between the
  processes
• using coupling.
• Will skip the details

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Conclusion and Open problems

• Weve seen that the mixing time of the
  pseudo-random next to random shuffle has the same
  mixing time as the random transposition shuffle.
• Proof is not that hard.
• Problem How general is the phenomenon?
• In particular
• Open problem Are there any sequences
  (deterministic/random) It, such that the It to
  random shuffle mixes in less than n log n time?

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