Timoteo Carletti & Duccio Fanelli3) Weak Collatz conjecture: all orbits are bounded Proof Assume...

www.unamur.be [email protected]

September the 22nd, 2016, Amsterdam, The Netherlands

On the Collatz conjecture: a contracting Markov walk

on a directed graph.

Timoteo Carletti & Duccio Fanelli


Acknowledgements

Several colleagues, among which:

Jean-Charle Delvenne, Craig Alan Feinstein, Steffen Kionke, Vassilis Papanicolaou and Cédric Villani

IAP$VII/19$)$DYSCO$


The Collatz map T (Lothar Collatz, 1937)

8n 2 N T (n) =

(n2 if n is even

3n+ 1 if n is odd.

1 ��!even

2/2 = 1��!odd

3⇥ 1 + 1 = 4��!even

4/2 = 2

��!odd

3⇥ 3 + 1 = 103 ��!even

10/2 = 5��!odd

5⇥ 3 + 1 = 16

16 ��!even

16/2 = 8 ��!even

8/2 = 4 ! 2 ! 1

! 22! 11! 34! 17! 52! 267 ! 13! 40! 20

! 10! 5! 16! 8 ! 4! 2! 1

Numerically verified up to 1.9x1017 T. Oliveira E Silva, Math. Comp. 68 No. 1 (1999), 371


The Collatz map T (Lothar Collatz, 1937)

Conjecture For any given positive integer the iterates of the Collatz map eventually converge to the period-3 cycle {1,2,4} as .

n0 2 Nnk = T �k(n)

k ! 1

J.C. Lagarias, "The Ultimate Challenge: the 3x+1 problem". American Mathematical Society, (2010).


Main ideas of the proof

1) Introduce the third iterate of the Collatz map and consider the equivalence classes of integer numbers modulo 8;

2) Define a finite state Markov chain with a suitable invariant probability measure, whose transition probabilities reflect the deterministic map;

3) Prove a weak version of the conjecture: diverging orbits cannot exist due to the bound on the average stationary distribution;

4) Prove the conjecture by excluding other periodic orbits.


1) The third iterate of the Collatz map S=To3

B(k, 8) := {n 2 N : 9m 2 N, n = 8m+ k} k 2 {0, . . . , 7}

The 3-cycle {1,2,4} becomes a fixed point

S(n) =

8>>>>>>>>>>>>><

>>>>>>>>>>>>>:

n8 if n 2 B(0, 8)6n+2

8 if n 2 B(1, 8)6n+4

8 if n 2 B(2, 8)36n+20

8 if n 2 B(3, 8)6n+8

8 if n 2 B(4, 8)6n+2

8 if n 2 B(5, 8)6n+4

8 if n 2 B(6, 8)36n+20

8 if n 2 B(7, 8) .

8n 2 N

contraction by 1/8contraction by 3/4contraction by 3/4

contraction by 3/4contraction by 3/4contraction by 3/4

expansion by 9/2

expansion by 9/2


2) A finite states Markov process

8 symbols alphabet: B(k, 8) k = 0, . . . , 7

Transition probabilities:q

⇤ij = P [S(x) 2 B(j, 8)|x 2 B(i, 8)]

q⇤ij :=µinv

⇥B(i, 8) \ S�1B(j, 8)

⇤

µinv [B(i, 8)]

invariant (probability) measureµinv

Transition prob of Sok is the kth power of the transition prob of S


2) Markov process. Transition probabilities: an intuition

Let thenn = 8k 2 B(0, 8) S(n) = n/8 = k

with probability 1/8, S(n) belongs to any one of the eight classes

with probability 1/2, S(n) belongs to

Let thenn = 3 + 8k 2 B(3, 8)S(n) = (36n+ 20)/8 = 16 + 36k = 8(4k + 2) + 4k

B(0, 8) B(4, 8)or to


2) Markov process. Invariant measure

Let , there exist and n 2 N s0, . . . , sk�1 2 {0, . . . , 7}k � 1

n = sk�18k�1 + · · ·+ s18 + s0

⌫(s0) =1

6if s0 = 0, 2, 4, 6 and ⌫(s0) =

1

12if s0 = 1, 3, 5, 7.

µinv(sk�1 . . . s1s0) =1

2k1

7

1

8k�2⌫(s0)

Note: on all integers.µinv > 0


2) Markov process. Transition probabilities

7

0

6

5

4

3

2

1

Q⇤ =

0

BBBBBBBBBB@

18

18

18

18

18

18

18

18

0 14 0 1

4 0 14 0 1

414 0 1

4 0 14 0 1

4 012 0 0 0 1

2 0 0 014 0 1

4 0 14 0 1

4 014 0 1

4 0 14 0 1

4 00 1

4 0 14 0 1

4 0 14

0 0 12 0 0 0 1

2 0

1

CCCCCCCCCCA

Q* is a stochastic matrix

8i = 0, . . . , 7X

j

q⇤ij = 1

q⇤ij :=µinv

⇥B(i, 8) \ S�1B(j, 8)

⇤

µinv [B(i, 8)]


2) Markov process. Some properties

The Markov process with stochastic matrix Q* is ergodic.

visiting frequencies of classes B(k, 8)Contracting/expanding property of S

and

fQ⇤ =

✓1

8

◆1/6 ✓3

4

◆2/3 ✓9

2

◆1/6

=3

4< 1

average contracting/expanding factor:

~PS = (1/6, 1/12, 1/6, 1/12, 1/6, 1/12, 1/6, 1/12)

The left eigenvector associated to is � = 1 ~PSQ⇤ = ~PS

µinv(B(k, 8)) = PS,k


3) Weak Collatz conjecture: all orbits are bounded

ProofAssume there exists a diverging orbit for S. Observe it is a possible realisation of the Markov chain.

Given any , then there exists a positive integer M(n0) such that for all k>0.

n0 2 NS�k(n0) M(n0)

limk!1

#{0 j k � 1 : S�j(n0) 2 B(i, 8)}k

= µinv (B(i, 8)) 8i 2 {0, . . . , 7}

Because of the ergodic theorem, for - almost allµinv n0 2 N

Contradiction with the unboundedness of the orbit.

Hence for this orbits it holds true: fQ⇤ < 1

µinvn0 2 N

The existence of zero measure sets can be ruled out because for allµinv(n0) > 0


4) Collatz cycle: {1,2,4} is the unique attracting cycle

Proof

Assume there exists a L-periodic orbit for S with L > 1. Observe it is a possible realisation of the Markov chain, hence it should infinitely visit the 8 classes.

No periodic orbits of period different from 1 exist for S.

Invoking the ergodic theorem this orbit should have fQ⇤ < 1

Contradicting the periodicity (that would require )fQ⇤ = 1

µinvThe existence of zero measure sets can be ruled out as previously done


4’) Markov process with absorbing state

Let us observe that once we have proved the weak Collatz conjecture we can conclude using the following finite state Markov process

8

7

0

6

5

4

3

2

1

8>>>>>>>>>>>>>>><

>>>>>>>>>>>>>>>:

C0 = B(0, 8)C1 = B(1, 8) \ {1}C2 = B(2, 8) \ {2}C3 = B(3, 8)C4 = B(4, 8) \ {4}C5 = B(5, 8)C6 = B(6, 8)C7 = B(7, 8)C8 = {1, 2, 4}


Help save Dmitry!

Vsevolod Salnikov is currently postdoc in the Namur Center for Complex Systems. His son

Dmitry has been diagnosed with acute lymphoblastic leukaemia . The best chances of beating the disease are given by a cutting edge CAR T-cell therapy, but the treatment is currently only accessible in Seattle (USA). The family is

currently trying to collect the sum for the treatment, as it is not covered by standard

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More information on : http://dmitry.be/ or ask the Namur people (Timoteo Carletti, Sarah de Nigris, Renaud Lambiotte, Michael

Schaub, etc.)


September the 19th, 2016, Amsterdam, The Netherlands

On the Collatz conjecture: a contracting Markov walk on a directed

graph.

Timoteo Carletti & Duccio Fanelli


Markov process. Invariant measure

µinv(B(�, 8)) =1

6if � = 0, 2, 4, 6,

µinv(B(�, 8)) =1

12if � = 1, 3, 5, 7 and µinv(N) = 1

µinv

⇥S�1B(j, 8m)

⇤= µinv [B(j, 8m)]

For any and m � 1 j = 0, . . . , 8m � 1

Probability measure

Invariance

Timoteo Carletti & Duccio Fanelli3) Weak Collatz conjecture: all orbits are bounded Proof Assume...

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