Coding for DNA Storage in Live Organisms - Unicamp · (Source:Wikipedia) Theorem(Perron-Frobenius(Partial)) IfG isaprimitivegraphthen: 1 = (AG) ,maxfj j: isaneigenvalueofAGg also

Coding for DNA Storagein Live Organisms

Moshe Schwartz

Electrical & Computer EngineeringBen-Gurion University

Israel

Based on joint works with: (alphabetically)

• Jehoshua Bruck – Caltech

• Ohad Elishco – Ben-Gurion University (now MIT)

• Farzad Farnoud (Hassanzadeh) – University of Virginia

• Siddharth Jain – Caltech

• Yonatan Yehezkeally – Ben-Gurion University

Introduction 2 / 79

Science fiction distant future dream?

Introduction 3 / 79

No – It’s just around the corner!

Introduction 4 / 79

DNA is a long string

Genetic information is stored in DNA, which is astring of nucleotides: Adenine, Cytosine, Guanine,and Thymine.

In E. coli bacteria, genetic information is stored inabout 4 · 106 base pairs.In humans, genetic information is stored in over3 · 109 base pairs.

Introduction 5 / 79

Why store information in DNA?

DNA is dense!

It stores information in the molecular level.

DNA can potentially hold 250 · 250 bytes (250 peta-byte) of informationin 1 gram of DNA.

If we were to use 8Tb hard-drives to store the same amount, we’ll need32000 hard-drives, with a total weight of about 25 tons!

Introduction 6 / 79

OK, but why in living organisms?

• Reading from DNA is destructive, hence we need several copies.Living organisms replicate and solve this problem.

• Data longevity is (potentially) better, due to replication oforganisms.

• The organism’s outer shell provides extra protection.

• Labeling organisms for biological studies.

• Watermarking genetically modified organisms (GMOs).

Main disadvantage:

Mutations!

We need error-correcting codes.

Introduction 7 / 79

Error-correcting codes – An age old story

An error-correcting code has two maincomponents:

1 An error ball: Its size and shape depend onthe kind of errors the channel induces.

2 A packing of error balls: Its density affectscommunication efficiency. Its structureaffects ease of encoding/decoding.

Introduction 8 / 79

What kinds of errors do we expect?

Insertion Duplication

Substitution Deletionu v w

u w

u v w

u v w

u v′ w

u v w

u w

u v v w

Which is the most common? Unknown yet, but…

Introduction 9 / 79

Repeated sequences are everywhere

More than 50% of human genome is repeated sequences!1

Repetitions were shown to be connected with diseases such as cancer,myotonic dystrophy, Huntington’s disease, and important phenomenasuch as chromosome fragility, expansion diseases, silencing genes, andrapid morphological variation.

Repetitions are common in other species as well, and are claimed to bea major evolutionary force during vertebrate evolution.1

1Lander et al., Nature 2001.Introduction 10 / 79

Duplication processes may repeat

ACTCA⇓

ACTACTCA⇓

ACTATACTCA⇓

ACTATACACTCA

It is conceivable that a substantial portion of the unique genome, thepart that is not known to contain repeated sequences, also has itsorigins in ancient repeated sequences that are no longer recognizabledue to change over time.2

2Lander et al., Nature 2001.Introduction 11 / 79

Duplication processes may differ

Palindromic Duplication Interspersed Duplication

End Duplication Tandem Duplicationu v w

u v w

u v w

u v w z

u v w v

u v vR w

u v v w

u v w v z

Introduction 12 / 79

A formal definition

Definition

Let Σ be a finite alphabet, s ∈ Σ∗ some string, and T ⊆ Σ∗Σ∗a set of

string-duplication rules. A string-duplication system, S, defined by thetuple (Σ, s, T ), is the reflexive transitive closure of T operating on s,namely, S ⊆ Σ∗ is the minimal set for which:

1 s ∈ S.

2 s′ ∈ S and T ∈ T imply T(s′) ∈ S.

We write S = S(Σ, s, T ).


End duplication - formally

Definition (End Duplication)

Tendi,k (x) =

{uvwv if x = uvw, |u| = i, |v| = k

x otherwise.

T endk =

{Tendi,k

∣∣∣ i > 0}.

The end-duplication system is defined as Sendk = S(Σ, s, T end

k ).

u v w

u v w v


Tandem duplication - formally

Definition (Tandem Duplication)

Ttani,k (x) =

{uvvw if x = uvw, |u| = i, |v| = k

x otherwise.

T tank =

{Ttani,k

∣∣∣ i > 0}.

The tandem-duplication system is defined as Stank = S(Σ, s, T tan

k ).

u v w

u v v w


How expressive is a duplication system?

Definition

The capacity of a string system S ⊆ Σ∗ is defined by

cap(S) = lim supn→∞

log2 |S ∩ Σn|n

.

Definition

Let S ⊆ Σ∗ be a string system. We shall say S is fully expressive if forevery v ∈ Σ∗ there exist u,w ∈ Σ∗ such that uvw ∈ S.


We are interested in:

• How does the capacity depend on the choice of duplication rules?

• How does the capacity depend on the choice of seed string?

• Which systems are fully expressive?

• What is the connection between capacity and full expressiveness?


Some related previous work exists

Tandem duplication was studied in the context of formal languages:

• Martín-Vide and Paun, Acta Cybernetica (1999):Where are tandem-duplication languages located in the Chomskyhierarchy?

• Dassow, Mitrana and Paun, Bull. of the EATCS (1999):Binary tandem-duplication languages are regular.

• Ming-Wei, Bull. of the EATCS (2000):Non-binary tandem-duplication languages are irregular.


More related previous work exists

Tandem duplication was studied in an algorithmic context:

• Main and Lorentz, J. Alg. (1984), Gusfield and Stoye, J. Comp. andSystems Sci. (2004):How to efficiently find tandem duplications in a string.

• Matroud, Hendy, and Tuffley, Nucleic Acids Research (2011):How to efficiently find nested tandem duplications.

• Elemento et al., Molecular Bio. and Evolution (2002), Lajoie et al.,J. Comp. Biology (2007), Brejová et al., Phil. Trans. R. Soc. A (2014):How to reconstruct the derivation process of a tandem-duplicatedstring.


End duplication has full capacity

Theorem

For Sendk = S(Σ, s, T end

k ), |s| > k,

cap(Sendk ) = log2 |Σ| .

AssumptionThe initial string s contains every symbols of Σ at least once.

End Duplication 20 / 79

End duplication has full capacity (Cont.)

Proof.

6: We obviously have,

cap(Sendk ) = lim sup

n→∞

log2∣∣Send

k ∩ Σn∣∣

n

6 lim supn→∞

log2 |Σn|n

= log2 |Σ| .



Proof.

>: We claim that starting with any string s ∈ Σ>k, with each symbolappearing at least once, and any w = w1w2 . . .wk ∈ Σk, we can derive astring y with w as a suffix.Step I: Duplicate prefix. Assume s = uv, |u| = k, then

s = uv ⇒ uvu = s′.

Observation: Every symbol of Σ appears in the beginning and end of ak-substring of s′.Step II: Force w1 at the end.

k

w1 w1⇒



Proof.

Step III: Force w1w2 at the end.

k

w2 w1 w1 w2⇒

and then

k

w1w2 w1w2⇒

Repeat Step III inductively to get w1w2 . . .wk as a suffix.



Proof.

Step IV: Repeat previous steps to get every k-word from Σk as asubstring.Thus, after at most 2k |Σ|k duplications we get a string s′′ containing allpossible k-substrings, |s′′| 6 2k2 |Σ|k.For any n = |s′′|+ tk we can now create |Σ|tk distinct strings. Hence,

cap(Sendk ) = lim sup

n→∞

log2∣∣Send

k ∩ Σn∣∣

n> lim sup

t→∞

log2(|Σ|tk)

|s′′|+ tk> log2 |Σ| .

Corollary

Sendk systems are fully expressive.


Tandem duplication behaves differently

But first…Main tool – φk-transform domain. We assume WLOG that Σ = Zq.

Definition

We define the transform φk : Z>kq → Zkq × Z∗

q by,

φk(x) = (Prefk(x), Suff|x|−k(x)− Pref|x|−k(x)),

as well as ζi,k : Zkq × Z∗

q → Zkq × Z∗

q,

ζi,k(x, y) =

{(x, u0kw) if y = uw, |u| = i

(x, y) otherwise,

where Prefi(x) and Suffi(x) are, respectively, the i-prefix and i-suffix of x.

Tandem Duplication 25 / 79

Main tool - φk-transform domain

Lemma

The following diagram commutes:

Z>kq

Ttani,k−−−−→ Z>kqyφk

yφk

Zkq × Z∗

qζi,k−−−−→ Zk

q × Z∗q

i.e., for every string x ∈ Z>kq ,

φk(Ttani,k (x)) = ζi,k(φk(x)).



ExampleAssume Σ = Z4. Starting with 02123 and letting i = 1 and k = 2 leads to

02123Ttan1,2−−−−→ 0212123yφ2

yφ2

(02, 102)ζ1,2−−−−→ (02, 10002)

where the inserted elements are underlined.



Theorem

For Stank = S(Σ, s, T tan

k ), |s| > k, cap(Stank ) = 0.

Proof.

In the φk-transform domain, φk(s) = (x, y), and tandem duplicationbecomes an insertion of 0k in the y-part. Thus, a tandem duplicationoperation is equivalent to throwing k balls into a bin. There are atmost |y| = |s| − k+ 1 bins. Thus, after t tandem-duplicated operations,there are at most

(|s|−k+tt

)6 (|s| − k+ t)|s|−k outcomes. Thus,

cap(Stank ) 6 lim sup

t→∞

log2((|s| − k+ t)|s|−k)

|s|+ tk= 0



Corollary

Stank systems are never fully expressive.

Proof.

If φk(s) = (x, y), then all possible mutations are limited (in theφk-transform domain) to (x, y′) with y′ being the same as y except forextra zeros. Thus, φ−1

k (x, y1) can never be obtained from s.


Were we too strict?

Definition (Tandem Duplication)

Ttani,k (x) =

{uvvw if x = uvw, |u| = i, |v| = k

x otherwise.

T tan>k =

{Ttani,k′

∣∣∣ i > 0, k′ > k}.

The lower-bounded tandem-duplication system is defined asStan>k = S(Σ, s, T tan

>k ).

u v w

u v v w


Yes, we were! Here’s full expressiveness:

Theorem

Stan>k is fully expressive.

Proof.

Employ a similar procedure to generate each substring as in the prooffor Send

k , only each time copy a suffix of the string (from the chosenstarting point, to the end).


What about full capacity?

Theorem

For any finite alphabet Σ, and s ∈ Σ∗, we have

cap(Stan>1 ) > log2(r+ 1),

where r is the largest (real) root of the polynomial

f(x) = x|Σ| −|Σ|−2∑i=0

xi.

Proof Strategy: Find a set S ⊆ Stan>1 for which we can calculate the

capacity. But how?


Regular languages to the rescue

Definition (Recipe for a regular language)

• A finite alphabet Σ

• A finite directed labeled graph G = (V, E, L), with E ⊆ V× V in themultiset sense, and L : E → Σ.

• A starting state s ∈ V and a set of accepting states A ⊆ V.

• If e1e2 . . . en is a directed path in G, it generates the wordL(e1)L(e2) . . . L(en).

• The language represented by G, denoted S(G), is defined as the setof all words generated by directed paths starting at s and endingin A.


A simple example for a regular language

ExampleConsider the following directed labeled graph G:

0

1

0

S(G) is the set of all binary strings where a 1 is followed by a 0.


Graphs have properties

Definition

Let G = (V, E, L) be a graph generating a regular language.

• G is irreducible if for every v1, v2 ∈ V, there is a directed pathv1 v2.

• G is primitive if it is irreducible and the gcd of all cycle lengths is 1.

• G is lossless if for every v1, v2 ∈ V, and every word w ∈ Σ∗, there isat most one path v1 v2 that generates w.


Counting paths is easy

Definition

For G = (V, E, L) define the adjacency matrix AG = (au,v) as the |V| × |V|matrix where au,v is the number of edges from u to v in G.

Observation

• The number of paths u v of length n is exactly (AnG)u,v.

• For a lossless graph G with one accepting state, i.e., A = {v}, wehave |S(G) ∩ Σn| = (AnG)s,v.

• Thus (with the above setting),

cap(S(G)) = lim supn→∞

log2((AnG)s,v)

n.


Enter Perron and Frobenius

O. Perron

G. Frobenius(Source: Wikipedia)

Theorem (Perron-Frobenius (Partial))

If G is a primitive graph then:

1 λ = λ(AG) , max {|µ| : µ is an eigenvalue of AG} alsocalled the spectral radius of G, is an eigenvalue of AG .

2 There exist y, x > 0, unique (up to scalar multiplication)left and right eigenvectors for λ.

3 If y · xT = 1, then

limn→∞

1

λnAnG = xT · y.

Corollary

For a primitive lossless graph G, cap(S(G)) = log2(λ(AG)).


Back to Stan>1

Proof.

Main Idea: Find a regular language that “resides” within Stan>1 and use

its capacity to lower bound cap(Stan>1 ).

Phase I: Denote the alphabet letters as a1, a2, . . . , a|Σ|. As in the proofof full expressiveness, assume we reach a string with a|Σ| . . . a2a1 as asuffix. From now on, we ignore everything except this suffix.Phase II: Run in iterations. In iteration i, where i = |Σ| , |Σ| − 1, . . . , 3, 2use tandem duplication only on strings of the form aiai−1 . . . a1. In thelast iteration, tandem duplicate single letters.It is easy to verify the resulting strings form the following regularlanguage,

S =

(a+|Σ|

(a+|Σ|−1

(. . .(a+2(a+1)+)+)+

)+)+

.


Proof by sub-language (Cont.)

Proof.

S =

(a+|Σ|

(a+|Σ|−1

(. . .(a+2(a+1)+)+)+

)+)+

.

a|Σ| a|Σ|−1 a2 a1

a|Σ| a|Σ|−1 a2

a1

a1a1


Proof by sub-language (Cont.)

Proof.

The graph is lossless, irreducible, and primitive. Its adjacency matrix is

AG =

1 1

1 11 1

. . .. . .1 1

1 1 1 . . . 1 1

,

Thus, the number of paths of length n from the starting vertex to theaccepting vertex grows exponentially as λn, where λ is the spectralradius of the graph, i.e., the largest root of

χAG (λ) = det(λI− AG) = (λ− 1)|Σ| −|Σ|−2∑i=0

(λ− 1)i.

Set x = λ− 1 and we obtain the result.


What do we have so far?

CapacityType System Zero Partial Full Full Expressiveness

EndSendk − − X XSend>k − − X X

TandemStank X − − −Stan>k − ? ? X

Open Question

Find cap(Stan>k ) or improve the bounds on it.


Full capacity⇔ full expressiveness?

Theorem

Let S be a string system over the alphabet Σ. If S has full capacity then Shas full expressiveness.

Proof.

Assume to the contrary S never contains w ∈ Σk as a substring.Partition every word x ∈ S into blocks of length k (and perhaps aremainder block of length at most k− 1). Each block has at most|Σ|k − 1 choices, since w is forbidden. Thus,

|S ∩ Σn| 6 (|Σ|k − 1)bn/kc · |Σ|k−1 .

Thencap(S) 6 log2(|Σ|

k − 1)

k< log2 |Σ| .


What about the other direction?

ExampleConsider the following string system,

S = {vv | v ∈ Σ∗} .

It is obvious that S is fully expressive, but

cap(S) =1

2.

Open Question

This example is not a string-duplication system. What is theconnection between full capacity and full expressiveness forstring-duplication systems?


A bit more on the big picture…

CapacityType System Zero Partial Full Full Expressiveness

EndSendk − − X XSend>k − − X X

TandemStank X − − −Stan>k − ? ? X

Palindromic Spalk − ? ? X

Interspersed Sintk,k′ X X ? X

Open Question

Complete the missing pieces in this table.


Let’s add probability to the mix

Why?

• Real biological processes are not always deterministic.

• Just like Shannon vs. Hamming: it is interesting!

Case study:

• Binary alphabet, Σ = {0, 1}, Duplication length k = 1.

• The position to duplicate is chosen independently and uniformly.• Two options:

• Stan1 – Tandem duplication: bit b becomes bb.

• Stan1 – Complement tandem duplication: bit b becomes bb.

Pólya String Models 45 / 79

Is this a Pólya urn model?

An urn contains B black balls and W white balls.At each step, a ball is extracted uniformly andindependently from the urn. The ball is returnedto the urn, together with another ball of thesame color. The process repeats.

Crucial difference:

There is no string structure in a Pólya urn model.


How would we define capacity?

Let S(i) denote the random variable whose value is the string after imutations, and S(0) = s the seed string.

Definition

The probabilistic capacity of the process S is defined as

capProb(S) = lim supn→∞

1

nH(S(n)),

where H(S(n)) is the entropy of S(n), i.e.,

H(S(n)) = −∑w∈Σ∗

Pr(S(n) = w) log2 Pr(S(n) = w).

The combinatorial capacity will be denoted by capComb.


Not everything is uniformly distributed

Assume Stan1 with S(0) = 0:

n = 0 :

n = 1 :

n = 2 :

n = 3 :

0ε

01

1

011

21

0111

321

0101

231

0110

213

010

12

0110

312

0100

132

0101

123

Thus, Pr(S(3) = 0110) = 13 but Pr(S(3) = 0111) = 1

6 .


One simple connection exists

Lemma

For S ∈{Stan1 , Stan

1

}, capProb(S) 6 capComb(S).

Proof.

H(S(n)) is maximized when S(n) is uniformly distributed,

H(S(n)) 6 log2

∣∣∣S ∩ Σ|S(0)|+n∣∣∣ .

Thus,

capProb(S) = lim supn→∞

1

nH(S(n))

6 lim supn→∞

1

nlog2

∣∣∣S ∩ Σ|S(0)|+n∣∣∣ = capComb(S).


So for tandem duplication…

Corollary

For any S(0) we havecapProb(S

tan1 ) = 0.

Proof.

We obviously have capProb(Stan1 ) > 0. Additionally,

capProb(Stan1 ) 6 capComb(S

tan1 ) = 0,

which we already proved.


Complement-tandem duplication is harder

Assume S(0) = 0 for simplicity. Let us record the history of mutationsin a string, whose ith position equals j if the jth mutation caused theith symbol.

Example 0 → 01 → 010 → 0110,ε → 1 → 12 → 312.

Observation

1 History is a permutation.

2 Each permutation is equally likely.


Here it is again


n = 0 :

n = 1 :

n = 2 :

n = 3 :

0ε

01

1

011

21

0111

321

0101

231

0110

213

010

12

0110

312

0100

132

0101

123

Some histories results in the same mutated string.


It’s all in the signature

Definition

The signature of a permutation π ∈ Sn, is a binary stringw = w1w2 . . .wn−1, where

wi =

{0 π(i) > π(i+ 1),

1 π(i) < π(i+ 1).

Theorem

Consider Stan1 with S(0) = 0. Then Pr(S(n) = 01w) is the same as the

probability of getting the signature w when choosing a permutation fromSn (uniformly).


It’s all in the signature – Proof

Proof.

Assuming w ∈ {0, 1}n−1, some notation first:

1 Π01w – The set of history permutations that lead to a mutatedstring 01w.

2 Ψw – The set of permutations from Sn with signature w.

3 For any string v ∈ {0, 1}`, the set of positions where 0 is precededby a 1 (including possible edges)

Tv = {i ∈ [`+ 1] : (vi−1 = 1 or i = 1) and (vi = 0 or i = `+ 1)}

Example: for v = 0011010 we have Tv = {1, 5, 7}.


It’s all in the signature – Proof (Cont.)

Proof.

Strategy: Prove |Π01w| = |Ψw| by showing both expressions have thesame recursion with the same starting conditions.

Starting conditions: Trivially |Π01ε| = |Ψε| = 1.

Recursion for Ψw: Given w ∈ {0, 1}n−1, we can recursively construct apermutation π ∈ Sn with signature w by picking π−1(n), which canonly be some i ∈ Tw. We then recursively construct two permutations,with signatures w1...i−2 and wi...n−1. Thus,

|Ψw| =∑i∈Tw

(n− 1

i− 1

) ∣∣Ψw1...i−2

∣∣ ∣∣Ψwi+1...n−1

∣∣ .


It’s all in the signature – Proof (Cont.)

Proof.

Recursion for Π01w: Given w ∈ {0, 1}n−1, consider a historypermutation π ∈ Sn resulting in the mutated sequence 01w. Obviouslyπ−1(1) is a position of a bit 1 in 01w which is last in a run, i.e.,followed by a 0 or last in the string. Thus, pick π−1(1), and constructthe rest of the permutation recursively using w1...i−2 and wi...n−1. Thus,

|Π01w| =∑i∈Tw

(n− 1

i− 1

) ∣∣Π01w1...i−2

∣∣ ∣∣Π10wi+1...n−1

∣∣ .


Last time I’m showing this slide


n = 0 :

n = 1 :

n = 2 :

n = 3 :

0ε

01

1

011

21

0111

321

0101

231

0110

213

010

12

0110

312

0100

132

0101

123

Open Question

Find a nice bijection between Π01w and Ψw.


And now, the capacity

Theorem

For Stan1 with S(0) = 0,

0.7213 ≈ log2(e)2

6 capProb(Stan1 ) 6 H2

(1

3

)≈ 0.9183,

where H2(x) , −x log2(x)− (1− x) log2(1− x) is the binary entropyfunction.


Proof of the bounds

Proof.

Consider the real random variables X1, X2, . . . , chosen i.i.d., uniformlyfrom [0, 1]. Sorting Xn1 , X1, X2, . . . , Xn generates a random permutation(due to symmetry, chosen uniformly from Sn).Define

Qi ,{1 Xi < Xi+1

0 Xi > Xi+1,

(except for a 0-measure undefined set). So Qn−11 is a signature of a

uniformly chosen random permutation from Sn.


Proof of the bounds (Cont.)

Proof.

We now havePr(S(n) = 01w) = Pr(Qn−1

1 = w).

and

capProb(Stan1 ) = lim sup

n→∞

1

nH(S(n)) = lim sup

n→∞

1

nH(Qn−1

1 )

= lim supn→∞

1

n

n−1∑i=1

H(Qi | Qi−11 ).



Proof.

Lower bound: Since Qi−11 → Xi → Qi we have

H(Qi | Qi−11 ) > H(Qi | Xi).

Furthermore, Pr(Qi = 0 | Xi = x) = x. Thus,


n→∞

1

n

n−1∑i=1

H(Qi | Qi−11 )

> lim supn→∞

1

n

n−1∑i=1

H(Qi | Xi)

= H(Q1 | X1) =∫ 1

0H2(x)dx =

log2(e)2

.



Proof.

Upper bound:


n→∞

1

n

n−1∑i=1

H(Qi | Qi−11 ) 6 lim sup

n→∞

1

n

n−1∑i=2

H(Qi | Qi−1)

= H(Q2 | Q1) =1

2(H(Q2 | Q1 = 0) + H(Q2 | Q1 = 1))

= H2

(1

3

),

since

Pr(Q2 = 0 | Q1 = 0) =

∫ 10 dx1

∫ x10 dx2

∫ x20 dx3∫ 1

0 dx1∫ x10 dx2

=1/6

1/2=

1

3,

and similarly for Pr(Q2 = 1 | Q1 = 1).


Probabilistic 6= Combinatorial

Observation

capProb(Stan1 ) 6 H2

(1

3

)< 1 = capComb(S

tan1 ).

Open Questions

1 Find capProb(Stan1 ).

2 We know nothing for duplication length > 2.


Moving on to error correction

An error-correcting code has two maincomponents:



3 A packing of error balls: Its density affectscommunication efficiency. Its structureaffects ease of encoding/decoding.

Error-Correcting Codes 64 / 79

Let us recall the scenario

• Information is stored in the DNA of some bacteria.

• The bacteria mutate over time.

• When the information is read, the DNA has gone through a(perhaps unbounded) number of duplications.

Goal

Protect information against duplication errors!

Case studyWe focus on Stan

k – tandem duplication with fixed duplication length k.


Some definitions are required

Definition

If v ∈ Stank (Σ, u, T tan

k ), we denote it as u=⇒∗k v. We

say u is an ancestor of v, and v is a descendant ofu. We define the descendant cone of u as

D∗k(u) =

{v ∈ Σ∗ : u

∗=⇒k

v},

and the ancestor cone as

A∗k(u) ={v ∈ Σ∗ : v

∗=⇒k

u}.

A∗k (u)

D∗k (u)

u

Time


Now we define a code

Definition

An (n,M; ∗)k code C is a subset C ⊆ Σn of size |C| = M, such that foreach u, v ∈ C, u 6= v,

D∗k(u) ∩ D∗

k(v) = ∅.

The decoding problem

Given an (n,M; ∗)k code C, and a (mutated) word v ∈ Σ∗, find

Decode(v) = A∗k(v) ∩ C.


Reminder – The φk-transform

We assume WLOG that Σ = Zq.

Definition

We define the transform φk : Z>kq → Zkq × Z∗

q by,

φk(x) = (Prefk(x), Suff|x|−k(x)− Pref|x|−k(x)),

as well as ζi,k : Zkq × Z∗

q → Zkq × Z∗

q,

ζi,k(x, y) =

{(x, u0kw) if y = uw, |u| = i

(x, y) otherwise,

where Prefi(x) and Suffi(x) are, respectively, the i-prefix and i-suffix of x.



Lemma

The following diagram commutes:

Z>kq

Ttani,k−−−−→ Z>kqyφk

yφk

Zkq × Z∗

qζi,k−−−−→ Zk

q × Z∗q

i.e., for every string x ∈ Z>kq ,

φk(Ttani,k (x)) = ζi,k(φk(x)).



ExampleAssume Σ = Z4. Starting with 02123 and letting i = 1 and k = 2 leads to

02123Ttan1,2−−−−→ 0212123yφ2

yφ2

(02, 102)ζ1,2−−−−→ (02, 10002)

where the inserted elements are underlined.


The ancestors are the key component

A∗k (u)

D∗k (u)

u

Time

Rk(u)

Definition

If A∗k(v) = {v} we say v is irreducible. The set ofirreducible words is denoted Irrk. The roots of v ∈ Σ∗

are defined by Rk(v) = A∗k(v) ∩ Irrk.

Lemma

For tandem duplication of length k, and every v ∈ Σ∗,|Rk(v)| = 1.

Already proved by Leupold et al. (2005). We give adifferent proof, using φk, enabling a code construction.


Proof of root uniqueness

Proof.

Denote φk = (x, y), and y = 0m0y10m1y20m2 . . . 0mt−1yt0mt , where yi 6= 0for all i. Any ancestor v′ ∈ A∗k(v) must be of the form,

φk(v′) = (x, 0m0−i0ky10m1−i1ky20m2−i2k . . . 0mt−1−it−1kyt0mt−itk),

and it is irreducible if and only if

φk(v′) = (x, 0m0 mod ky10m1 mod ky20m2 mod k . . . 0mt−1 mod kyt0mt mod k),

giving a unique root.


Disjoint descendant cones are simple

Corollary

D∗k(u) ∩ D∗

k(v) 6= ∅ if and only if Rk(u) = Rk(v).

Proof.

⇒: If w ∈ D∗k(u) ∩ D∗

k(v) then

Rk(u)∗=⇒k

u∗=⇒k

w and Rk(v)∗=⇒k

v∗=⇒k

w,

and since the root of w is unique, Rk(u) = Rk(v).


Disjoint cones proof (Cont.)

Proof.

⇐: If Rk(u) = Rk(v) then denote

φk(Rk(u)) = φk(Rk(v)) = (x, 0m0y10m1y20m2 . . . 0mt−1yt0mt).

Then,

φk(u) = (x, 0m′0y10m

′1y20m

′2 . . . 0m

′t−1yt0m

′t)

φk(v) = (x, 0m′′0 y10m

′′1 y20m

′′2 . . . 0m

′′t−1yt0m

′′t ).

Define w ∈ Σ∗ such that

φk(w) = (x, 0max(m′0,m

′′0 )y10max(m′

1,m′′1 ) . . . 0max(m′

t−1,m′′t−1)yt0max(m′

t,m′′t )),

which immediately shows u=⇒∗k w and v=⇒∗

k w.


Putting it all together

Theorem

• v ∈ Irrk iff φk(v) = (x, y) and y is (0, k− 1)-RLL.

• Irrk ∩Σn is an (n,M; ∗)k-code.• Decoding v ∈ Σ∗ may be done in linear time by:

1 Finding φk(v) = (x, y).2 Reducing runs of 0’s in y modulo k to obtain y′.3 Returning the answer φ−1

k (x, y′).

Observation

The code may be further enlarged (and made optimal!) by carefullyadding shorter RLL sequences.


Other results

• Stan63 : forms a regular language, unique root, positive (though notfull) capacity, not fully expressive.3

• Unique root exists in several other cases, enabling codeconstruction and decoding.3

Theorem

Let Σ 6= ∅ be an alphabet, and U ⊆ N, U 6= ∅, a set of tandem-duplicationlengths. Denote k = min(U). Then (Σ,U) is a unique-root pair if and onlyif it matches one of the following cases:

|Σ| = 1 U ⊆ kN

|Σ| = 2U = {k}U ⊇ {1, 2}

|Σ| > 3U = {k}U = {1, 2}U = {1, 2, 3}

3Jain et al., IEEE Trans. on Inform. Th. 2017.Conclusion 76 / 79

Other results

• What is the longest duplication distance to the root (inunbounded tandem duplication)? Apparently for length nsequences it is Θ(n) in the worst (and common!) case.4

• In the probabilistic models we know also the capacity of endduplication, as well as a mix duplication and complementduplication – but only for duplication length k = 1.5

• Tandem duplication with point-mutation (substitution) has morecapacity and expressiveness, but requires more care whenconstructing error-correcting codes.6

4Alon et al., ISIT 2016.5Elishco et al., ISIT 2016.6Jain et al., ISIT 2017.

Conclusion 77 / 79

Many open questions remain!

Open Questions

• Study error-correcting codes for duplication models other thantandem duplication.

• Find error-correcting codes for a probabilistic channel, correctingtypical errors.

• Study a mix of duplication and other mutations (substitutions,insertions/deletions).

• Study error models which are context sensitive.

• For the biologists: Find out the channel parameters in the realworld.

Conclusion 78 / 79

Thank You

Conclusion 79 / 79

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