1

On Prime Root-of-Unity Sequences withPerfect Periodic Correlation

Mojtaba Soltanalian* and Petre Stoica, Fellow, IEEE

Abstract—In this paper, Perfect Root-of-Unity Sequences(PRUS) with entries in αp = x ∈ C | xp = 1 (where p isa prime) are studied. A lower bound on the number of distinctphases that are used in PRUS over αp is derived. We showthat PRUS of length L ≥ p(p − 1) must use all phases in αp.Certain conditions on the lengths of PRUS are derived. Showingthat the phase values of PRUS must follow a given differencemultiset property, we derive a set of equations (which we callthe principal equations) that give possible lengths of a PRUSover αp together with their phase distributions. The usefulnessof the principal equations is discussed, and guidelines for efficientconstruction of PRUS are provided. Through numerical results,also contributions are made to the current state-of-knowledgeregarding the existence of PRUS. In particular, a combination ofthe developed ideas allowed us to numerically settle the problemof existence of PRUS with (L, p) = (28, 7) within about twoweeks— a problem whose solution (without using the ideas inthis paper) would likely take more than three million years on astandard PC.

Index Terms—Perfect sequences, Root-of-unity sequences, Pe-riodic autocorrelation, Phase distribution, Sequence construction

I. INTRODUCTION

Perfect Root-of-Unity Sequences (PRUS), also known asperfect N -phase [1], N -ary [2], or polyphase [3] sequences,are unimodular sequences with entries in αN = x ∈C | xN = 1 and the property that all their out-of-phaseperiodic autocorrelations are equal to zero [4]-[7]. Thesesequences are of interest in several applications includingfast startup equalization and channel estimation [1], as wellas communication schemes such as Direct-Sequence Spread-Spectrum Multiple-Access (DS/SSMA) and Frequency-Hopping Spread-Spectrum Multiple-Access (FH/SSMA) [8].They can also be used as key sequences in pulse compressionfor continuous-wave radars [4]-[8].

Due to implementation issues it is usually desirable thatthe entries of the sequence are from a small alphabet. Withthis fact in mind, it is interesting to note that 4, 6 and 11out of the first 8, 16 and 32 natural numbers, respectively,are prime. The study of PRUS with prime-size alphabets isimportant not only because of this relatively high densityof prime numbers in small alphabet sizes, but also becauseof the role of prime numbers as building blocks of natural

This work was supported in part by the European Research Council (ERC)under Grant #228044 and the Swedish Research Council. The authors arewith the Dept. of Information Technology, Uppsala University, Uppsala, SE75105, Sweden.

* Please address all the correspondence to Mojtaba Soltanalian,Phone: (+46) 18-471-3168; Fax: (+46) 18-511925; Email:mojtaba.soltanalian@it.uu.se

numbers. A similar building block property can be seen in thePRUS case: let n = mk where m and k are co-prime andassume that there exist PRUS u = (u0, · · · , um−1) and v =(v0, · · · , vk−1) with alphabet sizes m† and k† respectively;then w = (w0, · · · , wn−1) where wl = u(l mod m)v(l mod k)

is a PRUS with alphabet size n† = m†k†. This constructionis known as Chinese Remainder Theorem (CRT) constructionor simply as the direct product [11].

A general computational framework for designing se-quences with optimal correlation was proposed in [13]. It isknown that for lengths L that are square-free there exist aninfinite number of independent unimodular sequences withperfect periodic correlation, see [6]. A fast computationalmethod to find perfect unimodular sequences is proposed in[14]. Algebraic constructions for perfect unimodular sequencesof lengths p, 2p, 3p, pp′ and ps (where p and p′ are prime) wereintroduced and studied in [16]-[20]. When it comes to root-of-unity sequences (which correspond to the finite alphabet caseof the unimodular sequences), the problem appears to be morecomplicated. For example, it is not known whether there existsnone, a few or plenty of PRUS for some lengths or alphabetsizes. Besides construction methods, some publications (e.g.[1], [21]) have introduced and used the following necessarycondition on PRUS: if x = xlL−1

l=0 is a PRUS of length Lthen ∣∣∣∣∣

L−1∑l=0

xl

∣∣∣∣∣ = √L (1)

This necessary condition follows directly from the fact thatthe DFT of a PRUS has a constant magnitude (note that theDFT value at zero frequency is the sum of the sequence). In[2], several useful results are obtained which can be combinedwith the results in this paper. Namely, it was shown in [2] thatthe existence of PRUS of length L = mp (for a prime p) withentries in αp is connected to the existence of (L, p, L,m)-relative difference sets. Using some existence results of relativedifference sets, the authors in [2] prove for example that thereis no PRUS of length L = ps (for s ≥ 3), L = 2ps (fors ≥ 1), and L = pp′ (for prime p′ > p) with entries in αp.However, the strongest claim in the literature regarding theexistence (and construction) of PRUS is known as the Mow’sconjecture [8]:

Mow’s conjecture (for prime p): Let M(L,p) be the totalnumber of PRUS with length L over αp. Let L = sq2, wheres and q are both natural numbers and s is square-free. Then

M(L, p) =

q!sqΦq(s)pm, pmin = p,0, otherwise,

(2)

2

where

pmin =

2sq, for s even and m odd,sq, otherwise,

(3)

and the Euler totient function Φ(n) shows the number ofk ∈ Zn for which k and n are co-prime, and Φ(1) = 1 bydefinition. Moreover, all such PRUS can be constructed usinga unified approach, see [8].

Note that a proof of Mow’s conjecture would imply thatno PRUS of lengths other than L = p and L = p2 existsover αp. In this work1, PRUS with a prime-size alphabet arestudied. In particular, we study the phase distribution of suchsequences, and introduce a set of principal equations thatcan yield the possible phase distributions of PRUS for anygiven p and L. In general, the provided phase distributions cansignificantly reduce the size of search space for finding PRUS.Based on the obtained phase distributions, we also providepractical guidelines for construction of PRUS. A combinationof the ideas in this paper provides us, for example, with thepossibility to numerically settle the problem of the existenceof PRUS for (L, p) = (28, 7) within two weeks; a problem forwhich an exhaustive search of the associated search space isguaranteed to take more than three million years on a standardPC.

The rest of this paper is organized as follows. In SectionII, the phase distribution of PRUS over αp is discussed. Weshow that the phase values of PRUS over αp must follow aspecific difference multiset property. Furthermore, Section IIpresents the principal equations. Section III is devoted to thestudy of some special cases. A discussion on the usefulnessof the principal equations as well as guidelines for an efficientconstruction of perfect sequences along with some examplesare included in section IV. In Section V, we provide ournumerical results. Finally, Section VI concludes the paper.

Notation: We use bold lowercase letters for vec-tors/sequences and bold uppercase letters for matrices. (.)T

and (.)∗ denote the vector/matrix transpose and complex con-jugate respectively. The symbol ⊙ is used for the Hadamardelement-wise product of two matrices. x is a vector con-taining the same entries as x but in reversed order. 0n and1n are all zero and all one vectors of length n. e(n)l is thelth standard basis vector in Rn. ∥x∥n or the ln-norm of thevector x is defined as (

∑k |x(k)|n)

1n where x(k) are the

entries of x. N, Z, R and C represent the set of natural, integer,real and complex numbers respectively. Zn represents the set0, 1, · · · , n− 1. Considering the multiplicities of elements,we use the concept of multiset with the notation [·]. Finally,p denotes a prime number throughout the paper.

II. PHASE STUDY

In this section, we study the phase distribution of PRUSover the alphabet αp of prime size. Let x = xlL−1

l=0 =ej

2πp kl

L−1

l=0be an L-length PRUS over αp. All kl are in

1Some parts of this work were presented at the IEEE International Confer-ence on Acoustics, Speech and Signal Processing (ICASSP) 2011 [22].

Zp and we call them the integer phases of the sequence. Theperiodic autocorrelation of x at lag u ∈ ZL is defined as

Ru =L−1∑l=0

ej2πp (kl−kl+u)

=

L u = 00 u ∈ ZL − 0 (4)

where the indices of kl are used in a periodic manner (i.e.mod L). It is interesting to note that Ru is a summation ofterms which are also in αp. Theorem 1 paves the way forusing this observation (see Appendix A for a short proof):

Theorem 1. If∑p−1

k=0 akej 2π

p k = 0 for some ak ∈ Z, then allak must be identical.

Corollary 1. If there exists a PRUS of length L over αp thenp|L.

Proof: Let u ∈ ZL − 0. Then it follows from (4) andTheorem 1 that Ru = m

∑p−1k=0 e

j 2πp k = 0 where m = L/p

must be an integer.

The fact pointed out in Corollary 1 is already known in theliterature, see e.g. [2]. However, it is worthwhile to commenton the more general case of PRUS of length L over αN . Froma number theory perspective, the authors of [23] study thevanishing sums of roots-of-unity, viz. q1 + q2 + · · · qT = 0with ql ∈ αN (for all l), and general N ∈ N. In particular,they show that if N = pa1

1 · · · parr (with p1 < · · · < pr)

represent the prime factorization of N then a vanishing sumof T root-of-unity numbers ql (with ql ∈ αN ) can occuronly if there exist non-negative integers tk such that T canbe written as T = t1p1 + · · ·+ trpr. Interestingly, we can usethis result in the context of PRUS. Namely, the autocorrelationsums similar to that in (4) may become zero only if L can bewritten as

L = t1p1 + · · ·+ trpr (5)

where tk are non-negative integers. Therefore, satisfying(5) is a necessary condition for a PRUS of length L overαN . Considering (5), also typically known as the Frobeniuscoin problem [24]-[26], can be particularly useful for showingthe non-existence of PRUS when L is rather small. On thecontrary, it can be shown that if L ≥ (p1 − 1)(p2 − 1) then(5) always has a solution.

In the sequel we use the notation L = mp, m ∈ N, for thelength of PRUS over αp.

Corollary 2. Let x =ej

2πp kl

mp−1

l=0be a PRUS of length

L = mp over αp. Then for every s ∈ Zp and u ∈ ZL −0, there exist exactly m distinct integers l such that kl ≡kl+u + s (mod p).

Proof: We only need to observe that, according to The-orem 1, all sums in (4) for Ruu∈ZL−0 must have exactlym terms equal to ej

2πp s for every s ∈ Zp .

We note that in light of the above results, a general differ-ence set structure is obtained. Let D = [d0, d1, · · · , ds−1] be amultiset over a group G of order v. D is a (v, s, λ)-difference

3

multiset over G iff the multiset ∆D = [dk − dl : k, l ∈Zs, k = l] contains each element of G (0 included) exactlyλ times. By this definition, the multiset of integer phases[k0, k1, · · · , kL−1] of a PRUS of length L = mp over αp

is a (p,mp,m)-difference multiset. The next two definitionsappear to be essential in order to continue our study.

Definition 1. Let ξ0, · · · , ξt−1 be real numbers whose sum isa constant C. From the Cauchy-Schwarz inequality we havethat

∥ξ∥22 ≥(1Tt ξ)2/t (6)

where ξ is the vector with entries ξk. Therefore,∑t−1

k=0 ξ2k at-

tains its minimum value when the sum is uniformly distributedover ξk. We define

Γ(C, t) ,C2/t t > 00 otherwise

(7)

as the minimum value of the sum of squares of t real variableswith sum C.

Definition 2. We let Φx be the circulant matrix made fromthe integer phases kl of the sequence x, viz.

Φx ,

k0 k1 · · · kmp−1

kmp−1 k0 · · · kmp−2

......

. . ....

k1 k2 · · · k0

. (8)

For the lth column of Φx, consider the location of the entrieswhich are equal to kl (l = 0, · · · ,mp− 1). Considering theselocations for all columns, we build an mp×mp equivalencematrix Φe of x whose entries in the mentioned locations are1; otherwise they are 0. We also extend the definition of Φe toΦ(s)

e (with Φ(0)e = Φe), for s ∈ Zp as follows: by finding the

locations of the entries kl′ in the lth column of Φx such thatkl′ ≡ kl+ s (mod p), we represent these locations in Φ(s)

e by1, and by 0 otherwise.

There exist several construction methods for PRUS with thelength L = p; see Section III for details. Moreover, the caseof L = p presents some unique properties (see below) thatmakes its study relevant. We also study the PRUS of lengthL = mp, m > 1, and present our general results in sub-sectionII-B below.

A. The case of L = p (corresponding to m = 1)

Based on the above discussions, the integer phases of PRUSwith length L = p over αp have a (p, p, 1)-difference multisetstructure. Such PRUS are in close connection with prime-length binary sequences with optimal periodic correlation. Adetailed discussion revealing such close relationship is pro-vided in Appendix B. Theorem 2 studies the phase distributionof PRUS in this case.

Theorem 2. A PRUS of length L = p > 2 over αp hasexactly 1

2 (p + 1) distinct phases. Even more precisely, suchPRUS consists of a singleton and 1

2 (p− 1) equi-phase pairs.

Proof: Note that due to the (p, p, 1)-difference multisetstructure, all rows of Φe (in this case) have exactly one 1

except the first row whose all entries are 1. Now let µk bethe number of times that ej

2πp k (for k ∈ Zp) occurs in the

sequence, and let us assume that t of µk are nonzero. Wehave

p−1∑k=0

µk = p. (9)

As discussed above, by considering the rows of Φe, weconclude that there are (2p−1) ones in Φe. On the other hand,since for every integer phase k ∈ Zp we have µk columns withµk ones in each of them, the number of ones in Φe is equalto∑p−1

k=0 µ2k; hence

p−1∑k=0

µ2k = 2p− 1. (10)

Since the sum of µk is constant, we have

2p− 1 =

p−1∑k=0

µ2k ≥ Γ(p, t) =

p2

t, (11)

and as a result

t ≥⌈

p2

2p− 1

⌉=

1

2(p+ 1) (12)

where ⌈x⌉ denotes the smallest integer greater than or equalto x. Now let us suppose that t ≥ 1

2 (p + 1) + 1 = 12 (p + 3).

Also let v1 be the number of µk which are equal to one.Therefore

p =

p−1∑k=0

µk ≥ v1 + 2(t− v1) = 2t− v1. (13)

Note that (13) implies v1 ≥ 2t − p ≥ 3. This leads to acontradiction for p = 3 as all phases should be different; i.e.µ0 = µ1 = µ2 = 1 which yields

∑p−1k=0 µ

2k = 2(3) − 1. Next

we consider the case of p ≥ 5. Note that

p−1∑k=0

µ2k ≥ v1 + Γ

(p− v1,

p+ 3

2− v1

). (14)

By substituting (10) in (13) we get v1 ≤ 3 and as a resultv1 = 3. Now let µk∗ ≥ 2 for some k∗ ∈ Zp; then

p−1∑k=0

µ2k ≥ 3 + µ2

k∗+ Γ

(p− 3− µ2

k∗,p+ 3

2− 4

). (15)

Again by substituting (10) in (15) we obtain µk∗ ≤ 3. Thisshows that except for µk = 1, the only possible values of µk

are 2 and 3. Let us denote the number of them by v2 and v3,respectively. Then:

3 + 2v2 + 3v3 = p3 + 4v2 + 9v3 = 2p− 1

(16)

which is not feasible for integer numbers v2 and v3. Thanksto the latter contradiction, we conclude that the number ofdistinct phases is equal to t = 1

2 (p+ 1). In order to obtain acomplete picture of the phase distribution of x, let vk denotethe number of µk which are equal to k, for all k ∈ Zp.

4

Using the inequality in (13) we have that v1 ≥ 2t − p ≥ 1.On the other hand,

p−1∑k=0

µ2k ≥ v1 + Γ

(p− v1,

p+ 1

2− v1

)(17)

which implies v1 ≤ 1, and as a result v1 = 1. If µk∗ ≥ 2 forsome k∗ ∈ Zp then

p−1∑k=0

µ2k ≥ 1 + µ2

k∗+ Γ

(p− 1− µ2

k∗,p+ 1

2− 2

). (18)

As before, by substituting (10) in (18) we obtain µk∗ ≤ 2, andhence µk∗ = 2. This implies that v2 = 1

2 (p − 1) and vk = 0for k > 2 which completes the proof.

B. The case of L = mp (general case)

As indicated earlier, the integer phases of a PRUS of lengthL = mp over αp build a (p,mp,m)-difference multiset, whichimplies that for every u ∈ ZL − 0, there exist exactly mdistinct integers l such that kl = kl+u. Therefore, for anmp ×mp matrix Φe built as in Definition 2, the number ofones is equal to mp+m(mp−1). Let µk represent the numberof times that ej

2πp k occurs in the sequence. Then we have that

p−1∑k=0

µk = mp, (19)

p−1∑k=0

µ2k = mp+m(mp− 1). (20)

We assume t of µk are nonzero, which implies

m2p+m(p− 1) =

p−1∑k=0

µ2k

≥ Γ (mp, t) =(mp)2

t, (21)

and as a result

t ≥ mp2

(m+ 1)p− 1. (22)

The above lower bound shows that as m increases, a largernumber of phases from αp might be needed to build a PRUSof length L = mp. For sufficiently large values of m we needall phases:

Theorem 3. For m ≥ p − 1, all phase values must be usedin a PRUS.

Proof: This is a direct consequence of the lower boundin (22).

Now, for every s ∈ Zp − 0 we consider the matrixΦ(s)

e built as in Definition 2. Based on the difference multisetproperty, for every u ∈ ZL−0, there exist exactly m distinctintegers l such that kl+u ≡ kl + s (mod p). Therefore, thematrix Φ(s)

e has exactly m ones in each of its rows exceptfor the first row which is all zero. This implies that Φ(s)

e hasm(mp − 1) ones. On the other hand, the number of ones inΦ(s)

e is equal to∑p−1

k=0 µkµk+s as it equals the number of all

pairs with the property kl+u ≡ kl+s (mod p). Therefore, theout-of-phase correlations of the sequence µk are given by

p−1∑k=0

µkµk+s = m(mp− 1), s ∈ Zp − 0. (23)

Based on (19), (20) and (23), we conclude the following.

Theorem 4. Let µk denote the phase distribution of a PRUSwith length L = mp over αp. If we define rk , µk −m, thenrk satisfy the following set of principal equations:

∑p−1k=0 rk = 0∑p−1k=0 r

2k = m(p− 1)∑p−1

k=0 rkrk+s = −m, s ∈ Zp − 0.(24)

Solving the principal equations indicate the possible PRUSphase distributions for given L and p. It is interesting to notethat if rk is a solution to (24), then −rk, r−k andrk+l where l ∈ Zp are also valid solutions to (24). In otherwords, the set of principal equations induces a certain type ofequivalence class on its solutions. We note that the unimodularperfect sequences enjoy a similar set of equivalence properties:let x be a unimodular perfect sequence, then x∗ and ejϕx(where ϕ can be chosen arbitrarily) are also unimodular perfectsequences. This shows that given a solution rk to theprincipal equations, the solutions r−k and rk+l do notlead to new PRUS. In contrast, the solution −rk might leadto new sequences. As an aside remark, note that the secondequation of (24) can be viewed as a sum of squares problem,which has been widely studied for many years. More detailson this aspect are deferred to Appendix C. The followingdiscussion is devoted to a geometrical study of the problem.

Let r0 = (r0, · · · , rp−1)T and also let rk represent the

circularly shifted version of r0 by k ∈ Zp. The principalequations can be rephrased as follows over the vectors rk:

1Tp rk = 0

∥rk∥2 =√m(p− 1)

rTk rl = −m, k = l

(25)

The angle between each pair of vectors (rk, rl)k =l is givenby

θ = arccos

(rTk rl

∥rk∥2∥rl∥2

)= arccos

(−1

p− 1

)(26)

Therefore, rkk∈Zp , form a set of p vectors lying in a(p−1)-dimensional space which is the hyperplane orthogonalto 1p and the angle between each pair of them is the valuegiven in (26). We further note that the structure made byconnecting all vertices pointed by rk is a known multi-dimensional object called a regular simplex [29]. Such struc-tures are shown in Fig. 1 for one, two and three dimen-sions. The reference [29] suggests e(p)k (i.e. the standardbasis) as vertices of a regular simplex of edge

√2 lying

in the hyperplane 1Tp x = 1. It can be easily verified that

r0 =((p− 1)

√mp ,−

√mp , · · · ,−

√mp

)T∈ Rp together

with its circularly shifted versions (denoted by rk) satisfythe principal equations. It is also straightforward to verify that

5

Fig. 1. (a-c) Regular simplexes in one, two and three dimensional space.In n dimensions they can be characterized with n+1 vectors with the samel2-norm and also the same angle between them as described in Eq. (26).

these points can be obtained from the vectors e(p)k by ascaling and a translation. As every two regular simplexes withtheir center at 0p can be obtained from each other by a set ofrotations, we can find the vectors rk by rotations of rksuch that their end lie at the integral lattice. Note that as theregular simplex made by rk is in a (p − 1)-dimensionalspace, its rotation could be parametrized with (p − 2) anglesψ0, · · · , ψp−3 and as a result the vectors rk could be writtenas a function of sinψk and cosψk for k ∈ Zp−2. Taking intoconsideration the fact that rk are integral, this gives a closedform solution for m (and as a result a closed form solutionfor the sequence length) as well as the phase distribution. Wegive an example of the usage of such a geometrical approachin Section III. Nevertheless, for large values of L and p, anefficient numerical approach to tackle the principal equationsis discussed in Appendix D.

III. SPECIAL CASES

This section not only considers special cases associatedwith the principal equations but also aims to establish theconnections between the results in Section II and the existingliterature on PRUS. Several special cases for m and p arediscussed. The cases m = 1 and m = p are discussed becauseof the well-known constructions which give sequences forthese values. Additionally, we study the case of m = h2m′

where a solution for the principal equations can be found forlength L′ = m′p. The special cases p = 2 and p = 3 arealso discussed to give closed form solutions for the length andphase distribution of PRUS. The case p = 3 can be consideredas an example of using the geometrical approach based on theregular simplex to solve the principal equations.

A. Special Cases of m

• m = 1 and m = p: Sequences with m = 1 (i.e.length L = p) can be constructed for example by Zadoff,

Chu, Golomb polyphase, P3 and P4 methods [4]. Thesemethods are all based on quadratic integer phases and itis easy to verify that all of them follow the distributiongiven in Section II-A. Examples of construction methodsfor m = p (i.e. length L = p2) include Frank, P1, P2and Px methods [4]. Sequences of this length contain allphase values, see Theorem 3.

• m = h2m′: Let r(m′)

k be a solution of the principalequations for length L′ = m′p over αp. Then, onecan verify that r(m)

k = hr(m′)

k is a solution of theprincipal equations for the length L = mp. Interestingly,the Mow’s conjecture suggests that the latter consructionof solutions for the principal equations cannot lead tonew PRUS. Nevertheless, existence of such PRUS is notdisproved by the principal equations.

B. Special Cases of p

• p = 2: Solving the principal equations, viz.

µ20 + µ2

1 = 2m2 +m (27)2µ0µ1 = 2m2 −m (28)

for this case (which is the case of perfect binarysequences) yields µ0 = 1

2

(2m±

√2m)

and µ1 =12

(2m∓

√2m). Therefore m must be of the form 2h2

and as a result µ0 = 2h2 ± h and µ1 = 2h2 ∓ h. Thisenumeration of +1 and −1 in perfect binary sequencescan be obtained also by the necessary condition (1) andis mentioned in several publications including [30].

• p = 3: Here we use the geometrical approach discussed inSection II-B to solve the principal equations. For 3-phaseperfect sequences, the rk make a two dimensionalregular simplex orthogonal to 13, which has 3 vectorsand each two of them have an angle of 2π

3 . The structureof this regular simplex is shown in Fig. 1(b). Let R13

be the unitary rotation matrix which maps 13 to√3e

(3)3 .

Also let

r′k =√2m

cos(2kπ3 + ψ

)sin(2kπ3 + ψ

)0

(29)

for k ∈ Z3 and ψ ∈ [0, 2π). Therefore, rk is equal toR−1

13r′k for some ψ. This implies that rk ∈ Z3 is of the

form

√2m

√22 cos

(2kπ3 + ψ

)−

√66 sin

(2kπ3 + ψ

)√63 sin

(2kπ3 + ψ

)−

√22 cos

(2kπ3 + ψ

)−

√66 sin

(2kπ3 + ψ

)

for k ∈ Z3. As rk are the circularly shifted versionsof each other, it is sufficient to study one of them. Fork = 0, we infer that both h1 = 2

√m3 sinψ (which is

the second entry of r0) and h2 = 2√m cosψ (which is

the difference between the first and the third entry of r0)must be integers. We conclude that 3h21 + h22 = 4m and

r0 =1

2

(h2 − h1)2h1

−(h2 + h1)

. (30)

6

Therefore, the sequence length must be of the form

L =1

4

(9h21 + 3h22

)(31)

while its phase distribution is given by

1

4

(3h21 + h22

)13 +

1

2

(h2 − h1)2h1

−(h2 + h1)

, (32)

for integers h1 and h2.

IV. PRUS: FROM PHASE DISTRIBUTIONS TOCONSTRUCTION

In this section, we discuss the use of the ideas introducedearlier for an efficient search or construction of PRUS.

A. Are the Principal Equations Useful?First we show that if rk satisfy the principal equations,

then the necessary condition in (1) will be satisfied; i.e. theprincipal equations are more informative than (1). Let S bethe sum of entries of the PRUS:

S ,mp−1∑l=0

ej2πp kl =

p−1∑k=0

µkej 2π

p k. (33)

For rk satisfying the principal equations:p−1∑k=0

rkrk+s =

(p−1∑k=0

r2k

)−mp = −m, s ∈ Zp − 0. (34)

Therefore,

|S|2 =

∣∣∣∣∣p−1∑k=0

µkej 2π

p k

∣∣∣∣∣2

=

∣∣∣∣∣p−1∑k=0

rkej 2π

p k

∣∣∣∣∣2

=

p−1∑s=0

(p−1∑k=0

rkrk+s

)ej

2πp s = mp

which implies the satisfaction of (1).It is worth emphasizing that satisfaction of the principal

equations is necessary but not sufficient for a PRUS. Thenecessity induced by the principal equations guarantees thatif a sequence exists over αp then it will have a specific lengthand phase distribution; particularly, the number of sequenceswhich are needed to be checked for enummeration of PRUSof length L over αp reduces from

pL =∑

µk≥0p−1k=0:

∑p−1k=0 µk=L

(L

µ0, µ1, · · · , µp−1

)(35)

to ∑µkp−1

k=0 ∈ Ω

(L

µ0, µ1, · · · , µp−1

)(36)

where Ω represents the set of feasible phase distributions.Note that the expression (36) typically contains only a fewout of

(L+p−1p−1

)summation terms of Eq. (35). Therefore, the

principal equations can be used to show the impossibility ofsome lengths (in cases for which no feasible phase distributionexists) and to significantly reduce the size of search space ofPRUS in general; see Section V for some numerical evidenceon this aspect.

B. PRUS Construction: Guidelines and Examples

Once we have obtained the length and phase distribution ofa PRUS, an efficient method for its construction (or furtherelimination of non-suitable cases) is needed. Hereafter, weaim to explain how a PRUS could be constructed using theΦ(s)

e matrices introduced in Section II. Examples of such aconstruction for different scenarios are also provided.

For the Φe matrix (built based on the equality of phases)we need an all one first row and exactly m ones in anyother row. On the other hand, the matrices Φ(s)

e s=0 (builtbased on the inequality of phases) contain no ones in thefirst row and exactly m ones in the other rows. It is animportant observation that this condition is equivalent to theperfectness of the sequence and as a result all we need isto check whether assigning different phases to indices in thesequence preserves this condition. Let us assume that thevector χk = (χk(0), · · · , χk(µk − 1))

T contains the indicesthat are assigned to the kth phase, i.e. ej

2πp k. As an example,

the configuration of the matrix Φx and the correspondingvectors χk are shown in Fig. 2 for the Frank sequence oflength 9. As the Φx matrix has a circulant structure, we can

Fig. 2. Configuration of Φx and the vectors χk for the Frank sequenceof length 9.

observe that rows of Φ(s)e with a one in the χk(l)

th positionare given by

(χk(l)− χk(1))1µ(k+s)+ χ

(k+s). (37)

Fig. 3 depicts the construction of the matrices Φ(s)e for the

Frank sequence of length 9 based on (37).According to the above discussion, instead of the classical

method based on calculation of the autocorrelation for allsequences, one can make updates of the matrices Φ(s)

e foreach assignment of indices to phases in αp respectively, andcheck whether for each assignment the number of nonzeroentries in each row (except the first row) of the updatedmatrices Φ(s)

e does not exceed m. Note that since thematrices Φ(s)

e represent all distinct differences (s ∈ Zp

where s = 0 denotes the case of phase equality), they can

7

(a) (b) (c)

Fig. 3. (a-c) Construction of the matrices Φ(s)e for the Frank sequence of length 9 and integer phase differences equal to s = 0 (i.e. equality of phases),

1 and 2 respectively. The positions of 1s are shown by the corresponding phases that satisfy the difference s.

be viewed as complementary binary matrices; i.e. (i) thereexists no common positions for ones in the matrices Φ(s)

e (equivalently Φ(s1)

e ⊙ Φ(s2)e is an all zero matrix for any

s1 = s2, where s1, s2 ∈ Zp), and (ii) the sum of matricesΦ(s)

e s∈Zp is an all one matrix. Therefore, if by assigning allindices to elements of αp, there still exists no row violatingthe above condition, then this shows that the number of onesin all rows of Φ(s)

e (except the first row) is equal to m.It is important to note that by using the discussed idea, still

we need to check all(L

µ0, µ1, · · · , µp−1

)(38)

possible arrangements of entries of PRUS (with given phasedistribution µk). But the proposed method is also sensitive tounsuitable partial assignments of phases. A suitable phase ar-rangement is as shown in Fig. 2 corresponding to the Frank se-quence of length 9. The method assigns χ0 = (0, 1, 2, 3, 6)T ,χ1 = (5, 7)T and χ2 = (4, 8)T one after another and noneof the matrices Φ(s)

e violates the above rule about thenumber of ones in their rows. On the other hand, there areunsuitable phase arrangements which coincide with the phaseconfiguration of the Frank sequence. For example, supposethat the method already has assigned χ0 = (0, 1, 2, 3, 6)T .Now if the method assigns χ1 = (4, 5)T then by updating thematrices Φ(s)

e (as depicted in Fig. 4), it appears that sucha partial phase arrangement violates the expected number ofones in rows of Φ(s)

e .Note that by recognizing any partial assignment of phases as

unsuitable, the proposed construction approach avoids testinglots of sequences and thus is considerably more efficient thanthe classical approach. We refer the interested reader to afurther efficiency analysis of the proposed construction methodin Appendix E.

V. NUMERICAL RESULTS

We provide several numerical results that rely on the ideasdiscussed in the paper. Table I presents all feasible lengths

(less than or equal to 500) along with their correspondingphase distributions for p = 5 and 7. Using the equivalenceproperties of PRUS, the µk sequences are circularly shiftedsuch that µ0 take the maximum value among all µk. Thenon-existence results of [2] are used to omit some cases of(L, p) wihtout solving the principal equations. Note that byproviding the phase distributions we are able to significantlyreduce the size of the search space in all cases. The searchspace cardinality reduction induced by using the principalequations is also reported in Table I.

In order to contribute to the current state-of-knowledge re-garding the existence of PRUS, next we consider the unsolvedcases of PRUS of length L ≤ 50 in [2]; which are also shownin Table II. As in the previous example, the size reduction ofthe PRUS search space is reported when the phase distributionswere derived by solving the principal equations. Nevertheless,even after using the principal equations, the size of the searchspaces appears to be prohibitive for an exhaustive search. Tohelp the interpretation of the results in Table II, and to see howexpensive tackling such search problems can be, we considerthe following analysis for the case (L, p) = (28, 7):

• The initial size of the search space is 728 ≈ 4.60× 1023.Supposing that a standard PC can handle 5× 109 simplemath operations per second, we can see that a search forPRUS in this case would take more than

728

(3600× 24× 365)(5× 109)years (39)

i.e. approximately 3 million years.• Using the principal equations, we reduce the size of the

search space by a factor of 5.33 × 104. On the samestandard PC, an exhaustive search of PRUS for (L, p) =(28, 7) in this case will take more than

8.63× 1018

(3600× 24× 365)(5× 109)years (40)

i.e. approximately 55 years.• By employing the construction guidelines provided in

8

(a) (b) (c)

Fig. 4. (a-c) Updated matrices Φ(s)e for an unsuitable phase arrangement which coincides with the phase configuration of the Frank sequence (integer

phase differences are equal to s = 0, 1 and 2 for (a), (b) and (c) respectively). The positions of 1s are shown by the corresponding phases that satisfy thedifference s. χ1 = (4, 5)T is assigned after considering χ0 = (0, 1, 2, 3, 6)T and as a result the number of 1s in the second and ninth row of (a) are morethan m = 3.

TABLE IALL POSSIBLE LENGTHS (LESS THAN OR EQUAL TO 500) OF PRUS OVER αp FOR p = 5 AND 7 TOGETHER WITH PHASE DISTRIBUTIONS.

p Length (L) Phase distributions µkp−1k=0 Reduction in the size of search space induced by the

principal equations (, initial size / final size)5 (2, 1, 2, 0, 0) (2, 2, 0, 1, 0) 52.0825 (6, 6, 6, 6, 1) (9, 4, 4, 4, 4) 1.60× 103

180 (42, 36, 42, 30, 30) (42, 42, 30, 36, 30) 8.39× 104

5 220 (52, 46, 34, 46, 42) (52, 46, 46, 42, 34) 6.31× 104

(54, 42, 46, 36, 42) (54, 46, 42, 42, 36)275 (64, 54, 59, 44, 54) (64, 59, 54, 54, 44) 9.86× 104

(66, 56, 46, 56, 51) (66, 56, 56, 51, 46)

7 (2, 2, 1, 0, 0, 2, 0) 1.31× 103

28 (6, 6, 4, 2, 2, 6, 2) (7, 5, 5, 2, 5, 2, 2) 5.33× 104

49 (13, 6, 6, 6, 6, 6, 6) (8, 8, 8, 8, 8, 8, 1) 3.05× 105

56 (11, 10, 10, 5, 10, 5, 5) (11, 11, 6, 11, 6, 6, 5) 4.55× 105

112 (18, 18, 17, 18, 17, 17, 17) (20, 20, 14, 20, 14, 14, 10) 2.52× 106

(20, 20, 16, 12, 12, 20, 12) (22, 18, 18, 12, 18, 12, 12)(25, 15, 15, 14, 15, 14, 14)

7 196 (30, 30, 30, 30, 30, 30, 16) (33, 33, 26, 33, 26, 26, 19) 1.02× 107

(37, 30, 30, 23, 30, 23, 23) (40, 26, 26, 26, 26, 26, 26)224 (36, 36, 32, 36, 32, 32, 20) (37, 37, 35, 26, 26, 37, 26) 1.02× 107

(38, 36, 36, 26, 36, 26, 26) (38, 38, 26, 28, 28, 38, 28)(38, 38, 27, 38, 27, 27, 29) (44, 32, 32, 28, 32, 28, 28)

372 (62, 62, 55, 62, 55, 55, 41) (62, 62, 62, 48, 62, 48, 48) 8.23× 107

(64, 64, 50, 64, 50, 50, 50) (71, 57, 57, 50, 57, 50, 50)448 (68, 68, 66, 68, 66, 66, 46) (71, 71, 62, 71, 62, 62, 49) 7.02× 107

(72, 72, 56, 72, 56, 56, 64) (72, 72, 60, 72, 60, 60, 52)(76, 68, 68, 56, 68, 56, 56) (79, 66, 66, 57, 66, 57, 57)(82, 62, 62, 60, 62, 60, 60)

Section IV, we developed a MATLAB code1 to search forPRUS with (L, p) = (28, 7). We used two standard PCs,which dealt with the two possible phase distribtuions ofthe case (L, p) = (28, 7) (see Table II), in parallel. Usingthis approach, we were able to confirm the non-existenceof PRUS with (L, p) = (28, 7) in about 2 weeks.

Note that due to the exponential growth in the size of searchspace, the questions regarding the existence of PRUS in casesfor (L, p) = (33, 11) and (L, p) = (39, 13) remain open.

Although the study of PRUS over general root-of-unity

1The MATLAB code associated with this experiment is provided online:http://www.it.uu.se/katalog/mojso279/test-p7.rar

alphabet (i.e. αN with general N ∈ N) is beyond the scopeof this paper, it can be interesting to study the length/alphabetrestrictions imposed by (5) and the remarks after Corollary1. The eliminated cases of (L,N) via (5) are plotted in Fig.5 for 2 ≤ L,N ≤ 100. Interestingly, the prime values ofN appear to support rather smaller numbers of lengths Lthan the nonprime values do. Via the results of Fig. 5 weprove the non-existence of PRUS for 3443 cases of (L,N)for 2 ≤ L,N ≤ 100.

VI. CONCLUDING REMARKS

Perfect root-of-unity sequences with prime-size alphabetshave been studied. The results can be summarized as follows:

9

TABLE IIREMAINING UNSOLVED CASES FROM [2], L ≤ 50

p Length (L) Phase distributions µkp−1k=0 Reduction in the size of search space

induced by the principal equations (,initial size / final size)

Final size of thesearch space

Existence

7 28 (6, 6, 4, 2, 2, 6, 2) (7, 5, 5, 2, 5, 2, 2) 5.33× 104 8.63× 1018 Negative11 33 (8, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2) 2.68× 107 8.65× 1026 ?13 39 (5, 1, 4, 5, 5, 3, 3, 1, 4, 3, 4, 0, 1) 1.03× 109 2.69× 1034 ?

(5, 1, 1, 4, 4, 3, 5, 4, 5, 0, 3, 1, 3)(6, 3, 5, 3, 1, 5, 5, 2, 2, 3, 1, 2, 1)(6, 5, 1, 5, 2, 1, 1, 3, 3, 5, 2, 3, 2)

2 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

25

10

15

20

25

30

35

40

45

50

55

60

65

70

75

80

85

90

95

100

Length (L)

Alp

habet

Siz

e(N

)

Fig. 5. The eliminated PRUS cases (represented by black squares) of (L,N) by solving (5) for 2 ≤ L,N ≤ 100. The results in this figure prove thenon-existence of PRUS for 3443 cases of (L,N) for 2 ≤ L,N ≤ 100.

• The phase distribution of p-length PRUS over αp wasgiven for p > 2: it was shown that such sequences have12 (p+1) distinct phases with 1

2 (p−1) of them appearingin pairs and one of them being a singleton.

• A lower bound on the number of distinct phases whichmust be used in a PRUS over αp was derived. The lowerbound was used to show that for PRUS of length L ≥p(p−1) (i.e. m ≥ p−1) over αp, the sequence must useall phase values.

• Guidelines to find possible lengths (L) of PRUS over αp

were given. It was shown that integer phases of the se-quence must follow a specific difference multiset propertyand there should exist a sequence of µk (introduced

in Section II) satisfying the principal equations. For apossible length, the phase distribution is then given byµkp−1

k=0.• A geometrical analytical method to solve the principal

equations was introduced for a specific p using the regularsimplex. In particular, it was shown using the geometricalapproach that if there exists a perfect sequence over α3

(i.e. a 3-phase perfect sequence), its length must be ofthe form L = 1

4

(9h21 + 3h22

)for (h1, h2) ∈ Z2 and the

phase distribution of the sequence was also derived.• The usefulness of the principal equations was discussed.

Given the phase distribution, guidelines for efficientconstruction of PRUS (in comparison to the exhaustive

10

search) along with some examples were provided.• Numerical evidence was provided to show the potential

of using the principal equations and the constructionguidelines of Section IV in practice. Through numericalexamples, new contributions were made to the currentstate of knowledge regarding the existence of PRUS.

We conclude the paper with two remarks. First of all, whileTheorem 1 shows that a uniform distribution of phases leads toperfect sequences, for almost-perfect sequences we may focuson almost-uniform distributions. A clear possibility here canbe outlined as follows: in cases for which a perfect sequencedoes not exist, one can try to build sequences with a phasedistribution which approximately satisfies the principal equa-tions and approximately preserves the construction conditionsfor Φ(s)

e in Section D. Finally, we would like to emphasizethe possible connections between the study of PRUS and theSzemeredi theorem and related results which study the min-imal size and properties of subsets of Zn containing specificlength arithmetic progressions (see Appendix E). These resultsmight be usable to further examine the existence as well asthe construction for PRUS over αp.

APPENDIX APROOF OF THEOREM 1

We prove a more general form of the theorem by using someresults from the theory of algebraic numbers and minimalpolynomials: a number is called algebraic iff it is a root of apolynomial with rational coefficients. The minimal polynomialPmin(x) of an algebraic number x0 is the polynomial withrational coefficients, minimum degree and the leading coeffi-cient equal to 1 which satisfies Pmin(x0) = 0. It is known[34] that the minimal polynomial of the primitive nth root ofunity (ej

2πn ) is of degree d = ϕ(n) where ϕ (the Euler’s totient

function) shows the number of k ∈ Zn for which k and n areco-prime. For n = p, p prime, the minimal polynomial is of theunique form

∑p−1k=0 x

k [35]. We conclude that if P (ej2πp ) = 0

for a polynomial P (x) with rational coefficients and degreed = p − 1 then P (x) must be equal to w

∑p−1k=0 x

k for somerational scalar w. This implies that all coefficients of P (x)must be equal, which completes the proof.

APPENDIX BCONNECTIONS BETWEEN PRUS OF LENGTH L = p

OVER αp AND BINARY SEQUENCES WITHOPTIMAL PERIODIC CORRELATION

Let x be a PRUS of length p over αp. Let µk represent thenumber of times for which ej

2πp k occurs in x. It is shown in

Section III-B that the distribution of µk is given by (p− 1)/2, µk = 01, µk = 1(p− 1)/2, µk = 2

(41)

Since x is perfect, it has a constant magnitude of√p in the

discrete Fourier domain. Note that the value of the discreteFourier domain, at the frequency zero, represents the sum of

the sequence x; hence, for x =ej

2πp kl

p−1

l=0we have that∣∣∣∣∣

p−1∑l=0

ej2πp kl

∣∣∣∣∣ = √p, (42)

or equivalently ∣∣∣∣∣p−1∑k=0

µkej 2π

p k

∣∣∣∣∣ = √p. (43)

Now let rk = µk−1. Therefore, (43) can be rewritten basedon rk as ∣∣∣∣∣

p−1∑k=0

rkej 2π

p k

∣∣∣∣∣ = √p. (44)

The latter equality implies thatp−1∑k=0

(p−1∑l=0

rlrl+k

)ej

2πp k = p. (45)

By applying the result of Theorem 1 to (45), we obtain:(p−1∑l=0

r2l

)− p =

p−1∑l=0

rlrl+1 = · · · =p−1∑l=0

rlrl+p−1. (46)

On the other hand, it is interesting to note thatp−1∑l=0

r2l =

p−1∑l=0

(µl − 1)2= p− 1. (47)

Therefore, the sequence rk has in-phase autocorrelation ofp−1, and respectively, a constant out-of-phase autocorrelationof −1. Moreover, the distribution of rk is given by (p− 1)/2, rk = −1

1, rk = 0(p− 1)/2, rk = +1

(48)

which implies that rk is a balanced [36] punctured [37]binary sequence with only one zero. Note that replacing thezero element of the sequence with +1 or −1 can changethe out-of-phase correlation lags by 0, 2 or −2. We alsonote that all correlation values are congruent to the lengthof the sequence (i.e. p) modulo 4. Therefore, the out-of-phasecorrelation lags would turn to −1 for p ≡ 3, and to 1,−3for p ≡ 1 (mod 4).

The discussed idea of constructing binary sequences withoptimal periodic correlation from PRUS is summarized inTable III. As indicated earlier, there are several methods toconstruct a PRUS of length p over αp. However, we do notlimit our statements to the known construction methods as it isnot yet proven that the currently known construction methodscover all possible PRUS.

APPENDIX CSTUDY OF THE PRINCIPAL EQUATIONS IN (24) USING THE

SUM OF SQUARES PROBLEM

We note that the second equality in (24) may be viewed asa sum of squares problem. This approach can be applied inparticular to the case of p = 3. For p = 3, Gauss showed that a

11

TABLE IIICONSTRUCTION OF PRIME-LENGTH BINARY SEQUENCES WITH OPTIMAL

PERIODIC CORRELATION FROM PRUS

Step 0: Consider a PRUS of length L = p over αp.

Step 1: Let µk represent the number of times for which ej 2π

pk occurs

in the considered PRUS. Form the sequence rk = µk − 1.Step 2: Replace the only zero element in rk with +1 or −1.

natural number can be represented as the sum of three squaresiff it is not of the form 4k(8l−1), (k, l) ∈ Z2 [38]. For p > 3,one may note that according to a theorem by Lagrange everynatural number can be represented as the sum of four squares[38]. The latter result implies that every natural number canbe written as sum of p > 3 squares. We refer the interestedreader to [39]-[42] for additional information on cases p = 5,7, 11, and 13.

APPENDIX DEFFICIENT TEST METHOD FOR SPECIFIC LENGTHS

Herein we propose an efficient method for testing if a PRUSof a specific length might exist over αp and for determining itsphase distribution. This test method is useful for cases in whichthe length of the needed sequence is fixed or the derivationof closed form solutions of the principal equations for thesequence length and phase distribution over a desired alphabetis deemed to be expensive. Our test method is based on twoapproaches to reduce the size of search space for rk: (i)imposing adaptive bounds on rk and (ii) assigning rkto certain classes of residues for different integer values. Apreliminary bound on rk is given by the following lemma:

Lemma 1. Let r0, · · · , rp−1 be a solution to the principalequations; then

|rk| ≤ (p− 1)

√m

p, k ∈ Zp. (49)

Also if rk∗ has the maximum absolute value among all rk,then

|rk∗ | ≥(√

p− 1)√m

p. (50)

Proof: From the principal equations we have rk =−∑

l∈Zp−k rl, and as a result |rk| ≤∑

l∈Zp−k |rl|.Therefore,

m(p− 1) =

p−1∑l=0

r2l = r2k +∑

l∈Zp−k

r2l

≥ r2k + Γ

∑l∈Zp−k

|rl|, p− 1

≥ r2k + Γ (|rk|, p− 1)

=

(p

p− 1

)r2k (51)

which yields the inequality (49). Next, note that

m(p− 1) =

p−1∑k=0

r2k ≤ pr2k∗(52)

which implies the inequality (50).

Interestingly, similar bounds on rk could also be estab-lished in the case that some of rk are known. Let us assumethat we know the values of rkk∈Zq and let

S1 =∑q−1

k=0 rkS2 =

∑q−1k=0 r

2k

(53)

Therefore, for every k ∈ Zp − Zq ,

rk = −S1 −∑

l∈(Zp−Zq)−k

rl (54)

and as a result

m(p− 1)− S2 = r2k +∑

l∈(Zp−Zq)−k

r2l

≥ r2k + Γ (rk + S1, p− q − 1) . (55)

The above quadratic inequality implies that such an arrange-ment of rkk∈Zq might be possible only if

S21

p− q+ S2 ≤ m(p− 1), (56)

and that rkk∈Zp−Zq are bounded by

b± =−S1 ±

√(p− q − 1) ((p− q)(m(p− 1)− S2)− S2

1)

(p− q). (57)

These adaptive bounds help us make convenient successiveselections of rk.

In the following, we propose another useful idea to reducethe size of the search space, inspired by Minkowski-Hasseprinciple for quadratic forms [38]:

Minkowski-Hasse Principle. A quadratic form

Q (r0, · · · , rn−1) =∑

(k,l)∈Z2n

Qklrkrl (58)

of rank n with integral coefficients represents zero overthe rationals iff for any g ∈ Z − 0, the congruenceQ (r0, · · · , rn−1) ≡ 0 (mod g) has a primitive solution and inaddition Q represents zero over the reals, i.e. it is indefinite.

Let∑p−1

k=0 rk ≡ 0∑p−1k=0 r

2k ≡ m(p− 1)∑p−1

k=0 rkrk+s ≡ −m, s ∈ Zp − 0(59)

be a set of congruence mod g ∈ N. Note that the second andthird term of (59) are quadratic. It is also interesting to notethat, as rk are bounded according to Lemma 1 and all othervalues are known and finite, the necessity and sufficiency ofthe congruence are obvious. In fact, choosing a sufficientlylarge g turns the congruence into an equality by adding anintegral constant to rk. The second fact we need to take intoconsideration is that the sum

∑p−1k=0 rk, rk ∈ Zg , gets all

the residue values in Zg exactly gp−1 times. Therefore, bysearching over all rk ∈ Zg , the congruence set (59) mustreduce the search space at least by a factor of g. Starting from asmall g (say g = 2), we can omit at least (g−1)/g of the searchspace elements by testing at most gp elements. But, since many

12

of these elements are redundant for different g, it turns out thatfor sufficiently large values of g the number of newly omittedelements is less that the number of tested elements. Therefore,a combination of this method and the adaptive bounds in(57) appears to be more useful. Our proposed method canbe described as follows:

• 0: Consider the integral search space Ω bounded by theinequality (49). Without loss of generality, we assume thatr0 has the maximum value among rk and is boundedas in (50). Also let g = 2.

• 1: Solve the congruence set in (59) for g. This can bedone by a brute-force search over Zg .

• 2: Reduce the size of Ω by omiting elements whichbelong to residue classes not feasible for the congruenceset in (59). Let ∆g represent the number of omittedelements. If ∆g > gp, increase g by one and goto 1.

• 3: Consider all possible values of r0 in Ω and updateS1 and S2 for each of them. By considering all k =1, · · · , p− 1, respectively, do the following:

– 3.0: Establish the bounds b± as in (57).– 3.1: Consider all possible rk in Ω that follow the

bounds b± and their absolute value is at most equalto r0.

– 3.2: Update S1 and S2 for each rk considered in 3.1.• 4: Check whether the obtained rk satisfy the third part

of the principal equations (i.e. all out-of-phase correla-tions of rk are −m).

APPENDIX EFURTHER EFFICIENCY ASSESSMENT OF THE PROPOSED

CONSTRUCTION IN SECTION IV-B

To explain in more detail how the proposed approachcontributes an efficient construction scheme, consider thefollowing: let 1+ad(χk) be the length of the longest arithmeticprogression with common difference d in χk. For a PRUS, wemust have that

p−1∑k=0

ad(χk) ≤ m. (60)

On the other hand, ad(χk) are not independent for differentvalues of d as it can be checked (by construction) that

aqd(χk) ≥⌊1

qad(χk)

⌋(61)

for every q ∈ Zad(χk)− 0. This immediately shows that

a large ad(χk∗) (for a k∗ ∈ Zp) limits not only ad(χk)

where k ∈ Zp − k∗ but also ad(χk) for some other valuesof d. A preliminary result from (60) is that none of χkhas an arithmetic progression of length greater than m + 1.However, the number of elements of the set (A×) of sequencesfor which the assigned vectors χ0, · · · ,χl (l ∈ Zp) are notfeasible according to (60) and (61) is significantly larger thanthe number of sequences for which at least one of the elementsof the set χ0, · · · ,χl has an arithmetic progression of lengthgreater than m+1. We further note that A× is a subset of allsequences that the proposed construction approach identifiesas unsuitable for PRUS before assigning all χkk∈Zp .

ACKNOWLEDGEMENT

The authors are grateful to Prof. Matthew Geoffrey Parkerfor his detailed comments on an early version of this paper.

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