Table Of ContentOptimal One-shot Quantum Algorithm for
EQUALITY and AND
Andris Ambainis and J¯anis Iraids
Faculty of Computing, University of Latvia, Rain¸a bulv¯aris 19, Riga, LV-1586,
Latvia,[email protected],[email protected]
7
Abstract. Westudy the computation complexity of Boolean functions
1
0 in the quantum black box model. In this model our task is to compute
2 a function f : {0,1} → {0,1} on an input x ∈ {0,1}n that can be
accessed byqueryingtheblackbox.Quantumalgorithms areinherently
n
probabilistic; we are interested in the lowest possible probability that
a
J thealgorithmoutputsincorrectanswer(theerrorprobability)forafixed
numberofqueries.Weshowthatthelowestpossibleerrorprobabilityfor
4
2 ANDn and EQUALITYn+1 is 21 − n2n+1.
] Keywords: quantum query complexity; bounded error; total Boolean
h function; and; equality; single query
p
-
t
n 1 Introduction
a
u
q In this paper we study the computational complexity of Boolean functions in
[ the quantum black box model. It is a generalization of the decision tree model,
where we are computing an n-bit function f : {0,1}n → {0,1} on an input
1
x∈{0,1}n that canonly be accessedthrougha black box by queryingsome bit
v
2 xi of the input. In the quantum black box model the state of the computation
4 is described by a quantum state from the Hilbert space H ⊗H ⊗H where
Q W O
9 H ={|0i,|1i,...,|ni} is the query subspace, H is the working memory and
Q W
6
H = {|0i,|1i} is the output subspace. A computation using t queries consists
0 O
ofasequenceofunitarytransformationsU ·O ·U ·O ·...·O ·U followedbya
. t x t−1 x x 0
1
measurement,wheretheU ’sareindependentoftheinputandO =O ⊗I⊗I
i x Q,x
0
with
7
1 (−1)xi|ii=xˆi|ii, if i∈[n],
O |ii=
Q,x
v: (|0i, if i=0,
i
X isthe querytransformation,wherexi ∈{0,1}orequivalently,xˆi ∈{−1,1}.The
final measurement is a complete projective measurement in the computational
r
a basis and the output of the algorithm is the result of the last register, H .
O
For and 0 ≤ ǫ < 1 we denote by Q (f) the smallest number of queries for an
2 ǫ
quantumalgorithmoutputtingf(x)withprobabilityatleast1−ǫ.Usuallytheǫ
is omitted from Q (f) because it changes Q (f) by a constant factor, and Q(f)
ǫ ǫ
is called the bounded error quantum query complexity of f. This complexity
measure is widely studied as most computational problems can be expressed in
the query model. The most well known examples are by [3,8]. For the searching
problem Grover’s algorithm is exactly optimal as shown by [9].
However, if one is interested in computing functions with constant number
of inputs (for example, as a part of small circuit), then it may be useful to
fix the number queries and minimize the probability of an incorrect answer. In
this paper we will be concernedwith quantum algorithmsperforming at most 1
query, thus we introduce E(f).
Definition 1. Letf beaBoolean function.ThenletE(f)betheminimumerror
probability for a quantum algorithm that calculates f using just one query, i.e.,
E(f)= min maxPr[algorithm A does not output f(x)].
A:A performs1query x
We will be focusing on two Boolean functions defined as follows:
1, if x =x =...=x
EQUALITY (x)= 1 2 n
n
(0, otherwise
and
1, if x =x =...=x =1
AND (x)= 1 2 n .
n
(0, otherwise
In her doctoral thesis [5] gave quantum algorithms showing that
1 1
E(EQUALITY )≤ ;E(AND )≤ ;
3 10 2 10
1 1
E(EQUALITY )≤ ;E(AND )≤ ;
4 4 3 4
7 7
E(EQUALITY )≤ ;E(AND )≤ .
6 16 5 16
Our main result asserts that
Theorem 1.
1 n
E(AND )=E EQUALITY = − .
n n+1 2 n2+1
(cid:0) (cid:1)
The proof can be summarized in a series of three inequalities:
1 n 1 n
− ≤E(AND )≤E EQUALITY ≤ − .
2 n2+1 n n+1 2 n2+1
(cid:0) (cid:1)
The first inequality can be proven using a characterization of symmetric
sum-of-squares polynomials known as the Blekherman’s theorem.
Theorem 4 (Blekherman). Let q(xˆ) be the symmetrization of a polynomial
p2(xˆ)wherep(xˆ)isamultilinearpolynomialofdegreet≤ n andxˆ=(x ,...,x ).
2 1 n
Then, over the Boolean hypercube xˆ∈{−1,1}n,
t
q(xˆ)= p (|x|) (|x|−i)(n−|x|−i)
t−j
j=0 0≤i<j
X Y
where p is a univariate polynomial that is a sum of squares of polynomials of
t−j
degree at most t−j and |x| denotes the number of variables i:xˆ =−1.
i
Eventhoughitisanunpublishedresult,thereareproofs—see[4]orSection4
in this paper for a considerably shorter proof using representation theory.
The second inequality is trivial, since
AND (x ,...,x )=EQUALITY (x ,...,x ,1),
n 1 n n+1 1 n
and so we can use an algorithm for EQUALITY to calculate AND .
n+1 n
Thethird inequalitycanbe provedby constructingaquantumalgorithmfor
the function EQUALITY . Since the algorithm is very simple we present it
n+1
before the more involved proof of the first inequality.
If we compare E(f) with the classical analogue, let us call it EC(f), [5] has
shown that EC(EQUALITY )= 1 and EC(AND )= 1 − 1 .
n 2 n 2 4n−2
2 Algorithm for EQUALITY
Theorem 2.
1 n
E EQUALITY ≤ −
n+1 2 n2+1
(cid:0) (cid:1)
Proof. We will prove that the following algorithm has the claimed error proba-
bility:
Algorithm 1 Algorithm for EQUALITY
n+1
1: Statespace: |1i,|2i,...,|n+1i
2: Start in uniform superposition Pn+1 1 |ii
i=1 √n+1
3: Query:Pni=+11 √n1+1|ii−Q→Pni=+11 (√−n1+)x1i |ii
4: Perform quantum Fourier transform Fn+1|ii = Pnj=+11 ω(i−1)(jn−+11)(−1)xj |ii,ω =
en2+π1i:
Pni=+11 (√−n1+)x1i |ii−F−n−+→1 Pni=+11Pnj=+11 ω(i−1)(jn−+11)(−1)xj |ii
5: Perform a complete measurement
6: if the result is state |1i then
7: With probability 1 − n output 0; otherwise output 1
2 n2+1
8: else
9: Output0
First, let us consider the case when EQUALITY = 1. In that case the
n+1
state |1i will be measured with certainty and hence the probability to output
the incorrect answer 0 is 1 − n .
2 n2+1
If on the other hand the input is such that EQUALITY = 0, the algo-
n+1
rithm has an opportunity to answer incorrectly only in the case it measures
|1i. Denote by m := n+1x . The probability that the algorithm answers
i=1 i
2
1 is m · 1 + nP . The value of this expression is maximized when
n+1 2 n2+1
m =(cid:16)±(n−(cid:17) 1)(cid:16)and so the(cid:17)probability to answer 1 on the worst kind of input
(namely the input where only one bit is different from every other bit) is
n−1 2 1 n n−1 2 (n+1)2 n2+1−2n 1 n
+ = = = − .
n+1 2 n2+1 n+1 2(n2+1) 2(n2+1) 2 n2+1
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
⊓⊔
3 Lower Bound for AND
Theorem 3.
1 n
E(AND )≥ −
n 2 n2+1
Proof. First, we will restrict the domain of the inputs of the AND function to
n
bitlistswithHammingweightof0,n−1orn.Itturnsoutthatthispromiseprob-
lem has the same optimal error probability. Consider any quantum algorithm
computing AND with error probability ǫ. Following the familiar reasoning of
n
[1]wecanwritetheprobabilitythatthealgorithmoutputs1asasum-of-squares
polynomial of degree at most 2:
Pr[algorithm outputs 1]= p2(xˆ ,...,xˆ ).
i 1 n
i
X
FromBlekherman’stheoremweobtainthatbysymmetrizationtheremustexist
a degree at most 2 univariate polynomial of the form
p(s)= (a s+b )2+(n−s)s c2
i i j
i j
X X
such that
1−ǫ≤p(s)≤1
when s=|x|=n and
0≤p(s)≤ǫ
when s=|x| and s∈{0,n−1}.
A geometric representation of the potential regions where p(s) intersects
s=0,s=n−1ands=nisdepictedinFigure1.Clearly,adegree0polynomial
p — a constant, would produce an error probability ǫ = 1. Consider a degree
2
1 polynomial — a straight line. We will apply transformations to the line that
1
1−ǫ
1
2
ǫ
0
0 n−1 n
Fig.1. Regions where p(s) may intersect s=0, s=n−1 and s=n
do not increase (but may decrease) the error probability it achieves ǫ. First,
we stretch the line vertically with respect to the horizontal line y = 1 until it
2
passesthroughtheorigin.Thenwestretchthelineverticallywithrespecttothe
horizontal line y =0 until it passes through both (n−1,ǫ) and (n,1−ǫ). This
line has a slope ǫ = 1−ǫ and so ǫ= 1 − 1 .
n−1 n 2 4n−2
Finally, consider a degree 2 polynomial p — a parabola. If the parabola is
concave,we may reasonsimilarly as in the line case, except, the point (n−1,ǫ)
mustnowbeabovethe linepassingthrough(0,0)and(n,1−ǫ)andsothe error
probability is higher. If the parabola is convex, we consider further two cases.
a) If the vertex of the parabola has s ≤ 0, then we perform the same vertical
stretchings. Since the parabola now passes through (0,0) we can describe it
with an equation as2+bs where a>0. Since the vertex of the parabola has
s≤0,thecoefficientbmustbenon-negative.Thesmallestǫpossibleforsuch
parabolas can be described through the system
1−ǫ=maxan2+bn such that
a,b
an2+bn+a(n−1)2+b(n−1)=1
(b≥0
Fromtheequalitywecanexpressb= 1−a(n2+(n−1)2) andhencea≤ 1 .
2n−1 n2+(n−1)2
Plugging it all into the objective function we have that
n(1−a(n2+(n−1)2))
1−ǫ≤an2+ ≤
2n−1
n2 (n+1)2 1 n
≤ ≤ = +
n2+(n−1)2 2(n2+1) 2 n2+1
b) Ifthevertexoftheparabolahass≥0thenclearlythevertexhastobeinthe
interval s∈[0,n]. Therefore we use the property from Blekherman’s charac-
terization that the polynomial p(s) is non-negative in the interval s ∈ [0,n],
i.e., the term (a s+b )2 is non-negative everywhere and (n−s)s c2
i i i j j
is non-negative for s∈[0,n]. Now we stretch the parabola horizontally with
P P
respect to line s = n until p(0) = p(n −1). This will not increase ǫ and
preserve the non-negativity in the interval s ∈ [0,n]. Next we stretch the
parabola vertically with respect to line y = p(n) until p n−1 = 0. Again,
2
this step does not increase ǫ. Finally, we stretch vertically with respect to
(cid:0) (cid:1)
y =0 until p(0)=1−p(n). The last step preservedthe vertex at (n−1,0) so
2
the parabola has an equation
2
n−1
p(s)=a s− .
2
(cid:18) (cid:19)
But from the equation p(0)=1−p(n) we obtain
n−1 2 n+1 2
a =1−a ;
2 2
(cid:18) (cid:19) (cid:18) (cid:19)
2
a= .
n2+1
Consequently,
2 n−1 2 1 n
ǫ≥p(0)= = − .
n2+1 2 2 n2+1
(cid:18) (cid:19)
⊓⊔
Interestingly, the proof only really requires the sum-of-squares characterization
when n−1 is not an integer. The fact that the parabola p(s) is non-negative at
2
s= n−1 is sufficient.
2
4 Proof of Blekherman’s Theorem
In this section we prove Blekherman’s theorem.
Theorem 4 (Blekherman). Let q(xˆ) be the symmetrization of a polynomial
p2(xˆ)wherep(xˆ)isamultilinearpolynomialofdegreet≤ n andxˆ=(x ,...,x ).
2 1 n
Then, over the Boolean hypercube xˆ∈{−1,1}n,
t
q(xˆ)= p (|x|) (|x|−i)(n−|x|−i)
t−j
j=0 0≤i<j
X Y
where p is a univariate polynomial that is a sum of squares of polynomials of
t−j
degree at most t−j and |x| denotes the number of variables i:xˆ =−1.
i
Ourproofutilizesconceptsofrepresentationtheory.Foradescriptionofthecore
tools of representation theory that we require refer to the first two chapters of
[7].
4.1 Group representation
LetH beaHilbertspacewithbasisstatesxˆ (forallS ⊆[n])correspondingto
℘ S
monomials xˆ . Then, the vectors in H correspond to multilinear polyno-
i∈S i ℘
mialsinvariablesxˆ .Weconsideragrouprepresentationofthesymmetricgroup
i
S on H wQith transformations U defined by U xˆ = xˆ . The irreducible
n ℘ π π S π(S)
representations contained in H are well known:
℘
Let S (xˆ ,...,xˆ )= xˆ ...xˆ be the mth elementary symmetric
m 1 n i1,...,im i1 im
polynomial. We use S (xˆ ,...,xˆ ) to denote the constant 1.
0 1 n
P
Lemma 1. A subspaceH ⊆H is irreducible if andonlyifthereexistb andα
℘ m
−→
for m=0,1,...,n−2bsuchthat H is spannedbyvectors p corresponding
i1,...,jb
to polynomials p (for all choices of pairwise distinct i ,j ,...,i ,j ∈[n])
i1,...,jb 1 1 b b
where
n−2b
p (xˆ ,...,xˆ )=(xˆ −xˆ )...(xˆ −xˆ ) α S (xˆ′)
i1,...,jb 1 n i1 j1 ib jb m m
m=0
X
and xˆ′ ∈{−1,1}n−2b consists of all xˆ for i∈[n], i∈/ {i ,...,j }.
i 1 b
See [2] for a short proof of Lemma 1.
4.2 Decomposition of q(xˆ)
Let
p(xˆ ,...,xˆ )= a xˆ .
1 n S S
S:X|S|≤t
We associate p2(xˆ ,...,xˆ ) with the matrix (P ) with rows and columns
1 n S1,S2 −→
indexed by S ⊆ [n],|S| ≤ t defined by P = a a . Let x be a column
vector consisting of all xˆ for S :|S|≤t. TS1h,eSn2, p2(xˆS1,.S.2.,xˆ )=−→xTP−→x. This
S 1 n
means that P is positive semidefinite.
For a permutation π ∈S , let Pπ be the matrix defined by
n
Pπ =a a
S1,S2 π(S1) π(S2)
and let Q = 1 Pπ be the average of all Pπ. Then, q(xˆ) = −→xTQ−→x.
n! π∈Sn
Q is also positive semidefinite (as a linear combination of positive semidefinite
matrices Pπ withPpositive coefficients).
We decompose Q = λ Q with λ ranging over different non-zero eigen-
i i i i
values and Q being the projectors on the respective eigenspaces. Since Q is
i
P
positive semidefinite, we have λ >0 for all i.
i
WeinterprettransformationsUπaspermutationmatricesdefinedby(Uπ)S,S′ =
1 if S =π(S′) and (Uπ)S,S′ =0 otherwise. Then, we have
1 1 1
U QU† = U PτU† = Pπτ = Pτ =Q.
π π n! π π n! n!
τX∈Sn τX∈Sn τX∈Sn
Since we also have
U QU† = λ U Q U†,
π π i π i π
i
X
we must have Q = U Q U†. This means that Q is a projector to a subspace
i π i π i
H ⊆ H that is invariant under the action of S . If H is not irreducible, we
i ℘ n i
can decompose it into a direct sum of irreducible subspaces
H =H ⊕H ⊕...⊕H .
i i,1 i,2 i,mi
Then, we have Q = mi Q where Q is a projector to H and Q =
i j=1 i,j i,j i,j
λ Q . This means that we can decompose q(xˆ) = λ q (xˆ) where
q i,(jxˆ)i=i,j−→xTQ −→x anPd it suffices to show the theorem foir,joniei,pjolynomial
Pi,j i,j P
q (xˆ) instead of the whole sum q(xˆ).
i,j
4.3 Projector to one subspace.
LetH ⊆H beanirreducibleinvariantsubspace.Weclaimthattheprojection
℘,ℓ ℘
to the subspace H denoted by Π is of the following form:
℘,ℓ ℘,ℓ
Lemma 2.
Π =cρ where ρ = −→p −→pT
℘,ℓ ℘,ℓ ℘,ℓ i1,...,jb i1,...,jb
i1X,...,jb
for some constant c.
Proof. If we restrict to the subspace H , then Π is just the identity I.
℘,ℓ ℘,ℓ
On the right hand side, ρ is mapped to itself by any U (since any U
℘,ℓ π π
−→
permutes the vectors p in some way). Therefore, all U also map the
i1,...,jb π
eigenspaces of ρ to themselves. This means that, if ρ has an eigenspace
℘,ℓ ℘,ℓ
V ⊂ H , then U acting on V also form a representation of S but that
℘,ℓ π n
would contradict H being an irreducible representation. Therefore, the only
℘,ℓ
eigenspace of ρ is the entire H . This can only happen if ρ is cI for some
℘,ℓ ℘,ℓ ℘,ℓ
constant c. ⊓⊔
4.4 Final polynomial
From the previous subsection, it follows that q (xˆ) is a positive constant times
i,j
(xˆ −xˆ )2...(xˆ −xˆ )2S2(xˆ′)
i1 j1 ib jb
i1X,...,jb
where S(xˆ′) is a symmetric polynomial of degree at most t−b. Instead of the
sum, we consider the expected value of (xˆ −xˆ )2...(xˆ −xˆ )2S2(xˆ′) when
i1 j1 ib jb
i ,...,j are chosen randomly. (Since the sum and the expected value differ by
1 b
a constant factor, this is sufficient.)
Terms (xˆ −xˆ )2 are nonzero if and only if one of x and x is 1 and the
ik jk ik jk
other is −1. Then, for k =1, we have
2s(n−s)
Pr[{xˆ ,xˆ }={−1,1}]= ,
i1 j1 n(n−1)
since there are n(n−1) possible sets {xˆ ,xˆ } and s(n−s) of them contain one
2 i1 j1
1 and one −1. For k >1,
Pr[{xˆ ,xˆ }={−1,1}|{xˆ ,xˆ }={−1,1} for l∈[k−1]]
ik jk il jl
2(s−k+1)(n−s−k+1)
= ,
(n−2k+2)(n−2k+1)
since the condition {xˆ ,xˆ } = {−1,1} for l ∈ [k−1] means that, among the
il jl
remaining variables, there are s−k+1 variables xˆ = −1 and n−s−k+1
j
variables xˆ = 1 and n−2k+2 variables in total (and, given that, the k = 1
j
argument applies). Thus,
b 2bs(s−1)...(s−b+1)(n−s)...(n−s−b+1)
Pr (xˆ −xˆ )2 =1 = .
ik jk n(n−1)...(n−2b+1)
" #
k=1
Y
SinceS isasymmetricpolynomial,wehaveS(xˆ′)=S′(s′)whereS′ isapolyno-
mial of one variable s′, with s′ equal to the number of variables xˆ′ =−1. Since
j
there are b variables xˆ =−1 that do not appear in xˆ′, we have s′ =s−b. This
j
means that S′ can be rewritten as a polynomial in s (instead of s′).
5 Conclusion
In this paper we have shown that
1 n
E(AND )=E EQUALITY = − .
n n+1 2 n2+1
(cid:0) (cid:1)
There is a natural way to generalize E(f) to any fixed number of queries t. We
may denote it by E (f) and have
t
E (f)= min maxPr[algorithm A does not output f(x)].
t
A:Aperformstqueries x
From the numerical experiments of [6] it seems that the connection between
EQUALITY and AND goes much deeper.
n+1 n
Conjecture 1. For all positive integers t and n:
E EQUALITY =E (AND ).
t n+1 t n
(cid:0) (cid:1)
Acknowledgements
The research leading to these results has received funding from the European
UnionSeventhFrameworkProgramme(FP7/2007-2013)undergrantagreement
n◦ 600700(QALGO),ERCAdvancedGrantMQC,LatvianStateResearchpro-
gramme NexIT project No.1.
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