Adjacent vertices can be hard to find by quantum walks
AAdjacent vertices can be hard to find by quantumwalks ∗ Nikolajs Nahimovs and Raqueline A. M. Santos
Faculty of Computing, University of LatviaRaina bulv. 19, Riga, LV-1586, Latvia [email protected], [email protected]
Abstract
Quantum walks have been useful for designing quantum algorithms thatoutperform their classical versions for a variety of search problems. Mostof the papers, however, consider a search space containing a single markedelement only. We show that if the search space contains more than onemarked element, their placement may drastically affect the performance ofthe search. More specifically, we study search by quantum walks on generalgraphs and show a wide class of configurations of marked vertices, for whichsearch by quantum walk needs Ω( N ) steps, that is, it has no speed-up overthe classical exhaustive search. The demonstrated configurations occur forcertain placements of two or more adjacent marked vertices. The analysis isdone for the two-dimensional grid and hypercube, and then is generalizedfor any graph. Quantum walks are quantum counterparts of classical random walks [9]. Similarlyto classical random walks, there are two types of quantum walks: discrete-timequantum walks, first introduced by Aharonov et al. [1], and continuous-timequantum walks, introduced by Farhi et al. [5]. For the discrete-time version, thestep of the quantum walk is usually given by coin and shift operators, which areapplied repeatedly. The coin operator acts on the internal state of the walker andrearranges the amplitudes of going to adjacent vertices. The shift operator movesthe walker between the adjacent vertices. ∗ This work was supported by the European Union Seventh Framework Programme (FP7/2007-2013) under the QALGO (Grant Agreement No. 600700) project and the RAQUEL (GrantAgreement No. 323970) project, the Latvian State Research Programme NeXIT project No. 1,and the ERC Advanced Grant MQC. a r X i v : . [ qu a n t - ph ] O c t uantum walks have been useful for designing algorithms for a variety ofsearch problems [3, 6, 2]. To solve a search problem using quantum walks, weintroduce the notion of marked elements (vertices), corresponding to elements ofthe search space that we want to find. We perform a quantum walk on the searchspace with one transition rule at the unmarked vertices, and another transitionrule at the marked vertices. If this process is set up properly, it leads to a quantumstate in which marked vertices have higher probability than the unmarked ones.This method of search using quantum walks was first introduced in [11], whichdescribes a quantum search in the hypercube, and has been used many timessince then.Not many papers in the literature consider search by quantum walks withmultiple marked vertices. Wong [13] analyzed the spatial search problem solved bycontinuous-time quantum walk on the simplex of complete graphs and showed thatthe location of marked vertices can dramatically influence the required jumpingrate of the quantum walk. Wong and Ambainis [14] analysed the discrete-timequantum walk on the simplex of complete graphs and showed that if one of thecomplete graphs is fully marked then there is no speed-up over classical exhaustivesearch. Nahimovs and Rivosh [8] studied the dependence of the running time ofthe AKR algorithm [3] on the number and the placement of marked locations.They found some “exceptional configurations” of marked vertices, for which theprobability of finding any of the marked vertices does not grow over time. Anotherpreviously known exceptional configuration for the two-dimensional grid is the“diagonal construction” by Ambainis and Rivosh [4].In this paper, we extend the work of Nahimovs and Rivosh [7]. We studysearch by quantum walks on general graphs with multiple marked vertices andshow a wide class of configurations of marked vertices, for which the probability offinding any of the marked vertices does not grow over time. These configurationsoccur for certain placements of two and more adjacent marked vertices. We provethat for such configurations the state of the algorithm is close to a stationarystate.We start by reviewing the simple example of the two-dimensional grid from [7]by showing that any pair of adjacent marked vertices forms an exceptionalconfiguration. The same construction is valid for the hypercube. We extend theproof to general graphs by showing that any pair of adjacent marked verticeshaving the same degree d forms an exceptional configuration, for which theprobability of finding a marked vertex is limited by const · d /N . Then, weprove that any k -clique of marked vertices forms an exceptional configuration.Additionally, we formulate general conditions for a state to be stationary given aconfiguration of marked vertices. Our results greatly extend the class of knownexceptional configurations. 2 Two-dimensional grid
Consider a two-dimensional grid of size √ N × √ N with periodic (torus-like)boundary conditions. Let us denote n = √ N . The locations of the grid arelabeled by the coordinates ( x, y ) for x, y ∈ { , . . . , n − } . The coordinates definea set of state vectors, | x, y (cid:105) , which span the Hilbert space, H P , associated tothe position. Additionally, we define a 4-dimensional Hilbert space with the setof states {| c (cid:105) : c ∈ {← , → , ↑ , ↓}} , H C , associated with the direction. We referto it as the direction or the coin subspace. The quantum walk has associatedHilbert space H P ⊗ H C with basis states | x, y, c (cid:105) for x, y ∈ { , . . . , n − } and c ∈ {↑ , ↓ , ← , →} .The evolution of a state of the walk is driven by the unitary operator U = S · ( I ⊗ C ), where S is the flip-flop shift operator S | i, j, ↑(cid:105) = | i, j + 1 , ↓(cid:105) (1) S | i, j, ↓(cid:105) = | i, j − , ↑(cid:105) (2) S | i, j, ←(cid:105) = | i − , j, →(cid:105) (3) S | i, j, →(cid:105) = | i + 1 , j, ←(cid:105) , (4)and C is the coin operator, given by the Grover’s diffusion transformation C = 12 − − − − . (5)The spatial search algorithm uses the unitary operator U (cid:48) = S · ( I ⊗ C ) · ( Q ⊗ I ),where Q is the query transformation which flips the sign of marked vertices, thatis, Q | x, y (cid:105) = −| x, y (cid:105) , if ( x, y ) is marked and Q | x, y (cid:105) = | x, y (cid:105) , otherwise. Theinitial state of the algorithm is | ψ (0) (cid:105) = 1 √ N n − (cid:88) i,j =0 (cid:0) | i, j, ↑(cid:105) + | i, j, ↓(cid:105) + | i, j, ←(cid:105) + | i, j, →(cid:105) (cid:1) . (6)Note that | ψ (0) (cid:105) is a 1-eigenvector of U but not of U (cid:48) . If there are marked vertices,the state of the algorithm starts to deviate from | ψ (0) (cid:105) . In case of one markedvertex, after O ( √ N log N ) steps the inner product (cid:104) ψ ( t ) | ψ (0) (cid:105) becomes close to0. If we measure the state at this moment, we will find the marked vertex with O (1 / log N ) probability [3]. This gives the total running time of O ( √ N log N )steps with amplitude amplification. 3y analyzing the quantum search algorithm for a group of marked vertices ofsize √ k × √ k , Ref. [7] identified that the algorithm does not work as expectedwhen k is even, meaning that the overlap of the initial state with the state attime t , (cid:104) ψ (0) | ψ ( t ) (cid:105) , stays close to 1. Moreover, the same effect holds for any blockof size 2 k × l or k × l , with l and k being positive integers. The reason for suchbehavior is that blocks of marked vertices form stationary states, as we are goingto see below. Consider a two-dimensional grid with two marked vertices ( i, j ) and ( i, j + 1).Let | φ astat (cid:105) be a state having amplitudes of all basis states equal to a except for | i, j, →(cid:105) and | i + 1 , j, ←(cid:105) , which have amplitudes equal to − a (see Fig. 1), thatis, ( i + 1 , j )( i, j ) − a − aa aaaa a ...... ...... . . .. . . a aa a ...... . . .a aa a ...... . . . Figure 1: The amplitudes of the state | φ astat (cid:105) . The vertices ( i, j ) and ( i + 1 , j ) aremarked. | φ astat (cid:105) = n − (cid:88) x,y =0 (cid:88) c a | x, y, c (cid:105) − a | i, j, →(cid:105) − a | i + 1 , j, ←(cid:105) . (7)Then, this state is not changed by a step of the algorithm. Lemma 1.
Consider a grid of size √ N × √ N with two adjacent marked vertices ( i, j ) and ( i + 1 , j ) . Then the state | φ astat (cid:105) , given by Eq. (7), is not changed by thestep of the algorithm, that is, U (cid:48) | φ astat (cid:105) = | φ astat (cid:105) .Proof. Consider the effect of a step of the algorithm on | φ astat (cid:105) . The querytransformation changes the signs of all the amplitudes of the marked vertices.The coin transformation performs an inversion about the average: for unmarkedvertices, it does nothing, as all amplitudes are equal to a ; for marked vertices,the average is 0, so applying the coin results in sign flip. Thus, ( I ⊗ C )( Q ⊗ I )does nothing for the amplitudes of the non-marked vertices and twice flips thesign of the amplitudes of the marked vertices. Therefore, we have ( I ⊗ C )( Q ⊗ ) | φ astat (cid:105) = | φ astat (cid:105) . The shift transformation swaps the amplitudes of near-byvertices. For | φ astat (cid:105) , it swaps a with a and − a with − a . Thus, we have S ( I ⊗ C )( Q ⊗ I ) | φ astat (cid:105) = | φ astat (cid:105) .The initial state of the algorithm, given by Eq. (6), can be written as | ψ (0) (cid:105) = | φ astat (cid:105) + 4 a ( | i, j, →(cid:105) + | i + 1 , j, ←(cid:105) ) , (8)for a = 1 / √ N . Therefore, the only part of the initial state which is changed bythe step of the algorithm is2 √ N ( | i, j, →(cid:105) + | i + 1 , j, ←(cid:105) ) . (9)Let us establish an upper bound on the probability of finding a marked vertex, p M = (cid:104) ψ ( t ) | (cid:32) (cid:88) v ∈ M | v (cid:105)(cid:104) v | ⊗ I (cid:33) | ψ ( t ) (cid:105) , (10)where M is the set of marked vertices. Lemma 2.
Consider a grid of size √ N × √ N with two adjacent marked vertices ( i, j ) and ( i, j + 1) . Then for any number of steps, the probability of finding amarked vertex p M is O (cid:0) N (cid:1) .Proof. Follows from the proof of Theorem 2 by substituting d = 4 and m =2 N .Fig. 2 shows the absolute value of the overlap, |(cid:104) ψ (0) | ψ ( t ) (cid:105)| , and the probabilityof finding a marked vertex, p M , during the first 100 steps of the walk for a grid ofsize 50 ×
50 and two different sets of marked vertices. In the first case (solid line),we have two adjacent marked vertices, M = { (0 , , (0 , } and in the second case(dashed line), we have M = { (0 , , (0 , } . Clearly, one can see the effect of thestationary state on the evolution. If the two marked vertices are adjacent, theoverlap stays closes to 1 and the probability of finding a marked vertex stays closeto the probability in the initial state. If the two marked vertices are not adjacent,the quantum walk behaves as expected (as in the single marked vertex case).Note that if we have a block of marked vertices of size k × m we can constructa stationary state as long as we can tile the block by the blocks of size 1 × ×
1. For example, consider M = { (0 , , (0 , , (2 , , (3 , } for n ≥
3. Then thestationary state is given by | φ astat (cid:105) = n − (cid:88) x,y =0 (cid:88) c a | x, y, c (cid:105)− a | , , →(cid:105)− a | , , ←(cid:105)− a | , , ↑(cid:105)− a | , , ↓(cid:105) . (11)More details on alternative constructions of stationary states for blocks of markedvertices on the two-dimensional grid can be found in [7].5igure 2: Probability of finding a marked vertex, p M , and absolute value of theoverlap, |(cid:104) ψ (0) | ψ ( t ) (cid:105)| , for the first 100 steps of the quantum walk on a grid ofsize 50 ×
50. (Solid line) We have two adjacent marked vertices, (0 ,
0) and (0 , ,
0) and (0 , The n -dimensional hypercube is a graph with N = 2 n vertices where each vertexhas degree n . The discrete-time quantum walk has associated Hilbert space H n ⊗ H n . The evolution operator is given by U = S · ( I ⊗ C ), where the shiftoperator, S , acts in the following way S | (cid:126)v (cid:105)| c (cid:105) = | (cid:126)v ⊕ (cid:126)e c (cid:105)| c (cid:105) , (12)with (cid:126)v being the binary representation of v . This operator moves the walkerfrom state | (cid:126)v (cid:105)| c (cid:105) to | (cid:126)v ⊕ (cid:126)e c (cid:105)| c (cid:105) , where (cid:126)e c is the binary vector with 1 in the c -thposition. Note, that vertices are connected to each other if they differ in only onebit position. The coin transformation is the Grover coin C = 2 | s (cid:105)(cid:104) s | − I , where | s (cid:105) = √ n (cid:80) n − i =0 | i (cid:105) .Searching for marked vertices in the hypercube is done by using the unitaryoperator U (cid:48) = S · ( I ⊗ C ) · ( Q ⊗ I ), where Q is the query transformation, whichflips the sign of marked vertices. The initial state of the algorithm is given by | ψ (0) (cid:105) = 1 √ nN N − (cid:88) (cid:126)v =0 n − (cid:88) c =0 | (cid:126)v (cid:105)| c (cid:105) . (13)In case of SKW algorithm with one marked vertex [11], if we measure the stateof the quantum walk after O ( √ N ) time steps, we will find the marked vertex withprobability 1 / − O (1 /n ). Hence, we expect to repeat the algorithm a constantnumber of times, which gives the total running time of O ( √ N ) steps.6 .2 Stationary states for the hypercube Consider a hypercube with two adjacent marked vertices i and j . Without lossof generality, suppose (cid:126)i and (cid:126)j differ in the first bit. Let | φ astat (cid:105) be a state havingamplitudes of all basis states equal to a except for | (cid:126)i, (cid:105) and | (cid:126)j, (cid:105) which haveamplitudes equal to − ( n − a (see Fig. 3), that is, | φ astat (cid:105) = a N − (cid:88) v =0 n − (cid:88) c =0 | v, c (cid:105) − an (cid:16) | (cid:126)i, (cid:105) + | (cid:126)j, (cid:105) (cid:17) . (14) i j − ( n − a − ( n − a . . .. . .. . .. . .. . .. . . aaa ... aa a ... Figure 3: Amplitudes of the stationary state in an n -dimensional hypercube withtwo adjacent marked vertices i and j . Lemma 3.
Consider an n -dimensional hypercube with two adjacent markedvertices i and j . Then | φ astat (cid:105) , given by Eq. (14), is not changed by a step of thealgorithm, that is, U (cid:48) | φ astat (cid:105) = | φ astat (cid:105) .Proof. Similar to proof of Lemma 1.The initial state of the algorithm, given by Eq. (13), can be written as | ψ (0) (cid:105) = | φ astat (cid:105) + an (cid:16) | (cid:126)i, (cid:105) + | (cid:126)j, (cid:105) (cid:17) , (15)for a = 1 / √ nN . Therefore, the only part of the initial state, which is changed bya step of the algorithm is (cid:114) nN (cid:16) | (cid:126)i, (cid:105) + | (cid:126)j, (cid:105) (cid:17) . (16)From this fact, we can establish an upper bound for the probability of finding amarked vertex. Lemma 4.
Consider an n -dimensional hypercube with two adjacent markedvertices i and j . Then for any number of steps, the probability of finding a markedvertex p M is O (cid:16) n N (cid:17) .Proof. Follows from the proof of Theorem 2 by substituting d = n and m =( nN ) /
2. 7ig. 4 shows the probability of finding a marked vertex and the absolute valueof the overlap, |(cid:104) ψ (0) | ψ ( t ) (cid:105)| , for a hypercube of dimension n = 10 for the first 100steps of the algorithm. We consider two different sets of marked vertices. In thefirst case (solid line), we have two adjacent marked vertices M = { , } . In thiscase, the overlap stays close to 1 and the probability stays close to the probabilityin the initial state, because the quantum walk has a stationary state. In theFigure 4: Probability of finding a marked vertex, p M , and absolute value ofthe overlap, |(cid:104) ψ (0) | ψ ( t ) (cid:105)| , for 100 steps of the quantum walk on hupercube with N = 2 vertices. (Solid line) We have two adjacent marked vertices, 0 and 1.(Dashed line) We have two non-adjacent marked vertices, 0 and 3.second case (dashed line), we have two non-adjacent marked vertices M = { , } .As one can see, the behavior in the second case is very different from the behaviorin the first case.Note that any set of marked vertices which can be divided into unique blocksof two adjacent marked vertices will result in a stationary state. Consider a graph G = ( V, E ) with a set of vertices V and a set of edges E . Let n = | V | and m = | E | . The discrete-time quantum walk on G has associatedHilbert space H m with the set of basis states {| v, c (cid:105) : v ∈ V, ≤ c < d v } , where d v is the degree of vertex v . Note, that the state | v, c (cid:105) cannot be written as | v (cid:105) ⊗ | c (cid:105) unless G is regular.The evolution operator is given by U = SC. (17)8he coin transformation C is the direct sum of coin transformations for indi-vidual vertices, i.e. C = C d (cid:76) · · · (cid:76) C d n with C d i being the Grover diffusiontransformation of dimension d i . The shift operator S acts in the following way, S | v, c (cid:105) = | v (cid:48) , c (cid:48) (cid:105) , (18)where v and v (cid:48) are adjacent, c and c (cid:48) represent the directions that points v to v (cid:48) and v (cid:48) to v , respectively.Searching for marked vertices is done by using the unitary operator U (cid:48) = SCQ, (19)where Q is the query transformation, which flips the signs of the amplitudes atthe marked vertices, that is, Q = I − (cid:88) w ∈ M d w − (cid:88) c =0 | w, c (cid:105)(cid:104) w, c | , (20)with M being the set of marked vertices. The initial state of the algorithm is theequal superposition over all vertex-direction pairs: | ψ (0) (cid:105) = 1 √ m n − (cid:88) v =0 d v − (cid:88) c =0 | v, c (cid:105) . (21)It can be easily verified that the initial state stays unchanged by the evolutionoperator U , regardless of the number of steps.The running time of the algorithm depends on both the structure of thegraph as well as the placement of marked vertices. Next, we describe a class ofexceptional configurations of marked vertices, for which the probability of findinga marked vertex is limited. It is not difficult to see that the construction of stationary states for the two-dimensional grid and the hypercube can be generalized to any graph. First, weconsider only two adjacent marked vertices. Then, we show we can also constructa stationary state for three adjacent marked vertices. Later, we describe generalconditions for a state to be stationary.
Consider a graph G = ( V, E ) with two adjacent marked vertices i and j with thesame degree, that is, d i = d j = d . Let | φ astat (cid:105) be a state having all amplitudes9qual to a except of the amplitude of vertex i pointing to vertex j and amplitudeof vertex j pointing to vertex i , which are equal to − ( d − a . Fig. 5 shows theconfiguration of the amplitudes in the marked vertices. Then, this state is notchanged by a step of the algorithm. i j − ( d − a − ( d − a aaa ... aa a ... d − d − Figure 5: The amplitudes for the stationary state with two adjacent markedvertices i and j . Theorem 1.
Let G = ( V, E ) be a graph with two adjacent marked vertices i and j with d i = d j = d , and let | φ ai,j (cid:105) = − ad (cid:0) | i, c ( i,j ) (cid:105) + | j, c ( j,i ) (cid:105) (cid:1) , (22) where c ( i,j ) represents the direction which points vertex i to vertex j . Then, | φ astat (cid:105) = a n − (cid:88) v =0 d v − (cid:88) c =0 | v, c (cid:105) + | φ ai,j (cid:105) , (23) is not affected by a step of the algorithm, that is, U (cid:48) | φ astat (cid:105) = | φ astat (cid:105) .Proof. Consider the effect of a step of the algorithm. The query transformationchanges the sign of all amplitudes of the marked vertices. The coin flip performsan inversion about the average of these amplitudes: for unmarked vertices, it doesnothing as all amplitudes are equal to a ; for marked vertices, the average is 0, soit results in sign flip. Thus, CQ does nothing for the amplitudes of the unmarkedvertices and twice flips the sign of amplitudes of the marked vertices. Therefore,we have CQ | φ astat (cid:105) = | φ astat (cid:105) . (24)The shift transformation swaps amplitudes of adjacent vertices. For | φ astat (cid:105) , itswaps a with a and − ( d − a with − ( d − a . Thus, we have SCQ | φ astat (cid:105) = | φ astat (cid:105) . (25)The initial state of the algorithm | ψ (0) (cid:105) , given by Eq. (21), can be written as | ψ (0) (cid:105) = | φ astat (cid:105) − | φ ai,j (cid:105) , (26)10or a = 1 / √ m . Therefore, the only part of the initial state which is changed bya step of the algorithm is | φ ai,j (cid:105) . From this fact, we can establish an upper boundfor the probability of finding a marked vertex. Theorem 2.
Let G = ( V, E ) be a graph with two adjacent marked vertices i and j with d i = d j = d , and let the probability of finding a marked vertex be given by p M = (cid:104) ψ ( t ) | (cid:32) (cid:88) v ∈ M d v − (cid:88) c =0 | v, c (cid:105)(cid:104) v, c | (cid:33) | ψ ( t ) (cid:105) , (27) where | ψ ( t ) (cid:105) = U t | ψ (0) (cid:105) . Then, p M = O (cid:16) d m (cid:17) , where m is the number of edges ofthe graph.Proof. The only part of the initial state | ψ (0) (cid:105) which is changed by the step ofthe algorithm is | φ ai,j (cid:105) = − ad (cid:0) | i, c ( i,j ) (cid:105) + | j, c ( j,i ) (cid:105) (cid:1) , for a = 1 / √ m . Since theevolution is unitary, this part will keep its norm unchanged. In this way, we wantto find how big amplitudes can get in order to maximize the value of p M . Thismeans we want to maximize the function2( d − a + x ) + 2( − ( d − a − x ) , (28)subject to 2( d − x + 2 x = ||| φ ai,j (cid:105)|| = 2 a d . Note that x represents theamplitudes going from the marked vertices to unmarked vertices and x representsthe amplitudes going from one marked vertex to the other. Then, we obtain p M ≤ a (2 (cid:112) ( d − d + d (2 d − O (cid:18) d m (cid:19) . (29)One of corollaries of Theorem 2 is that if the degree of the marked vertices isconstant or if it does not grow as a function of n , then for large n , the probabilityof finding a marked vertex will stay close to the initial probability.Consider the example of the graphs in Fig. 6. Intuitively, one might expectthat the probability of finding a marked vertex in graph 6a would be bigger thanin 6b. However, Fig. 7 shows something different. We calculate the probability offinding a marked vertex and the absolute value of the overlap, |(cid:104) ψ (0) | ψ ( t ) (cid:105)| , for100 steps of the evolution and when the size of the two complete graphs attachedto the marked vertices is 10. The graph 6a produces a stationary state, and thatis why its overlap stays very close to 1 and the probability of finding a markedvertex is smaller than in the other graph.11 j (a) Two adjacent marked vertices withthe same degree. i j (b) Two adjacent marked vertices withdifferent degrees. Figure 6: Graphs with two adjacent marked vertices i and j connected to twocomplete graphs of size 5. ( a ) The marked vertices have the same degree. ( b ) Themarked vertices have different degrees.Figure 7: Probability of finding a marked vertex, p M , and absolute value of theoverlap, |(cid:104) ψ (0) | ψ ( t ) (cid:105)| , for 100 steps of the quantum walk. (Solid line) Graph 6awith two adjacent marked vertices with the same degree and complete graphs ofsize 10. (Dashed line) Graph 6b with two adjacent marked vertices with differentdegree and complete graphs of size 10. Corollary 1.
Suppose the set of marked vertices, M , can be divided in groups oftwo adjacent marked vertices without intersecting each other. Let M (cid:48) be the setof such pairs. Then | φ astat (cid:105) = n − (cid:88) v =0 d v − (cid:88) c =0 a | v, c (cid:105) + (cid:88) ( i,j ) ∈ M (cid:48) | φ ai,j (cid:105) (30) is not changed by a step of the algorithm, that is, U (cid:48) | φ astat (cid:105) = | φ astat (cid:105) .Proof. Follows from the proof of Theorem 1.
Now, consider a graph G = ( V, E ) with three adjacent marked vertices i , j and k , that is, a marked triangle. The stationary state for this case will have the12mplitudes in the marked vertices as depicted in Fig. 8. i jk − al ij − al ik − al jk aaa ... aa a ... aaa · · · d j − d k − d i − Figure 8: Sketch of amplitudes for the stationary state with three adjacent markedvertices i , j and k .Note that in order to have a stationary state the sum of amplitudes of eachmarked vertex should be 0, so the action of the coin operator will be a sign flip.By solving the following system of equations: l ij + l ik = d i − l ij + l jk = d j − l ik + l jk = d k − l ij = d i + d j − d k − , l ik = d i + d k − d j − l jk = d j + d k − d i − . (32) Theorem 3.
Let G = ( V, E ) be a graph with three adjacent marked vertices i , j and k ; and let | φ ai,j,k (cid:105) = − a ( l ij + 1) (cid:0) | i, c ( i,j ) (cid:105) + | j, c ( j,i ) (cid:105) (cid:1) − a ( l ik + 1) (cid:0) | i, c ( i,k ) (cid:105) + | k, c ( k,i ) (cid:105) (cid:1) −− a ( l jk + 1) (cid:0) | i, c ( j,k ) (cid:105) + | k, c ( k,j ) (cid:105) (cid:1) , (33) where l ij , l ik , and l jk are defined in (32). Then, | φ astat (cid:105) = a n − (cid:88) v =0 d v − (cid:88) c =0 | v, c (cid:105) + | φ ai,j,k (cid:105) , (34) is not affected by a step of the quantum walk on G .Proof. Similar to Theorem 1. 13ig. 9 shows two graphs with three adjacent marked vertices i, j and k connected to three complete graphs. Remember, that for three adjacent markedvertices it doesn’t matter if their degrees are different or equal. Both cases willresult in a stationary state. i jk (a) Three adjacent marked vertices with d i = 4 , d j = 3 and d k = 5. ki j (b) Three adjacent marked vertices with d i = 4 , d j = 3 and d k = 6. Figure 9: Graphs with three marked vertices connected to three complete graphsof size 5.In Fig. 10, we can see how the probability of finding a marked vertex dependson the degree of the marked vertices. Although, both graphs have a stationarystate, the maximum probability achieved by 9b is bigger. Moreover, if we considerthat the size of the complete graphs can grow, then for graph 9a the probabilityof finding a marked vertex will stay close to the initial probability (the degree ofthe marked vertices does not scales as a function of the number of vertices), whilefor graph 9b it will increase ( d k increases with the size of the complete graph). k -clique of marked vertices Next, we generalize the previous result for any complete subgraph of markedvertices.
Theorem 4.
Let G = ( V, E ) be a graph with k marked vertices v , . . . , v k forminga k -clique. Then it forms an exceptional configuration.Proof. Let d v j = ( k −
1) + d (cid:48) j , where d (cid:48) j is the number of edges of v j outside theclique. To construct a stationary state, we need to assign amplitudes to internaledges of the clique, so that the amplitudes in vertex v j sum up to d (cid:48) j .Without a loss of generality let d (cid:48) < d (cid:48) < · · · < d (cid:48) k . We set the amplitude ofthe edge ( v , v ) to − ad (cid:48) and amplitudes of other edges within the clique outgoingfrom v to 0. By this, we have satisfied the condition for the vertex v and reduced14igure 10: Probability of finding a marked vertex, p M , and absolute valueof the overlap, |(cid:104) ψ (0) | ψ ( t ) (cid:105)| , for 100 steps of the quantum walk. (Solid line)Graph 9a with complete graphs of size 20 and three adjacent marked verticeswith d i = 4 , d j = 3 and d k = 5. (Dashed line) Graph 9b with complete graphs ofsize 20 and three adjacent marked vertices with d i = 4 , d j = 3 and d k = 21.the problem from size k to k −
1. I.e. now we have a ( k − d (cid:48) − d (cid:48) ) , d (cid:48) , . . . , d (cid:48) k . Next, we recursively repeat the previous step until we get a3-clique, which always have an assignment. In this way, we have constructed astationary state for a k -clique of marked vertices.Note that any set of marked vertices which can be divided into unique blocksof two adjacent marked vertices with the same degree and/or k -clique markedvertices will result in a stationary state. In this section, we describe general conditions in which a state is stationary.
Theorem 5 (General conditions) . Let | ψ (cid:105) be a state with the following properties:1 All amplitudes of the unmarked vertices are equal;2 The sum of the amplitudes of any marked vertex is 0;3 The amplitudes of two adjacent vertices pointing to each other are equal.Then, | ψ (cid:105) is not changed by a step of the quantum walk, that is, U | ψ (cid:105) = | ψ (cid:105) .Proof. Item 1 is required in order for the coin transformation to have no effect onthe unmarked vertices. It is easy to see that C n | u (cid:105) = | u (cid:105) , where | u (cid:105) = a (cid:80) n − i =0 | i (cid:105) for some constant a . 15tem 2 is necessary so the coin transformation can flip the signs of the ampli-tudes in the marked vertices. Note that previously the sign of these amplitudeswere inverted by the query transformation.Item 3 is necessary for the shift transformation to have no effect on thestate.Note that the aforementioned conditions are established for the case CQ | ψ (cid:105) = | ψ (cid:105) and S | ψ (cid:105) = | ψ (cid:105) . There might be even more general conditions for the case U (cid:48) | ψ (cid:105) = | ψ (cid:105) . In this paper, we have demonstrated a wide class of exceptional configurations ofmarked vertices for the quantum walk based search on various graphs. The abovephenomenon is purely quantum. Classically, additional marked vertices resultin the decrease of the number of steps of the algorithm and the increase of theprobability of finding a marked location. Quantumly, the addition of a markedvertex can drastically drop the probability of finding a marked location.An open question is whether the found phenomenon has analogs for othermodels of quantum walks (continuous time quantum walks [5], Szegedy’s quantumwalk [12], staggered quantum walks [10], etc.) as well as for alternative coinoperators.Another open question is algorithmic applications of the found effect. Forexample, in case of two-dimensional grid the search algorithm can “distinguish”between odd-times-odd and even-times-even groups of marked locations. More-over, if there are multiple odd-times-odd and even-times-even groups of markedlocations, the algorithm will find only odd-times-odd groups and will “ignore”even-times-even groups. Nothing like this is possible for classical random walkswithout adding additional memory resources and complicating the algorithm.Another example is the general graphs where the algorithm “ignores” marked tri-angles. We think that the found phenomenon should have algorithmic applicationswhich would be very interesting to find.
Acknowledgements.
The authors thank Andris Ambainis for helpful ideasand suggestions, which was a great help during this research, and Tom Wong foruseful comments and careful revision of the manuscript.
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