Lecture 8: Period Finding: Simon s Problem over Z N

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1 Quantum Computation (CMU 8-859BB, Fall 205) Lecture 8: Period Finding: Simon Problem over Z October 5, 205 Lecturer: John Wright Scribe: icola Rech Problem A mentioned previouly, period finding i a rephraing of Simon algorithm, but intead of uing the Fourier tranform over Z n 2, we will ue the Fourier tranform over Z. Moreover, an efficient quantum algorithm for period finding will eentially give u Shor algorithm [Sho97] for efficient factorization the detail will be preented next lecture. Definition. (Period-Finding Problem). Given i ome f : Z color (i.e. the image of f i ome untructured et). Pictorially, f can be imagined a an array: R G B Y R G B Y } {{ } length A uual, we have oracle acce to f, and we denote the oracle by O f. For thi problem, it i mot convenient to work with the oracle which behave a follow: O f ( x b ) = x b f(x), where b i an m-qubit tring. We have the promie that f i periodic; namely for ome Z \ {0}, f(x) = f(x + ) for all x Z. Otherwie, all of f value are aumed to be ditinct: that i, we never have f(x) = f(y) if x and y don t differ by a multiple of. The goal of the problem i to find. Remark.2. Claically, we can actually olve thi problem very efficiently. ote that the condition on implie that divide. Auming = 2 n, then mut lie in the et {, 2, 4,..., }. So we obtain an efficient claical algorithm by imply teting if = i f period, then if = 2 i f period, etc. Thi require u to tet n = log value of, o the query complexity, and run-time, i O(n). So, why do we care about olving thi quantumly? Shor algorithm [Sho97] for factoring will actually look at a variant where f i almot-periodic. So we will not necearily know that divide. However, the quantum algorithm we develop today will generalize to account for thi cae.

2 2 The Algorithm Here i the quantum algorithm that we will ue to olve thi problem: Prepare the tate x Z x, i.e. the uniform uperpoition of all the ket x. ( Attach the tate 0 m, thereby obtaining x Z x ) 0 m = x Z x 0 m. Query the oracle O f on the current input. The tate of the ytem will now be x Z x f(x). () Meaure the color regiter, i.e. the regiter correponding to the 0 m f(x) part of the output of O f. Apply F to the remaining n qubit. Meaure the remaining n qubit. Then, we do a little bit of claical computation with the output, which will be made clear later on. To aid with the analyi of the algorithm, we will ue the following notation: otation 2.. For each color c, define f c : Z {0, } by { iff(x) = c, f c (x) = 0 otherwie. Thu, f c i the indicator function for the event that f(x) = c. Thu, () can be rewritten a color c x Z (f c (x) x ) f(x). (2) Indeed, if f c (x) = 0 then the term x f(x) diappear from the um, o we only count x f(x) for the one color c uch that f(x) = c. Moreover, we comment at that in the ummation (2), the number of nonzero term of the form (f c (x) x ) i preciely, which i the number of ditinct value x Z that map to the color c under f. 2

3 3 Analyi The firt quetion that we hould ak i: what do we get when we perform the color meaurement? That i, with what probability do we meaure a fixed color c? Uing (2), we ee that thi probability i preciely ( ) 2 f c (x) = f c (x) 2 = f c (x) x Z x Z x Z = E [f c (x)] = Pr [f(x) = c] x R Z x R Z = fraction of f output which have color c =. In the above computation, we write x R Z to denote the ditribution in which x i ampled from Z uniformly at random. We ued the fact that f c (x) 2 = f c (x) for all x ince f c i {0, }- valued. We alo oberved that E x R Z [f c (x)] = Pr x R Z [f c (x) = ] ince f c (x) for random x i a {0, }-valued random variable, and then the imple fact that f c (x) = f(x) = c by the definition of f c to conclude Pr x R Z [f c (x) = ] = Pr x R Z [f(x) = c]. Thu, we obtain a uniformly random color a our anwer! Given that we have oberved a color c, what doe the tate (2) collape to? Well, it the ubtate that i conitent with the color c. Thi i the tate ( ) f c (x) x c normalized. x Z The normalizing factor i, o the tate i ( ) f c (x) x x Z It will be mot convenient for u to rewrite thi tate a c x Z fc (x) x. (3) At thi point, we will ue the powerful trick that alway eem to work when we analyze quantum algorithm. Thi trick i, of coure, the quantum Fourier tranform. More preciely, we will: Apply the Quantum Fourier Tranform over Z and meaure. Let u briefly recall what generally happen when we ue thi trick. If we have a function g : G C uch that E x R G[ g(x) 2 ] =, where G i either Z n 2 or Z, then the following i a valid quantum tate: g(x) x. x G 3

4 Figure : The function g = 2 {,5,9,3,...,} If we apply the Fourier tranform, which correpond to applying H if G = Z n 2 and F if G = Z, then the tate we obtain i ĝ(γ) γ. Thu, when we meaure, we oberve γ G with probability ĝ(γ) 2. γ G Remark 3.. Thi procedure i called pectral ampling. The et {ĝ(γ)} γ G i called the Fourier pectrum of g. Thi i *almot* alway how exponential peed-up are obtained in quantum computing. In our cae, recalling (3), we hould put g = f c, a then x Z g(x) x i the quantum tate after meauring the color regiter. Remark 3.2. For Simon algorithm, if we define { x = y, y (x) = 0 x y, we had g = 2( y + y+ ), where f (c) = {y, y + } for ome color c. Before continuing, let define the following notation which we will ue throughout the analyi: otation 3.3. For S Z, we define S : Z {0, } by { if x S, S (x) = 0 if x / S. Example 3.4. Say = 4, c i Green, and f i Green on, 5, 9, 3,.... Then g = f c = 2 {,5,9,3,...}. Figure demontrate what thi function look like. 4

5 Figure 2: The function f Green = {2,6,0,...} Figure 3: The function f Red = {0,4,8,...} Our algorithm output γ Z with probability ĝ(γ) 2 = f c (γ) 2. So, the quetion now i: what are thee Fourier coefficient? We have a periodic pike function. It turn out that the Fourier tranform of uch a function i alo a periodic pike function. To implify our analyi, we would like to how that f c (γ) 2 i independent of c. That way, it will uffice to compute f c (γ) 2 for one nice choice of c. Hence, we will prove the following claim: Claim 3.5. Let g : Z C and let t Z. Define g +t : Z C by g +t (x) = g(x + t). Then g and g +t have eentially the ame Fourier coefficient. That i, they differ by a multiplicative factor of magnitude, o they have the ame magnitude. Why i thi helpful? Well, ay f Green = {2,6,0,...,} and f Red = {0,4,8,...,}. Then, thee two function are hift of each other, a f Green = f +2 Red. It therefore uffice to tudy f c for a ingle value of c, a deired. See figure 2 and 3. Proof. Let ω = e 2πi, o that χ γ (x) = ω γ x. We compute: ĝ +t (γ) = E [g +t (x)χ γ (x) ] x R Z = E x R Z [g(x + t)χ γ (x) ] ( ) 5

6 At thi point, we make the change of variable y = x + t. For fixed t, if x i elected from Z uniformly at random, o i y. Thu, ( ) = E [g(y)χ γ (y t) ] y R Z = E x R Z [g(y)χ γ (y) χ γ ( t) ] = χ γ ( t) E y Z [g(y)χ γ (y) ] = ω γt ĝ(γ). Recalling that ω γt i an -th root of, we conclude that it i a complex number of magnitude. In the above computation, we ued the important fact that χ γ (x +y) = χ γ (x)χ γ (y) for all x, y Z, a well a the obervation that χ γ ( t) doe not depend on the randomly elected y R Z. Thi immediately yield the following corollary: Corollary 3.6. ĝ+t (γ) 2 = ω γt 2 ĝ(γ) 2 = ĝ(γ) 2. Thu, the probability of ampling γ Z i independent of t. In our cae, thi mean that when we do the pectral ampling at the end of the algorithm, it doe not matter what color we meaured earlier. It i therefore no lo of generality to aume that, if c i the color we ampled, f(0) = c, from whence it follow that f c = {0,,2,3,...}. What i o pecial about the et {0,, 2,...}? It preciely the ubgroup of Z generated by the number! Remark 3.7. For Simon algorithm, we would tudy {0,}, a {0, } i the ubgroup of Z n 2 generated by. We are now prepared to analyze the Fourier coefficient of g. Propoition 3.8. Let H = {0,, 2,...} Z and let h = H. Then { if γ {0, ĥ(γ) =, 2, 3,...}, 0 otherwie. Remark 3.9. Oberve that {0,, 2, 3,...} =. To ee thi, recall that = = 0 mod, o the ize of the et i at mot becaue a = (a+) for all value of a. But if r < then r 0 mod, o if a, b < with a < b then a b a otherwie (b a) = 0 mod even though b a b <. Thu, we conclude that there are preciely value of γ for which ĥ(γ) 0. Auming the propoition, what do we conclude? Well, the tate change a follow. Recalling that the tate prior to applying the F gate i (3), we have F h(x) x ĥ(γ) γ x Z γ Z meaure γ with probability ĥ(γ) 2. 6

7 Since ĥ(γ) = iff γ H, we meaure γ / H with probability 0 and γ H with probability 2 =. That i, we ample a uniformly random γ H. Recalling that H = {0,, 2,...}, any γ H atifie γ = 0 mod. Thu, we conclude that we ample a uniformly random γ uch that γ = 0 mod. Thi i jut like what happened in Simon algorithm! There, we ampled a uniformly random γ Z n 2 uch that γ = 0, o the only difference i that now we conider multiplication modulo intead of the dot product of two length n tring modulo 2. ow, we will prove Propoition 3.8. Proof. We compute ĥ(γ) = E [h(x) χ γ (x) ] = x R Z E [χ γ(x) ] = ( ). x R H The econd equality follow from the fact that h(x) i only ever nonzero on the et H which ha ize. We conider two cae: Cae. Suppoe γ {0,, 2,...}. Then χ γ (x) = ω γ x ow, recall that x i a multiple of, ince i aumed to be ampled from H. Similarly, if γ i in {0,, 2,...}, γ i a multiple of. Thu, χ γ (x) = ω (multiple of ) (multiple of ) = ω (multiple of ) =, uing that ω i an -th root of unity. Thu, χ γ (x) = for all x H, o ( ) = E x H [] =. Cae 2. We could, uing ome elementary arithmetic, directly compute E x R H[χ γ (x) ] to how that it i 0. However, we ll be a bit more lick. We ve already hown that our algorithm output γ {0,, 2 } with probability ( ) 2 =. Since {0,, 2,..., } =, there i no probability left to give to the h(γ)! That i, = ĥ(γ) 2 = ĥ(γ) 2 + ĥ(γ) 2 = + ĥ(γ) 2 γ Z γ H γ / H γ / H o implying ĥ(γ) = 0 for all γ / H. ĥ(γ) 2 = 0, γ / H 7

8 4 Summary Let reviewed what we ve accomplihed. We know that f i promied to be of the form R G B Y R G B Y } {{ } length That i, f i -periodic. We have hown that, with one query to O f and O(n 2 ) gate along with a couple meaurement, we get a uniformly random γ Z uch that γ = 0 mod (or, equivalently, uch that γ {0,, 2,...}). Well, what hould we do? A i uually the cae, we ll want to repeat the algorithm ome number of time. But how many time how we repeat the algorithm to find? Firt of all, we oberve that to find, it actually uffice to find. After that, we can divide through by and take the reciprocal to compute. So, how do we find? For the time being, forget that and are power of 2. Let m =, o our algorithm ample a random integer multiple of m. Suppoe we have two random ample, which we can write am and bm. ote that gcd(am, bm) = gcd(a, b)m. So, if gcd(a, b) =, by taking the gcd of the two output am and bm, we will have found m! Thu, the quetion become the following: If a and b are ampled from {0,,..., } uniformly at random and independently, what the probability that gcd(a, b) =? Claim 4.. If a and b are ampled a above, Pr[gcd(a, b) = ] Ω(). Proof. Firt condition on the event that a and b are not equal to zero. Thi event occur with large probability: ( ) 2 which i at leat /4. Fix a prime p. Oberve that Pr[p divide a and b] = Pr[p divide a] Pr[p divide b] ince a and b are independent = Pr[p divide a] 2 ince a and b are identically ditributed ote that at mot a /p fraction of the element in the et {, 2,..., } are multiple of p. Since we are conditioning on the event that a 0, we conclude that Pr[p divide a] /p. Therefore Pr[p divide a and b] p 2. Thu, Pr[gcd(a, b) > ] = Pr[a and b hare a prime factor] p 2 prime p n 2 n 2 = π , Union Bound 8

9 where we have ued the famou number-theoretic fact that n where ζ denote the Riemann zeta function. Remark 4.2. Computing prime p Reference p 2 π2 = ζ(2) = n2 6, exactly i an open problem. [Sho97] Peter Shor. Polynomial-time algorithm for prime factorization and dicrete logarithm on a quantum computer. SIAM journal on computing, 26(5): ,