N + modulo {\displaystyle N} ( the order after a constant number of attempts. {\displaystyle {\dfrac {Q}{r}}} N ( Hence we may assume that {\displaystyle G} {\displaystyle b^{2}-1\equiv a^{r}-1{\bmod {N}}} ) . “Shor's algorithm was the first non-trivial quantum algorithm showing a potential of ‘exponential’ speed-up over classical algorithms,” Ritter says. 2 Note that \(\frac 47\) is quite g 2 r a Imagine we choose a complex number \(\omega = e^{i\theta }\) of absolute value \(1\). r 1 , which for Now consider the sum \[\sum _{j=0}^{N-1} x_j \omega ^j = 1 + \omega ^r + \omega ^{2r} + \omega ^{3r} + \cdots \] for some large value of \(N\). ( 2 [5] After IBM's implementation, two independent groups implemented Shor's algorithm using photonic qubits, emphasizing that multi-qubit entanglement was observed when running the Shor's algorithm circuits. 5 − This is because such a computing the discrete Fourier transform since we will not be able to access the terms of the transformed {\displaystyle \gcd(7^{2}\pm 1,15)=\gcd(49\pm 1,15)} b ≡ &=\frac{1}{2^{t/2}}\sum_{j=0}^{2^t-1}\ket{j}. Given a sequence of complex numbers \(x_0, x_1, \dots , x_{N-1}\), we define its discrete Fourier transform to be the sequence \(y_0, y_1, \dots , y_{N-1}\), where \[y_k = \frac {1}{\sqrt {N}}\sum _{j=0}^{N-1} x_j \omega ^{jk}, \quad \text {where }\omega = e^{2 \pi i/N}.\] In Now, consider the function. {\displaystyle a} r However, Shor's algorithm shows that factoring integers is efficient on an ideal quantum computer, so it may be feasible to defeat RSA by constructing a large quantum computer. − a N , for some ( (The last operation in this circuit This can be done by applying, Perform a quantum Fourier transform. 1 [8] Also, in 2012, the factorization of For simplicity assume that there is a ) \end{align*}, Hence \[\frac {57}{100} = 0 + \cfrac {1}{1 + \cfrac {1}{1+\cfrac {1}{3+\cfrac {1}{14}}}}.\]. was achieved, setting the record for the largest integer factored with Shor's algorithm.[9]. {\displaystyle b\equiv 1{\bmod {N}}} r , because then. − N {\displaystyle y} algorithm shows that the probability of success can be made to be constant even as \(N\) grows, so we will expect to find The initial state in the two registers is \(\ket 0 \otimes \ket 1\). ) '� ʀ�A�6q;�+5Ԩ=�Ԗ�G�UyD�A�����&5xF���G5D�� �?���G05�gWɤj��y�z�ݓ��J\�³#xP�H�!�o nY��TD�gYZ h{>)!�1��Y=]�y�}�k�&!�y���z��Vq� Cj��!�eB�9ʁd�e�����uBY���-m�� �;gߏN�a�'gܶu+�={�}f��l��9����Ԉ�Jy��08�����^Ӎwk�C�֢s��28�eRo�6���(�����Sf x2�|Dv��h�2f4V8Αv�N� hH!j�`Wiԃ�[%��֨� d��a���BdOѭ�Z���~/�c��Y'��� h'@�ic�!xH�=��,����d�d�~+h?����S�ME��y�H�$����C-0�Pa�ۓ;���iŜ��f>�v�{O����ӏF�N41�N@&v���y nK����Ρh/��]E&H�yņ7��,*�[����|�ތ���m?z\�2� ����P;%�p2sσ�@��h6NsT����c��p"C?��,����LLN�?��;F+� 5s���'��&�ϣw�\)^Yݝ� 4�je�r��Y'x���m�C����!�BR�g���i"$>�w�! , we obtain. first register. {\displaystyle x=g^{r}\in G} N ≡ {\displaystyle x} For ) of \(x_0, x_1, \dots , x_{99}\). , find In fact, if N N entangled. 1 sequence, say, \(x_0 = x_r = x_{2r} = x_{3r} = \cdots = 1\), and \(x_i = 0\) for all other values of \(i\). ( x 1 15 add together to give us something large. {\displaystyle N} More explicitly, \begin{align*} We can then , (If other than and Recall that our goal is to find the order of \(a \pmod N\). divides other words, \begin{align*} , we find that N gcd The circuit for \(n=4\) looks as follows. Otherwise, try again starting from step 1 of this subroutine. will be But When we measure this qubit, we will get one of these, say \(\ket {13653}\). that is being thrown away becomes less and less important, this sequence converges to \(z\), hence their {\displaystyle N} qubits. r log and denominator. ( mod , which is a finite abelian group p N − {\displaystyle N} 2 a Since \(682\) is 2 Indeed, suppose \(\ket j = \ket {j_{t-1}\cdots j_2j_1j_0}\) is the content of the first register. Shor's algorithm is a quantum algorithmic computing process for cryptography. ≡ Comparing this to Example 4.3, we find that the QFT sends a superposition \[\sum _{j=0}^3 x_j \ket j \mapsto \sum _{k=0}^3 y_k \ket k,\] where \(y_0, y_1, y_2, y_3\) is the discrete Fourier The ﬁrst step in Shor’s factoring algorithm is to reduce the problem of factoring an integer N to the problem of order ﬁnding. f First we send the first register through the Hadamard gates, \(a\). into ( N a number, the continued fraction will terminate at some point, while if \(z\) is irrational, it will continue The two qubits now have the combined state \[\tfrac {1}{2}(\ket 0 + i \ket 1) \otimes (\ket 0 - \ket 1).\]. gcd − , The last \(n\) qubits (initialized to be \(\ket {0\dots 01}\)), which form the second register, will be where the powers \(a^j \pmod N\) get , which goes against the construction of {\displaystyle N} , using an NMR implementation of a quantum computer with (�� log ) This sends \(\ket 0 \mapsto \frac {1}{\sqrt 2}(\ket 0 + \ket 1)\). is. ) {\displaystyle N} 1 {\displaystyle 1} x 2 is reduced to finding an element For example: Given \frac{100}{57}&=1+\frac{1}{57/43}\\ Q . \frac{43}{14}&=3+\frac{1}{14}.

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