# shor algorithm

November 7, 2020Uncategorized

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�2f4V8Αv�N� hH!j�`Wiԃ�[%��֨� d��a���BdOѭ�Z���~/�c��Y'��� h'@�ic�!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}.