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NTT using verilog

₹600-1500 INR

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Posted almost 3 years ago

₹600-1500 INR

Paid on delivery
I need Number theoretic transform implementation (NTT) and Inverse NTT using Verilog. A reference file is uploaded below. Show the results using an example n degree polynomial.
Project ID: 30193205

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Hi, I am a senior digital design engineer, I have a broad knowledge of digital design in ASIC and FPGA using both VHDL and Verilog. I am using Vivado, ISE, and Quartise for FPGA, using DC, ICC, and prime-time for ASIC. I will provide you a professional report about your project with citation and scientific formatting. Please contact me to know more about your needs. Regards, moaaz.
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3 freelancers are bidding on average ₹2,367 INR for this job
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Dear sir, I am a digital design engineer expert in FPGA and ASIC design flows using Verilog and VHDL programming. Also, I am experienced with Vivado, ISE, Vivado IPs, SDK, Quartus, Design Compiler, IC Compiler and others. I have done generic polynomial evaluator in VHDL with its testbench here: https://www.freelancer.com/projects/electronics/Vhdl-coding-for-profvip/reviews Please contact me to discuss more about this project. Kindest regards.
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Algorithm 1 INTT based on Gentleman-Sande butterfly [21] Input: a = (a[0], a[1], ··· , a[N − 2], a[N − 1]), p, Ψ−1 rev = (Ψ−1 rev[0], Ψ−1 rev[1], ··· , Ψ−1 rev[N − 1]) Output: a ← INTT(a) 1: t = 1 2: for (m = N; m > 1; m = m/2) do 3: j1 = 0 4: h = m/2 5: for (i = 0; i<h; i = i + 1) do 6: j2 = j1 + t − 1 7: W = Ψ−1 rev[h + i] 8: for (j = j1; j ≤ j2; j = j + 1) do 9: ButterflyUnit(a[j], a[j + t], W, p) 10: end for 11: j1 = j1 + 2 × t 12: end for 13: t = 2 × t 14: end for 15: return a Procedure for the NTT Suppose the input vector is a sequence of n non-negative integers. Choose a minimum working modulus M such that 1≤n<M and every input value is in the range [0,M). Select some integer k≥1 and define N=kn+1 as the working modulus. We require N≥M, and N to be a prime number. Dirichlet’s theorem guarantees that for any n and M, there exists some choice of k to make N be prime. Because N is prime, the multiplicative group of ZN has size φ(N)=N−1=kn. Furthermore, the group must have at least one generator g, which is also a primitive (N−1)th root of unity. Define ω≡gk mod N. We have ωn=gkn=gN−1=gφ(N)≡1 mod N due to Euler’s theorem. Furthermore because g is a generator, we know that ωi=gik≢1 for 1≤i<n, because ik<nk=N−1. Hence ω is a primitive nth root of unity, as required by the DFT of length n. Thanks you
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