![]() Magnetic resonance imaging (MRI) is a safe, non-ionizing, and non-invasive imaging modality that provides high resolution and excellent contrast of soft tissues. Our analyses (both quantitative as well as qualitative/visual analysis) establish that the proposed method is robust and reduces several-fold the computational time reported by the current state-of-the-art technique. The proposed method has been rigorously evaluated on synthetically corrupted data on varying degrees of motion, numbers of shots, and encoding trajectories. First CG-SENSE reconstruction is employed to reconstruct an image from the motion-corrupted k-space data and then the GAN-based proposed framework is applied to correct the motion artifacts. ![]() In this paper, we propose a novel generative adversarial network (GAN)-based conjugate gradient SENSE (CG-SENSE) reconstruction framework for motion correction in multishot MRI. Numerous efforts have focused on addressing this issue however, all of these proposals are limited in terms of how much motion they can correct and require excessive computational time. The downside of multishot MRI is that it is very sensitive to subject motion and even small levels of motion during the scan can produce artifacts in the final magnetic resonance (MR) image, which may result in a misdiagnosis. As it is also non-empty, by connectedness of $V$ we have that im $(f)$ is the whole of $V$.Multishot Magnetic Resonance Imaging (MRI) is a promising data acquisition technique that can produce a high-resolution image with relatively less data acquisition time than the standard spin echo. Then by continuity, $f(X)=Y$, so im $(f)$ is closed.īy invariance of domain we also know that im $(f)$ is open. Then we have a sequence of vectors $X_i\in V$ with $f(X_i)\to Y$ and $f(X_i)$ Cauchy, so $X_i$ is Cauchy. Let $Y$ be an accumulation point of im $(f)$. Then $f$ is clearly injective and continuous, and im $(f)$ is non-empty. Suppose that $f$ is a rigid motion of $V$. Let $V$ be a finite dimensional real vector space with Euclidean norm. The cases $\lambda\leq0$ and $\lambda\in (0,1)$ follow analogously. Then $Y=\lambda g(X)$ is the unique vector satisfying $||Y||=\lambda||X||$ and $g(X)$ lies on the line segment $OY$.Īs $X$ lies on the line segment $O(\lambda X)$, we know $g(X)$ lies on the line segment $O(g(\lambda X))$. The vectors $Z$ in the line segment $XY$ are characterized by the condition $||X-Y|| =||X-Z|| + ||Z-Y||$, so $g$ preserves line segments. Bijectivity of an arbitrary rigid motion follows. ![]() Suppose that $g$ is a rigid motion of $V$, fixing the origin $O$. Solution $2$ (Geometric): Let $V$ be a finite dimensional real vector space with Euclidean norm. In my view the one by is the best of these 3, as it is the most succinct and self-contained, but the others are interesting in their own right. Variants of this question have been asked so many times I expect someone has also posted the geometric solution too, at some point. ![]() Credit for the topological solution to for the comment here. However it uses invariance of domain instead, which is actually less trivial to prove than linearity. The topological one is noteworthy as it takes a different route to most solutions, by not showing that any maps are linear. Here are two alternative arguments: one geometric and one topological. If $\ g(x)=f(x)-f(0)\ $ then $\ g\ $ must be an orthogonal linear transformation. ![]()
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