<table cellspacing="0" cellpadding="0" border="0"><tr><td valign="top" style="font: inherit;"><p>Hi Xianjun,</p>
<p>it may not be right but it is just what I am thinking.<br>
Could it be the reverse transform if you take the opposite offsets, i.e reverse_offset = -offset ?</p>
<p>Best,<br>
Willi</p>
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<span style="font-weight:bold;">From:</span>
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xianjun.sun.ext_CRC <xianjun.sun.ext@united-imaging.com>; <br>
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<span style="font-weight:bold:">To:</span>
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insight-users@itk.org <insight-users@itk.org>; <br>
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<span style="font-weight:bold:">Subject:</span>
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[Insight-users] How to do reverse transformation of cubic Bspline-based FFD? <br>
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<span style="font-weight:bold;">Sent:</span>
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Thu, Nov 1, 2012 3:14:27 AM <br>
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<p class="MsoNormal"><span lang="EN-US" style='font-size:14.0pt;'>Hello everyone,</span></p>
<p class="MsoNormal"><span lang="EN-US" style='font-size:14.0pt;'> </span></p>
<p class="MsoNormal" style='margin-left:21.0pt;'><span lang="EN-US"
style='font-size:14.0pt;'>I got offsets of the control points employed to
transform image I1 to image I2, now I want to get the reverse transformation.
How can I do it? Any hints or reference paper?</span></p>
<p class="MsoNormal" style='margin-left:21.0pt;'><span lang="EN-US"
style='font-size:14.0pt;'>Thanks a lot.</span></p>
<p class="MsoNormal"><span lang="EN-US" style='font-size:14.0pt;'> </span></p>
<p class="MsoNormal"><span lang="EN-US" style='font-size:14.0pt;'> </span></p>
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