in reply to Re: Finding the simplest combination of descrete thicknesses that sums nearest to the target thickness.
in thread Finding the simplest combination of descrete thicknesses that sums nearest to the target thickness.

Thank you Scott,

I did try to just get to the basics of what I need and left out some important info. Here is the totality of what I am trying to accomplish:

I am a Medical Physicist (deals with treating cancer with radiation). When we irradiate the entire body (this is done to supress the immune system before a transplant, i.e bone marrow), we want the entire body to receive uniform dose. If you look at the body from the side, and consider its verying thicknesses (neck, shoulder, hips, knees,...), it becomes clear that we need to attenuate some of the radiation at different areas. For example, at the midline of the patient, the radiation has to travel though more tissue at the shoulders then it does at the neck, so we place a lead filter of a certain thickness for only the radiation that will go though the neck so that the dose at the midline of the patient will be the same. Ideally we want to build up these filters, if you just place the filters next to eachother, then there is a chance that radiation can escape unfiltered between the filters leading to hotspots.

Now, when filling out the info the width/thickness of the filters would start from the top of the patient and would continue in series towards the patients feet.

i.e. 3,2
5,4

Would indicate that the first filter would be 3cm wide and 2mm thick and the second filter would be 5cm wide (starting after the first filter) and 4mm thick.

This would be built using an 8cm piece of 1.6mm lead, an 8cm piece of 0.4mm lead, a 5cm piece of 1.6mm lead, and a 5cm piece of 0.4mm lead and would look like the following:

____________________ |____________________| 0.4mm | | ___________|____________________| 1.6mm |________________________________| 0.4mm | | |________________________________| 1.6mm
Let me know if I can clear anything up on this.

Thanks again everyone!
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