Finding the path between two points

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Zena Assaad
Zena Assaad 2015 年 7 月 2 日
コメント済み: Zena Assaad 2015 年 7 月 6 日
Hi all,
I have two x y z coordinates and I need the path between these two points. It does not necessarily need to be the shortest distance path. I am working with a (3x30) position matrix. The first row represents x, the second row is y and the third row is z.
Point 1: (x1,y1,z1)
Point 30: (x30,y30,z30)
Position matrix: [x1 .......... x30
y1 .......... y30
z1 .......... z30]
Point 1 and point 30 are given. I need something simple that will give me the remaining 28 points within the matrix.
Thanks in advance!
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James Tursa
James Tursa 2015 年 7 月 2 日
編集済み: James Tursa 2015 年 7 月 2 日
Any restrictions on the path? E.g., is a diagonal jump (two indexes changing at the same time) OK? Can indexes only change a max of 1 for each point?
Zena Assaad
Zena Assaad 2015 年 7 月 2 日
No, there are no restrictions on the path.

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James Tursa
James Tursa 2015 年 7 月 2 日
編集済み: James Tursa 2015 年 7 月 2 日
E.g., a somewhat "linear" path allowing "diagonal" jumps (more than one index can change at a time):
Point1 = whatever;
Point30 = whatever;
Pmatrix = [round(linspace(Point1(1),Point30(1),30));
round(linspace(Point1(2),Point30(2),30));
round(linspace(Point1(3),Point30(3),30))];
If the points do not need to be integers, then skip the rounding.
  7 件のコメント
James Tursa
James Tursa 2015 年 7 月 2 日
編集済み: James Tursa 2015 年 7 月 2 日
They are not zero ... they are just displaying as zero because of the display format you are using (probably short) and the wide range of values in the result. Try this:
format long g
Then display the result again.
Or just look at the last row by itself:
Pmatrix(3,:)
Zena Assaad
Zena Assaad 2015 年 7 月 6 日
That worked perfectly. Thank you very much for all your help.

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Image Analyst
Image Analyst 2015 年 7 月 2 日

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