Spiralling Algorithm in C++

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  • Thread starter ron_jay
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  • #1
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Hey, I got an assignment to write and algorithm in C++ to print this pattern:

35 34 33 32 31 30
16 15 14 13 12 29
17 04 03 02 11 28
18 05 00 01 10 27
19 06 07 08 09 26
20 21 22 23 24 25

As you can see, the number starts near the center and spirals out... I am stumped. Any help?
 

Answers and Replies

  • #2
mgb_phys
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Simple way for small size is to create a 2d array and fill it in starting from the centre (just need a couple of loops) and then print it.
Generating one line-by-line for an arbitrary size isn't too hard but requires thinking about the loop counters a bit!
 
  • #3
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This one is a little tricky because the numbering starts from the center. I have already figured out how to do this when the numbering starts from the top left hand corner (recursion) using
"virtual motion analogy" (that's what I like to call it).However,looping with for or while is clumsy. Recursion should solve it elegantly. One can assume the array matrix as a 2D "field" and set them to 0. By, recursion, we can then set the up, down, left and right freedom of movement. However, this is the part which is slippery...how do the axes change to suit the situation? Its like a virtual loop that keeps on depositing a solid layer around it as it rotates on its outer periphery being pulled by the central force - just an analogy...if this could be interpreted in the terms of the matrix....
 
  • #4
rcgldr
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Use an offset value for the center, in this case it's 3,3 (using y,x as index order)

Then the starting relative index will be -1,-1, followed by -1,+1, +1,+1, +1,-1.

Each time the index pattern is +n, -n, your at end of the current box, and ready to start the next box.

Don't forget to handle the case where the size of the array is an odd number of columns and rows.
You start at the middle, go right, fill in the 8 elements counter clockwise, then step out to the right
again.
 
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