Mailman 3 A bit long, but would appreciate anyone's help, if time permits! - NumPy-Discussion

July 23, 2004

      Hi.  Like my previous post, my question is not directly related to 
Numpy, but I couldn't help posting it since many people here deal with 
numbers.  I have a question that requires a bit of explanation.  I 
would highly appreciate it if anyone could read this and offer any 
suggestions, whenever time permits.

I'm trying to write a program that 1) gives all possible rotations of 
an ordered list, 2) chooses the ordering that has the smallest 
difference from first to last element of the rotation, and 3) continues 
to compare the difference from first to second-to-last element, and so 
on, if there was a tie in step 2.

The following is the output of a function I wrote.  The first 6 lines 
are all possible rotations of [0,1,3,6,7,10], and this takes care of 
step 1 mentioned above.  The last line provides the differences (mod 
12).  If the last line were denoted as r, r[0] lists the differences 
from first to last element of each rotation (p0 through p5), r[1] the 
differences from first to second-to-last element, and so on.
...
...
...
from normal import normal
normal([0,1,3,6,7,10])
[0, 1, 3, 6, 7, 10]	#p0
[1, 3, 6, 7, 10, 0]	#p1
[3, 6, 7, 10, 0, 1]	#p2
[6, 7, 10, 0, 1, 3]	#p3
[7, 10, 0, 1, 3, 6]	#p4
[10, 0, 1, 3, 6, 7]	#p5
[[10, 11, 10, 9, 11, 9], [7, 9, 9, 7, 8, 8], [6, 6, 7, 6, 6, 5], [3, 5, 
4, 4, 5, 3], [1, 2, 3, 1, 3, 2]]     #r

Here is my question.  I'm having trouble realizing step 2 (and 3, if 
necessary).  In the above case, the smallest number in r[0] is 9, which 
is present in both r[0][3] and r[0][5].  This means that p3 and p5 and 
only p3 and p5 need to be further compared.  r[1][3] is 7, and r[1][5] 
is 8, so the comparison ends here, and the final result I'm looking for 
is p3, [6,7,10,0,1,3] (the final 'n' value for 'pn' corresponds to the 
final 'y' value for 'r[x][y]').

How would I find the smallest values of a list r[0], take only those 
values (r[0][3] and r[0][5]) for further comparison (r[1][3] and 
r[1][5]), and finally print a p3?

Thanks again for reading this.  If there is anything unclear, please 
let me know.

Best,
Kye

My code begins here:
#normal.py
def normal(s):
	s.sort()
	r = []
	q = []
	v = []

	for x in range(0, len(s)):
		k = s[x:]+s[0:x]
		r.append(k)

	for y in range(0, len(s)):
		print r[y], '\t'
		d = []
		for yy in range(len(s)-1, 0, -1):
			w = (r[y][yy]-r[y][0])%12
			d.append(w)
		q.append(d)

	for z in range(0, len(s)-1):
		d = []
		for zz in range(0, len(s)):
			w = q[zz][z]
			d.append(w)
		v.append(d)
	print '\n', v

A bit long, but would appreciate anyone's help, if time permits!

Hee-Seng Kye

Colin J. Williams

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