Closed-form heel formula

In my notes on modifying the heel from my Widdershins pattern, I gave a recursive formula to determine the number of stitches needed for the heel flap. In the comments on that entry, Carol posted a closed form:

Embarrassingly, I had to ask my son, who's studying math at university what k was. He told me to think about it. It finally clicked.
I still couldn't figure out how to program the formula but did get it into closed form.
I think it reduces to
h(n) = n – 2 – 2 * int(n/9),
where int(n/9) stands for the integer part of n/9.
btw. I really like the heel.

Of course, I had to try this out. Being a bit to lazy to do a real proof, I just coded it up quickly and compared the two functions for values up to 150. If anyone is planning on knitting a sock over 300 stitches around, I could go higher.

>>> def h1(n):
...   def h(k):
...     if k < n/3:
...       return k
...     else:
...       return 4 + h(k-6)
...   return h(n)
...
>>> h1(33)
25
>>> def h2(n): return n-2-2*(n/9)
...
>>> h2(33)
25
>>> for i in range(150):
...   val1 = h1(i)
...   val2 = h2(i)
...   if val1 != val2:
...     print i, val1, val2
...
8 4 6
17 11 13
26 18 20
35 25 27
44 32 34
53 39 41
62 46 48
71 53 55
80 60 62
89 67 69
98 74 76
107 81 83
116 88 90
125 95 97
134 102 104
143 109 111

For some numbers, my formula suggests one more pair of turning rows than Carol's formula, but given that my original n/3 came from me thinking, "Oh,one-third of the stitches in the center sounds reasonable." I don't think it'll make too much difference unless you are really picky about how pointy you like your sock heels. Thank you, Carol! I was feeling a bit guilty about not providing a closed form, but the guilt didn't motivate me quite enough to get out my college textbooks and remember how to solve it myself. Now I don't have to!
Categories: knitting, socks, Widdershins, python