Comments on: Determining List Steepness in a Homomorphism

By: Mark Essel

Mark Essel — Tue, 14 Jul 2009 18:37:17 +0000

Just had time to scan the top few paragraphs, I’m interested enough to come back and do some homework. In a world with no free time, we still find time to have some fun :D

By: cmcq

cmcq — Sat, 11 Jul 2009 09:03:40 +0000

1) No, because s>=(2^n-1)c where n is the list length. For negative c this puts a lower bound on n. For example [-8,-4,-2,-1] has s=-15, c=-1, so -15>=(2^n-1)*(-1), which gives n>=4.

A technique based on fixed-length inverses is very restricted: they can’t compute length!

2) You don’t need rationals ever: for sumcaps of two element lists you can use g(s,c)=[s-c,c]. Though personally I would not hesitate to extend the ring of numbers if it gave a simpler proof! Isn’t that what proof irrelevance is all about?

By: Shin

Shin — Sat, 11 Jul 2009 04:30:18 +0000

Thanks for the encouragement! I've been relying on WordPress to generate the RSS (was this the link you tried?). What error message did you see? I'll check whether other WordPress users had the same problem.

By: Shin

Shin — Sat, 11 Jul 2009 04:27:44 +0000

Thanks for the interesting comments, cmcq. Indeed it is harder than it looks. The first guess of mine (and some students here) was: g (s, c) = [(s+c)/2, (s-c)/2]. Not only was that wrong (it fails for, e.g, (8,8), output of sumcap [8]), but also forces us to move from integer to rational (that was why I was vague about the type of "numbers" in the post). I've been thinking: 1) is it possible to find an inverse that produces a constant-length list? 2) Can we stay within Int, if the input is Int? Goal 1) is essential for deriving ⊚, and 2) is desirable because the ⊚ that eventually worked does operate on integers only. You seem to have shown that goal 1) is hard to achieve. Well, perhaps the construction of list homomorphism by inverses is just not as widely applicable as we've expected.

By: Matías Giovannini

Matías Giovannini — Sat, 11 Jul 2009 00:35:40 +0000

Great post! Shin-Cheng, your RSS feed is broken. I’d love to subscribe, but unfortunately the XML parser in Firefox 3.0.x complains.

By: cmcq

cmcq — Fri, 10 Jul 2009 11:35:26 +0000

Wordpress ate my definition of g:
For s<3c<0, pick n such that s<=(2^n-1)c and use g(s,c)=[s+2^(n-1)c, 2^(n-2)c, ..., 2c, c].

By: cmcq

cmcq — Fri, 10 Jul 2009 11:33:18 +0000

Sorry, that’s wrong; both those expressions only work for s>=3c. In a list of length n with rcap c, the last element must be at least c, so the second-to-last element must be at least 2c, etc. The sum is therefore at least (2^n-1)c. For c>=0 this gives g(s,s)=[s] and g(s,c)=[s-c,c] for s>=3c. For s<3c=(2^n-1)c and use g(s,c)=[s+2^(n-1)c, 2^(n-2)c, ..., 2c, c].

Surely the definition of sumcap won’t come from this inverse though!

By: cmcq

cmcq — Fri, 10 Jul 2009 09:06:07 +0000

If you guess g(s,c)=[a,b] then a bit of algebra gives:
a = max(s-c, (s+c)/2)
b = min(c, s-c/2)

If s>=3c this gives [s-c,c]
If s<=3c this gives [(s+c)/2, (s-c)/2]