two's complement integers

[Michael Shulman]

Has anyone considered the following HIT definition of the integers?

data ℤ : Type where
zero : ℤ -- 0
negOne : ℤ -- -1
dbl : ℤ → ℤ -- n ↦ 2n
sucDbl : ℤ → ℤ -- n ↦ 2n+1
dblZero : dbl zero ≡ zero -- 2·0 = 0
sucDblNegOne : sucDbl negOne ≡ negOne -- 2·(-1)+1 = -1

The idea is that it's an arbitrary-precision version of little-endian
two's-complement, with sucDbl and dbl representing 1 and 0
respectively, and negOne and zero representing the highest-order sign
bit 1 and 0 respectively. Thus for instance

sucDbl (dbl (sucDbl (sucDbl zero))) = 01101 = 13
dbl (sucDbl (dbl (dbl negOne))) = 10010 = -14

The two path-constructors enable arbitrary expansion of the precision, e.g.

01101 = 001101 = 0001101 = ...
10010 = 110010 = 1110010 = ...

This seems like a fairly nice definition: it should already be a set
without any truncation, it's binary rather than unary, and the
arithmetic operations can be defined in the usual digit-by-digit way
without splitting into cases by sign. Mathematically speaking, it
represents integers by their images in the 2-adic integers, with zero
and negOne representing infinite tails of 0s and 1s respectively. (An
arbitrary 2-adic integer, of course, is just a stream of bits.)

Best,
Mike

[Martin Escardo]

Nice. I haven't seen this. But I have considered two similar things:

(1) The natural numbers defined by

0
2x+1
2x+2

This is not a HIT - just a normal inductive types.

(2) A HIT for the dyadics numbers D in the interval [0,1] with constructors

data 𝔹 : Type₀ where
L R : 𝔹 -- 0 and 1
l r : 𝔹 → 𝔹 -- x ↦ x/2 and x ↦ (x+1)/2
eqL : L ≡ l L -- 0 = 0/2
eqM : l R ≡ r L -- 1/2 = (0+1)/2
eqR : R ≡ r R -- 1 = (1+1)/2

Then, like your HIT, this is automatically a set without the need of
truncation. But I didn't find an simple proof of this. Together with
Alex Rice, we proved this by showing it to be equivalent to another type
which is easily seen to be a aset:

data 𝔻 : Type₀ where
middle : 𝔻
left : 𝔻 → 𝔻
right : 𝔻 → 𝔻

data 𝔹' : Type₀ where
L' : 𝔹'
R' : 𝔹'
η : 𝔻 → 𝔹'

l' : 𝔹' → 𝔹'
l' L' = L'
l' R' = η middle
l' (η x) = η (left x)

r' : 𝔹' → 𝔹'
r' L' = η middle
r' R' = R'
r' (η x) = η (right x)

eqL' : L' ≡ l' L'
eqM' : l' R' ≡ r' L'
eqR' : R' ≡ r' R'

eqL' = refl
eqM' = refl
eqR' = refl

Then the types 𝔹 and 𝔹' are equivalent, and clearly 𝔹' is a set
because it has decidable equality.

Moreover, the "binary systems" (𝔹,L,R,l,r,eqL,eqM,eqR) and
(𝔹',L',R',l',r',eqL',eqM',eqR') are isomorphic.

(And we implemented in this in Cubical Agda.)

I wonder if your definition of the integers with the HIT is isomorphic
to an inductively defined version without a HIT as above, and this is
how you proved it to be a set.

Martin

[Michael Shulman]

On Thu, Mar 4, 2021 at 1:11 PM Martin Escardo <escardo...@gmail.com> wrote:
> I wonder if your definition of the integers with the HIT is isomorphic
> to an inductively defined version without a HIT as above, and this is
> how you proved it to be a set.

No, I just did an encode-decode. I formalized most of it, up to some
obviously true lemmas about squares and cubes.

I thought for a little bit about whether there is an equivalent
"lower" inductive version, but nothing immediately occurred to me. Do
you have a suggestion?

[Martin Escardo]

On 04/03/2021 22:05, Michael Shulman wrote:

For example, take ℤ' = ℕ + ℕ and define the functions you have in your
hit and prove the equations you have in your HIT.

Then using the HIT universal property you get a function from your ℤ to
this ℤ', and conversely you define a function from ℤ' to ℤ "manually".

This is what Alex Rice and I did to prove that our 𝔹 and 𝔹' are
equivalent. The cubes don't disappear completely, of course. But there
were much fewer cubes than when we tried to prove directly that 𝔹 is a
set, and we gave up because the code was getting too long and
unpleasant, with too many cases to check. The indirect version was much
shorter and pleasing.

Martin

[Nicolai Kraus]

On 04/03/2021 22:05, Michael Shulman wrote:

I thought for a little bit about whether there is an equivalent

"lower" inductive version, but nothing immediately occurred to me.  Do
you have a suggestion?

I'm not sure what the precise thing is that you're looking for because, without further specification, any standard definition of Z would qualify :-)

The HIT is neat, but wouldn't it in practice behave pretty similar to a standard representation via binary lists? E.g. something like Unit + Bool * List(Bool), where inl(*) is zero, the first Bool is the sign, and you add a 1 in front of the list in order to get a positive integer. What's the advantage of the HIT - maybe one can avoid case distinctions?

Nicolai

[Michael Shulman]

On Thu, Mar 4, 2021 at 3:16 PM Nicolai Kraus <nicola...@gmail.com> wrote:
> I'm not sure what the precise thing is that you're looking for because, without further specification, any standard definition of Z would qualify :-)

Yes, that seems to be what Martin suggested too with ℕ + ℕ. It seemed
to me as though the distance between ℕ + ℕ and my ℤ is greater than
the distance between his 𝔹 and 𝔹', but maybe not in any important
way.

> The HIT is neat, but wouldn't it in practice behave pretty similar to a standard representation via binary lists? E.g. something like Unit + Bool * List(Bool), where inl(*) is zero, the first Bool is the sign, and you add a 1 in front of the list in order to get a positive integer. What's the advantage of the HIT - maybe one can avoid case distinctions?

Is there a non-HIT binary representation that can be interpreted as
two's-complement (thereby avoiding case distinctions on sign)? I
haven't been able to figure out a way to do that with mere lists of
booleans.

[Jason Gross]

Note that the Coq standard library Z is a binary representation of integers and Z.testbit gives access to the infinite twos-compliment representation.  Of course, it makes use of case-distinctions on sign to define it, but you go bit-by-bit; if the number is negative, you just invert the bit before returning it.  Is there something you're after by having the representation not encode sign bits separately?

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[Michael Shulman]

I don't have a particular application in mind at the moment. It just
struck me how 2-adic integers have such simple coinductive definitions
of the arithmetic operations, without any case distinctions at all,
and I wondered whether there was a definition of plain integers with a
similar property. It seems annoying to have to constantly case-split
on a sign bit.

I guess a lower-inductive representation could take advantage of the
fact that the 2-adic representation of an integer other than 0 or -1
has a unique largest digit that differs from all larger digits, i.e.
it's either

...0000 (i.e. 0)
...1111 (i.e. -1)
...0001 + arbitrary bit string
...1110 + arbitrary bit string

This involves more case splits when defining arithmetic, but it would
probably be an easy set to show that the HIT one is equivalent to.

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