### Re: using Induction.Nat

Using Data.Bin was also my hunch. Here is a partial solution to your
problem:
open import Function
open import Data.Bool
open import Data.List
open import Data.Nat
open import Data.Fin using (zero; suc)
open import Data.Bin using (Bin⁺; 0#; _1#; fromℕ)
open import Relation.Binary.PropositionalEquality
-- Binary exponentiation
module Power {A : Set} (_•_ : A → A → A) (x : A) where
binPower : Bin⁺ → A
binPower [] = x
binPower (b ∷ bs) = case b of
λ { zero → r²
; (suc zero) → x • r²
; (suc (suc ()))
}
where
r = binPower bs
r² = r • r
posPower : (n : ℕ) → n > 0 → A
posPower 0 ()
posPower (suc n) _ with fromℕ (1 + n)
... | bs 1# = binPower bs
... | 0# = x -- this case is impossible; showing this is extra
work...
On 2013-10-08 10:13, Daniel Peebles wrote:
> My first inclination for your repeated halving is to switch to a
> representation where your work is structural Data.Bin comes to mind,
> and you could write your exponentiation algorithm on Bin values, then
> convert back and forth on the way in and out as a wrapper. It'd
> probably be a lot more efficient, too.
>
> Alternately, since the conversion to Data.Bin is doing well-founded
> recursion of the sort you need, you could use it as an example. The
> real work is done in Data.Digits, and you can find more
> at http://www.cse.chalmers.se/~nad/listings/lib/Data.Digit.html#2612
> [2].
>
> On Mon, Oct 7, 2013 at 10:30 AM, Sergei Meshveliani
> <mechvel@...> wrote:
>
>> People,
>> I have this program for powering x^(n+1)
>> for x : Carrier of a Semigroup:
>>
>> half : ℕ → ℕ
>> half = ⌊_/2⌋
>>
>> powerTo-suc : C → (e : ℕ) → C -- x^(e+1)
>> powerTo-suc x 0 = x
>> powerTo-suc x (suc e) = case even? e of
>> { (yes _) → p ∙ p
>> ; _ → x ∙ (p ∙ p) }
>> where
>> h = half e
>> p = powerTo-suc x h
>>
>> The checker cannot prove its termination.
>> I can help this proof by adding the counter argument cnt for the
>> number of the `half' calls.
>> But this complicates various further proofs for powerTo-suc.
>>
>> So, I think of trying _inductive families_ for ℕ provided by
>> Induction, Induction.Nat ... in the standard library
>> (the question of whether this will make further proofs easier is
>> open
>> for me).
>>
>> But it is difficult to understand the idea, some introductory text
>> is
>> needed, with a couple of simple examples.
>>
>> Induction.Nat gives an example for `half'. It is too simple,
>> because
>> it is for the step from n to suc n.
>> And in my example, it is the step from half (suc e) to suc e,
>> assuming that we have a proof for
>> half (suc e) < suc e.
>>
>> Probably cRec needs here to be given something additional, maybe
>> some
>> builder needs to be defined.
>>
>> May be you can demonstrate this Agda induction right on this example
>> of
>> powerTo-suc ?
>>
>> Thank you in advance for explanation.
>>
>> ------
>> Sergei
>>
>> _______________________________________________
>> Agda mailing list
>> Agda@...
>> https://lists.chalmers.se/mailman/listinfo/agda [1]
>
>
>
> Links:
> ------
> [1] https://lists.chalmers.se/mailman/listinfo/agda
> [2] http://www.cse.chalmers.se/~nad/listings/lib/Data.Digit.html#2612
>
> _______________________________________________
> Agda mailing list
> Agda@...
> https://lists.chalmers.se/mailman/listinfo/agda
--
Andreas Abel <>< Du bist der geliebte Mensch.
Theoretical Computer Science, University of Munich
http://www.tcs.informatik.uni-muenchen.de/~abel/

module BinPowerStdLib where
open import Function
open import Data.Bool
open import Data.List
open import Data.Nat
open import Data.Fin using (zero; suc)
open import Data.Bin using (Binâº; 0#; _1#; fromâ)
open import Relation.Binary.PropositionalEquality
-- Binary exponentiation
module Power {A : Set} (_â¢_ : A â A â A) (x : A) where
binPower : Binâº â A
binPower [] = x
binPower (b â· bs) = case b of
Î» { zero â rÂ²
; (suc zero) â x â¢ rÂ²
; (suc (suc ()))
}
where
r = binPower bs
rÂ² = r â¢ r
posPower : (n : â) â n > 0 â A
posPower 0 ()
posPower (suc n) _ with fromâ (1 + n)
... | bs 1# = binPower bs
... | 0# = x -- this case is impossible; showing this is extra work...

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