Constructing List Homomorphisms from Proofs

Constructing List Homomorphisms from Proofs

Yun-Yan Chi Shin-Cheng Mu

IIS, Academia Sinica, Taiwan

September 17, 2012

Yun-Yan Chi, Shin-Cheng Mu APLAS 2011 1/ 25

Prelude
Preliminaries
The Way To Go

List Homomorphism

▸ A function h on lists is called a list homomorphism if it
satisﬁes
h (xs + ys) = h xs h ys,
+
for some associative operator ( )
▸ E.g. sum (xs + ys) = sum xs + sum ys
+


Prelude
Preliminaries
The Way To Go

List Homomorphism

▸ A function h on lists is called a list homomorphism if it
satisﬁes
h (xs + ys) = h xs h ys,
+
for some associative operator ( )
▸ E.g. sum (xs + ys) = sum xs + sum ys
+
▸ Potential chances of parallelisation
▸ compute h xs and h ys in parallel
▸ combine the results using ( )
▸ Can a list homomorphism be mechanically constructed?


Prelude
Preliminaries
The Way To Go

A clue

▸ The well-known third list-homomorphism theorem:
▸ h is a list homomorphism
▸ if h can be foldr (⊲) e and foldl(⊳) e
for some (⊲), (⊳) and e
▸ E.g.

sum ([3, 5, 7, 9]) = 3 + sum [5, 7, 9]
= sum [3, 5, 7] + 9
= sum [3, 5] + sum [7, 9]


Prelude
Preliminaries
The Way To Go

But, how?

▸ Plenty of previous work was devoted to the construction of
( ) from the definitions of (⊲) and (⊳)
▸ Practically, efforts are needed to prove
h = foldr (⊲) e = foldl (⊳) e
▸ This occurs often that one of (⊲) or (⊳) is picked as definition
of h, while the other is much harder to find


Prelude
Preliminaries
The Way To Go

But, how?

▸ We may have a good guess of ( ) by mixing (⊲) and (⊳)
▸ The proof of the correctness of ( ) is very similar to the proof
of h = foldr (⊲) e = foldl (⊳) e, which we have to provide
anyway
▸ Our idea: transform the proof of the correctness of ( ) from
the proof of foldr = foldl, after assembling a possible ( )


Prelude Fold
Preliminaries Tupling
The Way To Go Foldr-Fusion Law

Prelude

Preliminaries
Fold
Tupling
Foldr-Fusion Law

The Way To Go
Proof by Fusion
Proof Generalisation


Prelude Fold

Foldr
▸ A function h is a instance of foldr (⊲) e if h can be deﬁned as
▸ h [] = e
▸ h (x xs) = x ⊲ h xs
for some e and (⊲)
▸ E.g.

foldr (⊲) e xs
= foldr (⊲) e (x1 (x2 (x3 [ ])))
= x1 ⊲ foldr (⊲) e (x2 (x3 [ ]))
= ...
= x1 ⊲ (x2 ⊲ (x3 ⊲ e))


Prelude Fold

Foldl
▸ Symmetrically, h is a foldl(⊳) e if it can be deﬁned as
▸ h [] = e
▸ h (xs + [x]) = h xs ⊳ x,
+
for some (⊳) and e
▸ E.g.

foldl (⊳) e xs
= foldl (⊳) e ((([ ] + [x1 ]) + [x2 ]) + [x3 ])
+ + +
= foldl (⊳) e (([ ] + [x1 ]) + [x2 ]) ⊳ x3
+ +
= ...
= ((e ⊳ x1 ) ⊳ x2 ) ⊳ x3


Prelude Fold

Tupling

▸ Not all functions can be a fold
▸ Tupling:
for h, ﬁnd a k such that ⟨h, k⟩ is a fold
▸ ⟨h, k⟩ x = (h x, k x)


Prelude Fold

Foldr-Fusion Law

▸ One can fuse f and foldr into another foldr


Prelude Fold

Foldr-Fusion Law

▸ One can fuse f and foldr into another foldr

(f ○ foldr (⊲) e) xs
= (f ○ foldr (⊲) e) (x1 (x2 (x3 ... [])))
= f (x1 ⊲ (x2 ⊲ (x3 ⊲ ... ⊲ e)))
= { f (x ⊲ z) = x ⊕ f z }
x1 ⊕ (f (x2 ⊲ (x3 ⊲ ... ⊲ e)))
= ...
= x1 ⊕ (x2 ⊕ (x3 ⊕ ... ⊕ (f e)))
= foldr (⊕) (f e) xs


Prelude
Proof by Fusion
Preliminaries
The Way To Go

Return to our approach

▸ Since we try to transform the proof of
▸ h = foldr (⊲) e = foldl (⊳) e
to the proof of
▸ the correctness of ( ),
▸ we want to know how to
1. prove that h = foldr (⊲) e = foldl (⊳) e
2. prove that ( ) do deﬁne a list homomorphism
3. transform the former to the latter


Prelude
Proof by Fusion
Preliminaries
The Way To Go


▸ Let h = foldr (⊲) e
▸ To prove that h = foldl (⊳) e, we have to show
▸ h[] = e
▸ h (xs + [z]) = h xs ⊳ z
+
▸ In point-free style: h ○ (+
+[z]) = (⊳ z) ○ h

h ○ (+
+[z])
= { foldr -fusion, since (+
+[z]) = foldr ( ) [z] }
foldr (⊲) (h [z])
= { foldr -fusion (backwards) }
(⊳ z) ○ foldr (⊲) e
= (⊳ z) ○ h


Prelude
Proof by Fusion
Preliminaries
The Way To Go


▸ For the second foldr -fusion
▸ z ⊲e=e⊳z
▸ (x ⊲ y ) ⊳ z = x ⊲ (y ⊳ z) - the associativity of (⊲) and (⊳)

▸ We will have the proof of h = foldr (⊲) e = foldl (⊳) e
if we have the proof of above fusion conditions


Prelude
Proof by Fusion
Preliminaries
The Way To Go

h is list homomorphism

▸ To prove that h is a list homomorphism, we have to show
▸ h (xs + ys) = h xs
+ h ys
▸ In point-free style: h ○ (+
+ys) = ( h ys) ○ h

h ○ (+
+ys)
= { foldr -fusion, since (+
+ys) = foldr ( ) ys }
foldr (⊲) (h ys)
= { foldr -fusion (backwards) }
( h ys) ○ foldr (⊲) e
= ( h ys) ○ h


Prelude
Proof by Fusion
Preliminaries
The Way To Go

h is list homomorphism

▸ For the second foldr -fusion
▸ h ys = e h ys
▸ (x ⊲ y ) h ys = x ⊲ (y h ys)

▸ If we have the proof of those fusion conditions, we will have
the proof of ( ) do deﬁne a list homomorphism


Prelude
Proof by Fusion
Preliminaries
The Way To Go

Generalisation

▸ To transform the proof of
▸ h = foldr (⊲) e = foldl (⊳) e

to the proof of
▸ h (xs + ys) = h xs
+ h ys


Prelude
Proof by Fusion
Preliminaries
The Way To Go

Generalisation

▸ z ⊲e=e⊳z
▸ (x ⊲ y ) ⊳ z = x ⊲ (y ⊳ z)
to the proof of
▸ h ys = e h ys
▸ (x ⊲ y ) h ys = x ⊲ (y h ys)


Prelude
Proof by Fusion
Preliminaries
The Way To Go

Generalisation

▸ z ⊲e=e⊳z
▸ (x ⊲ y ) ⊳ z = x ⊲ (y ⊳ z)
to the proof of
▸ h ys = e h ys
▸ (x ⊲ y ) h ys = x ⊲ (y h ys)

▸ To come up with ( ) and its correctness proof
▸ Generalise the former proof to the latter by replacing the
occurrences of z in (⊳) by metavariables


Setup
Example: Steep
Proving foldr (⊲) e = foldl (⊳) e
Conclusions
Constructing ( )

Example: Steep
Setup
Constructing ( )

Conclusions


Setup
Example: Steep
Conclusions
Constructing ( )

Steep
▸ A list of numbers is said to be steep if each number is larger
than the sum of the numbers to its right.
▸ E.g. steep [20, 10, 5, 2]


Setup
Example: Steep
Conclusions
Constructing ( )

Steep
▸ Can steep be a foldr ?


Setup
Example: Steep
Conclusions
Constructing ( )

Steep
▸ Can steep be a foldl ?


Setup
Example: Steep
Conclusions
Constructing ( )

Cap
▸ cap xs, upper-bound of value we can attach to the right of xs
▸ cap can be a foldr together with sum


Setup
Example: Steep
Conclusions
Constructing ( )

Cap
▸ cap xs, upper-bound of value we can attach to the right of xs
▸ cap can be a foldl


Setup
Example: Steep
Conclusions
Constructing ( )

Steep as Fold

▸ We can compute steep if we can compute cap
▸ ⟨cap, sum⟩ can be foldr (⊲) (∞, 0) and foldl (⊳) (∞, 0),
where
▸ x ⊲ (c2 , s2 ) = ((x − s2 ) ↓ c2 , x + s2 )
▸ (c1 , s1 ) ⊳ z = ((c1 − z) ↓ z, s1 + z)
▸ It is not so obvious that foldr = foldl


Setup
Example: Steep
Conclusions
Constructing ( )

Example: Steep
Setup
Constructing ( )

Conclusions


Setup
Example: Steep
Conclusions
Constructing ( )

Fusion Conditions

▸ z ⊲ (∞, 0) = (∞, 0) ⊳ z
▸ (x ⊲ y ) ⊳ z = x ⊲ (y ⊳ z)
▸ The former condition trivially holds:

z ⊲ (∞, 0)
= { deﬁnition of (⊲) }
((z − 0) ↓ ∞, z + 0)
= { arithmetics }
((∞ − z) ↓ z, 0 + z)
= { deﬁnition of (⊳) }
(∞, 0) ⊳ z.


Setup
Example: Steep
Conclusions
Constructing ( )

Proof of Associativity

(x ⊲ (c, s)) ⊳ z
((x − s) ↓ c, x + s) ⊳ z
((((x − s) ↓ c) − z) ↓ z, x + s + z)
= { (−z) distributes over (↓) }
(((x − s − z) ↓ (c − z)) ↓ z, x + s + z)
= { arithmetics }
(((x − (s + z)) ↓ ((c − z) ↓ z), x + s + z)
x ⊲ ((c − z) ↓ z, s + z)
x ⊲ ((c, s) ⊳ z)


Setup
Example: Steep
Conclusions
Constructing ( )

Example: Steep
Setup
Constructing ( )

Conclusions


Setup
Example: Steep
Conclusions
Constructing ( )

The Aim now is

1. Generalise the proof of
▸ (x ⊲ y ) ⊳ z = x ⊲ (y ⊳ z)
to a proof of
▸ (x ⊲ y ) (c2 , s2 ) = x ⊲ (y (c2 , s2 )).
2. Construct a deﬁnition of ( )


Setup
Example: Steep
Conclusions
Constructing ( )

Generalise The Proof
▸ Copy the proof of associativity
(x ⊲ (c, s)) ⊳ z
((x − s) ↓ c, x + s) ⊳ z
((((x − s) ↓ c) − z) ↓ z, x + s + z)
= { -z distributes over (↓) }
(((x − s − z) ↓ (c − z)) ↓ z, x + s + z)
= { arithmetics }
(((x − (s + z)) ↓ ((c − z) ↓ z), x + s + z)
x ⊲ ((c − z) ↓ z, s + z)
x ⊲ ((c, s) ⊳ z)


Setup
Example: Steep
Conclusions
Constructing ( )

▸ Generalise ⊳ z to (c2 , s2 )
(x ⊲ (c, s)) ⊳ z
((x − s) ↓ c, x + s) ⊳ z
((((x − s) ↓ c) − z) ↓ z, x + s + z)
(((x − s − z) ↓ (c − z)) ↓ z, x + s + z)
= { arithmetics }
(((x − (s + z)) ↓ ((c − z) ↓ z), x + s + z)
x ⊲ ((c − z) ↓ z, s + z)
x ⊲ ((c, s) ⊳ z)


Setup
Example: Steep
Conclusions
Constructing ( )

▸ Generalise ⊳ z to (c2 , s2 )
(x ⊲ (c, s)) (c2 , s2 )
((x − s) ↓ c, x + s) (c2 , s2 )
= { deﬁnition of ( ) }
((((x − s) ↓ c) − z) ↓ z, x + s + z)
(((x − s − z) ↓ (c − z)) ↓ z, x + s + z)
= { arithmetics }
(((x − (s + z)) ↓ ((c − z) ↓ z), x + s + z)
x ⊲ ((c − z) ↓ z, s + z)
x ⊲ ((c, s) (c2 , s2 ))


Setup
Example: Steep
Conclusions
Constructing ( )

▸ Replace z by metavariable Xi
(x ⊲ (c, s)) (c2 , s2 )
((x − s) ↓ c, x + s) (c2 , s2 )
((((x − s) ↓ c) − z) ↓ z, x + s + z)
(((x − s − z) ↓ (c − z)) ↓ z, x + s + z)
= { arithmetics }
(((x − (s + z)) ↓ ((c − z) ↓ z), x + s + z)
x ⊲ ((c − z) ↓ z, s + z)
x ⊲ ((c, s) (c2 , s2 ))


Setup
Example: Steep
Conclusions
Constructing ( )

▸ Replace z by metavariable Xi
(x ⊲ (c, s)) (c2 , s2 )
((x − s) ↓ c, x + s) (c2 , s2 )
((((x − s) ↓ c) − X1 ) ↓ X2 , x + s + X3 )
= { -X1 distributes over (↓) }
(((x − s − X1 ) ↓ (c − X1 )) ↓ X2 , x + s + X3 )
= { arithmetics }
(((x − (s + X1 )) ↓ ((c − X1 ) ↓ X2 ), x + s + X3 )
x ⊲ ((c − X1 ) ↓ X2 , s + X1 )
x ⊲ ((c, s) (c2 , s2 ))


Setup
Example: Steep
Conclusions
Constructing ( )

▸ (s + X1 ) and (s + X3 ) have to be the same term
(x ⊲ (c, s)) (c2 , s2 )
((x − s) ↓ c, x + s) (c2 , s2 )
((((x − s) ↓ c) − X1 ) ↓ X2 , x + s + X3 )
(((x − s − X1 ) ↓ (c − X1 )) ↓ X2 , x + s + X3 )
= { arithmetics }
(((x − (s + X1 )) ↓ ((c − X1 ) ↓ X2 ), x + s + X3 )
x ⊲ ((c − X1 ) ↓ X2 , s + X1 )
x ⊲ ((c, s) (c2 , s2 ))


Setup
Example: Steep
Conclusions
Constructing ( )

▸ (s + X1 ) and (s + X3 ) have to be the same term
(x ⊲ (c, s)) (c2 , s2 )
((x − s) ↓ c, x + s) (c2 , s2 )
((((x − s) ↓ c) − X1 ) ↓ X2 , x + s + X1 )
(((x − s − X1 ) ↓ (c − X1 )) ↓ X2 , x + s + X1 )
= { arithmetics }
(((x − (s + X1 )) ↓ ((c − X1 ) ↓ X2 ), x + s + X1 )
x ⊲ ((c − X1 ) ↓ X2 , s + X1 )
x ⊲ ((c, s) (c2 , s2 ))


Setup
Example: Steep
Conclusions
Constructing ( )

▸ Proof of (x ⊲ y ) (c2 , s2 ) = x ⊲ (y (c2 , s2 ))
(x ⊲ (c, s)) (c2 , s2 )
((x − s) ↓ c, x + s) (c2 , s2 )
((((x − s) ↓ c) − X1 ) ↓ X2 , x + s + X1 )
(((x − s − X1 ) ↓ (c − X1 )) ↓ X2 , x + s + X1 )
= { arithmetics }
(((x − (s + X1 )) ↓ ((c − X1 ) ↓ X2 ), x + s + X1 )
x ⊲ ((c − X1 ) ↓ X2 , s + X1 )
x ⊲ ((c, s) (c2 , s2 ))


Setup
Example: Steep
Conclusions
Constructing ( )

Reﬁning The ( )

▸ (c1 , s1 ) (c2 , s2 ) = ((c1 − X1 ) ↓ X2 , s1 + X1 )
▸ Satisﬁes that (c2 , s2 ) = (∞, 0) (c2 , s2 )

(c2 , s2 ) = ((∞ − X1 ) ↓ X2 , 0 + X1 )
≡ (c2 , s2 ) = (∞ ↓ X2 , X1 )
≡ (c2 , s2 ) = (X2 , X1 )


Setup
Example: Steep
Conclusions
Constructing ( )

Reﬁning The ( )

▸ (c1 , s1 ) (c2 , s2 ) = ((c1 − X1 ) ↓ X2 , s1 + X1 )
▸ Satisﬁes that (c2 , s2 ) = (∞, 0) (c2 , s2 )

(c2 , s2 ) = ((∞ − X1 ) ↓ X2 , 0 + X1 )
≡ (c2 , s2 ) = (∞ ↓ X2 , X1 )
≡ (c2 , s2 ) = (X2 , X1 )
▸ We have thus discovered that
▸ (c1 , s1 ) (c2 , s2 ) = ((c1 − s2 ) ↓ c2 , s1 + s2 )
▸ This ( ) has got to be correct, because we have the proof
already!


Example: Steep
Conclusions

Conclusions

▸ We have proposed and demonstrated a novel approach to
constructing ( ).
▸ Starting with a trivial generalisation of either (⊲) or (⊳), we
exploit the constraint enforced by the proof of associativity to
reﬁne ( ).
▸ Once we have constructed ( ), we have its correctness proof
too.


Constructing List Homomorphisms from Proofs

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