Combinatorial programming with functions

Jo Devriendt

slides.com/jod/manyworlds-functions

ManyWorlds in a nutshell

Extremely accessible language

With blistering propagation performance

To solve combinatorial problems

Accessible language

3 core ideas

First-class functions

Everything must be finite

bool, int, string primitive types

Online editor @ manyworlds.site

declare S,E,N,D,M,O,R,Y: -> {0..9}.

              1000*S() + 100*E() + 10*N() + D()
            + 1000*M() + 100*O() + 10*R() + E()
//---------------------------------------------
= 10000*M() + 1000*O() + 100*N() + 10*E() + Y().

M() > 0.

distinct(S(),E(),N(),D(),M(),O(),R(),Y()).

Accessible language

declare some 0-arity functions

function that takes arbitrary number of arguments

My mother can understand this. Yours too! :p

write constraints using normal operators

functions

Accessible language

= != > < >= =< + - * / %
not and or xor implies

abs min max count

distinct same if-then-else

Strongly typed and total along bool, int, string

to-do

overloading for multiple types

Builtin functions

Accessible language

declare <name>: <type>, <type>, ... -> <finite range>.

User functions

declare prime: int -> {true, false}.

declare distance: string, string -> {0..1000}.

declare cities: string -> bool.

declare state: int -> {"busy", "ready", "down"}.

declare edge_color: string, string -> {"r", "g", "b"}.

signature

all ranges have a clear type, but are finite no int or string

declare n: -> {0..9}.

Accessible language

declare color: string -> {"r", "g", "b", "y"}.

color("NL") != color("BE").
color("NL") != color("DE").
color("BE") != color("LU").
color("BE") != color("DE").
color("BE") != color("FR").
color("FR") != color("LU").
color("FR") != color("DE").
color("LU") != color("DE").

Map coloring 1

There must be a better way...

Accessible language

declare color: string -> {"r", "g", "b", "y"}.

declare border: string, string -> bool.

define border as {("NL","BE"), ("NL","DE"),
("BE","LU"), ("BE","DE"), ("BE","FR"),
("FR","LU"), ("FR","DE"), ("LU","DE")} else false.

all [ color(x)!=color(y) for x,y where border(x,y) ].

Map coloring 2

definition fixes meaning of function

FMF expression

total definition

compiling / flattening / unrolling / grounding / instantiating
yields previous disequalities

Accessible language

FMF expressions

max(f(1), f(2), f(3))
all
any
none
count
sum
product
min
max
distinct
same
odd
even
and
or
not( or )
count
+
*
min
max
distinct
same
xor
not( xor )
max [ f(x) for x where x in {1..3} ]

FMF

corresponding

n-ary function

"scoped builtins"

Accessible language

all [ color(x)!=color(y) for x,y where border(x,y) ].

FMF expressions

Filter: select all x,y where border(x,y) holds

Fold: reduce those color(x)!=color(y) to true iff all are true

Map: map those x,y to color(x)!=color(y)

Accessible language

all [ color(x)!=color(y) for x,y where border(x,y) ].

FMF expressions

list comprehension

aggregate function converting list to one expression

Accessible language

FMF expressions

Flexible!

sum [ 
  weight(x) 
  for x where Item(x) and inKnapsack(x) 
] <= capacity().

Captures global constraints and global functions.

distinct [ 
  color(x) 
  for x where Country(x) and not color(x)="r" 
].

"allDiff except 0"

Accessible language

decdef num as {1..9}.

declare cell: int,int -> num.

all [ 
    distinct [ cell(r,c) for c where num(c) ] 
for r where num(r) ].
all [ 
    distinct [ cell(r,c) for r where num(r) ] 
for c where num(c) ].

declare square: int,int,int -> bool.

define square as ...

all [ 
    distinct [ cell(r,c) for r,c where square(r,c,s)]
for s where num(s) ].

define cell as ...

FMF can be arbitrarily nested

Sudoku

Accessible language

FMF expressions

Write your own decomposition

Circuit constraint in TSP

decdef City as {"city_" 1..n}.

declare next: string -> City.

declare order: string -> {1..n}.

order("city_1") = 1.

all [ order(next(x)) = order(x) + 1 
    for x where City(x) and next(x) != "city_1" ].

Accessible language

decdef City as {"city_" 1..n}.

declare next: string -> City.

declare order: string -> {1..n}.

order("city_1") = 1.

all [ order(next(x)) = order(x) + 1 
    for x where City(x) and next(x) != "city_1" ].

FMF caveat: compiler must derive finite instantiation from "where" block

City limits instantations of x as it is a finite set of strings.

Accessible language

3 core ideas

First-class functions

Everything must be finite

bool, int, string primitive types

Online editor @ manyworlds.site

Solve combinatorial problems

find a consistent world

FOUND WORLD

color as {("DE","b"), ("BE","g"), ("FR","r"), ("LU","y"), ("NL","r")}.

declare color: string -> {"r", "g", "b", "y"}.

declare border: string, string -> bool.

define border as {("NL","BE"), ("NL","DE"),
("BE","LU"), ("BE","DE"), ("BE","FR"),
("FR","LU"), ("FR","DE"), ("LU","DE")} else false.

all [ color(x)!=color(y) for x,y where border(x,y) ].

Solve combinatorial problems

debug no consistent world

FOUND BLOCKERS
Line 9: not color("BE")=color("DE")
Line 9: not color("BE")=color("FR")
Line 9: not color("BE")=color("LU")
Line 9: not color("DE")=color("FR")
Line 9: not color("DE")=color("LU")
Line 9: not color("FR")=color("LU")
declare color: string -> {"r", "g", "b", "y"}.

declare border: string, string -> bool.

define border as {("NL","BE"), ("NL","DE"),
("BE","LU"), ("BE","DE"), ("BE","FR"),
("FR","LU"), ("FR","DE"), ("LU","DE")} else false.

all [ color(x)!=color(y) for x,y where border(x,y) ].

NL is not involved, as it only borders two other countries

Solve combinatorial problems

optimize over all worlds

minimize sum [ distance(x,next(x)) for x where City(x) ].

find a world now returns

  • optimal world
  • optimal objective value
  • "optimization blockers"

Solve combinatorial problems

declare man: string -> bool.
declare mortal: string -> bool.

all [ man(x) implies mortal(x) 
      for x where 
      x in {"Socrates","Athens","poison cup","Zeus"} 
    ].

man("Socrates").
not mortal("Zeus").
FOUND INTERSECTION

man as {("Socrates",true), ("Zeus",false)}.

mortal as {("Socrates",true), ("Zeus",false)}.

intersect all worlds

Solve combinatorial problems

Planned:

count all worlds

find relevant unknowns

Performance

BIG to-do

Goal

Means

  • C++
  • Exact
  • multithreaded propagation
  • strong compilation

Fastest "intersect" on the planet

Compilation?

Pipeline

  1. fix parsing
  2. desugar
  3. simplify
  4. instantiate
  5. simplify
  6. unnest 1
  7. simplify
  8. merge
  9. unnest 2
  10. add constraints to solver

Bad: in each step, full expression tree is rebuilt...

Good: minimal set of compiled constraints

Performance

ManyWorlds in a nutshell

Extremely accessible language

With blistering propagation performance

To solve combinatorial problems

Motivation

Combinatorial programming is simpler than imperative programming. Why are the languages so hard?

Successful when a 12-year old can do their math homework with ManyWorlds and when a lawyer can recognize ManyWorld-encoded laws.

Caveat: no silver bullet!
But simple problems should have simple solutions.

Problem domain description should match code.

Usability

  • No order on expressions
    • e.g., use a function before you declare it
  • Deterministic
  • Helpful debug information
    • syntax mistakes
    • compilation errors
    • blockers
  • Online editor
  • No limit on integer type
  • Syntax highlighting (to-do)
  • Good documentation and tutorials (... to-do)
  • ...

Relation to CPMpy

Both focus on

  • ease of use
  • flexibility
  • solver independence
  • explanations

Differences

  • library vs system
  • expressivity of language
  • access to solving technology

Summer 2024:

Stateful ManyWorlds with Python interface. New backend? ;)

Little directionality

sum [ distance(x,next(x)) for x where City(x) ]
  • order of expressions does not matter
  • (almost) no distinction between parameters and variables

     
  • even definitions can be used backwards
declare fib: int -> {0..1e21}.


define fib(x) as
  0   if x=0 else
  1   if x=1 else
  fib(x-1) + fib(x-2)
where x in {0..100} else 0.
declare fib: int -> {0..1e21}.
declare a,b: -> {0..10}.

define fib(x) as
  a() if x=0 else
  b() if x=1 else
  fib(x-1) + fib(x-2)
where x in {0..100} else 0.

fib(10)=144.

Thanks for your attention!

ManyWorlds - Combinatorial programming with functions

By Jo Devriendt

ManyWorlds - Combinatorial programming with functions

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