Measuring With Jugs

A Solution in

Finite Domain

Outline

The Puzzle

The Algorithm

Introduction to Oz3

Implementation in Oz3

The Puzzle

Given a set of jugs, and an unlimited supply of water, how would you measure out a quantity of water.

Rules Unlimited Supply of water When a jug is poured into, it must be

filled to capacity( unless there is not enough to fill it)

Any jug may be emptied.

Example Given a 5 liter jug, a 3 liter jug, how do

you measure out exactly 4 liters?

Solution: Fill the 5 liter jug, empty it into the 3 liter jug leaving the 5 liter with 2 liters. Empty the 3 liter jug. Pour 2 liters from the 5 liter jug into the 3 liter jug. Fill the 5 liter jug. Pour 1 liter into the 3 liter jug to fill it. % liter jug now has 4 liters!

A Bit of History

“Classic Mind Teaser” Puzzle was explored in “Measuring

With Jugs” by Boldi, Santini and Vigna.

In “Die Hard III”

The Algorithm

This can be broken down fairly since the only

important quantities are the capacities of the

jugs and the final volume. So…

Number of filling operations-Number of

emptying operations=Final

CapA*TimesA-CapB*TimesB=Final

Constraint Programming

Constraint programming applies

constraints given by the programmer to

limit choices …

Intro to Oz

variables -- a set of alphanumeric characters starting with a CAPITOL letter

atoms -- set of alphanumeric characters

beginning with a lower case letter

(anAtom in Java would be "anAtom")

keywords – declare, fun, Browse, Search

Other Neat Stuff

Virtual string: a#b#C Virtual string: a#b#C

Record: root(a b e o p)Record: root(a b e o p)

List: [a 1 r B 24.6]List: [a 1 r B 24.6]

Anonymous function:Anonymous function: fun{$ A B} ……. endfun{$ A B} ……. end

Oz Program declaredeclare fun{Factors Number}fun{Factors Number} fun{FactsIter Fact Number}fun{FactsIter Fact Number} if Fact == Number then [Fact]if Fact == Number then [Fact] elseif (Number mod Fact==0) then Fact|{FactsIter 2 (Number elseif (Number mod Fact==0) then Fact|{FactsIter 2 (Number

div Fact)}div Fact)} else {FactsIter Fact+1 Number}else {FactsIter Fact+1 Number} endend endend inin {FactsIter 2 Number}{FactsIter 2 Number} endend

{Browse {Factors 740}}{Browse {Factors 740}}

Constraint Programming Applied

CapA*TimesA-CapA*TimesANeg Each jug is represented by a virtual

string

CapX#XTimesX#Conts show output explain output numbers

Numbers to Operations

Introduction

Structure of functions

Code bits

Final output

Overview

Each jug is represented by a virtual string CapX#XTimesX#Conts

Series of operations is represented by a List

Main Functions in JugsExten.oz DoSomething - all of the emptying and filling

operations are performed in this function

Combine2 – combines the contents of a list of jugs and writes a path that pours all into largest

Combine -- combines all jugs into largest and writes a path

WritePath -- Uses previous functions to Write a path describing all operations

Jugs -- finite domain portion of program to find numbers

Bits Of Code -- Combine

%% Combines contents of all of the jugs into the largest

fun{Combine Jugs} X List Total in {GCap Jugs List X _} Total={SumList3 X|List _}

{Combine2 List X.1}|[to get Total] end

Output from Jugs7.oz

CapA=6

CapB=10 CapC=15

JugsExten.oz Output

Conclusion

This shows that Finite Domain Constraint Programming is useful for areas where it might not at first seem useful.