Research Plan

1
Logical and Algebraic Formulation of Origami
Axioms
Advisor Tetsuo Ida Graduate School of Systems
and Information Engineering Student Fadoua
Ghourabi 200621004 First year
November 16, 2006
2
Outline

• Introduction
• Logical Formulation of Huzitas Axioms
• Algebraic Interpretation
• Applications
• Conclusion and Further Research

3
Introduction
Origami (???) is the Japanese paper fold art

• Power of Origami
• Origami Mathematics
• Computational Origami
• Computer assisted construction of geometrical
  objects by means of paper folds

4
Introduction

• Definition of origami construction by axioms
• Given points and lines in an origami paper, what
  kind of folds can we make ?

Huzitas axioms (HA) Six basic folds are
possible.
5
Introduction

• Huzitas axioms Basic folding operations
• Axiom (O1)
• Axiom (O3)
• Axiom (O4)
• Axiom (O6)

n
m
P

m
n
m
6
Logical Formulation of HA

• Syntax
• Point is sort of points
• Line is sort of lines
• Variables V VPoint ? VLine
• ? is a sorted signature ? ( P, F )
• where P is set of predicates and F is the
  set of functions
• L the first order predicate logic language over
  the sorted signature ? ( P, F )
• P OnLine, Perpendicular,
• F SymmetricPoint,

7
Logical Formulation of HA
t P l f(t,..., t)

• Terms are defined by
• where P ?VPoint, l ? VLine and f ? F
• Formulas are in prenex normal form
• where Qi ??, ?, F is quantifier free formula, A
  is an atomic formula

? Q1x1 Qnxn Fx1xn F A F ? F F ? F
F
8
Logical Formulation of HA

• Axiom (O1)

Given two points P and Q, we can make a fold
along the line that passes through P and Q
? P, Q ? Point ? k ? Line OnLineP, k ?
OnLineQ, k
9
Logical Formulation of HA
Axiom (O3)
Given two lines m and n, we can make a fold to
superpose m and n.
? m, n ? Line ? k ? Line ? P ? Point OnLineP, k
? DistanceP, n DistanceP, m
10
Logical Formulation of HA
Axiom (O4)
Given a point P and a line m, we can make a fold
along the line that is perpendicular to m and
passes through P
? P ? Point ? m ? Line ? k ? Line OnLineP, k ?
Perpendiculark, m
11
Logical Formulation of HA
Axiom (O6)
Given two points P and Q and two lines m and n,
we can make a fold to superpose P and m, and Q
and n, simultaneously
? P, Q ? Point ? m, n ? Line ? k ?
Line OnLineSymmetricPointP, k,
m? OnLineSymmetricPointQ, k, n
12
Algebraic Interpretation
Logical formulas
Algebraic forms Polynomial equalities,
inequalities and disequalities

• Why?

Further applications

• Realization of origami construction
• Proof of the correctness of the origami
  construction

13
Algebraic Interpretation

• Algebraic interpretation A .
• APoint ?2
• ALine (a, b, c) ? ?3 a2b21
• A OnLineP, m am xp bm yp cm0,
  am2bm2-10
• where A P(xP,yP) and A m(am, bm, cm)
• A Perpendicularm, n am an bm bn 0
  ,
• am2bm2 -1 0, an2bn2 -1 0
• where A m (am, bm, cm) and A n(an, bn,
  cn)

14
Algebraic Interpretation
A over formulas
1/ Conjunction
A ? i?1,...,n Fi ? i?1,...,n A Fi
Example
A F1 p10 A F2 q10, q20 A
F1 ? F2 p10 ? q10, q20 p10,
q10, q20
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Algebraic Interpretation
2/ Disjunction
A ? i?1,...,n Fi p1 pn 0 pi0?
AFi , 1 ? i ? n
Example
A F1 p10 A F2 q10, q20 A
F1 ? F2 p1 q1 0, p1 q20
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Algebraic Interpretation
3/ Negation
A ?F ? p0?AF (p ?p - 1)0
Example
A F p0 A ?F p ?p - 1 0
17
Applications

• Origami construction as constraint solving

Stepwise construction where each step is a fold
operation that satisfies one of Huzitas
axioms Geometric constraints are defined to
specify origami geometric properties
formulate origami axioms
18
Applications
Example trisecting the angle ?FEG
(O1) Construction of the edges of ?FEG c
Constraintk ? Line, ThruQE, G, k s
SolveConstraintc Solution k ?Line-1, 1,
-1
19
Applications

• Trisection of ?FEG
• s SolveConstraintConstraintx ? Line,
  y?Line, yBringLineQEF, x?xBringLineQEG,
  y
• Solution Three numeric solutions

Second case is trisection of the internal angle
?FEG.
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Axiom (O3)
Given two lines m and n, we can make a fold to
superpose m and n.
(? m, n ? Line ? k ? Line ? P ? Point OnLineP,
k ? DistanceP, n DistanceP, m)
21
Applications

• Automated Theorem Proving

• Collect logical constraints of the origami
  construction
• Translate logical constraints into polynomials
  and then into algebraic equations Premises P
• Translate conclusion into algebraic form
  Conclusion C
• Prove P gt C using theorem provers based on GB,
  CAD

22
Conclusion and further research

• Logical and algebraic formulation of Huzitas
  axioms
• Constraint solving Theorem proving
• Future improvements

Generalization of logical formulation
A proof that reflects the geometric nature of the
construction problems
23
Thank you for your attention
24
Algebraic Interpretation
Example Axiom (O2)
A ?P, Q ? Point ?k ? Line SymmetricPointP, k
Q