Propositional Logic Computer exercises€¦ · Propositional Logic Computer exercises Mario Alviano University of Calabria, Italy A.Y. 2017/2018 1/40

Propositional LogicComputer exercises

Mario Alviano

University of Calabria, Italy

A.Y. 2017/2018

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Outline

1 DIMACS format

2 Setup and simple tests

3 Pigeon Hole Problem

4 Chess Piece Independence

5 Hardware correctness

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Outline

1 DIMACS format





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SAT solvers input: DIMACS format

Standard input format for SAT solversRepresentation of CNF formulas stemming from achallenge run by DIMACS (Center for DiscreteMathematics and Theoretical Computer Science) in 1992

Example

(x1 ∨ x2 ∨ ¬x3) ∧ (¬x2) ∧ (x4 ∨ ¬x3)

DIMACS encoding

c This is a CNF in DIMACScp cnf 4 31 2 -3 0

-2 04 -3 0

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SAT solvers input: DIMACS format

Standard input format for SAT solversRepresentation of CNF formulas stemming from achallenge run by DIMACS (Center for DiscreteMathematics and Theoretical Computer Science) in 1992

Example

(x1 ∨ x2 ∨ ¬x3) ∧ (¬x2) ∧ (x4 ∨ ¬x3)

DIMACS encoding

c This is a CNF in DIMACScp cnf 4 31 2 -3 0

-2 04 -3 0

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DIMACS format in BNF grammar

BNF grammar

<input> ::= <preamble> <formula> EOF

<preamble> ::= [<commentlines>] <problemline><commentlines> ::= <commentline> <commentlines> | <commentline><commentline> ::= c <text> EOL<problemline> ::= p cnf <pnum> <pnum> EOL

<formula> ::= <clauselist><clauselist> ::= <clause> <clauselist> | <clause><clause> ::= <literal> <clause> | <literal> 0<literal> ::= <num>

<text> ::= A sequence of non-special ASCII characters<pnum> ::= A signed integer greater than 0<num> ::= A signed integer different from 0

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Outline

1 DIMACS format





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Setup

1 (I suggest to) Boot Linux

2 Download one of the solvers athttp://minisat.se/MiniSat.html

MiniSat 2.2.0: only sourcesMiniSat 1.14: sources and binary (if you are lazy)

3 Build and have a look at the help$ minisat --help

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http://minisat.se/MiniSat.html

Setup

1 (I suggest to) Boot Linux2 Download one of the solvers athttp://minisat.se/MiniSat.html



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Setup


MiniSat 2.2.0: only sources

MiniSat 1.14: sources and binary (if you are lazy)


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Setup




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Setup




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Simple test (1)

(x1 ∨ x2) ∧ (x1 ∨ ¬x2) ∧ (¬x1 ∨ ¬x2)

c This is a CNF in DIMACScp cnf 2 31 2 01 -2 0

-1 -2 0

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Simple test (1)

(x1 ∨ x2) ∧ (x1 ∨ ¬x2) ∧ (¬x1 ∨ ¬x2)

c This is a CNF in DIMACScp cnf 2 31 2 01 -2 0

-1 -2 0

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Simple test (2)

(x1 ∨ x2) ∧ (x1 ∨ ¬x2) ∧ (¬x1 ∨ x2) ∧ (¬x1 ∨ ¬x2)

p cnf 2 41 2 01 -2 0

-1 2 0-1 -2 0

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Simple test (2)

(x1 ∨ x2) ∧ (x1 ∨ ¬x2) ∧ (¬x1 ∨ x2) ∧ (¬x1 ∨ ¬x2)

p cnf 2 41 2 01 -2 0

-1 2 0-1 -2 0

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More tests

Example

1 (x1 ∨ x2 ∨ ¬x3) ∧ (¬x2) ∧ (x4 ∨ ¬x3)

2 (x1 ∨ x2 ∨ ¬x3 ∨ ¬x4) ∧ (¬x2) ∧ (x4 ∨ ¬x3)

3 {x1 ∨ x2 ∨ x3, x1 ∨ x2 ∨ ¬x3, x1 ∨ ¬x2 ∨ x3, x1 ∨ ¬x2 ∨ ¬x3,¬x1 ∨ x4, x1 ∨ ¬x4 ∨ ¬x5 ∨ x6, ¬x1 ∨ x7}

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More tests

Example

1 (x1 ∨ x2 ∨ ¬x3) ∧ (¬x2) ∧ (x4 ∨ ¬x3)

2 (x1 ∨ x2 ∨ ¬x3 ∨ ¬x4) ∧ (¬x2) ∧ (x4 ∨ ¬x3)


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More tests

Example

1 (x1 ∨ x2 ∨ ¬x3) ∧ (¬x2) ∧ (x4 ∨ ¬x3)

2 (x1 ∨ x2 ∨ ¬x3 ∨ ¬x4) ∧ (¬x2) ∧ (x4 ∨ ¬x3)


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Outline

1 DIMACS format





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Pigeons and Holes

Pigeons in their Holes

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Pigeon Hole Problem

Pigeon Hole Problem — PHP

The problem PHP(m,n) is whether m pigeons can enter into npigeon holes

Task

Find a family of formulas which are satisfiablewhen PHP(m,n) is

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Pigeon Hole Problem

Pigeon Hole Problem — PHP

The problem PHP(m,n) is whether m pigeons can enter into npigeon holes

Task

Find a family of formulas which are satisfiablewhen PHP(m,n) is

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PHP — Modellation

Modellation

m × n propositional variables:xi,j , where i ≤ m and j ≤ nxi,j means that pigeon i is put into hole j

The formula should express:

1 Each pigeon is in some hole2 Each pigeon is in at most one hole3 In each hole there is at most one pigeon

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PHP — Modellation

Modellation



1 Each pigeon is in some hole2 Each pigeon is in at most one hole3 In each hole there is at most one pigeon

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PHP — Modellation

Modellation


The formula should express:1 Each pigeon is in some hole

2 Each pigeon is in at most one hole3 In each hole there is at most one pigeon

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PHP — Modellation

Modellation


The formula should express:1 Each pigeon is in some hole2 Each pigeon is in at most one hole

3 In each hole there is at most one pigeon

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PHP — Modellation

Modellation


The formula should express:1 Each pigeon is in some hole2 Each pigeon is in at most one hole3 In each hole there is at most one pigeon

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Example: PHP(3,2)

1 Each pigeon is in some hole

x1,1 ∨ x1,2, x2,1 ∨ x2,2, x3,1 ∨ x3,2

2 Each pigeon is in at most one hole

x1,1 → ¬x1,2x1,2 → ¬x1,1x2,1 → ¬x2,2x2,2 → ¬x2,1x3,1 → ¬x3,2x3,2 → ¬x3,1


x1,1 → ¬x2,1 ∧ ¬x3,1x2,1 → ¬x1,1 ∧ ¬x3,1x3,1 → ¬x1,1 ∧ ¬x2,1x1,2 → ¬x2,2 ∧ ¬x3,2x2,2 → ¬x1,2 ∧ ¬x3,2x3,2 → ¬x1,2 ∧ ¬x2,2

≡

x1,1 → ¬x2,1, x1,1 → ¬x3,1x2,1 → ¬x1,1, x2,1 → ¬x3,1x3,1 → ¬x1,1, x3,1 → ¬x2,1x1,2 → ¬x2,2, x1,2 → ¬x3,2x2,2 → ¬x1,2, x2,2 → ¬x3,2x3,2 → ¬x1,2, x3,2 → ¬x2,2

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Example: PHP(3,2)

1 Each pigeon is in some holex1,1 ∨ x1,2, x2,1 ∨ x2,2, x3,1 ∨ x3,2

2 Each pigeon is in at most one hole

x1,1 → ¬x1,2x1,2 → ¬x1,1x2,1 → ¬x2,2x2,2 → ¬x2,1x3,1 → ¬x3,2x3,2 → ¬x3,1



≡


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Example: PHP(3,2)


2 Each pigeon is in at most one holex1,1 → ¬x1,2x1,2 → ¬x1,1

x2,1 → ¬x2,2x2,2 → ¬x2,1x3,1 → ¬x3,2x3,2 → ¬x3,1



≡


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Example: PHP(3,2)


2 Each pigeon is in at most one holex1,1 → ¬x1,2x1,2 → ¬x1,1x2,1 → ¬x2,2x2,2 → ¬x2,1

x3,1 → ¬x3,2x3,2 → ¬x3,1



≡


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Example: PHP(3,2)


2 Each pigeon is in at most one holex1,1 → ¬x1,2x1,2 → ¬x1,1x2,1 → ¬x2,2x2,2 → ¬x2,1x3,1 → ¬x3,2x3,2 → ¬x3,1



≡


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Example: PHP(3,2)



3 In each hole there is at most one pigeonx1,1 → ¬x2,1 ∧ ¬x3,1x2,1 → ¬x1,1 ∧ ¬x3,1x3,1 → ¬x1,1 ∧ ¬x2,1

x1,2 → ¬x2,2 ∧ ¬x3,2x2,2 → ¬x1,2 ∧ ¬x3,2x3,2 → ¬x1,2 ∧ ¬x2,2

≡


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Example: PHP(3,2)



3 In each hole there is at most one pigeonx1,1 → ¬x2,1 ∧ ¬x3,1x2,1 → ¬x1,1 ∧ ¬x3,1x3,1 → ¬x1,1 ∧ ¬x2,1x1,2 → ¬x2,2 ∧ ¬x3,2x2,2 → ¬x1,2 ∧ ¬x3,2x3,2 → ¬x1,2 ∧ ¬x2,2

≡


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Example: PHP(3,2)




≡


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Example: PHP(3,2)




≡

x1,1 → ¬x2,1, x1,1 → ¬x3,1x2,1 → ¬x1,1, x2,1 → ¬x3,1x3,1 → ¬x1,1, x3,1 → ¬x2,1

x1,2 → ¬x2,2, x1,2 → ¬x3,2x2,2 → ¬x1,2, x2,2 → ¬x3,2x3,2 → ¬x1,2, x3,2 → ¬x2,2

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Example: PHP(3,2)




≡


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PHP(m,n) — Formula

m∧i=1

n∨j=1

xi,j

∧m∧

i=1

n∧j=1

(xi,j →n∧

k=1k 6=j

¬xi,k ) ∧m∧

i=1

n∧j=1

(xi,j →m∧

k=1k 6=i

¬xk ,j)

CNF Formula

m∧i=1

n∨j=1

xi,j

∧m∧

i=1

n∧j=1

n∧k=1k 6=j

(¬xi,j ∨ ¬xi,k ) ∧m∧

i=1

n∧j=1

m∧k=1k 6=i

(¬xi,j ∨ ¬xk ,j)

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PHP(m,n) — Formula

m∧i=1

n∨j=1

xi,j

∧m∧

i=1

n∧j=1

(xi,j →n∧

k=1k 6=j

¬xi,k ) ∧m∧

i=1

n∧j=1

(xi,j →m∧

k=1k 6=i

¬xk ,j)

CNF Formula

m∧i=1

n∨j=1

xi,j

∧m∧

i=1

n∧j=1

n∧k=1k 6=j

(¬xi,j ∨ ¬xi,k ) ∧m∧

i=1

n∧j=1

m∧k=1k 6=i

(¬xi,j ∨ ¬xk ,j)

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Example: PHP(3,2) in CNF


2 Each pigeon is in at most one hole¬x1,1 ∨ ¬x1,2¬x1,2 ∨ ¬x1,1¬x2,1 ∨ ¬x2,2¬x2,2 ∨ ¬x2,1¬x3,1 ∨ ¬x3,2¬x3,2 ∨ ¬x3,1

3 In each hole there is at most one pigeon¬x1,1 ∨ ¬x2,1, ¬x1,1 ∨ ¬x3,1¬x2,1 ∨ ¬x1,1, ¬x2,1 ∨ ¬x3,1¬x3,1 ∨ ¬x1,1, ¬x3,1 ∨ ¬x2,1¬x1,2 ∨ ¬x2,2, ¬x1,2 ∨ ¬x3,2¬x2,2 ∨ ¬x1,2, ¬x2,2 ∨ ¬x3,2¬x3,2 ∨ ¬x1,2, ¬x3,2 ∨ ¬x2,2

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PHP in DIMACS

We must convert the variables to positive integers

x1,1 7→ 1...

xm,1 7→ m

x1,2 7→ m + 1...

xm,2 7→ 2×m...

x1,n 7→ (n − 1)×m + 1...

xm,n 7→ n ×m

Therefore, xi,j will be represented by (j − 1)×m + i

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PHP in DIMACS


x1,1 7→ 1...

xm,1 7→ mx1,2 7→ m + 1

...xm,2 7→ 2×m

...x1,n 7→ (n − 1)×m + 1

...xm,n 7→ n ×m


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PHP in DIMACS


x1,1 7→ 1...

xm,1 7→ mx1,2 7→ m + 1

...xm,2 7→ 2×m

...x1,n 7→ (n − 1)×m + 1

...xm,n 7→ n ×m


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PHP in DIMACS


x1,1 7→ 1...

xm,1 7→ mx1,2 7→ m + 1

...xm,2 7→ 2×m

...x1,n 7→ (n − 1)×m + 1

...xm,n 7→ n ×m


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Example: PHP(3,2) in DIMACS

1 x1,1∨x1,2, x2,1∨x2,2, x3,1∨x3,2

2 ¬x1,1 ∨ ¬x1,2¬x1,2 ∨ ¬x1,1¬x2,1 ∨ ¬x2,2¬x2,2 ∨ ¬x2,1¬x3,1 ∨ ¬x3,2¬x3,2 ∨ ¬x3,1

3 ¬x1,1 ∨ ¬x2,1, ¬x1,1 ∨ ¬x3,1¬x2,1 ∨ ¬x1,1, ¬x2,1 ∨ ¬x3,1¬x3,1 ∨ ¬x1,1, ¬x3,1 ∨ ¬x2,1¬x1,2 ∨ ¬x2,2, ¬x1,2 ∨ ¬x3,2¬x2,2 ∨ ¬x1,2, ¬x2,2 ∨ ¬x3,2¬x3,2 ∨ ¬x1,2, ¬x3,2 ∨ ¬x2,2

p cnf 6 211 4 0 -1 -2 02 5 0 -1 -3 03 6 0 -4 -5 0

-1 -4 0 -4 -6 0-4 -1 0 -2 -1 0-2 -5 0 -2 -3 0-5 -2 0 -5 -4 0-3 -6 0 -5 -6 0-6 -3 0 -3 -1 0-3 -2 0-6 -4 0-6 -5 0

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PHP — Why?

Hard problem for SAT solvers

PHP(m,n) is unsatisfiable if and only if m > n

SAT solvers have an exponential behaviorSymmetries tend to mess up backtracking algorithmsTry it with the PHP(m,n) formula generator!

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PHP — Why?


PHP(m,n) is unsatisfiable if and only if m > nSAT solvers have an exponential behavior

Symmetries tend to mess up backtracking algorithmsTry it with the PHP(m,n) formula generator!

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PHP — Why?


PHP(m,n) is unsatisfiable if and only if m > nSAT solvers have an exponential behaviorSymmetries tend to mess up backtracking algorithms

Try it with the PHP(m,n) formula generator!

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PHP — Why?


PHP(m,n) is unsatisfiable if and only if m > nSAT solvers have an exponential behaviorSymmetries tend to mess up backtracking algorithmsTry it with the PHP(m,n) formula generator!

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Outline

1 DIMACS format





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Rook Independence Problem

Rook Independence Problem — RIP(m,n)

Place m rooks on an n × n chessboard so that they do notthreaten each other

Reminder: Rook moves

8 0Z0Z0Z0Z7 Z0Z0Z0Z06 0Z0Z0Z0Z5 Z0Z0Z0Z04 0Z0ZrZ0Z3 Z0Z0Z0Z02 0Z0Z0Z0Z1 Z0Z0Z0Z0

1 2 3 4 5 6 7 8

Example. RIP(2,2)

2 rZ1 Zr

1 2

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1 2 3 4 5 6 7 8

Example. RIP(2,2)

2 rZ1 Zr

1 2

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1 2 3 4 5 6 7 8

Example. RIP(2,2)

2 rZ1 Zr

1 2

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RIP — Modellation

Modellation

m × n × n propositional variables:xi,j,k , where i ≤ n, j ≤ n, k ≤ mxi,j,k means that rook k is put onto field (i , j)


1 Each rook is placed on some field2 Each rook is placed on at most one field3 Each field holds at most one rook4 No rook threatens another rook

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RIP — Modellation

Modellation



1 Each rook is placed on some field2 Each rook is placed on at most one field3 Each field holds at most one rook4 No rook threatens another rook

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RIP — Modellation

Modellation


The formula should express:1 Each rook is placed on some field

2 Each rook is placed on at most one field3 Each field holds at most one rook4 No rook threatens another rook

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RIP — Modellation

Modellation


The formula should express:1 Each rook is placed on some field2 Each rook is placed on at most one field

3 Each field holds at most one rook4 No rook threatens another rook

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RIP — Modellation

Modellation


The formula should express:1 Each rook is placed on some field2 Each rook is placed on at most one field3 Each field holds at most one rook

4 No rook threatens another rook

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RIP — Modellation

Modellation


The formula should express:1 Each rook is placed on some field2 Each rook is placed on at most one field3 Each field holds at most one rook4 No rook threatens another rook

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Example: RIP(3,2) (1)

1 Each rook is placed on some field

x1,1,1 ∨ x1,2,1 ∨ x2,1,1 ∨ x2,2,1x1,1,2 ∨ x1,2,2 ∨ x2,1,2 ∨ x2,2,2x1,1,3 ∨ x1,2,3 ∨ x2,1,3 ∨ x2,2,3

2 Each rook is placed on at most one fieldx1,1,1 → ¬x1,2,1 ∧ ¬x2,1,1 ∧ ¬x2,2,1x1,2,1 → ¬x1,1,1 ∧ ¬x2,1,1 ∧ ¬x2,2,1x2,1,1 → ¬x1,1,1 ∧ ¬x1,2,1 ∧ ¬x2,2,1x2,2,1 → ¬x1,1,1 ∧ ¬x1,2,1 ∧ ¬x2,1,1

...x1,1,3 → ¬x1,2,3 ∧ ¬x2,1,3 ∧ ¬x2,2,3x1,2,3 → ¬x1,1,3 ∧ ¬x2,1,3 ∧ ¬x2,2,3x2,1,3 → ¬x1,1,3 ∧ ¬x1,2,3 ∧ ¬x2,2,3x2,2,3 → ¬x1,1,3 ∧ ¬x1,2,3 ∧ ¬x2,1,3

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1 Each rook is placed on some fieldx1,1,1 ∨ x1,2,1 ∨ x2,1,1 ∨ x2,2,1x1,1,2 ∨ x1,2,2 ∨ x2,1,2 ∨ x2,2,2x1,1,3 ∨ x1,2,3 ∨ x2,1,3 ∨ x2,2,3



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3 Each field holds at most one rook

x1,1,1 → ¬x1,1,2 ∧ ¬x1,1,3x1,1,2 → ¬x1,1,1 ∧ ¬x1,1,3x1,1,3 → ¬x1,1,1 ∧ ¬x1,1,2

x1,2,1 → ¬x1,2,2 ∧ ¬x1,2,3x1,2,2 → ¬x1,2,2 ∧ ¬x1,2,3x1,2,3 → ¬x1,2,1 ∧ ¬x1,2,2

x2,1,1 → ¬x2,1,2 ∧ ¬x2,1,3x2,1,2 → ¬x2,1,1 ∧ ¬x2,1,3x2,1,3 → ¬x2,1,1 ∧ ¬x2,1,2

x2,2,1 → ¬x2,2,2 ∧ ¬x2,2,3x2,2,2 → ¬x2,2,1 ∧ ¬x2,2,3x2,2,3 → ¬x2,2,1 ∧ ¬x2,2,2

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3 Each field holds at most one rookx1,1,1 → ¬x1,1,2 ∧ ¬x1,1,3x1,1,2 → ¬x1,1,1 ∧ ¬x1,1,3x1,1,3 → ¬x1,1,1 ∧ ¬x1,1,2

x1,2,1 → ¬x1,2,2 ∧ ¬x1,2,3x1,2,2 → ¬x1,2,2 ∧ ¬x1,2,3x1,2,3 → ¬x1,2,1 ∧ ¬x1,2,2

x2,1,1 → ¬x2,1,2 ∧ ¬x2,1,3x2,1,2 → ¬x2,1,1 ∧ ¬x2,1,3x2,1,3 → ¬x2,1,1 ∧ ¬x2,1,2

x2,2,1 → ¬x2,2,2 ∧ ¬x2,2,3x2,2,2 → ¬x2,2,1 ∧ ¬x2,2,3x2,2,3 → ¬x2,2,1 ∧ ¬x2,2,2

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x1,2,1 → ¬x1,2,2 ∧ ¬x1,2,3x1,2,2 → ¬x1,2,2 ∧ ¬x1,2,3x1,2,3 → ¬x1,2,1 ∧ ¬x1,2,2

x2,1,1 → ¬x2,1,2 ∧ ¬x2,1,3x2,1,2 → ¬x2,1,1 ∧ ¬x2,1,3x2,1,3 → ¬x2,1,1 ∧ ¬x2,1,2

x2,2,1 → ¬x2,2,2 ∧ ¬x2,2,3x2,2,2 → ¬x2,2,1 ∧ ¬x2,2,3x2,2,3 → ¬x2,2,1 ∧ ¬x2,2,2

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x1,2,1 → ¬x1,2,2 ∧ ¬x1,2,3x1,2,2 → ¬x1,2,2 ∧ ¬x1,2,3x1,2,3 → ¬x1,2,1 ∧ ¬x1,2,2

x2,1,1 → ¬x2,1,2 ∧ ¬x2,1,3x2,1,2 → ¬x2,1,1 ∧ ¬x2,1,3x2,1,3 → ¬x2,1,1 ∧ ¬x2,1,2

x2,2,1 → ¬x2,2,2 ∧ ¬x2,2,3x2,2,2 → ¬x2,2,1 ∧ ¬x2,2,3x2,2,3 → ¬x2,2,1 ∧ ¬x2,2,2

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x1,2,1 → ¬x1,2,2 ∧ ¬x1,2,3x1,2,2 → ¬x1,2,2 ∧ ¬x1,2,3x1,2,3 → ¬x1,2,1 ∧ ¬x1,2,2

x2,1,1 → ¬x2,1,2 ∧ ¬x2,1,3x2,1,2 → ¬x2,1,1 ∧ ¬x2,1,3x2,1,3 → ¬x2,1,1 ∧ ¬x2,1,2

x2,2,1 → ¬x2,2,2 ∧ ¬x2,2,3x2,2,2 → ¬x2,2,1 ∧ ¬x2,2,3x2,2,3 → ¬x2,2,1 ∧ ¬x2,2,2

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4 No rook threatens another rook

x1,1,1 → ¬x1,2,2 ∧ ¬x1,2,3 ∧ ¬x2,1,2 ∧ ¬x2,1,3x1,1,2 → ¬x1,2,1 ∧ ¬x1,2,3 ∧ ¬x2,1,1 ∧ ¬x2,1,3x1,1,3 → ¬x1,2,1 ∧ ¬x1,2,2 ∧ ¬x2,1,1 ∧ ¬x2,1,2

x2,1,1 → ¬x2,2,2 ∧ ¬x2,2,3 ∧ ¬x1,1,2 ∧ ¬x1,1,3...

x1,2,1 → ¬x1,1,2 ∧ ¬x1,1,3 ∧ ¬x2,2,2 ∧ ¬x2,2,3...

x2,2,1 → ¬x2,1,2 ∧ ¬x2,1,3 ∧ ¬x1,2,2 ∧ ¬x1,2,3...

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4 No rook threatens another rookx1,1,1 → ¬x1,2,2 ∧ ¬x1,2,3 ∧ ¬x2,1,2 ∧ ¬x2,1,3x1,1,2 → ¬x1,2,1 ∧ ¬x1,2,3 ∧ ¬x2,1,1 ∧ ¬x2,1,3x1,1,3 → ¬x1,2,1 ∧ ¬x1,2,2 ∧ ¬x2,1,1 ∧ ¬x2,1,2

x2,1,1 → ¬x2,2,2 ∧ ¬x2,2,3 ∧ ¬x1,1,2 ∧ ¬x1,1,3...

x1,2,1 → ¬x1,1,2 ∧ ¬x1,1,3 ∧ ¬x2,2,2 ∧ ¬x2,2,3...

x2,2,1 → ¬x2,1,2 ∧ ¬x2,1,3 ∧ ¬x1,2,2 ∧ ¬x1,2,3...

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x2,1,1 → ¬x2,2,2 ∧ ¬x2,2,3 ∧ ¬x1,1,2 ∧ ¬x1,1,3...

x1,2,1 → ¬x1,1,2 ∧ ¬x1,1,3 ∧ ¬x2,2,2 ∧ ¬x2,2,3...

x2,2,1 → ¬x2,1,2 ∧ ¬x2,1,3 ∧ ¬x1,2,2 ∧ ¬x1,2,3...

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x2,1,1 → ¬x2,2,2 ∧ ¬x2,2,3 ∧ ¬x1,1,2 ∧ ¬x1,1,3...

x1,2,1 → ¬x1,1,2 ∧ ¬x1,1,3 ∧ ¬x2,2,2 ∧ ¬x2,2,3...

x2,2,1 → ¬x2,1,2 ∧ ¬x2,1,3 ∧ ¬x1,2,2 ∧ ¬x1,2,3...

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x2,1,1 → ¬x2,2,2 ∧ ¬x2,2,3 ∧ ¬x1,1,2 ∧ ¬x1,1,3...

x1,2,1 → ¬x1,1,2 ∧ ¬x1,1,3 ∧ ¬x2,2,2 ∧ ¬x2,2,3...

x2,2,1 → ¬x2,1,2 ∧ ¬x2,1,3 ∧ ¬x1,2,2 ∧ ¬x1,2,3...

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RIP(m,n) — Formula

1

m∧k=1

n∨i=1

n∨j=1

xi,j,k

2

n∧i=1

n∧j=1

m∧k=1

(xi,j,k →

nn∧

l=1h=1

(l,h)6=(i,j)

¬xl,h,k )

3

n∧i=1

n∧j=1

m∧k=1

(xi,j,k →m∧

l=1l 6=k

¬xi,j,l)

4

n∧i=1

n∧j=1

m∧k=1

(xi,j,k →

mn∧

l=1h=1

(l,h)6=(i,k)

¬xl,j,h ∧

mn∧

l=1h=1

(l,h)6=(j,k)

¬xi,l,h)

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RIP — CNF Formula

Your turn!

1 Figure out CNF2 Use variable enumeration as in the following slide

Or trick with some hashmap, associative array, dictionary, ...

3 Copy php.pl to rip.pl and modify accordingly

Or start with php.py if you like pythonic solutions!

4 Try it with MiniSat for some m,n

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RIP — CNF Formula

Your turn!

1 Figure out CNF

2 Use variable enumeration as in the following slide





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RIP — CNF Formula

Your turn!


Or trick with some hashmap, associative array, dictionary, ...3 Copy php.pl to rip.pl and modify accordingly



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RIP — CNF Formula

Your turn!






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RIP — CNF Formula

Your turn!





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RIP — CNF Formula

Your turn!





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RIP — CNF Formula

Your turn!





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RIP — Variable enumeration

x1,1,1 7→ 1...

xn,1,1 7→ nx1,2,1 7→ n + 1

...xn,2,1 7→ 2× n

...x1,n,1 7→ (n − 1)× n + 1

...xn,n,1 7→ n × n

x1,1,2 7→ n × n + 1...

x1,n,2 7→ n × n + (n − 1)× n + 1...

xn,n,2 7→ 2× n × n...

xn,n,m 7→ m × n × n

xi,j,k 7→ (k − 1)× n × n + (j − 1)× n + i

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Queen Independence Problem

Queen Independence Problem — QIP(m,n)

Place m queens on an n × n chessboard so that they do notthreaten each other

Reminder: Queen moves

8 0Z0Z0Z0Z7 Z0Z0Z0Z06 0Z0Z0Z0Z5 Z0Z0Z0Z04 0Z0ZqZ0Z3 Z0Z0Z0Z02 0Z0Z0Z0Z1 Z0Z0Z0Z0

1 2 3 4 5 6 7 8

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Queen Independence Problem: Diagonals (1)

4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,4 1,3 1,2

1,1

2,1 3,1 4,12,4 2,3

2,2

3,2 4,23,4

3,3

4,3

4,4

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,4 1,3

1,2 1,1

2,1 3,1 4,12,4

2,3 2,2

3,2 4,2

3,4 3,3

4,3

4,4

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,4

1,3 1,2 1,1

2,1 3,1 4,1

2,4 2,3 2,2

3,2 4,2

3,4 3,3

4,3

4,4

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,4 1,3 1,2 1,1

2,1 3,1 4,1

2,4 2,3 2,2

3,2 4,2

3,4 3,3

4,3

4,4

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,4 1,3 1,2 1,1 2,1

3,1 4,1

2,4 2,3 2,2 3,2

4,2

3,4 3,3 4,34,4

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,4 1,3 1,2 1,1 2,1 3,1

4,1

2,4 2,3 2,2 3,2 4,23,4 3,3 4,3

4,4

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,4 1,3 1,2 1,1 2,1 3,1 4,12,4 2,3 2,2 3,2 4,2

3,4 3,3 4,34,4

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,1 1,2 1,3

1,4

2,4 3,4 4,42,1 2,2

2,3

3,3 4,33,1

3,2

4,2

4,1

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,1 1,2

1,3 1,4

2,4 3,4 4,42,1

2,2 2,3

3,3 4,3

3,1 3,2

4,2

4,1

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,1

1,2 1,3 1,4

2,4 3,4 4,4

2,1 2,2 2,3

3,3 4,3

3,1 3,2

4,2

4,1

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,1 1,2 1,3 1,4

2,4 3,4 4,4

2,1 2,2 2,3

3,3 4,3

3,1 3,2

4,2

4,1

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,1 1,2 1,3 1,4 2,4

3,4 4,4

2,1 2,2 2,3 3,3

4,3

3,1 3,2 4,24,1

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,1 1,2 1,3 1,4 2,4 3,4

4,4

2,1 2,2 2,3 3,3 4,33,1 3,2 4,2

4,1

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4 0Z0Z3 Z0Z02 0Z0Z1 Z0Z0

1 2 3 4

1,1 1,2 1,3 1,4 2,4 3,4 4,42,1 2,2 2,3 3,3 4,3

3,1 3,2 4,24,1

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Homework (1)

1 Write a CNF formula generator for the BishopIndependence Problem

2 Write a CNF formula generator for the King IndependenceProblem

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Homework (2)

Latin Square

http://mathworld.wolfram.com/LatinSquare.html

1 Model the Latin Square problemStart with a few examplesThen, work out a general formulaFinally, transform the formula in CNF

2 Write a CNF generator for Latin Square3 Test your CNF with a SAT solver!

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Homework (2)

Latin Square



2 Write a CNF generator for Latin Square

3 Test your CNF with a SAT solver!

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Homework (2)

Latin Square



2 Write a CNF generator for Latin Square3 Test your CNF with a SAT solver!

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Homework (3)

1 Represent a 4× 4 Sudoku with 2× 2 squares using apropositional logic formula. This means that there is a4× 4 grid of fields, each of which should be filled withexactly one number between 1 and 4. The grid is dividedinto 4 non-overlapping regions of dimension 2× 2.

2 Generalize the formula for grids of n2 × n2 fields, into eachof which a number between 1 and n2 should be written.The grid is divided into n2 non-overlapping square regionsof dimension n × n. Here n is an arbitrary positive integer(so the previous example was a special case for n = 2).

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Outline

1 DIMACS format





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Full Adder

Standard circuit

Alternative circuit

Full Adder Equivalence Problem

Do these two circuits implement the same functionality?

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Full Adder

Standard circuit Alternative circuit



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Full Adder

Standard circuit Alternative circuit



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Homework (1)Represent the following Boolean circuit using a set of wffs:

Each component of the circuit has inputs at its left-hand side andoutput at its right-hand side. The output of a component is 1 if andonly if the sum of its inputs satisfies the condition written inside thecomponent; so if the sum of the inputs does not satisfy the condition,the output is 0. The inputs and outputs of the circuit (A, B, C, D, E, F)may have values 1 or 0. Bifurcations are indicated using a dot;crossing wires without a dot.The modelled formula must have models that correspond exactly tothe admissible input and output values of the circuit. 38 / 40

Homework (2)

1 In the previous exercise, how can you find out whether thevalue of F can ever be 1 in an admissible state of thecircuit?

2 Whether the value of E can ever be equal to the value of A?3 Whether the value of F is 1 if and only if the value of D is 0?

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END OF THELECTURE

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Documents

Propositional Logic Computer exercises€¦ · Propositional Logic Computer exercises Mario Alviano University of Calabria, Italy A.Y. 2017/2018 1/40