smtlecture.6

Satisﬁability Modulo Theories
Lecture 6 - A Theory Solver for LRA
(slides revision: Saturday 14th
March, 2015, 11:47)
Roberto Bruttomesso
Seminario di Logica Matematica
(Corso Prof. Silvio Ghilardi)
24 Novembre 2011
R. Bruttomesso (SMT) T -solver for LRA 24 Novembre 2011 1 / 21

Outline
1 Basic Solving
Introduction
Preprocessing
Solving
2 Improvement for T -solver
T -solver features
Strict inequalities
Integers

Simplex Algorithm
Invented by Tobias Dantzig around 1950
Used to solve optimization problems in linear programming
Greg Nelson was the first to employ it for constraint solving, AFAIK,
around 1980
Difference is that in linear programming input problem is feasible and
one looks for optimum. In constraint solving problem can be infeasible,
and we are interested in finding any solution

Introduction
Linear Rational Arithmetic LRA consists in solving Boolean
combinations of atoms of the form
n
j=1
ajxj ≤ b
n
j=1
ajxj ≥ b
where aj are constants (coeﬃcients), xj are variables, and b is a
constant (bound). The domain of aj, xj, b is that of rationals.

Introduction
Linear Rational Arithmetic LRA consists in solving Boolean
combinations of atoms of the form
n
j=1
ajxj ≤ b
n
j=1
ajxj ≥ b
where aj are constants (coeﬃcients), xj are variables, and b is a
constant (bound). The domain of aj, xj, b is that of rationals.
Notice that the following translations hold
n
j=1 ajxj = b =⇒ ( n
j=1 ajxj ≤ b) ∧ ( n
j=1 ajxj ≥ b)
n
j=1 ajxj < b =⇒ see later
n
j=1 ajxj > b =⇒ see later
n
j=1 ajxj = b =⇒ ( n
j=1 ajxj < b) ∨ ( n
j=1 ajxj > b)

Preprocessing
In the SMT setting, we are given a formula ϕ like
(x ≥ 0) ∧ ((x + y ≤ 2) ∨ (x + 2y − z ≥ 6)) ∧ ((x + y ≥ 2) ∨ (2y − z ≤ 4))
We perform a preprocessing step, in order to separate the formula
into a set of equations and a set of simple bounds. This is done by
introducing fresh variables.

Preprocessing
(x ≥ 0) ∧ ((x + y ≤ 2) ∨ (x + 2y − z ≥ 6)) ∧ ((x + y ≥ 2) ∨ (2y − z ≤ 4))
The formula above ϕ is equivalent to (the conjunction of)
(x ≥ 0) ∧ ((s1 ≤ 2) ∨ (s2 ≥ 6)) ∧ ((s1 ≥ 2) ∨ (s3 ≤ 4))
(s1 = x + y)
(s2 = x + 2y − z)
(s3 = 2y − z)

Preprocessing
(x ≥ 0) ∧ ((x + y ≤ 2) ∨ (x + 2y − z ≥ 6)) ∧ ((x + y ≥ 2) ∨ (2y − z ≤ 4))
The formula above ϕ is equivalent to (the conjunction of)
(x ≥ 0) ∧ ((s1 ≤ 2) ∨ (s2 ≥ 6)) ∧ ((s1 ≥ 2) ∨ (s3 ≤ 4)) ϕ
(s1 = x + y)
(s2 = x + 2y − z) Ax = 0
(s3 = 2y − z)

Preprocessing
In general, from a formula ϕ, we end up in a rewritten formula of the
kind
ϕ ∧ Ax = 0
where ϕ is a Boolean combination of bounds, while Ax = 0 is a
system of equations of the form
a11x1 + . . . + a1nxn = 0
a21x1 + . . . + a2nxn = 0
. . .
ai1x1 + . . . + ainxn = 0
. . .
am1x1 + . . . + amnxn = 0

Preprocessing
In general, from a formula ϕ, we end up in a rewritten formula of the
kind
ϕ ∧ Ax = 0
where ϕ is a Boolean combination of bounds, while Ax = 0 is a
system of equations of the form
a11x1 + . . . + a1nxn = 0
a21x1 + . . . + a2nxn = 0
. . .
ai1x1 + . . . + ainxn = 0
. . .
am1x1 + . . . + amnxn = 0
Now we detach Ax = 0 from the formula, and we store it into the
T -solver permanently. The SAT-solver will work only on ϕ . Therefore
the constraints that are pushed into and popped from the
T -solver are just bounds

The Tableau
The equations Ax = 0 are kept in a tableau, the most important structure of the Simplex
The variables are partitioned into the set of non-basic N and basic B variables
E.g., B = {x1, x3, x4}, N = {x2, x5, x6}
x1 = 4x2 + x5
x3 = 5x2 + 3x6
x4 = x5 − x6
non-basic variables can be considered as independent, while basic variables assume values
forced by the non-basic ones. E.g., in the row
x1 = 4x2 + x5
suppose that x2 = 2, x5 = 1, then we set x1 = 9

Solving
The T -solver stores
the Tableau (does not grow/shrink)
the active bounds on variables (initially none)
the current model µ (initially all 0, but could be chosen diﬀerently)
Tableau
x1 = a11xm+1 . . . + a1nxn
x2 = a21xm+1 . . . + a2nxn
. . .
xi = ai1xm+1 . . . + ainxn
. . .
xm = am1xm+1 . . . + amnxn
lb Bounds ub
−∞ ≤ x1 ≤ ∞
−∞ ≤ x2 ≤ ∞
. . .
−∞ ≤ xi ≤ ∞
. . .
−∞ ≤ xm ≤ ∞
. . .
−∞ ≤ xn ≤ ∞
µ
x1 → 0
x2 → 0
. . .
xi → 0
. . .
xm → 0
. . .
xn → 0
The T -solver is in a consistent state if the model (i) respects the tableau and (ii) satisﬁes
the bounds (for all x, lb(x) ≤ µ(x) ≤ ub(x))

Asserting a bound on a non-basic variable
Asserting a bound x ≤ c (resp. x ≥ c), x ∈ N may result in

unsatisﬁability, if c < lb(x) (resp. c > ub(x))

nothing, if c > ub(x) (resp. c < lb(x))

bound tightening, model not aﬀected if µ(x) ≤ c (resp. µ(x) ≥ c)

bound tightening, model aﬀected if µ(x) > c (resp. µ(x) < c)

The last case is “problematic” because we need to
(i) adjust µ(x): µ(x) is set to c
(ii) adjust the values of basic variables
We assume we have a function Update(x, c) that implements (i) − (ii)

Tableau
x1 = −x3 + x4
x2 = x3 + x4
lb Bounds ub
−∞ ≤ x1 ≤ ∞
−∞ ≤ x2 ≤ ∞
− ∞ ≤ x3 ≤ ∞
−∞ ≤ x4 ≤ ∞
µ
x1 → 0
x2 → 0
x3 → 0
x4 → 0
T -solver stack:

Tableau
x1 = −x3 + x4
x2 = x3 + x4
lb Bounds ub
−∞ ≤ x1 ≤ ∞
−∞ ≤ x2 ≤ ∞
− ∞ ≤ x3 ≤ − 4
−∞ ≤ x4 ≤ ∞
µ
x1 → 0
x2 → 0
x3 → 0
x4 → 0
T -solver stack:
x3 ≤ −4 (tighten ub(x3), aﬀects other values)

Tableau
x1 = −x3 + x4
x2 = x3 + x4
lb Bounds ub
−∞ ≤ x1 ≤ ∞
−∞ ≤ x2 ≤ ∞
− ∞ ≤ x3 ≤ − 4
−∞ ≤ x4 ≤ ∞
µ
x1 → 4
x2 → − 4
x3 → − 4
x4 → 0
T -solver stack:

Tableau
x1 = −x3 + x4
x2 = x3 + x4
lb Bounds ub
−∞ ≤ x1 ≤ ∞
−∞ ≤ x2 ≤ ∞
− 8 ≤ x3 ≤ − 4
−∞ ≤ x4 ≤ ∞
µ
x1 → 4
x2 → − 4
x3 → − 4
x4 → 0
T -solver stack:
x3 ≥ −8 (tighten lb(x3), does not aﬀect other values)

Tableau
x1 = −x3 + x4
x2 = x3 + x4
lb Bounds ub
−∞ ≤ x1 ≤ ∞
−∞ ≤ x2 ≤ ∞
− 8 ≤ x3 ≤ − 4
−∞ ≤ x4 ≤ ∞
µ
x1 → 4
x2 → − 4
x3 → − 4
x4 → 0
T -solver stack:
x3 ≥ −8 (tighten lb(x3), does not aﬀect other values)
x3 ≤ 0 (does not tighten ub(x3))

Asserting a bound on a basic variable
Asserting a bound x ≤ c (resp. x ≥ c), x ∈ B may result in the same 4 cases as before
however, since a basic variable is dependent from non-basic variables in the tableau, we
cannot use function Update directly. Before we need to turn x into a non-basic variables.

Asserting a bound on a basic variable
Asserting a bound x ≤ c (resp. x ≥ c), x ∈ B may result in the same 4 cases as before
however, since a basic variable is dependent from non-basic variables in the tableau, we
cannot use function Update directly. Before we need to turn x into a non-basic variables.
So if µ(x) > c or µ(x) < c we have to
(i) turn x into non-basic (another non-basic variable will become basic instead)
(ii) adjust µ(x): µ(x) is set to c
(iii) adjust the values of basic variables
Step (i) is performed by a function Pivot(x, y) (see next slide). Therefore asserting a bound
on a basic variable x consists in executing Pivot(x, y) and then Update(x, c)

Pivoting
Pivot(x, y) is the operation of swapping a basic variable x with a non-basic variable y (how
to choose y ? See next slide)
It consists of the following steps:
1 Take the row x = ay + R in the tableau (R = rest of the polynome)
2 Rewrite it as y = x−R
a
3 Substitute y with x−R
a
in all the other rows of the tableau (and simplify)

Pivoting
a
a
Example for Pivot(x1, x3)
Step 1
x1 = 3x2 + 4x3 − 5x4
x5 = −x2 − x3
x6 = 10x2 + 5x4
Step 2
x3 = −3
4
x2 + 1
4
x1 + 5
4
x4
x5 = −x2 − x3
x6 = 10x2 + 5x4
Step 3
x3 = −3
4
x2 + 1
4
x1 + 5
4
x4
x5 = −1
4
x2 − 1
4
x1 − 5
4
x4
x6 = 10x2 + 5x4

Pivoting
a
a
Example for Pivot(x1, x3)
Step 1
x1 = 3x2 + 4x3 − 5x4
x5 = −x2 − x3
x6 = 10x2 + 5x4
Step 2
x3 = −3
4
x2 + 1
4
x1 + 5
4
x4
x5 = −x2 − x3
x6 = 10x2 + 5x4
Step 3
x3 = −3
4
x2 + 1
4
x1 + 5
4
x4
x5 = −1
4
x2 − 1
4
x1 − 5
4
x4
x6 = 10x2 + 5x4
we moved from B = {x1, x5, x6} to B = {x3, x5, x6}

Choosing Pivoting Variable
Consider the following situation
Tableau
. . .
x1 = 3x2 − 4x3 + 2x4 − x5
. . .
lb Bounds ub
−4 ≤ x1 ≤ 10
1 ≤ x2 ≤ 3
−4 ≤ x3 ≤ −1
1 ≤ x4 ≤ 2
−1 ≤ x5 ≤ 10
µ
x1 → 12
x2 → 1
x3 → −1
x4 → 2
x5 → −1
which among N = {x2, x3, x4} do I choose for pivoting ? Clearly, the value of µ(x1) is too
high, I have to decrease it by playing with the values of N:

Tableau
. . .
x1 = 3x2 − 4x3 + 2x4 − x5
. . .
lb Bounds ub
−4 ≤ x1 ≤ 10
1 ≤ x2 ≤ 3
−4 ≤ x3 ≤ −1
1 ≤ x4 ≤ 2
−1 ≤ x5 ≤ 10
µ
x1 → 12
x2 → 1
x3 → −1
x4 → 2
x5 → −1
3x2 cannot decrease, as µ(x2) = lb(x2) and cannot be moved down

Tableau
. . .
x1 = 3x2 − 4x3 + 2x4 − x5
. . .
lb Bounds ub
−4 ≤ x1 ≤ 10
1 ≤ x2 ≤ 3
−4 ≤ x3 ≤ −1
1 ≤ x4 ≤ 2
−1 ≤ x5 ≤ 10
µ
x1 → 12
x2 → 1
x3 → −1
x4 → 2
x5 → −1
−4x3 cannot decrease, as µ(x3) = ub(x3) and cannot be moved up

Tableau
. . .
x1 = 3x2 − 4x3 + 2x4 − x5
. . .
lb Bounds ub
−4 ≤ x1 ≤ 10
1 ≤ x2 ≤ 3
−4 ≤ x3 ≤ −1
1 ≤ x4 ≤ 2
−1 ≤ x5 ≤ 10
µ
x1 → 12
x2 → 1
x3 → −1
x4 → 2
x5 → −1
2x4 can decrease, as µ(x4) = ub(x4), and can be moved down

Tableau
. . .
x1 = 3x2 − 4x3 + 2x4 − x5
. . .
lb Bounds ub
−4 ≤ x1 ≤ 10
1 ≤ x2 ≤ 3
−4 ≤ x3 ≤ −1
1 ≤ x4 ≤ 2
−1 ≤ x5 ≤ 10
µ
x1 → 12
x2 → 1
x3 → −1
x4 → 2
x5 → −1
−x5 can decrease, as µ(x5) = lb(x5), and can be moved up

Tableau
. . .
x1 = 3x2 − 4x3 + 2x4 − x5
. . .
lb Bounds ub
−4 ≤ x1 ≤ 10
1 ≤ x2 ≤ 3
−4 ≤ x3 ≤ −1
1 ≤ x4 ≤ 2
−1 ≤ x5 ≤ 10
µ
x1 → 12
x2 → 1
x3 → −1
x4 → 2
x5 → −1
−x5 can decrease, as µ(x5) = lb(x5), and can be moved up
both x4 and x5 are therefore good candidates for pivoting. To avoid loops, choose variable
with smallest subscript (Bland’s Rule). This rule is not necessarily eﬃcient, though

Detect Unsatisﬁability
There might be cases in which no suitable variable for pivoting can be found. This
indicates unsatisﬁability.

Detect Unsatisfiability
There might be cases in which no suitable variable for pivoting can be found. This
indicates unsatisfiability. Consider the following where we have just asserted x1 ≤ 9
Tableau
. . .
x1 = 3x2 − 4x3 + 2x4 − x5
. . .
lb Bounds ub
−4 ≤ x1 ≤ 9
1 ≤ x2 ≤ 3
−4 ≤ x3 ≤ −1
2 ≤ x4 ≤ 2
−1 ≤ x5 ≤ −1
µ
x1 → 12
x2 → 1
x3 → −1
x4 → 2
x5 → −1
no variable among N = {x2, x3, x4} can be chosen for pivoting. This is because (due to
tableau)
x2 ≥ 3 ∧ x3 ≤ −1 ∧ x4 ≥ 4 ∧ x5 ≤ −1 ⇒ x1 ≥ 12 ⇒ ¬(x1 ≤ 9)
Therefore
{x2 ≥ 3, x3 ≤ −1, x4 ≥ 4, x5 ≤ −1, ¬(x1 ≤ 9)}
is a T -conflict (modulo the tableau)

Solving
1 while( true )
2 pick first xi ∈ B such that µ(xi) < lb(xi) or µ(xi) > ub(xi)
3 if ( there is no such xi ) return sat
4 xj = ChoosePivot(xi)
5 if ( xj == undef ) return unsat
6 Pivot(xi, xj)
7 if ( µ(xi) < lb(xi) )
8 Update(xi, lb(xi))
9 if ( µ(xi) > ub(xi) )
10 Update(xi, ub(xi))
11 end
xj = ChoosePivot(xi): returns variable to use for pivoting with xi, or undef if conflict
pick first xi . . . : it’s again Bland’s rule

An Example
Consider the following problem
x1 − x2 ≤ 8 ∧ x2 − x3 ≤ −1 ∧ x3 − x4 ≤ 2 ∧ x4 − x1 ≤ −10

An Example
x5 ≤ 8 ∧ x6 ≤ −1 ∧ x7 ≤ 2 ∧ x8 ≤ −10
Tableau
x5 = x1 − x2
x6 = x2 − x3
x7 = x3 − x4
x8 = x4 − x1
lb Bounds ub
−∞ ≤ x5 ≤ ∞
−∞ ≤ x6 ≤ ∞
−∞ ≤ x7 ≤ ∞
−∞ ≤ x8 ≤ ∞
µ
x1 → 0
x2 → 0
x3 → 0
x4 → 0
x5 → 0
x6 → 0
x7 → 0
x8 → 0

An Example
x5 ≤ 8 ∧ x6 ≤ −1 ∧ x7 ≤ 2 ∧ x8 ≤ −10
Tableau
x5 = x1 − x2
x6 = x2 − x3
x7 = x3 − x4
x8 = x4 − x1
lb Bounds ub
−∞ ≤ x5 ≤ 8
−∞ ≤ x6 ≤ ∞
−∞ ≤ x7 ≤ ∞
−∞ ≤ x8 ≤ ∞
µ
x1 → 0
x2 → 0
x3 → 0
x4 → 0
x5 → 0
x6 → 0
x7 → 0
x8 → 0
Receiving x5 ≤ 8: OK

An Example
x5 ≤ 8 ∧ x6 ≤ −1 ∧ x7 ≤ 2 ∧ x8 ≤ −10
Tableau
x5 = x1 − x2
x6 = x2 − x3
x7 = x3 − x4
x8 = x4 − x1
lb Bounds ub
−∞ ≤ x5 ≤ 8
−∞ ≤ x6 ≤ − 1
−∞ ≤ x7 ≤ ∞
−∞ ≤ x8 ≤ ∞
µ
x1 → 0
x2 → 0
x3 → 0
x4 → 0
x5 → 0
x6 → 0
x7 → 0
x8 → 0
Receiving x6 ≤ −1: µ(x6) out of bounds

An Example
x5 ≤ 8 ∧ x6 ≤ −1 ∧ x7 ≤ 2 ∧ x8 ≤ −10
Tableau
x5 = x1 − x2
x3 = x2 − x6
x7 = x2 − x6 − x4
x8 = x4 − x1
lb Bounds ub
−∞ ≤ x5 ≤ 8
−∞ ≤ x6 ≤ − 1
−∞ ≤ x7 ≤ ∞
−∞ ≤ x8 ≤ ∞
µ
x1 → 0
x2 → 0
x3 → 1
x4 → 0
x5 → 0
x6 → − 1
x7 → 1
x8 → 0
Pivot(x6, x3), Update(x6, −1). Values of x3 and x7 change as well

An Example
x5 ≤ 8 ∧ x6 ≤ −1 ∧ x7 ≤ 2 ∧ x8 ≤ −10
Tableau
x5 = x1 − x2
x3 = x2 − x6
x7 = x2 − x6 − x4
x8 = x4 − x1
lb Bounds ub
−∞ ≤ x5 ≤ 8
−∞ ≤ x6 ≤ − 1
−∞ ≤ x7 ≤ 2
−∞ ≤ x8 ≤ ∞
µ
x1 → 0
x2 → 0
x3 → 1
x4 → 0
x5 → 0
x6 → − 1
x7 → 1
x8 → 0
Receiving x7 ≤ 2: OK

An Example
x5 ≤ 8 ∧ x6 ≤ −1 ∧ x7 ≤ 2 ∧ x8 ≤ −10
Tableau
x5 = x1 − x2
x3 = x2 − x6
x7 = x2 − x6 − x4
x8 = x4 − x1
lb Bounds ub
−∞ ≤ x5 ≤ 8
−∞ ≤ x6 ≤ − 1
−∞ ≤ x7 ≤ 2
−∞ ≤ x8 ≤ − 10
µ
x1 → 0
x2 → 0
x3 → 1
x4 → 0
x5 → 0
x6 → − 1
x7 → 1
x8 → 0
Receiving x8 ≤ −10: µ(x8) out of bounds

An Example
x5 ≤ 8 ∧ x6 ≤ −1 ∧ x7 ≤ 2 ∧ x8 ≤ −10
Tableau
x5 = x4 − x8 − x2
x3 = x2 − x6
x7 = x2 − x6 − x4
x1 = x4 − x8
lb Bounds ub
−∞ ≤ x5 ≤ 8
−∞ ≤ x6 ≤ − 1
−∞ ≤ x7 ≤ 2
−∞ ≤ x8 ≤ − 10
µ
x1 → 10
x2 → 0
x3 → 1
x4 → 0
x5 → 10
x6 → − 1
x7 → 1
x8 → − 10
Pivot(x8, x1), Update(x8, −10). Values of x5 and x1 change as well. µ(x5) out of bounds

An Example
x5 ≤ 8 ∧ x6 ≤ −1 ∧ x7 ≤ 2 ∧ x8 ≤ −10
Tableau
x2 = x4 − x8 − x5
x3 = x4 − x8 − x5 − x6
x7 = − x5 − x8 − x6
x1 = x4 − x8
lb Bounds ub
−∞ ≤ x5 ≤ 8
−∞ ≤ x6 ≤ − 1
−∞ ≤ x7 ≤ 2
−∞ ≤ x8 ≤ − 10
µ
x1 → 10
x2 → 2
x3 → 1
x4 → 0
x5 → 8
x6 → − 1
x7 → 3
x8 → − 10
Pivot(x5, x2), Update(x5, 8). Values of x2 and x7 change as well. µ(x7) out of bounds

An Example
x5 ≤ 8 ∧ x6 ≤ −1 ∧ x7 ≤ 2 ∧ x8 ≤ −10
Tableau
x2 = x4 − x8 − x5
x3 = x4 − x8 − x5 − x6
x7 = − x5 − x8 − x6
x1 = x4 − x8
lb Bounds ub
−∞ ≤ x5 ≤ 8
−∞ ≤ x6 ≤ − 1
−∞ ≤ x7 ≤ 2
−∞ ≤ x8 ≤ − 10
µ
x1 → 10
x2 → 2
x3 → 1
x4 → 0
x5 → 8
x6 → − 1
x7 → 3
x8 → − 10
Cannot ﬁnd suitable variable for pivoting (ChoosePivot(x7) == undef). Return unsat

Outline
1 Basic Solving
Introduction
Preprocessing
Solving
2 Improvement for T -solver
T -solver features
Strict inequalities
Integers

T -solver features
Incrementality: it comes for free, as we keep µ updated

T -solver features
Backtrackability: use the same trick as for BF: update a model µ ,
if satisﬁable set µ = µ , otherwise forget µ . When backtracking
don’t change µ

T -solver features
don’t change µ
Minimal T -conﬂicts: seen already

T -solver features
don’t change µ
Theory Propagation:
Bound propagation (cheap): if x ≤ c has been asserted, then all
other inactive x ≤ c with c ≤ c are implied (similar for ≥)

T -solver features
don’t change µ
Theory Propagation:
Bound propagation (cheap): if x ≤ c has been asserted, then all
other inactive x ≤ c with c ≤ c are implied (similar for ≥)
Tableau+Bound propagation: use a row x1 = a2x2 + a3x3 + . . . and
bounds on x2, x3, . . . to derive bounds on x1 (similar idea as to ﬁnd
conﬂicts)

Strict inequalities
So far we have used bounds like x ≤ 3, but never strict ones like x < 3

Strict inequalities
Since we are on the rationals x < 3 is same as x ≤ (3 − δ), where δ is a
symbolic positive parameter (δ > 0)

Strict inequalities
We do the following: instead of using Q numbers of form n, we use Qδ
numbers of form (n, k) to be intended as n + kδ

Strict inequalities
Addition in Qδ: (n1, k1) + (n2, k2) = (n1 + n2, k1 + k2)

Strict inequalities
Scalar multiplication in Qδ: c(n, k) = (cn, ck)

Strict inequalities
Comparison in Qδ: (n1, k1) ≤ (n2, k2) iﬀ (n1 < n2) or (n1 = n2 and
k1 ≤ k2)

Strict inequalities
Comparison in Qδ: (n1, k1) ≤ (n2, k2) iﬀ (n1 < n2) or (n1 = n2 and
k1 ≤ k2)
If constraints are satisﬁable, it is always possible to compute a rational
value for δ to translate Qδ numbers into Q numbers

Strict inequalities
Example: situation after receiving
x3 < −2, x4 > 0
and after Update(x3, (−2, −1)), Update(x4, (0, 1))
Tableau
x1 = −x3 + x4
x2 = x3 + x4
lb Bounds ub
−∞ ≤ x1 ≤ ∞
−∞ ≤ x2 ≤ ∞
−∞ ≤ x3 ≤ (−2, −1)
(0, 1) ≤ x4 ≤ ∞
µ
x1 → (2, 2)
x2 → (−2, 0)
x3 → (−2, −1)
x4 → (0, 1)
x1 = −(−2, −1) + (0, 1) = (2, 1) + (0, 1) = (2, 2)
x2 = (−2, −1) + (0, 1) = (−2, 0)

Solving LIA with LRA
The Simplex can be used also to reason (in a complete way) about the
integers (LIA), e.g., using known techniques in linear programming
Given a set of LIA constraints S
If S is unsatisfiable on LRA1 then it is also unsatisfiable on LIA
If S is satisfiable on LRA, then we have to check if there is an
integer solution
1
This means that we allow variables to assume values in Q instead of Z.

integer solution
For the latter case the convex polytope on Q is explored sistematically.
However, in general, search is necessary: LIA is NP-Complete, like
SAT
in SAT we split a and ¬a, in LIA we split x ≤ c and x ≥ c + 1
in SAT we learn clauses, in LIA we learn new tableau rows (new
constraints)
1

integer solution
For the latter case the convex polytope on Q is explored sistematically.
However, in general, search is necessary: LIA is NP-Complete, like
SAT
in SAT we split a and ¬a, in LIA we split x ≤ c and x ≥ c + 1
in SAT we learn clauses, in LIA we learn new tableau rows (new
constraints)
When on the integers, δ is set to 1 (Qδ numbers are not used)
1

Exercizes
1 Show that the T -conﬂicts generated by the Simplex are minimal
2 Suppose that (0, 1) ≤ x ≤ (1, −1). Compute the biggest possible
value for δ
3 Find the candidate for pivoting in this row (for simplicity we use
normal numbers)
x1 = 3x2 − 9x3 − 7x4
given these bounds −∞ ≤ x1 ≤ 1, 0 ≤ x2 ≤ ∞, −∞ ≤ x3 ≤ −1,
1 ≤ x4 ≤ 2 and this assignment
µ = {x1 → 2, x2 → 0, x3 → −1, x4 → 1}
4 Using the row of previous exercize, change the bounds so that the
only candidate for pivoting becomes x2
5 Now change the bounds so that no pivoting is possible. Compute
the conﬂict

smtlecture.6

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