hardness of multiple choice problems

Problem Statement and MotivationA New Satisfiability Result

Rainbow Minmax GapExact Coverage of Color Classes

......Hardness of Multiple Choice Problems

Aritra Banik1

Indian Institute of Technology Jodhpur

NMI Workshop on Complexity Theory at IIT Gandhinagar

1Joint work with Esther M. Arkin, Paz Carmi, Gui Citovsky, Matthew J.Katz, Joseph S. B. Mitchell, Marina Simakov

Aritra Banik Hardness of Multiple Choice Problems 1/31

.. Outline

...1 Problem Statement and Motivation

...2 A New Satisfiability Result

...3 Rainbow Minmax Gap

...4 Exact Coverage of Color Classes

.. Multiple choice problem

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A multiple choice problem consists of a set of color classesP = {C1,C2, . . . ,Cn}, where each color class Ci consists of apair of points.

Given a set P of color classes, the objective is to choose onepoint from each color class such that certain optimalitycriteria are satisfied.

We define a Rainbow to be a set of items containing one itemfrom each color class.

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MAX GAP

We define MAXGAP of a Rainbow Q to be equals to themaximum distance between a pair of consecutive points in Q.

Rainbow minmax gap (decision version): Given a set P ofn point color classes and a value d > 0, determine whetherthere exists a rainbow Q of size n with max gap at most d .

In a recent paper Consuegra and Narasimhan, present a2-approximation algorithm for rainbow minmax gap, but leavethe question whether the problem is NP-hard or not, open.

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MAX GAP

We define MAXGAP of a Rainbow Q to be equals to themaximum distance between a pair of consecutive points in Q.

Rainbow minmax gap (decision version): Given a set P ofn point color classes and a value d > 0, determine whetherthere exists a rainbow Q of size n with max gap at most d .

In a recent paper Consuegra and Narasimhan, present a2-approximation algorithm for rainbow minmax gap, but leavethe question whether the problem is NP-hard or not, open.

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Rainbow piercing: Given P and a set of intervals I,determine whether there exists a rainbow Q that is a piercingset for I.

Rainbow covering: Given a set P of interval color classes onthe x-axis and a set of points S on the x-axis, determinewhether there exists a rainbow Q that covers S (i.e., eachpoint in S is covered by at least one interval in Q).

Rainbow piercing: Given P and a set of intervals I,determine whether there exists a rainbow Q that is a piercingset for I.Rainbow covering: Given a set P of interval color classes onthe x-axis and a set of points S on the x-axis, determinewhether there exists a rainbow Q that covers S (i.e., eachpoint in S is covered by at least one interval in Q).

.. A New Satisfiability Result

We show the hardness of rainbow problems by showinghardness of a new and simple SAT instance which we callLSAT.

A SAT formula is an LSAT formula if each clause (viewed as aset of literals) intersects at most one other clause, and,moreover, if two clauses intersect, then they have exactly oneliteral in common.

We show the hardness of rainbow problems by showinghardness of a new and simple SAT instance which we callLSAT.

(z1 ∨ z2 ∨ z3)

z1 z2 z3 z4 z5z1 z2 z3z4 z5 z6 z7z6 z7

(z1 ∨ z4)(z4 ∨ z2) (z4 ∨ z5 ∨ z3) (z6 ∨ z7 ∨ z5) (z5 ∨ z6 ∨ z7)

Clause ClauseLiteral

A SAT formula is an LSAT formula if each clause (viewed as aset of literals) intersects at most one other clause, and,moreover, if two clauses intersect, then they have exactly oneliteral in common.

(z1 ∨ z2 ∨ z3)

z1 z2 z3 z4 z5z1 z2 z3z4 z5 z6 z7z6 z7

(z1 ∨ z4)(z4 ∨ z2) (z4 ∨ z5 ∨ z3) (z6 ∨ z7 ∨ z5) (z5 ∨ z6 ∨ z7)

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Let F be an LSAT formula and let T be its corresponding setof literals.

One can order the literals in T , such that

(i) each clause of F corresponds to at most three consecutiveliterals in the sorted list.

(ii) each clause may share at most one of its literals with anotherclause, in which case this literal is extreme in both clauses.

(z1 ∨ z2 ∨ z3)

z1 z2 z3 z4 z5z1 z2 z3z4 z5 z6 z7z6 z7

(z1 ∨ z4)(z4 ∨ z2) (z4 ∨ z5 ∨ z3) (z6 ∨ z7 ∨ z5) (z5 ∨ z6 ∨ z7)

C1 C2 C3 C4 C5 C6

(z1 ∨ z2 ∨ z3)

z1 z2 z3 z4 z5z1 z2 z3z4 z5 z6 z7z6 z7

(z1 ∨ z4)(z4 ∨ z2) (z4 ∨ z5 ∨ z3) (z6 ∨ z7 ∨ z5) (z5 ∨ z6 ∨ z7)

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.. If the clauses are disjoint

Consider a 3 SAT formula where the clauses are pairwisedisjoint, that is no two clause share a literal.(x1 ∨ x3 ∨ x4) ∧ (x2 ∨ x3 ∨ x4) ∧ (x1 ∨ x5 ∨ x6) ∧ (x2 ∨ x5 ∨ x6)

˜˜˜̃

This can be solved in polynomial time.

˜˜˜̃

This can be solved in polynomial time.

˜˜˜̃

This can be solved in polynomial time.Aritra Banik Hardness of Multiple Choice Problems 8/31

.. Outline

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

In the boolean satisfiability problem (SAT), one is given aformula in conjunctive normal form and the goal is todetermine whether it is satisfiable or not.

Consider a CNF formula F = (X , C) where each clausecontains three literals and each literal can appear at mosttwice.

Our aim is to construct a LSAT formula FL = (XL, CL) fromF , and show that there is a truth assignment for X such thateach clause in C is satisfied if and only if there is a truthassignment for XL such that each clause in CL is satisfied.

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

For each variable xi we add the following clauses to CL

(yi1 ∨ xi )(zi1 ∨ xi )

(xi ∨ yi2)(xi ∨ zi2)

yi1 xi yi2 zi1 xi zi2

Now, for each clause Ci ∈ C, we add a clause to CL as follows.

For each variable xi that appears in Ci , if xi appearsunnegated,

we replace it by yi1 or yi2

If xi appears negated,we replace it by zi1 or zi2

For each variable xi we add the following clauses to CL(yi1 ∨ xi )(zi1 ∨ xi )

We change the following clause(x1 ∨ x2 ∨ x3) ∧ (x2 ∨ x3 ∨ x4) ∧ (x3 ∨ x1 ∨ x4) ∧ (x1 ∨ x3 ∨ x4)

(y11∨y21∨y31)∧(y22∨z31∨z41)∧(y32∨z11∨y41)∧(y12∨z32∨z42)

.Theorem..

......F is satisfiable if and only if FL is satisfiable.

(y11∨y21∨y31)∧(y22∨z31∨z41)∧(y32∨z11∨y41)∧(y12∨z32∨z42)

.Theorem..

(y11∨y21∨y31)∧(y22∨z31∨z41)∧(y32∨z11∨y41)∧(y12∨z32∨z42)

.Theorem..

.Claim........F is satisfiable ⇒ FL is satisfiable

Assume F is satisfiable, that is, there exists a truthassignment for X such that each clause in C is satisfied.

If xi ∈ X was assigned FALSE (i.e., 0)

01 1 10 0

If xi ∈ X was assigned TRUE (i.e., 1)

0 111 00

01 1 10 0

0 111 00

xi ← 0⇒ yij ← 0 and zij ← 1

We have replaced xi by yij and xi by zij

Hence if F is satisfiable, then so is FL

01 1 10 0

0 111 00

01 1 10 0

0 111 00

01 1 10 0

0 111 00

.Claim........FL is satisfiable ⇒ F is satisfiable

Suppose FL is satisfiable, we will show that F is alsosatisfiable.

Consider any truth assignment for XL that satisfies FL. Thistruth assignment (restricted to X ) also satisfies F .

01 1 10 0

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01 1 10 0

0 111 00

01 1 10 0

0 111 00

01 1 10 0

0 111 00

Suppose the clause (x2 ∨ x3 ∨ x4) is not satisfied.

But in FL the clause (y22 ∨ z31 ∨ z41) is satisfied.

y22 = 1 =⇒ x2 = 1, z31 = 1 =⇒ x3 = 1 andz41 = 1 =⇒ x4 = 1.

Hence if FL is satisfiable, then so is F

01 1 10 0

0 111 00

y22 = 1 =⇒ x2 = 1, z31 = 1 =⇒ x3 = 1 andz41 = 1 =⇒ x4 = 1.

01 1 10 0

0 111 00

y22 = 1 =⇒ x2 = 1, z31 = 1 =⇒ x3 = 1 andz41 = 1 =⇒ x4 = 1.

01 1 10 0

0 111 00

y22 = 1 =⇒ x2 = 1, z31 = 1 =⇒ x3 = 1 andz41 = 1 =⇒ x4 = 1.

.. Outline

.. Rainbow minmax gap

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MAX GAP

P be a set of n color classes, where each color class Ci is apair of points {pi , pi} on the x-axis, and let d > 0.

We prove that the decision version of rainbow minmax gap isNP-complete, that is, it is NP-complete to determine whetherthere exists a rainbow Q ⊂ ∪ni=1Ci of size n, such that themaximum gap between a pair of consecutive points in Q is atmost d .

We present a reduction from LSAT.

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MAX GAP

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MAX GAP

(x1 ∨ x2 ∨ x3)

x1 x2 x3 x4 x5x1 x2 x3x4 x5 x6 x7x6 x7

(x1 ∨ x4)(x4 ∨ x2) (x4 ∨ x5 ∨ x3) (x6 ∨ x7 ∨ x5) (x5 ∨ x6 ∨ x7)

Given a LSAT formula we create the color classes as follows.

Place the points q1, q2, . . . , qk on the x-axis, from left toright, between any two disjoint clause set.

We place the points q1, . . . , qk at the end such that distancebetween any two consecutive points, including the distancebetween qk+l+1 and q1, is d + ϵ, for some ϵ > 0.

q1 q2 q3 q4 q1 q2 q3 q4

d+ ε d+ ε d+ ε d+ ε

(x1 ∨ x2 ∨ x3)

x1 x2 x3 x4 x5x1 x2 x3x4 x5 x6 x7x6 x7

(x1 ∨ x4)(x4 ∨ x2) (x4 ∨ x5 ∨ x3) (x6 ∨ x7 ∨ x5) (x5 ∨ x6 ∨ x7)

q1 q2 q3 q4 q1 q2 q3 q4

(x1 ∨ x2 ∨ x3)

x1 x2 x3 x4 x5x1 x2 x3x4 x5 x6 x7x6 x7

(x1 ∨ x4)(x4 ∨ x2) (x4 ∨ x5 ∨ x3) (x6 ∨ x7 ∨ x5) (x5 ∨ x6 ∨ x7)

q1 q2 q3 q4 q1 q2 q3 q4

(x1 ∨ x2 ∨ x3)

x1 x2 x3 x4 x5x1 x2 x3x4 x5 x6 x7x6 x7

(x1 ∨ x4)(x4 ∨ x2) (x4 ∨ x5 ∨ x3) (x6 ∨ x7 ∨ x5) (x5 ∨ x6 ∨ x7)

q4 q5x5 x6 x7x6 x7

(x6 ∨ x7 ∨ x5) ∨ (x5 ∨ x6 ∨ x7)

Our aim is to choose the distance between qi ,pa, pb, pc , pd , pe and qi+1 so that

If (xa ∨ xb ∨ xc) and (xc ∨ xd ∨ xe) is satisfied then distancebetween any two consecutive point is at most d .If the distance between any two consecutive point is at most dthen both (xa ∨ xb ∨ xc) and (xc ∨ xd ∨ xe) is satisfied.

We choose the distances as follows.

We assign distances for other cases similarly.

q4 q5x5 x6 x7x6 x7

(x6 ∨ x7 ∨ x5) ∨ (x5 ∨ x6 ∨ x7)

If (xa ∨ xb ∨ xc) and (xc ∨ xd ∨ xe) is satisfied then distancebetween any two consecutive point is at most d .

If the distance between any two consecutive point is at most dthen both (xa ∨ xb ∨ xc) and (xc ∨ xd ∨ xe) is satisfied.

q4 q5x5 x6 x7x6 x7

(x6 ∨ x7 ∨ x5) ∨ (x5 ∨ x6 ∨ x7)

d/2 d/4 d/4 d/4 d/4 d/2

q4 q5x5 x6 x7x6 x7

(x6 ∨ x7 ∨ x5) ∨ (x5 ∨ x6 ∨ x7)

qi xaxbxc qi+1

d/2 d/4 d/2

(xa ∨ xb ∨ xc)

qi xa xb xc qi+1

(xa ∨ xb) ∧ (xb ∨ xc)

d/2 d/2 d/2 d/2

Hence we have the following result.

.Theorem........The decision version of rainbow minmax gap is NP-complete.

We also proved the following results.

.Theorem........Rainbow piercing is NP-complete.

.Theorem........Rainbow covering is NP-complete.

Hence we have the following result.

.Theorem........The decision version of rainbow minmax gap is NP-complete.

We also proved the following results.

.Theorem........Rainbow piercing is NP-complete.

.Theorem........Rainbow covering is NP-complete.

.. Outline

.. Covering Exactly One

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Given set of color classes P = {C1,C2, . . . ,Cn}, where eachcolor class Ci consists of a pair of points.

Find a minimum-cardinality set I of intervals, such thatexactly one point from each color class is covered by aninterval in I.We showed the following problems to be NP-hard.

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Find a minimum-cardinality set I of intervals, such thatexactly one point from each color class is covered by aninterval in I.

We showed the following problems to be NP-hard.

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Find a minimum-cardinality set I of intervals, such thatexactly one point from each color class is covered by aninterval in I.We showed the following problems to be NP-hard.

.. Exact Coverage of Color Classes

Covering color classes with unit length intervals: Decidewhether or not there exists a set of unit length intervals, I,such that exactly one point from each color class is covered byan interval in I.

Covering color classes with the fewest unit length intervals:Find a minimum-cardinality set I of unit length intervals(assuming a feasible solution exists), such that exactly onepoint from each color class is covered by an interval in I.Covering color classes with intervals of arbitrary length: Finda minimum-cardinality set I of intervals of arbitrary length,such that exactly one point from each color class is covered byan interval in I.

Covering color classes with unit length intervals: Decidewhether or not there exists a set of unit length intervals, I,such that exactly one point from each color class is covered byan interval in I.Covering color classes with the fewest unit length intervals:Find a minimum-cardinality set I of unit length intervals(assuming a feasible solution exists), such that exactly onepoint from each color class is covered by an interval in I.

Covering color classes with intervals of arbitrary length: Finda minimum-cardinality set I of intervals of arbitrary length,such that exactly one point from each color class is covered byan interval in I.

Covering color classes with unit length intervals: Decidewhether or not there exists a set of unit length intervals, I,such that exactly one point from each color class is covered byan interval in I.Covering color classes with the fewest unit length intervals:Find a minimum-cardinality set I of unit length intervals(assuming a feasible solution exists), such that exactly onepoint from each color class is covered by an interval in I.Covering color classes with intervals of arbitrary length: Finda minimum-cardinality set I of intervals of arbitrary length,such that exactly one point from each color class is covered byan interval in I.

.. Conflict Free Covering

We call an interval on the x-axis that contains at most onepoint from each color class a conflict-free interval (orCF-interval for short).

Covering color classes with a set of given CF-intervals:Given a set P of point color classes on a line where each colorclass consists of a pair and a set of CF-intervals I, find aminimum-cardinality set I ′ ⊂ I of CF-intervals, such that atleast one point from each color class is covered by an intervalin I ′.

.. Conflict Free Covering

We call an interval on the x-axis that contains at most onepoint from each color class a conflict-free interval (orCF-interval for short).

Covering color classes with a set of given CF-intervals:Given a set P of point color classes on a line where each colorclass consists of a pair and a set of CF-intervals I, find aminimum-cardinality set I ′ ⊂ I of CF-intervals, such that atleast one point from each color class is covered by an intervalin I ′.

.. Covering color classes with a given set of CF-intervals

We present an 4-approximation for this problem.

From the set of given intervals I we create the following setof intervals J .By construction this new set of intervals are disjoint and henceeach color class can be contained in at most two intervals.

We can conver this to a vertex cover problem.

From the set of given intervals I we create the following setof intervals J .

By construction this new set of intervals are disjoint and henceeach color class can be contained in at most two intervals.

Each interval is a vertex and each color class represent anedge.

We have a 2-approximation for vertex cover.

Hence we have a 2-approximation when the intervals aredisjoint.

When the intervals are not disjoint.

OPT (J ) < 2OPT (I) =⇒ OURSOL < 4OPT (I).

.. Conclusion

Concluding remarks:

We also proved some variant of problem to be NP hard in theplane.

We also proved the problem to be NP hard when one needs tochoose at least one point from each color class.

We also have some approximation results.

Some open problems:

For exact cover finding a ”good” approximation.

More points in a color class.

Some other ”geometric object”??

.. Conclusion

Concluding remarks:

Some open problems:

.. Conclusion

Concluding remarks:

Some open problems:

.. Conclusion

Concluding remarks:

Some open problems:

.. Conclusion

Concluding remarks:

Some open problems:

.. Conclusion

Concluding remarks:

Some open problems:

Thank You

hardness of multiple choice problems

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