1. How Hard Can It Be?

2. Tractable versus Intractability
Given an algorithm A that solves a problem.
If the worst case time efficiency of A is O(p(n)), for some polynomial p, then A is said to solve the problem in polynomial time.
If a problem can be solved in polynomial time it is said to be tractable.
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3. P
A decision problem is a problem with one of two answers, e.g., yes/no, 1/0, or true/false.
P = {decision problems solvable by a deterministic algorithm in O(p(n)) time} = {tractable decision problems}
What problem would you be solving if it had a best algorithm with the largest degree polynomial you have experienced?
How “tractable” is a problem if the degree of p(n) is large?
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4. Decidability Are all decision problems solvable (decidable)?
A possibly undecidable problem: Does the Collatz sequence, a.k.a., the 3 n + 1 sequence always have finite length? Does the sequence, C(t) = 3*C(t-1)+1, if C(t-1) is odd, and C(t) = C(t-1)/2, if C(t-1) is even, C(t0)an integer such that C(t0)>1, always reach C(t) = 1 in a finite number of steps?
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5. Existence of Functions that are Not Turing Computable
Recall from the theory of computation that:
Computability of a function means one can find a Turing machine that computes the function
The set Turing machines is countable; thus, they can be put in 1-1 correspondence with the counting numbers, N¥
Consider the collection of functions f:N→N
How many such functions are there? 2|N|>>|N|
Thus, there are functions not Turing computable
¥ See, e.g., Dennis, Denning and Qualitz, Machines, Languages, and Computation, Prentice-Hall, 1978, pp. 497.
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6. An Example: The Halting Problem
The halting problem: Find an algorithm A such that:
Given any program P and any input I for P,
The algorithm decides whether P will finish processing with I in finite time
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7. A Proof By Contradiction that the Halting Problem is Undecidable
Let P be any program, I any input for P, and suppose A is an algorithm that can decide if P halts on I: 𝐴(𝑃, 𝐼)= 1, 𝑖𝑓 𝑃 ℎ𝑎𝑙𝑡𝑠 𝑜𝑛 𝐼 , 𝑖𝑓 𝑃 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑙𝑡 𝑜𝑛 𝐼
Let A operate on P with P as input in order to define a program Q: Q P = ℎ𝑎𝑙𝑡𝑠, 𝑖𝑓 𝐴 𝑃,𝑃 =0, 𝑖.𝑒., 𝑃 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑙𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑙𝑡, 𝑖𝑓 𝐴 𝑃,𝑃 =1, 𝑖.𝑒., 𝑃 ℎ𝑎𝑙𝑡𝑠 𝑜𝑛 𝑃
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8. Example Program Properties of Program Q Halts if A returns False;
Does not halt if A returns = True
#Program Q
import random
TIMELIMIT = 5
def A(): #Mimics an algorithm that decides whether a program terminates given some input
if random.random()<0.5:
return True
else:
return False
result = A()
if result:
print('Algorithm A returned "HALTS"')
print('Algorithm A returned "DOES NOT HALT" and program Q HALTS')
elapsedtime=0
while result and (elapsedtime< TIMELIMIT):
elapsedtime+=1
print(elapsedtime, 'Q does not HALT')
if elapsedtime >= TIMELIMIT:
print('Timed out')
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9. Proof (Continued)
On the previous slide is a program Q: Q P = ℎ𝑎𝑙𝑡𝑠, 𝑖𝑓 𝐴 𝑃,𝑃 =0, 𝑖.𝑒., 𝑃 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑙𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑙𝑡, 𝑖𝑓 𝐴 𝑃,𝑃 =1, 𝑖.𝑒., 𝑃 ℎ𝑎𝑙𝑡𝑠 𝑜𝑛 𝑃
What happens with Q(Q)?
Q Q = ℎ𝑎𝑙𝑡𝑠, 𝑖𝑓 𝐴 𝑄,𝑄 =0, 𝑖.𝑒., 𝑄 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑙𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 ℎ𝑎𝑙𝑡, 𝑖𝑓 𝐴 𝑄,𝑄 =1, 𝑖.𝑒., 𝑄 ℎ𝑎𝑙𝑡𝑠 𝑜𝑛 𝑄
Thus, Q must halt when Q does not halt, and Q does not halt when Q halts—a contradiction.
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10. An Analogy Consider the following statements:
p: The statement q is true.
q: The statement p is false.
What can be said about the truth of the statements p and q?
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11. Nondeterministic Algorithms
Nondeterministic algorithms possess two properties:
Candidate solutions may be generated by random guessing
Verification of validity and evaluation of quality are of O(p(n)) for some polynomial p
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12. Example: A TSP Algorithm
Repeat:
Generate a random permutation of the city labels (choose a trip by guessing)
Evaluate the length of the trip (validate and evaluate)
How hard is this?
This is not a decision problem but could be used to derive a decision problem.
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13. NP
Consider the nondeterministic algorithms with worst case run time O(p(n)) for some polynomial p.
NP = {decision problems solvable by a nondeterministic polynomial algorithm}
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14. Polynomially Reducible
Given two decision problems D1 and D2.
Suppose there exists a transformation t of instances of D1 into instances of D2 such that:
All “yes” instances of D1 are mapped to instances of D2 that produce “yes” responses
All “no” instances of D1 are mapped to instances of D2 that produce “no” responses
t(.) is O(p(n)) for some polynomial
Then D1 is said to be polynomially reducible to D2
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15. NP-Complete A decision problem D is said to be NP-Complete if:
D is in NP
Every other problem in NP is polynomially reducible to D
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16. Satisfiability
Consider a logic expression in conjunctive normal form (CNF), e.g., (x1 ∨ ¬x5 ∨ x4) ∧ (¬x1 ∨ x5 ∨ x3 ∨ x4)…
Does there exist an assignment of truth values for the propositions x1, x2, x3…, that makes the expression true?
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17. NP-Hard A problem is NP-Hard, if satisfiability reduces to it
Any problem in NP is thus at least as hard as an NP-complete problem.
Many optimization problems
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18. The Halting Problem, Again
The Halting Problem is
Not in NP: It is undecidable, so there is no algorithm, no matter what the complexity, to solve it.
NP-hard: Because satisfiability reduces to it:
Construct an algorithm A with a propositional formula X as input, X having n logic variables
A searches all 2n combinations of logic values to determine if X is satisfiable
If so, A stops; otherwise A enters an infinite loop
A halts if and only if X is satisfiable
Note: If there were a polynomial algorithm for the halting problem, the mapping would allow satisfiability to be solved using A and X as its inputs.
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