halting problem and tsp wednesday, week 8. background - halting problem common error: program goes...
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Halting Problem and TSP
Wednesday, Week 8
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Background - Halting Problem
Common error: Program goes into an infinite loop.
Wouldn’t it be nice to have a tool that would warn us that our program has this bug? (We could make a lot of money if we could develop
such a tool!)
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Example
Infinite loop - we get into a loop but for some reason we never get out of it.
for (i = 0; i < 10; i--)
{
document.writeln(“Help, I’m in an infinite loop!<BR>”);
}
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Example Output
Help, I’m in an infinite loop!
Help, I’m in an infinite loop!
Help, I’m in an infinite loop!
Help, I’m in an infinite loop!
Help, I’m in an infinite loop!
Help, I’m in an infinite loop!
…
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Definition
Design a program, H, that will do the following:
1. Take as its input a description of a program P (in binary, say).
2. Halt eventually and answer "yes" if P will eventually halt, or halt and answer "no" if P will run forever.
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Halting Problem
In other words,Program H will always halt and give the
correct answer no matter what program has been given as input.
Hprogram Halts? Yes or No
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Proving the Halting Problem
We want to prove that the halting problem is non-computable.
In other words, there is no algorithm that we can use to tell whether or not a program will halt.
We will use proof by contradiction.
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Proof by Contradiction
Prove: if x+y >= 2, then x>=1 or y>=1.– Assume that x+y>=2 and that x<1 and y<1.– Then, x+y < 1+1 = 2. – This is a contradiction since x+y >=2!– Thus our assumption that x<1 and y<1 is false.
Note: NOT(x>=1 or y>=1) = (x<1 and y<1) by DeMorgan’s Laws.
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Proof by Contradiction:
We’ll assume that the Halting Problem CAN be computed.
We’ll develop another program that uses the Halting Problem function.
We’ll find ourselves caught in a paradox (the contradiction).
We’ll have proven that our original assumption is false.
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Assume Halting Problem OK
Let H be a program (or sub-program) that determines whether a program will halt.
Hprogram Halts? Yes or No
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Let’s Build Another Program
Let P be a program that uses H.
For any given program, P will call H and pass it the given program.
Hprogram
program
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What does P do?
Now, P acts as follows:– P takes a program as input and feeds the program
to H as input.– If H answer yes, then P will enter an infinite loop
and run forever. – If H answer no, then P will stop.
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The program P
function P (program){
var halts = H(program);if (halts == True)
infinite loop;else
stop;}
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What Can We Do With P?
Let’s give P a copy of itself as its input.
HYes
No
program:
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What If P Halts?
HYes
No
program:
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What If P Loops Indefinitely?
HYes
No
program:
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The Paradox
If P is a program that halts when given itself as its input,
then, when given itself as input, P will go into an infinite loop.
If P is a program that loops indefinitely when given itself as its input,
then , when given itself as input, P will halt immediately.
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What Went Wrong?
There’s nothing wrong with P, itself.The problem must be with the assumption that we could write H.
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So, Proof by Contradiction:
We assumed that the Halting Problem COULD be computed.We developed another program that used the Halting Problem function.We found ourselves caught in a paradox (the contradiction).We proved that the Halting Problem is not computable.
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Conclusions
Self-referentiality is a real problem with programsRice’s theorem says that “Any nontrivial property of programs is undecidable.”Thus, we can’t write programs to answer questions about programs.
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Intractability
The Traveling Salesperson Problem is intractable.Proving it is beyond the scope of this class.We will just understand the problem and how hard it is.
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Traveling Sales Problem
A salesperson has a group of cities that he/she needs to visit. There are a bunch of distances between the cities.Our salesperson has to visit all the cities by following a path with the least distance (or cost).
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TSP Example #1
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TSP Answer #1
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TSP Example #2
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TSP Answer #2
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Traveling Sales Problem
Conclusions:– The only correct algorithm we could come up with
had to examine all possible paths.– The only correct algorithm that anyone has come up
with has to examine all possible paths.– Thus this algorithm is O(n!) and intractable.