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presents. Reducibility. w ith Dr. Bunsen Honeydew. Hello, I’m Dr. Bunsen Honeydew. Today we’re going to talk about Reducibility . We’ve been having some…issues here around the studio. .

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with Dr. Bunsen Honeydew


Hello, I’m Dr. Bunsen Honeydew. Today we’re going to talk about Reducibility.


We’ve been having some…issues here around the studio.


Beaker started experimenting with Turing Machines and…well, he found about about the Halting Problem the hard way.


He’s still looping.


Now, on to our lesson


Suppose we have a context-free grammar, EQCFG which we want to prove is undecidable. We use reducability!


Determining that EQCFG is undecidable is easy when we remember Corollary 5.23, which some incredibly smart person figured out for us. . .


Corollary 5.23 says:

If A ≤m B and A is undecidable, then B is undecidable


That means that if you have something that is undecidable, you can reduce it to a second object to prove that it is undecidable as well.


So let’s use this corollary to prove that EQCFG is undecidable. Do we know of anything that is already undecidable?


We already know that ALLCFG is undecidable!


Indeed. So if ALLCFG is undecidable, we should follow Corolary 5.23 and reduce it to EQCFG show that EQCFG is also undecidable


Let’s go to the chalkboard. . .


Goal: reduce ALLCFG to EQCFG to show that EQCFG is undecidable


Set up our contradiction proof

First, let’s just assume that EQCFG is actually decidable. Then we’re going to create a decider for it named E.

Now, let’s make a decider which will decide ALLCFG an we’ll call

it A. A does something special though. A is going to use E like a subroutine, giving it the ability to decide EQCFG

as well as ALLCFG

Since A is our decider,

if ALLCFG is decidable, A will accept

else reject

So now that we have our deciders, let’s build our machine!


For our decider A, we need to define certain actions for certain inputs, just like if we were writing code for A . . .

Now, for the input, let’s use an arbitrary name, G1. Now, ALLCFG is defined as:

ALLCFG = {<G1> | G1 is a CFG and L(G1)=Σ*} so G1 is a Context Free Grammar with the language Σ*. So write our definition for A as follows:

A: “on input <G1> (a CFG with the language defined above)

1. Change <G1> into <G2,G3>

2. Return E(<G2>,<G3>) (E is our decider for EQCFG )

You’ll notice that in number 1, we say we’re going to CHANGE G1.



  • This is part of Reducing something. We have input which is of ALLCFG. If we can change that input to be like EQCFG, then it means that ALLCFG and EQCFG are the same and therefore, both undecidable. Right?

  • So let’s skip ahead and look at what our final output will look like:

  • What will the change need to satisfy?? (This is part of phase 3)

    • <G1> ∈ ALLCFG => <G2, G3> ∈ EQCFG

      • This means that some G1, which is a member of ALLCFG will need to change into G2,G3, which are members of EQCFG

    • <G1> ∉ ALLCFG => <G2, G3> ∉ EQCFG

      • This means that some G1, which is not a member of ALLCFG will need to change into G2,G3, which are not members of EQCFG


Now that we have a little preview of our goal, let’s skip ahead and look at the second step of our reduction proof:


Now let’s actually change G1, like we were just talking about

“On input <G1>

1. Let some G2 = G1 (so we now have a CFG G2 which is

exactly the same as G1

2. Build G3 with L(G3) = Σ* (so the language is Σ*)

Output <G2, G3>”

Now that we have effectively changed, G1 into G2,G3. Now let’s see if it met our goals that we just talked about.



<G1> ∈ ALLCFG and the language of G1 is as follows: L(G1) = Σ*

We created G2 as a copy of G1 so L(G2) is like L(G1) and = Σ*

When we created G3, we defined its language as Σ*; therefore, L(G3)

also = Σ*

Let’s review that EQ = { <G2,G3>|G2 and G3 are CFGs and L(G2) = L(G3) }

Are the languages of G2 and G3 the same?


Therefore, we can conclude that:

<G2, G3> ∈ EQCFG because L(G2) = L(G3)


Phase3 Cont’d

That proves that the languages are the same, but can we show the reverse?

<G1> ∉ ALLCFGthen L(G1) ≠ Σ* (common sense if we look at the definition)

So L(G2) – and remember that G2 is a copy of G1 – L(G2) ≠ Σ*

Therefore, L(G2) ≠ L(G3) = Σ* (we defined the language of G3 as being Σ*!)

So we can conclude that

<G2, G3> ∉ EQCFG


So back to our decider. Does A accept?


Heavens no. The contradiction lies in the undecidability of ALLCFG, not EQCFG

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