Mapping Reductions: L1mL2

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Mapping Reductions L1?mL2

• An eight step process to help manufacture
  inspiration

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Step 1 Draw picture and look at language being
reduced to. Note input and output and what
output means.(Ex. Show ETM is undecidable)
ETM Input T.M. N, Output 1 L(N)? 0
???L(N)
1 L(N)? 0 ???L(N)
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Step 2 Choose language to reduce from. Note
input and output. You now have input and output
of Mf.(Ex. We will try to reduce from ATM)
ATM Input T.M. M, string w, Output 1
w?L(M) 0 w?L(M)
1 L(N)? 0 ?w?L(N)
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Step 3 Relate output of L1 to output of L2.
ETM ATM 1 L(N)? w?L(M) 0 ???L(N) w?L(M)
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Step 4 Come up with very basic inputs to second
machine that will produce accept and reject from
L2.
ETM ATM 1 L(N)? w?L(M) 0 ???L(N) w?L(M)
Very basic inputs 1 L(N)? 0 L(N)1
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Step 5 Describe clearly Mf and its output.This
is the tricky step, i.e. inspiration is
neededDont worry if the behavior is roughly
opposite of what you want. This is what step 7
is for.

• Mf On input ltM,wgt
• Create and output Turing Machine N
• N On input x
• If x1
• simulate M on input w
• else
• reject

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Hint The output of Mf is usually of the form
On input x reject/accept
On input x if x(some string) simulate
input to L1 (Possibly muck with output)
else reject/accept
-or-
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Step 6 Relate the behavior of ML2 acting on the
output of Mf to the implied output of L1.
ETM ATM VBI 1 L(N)? w?L(M) L(N) ? 0
???L(N) w?L(M) L(N)1
ETM(VBI) ? ATM 1 L(N) ? ? w?L(M) 0
L(N)1 ? w?L(M)
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Explanation we get the following implied output
of ATM since according to the behavior of N, the
language of N1 iff M accepts w. In all other
cases, the language of N has to be empty.
N On input x If x1 simulate M on input
w else reject
ETM(VBI) ATM 1 L(N) ? ? w?L(M) 0 L(N)1
? w?L(M)
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Step 7 Look for conflicts that can be remedied
by complimenting L1 or L2.
ETM ATM VBI ? ATM 1 L(N)? w?L(M)
L(N) ? ? w?L(M) 0 ???L(N) w?L(M)
L(N)1 ? w?L(M)
This can be fixed by reducing from ATM
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After modifying our notes to reflect the change
to ATM, we have
ETM ATM VBI ? ATM 1 L(N)? w?L(M)
L(N) ? ? w?L(M) 0 ???L(N) w?L(M)
L(N)1 ? w?L(M)
See. They match now!
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Step 8 Bring it on home! (i.e. actually state
that you proved something)

• Assuming ETM is decidable
• N?ETM ? L(N)f ? w?L(M) ? ltM,wgt?ATM
• N?ETM ? L(N)f ? w?L(M) ? ltM,wgt?ATM
• This implies ATM is a decidable language.
  However, we know ATM is undecidable. Therefore,
  our assumption is wrong so ETM must be
  undecidable.