Reducibility (I)
CS 301. Languages and Automata
© 2025 by Gonzalo A. Bello
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CS 301 - Lecture 24: Reducibility (I)
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Reducibility
Introduction (I)
• A reduction is a way of converting a problem 𝐴 into a problem 𝐵
such that a solution to problem 𝐵 can be used to solve problem 𝐴.
Reduces to
𝑨
𝑩
Solves
• Example:
• Let 𝐴 be the problem of finding your way around a city.
• Let 𝐵 be the problem of obtaining a map of the city.
• If I can obtain a map of the city (i.e., if I can solve problem 𝑩), then I can use
this map to find my way around the city (i.e., then I can solve problem 𝑨).
• Therefore, 𝐴 reduces to 𝐵 or 𝐴 is reducible to 𝐵.
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CS 301 - Lecture 24: Reducibility (I)
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Reducibility
Introduction (II)
• Another example:
Suppose Mary wants
a pet dragon
If Mary finds a dragon
shop, she can get
a pet dragon
Therefore, the problem of
getting a pet dragon
reduces to the problem
of finding a dragon shop
Source: Undecidable Problems (2020)
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CS 301 - Lecture 24: Reducibility (I)
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Reducibility
Introduction (III)
• Suppose 𝐴 is reducible to 𝐵.
• If 𝐵 is decidable, then 𝐴 is also decidable.
• Example:
• If I can obtain a map of the city (𝐵 is decidable), then I can find my way
around the city (𝐴 is decidable).
• How can we prove that a problem is decidable?
• Show that it reduces to a problem already known to be decidable.
To prove that 𝑷𝒏𝒆𝒘 is decidable,
show that 𝑷𝒏𝒆𝒘 reduces to 𝑷𝒌𝒏𝒐𝒘𝒏
where 𝑷𝒌𝒏𝒐𝒘𝒏 is decidable
• Example:
We proved that 𝐴𝑁𝐹𝐴 is decidable by showing that it reduces to 𝐴𝐷𝐹𝐴 ,
which was already known to be decidable.
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CS 301 - Lecture 24: Reducibility (I)
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Reducibility
Introduction (IV)
• Suppose 𝐴 is reducible to 𝐵.
• If 𝐴 is undecidable, then 𝐵 is also undecidable.
• Example:
• If I cannot find my way around the city (𝐴 is undecidable), then I cannot
obtain a map of the city (𝐵 is undecidable).
• How can we prove that a problem is undecidable?
• Show that a problem already known to be undecidable reduces to it.
To prove that 𝑷𝒏𝒆𝒘 is undecidable,
show that 𝑷𝒌𝒏𝒐𝒘𝒏 reduces to 𝑷𝒏𝒆𝒘
where 𝑷𝒌𝒏𝒐𝒘𝒏 is undecidable
Which problem is known to be undecidable?
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CS 301 - Lecture 24: Reducibility (I)
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Reducibility
Introduction (V)
• Another example:
If Mary finds a dragon shop,
she can get a pet dragon
Dragons do not exist (i.e., the
problem of getting a pet
dragon is undecidable).
Therefore, dragon shops can
also not exist (i.e., the problem
of finding a dragon shop is also
undecidable).
Source: Undecidable Problems (2020)
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CS 301 - Lecture 24: Reducibility (I)
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Reducibility
Exercise 24.1
If 𝑨 is reducible to 𝑩 and 𝑨 is undecidable,
we can conclude that 𝑩 is also undecidable.
A.
B.
True
False
If 𝑨 is reducible to 𝑩 and 𝑨 is decidable,
we can conclude that 𝑩 is also decidable.
A.
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B.
True
CS 301 - Lecture 24: Reducibility (I)
False
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Reducibility
Example: 𝑯𝑨𝑳𝑻𝑻𝑴 (I)
• Example: 𝐻𝐴𝐿𝑇𝑇𝑀 =
𝑀, 𝑤
𝑀 is a TM that halts on input 𝑤}.
• The halting problem is the problem of determining whether a TM halts (by
accepting or rejecting) on a given input.
How can we prove that 𝑯𝑨𝑳𝑻𝑻𝑴 is undecidable?
• Proof by contradiction:
• Assume that 𝐻𝐴𝐿𝑇𝑇𝑀 is decidable.
• Show that 𝐴 𝑇𝑀 is reducible to 𝐻𝐴𝐿𝑇𝑇𝑀 .
• Since 𝐻𝐴𝐿𝑇𝑇𝑀 is decidable and 𝐴 𝑇𝑀 is reducible to 𝐻𝐴𝐿𝑇𝑇𝑀 , then 𝐴 𝑇𝑀
is also decidable.
• But 𝐴 𝑇𝑀 is undecidable. This is a contradiction.
• Therefore, 𝐻𝐴𝐿𝑇𝑇𝑀 must be undecidable.
How can we reduce 𝑨𝑻𝑴 to 𝑯𝑨𝑳𝑻𝑻𝑴 ?
• Assume that we have a TM 𝑅 that decides 𝐻𝐴𝐿𝑇𝑇𝑀 .
• Use 𝑅 to construct a TM 𝑆 that decides 𝐴 𝑇𝑀 .
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CS 301 - Lecture 24: Reducibility (I)
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Reducibility
Example: 𝑯𝑨𝑳𝑻𝑻𝑴 (II)
• Example: 𝐻𝐴𝐿𝑇𝑇𝑀 =
𝑀, 𝑤
𝑀 is a TM that halts on input 𝑤}.
How can we reduce 𝑨𝑻𝑴 to 𝑯𝑨𝑳𝑻𝑻𝑴 ?
• Use 𝑅 to construct a TM 𝑆 that decides 𝐴 𝑇𝑀 .
𝑺 = “On input 𝑴, 𝒘 :
Is 𝑺 a decider?
1. Run TM 𝑹 that decides 𝑯𝑨𝑳𝑻𝑻𝑴 on input 𝑴, 𝒘 .
2. If 𝑹 rejects, _____________________
3. If 𝑹 accepts, run 𝑴 on input 𝒘.
4. If 𝑴 accepts, ____________________
5. If 𝑴 rejects, _____________________”
𝑆
𝑀
𝑤
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𝑅
Accept
𝑀
Reject
CS 301 - Lecture 24: Reducibility (I)
Accept
Reject
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Reducibility
Example: 𝑯𝑨𝑳𝑻𝑻𝑴 (III)
• Example: 𝐻𝐴𝐿𝑇𝑇𝑀 =
𝑀, 𝑤
𝑀 is a TM that halts on input 𝑤}.
How can we prove that 𝑯𝑨𝑳𝑻𝑻𝑴 is undecidable?
• Proof (by contradiction):
• Assume that 𝐻𝐴𝐿𝑇𝑇𝑀 is decidable. Let 𝑅 be a TM that decides 𝐻𝐴𝐿𝑇𝑇𝑀 .
• Construct a TM 𝑆 that decides 𝐴 𝑇𝑀 as follows.
𝑺 = “On input 𝑴, 𝒘 :
1. Run TM 𝑹 that decides 𝑯𝑨𝑳𝑻𝑻𝑴 on input 𝑴, 𝒘 .
2. If 𝑹 rejects (𝑴 does not halt on 𝒘), reject.
3. If 𝑹 accepts (𝑴 halts on 𝒘), run 𝑴 on input 𝒘.
4. If 𝑴 accepts, accept. If 𝑴 rejects, reject.”
• If 𝑴 accepts 𝒘, then 𝑀 halts on 𝑤, so 𝑅 and 𝑀 accept and 𝑺 accepts.
• If 𝑴 does not accept 𝒘, then either 𝑀 does not halt on 𝑤 or 𝑀 rejects 𝑤, so either
𝑅 or 𝑀 rejects and 𝑺 rejects.
• 𝑆 always halts because 𝑅 always halts and we only run 𝑀 if it halts.
• Then, 𝐴 𝑇𝑀 is decidable. But 𝐴 𝑇𝑀 is undecidable. This is a contradiction.
• Therefore, 𝐻𝐴𝐿𝑇𝑇𝑀 must be undecidable.
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Reducibility
References
• Michael Sipser. Introduction to the Theory of Computation.
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CS 301 - Lecture 24: Reducibility (I)
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