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- Formally modeling running time, and in particular notions such as
\(O(n)\) or \(O(n^3)\) time algorithms.
- The classes \(\mathbf{P}\) and \(\mathbf{EXP}\) modelling polynomial and
exponential time respectively.
- The
*time hierarchy theorem*, that in particular says that for every \(k \geq 1\) there are functions we*can*compute in \(O(n^{k+1})\) time but*can not*compute in \(O(n^k)\) time. - The class \(\mathbf{P_{/poly}}\) of
*non uniform*computation and the result that \(\mathbf{P} \subseteq \mathbf{P_{/poly}}\)

“When the measure of the problem-size is reasonable and when the sizes assume values arbitrarily large, an asymptotic estimate of … the order of difficulty of [an] algorithm .. is theoretically important. It cannot be rigged by making the algorithm artificially difficult for smaller sizes”, Jack Edmonds, “Paths, Trees, and Flowers”, 1963

“The computational complexity of a sequence is to be measured by how fast a multitape Turing machine can print out the terms of the sequence. This particular abstract model of a computing device is chosen because much of the work in this area is stimulated by the rapidly growing importance of computation through the use of digital computers, and all digital computers in a slightly idealized form belong to the class of multitape Turing machines.”, Juris Hartmanis and Richard Stearns, “On the computational complexity of algorithms”, 1963.

In chapefficient we saw examples of efficient algorithms, and
made some claims about their running time, but did not give a
mathematically precise definition for this concept. We do so in this
chapter, using the NAND++ and NAND<< models we have seen before.*dependence* of the number of steps the
program takes on the length of the input. That is, for any program \(P\),
we will be interested in the maximum number of steps that \(P\) takes on
inputs of length \(n\) (which we often denote as \(T(n)\)).*maximum* number of steps for
inputs of a given length, this concept is often known as *worst case
complexity*. The *minimum* number of steps (or “best case”
complexity) to compute a function on length \(n\) inputs is typically
not a meaningful quantity since essentially every natural problem
will have some trivially easy instances. However, the *average case
complexity* (i.e., complexity on a “typical” or “random” input) is
an interesting concept which we’ll return to when we discuss
*cryptography*. That said, worst-case complexity is the most
standard and basic of the complexity measures, and will be our focus
in most of this course.

We start by defining running time separately for both NAND<< and NAND++ programs. We will later see that the two measures are closely related. Roughly speaking, we will say that a function \(F\) is computable in time \(T(n)\) there exists a NAND<< program that when given an input \(x\), halts and outputs \(F(x)\) within at most \(T(|x|)\) steps. The formal definition is as follow:

Let \(T:\N \rightarrow \N\) be some function mapping natural numbers to
natural numbers. We say that a function
\(F:\{0,1\}^* \rightarrow \{0,1\}\) is *computable in \(T(n)\) NAND<<
time* if there exists a NAND<< program \(P\) such that for every every
sufficiently large \(n\) and every \(x\in \{0,1\}^n\), when given input \(x\),
the program \(P\) halts after executing at most \(T(n)\) lines and outputs
\(F(x)\).

Similarly, we say that \(F\) is *computable in \(T(n)\) NAND++ time* if
there is a NAND++ program \(P\) computing \(F\) such that on every
sufficiently large \(n\) and \(x\in \{0,1\}^n\), on input \(x\), \(P\) executes
at most \(T(n)\) lines before it halts with the output \(F(x)\).

We let \(TIME_{<<}(T(n))\) denote the set of Boolean functions that are computable in \(T(n)\) NAND<< time, and define \(TIME_{++}(T(n))\) analogously. The set \(TIME(T(n))\) (without any subscript) corresponds to \(TIME_{<<}(T(n))\).

time-def is not very complicated but is one of the most
important definitions of this book. Please make sure to stop, re-read
it, and make sure you understand it. Note that although they are defined
using computational models, \(TIME_{<<}(T(n))\) and \(TIME_{++}(T(n))\) are
classes of *functions*, not of *programs*. If \(P\) is a NAND++ program
then a statement such as “\(P\) is a member of \(TIME_{++}(n^2)\)” does not
make sense.

In the definition of time complexity, we count the number of times a
line is *executed*, not the number of lines in the program. For example,
if a NAND++ program \(P\) has 20 lines, and on some input \(x\) it takes a
1,000 iterations of its loop before it halts, then the number of lines
executed on this input is 20,000.

To make this count meaningful, we use the “vanilla” flavors of NAND++
and NAND<<, “unpacking” any syntactic sugar. For example, if a NAND++
program \(P\) contains a line with the syntactic sugar `foo = MACRO(bar)`

where `MACRO`

is some macro/function that is defined elsewhere using 100
lines of vanilla NAND++, then executing this line counts as executing
100 steps rather than a single one.

Unlike the notion of computability, the exact running time can be a
function of the model we use. However, it turns out that if we only care
about “coarse enough” resolution (as will most often be the case) then
the choice of the model, whether it is NAND<<, NAND++, or Turing or
RAM machines of various flavors, does not matter. (This is known as the
*extended* Church-Turing Thesis). Nevertheless, to be concrete, we will
use NAND<< programs as our “default” computational model for measuring
time, which is why we say that \(F\) is computable in \(T(n)\) time without
any qualifications, or write \(TIME(T(n))\) without any subscript, we mean
that this holds with respect to NAND<< machines.

Given that so far we have emphasized NAND++ as our “main” model of
computation, the reader might wonder why we use NAND<< as the default
yardstick where running time is involved. As we will see, this choice
does not make much difference, but NAND<< or *RAM Machines* correspond
more closely to the notion of running time as discussed in algorithms
text or the practice of computer science.

The class \(TIME(T(n))\) is defined in time-def with respect to
a *time* measure function \(T(n)\). When considering time bounds, we will
want to restrict our attention to “nice” bounds such as \(O(n)\),
\(O(n\log n)\), \(O(n^2)\), \(O(2^{\sqrt{n}})\), \(O(2^n)\), and so on. We do so
to avoid pathological examples such as non-monotone functions (where the
time to compute a function on inputs of size \(n\) could be smaller than
the time to compute it on shorter inputs) or other degenerate cases such
as running time that is not sufficient to read the input or cases where
the running time bound itself is hard to compute. Thus we make the
following definition:

A function \(T:\N \rightarrow \N\) is a *nice time bound function* (or
nice function for short) if:

- \(T(n) \geq n\)
- \(T(n) \geq T(n')\) whenever \(n \geq n'\)
- There is a NAND<< program that on input numbers \(n,i\), given in binary, can compute in \(O(T(n))\) steps the \(i\)-th bit of a prefix-free representation of \(T(n)\) (represented as a string in some prefix-free way).

All the “normal” time complexity bounds we encounter in applications
such as \(T(n)= 100 n\), \(T(n) = n^2 \log n\),\(T(n) = 2^{\sqrt{n}}\), etc.
are “nice”. Hence from now on we will only care about the class
\(TIME(T(n))\) when \(T\) is a “nice” function. Condition 3. of
nice-def means that we can compute the binary representation
of \(T(n)\) in time which itself is roughly \(T(n)\). This condition is
typically easily satisfied. For example, for arithmetic functions such
as \(T(n) = n^3\) or \(T(n)= \floor n^{1.2}\log n \rfloor\) we can typically
compute the binary representation of \(T(n)\) much faster than that: we
can do so in time which is polynomial in the *number of bits* in this
representation. Since the number of bits is \(O(\log T(n))\), any quantity
that is polynomial in this number will be much smaller than \(T(n)\) for
large enough \(n\).

The two main time complexity classes we will be interested in are the following:

**Polynomial time:**A function \(F:\{0,1\}^* \rightarrow \{0,1\}\) is*computable in polynomial time*if it is in the class \(\mathbf{P} = \cup_{c\in \{1,2,3,\ldots \}} TIME(n^c)\). That is, \(F\in \mathbf{P}\) if there is an algorithm to compute \(F\) that runs in time at most*polynomial*(i.e., at most \(n^c\) for some constant \(c\)) in the length of the input.**Exponential time:**A function \(F:\{0,1\}^* \rightarrow \{0,1\}\) is*computable in exponential time*if it is in the class \(\mathbf{EXP} = \cup_{c\in \{1,2,3,\ldots \}} TIME(2^{n^c})\). That is, \(F\in \mathbf{EXP}\) if there is an algorithm to compute \(F\) that runs in time at most*exponential*(i.e., at most \(2^{n^c}\) for some constant \(c\)) in the length of the input.

In other words, these are defined as follows:

Let \(F:\{0,1\}^* \rightarrow \{0,1\}\). We say that \(F\in \mathbf{P}\) if there is a polynomial \(p:\N \rightarrow \R\) and a NAND<< program \(P\) such that for every \(x\in \{0,1\}^*\), \(P(x)\) runs in at most \(p(|x|)\) steps and outputs \(F(x)\).

We say that \(F\in \mathbf{EXP}\) if there is a polynomial \(p:\N \rightarrow \R\) and a NAND<< program \(P\) such that for every \(x\in \{0,1\}^*\), \(P(x)\) runs in at most \(2^{p(|x|)}\) steps and outputs \(F(x)\).

Please make sure you understand why PandEXPdef and the bullets above define the same classes.

Since exponential time is much larger than polynomial time,
\(\mathbf{P}\subseteq \mathbf{EXP}\). All of the problems we listed in
chapefficient are in \(\mathbf{EXP}\),*non
Boolean* functions, but we will sometimes “abuse notation” and refer
to non Boolean functions as belonging to \(\mathbf{P}\) or
\(\mathbf{EXP}\). We can easily extend the definitions of these
classes to non Boolean and partial functions. Also, for every
non-Boolean function \(F:\{0,1\}^* \rightarrow \{0,1\}^*\), we can
define a Boolean variant \(Bool(F)\) such that \(F\) can be computed in
polynomial time if and only if \(Bool(F)\) is. See boolex

\(\mathbf{P}\) | \(\mathbf{EXP}\) (but not known to be in \(\mathbf{P}\)) |
---|---|

Shortest path | Longest Path |

Min cut | Max cut |

2SAT | 3SAT |

Linear eqs | Quad. eqs |

Zerosum | Nash |

Determinant | Permanent |

Primality | Factoring |

Most of the time we will restrict attention to computing functions that
are *total* (i.e., defined on every input) and *Boolean* (i.e., have a
single bit of output). However, time-def naturally extends to
non Boolean and to partial functions as well. We now describe this
extension, although we will try to stick to Boolean total functions to
the extent possible, so as to minimize the “cognitive overload” for the
reader and amount of notation they need to keep in their head.

We will define \(\overline{TIME}_{<<}(T(n))\) and \(\overline{TIME}_{++}(T(n))\) to be the generalization of \(TIME_{<<}(T(n))\) and \(TIME_{++}(T(n))\) to functions that may be partial or have more than one bit of output, and define \(\overline{\mathbf{P}}\) and \(\overline{\mathbf{EXP}}\) similarly. Specifically the formal definition is as follows:

Let \(T:\N \rightarrow \N\) be a nice function, and let \(F\) be a (possibly partial) function mapping \(\{0,1\}^*\) to \(\{0,1\}^*\). We say that \(F\) is in \(\overline{TIME}_{<<}(T(n))\) (respectively \(\overline{TIME}_{++}(T(n))\)) if there exists a NAND<< (respectively NAND++) program \(P\) such that for every sufficiently large \(n\) and \(x\in \{0,1\}^n\) on which \(F\) is defined, on input \(x\) the program \(P\) halts after executing at most \(T(n)\) steps and outputs \(F(x)\).

We let \(\overline{TIME}(T(n))\) (without a subscript) denote the set \(\overline{TIME}_{<<}(T(n))\) and let \(\overline{\mathbf{P}} = \cup_{c\in \{1,2,3,\ldots \}} \overline{TIME}(n^c)\) and \(\overline{\mathbf{EXP}} = \cup_{c\in \{1,2,3,\ldots \}} \overline{TIME}(2^{n^c})\).

We have seen in RAMTMequivalencethm that for every NAND<< program \(P\) there is a NAND++ program \(P'\) that computes the same function as \(P\). It turns out that the \(P'\) is not much slower than \(P\). That is, we can prove the following theorem:

There are absolute constants \(a,b\) such that for every function \(F\) and nice function \(T:\N \rightarrow \N\), if \(F \in TIME_{<<}(T(n))\) then there is a NAND++ program \(P'\) that computes \(F\) in \(T'(n)=a\cdot T(n)^b\). That is, \(TIME_{<<}(T(n)) \subseteq TIME_{++}(aT(n)^b)\)

The constant \(b\) can be easily shown to be at most five, and with more
effort can be optimized further. polyRAMTM-thm means that the
definition of the classes \(\mathbf{P}\) and \(\mathbf{EXP}\) are robust to
the choice of model, and will not make a difference whether we use
NAND++ or NAND<<. The same proof also shows that *Turing Machines* can
simulate NAND<< programs (and hence RAM machines). In fact, similar
results are known for many models including cellular automata,
C/Python/Javascript programs, parallel computers, and a great many other
models, which justifies the choice of \(\mathbf{P}\) as capturing a
technology-independent notion of tractability. As we discussed before,
this equivalence between NAND++ and NAND<< (as well as other models)
allows us to pick our favorite model depending on the task at hand
(i.e., “have our cake and eat it too”). When we want to *design* an
algorithm, we can use the extra power and convenience afforded by
NAND<<. When we want to *analyze* a program or prove a *negative
result*, we can restrict attention to NAND++ programs.

The idea behind the proof is simple. It follows closely the proof of
RAMTMequivalencethm, where we have shown that every function
\(F\) that is computable by a NAND<< program \(P\) is computable by a
NAND++ program \(P'\). To prove polyRAMTM-thm, we follow the
exact same proof but just check that the overhead of the simulation of
\(P\) by \(P'\) is polynomial. The proof has many details, but is not deep.
It is therefore much more important that you understand the *statement*
of this theorem than its proof.

As mentioned above, we follow the proof of RAMTMequivalencethm (simulation of NAND<< programs using NAND++ programs) and use the exact same simulation, but with a more careful accounting of the number of steps that the simulation costs. Recall, that the simulation of NAND<< works by “peeling off” features of NAND<< one by one, until we are left with NAND++.

We will not provide the full details but will present the main ideas used in showing that every feature of NAND<< can be simulated by NAND++ with at most a polynomial overhead:

- Recall that every NAND<< variable or array element can contain an
integer between \(0\) and \(T\) where \(T\) is the number of lines that
have been executed so far. Therefore if \(P\) is a NAND<< program
that computes \(F\) in \(T(n)\) time, then on inputs of length \(n\), all
integers used by \(P\) are of magnitude at most \(T(n)\). This means
that the largest value
`i`

can ever reach is at most \(T(n)\) and so each one of \(P\)’s variables can be thought of as an array of at most \(T(n)\) indices, each of which holds a natural number of magnitude at most \(T(n)\), which can be represented using \(O(\log T(n))\) bits. - As in the proof of RAMTMequivalencethm, we can think of
the integer array
`Foo`

as a two dimensional*bit*array`Foo_2D`

(where`Foo_2D[`

\(i\)`][`

\(j\)`]`

is the \(j\)-th bit of`Foo[`

\(i\)`]`

) and then encode the latter as a*one dimensional*bit array`Foo_1D`

by mapping`Foo_2D[`

\(i\)`][`

\(j\)`]`

to`Foo_1D[`

\(embed(i,j)\)`]`

where \(embed:\N \times N \rightarrow \N\) is some one-to-one function that embeds the two dimensional space \(\N\times \N\) into the one dimensional \(\N\). Specifically, if we use \(embed(i,j)= \tfrac{1}{2}(i+j)(i+j+1) + j\) as in pair-ex, then we can see that if \(i,j \leq O(T(n))\), then \(embed(i,j) \leq O(T(n)^2)\). We also use the fact that \(embed\) is itself computable in polynomial time in the length of its input. - All the arithmetic operations on integers use the grade-school algorithms, that take time that is polynomial in the number of bits of the integers, which is \(poly(\log T(n))\) in our case.
- In NAND++ one cannot access an array at an arbitrary location but
rather only at the location of the index variable
`i`

. Nevertheless, if`Foo`

is an array that encodes some integer \(k\in \N\) (where \(k \leq T(n)\) in our case), then, as we’ve seen in the proof of RAMTMequivalencethm, we can write NAND++ code that will set the index variable`i`

to \(k\). Specifically, using enhanced NAND++ we can write a loop that will run \(k\) times (for example, by decrementing the integer represented by`Foo`

until it reaches \(0\)), to ensure that an array`Marker`

that will equal to \(0\) in all coordinates except the \(k\)-th one. We can then decrement`i`

until it reaches \(0\) and scan`Marker`

until we reach the point that`Marker[i]`

\(=1\). - To transform the above from
*enhanced*to*standard*(i.e., “vanilla”) NAND++, all that is left is to follow our proof of enhancednandequivalence and show we can simulate`i+= foo`

and`i-= bar`

in vanilla NAND++ using our “breadcrumbs” and “wait for the bus” techniques. To simulate \(T\) steps of enhanced NAND++ we will need at most \(O(T^2)\) steps of vanilla NAND++ (see obliviousfig). Indeed, suppose that the largest point that the index`i`

reached in the computation so far is \(R\), and we are in the worst case where we are trying, for example, to increment`i`

while it is in a “decreasing” phase. Within at most \(2R\) steps we will be back in the same position at an “increasing” phase. Using this argument we can see that in the worst case, if we need to simulate \(T\) steps of enhanced NAND++ we will use \(O(1 + 2 + \cdots + T) = O(T^2)\) steps of vanilla NAND++.

Together these observations imply that the simulation of \(T\) steps of NAND<< can be done in \(O(T^a)\) steps of vanilla NAND++ for some constant \(a\), i.e., time polynomial in \(T\). (A rough accounting can show that this constant \(a\) is at most five; a more careful analysis can improve it further though this does not matter much.)

If we follow the equivalence results between NAND++/NAND<< and other
models, including Turing machines, RAM machines, Game of life, \(\lambda\)
calculus, and many others, then we can see that these results also have
at most a polynomial overhead in the simulation in each way.

polyRAMTM-thm shows that the classes \(\mathbf{P}\) and
\(\mathbf{EXP}\) are *robust* with respect to variation in the choice of
the computational model. They are also robust with respect to our choice
of the representation of the input. For example, whether we decide to
represent graphs as adjacency matrices or adjacency lists will not make
a difference as to whether a function on graphs is in \(\mathbf{P}\) or
\(\mathbf{EXP}\). The reason is that changing from one representation to
another at most squares the size of the input, and a quantity is
polynomial in \(n\) if and only if it is polynomial in \(n^2\).

More generally, for every function \(F:\{0,1\}^* \rightarrow \{0,1\}\), the answer to the question of whether \(F\in \mathbf{P}\) (or whether \(F\in \mathbf{EXP}\)) is unchanged by switching representations, as long as transforming one representation to the other can be done in polynomial time (which essentially holds for all reasonable representations).

We have seen in univnandppnoneff the “universal program” or
“interpreter” \(U\) for NAND++. Examining that proof, and combining it
with polyRAMTM-thm , we can see that the program \(U\) has a
*polynomial* overhead, in the sense that it can simulate \(T\) steps of a
given NAND++ (or NAND<<) program \(P\) on an input \(x\) in \(O(T^a)\) steps
for some constant \(a\). But in fact, by directly simulating NAND<<
programs, we can do better with only a *constant* multiplicative
overhead:

There is a NAND<< program \(U\) that computes the partial function
\(TIMEDEVAL:\{0,1\}^* \rightarrow \{0,1\}^*\) defined as follows: \[
TIMEDEVAL(P,x,1^T)=P(x)
\] if \(P\) is a valid representation of a NAND<< program which produces
an output on \(x\) within at most \(T\) steps. If \(P\) does not produce an
output within this time then \(TIMEDEVAL\) outputs an encoding of a
special `fail`

symbol. Moreover, for every program \(P\), the running time
of \(U\) on input \(P,x,1^T\) is \(O(T)\). (The hidden constant in the
\(O\)-notation can depend on the program \(P\) but is at most polynomial in
the length of \(P\)’s description as a string.).

The function \(TIMEDEVAL\) has a curious feature - its third input has the form \(1^T\). We use this notation to indicate a string of \(T\) ones. (For example, if we write \(1^5\) we mean the string \(11111\) rather using this to mean the integer one to the fifth power, which is a cumbersome way to write one; it will be always clear from context whether a particular input is an integer or such a string.)

Why don’t we simply assume that the function \(TIMEDEVAL\) gets the
integer \(T\) in the binary representation? The reason has to do with how
we define time as a function of the input length. If we represent \(T\) as
a string using the binary basis, then its length will be \(O(\log T)\),
which means that we could not say that a program that takes \(O(T)\) steps
to compute \(TIMEDEVAL\) would actually be considered as running in time
*exponential* in its input. Thus, when we want to allow programs to run
in time that is polynomial (as opposed to logarithmic) in some parameter
\(m\), we will often provide them with input of the form \(1^m\) (i.e., a
string of ones of length \(m\)).

There is nothing deep about representing inputs this way: this is merely a convention. However, it is a widely used one, especially in cryptography, and so is worth getting familiar with.

Before reading the proof of univ-nandpp, try to think how you
would compute \(TIMEDEVAL\) using your favorite programming language. That
is, how you would write a program `Timed_Eval(P,x,T)`

that gets a
NAND<< program `P`

(represented in some convenient form), a string
`x`

, and an integer `T`

, and simulates `P`

for `T`

steps. You will
likely find that your program requires \(O(T)\) steps to perform this
simulation. As in the case of polyRAMTM-thm, the proof of
univ-nandpp is not very deep and it more important to
understand its *statement*. If you understand how you would go about
writing an interpreter for NAND<< using a modern programming language
such as Python, then you know everything you need to know about this
theorem.

To present a universal NAND<< program in full we would need to describe a precise representation scheme, as well as the full NAND<< instructions for the program. While this can be done, it is more important to focus on the main ideas, and so we just sketch the proof here. A specification of NAND<< is given in the Appendix, and for the purposes of this simulation, we can simply use the representation of the code NAND<< as an ASCII string.

The program \(U\) gets as input a NAND<< program \(P\), an input \(x\), and a time bound \(T\) (given in the form \(1^T\)) and needs to simulate the execution of \(P\) for \(T\) steps. To do so, \(U\) will do the following:

- \(U\) will maintain variables
`current_line`

, and`number_steps`

for the current line to be executed and the number of steps executed so far. - \(U\) will scan the code of \(P\) to find the number \(t\) of unique
variable names that \(P\) uses. If we denote these names by
\(var_0,\ldots,var_{t-1}\) then \(U\) maintains an array
`Var_numbers`

that contains a list of pairs of the form \((var_s,s)\) for \(s\in [t]\). Using`Var_numbers`

we can translate the name of a variable to a number in \([t]\) that corresponds to its index. - \(U\) will maintain an array
`Var_values`

that will contain the current values of all \(P\)’s variables. If the \(s\)-th variable of \(P\) is a scalar variable, then its value will be stored in location`Var_values[`

\(s\)`]`

. If it is an array variable then the value of its \(i\)-th element will be stored in location`Var_values[`

\(t\cdot i + s\)`]`

. - To simulate a single step of \(P\), the program \(U\) will recover the
line corresponding to
`line_counter`

and execute it. Since NAND<< has a constant number of arithmetic operations, we can simulate choosing which operation to execute with a sequence of a constantly many if-then-else’s. When executing these operations, \(U\) will use the variable`step_counter`

that keeps track of the iteration counter of \(P\).

Simulating a single step of \(P\) will take \(O(|P|)\) steps for the program \(U\) where \(|P|\) is the length of the description of \(P\) as a string (which in particular is an upper bound on the number \(t\) of variable \(P\) uses). Hence the total running time will be \(O(|P|T)\) which is \(O(T)\) when suppressing constants that depend on the program \(P\).

To be a little more concrete, here is some “pseudocode” description of
the program \(U\):

We have seen that there are uncomputable functions, but are there
functions that can be computed, but only at an exorbitant cost? For
example, is there a function that *can* be computed in time \(2^n\), but
*can not* be computed in time \(2^{0.9 n}\)? It turns out that the answer
is **Yes**:

For every nice function \(T\), there is a function
\(F:\{0,1\}^* \rightarrow \{0,1\}\) in
\(TIME(T(n)\log n) \setminus TIME(T(n))\).

Note that in particular this means that \(\mathbf{P}\) is *strictly
contained* in \(\mathbf{EXP}\).

In the proof of halt-thm (the uncomputability of the Halting problem), we have shown that the function \(HALT\) cannot be computed in any finite time. An examination of the proof shows that it gives something stronger. Namely, the proof shows that if we fix our computational budget to be \(T\) steps, then not only we can’t distinguish between programs that halt and those that do not, but cannot even distinguish between programs that halt within at most \(T'\) steps and those that take more than that (where \(T'\) is some number depending on \(T\)). Therefore, the proof of time-hierarchy-thm follows the ideas of the uncomputability of the halting problem, but again with a more careful accounting of the running time.

If you fully understand the proof of halt-thm, then reading the following proof should not be hard. If you don’t, then this is an excellent opportunity to review this reasoning.

Recall the Halting function \(HALT:\{0,1\}^* \rightarrow \{0,1\}\) that was defined as follows: \(HALT(P,x)\) equals \(1\) for every program \(P\) and input \(x\) s.t. \(P\) halts on input \(x\), and is equal to \(0\) otherwise. We cannot use the Halting function of course, as it is uncomputable and hence not in \(TIME(T'(n))\) for any function \(T'\). However, we will use the following variant of it:

We define the *Bounded Halting* function \(HALT_T(P,x)\) to equal \(1\) for
every NAND<< program \(P\) such that \(|P| \leq \log \log |x|\), and such
that \(P\) halts on the input \(x\) within \(100 T(|x|)\) steps. \(HALT_T\)
equals \(0\) on all other inputs.

time-hierarchy-thm is an immediate consequence of the following two claims:

**Claim 1:** \(HALT_T \in TIME(T(n)\ log n)\)

and

**Claim 2:** \(HALT_T \not\in TIME(T(n))\).

Please make sure you understand why indeed the theorem follows directly from the combination of these two claims. We now turn to proving them.

**Proof of claim 1:** We can easily check in linear time whether an
input has the form \(P,x\) where \(|P| \leq \log\log |x|\). Since \(T(\cdot)\)
is a nice function, we can evaluate it in \(O(T(n))\) time. Thus, we can
perform the check above, compute \(T(|P|+|x|)\) and use the universal
NAND<< program of univ-nandpp to evaluate \(HALT_T\) in at
most \(poly(|P|) T(n)\) steps.

**Proof of claim 2:** This proof is the heart of
time-hierarchy-thm, and is very reminiscent of the proof that
\(HALT\) is not computable. Assume, toward the sake of contradiction, that
there is some NAND<< program \(P^*\) that computes \(HALT_T(P,x)\) within
\(T(|P|+|x|)\) steps. We are going to show a contradiction by creating a
program \(Q\) and showing that under our assumptions, if \(Q\) runs for less
than \(T(n)\) steps when given (a padded version of) its own code as input
then it actually runs for more than \(T(n)\) steps and vice versa. (It is
worth re-reading the last sentence twice or thrice to make sure you
understand this logic. It is very similar to the direct proof of the
uncomputability of the halting problem where we obtained a contradiction
by using an assumed “halting solver” to construct a program that, given
its own code as input, halts if and only if it does not halt.)

We will define \(Q\) to be the program that on input a string \(z\) does the following:

- If \(z\) does not have the form \(z=P1^m\) where \(P\) represents a NAND<< program and \(|P|< 0.1 \log\log m\) then return \(0\).
- Compute \(b= P^*(P,z)\) (at a cost of at most \(T(|P|+|z|)\) steps, under our assumptions).
- If \(b=1\) then \(Q\) goes into an infinite loop, otherwise it halts.

We chose \(m\) sufficiently large so that \(|Q| < 0.001\log\log m\) where \(|Q|\) denotes the length of the description of \(Q\) as a string. We will reach a contradiction by splitting into cases according to whether or not \(HALT_T(Q,Q1^m)\) equals \(0\) or \(1\).

On the one hand, if \(HALT_T(Q,Q1^m)=1\), then under our assumption that
\(P^*\) computes \(HALT_T\), \(Q\) will go into an infinite loop on input
\(z=Q1^m\), and hence in particular \(Q\) does *not* halt within
\(100 T(|Q|+m)\) steps on the input \(z\). But this contradicts our
assumption that \(HALT_T(Q,Q1^m)=1\).

This means that it must hold that \(HALT_T(Q,Q1^m)=0\). But in this case, since we assume \(P^*\) computes \(HALT_T\), \(Q\) does not do anything in phase 3 of its computation, and so the only computation costs come in phases 1 and 2 of the computation. It is not hard to verify that Phase 1 can be done in linear and in fact less than \(5|z|\) steps. Phase 2 involves executing \(P^*\), which under our assumption requires \(T(|Q|+m)\) steps. In total we can perform both phases in less than \(10 T(|Q|+m)\) in steps, which by definition means that \(HALT_T(Q,Q1^m)=1\), but this is of course a contradiction. This completes the proof of Claim 2 and hence of time-hierarchy-thm.

The time hierarchy theorem tells us that there are functions we can
compute in \(O(n^2)\) time but not \(O(n)\), in \(2^n\) time, but not
\(2^{\sqrt{n}}\), etc.. In particular there are most definitely functions
that we can compute in time \(2^n\) but not \(O(n)\). We have seen that we
have no shortage of natural functions for which the best *known*
algorithm requires roughly \(2^n\) time, and that many people have
invested significant effort in trying to improve that. However, unlike
in the finite vs. infinite case, for all of the examples above at the
moment we do not know how to rule out even an \(O(n)\) time algorithm. We
will however see that there is a single unproven conjecture that would
imply such a result for most of these problems.

The time hierarchy theorem relies on the existence of an efficient universal NAND<< program, as proven in univ-nandpp. We have mentioned that other models, such as NAND++ programs and Turing machines, are polynomially

We have now seen two measures of “computation cost” for functions. For a
finite function \(G:\{0,1\}^n \rightarrow \{0,1\}^m\), we said that
\(G\in SIZE(T)\) if there is a \(T\)-line NAND program that computes \(G\). We
saw that *every* function mapping \(\{0,1\}^n\) to \(\{0,1\}^m\) can be
computed using at most \(O(m2^n)\) lines. For infinite functions
\(F:\{0,1\}^* \rightarrow \{0,1\}^*\), we can define the “complexity” by
the smallest \(T\) such that \(F \in TIME(T(n))\). Is there a relation
between the two?

For simplicity, let us restrict attention to Boolean (i.e., single-bit output) functions \(F:\{0,1\}^* \rightarrow \{0,1\}\). For every such function, define \(F_n : \{0,1\}^n \rightarrow \{0,1\}\) to be the restriction of \(F\) to inputs of size \(n\). We have seen two ways to define that \(F\) is computable within a roughly \(T(n)\) amount of resources:

- There is a
*single algorithm*\(P\) that computes \(F\) within \(T(n)\) steps on all inputs of length \(n\). In such a case we say that \(F\) is*uniformly*computable (or more often, simply “computable”) within \(T(n)\) steps. - For every \(n\), there is a \(T(n)\) NAND program \(Q_n\) that computes
\(F_n\). In such a case we say that \(F\) has can be computed via a
*non uniform*\(T(n)\) bounded sequence of algorithms.

Unlike the first condition, where there is a single algorithm or “recipe” to compute \(F\) on all possible inputs, in the second condition we allow the restriction \(F_n\) to be computed by a completely different program \(Q_n\) for every \(n\). One can see that the second condition is much more relaxed, and hence we might expect that every function satisfying the first condition satisfies the second one as well (up to a small overhead in the bound \(T(n)\)). This indeed turns out to be the case:

There is some \(c\in \N\) s.t. for every nice \(T:\N \rightarrow \N\) and \(F:\{0,1\}^* \rightarrow \{0,1\}\) in \(TIME_{++}(T(n))\) and every sufficiently large \(n\in N\), \(F_n\) is in \(SIZE(c T(n))\).

To prove non-uniform-thm we use the technique of “unraveling the loop”. That is, we can use “copy paste” to replace a program \(P\) that uses a loop that iterates for at most \(T\) times with a “loop free” program that has about \(T\) times as many lines as \(P\).

Let us give an example using C-like syntax. Suppose we had a program of the form:

and we had the guarantee that the program would iterate the loop for at most \(4\) times before it breaks.

Then we could change it to an equivalent loop-free program of the following form:

That is all there is to the proof of non-uniform-thm

The proof follows by the argument of “unraveling the loop”. If \(P\) is a NAND++ program of \(L\) lines and \(T:\N \rightarrow \N\) is a function such that for every input \(x\in \{0,1\}^n\), \(P\) halts after executing at most \(T(n)\) lines (and hence iterating at most \(\floor{T(n)/L}\) times) then we can obtain a NAND program \(Q\) on \(n\) inputs as follows:

where for every number \(j\), we denote by `P〈i<-`

\(j\)`〉`

the NAND
program that is obtained by replacing all references of the form
`Foo[i]`

(which are allowed in NAND++, but illegal in NAND that has no
index variable `i`

) with references of the form `Foo[`

\(j\)`]`

(which are
allowed in NAND, since \(j\) is simply a number). Whenever we see a
reference to the variable `Xvalid[`

\(i\)`]`

in the program we will replace
it with `one`

or `zero`

depending on whether \(i<n\). Similarly, we will
replace all references to `X[`

\(i\)`]`

for \(i \geq n\) with `zero`

. (We can
use our standard syntactic sugar to create the constant `zero`

and `one`

variables.)

We simply repeat the lines of the form `IF (loop) P〈i<-`

\(j\)`〉`

for
\(\floor{T(n)/L}-1\) times, replacing each time \(j\) by \(0,1,0,1,2,\ldots\)
as in the definition of (standard or “vanilla”) NAND++ in
#vanillanandpp. We replace `IF`

with the appropriate
syntactic sugar, which will incur a multiplicative overhead of at most
\(4\) in the number of lines. After this replacement, each line of the
form `IF (loop) P〈i<-`

\(j\)`〉`

corresponds to at most \(4L\) lines of
standard sugar-free NAND. Thus the total cost is at most
\(4L \cdot (\tfrac{T(n)}{L}) \leq 4 \cdot T(n)\) lines.

By combining non-uniform-thm with polyRAMTM-thm, we get that if \(F\in TIME(T(n))\) then there are some constants \(a,b\) such that for every large enough \(n\), \(F_n \in SIZE(aT(n)^b)\). (In fact, by direct inspection of the proofs we can see that \(a=b=5\) would work.)

The proof of non-uniform-thm is *algorithmic*, in the sense
that the proof yields a polynomial-time algorithm that given a NAND++
program \(P\) and parameters \(T\) and \(n\), produces a NAND program \(Q\) of
\(O(T)\) lines that agrees with \(P\) on all inputs \(x\in \{0,1\}^n\) (as
long as \(P\) runs for less than \(T\) steps these inputs.) Thus the same
proof gives the following theorem:

There is an \(O(n)\)-time NAND<< program \(COMPILE\) such that on input a NAND++ program \(P\), and strings of the form \(1^n,1^m,1^T\) outputs a NAND program \(Q_P\) of at most \(O(T)\) lines with \(n\) bits of inputs and \(m\) bits of output satisfying the following property.

For every \(x\in\{0,1\}^n\), if \(P\) halts on input \(x\) within fewer than \(T\) steps and outputs some string \(y\in\{0,1\}^m\), then \(Q_P(x)=y\).

We omit the proof of the nand-compiler since it follows in a
fairly straightforward way from the proof of non-uniform-thm.
However, for the sake of concreteness, here is a *Python* implementation
of the function \(COMPILE\). (The reader can feel free to skip it.)

For starters, let us consider an imperfect but very simple program that
unrolls the loop. The following program will work correctly for the case
that \(m=1\) and that the underlying NAND++ program had the property that
it only modifies the value of the `Y[0]`

variable once. (A property that
can be ensured by adding appropriate flag variables and some `IF`

syntactic sugar.)

The `index`

function takes a number \(t\) and returns the value of the
index variable `i`

in the \(t\)-th iteration. Recall that this value in
NAND++ follows the sequence \(0,1,0,1,2,1,0,1,2,\ldots\) and it can be
computed in Python as follows:

Below is a more “robust” implementation of `COMPILE`

, that works for an
arbitrarily large number of outputs, and makes no assumptions on the
structure of the input program.

Since NAND<< programs can be simulated by NAND++ programs with polynomial overhead, we see that we can simulate a \(T(n)\) time NAND<< program on length \(n\) inputs with a \(poly(T(n))\) size NAND program.

To make sure you understand this transformation, it is an excellent exercise to verify the following equivalent characterization of the class \(\mathbf{P}\) (see Palternativeex). Prove that for every \(F:\{0,1\}^* \rightarrow \{0,1\}\), \(F\in \mathbf{P}\) if and only if there is a polynomial-time NAND++ (or NAND<<, it doesn’t matter) program \(P\) such that for every \(n\in \N\), \(P(1^n)\) outputs a description of an \(n\) input NAND program \(Q_n\) that computes the restriction \(F_n\) of \(F\) to inputs in \(\{0,1\}^n\). (Note that since \(P\) runs in polynomial time and hence has an output of at most polynomial length, \(Q_n\) has at most a polynomial number of lines.)

We can define the “non uniform” analog of the class \(\mathbf{P}\) as follows:

For every \(F:\{0,1\}^* \rightarrow \{0,1\}\), we say that \(F\in \mathbf{P_{/poly}}\) if there is some polynomial \(p:\N \rightarrow \R\) such that for every \(n\in \N\), \(F_n \in SIZE(p(n))\) where \(F_n\) is the restriction of \(F\) to inputs in \(\{0,1\}^n\).

non-uniform-thm implies that \(\mathbf{P} \subseteq \mathbf{P_{/poly}}\).

Please make sure you understand why this is the case.

Using the equivalence of NAND programs and Boolean circuits, we can also define \(P_{/poly}\) as the class of functions \(F:\{0,1\}^* \rightarrow \{0,1\}\) such that the restriction of \(F\) to \(\{0,1\}^n\) is computable by a Boolean circuit of \(poly(n)\) size (say with gates in the set \(\wedge,\vee,\neg\) though any universal gateset will do); see Ppolyfig.

The notation \(\mathbf{P_{/poly}}\) is used for historical reasons. It was
introduced by Karp and Lipton, who considered this class as
corresponding to functions that can be computed by polynomial-time
Turing Machines (or equivalently, NAND++ programs) that are given for
any input length \(n\) a polynomial in \(n\) long *advice string*. That this
is an equivalent characterization is shown in the following theorem:

Let \(F:\{0,1\}^* \rightarrow \{0,1\}\). Then \(F\in\mathbf{P_{/poly}}\) if and only if there exists a polynomial \(p:\N \rightarrow \N\), a polynomial-time NAND++ program \(P\) and a sequence \(\{ a_n \}_{n\in \N}\) of strings, such that for every \(n\in \N\):

- \(|a_n| \leq p(n)\)
- For every \(x\in \{0,1\}^n\), \(P(a_n,x)=F(x)\).

We only sketch the proof. For the “only if” direction, if \(F\in \mathbf{P_{/poly}}\) then we can use for \(a_n\) simply the description of the corresponding NAND program \(Q_n\), and for \(P\) the program that computes in polynomial time the \(NANDEVAL\) function that on input an \(n\)-input NAND program \(Q\) and a string \(x\in \{0,1\}^n\), outputs \(Q(n)\)>

For the “if” direction, we can use the same “unrolling the loop” technique of non-uniform-thm to show that if \(P\) is a polynomial-time NAND++ program, then for every \(n\in \N\), the map \(x \mapsto P(a_n,x)\) can be computed by a polynomial size NAND program \(Q_n\).

To make sure you understand the definition of \(\mathbf{P_{/poly}}\), I highly encourage you to work out fully the details of the proof of ppolyadvice.

non-uniform-thm shows that every function in \(TIME(T(n))\) is
in \(SIZE(poly(T(n)))\). One can ask if there is an inverse relation.
Suppose that \(F\) is such that \(F_n\) has a “short” NAND program for every
\(n\). Can we say that it must be in \(TIME(T(n))\) for some “small” \(T\)?
The answer is an emphatic **no**. Not only is \(\mathbf{P_{/poly}}\) not
contained in \(\mathbf{P}\), in fact \(\mathbf{P_{/poly}}\) contains
functions that are *uncomputable*!

There exists an *uncomputable* function
\(F:\{0,1\}^* \rightarrow \{0,1\}\) such that \(F \in \mathbf{P_{/poly}}\).

Since \(\mathbf{P_{/poly}}\) corresponds to non uniform computation, a function \(F\) is in \(\mathbf{P_{/poly}}\) if for every \(n\in \N\), the restriction \(F_n\) to inputs of length \(n\) has a small circuit/program, even if the circuits for different values of \(n\) are completely different from one another. In particular, if \(F\) has the property that for every equal-length inputs \(x\) and \(x'\), \(F(x)=F(x')\) then this means that \(F_n\) is either the constant function zero or the constant function one for every \(n\in \N\). Since the constant function has a (very!) small circuit, such a function \(F\) will always be in \(\mathbf{P_{/poly}}\) (indeed even in smaller classes). Yet by a reduction from the Halting problem, we can obtain a function with this property that is uncomputable.

Consider the following “unary halting function”
\(UH:\{0,1\}^* \rightarrow \{0,1\}\) defined as follows. We let
\(S:\N \rightarrow \{0,1\}^*\) be the function that on input \(n\in \N\),
outputs the string that corresponds to the binary representation of the
number \(n\) without the most significant \(1\) digit. Note that \(S\) is
*onto*. For every \(x\in \{0,1\}\), we define \(UH(x)=HALTONZERO(S(|x|))\).
That is, if \(n\) is the length of \(x\), then \(UH(x)=1\) if and only if the
string \(S(n)\) encodes a NAND++ program that halts on the input \(0\).

\(UH\) is uncomputable, since otherwise we could compute \(HALTONZERO\) by
transforming the input program \(P\) into the integer \(n\) such that
\(P=S(n)\) and then then running \(UH(1^n)\) (i.e., \(UH\) on the string of
\(n\) ones). On the other hand, for every \(n\), \(UH_n(x)\) is either equal
to \(0\) for all inputs \(x\) or equal to \(1\) on all inputs \(x\), and hence
can be computed by a NAND program of a *constant* number of lines.

The issue here is of course *uniformity*. For a function
\(F:\{0,1\}^* \rightarrow \{0,1\}\), if \(F\) is in \(TIME(T(n))\) then we
have a *single* algorithm that can compute \(F_n\) for every \(n\). On the
other hand, \(F_n\) might be in \(SIZE(T(n))\) for every \(n\) using a
completely different algorithm for every input length. For this reason
we typically use \(\mathbf{P_{/poly}}\) not as a model of *efficient*
computation but rather as a way to model *inefficient computation*. For
example, in cryptography people often define an encryption scheme to be
secure if breaking it for a key of length \(n\) requires more then a
polynomial number of NAND lines. Since
\(\mathbf{P} \subseteq \mathbf{P_{/poly}}\), this in particular precludes
a polynomial time algorithm for doing so, but there are technical
reasons why working in a non uniform model makes more sense in
cryptography. It also allows to talk about security in non asymptotic
terms such as a scheme having “\(128\) bits of security”.

While it can sometimes be a real issue, in many natural settings the difference between uniform and non-uniform computation does not seem to so important. In particular, in all the examples of problems not known to be in \(\mathbf{P}\) we discussed before: longest path, 3SAT, factoring, etc., these problems are also not known to be in \(\mathbf{P_{/poly}}\) either. Thus, for “natural” functions, if you pretend that \(TIME(T(n))\) is roughly the same as \(SIZE(T(n))\), you will be right more often than wrong.

To summarize, the two models of computation we have described so far are:

- NAND programs, which have no loops, can only compute finite
functions, and the time to execute them is exactly the number of
lines they contain. These are also known as
*straightline programs*or*Boolean circuits*. - NAND++ programs, which include loops, and hence a single program can
compute a function with unbounded input length. These are equivalent
(up to polynomial factors) to
*Turing Machines*or (up to polylogarithmic factors) to*RAM machines*.

For a function \(F:\{0,1\}^* \rightarrow \{0,1\}\) and some nice time bound \(T:\N \rightarrow \N\), we know that:

- If \(F\) is computable in time \(T(n)\) then there is a sequence \(\{ P_n \}\) of NAND programs with \(|P_n| = poly(T(n))\) such that \(P_n\) computes \(F_n\) (i.e., restriction of \(F\) to \(\{0,1\}^n\)) for every \(n\).
- The reverse direction is not necessarily true - there are examples of functions \(F:\{0,1\}^n \rightarrow \{0,1\}\) such that \(F_n\) can be computed by even a constant size NAND program but \(F\) is uncomputable.

This means that non uniform complexity is more useful to establish
*hardness* of a function than its *easiness*.

We have mentioned the Church-Turing thesis, that posits that the
definition of computable functions using NAND++ programs captures the
definition that would be obtained by all physically realizable computing
devices. The *extended* Church Turing is the statement that the same
holds for *efficiently computable* functions, which is typically
interpreted as saying that NAND++ programs can simulate every physically
realizable computing device with polynomial overhead.

In other words, the extended Church Turing thesis says that for every
*scalable computing device* \(C\) (which has a finite description but can
be in principle used to run computation on arbitrarily large inputs),
there are some constants \(a,b\) such that for every function
\(F:\{0,1\}^* \rightarrow \{0,1\}\) that \(C\) can compute on \(n\) length
inputs using an \(S(n)\) amount of physical resources, \(F\) is in
\(TIME(aS(n)^b)\).

All the current constructions of scalable computational models and programming language conform to the Extended Church-Turing Thesis, in the sense that they can be with polynomial overhead by Turing Machines (and hence also by NAND++ or NAND<< programs). consequently, the classes \(\mathbf{P}\) and \(\mathbf{EXP}\) are robust to the choice of model, and we can use the programming language of our choice, or high level descriptions of an algorithm, to determine whether or not a problem is in \(\mathbf{P}\).

Like the Church-Turing thesis itself, the extended Church-Turing thesis
is in the asymptotic setting and does not directly yield an
experimentally testable prediction. However, it can be instantiated with
more concrete bounds on the overhead, which would yield predictions such
as the *Physical Extended Church-Turing Thesis* we mentioned before,
which would be experimentally testable.

In the last hundred+ years of studying and mechanizing computation, no
one has yet constructed a scalable computing device (or even gave a
convincing blueprint) that violates the extended Church Turing Thesis.
However, *quantum computing*, if realized, will pose a serious challenge
to this thesis.*proof* that they would offer
super polynomial advantage over “classical” computing devices.
However, there seems to be no fundamental physical obstacle to
constructing them, and there are strong reasons to conjecture that
they do in fact offer such an advantage.*not* a challenge to the (non extended)
Church Turing thesis, as a function is computable by a quantum
computer if and only if it is computable by a “classical” computer
or a NAND++ program. It is only the running time of computing the
function that can be affected by moving to the quantum model.

- We can define the time complexity of a function using NAND++ programs, and similarly to the notion of computability, this appears to capture the inherent complexity of the function.
- There are many natural problems that have polynomial-time algorithms, and other natural problems that we’d love to solve, but for which the best known algorithms are exponential.
- The definition of polynomial time, and hence the class \(\mathbf{P}\), is robust to the choice of model, whether it is Turing machines, NAND++, NAND<<, modern programming languages, and many other models.
- The time hierarchy theorem shows that there are
*some*problems that can be solved in exponential, but not in polynomial time. However, we do not know if that is the case for the natural examples that we described in this lecture. - By “unrolling the loop” we can show that every function computable in time \(T(n)\) can be computed by a sequence of NAND programs (one for every input length) each of size at most \(poly(T(n))\)

Most of the exercises have been written in the summer of 2018 and haven’t yet been fully debugged. While I would prefer people do not post online solutions to the exercises, I would greatly appreciate if you let me know of any bugs. You can do so by posting a GitHub issue about the exercise, and optionally complement this with an email to me with more details about the attempted solution.

Prove that the classes \(\mathbf{P}\) and \(\mathbf{EXP}\) defined in PandEXPdef are equal to \(\cup_{c\in \{1,2,3,\ldots \}} TIME(n^c)\) and \(\cup_{c\in \{1,2,3,\ldots \}} TIME(2^{n^c})\) respectively. (If \(S_1,S_2,S_3,\ldots\) is a collection of sets then the set \(S = \cup_{c\in \{1,2,3,\ldots \}} S_c\) is the set of all elements \(e\) such that there exists some \(c\in \{ 1,2,3,\ldots \}\) where \(e\in S_c\).)

For every function \(F:\{0,1\}^* \rightarrow \{0,1\}^*\), define \(Bool(F)\) to be the function mapping \(\{0,1\}^*\) to \(\{0,1\}\) such that on input a (string representation of a) triple \((x,i,\sigma)\) with \(x\in \{0,1\}^*\), \(i \in \N\) and \(\sigma \in \{0,1\}\),

\[ Bool(F)(x,i,\sigma) = \begin{cases} F(x)_i & \sigma =0, i<|F(x)| \\ 1 & \sigma = 1,i<|F(x)| \\ 0 & \text{otherwise} \end{cases} \] where \(F(x)_i\) is the \(i\)-th bit of the string \(F(x)\).

Prove that \(F \in \overline{\mathbf{P}}\) if and only if \(Bool(F) \in \mathbf{P}\).

Prove that if \(F,G:\{0,1\}^* \rightarrow \{0,1\}^*\) are in
\(\overline{\mathbf{P}}\) then their *composition* \(F\circ G\), which is
the function \(H\) s.t. \(H(x)=F(G(x))\), is also in
\(\overline{\mathbf{P}}\).

Prove that there is some \(F,G:\{0,1\}^* \rightarrow \{0,1\}^*\) s.t.
\(F,G \in \overline{\mathbf{EXP}}\) but \(F\circ G\) is not in
\(\mathbf{EXP}\).

We say that a NAND++ program \(P\) is oblivious if there is some functions
\(T:\N \rightarrow \N\) and \(i:\N\times \N \rightarrow \N\) such that for
every input \(x\) of length \(n\), it holds that:

* \(P\) halts when given input \(x\) after exactly \(T(n)\) steps.

* For \(t\in \{1,\ldots, T(n) \}\), after \(P\) executes the \(t^{th}\) step
of the execution the value of the index `i`

is equal to \(t(n,i)\). In
particular this value does *not* depend on \(x\) but only on its
length.*oblivious* NAND++ program \(P'\)
that computes \(F\) in time \(O(T^2(n) \log T(n))\).

Prove that for every \(F:\{0,1\}^* \rightarrow \{0,1\}\), \(F\in \mathbf{P}\) if and only if there exists a polynomial time NAND++ program \(P\) such that \(P(1^n)\) outputs a NAND program \(Q_n\) that computes the restriction of \(F\) to \(\{0,1\}^n\).

Some topics related to this chapter that might be accessible to advanced students include: (to be completed)

Copyright 2018, Boaz Barak.

This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.

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