# Steb CT Quick Notes by wuzhengqin

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```									Computational Theory Quick Notes
1. Complexity
When picking an algorithm there are 2 contradicting goals. 1. Ease of understanding, debugging and
coding. 2. Efficiency.

Run Time
Consider: 1. Input to the program. 2. Quality of code generated by the compiler. 3. Nature and speed
of machine code. 4. Time complexity of the algorithm.

RUNNING TIME: T(n)
This is not on the input itself but the size of the input.

SPACE REQUIREMENTS: S(n)
Linear Search worst case T(n) = n. Binary Search Worst Case on sorted input T(n) = 4.

Big O-Notation
This places an upper bound on the running time of an algorithm, stated as a function of its input.

When doing this, always ask what happens as n approaches infinity.

The M is a constant. N0 is the threshold. The most interesting part of this is the f(n) part as it governs
growth. The lower the order the better. If A is O(f(n)) and B is O(g(n)) then A is considered to be
lower. Lowest order will not give you exactly how much time or space is needed to solve a problem
but shows us, for problems over a certain size, the lower order algorithm will take less time or space
than a higher order algorithm. Must be careful as it’s an upper bound and the quantity in question
might be much lower, input that causes the worst case may be unlikely to occur in practise, M and n0
are unknown and need not be small.

Linear Search: T(n) = n

O(n) = m.n, ∀n > n0

Ignore m & n0 as constant factors are of no interest. Larger constant factor may change
graph but not the characteristics of an equation. In some cases algorithms are sensitive to
the permutations of input ordering.

The statement that the running time of an algorithm is O(f (n))does not imply that the algorithm
ever takes that long: it only says that the analyst has been able to show that it never takes longer.
The actual running time might always be much lower. Even if the worse-case input is known, it can
be the case that the inputs encountered in practice lead to much lower running times.
Many extremely useful algorithms have a bad worst-case. For example, Quicksort has a running time
of O(n2), but it is possible to arrange things so that the running time of inputs encountered in
practice is closer to O(n log n). The constants n0 and M, implicit in the O-notation, often hide
implementation details that are important in practice. An algorithm with running time O(f (n)) says
nothing about the running time if n < n0, and M might be hiding a large amount of overhead
designed to avoid a bad worst-case. We would prefer an algorithm using N2 nanoseconds rather
than log N centuries, but we cannot make that choice based on O-notation.

Analysing an Algorithm
Various runtime properties of algorithms: frequency of instruction execution

1. Executed Frequency (Loops)
2. Executed Once (Initialization Time)
3. Never Executed (Branching [if/else])

Definition: (i) It is central to the function of the algorithm, and its behaviour typifies the overall
behaviour of the algorithm (ii) It is inside the main loops of the algorithm, and is executed as often as
any other part of the algorithm.

Sequential Search:
Algorithm if O(n). And so is of Linear Complexity

As input Size Doubles so does the worst case scenario

Binary Search:

The critical operation is the
comparison (L[mid] = key)
In the worst case, the max
number of comparisons will be
k, where k is the first integer
such that 2k _ n
In other words k = |log2n|, i.e.
the complexity of Binary
search is O(log2n). It has a
logarithmic complexity
Since log2n grows more slowly
than n, we know that above a
certain size of list, Binary
search will always run faster.
For Binary Search, data must be sorted for it to work.

Comparison of O(n) and O(log2 n)
A constant time algorithm, written O(1)

Bubble Sort

Selection Sort
First find smallest element then swap it with element in first position. Then repeat swapping the
elements with that in the next lowest position (1st time in 1st position, then 2nd, 3rd .....)
void selectionSort(int arr[], int n) {
int i, j, minIndex, tmp;
for (i = 0; i < n - 1; i++) {
minIndex = i;
for (j = i + 1; j < n; j++) {
if (arr[j] < arr[minIndex]) {
minIndex = j;
}
}
if (minIndex != i) {
tmp = arr[i];
arr[i] = arr[minIndex];
arr[minIndex] = tmp;
}
}
}
Outer Loop = n
Inner Loop = n – i
Selection Sort = T(n)
In the worst case we can expect every element will need to be swapped which is (n - 1) swaps.

To analyse an algorithm first identify looping bounds and the number of loops.
Then derived a math. Estimate for the work in each of these loops:
n
∑ (n - i)
i=1

Derived T(n) estimate (actual Work) : T(n) = n2/2 – n/2

Hamiltonian Cycle Problem (HCP0)in G
Each node can only be visited once.
Enumerate of possible permutations of G. Evaluate each permutation to see if a H.C. exists along the
path.
V1-V2-V3- so on.
By Definition a H.C in G visits each node and returns to starting node. Each node must have at least
one edge. Complete graph with n nodes is one in which each node is connected to every other node.
As there are only 2 routes to each node and you can only move along an unused route you only have
one usable route. As you only have one route each time you must eventually return to the root
node.

Travelling Salesman (TSP)
Enumerate all possible paths in G. Examine each path for optimization criteria. (max length/ min
length/ lowest cost)
Permutation Problem: Find all possible permutations of elements in set. Complexity of n!
Simplifications of complexity.
1 Graph symmetric in nature e.g. cost (b-a) = cost (a-b). 123 = 312
2 Starting node is not important when considering a given path in G.

Models of Computation
Not all problems can be solved by computers and not all computations are equal in complexity.
Finite State Machines

M= [S, I, O, Fs, Fo]

S = set of states

I = Input Alphabet

O = Output Alphabet

Fs                                    Fo

By Default machine boots into state 0. Ignore first output as it’s the default output of the boot.
There must be no unlabeled transitions, can be no 2 or more transitions from state Sn with the same
input. We look to minimize number of states in machine. FSM’s mostly contain cycles. A FSM
without a cycle is limited to input sequences with length equal to its longest path. A regular set is a
compact way to describe the sets.

Kleene’s Theorem

Any set recognised by a FSM is regular, and any regular set can be recognised by some FSM
Regular sets are exactly the sets FSMs are capable of recognising It follows that if a set is not regular,
there is no FSM which can recognise it Ex: The set S = {0n1n|n _ 0}, where 0n is the string
containing n 0’s. So this is the set of strings which have a specific number of 0’s followed by the
same number of 1’s.
Infinite -> FSM has a looping structure if we id a test for regularity we would be able to test if a given
string can be accepted by some FSM. FSM fail to accept strings which are not regular. FSM model of
computation is therefore deficient. We need some test to determine regularity.

Our Deterministic Finite Automaton (DFA) is denoted by a quintuple A = (Q,P, _, s, F) where
Q is a finite set of states; P is a finite input alphabet; _ is a transition function from Q × P to Q;
s 2 Q is the initial state of the automaton; F _ Q is the set of favourable states. If the automaton,
being in state q, has read the symbol a 2 P, it enters the state q′ = _(q, a). The new state is
completely determined by the content of the cell and the internal state of the automaton.
At some moment the reading head reaches the end of the input word a If at this moment the
automaton is in a favourable state q 2 F, the input word is said to be accepted by the
automaton. Otherwise the input word is not accepted. The set of all input words accepted by the
automaton A is called the language accepted by A .We denote this language L(A). As we have seen, a
convenient way to represent the finite automaton is a finite state diagram. Favourable states are
circled The transition diagram shown in Fig. 1 accepts the language containing all the strings anbm
for n = 0, 1, 2, ... and m = 1, 2, ....

The Pumping Lemma

Some languages are not regular and not amenable to the solution via FSM. How do we determine
the regularity of a language? Using the P.L. we can demonstrate that if a regular language containing
long strings it must contain an infinite set of strings of a particular form. If a language can be shown
not to posses strings of this form we there by demonstrate the language is not regular.
pigeonhole principle: the pigeonhole principle states that if n items are put into m pigeonholes with
n > m, then at least one pigeonhole must contain more than one item. Let L be a regular language.
There exists an integer n > 0 such that any string w 2 L, with length |w| _ n can be represented as
the concatenation xyz such that, The substring y is non-empty. |xy| _ n, and xyiz 2 L for each i _ 0. If
L is finite, then choose any n longer than the longest word in L and the theorem follows since there
are no words of length at least n. Suppose then that L is infinite. Let A be a finite automaton that
accepts L. Let n be the number of states in A. Consider any string w = w1w2...wm 2 L that has the
length m _ n. Consider a computation of A on the initial segment w1w2...wn of the string w. For any
k, 0 _ k _ n, let (qk,wk+1wk+2...wn) be the configuration of A after k steps of the computation. Since
A only has n states, and there are more than n configurations in the above fragment, by the
pigeonhole principle, there exist r and j, 0 _ r < j _ n, such that such that qr = qj. This means that the
string y = wr+1wr+2...wj brings the automaton A from state qr back to the same state. Note that the
string is non-empty, since r < j. Now, if we remove string y from the original string w, or insert yi
instead of y for any i, we get a string that will still be accepted by A. Thus, any string xyiz, where x =
w1w2...wr, y = wr+1wr+2...wj , and z = wj+1wj+2...wm, is accepted. Moreover, the total length of the
prefix x and the substring y does not exceed n.

Turing Machines
FSM are insufficient models of computation. Some algorithms that can’t be implemented using FSM.
For example recognising strings in L = {0n1n|n>=0}. To solve the problem or accepting strings in L the
computing model wound need to store at least the first half of strings in L. FSM cannot store their
input. Turing concluded FSM were inadequate and furthermore any model of computation must. A,
Erase & Write to some storage medium. B, Decision as to which character is read or written is based
on configuration of the machine. This leads to 2 main differences between FSM and TM. A, TM can
write & overwrite its input contents. B, Reread inputs already seen (rewind) or backup. In designing
we normally move the head right or left there is no concept of staying where you are.
All programming languages support 3 key features: A, Sequence Instructions. B, Selection (if). C,
Iteration (If, Else). Here the 1, 0 is the then.
(0, 0, 1, 0, R)
(0, 1, 1, 0, R)

TM: quintuple. To iterate just set the next state to the same as the current state.
As Function Computers 0 is encoded as 1
1 is encoded as 11, 2 is 111 .....

Church Turing
Any algorithmic procedure that can be carried out by a human being or by a computer can be carried
out by a Turing Machine. It is conceivable that it could be discarded in the future if someone can
identify a legitimate algorithm that cannot be implemented by a Turing machine. In the 70 or so
years since the thesis was postulated, this has not happened. Implementation 1. According to the
principle, if we have managed to derive even an informal or verbal description of a computational
procedure, this procedure can be implemented as a Turing Machine, and thus, as a computer
program. 2. Thus, to show that some function is computable, we need not write a full set of
transitions for our Turing Machine, it merely suffices to derive a clear and unambiguous description
of the algorithm acceptable to the theoretical computer science community.
Implementation 2. 1. Another implication of the thesis is perhaps even more important. 2. A Turing
machine is a mathematically precise model. 3. It opens a possibility to show that some problems
cannot be solved by any algorithm. 4. What we have to show is that there is no Turing machine that
can solve the problem in question.

Universal Turing Machine:
A universal Turing machine is a Turing machine that can simulate an arbitrary Turing machine on
arbitrary input. The universal machine essentially achieves this by reading both the description of
machine to be simulated as well as the input thereof from its own tape. A good example of a UTM is
a computer as it is the UTM in which other TM’s (applications and programs) run in.

The Halting Problem:
Can we design an algorithm halt that determines if a program will halt or not based on an input x?
No. If such an algorithm did exist it could be implemented as a program called, let’s say, HALT,
implemented in the same programming language as any other program P. This program, HALT(P,X)
would then take two parameters, P and X, and outputs 1 if P halted on X and 0 otherwise. We can
now slightly modify HALT to build another program which we call DIAGONAL(P). DIAGONAL(P) takes
one parameter the text of a program called P, and then calls HALT on input HALT(P, P). That is,
DIAGONAL calls HALT to see if the program P halts on input of program P. Finally, we build another
machine, which we call CONTR(P) that calls DIAGONAL as a module and does not halt if and only if
DIAGONAL(P) = 1 (that is, HALT determines in this case whether P halts on the input being its own
text).