Halting Problem--Week 11

This week we discuss that there are some problems in the world that can not solved by algorithms. That is there are uncomputable problems existed for any reasonable computational model. Consider the following halting problem:

def half(f,i) :
      '''return True iff f(i) eventually halts''

we are told that this halt function can not to be implemented. Here is the proof:

assume halt exist.and there are such function exist below:

def confuse(f):
      def halt(f,i):
           #some code here
      if halt(f,f):
          while True:
                    pass
     else:
          return False

we first assume confuse(confuse) halts, then halt(confuse,confuse) is True, then we enter the infinite while loop.
Therefore, we assume confuse(confuse) halts , but got the result of not half

we then assume confuse(confuse) doesn't halts, then we enter the else, which return False.
once again we assume confuse(confuse) doesn't halts but it halts.

so no matter which case we consider, we got the contradiction, that is halt doesn't exist.

Disprove Big-Oh ---Week 10

last week, we introduce the big-Oh prove, this week we disprove the big-Oh.
Here is one of the examples:

Disprove n^3  O(3n^2)

we first negate  cR+,  BN, nN, n>=B, n^3<=c(3n^2)

  cR+,  BN, nN, n>=B^( n^3)>c(3n^2)


proof:

assume c belongs to R+ and B belongs to N, (Generic c,B)
       pick n=B+ceil(3c)+1  (then n belongs to N)
       then n>=b
     
       then n>=B
       then n^3=n*n^2
       then n^3>3c*n^2 (since n>3c)
       then (n>=B)^ n^3>3cn^2

then  cR+,  BN, nN, n>=B^( n^3)>c(3n^2)

Big-oh of polynomials--Week 9

This week we introduce the big-Oh of polynomials. First, we introduce what is big-Oh O(n^2)? It is a polynomial function that  take a natural number and return a positive real number such that
 cR+,  BN, N, n>=B, 3n^2+2n<=c(n^2). This may look hard to understand, so we can give a simple example of big-Oh proof.

Prove 3n^2+2n O(n^2)

 cR+,  BN, nN, n>=B, 3n^2+2n<=c(n^2)

Pick c=5 ,then CR+
pick b=0, then  BN
assume nN,
    assume n>=B
         then 3n^2+2n<= 3n^2+2n^2
                                 =5n^2
                                =cn^2



     then n>=B ⇒ 3n^2+2n<=cn^2
then BN, then n>=B ⇒ 3n^2+2n<=cn^2
then  cR+,  BN, nN, n>=B, 3n^2+2n<=c(n^2)

Counting Time Complexity--Week 8

This week we start to learn the interesting and harder staff-----Time Complexity.

I will give a simple linear search python function to talk about the time complexity.

ex:

def  LS(A,x):
       '''Return an index i such that x==L[i]. Otherwise, return -1.'''
       i=0                                 (Line 1)
       while i<len(A):             (Line 2)
              if A[i]==x:             (Line 3)
                    return i             (Line 4)  
              i=i+1                      (Line 5)
       return -1                        (Line 6)

so, if we give an input (A=[2,4,6,8], x=4), what will the time complexity be:

we can trace our code like this:

Line 1: 1 step (i=0)
Line 2: 1 step (0<4)
Line 3: 1 step (A[0]==4)
Line 5: 1 step (i=1)

then again, we go to while loop

Line 2: 1 step (1<4)
Line 3:1 step (A[1]==4)
Line 4: 1 step ( return 1)

so our python function ends here, if we count the total time complexity, it will be 1+1+1+1+1+1+1=7 steps

We can even turn this question into a harder question: what is the worst case in which x isn't in A?

then we can think like this :
line 1 : 1 step (i=0)

while loop : 3* n   (line 2,3,5 executes once for each 0 to len(A)-1 , which line 2,3,5 executes once for n times)

then we have 2 more steps to add, which is last while loop check and the last line 'return -1')

so in worst case, the time complexity is  3n+1.

Inference Rules-Week 7

Up to today, we have learned some rules of inference to allow us to reach conclusions about certain situations. Here are some of them:

Conjunction Elimination: if we know A^B, we can conclude A and B are both True.

Existential Elimination: if we know that there exist a belongs to X, P(a), then we can certainly pick an element with that property. Pick a' belongs to X, then p(a').

Disjunction Elimination: if we know that A V B, and we know that not(A), we can conclude that B must be True.

Implication Elimination: if we know A implies B, and we also know that A is True, we can conclude that B is also True. On the other hand, if we know not(B), we must conclude that not(A) must be True.

Universal Elimination: if we know that for all x belongs to X, P(x), then if we choose any a belongs to X, we can conclude that P(a) must be True.

Implication introduction:if we assume A and then B follows, then we can conclude that A implies B

Universal Introduction: if we assume that a is generic element of D, and we derive P(a), then we can conclude that for all a belongs to D, P(a).

Existential Introduction:if we can show that x belongs to X, P(x), then we can conclude that there exist x belongs to X, such that P(x).

Conjunction introduction:if we know that A and we know B, then we can conclude A^B

Disjunction introduction: If we know A, we can immediately conclude that A v B