(5,16,22,45,2,10,18,30,50,12,1)
Consider the insertion of items with this set of keys, in the order given, into :
a. An initially empty (2,4) tree T’
b. An initially empty red black tree T’’
(5,16,22,45,2,10,18,30,50,12,1)
Consider the insertion of items with this set of keys, in the order given, into :
a. An initially empty (2,4) tree T’
b. An initially empty red black tree T’’
As stated in the question one of the solutions is 01000110.
So assuming this is the first representation of 3/8 in floating –point format.
3/8 can be written as 0.011 in binary having decimal and integer part.
01000110 is 0110.0 × 2-4
So other solutions will be
01000011 is 0011.0 × 2-3
01100001
01110000
The last two solutions are not precise due to fewer bits in Mantissa. This is the limitation of floating point numbers. Preciseness can be improve by having representation in more bits .
Ans. If the Language L = an bcba2n is regular than it can be converted into string w where w = xyz . so we have that xyz = an bcba2n . Now since y ≠ € , there can be three cases .
Case 1 : Now assume y is having an instance of a and z will have the rest of bcba2n . Now according to pumping lemma we will have xyiz where i ≥ 0 , will also lies in L but if i = k , then we can increase the instances of a without increasing the instance of a present in substring z . So it can’t be a regular language .
Case 2: Now assume that y = bcb , in that case . xyiz when i = 0 will not have any instance of bcb . so the resulting language would be falling in L . Hence L can’t be a regular language.
Case 3: Now assume that y is having an instance of a present after bcb . in that case x = anbcb , y = ak , z = a2n-k . Now according to pumping lemma xyiz where i≥0 , should lie in L . Now if we increase the value of i , we can have more instances of a in the later part which violates the property of L having exactly double number of instances of a present in the substring x . Hence L can’t be a regular language.
As none of the case is able to provide valid proof of pumping lemma L can’t be a regular language.