Answer:
a = int(input("Enter first number: "))
b = int(input("Enter second number: "))
c = int(input("Enter third number: "))
a,b = a+b,b+c
print(a,b,c)
Explanation:
Assignments like this are easy in python, in the sense that no helper variables are needed!
Answer:
He can define pas follows: Let P be defined on the set of languages accepted by some Turing machine M. Let it be True if 10 ] is 5 and False otherwise.
Explanation:
The domain of P is the SD languages since it is those languages that are accepted by some Turing machine H. P is nontrivial since P({ a, aa, aaa, aaaa, aaaaa, aaaaaa, b, bb, bbb, bbbb, bbbbb, bbbbbb } ) is True and P ( 5 ) is False.
Thus {< M> is a Turing machine and I L I - 5 and I L I - 16 }
Answer: Applications for desktop or laptop computers are sometimes called desktop applications, while those for mobile devices are called mobile apps. When you open an application, it runs inside the operating system until you close it
Explanation:
Answer: provided in the explanation section
Explanation:
Given that:
Assume D(k) =║ true it is [1 : : : k] is valid sequence words or false otherwise
now the sub problem s[1 : : : k] is a valid sequence of words IFF s[1 : : : 1] is a valid sequence of words and s[ 1 + 1 : : : k] is valid word.
So, from here we have that D(k) is given by the following recorance relation:
D(k) = ║ false maximum (d[l]∧DICT(s[1 + 1 : : : k]) otherwise
Algorithm:
Valid sentence (s,k)
D [1 : : : k] ∦ array of boolean variable.
for a ← 1 to m
do ;
d(0) ← false
for b ← 0 to a - j
for b ← 0 to a - j
do;
if D[b] ∧ DICT s([b + 1 : : : a])
d (a) ← True
(b). Algorithm Output
if D[k] = = True
stack = temp stack ∦stack is used to print the strings in order
c = k
while C > 0
stack push (s [w(c)] : : : C] // w(p) is the position in s[1 : : : k] of the valid world at // position c
P = W (p) - 1
output stack
= 0 =
cheers i hope this helps !!!
