WRITE THIS FUNCTION: CODE SHOULD BE IN PYTHON: import random def value_equality(e, f, num_samples=1000, tolerance=1e-6): """Return True if the two expressions self and other are numerically equivalent. Equivalence is tested by generating num_samples assignments, and checking that equality holds for all of them. Equality is checked up to tolerance, that is, the values of the two expressions have to be closer than tolerance. It can be done in less than 10 lines of code.""" # YOUR CODE HERE Should pass these test cases: assert value_equality('x', 'x') assert value_equality(3, 3) assert not value_equality(3, 'x') e1 = ('+', ('*', 'x', 1), ('*', 'y', 0)) e2 = 'x' assert value_equality(e1, e2) e3 = ('/', ('*', 'x', 'x'), ('*', 'x', 1)) e3b = ('/', ('*', 'x', 'y'), ('*', 'x', 1)) assert value_equality(e1, e3) assert not value_equality(e1, e3b) e4 = ('/', 'y', 2) assert not value_equality(e1, e4) assert not value_equality(e3, e4) e5 = ("+", "cat", ("-", "dog", "dog")) assert value_equality(e5, "cat")
WRITE THIS FUNCTION: CODE SHOULD BE IN PYTHON: import random def value_equality(e, f, num_samples=1000, tolerance=1e-6): """Return True if the two expressions self and other are numerically equivalent. Equivalence is tested by generating num_samples assignments, and checking that equality holds for all of them. Equality is checked up to tolerance, that is, the values of the two expressions have to be closer than tolerance. It can be done in less than 10 lines of code.""" # YOUR CODE HERE Should pass these test cases: assert value_equality('x', 'x') assert value_equality(3, 3) assert not value_equality(3, 'x') e1 = ('+', ('*', 'x', 1), ('*', 'y', 0)) e2 = 'x' assert value_equality(e1, e2) e3 = ('/', ('*', 'x', 'x'), ('*', 'x', 1)) e3b = ('/', ('*', 'x', 'y'), ('*', 'x', 1)) assert value_equality(e1, e3) assert not value_equality(e1, e3b) e4 = ('/', 'y', 2) assert not value_equality(e1, e4) assert not value_equality(e3, e4) e5 = ("+", "cat", ("-", "dog", "dog")) assert value_equality(e5, "cat")
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
Related questions
Question
WRITE THIS FUNCTION:
CODE SHOULD BE IN PYTHON:
import random
def value_equality(e, f, num_samples=1000, tolerance=1e-6):
"""Return True if the two expressions self and other are numerically
equivalent. Equivalence is tested by generating
num_samples assignments, and checking that equality holds
for all of them. Equality is checked up to tolerance, that is,
the values of the two expressions have to be closer than tolerance.
It can be done in less than 10 lines of code."""
# YOUR CODE HERE
Should pass these test cases:
assert value_equality('x', 'x')
assert value_equality(3, 3)
assert not value_equality(3, 'x')
e1 = ('+', ('*', 'x', 1), ('*', 'y', 0))
e2 = 'x'
assert value_equality(e1, e2)
e3 = ('/', ('*', 'x', 'x'), ('*', 'x', 1))
e3b = ('/', ('*', 'x', 'y'), ('*', 'x', 1))
assert value_equality(e1, e3)
assert not value_equality(e1, e3b)
e4 = ('/', 'y', 2)
assert not value_equality(e1, e4)
assert not value_equality(e3, e4)
e5 = ("+", "cat", ("-", "dog", "dog"))
assert value_equality(e5, "cat")
![[2]
Os
class IllegalOperator(Exception):
pass
Let us define a helper function calc,which takes as argument an operator and two numbers, and computes the required operation. It will make
it easier to write the rest of the code.
def calc(op, left, right):
if op == "+":
return left + right
elif op ==
"_":
return left -
right
elif op ==
"*":
return left * right
elif op == "/":
return left / right
else:
raise IllegalOperator(op)
With this, we can write our compute method as follows.
[ 4] def compute(e):
if isinstance(e, tuple):
# We have an expression.
op, 1, r = e
# We compute the subexpressions.
1l = compute(1)
rr = compute(r)
# And on the basis of those, the whole expression.
return calc (op, 11, rr)
else:
# base expression; just return the number.
return e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fecf10a40-9173-4876-b1dc-0585f27b87b2%2F3e52b1b8-a38b-4352-8357-6caf6025ace3%2F6186z5r_processed.png&w=3840&q=75)
Transcribed Image Text:[2]
Os
class IllegalOperator(Exception):
pass
Let us define a helper function calc,which takes as argument an operator and two numbers, and computes the required operation. It will make
it easier to write the rest of the code.
def calc(op, left, right):
if op == "+":
return left + right
elif op ==
"_":
return left -
right
elif op ==
"*":
return left * right
elif op == "/":
return left / right
else:
raise IllegalOperator(op)
With this, we can write our compute method as follows.
[ 4] def compute(e):
if isinstance(e, tuple):
# We have an expression.
op, 1, r = e
# We compute the subexpressions.
1l = compute(1)
rr = compute(r)
# And on the basis of those, the whole expression.
return calc (op, 11, rr)
else:
# base expression; just return the number.
return e
![[8] def simplify(e):
if isinstance(e, tuple):
op, l, r = e
# We simplify the children expressions.
11 = simplify( Add text cell
simplify(r)
rr =
# We compute the expression if we can.
if isnumber(ll) and isnumber(rr) :
return calc(op, ll, rr)
else:
return (op, l1, rr)
else:
# Leaf. No simplification is possible.
return e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fecf10a40-9173-4876-b1dc-0585f27b87b2%2F3e52b1b8-a38b-4352-8357-6caf6025ace3%2Foi1eawl_processed.png&w=3840&q=75)
Transcribed Image Text:[8] def simplify(e):
if isinstance(e, tuple):
op, l, r = e
# We simplify the children expressions.
11 = simplify( Add text cell
simplify(r)
rr =
# We compute the expression if we can.
if isnumber(ll) and isnumber(rr) :
return calc(op, ll, rr)
else:
return (op, l1, rr)
else:
# Leaf. No simplification is possible.
return e
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