61A Lecture 21
Monday, October 17
Monday, October 17, 2011
Space Consumption
2
Monday, October 17, 2011
Space Consumption
Which environment frames do we need to keep during evaluation?
2
Monday, October 17, 2011
Space Consumption
Which environment frames do we need to keep during evaluation?
Each step of evaluation has a set of active environments.
2
Monday, October 17, 2011
Space Consumption
Which environment frames do we need to keep during evaluation?
Each step of evaluation has a set of active environments.
Values and frames referenced by active environments are kept.
2
Monday, October 17, 2011
Space Consumption
Which environment frames do we need to keep during evaluation?
Each step of evaluation has a set of active environments.
Values and frames referenced by active environments are kept.
Memory used for other values and frames can be reclaimed.
2
Monday, October 17, 2011
Space Consumption
Which environment frames do we need to keep during evaluation?
Each step of evaluation has a set of active environments.
Values and frames referenced by active environments are kept.
Memory used for other values and frames can be reclaimed.
Active environments:
2
Monday, October 17, 2011
Space Consumption
Which environment frames do we need to keep during evaluation?
Each step of evaluation has a set of active environments.
Values and frames referenced by active environments are kept.
Memory used for other values and frames can be reclaimed.
Active environments:
• The environment for the current expression being evaluated
2
Monday, October 17, 2011
Space Consumption
Which environment frames do we need to keep during evaluation?
Each step of evaluation has a set of active environments.
Values and frames referenced by active environments are kept.
Memory used for other values and frames can be reclaimed.
Active environments:
• The environment for the current expression being evaluated
• Environments for calls that depend upon the value of the
current expression
2
Monday, October 17, 2011
Space Consumption
Which environment frames do we need to keep during evaluation?
Each step of evaluation has a set of active environments.
Values and frames referenced by active environments are kept.
Memory used for other values and frames can be reclaimed.
Active environments:
• The environment for the current expression being evaluated
• Environments for calls that depend upon the value of the
current expression
• Environments associated with functions referenced by active
environments
2
Monday, October 17, 2011
Fibonacci Environment Diagram
fib:
fib(n):
...
n: 3
fib
fib(3)
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
3
Monday, October 17, 2011
Fibonacci Environment Diagram
fib:
fib(n):
...
n: 3
fib
fib(3)
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
n: 1
fib
fib(n-2)
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
3
Monday, October 17, 2011
Fibonacci Environment Diagram
fib:
fib(n):
...
n: 3
fib
fib(3)
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
n: 1
fib
fib(n-2)
0
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
3
Monday, October 17, 2011
Fibonacci Environment Diagram
fib:
fib(n):
...
n: 3
fib
fib(3)
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
n: 1
fib
fib(n-2)
0
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
4
Monday, October 17, 2011
Fibonacci Environment Diagram
fib:
fib(n):
...
n: 3
n: 2
fib
fib
fib(3)
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
n: 1
fib
fib(n-2)
0
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
fib(n-1)
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
4
Monday, October 17, 2011
Fibonacci Memory Consumption
fib(6)
fib(4)
fib(2)
1
fib(5)
fib(3)
fib(1)
0
fib(3)
fib(2)
1
fib(4)
fib(1)
fib(2)
fib(2)
0
1
1
fib(3)
fib(1)
fib(2)
0
1
5
Monday, October 17, 2011
Fibonacci Memory Consumption
fib(6)
fib(4)
fib(2)
1
fib(5)
fib(3)
fib(1)
0
fib(3)
fib(2)
1
fib(4)
fib(1)
fib(2)
fib(2)
0
1
1
Assume we have
reached this step
fib(3)
fib(1)
fib(2)
0
1
5
Monday, October 17, 2011
Fibonacci Memory Consumption
fib(6)
fib(4)
fib(2)
1
fib(5)
fib(3)
fib(1)
0
fib(3)
fib(2)
1
fib(4)
fib(1)
fib(2)
fib(2)
0
1
1
Assume we have
reached this step
fib(3)
fib(1)
fib(2)
0
1
6
Monday, October 17, 2011
Fibonacci Memory Consumption
Has an active environment
fib(6)
fib(4)
fib(2)
1
fib(5)
fib(3)
fib(1)
0
fib(3)
fib(2)
1
fib(4)
fib(1)
fib(2)
fib(2)
0
1
1
Assume we have
reached this step
fib(3)
fib(1)
fib(2)
0
1
6
Monday, October 17, 2011
Fibonacci Memory Consumption
Has an active environment
Can be reclaimed
fib(6)
fib(4)
fib(2)
1
fib(5)
fib(3)
fib(1)
0
fib(3)
fib(2)
1
fib(4)
fib(1)
fib(2)
fib(2)
0
1
1
Assume we have
reached this step
fib(3)
fib(1)
fib(2)
0
1
6
Monday, October 17, 2011
Fibonacci Memory Consumption
Has an active environment
Can be reclaimed
Hasn't yet been created
fib(6)
fib(4)
fib(2)
1
fib(5)
fib(3)
fib(1)
0
fib(3)
fib(2)
1
fib(4)
fib(1)
fib(2)
fib(2)
0
1
1
Assume we have
reached this step
fib(3)
fib(1)
fib(2)
0
1
6
Monday, October 17, 2011
Active Environments for Returned Functions
make_adder:
make_adder(n):
...
def make_adder(n):
def adder(k):
return k + n
return adder
add1 = make_adder(1)
7
Monday, October 17, 2011
Active Environments for Returned Functions
make_adder:
make_adder(n):
...
def make_adder(n):
def adder(k):
return k + n
return adder
add1 = make_adder(1)
make_adder(1)
7
Monday, October 17, 2011
Active Environments for Returned Functions
make_adder:
make_adder(n):
...
def make_adder(n):
def adder(k):
return k + n
return adder
add1 = make_adder(1)
n: 1
adder:
make_adder
adder(k):
return k + n
make_adder(1)
def adder(k):
return k + n
return adder
7
Monday, October 17, 2011
Active Environments for Returned Functions
make_adder:
make_adder(n):
...
def make_adder(n):
def adder(k):
return k + n
return adder
add1 = make_adder(1)
n: 1
adder:
make_adder
adder(k):
return k + n
make_adder(1)
def adder(k):
return k + n
return adder
7
Monday, October 17, 2011
Active Environments for Returned Functions
make_adder:
add1:
make_adder(n):
...
def make_adder(n):
def adder(k):
return k + n
return adder
add1 = make_adder(1)
n: 1
adder:
make_adder
adder(k):
return k + n
make_adder(1)
def adder(k):
return k + n
return adder
7
Monday, October 17, 2011
Active Environments for Returned Functions
make_adder:
add1:
make_adder(n):
...
def make_adder(n):
def adder(k):
return k + n
return adder
add1 = make_adder(1)
n: 1
adder:
make_adder
adder(k):
return k + n
Associated with
an environment
make_adder(1)
def adder(k):
return k + n
return adder
7
Monday, October 17, 2011
Active Environments for Returned Functions
make_adder:
add1:
make_adder(n):
...
def make_adder(n):
def adder(k):
return k + n
return adder
add1 = make_adder(1)
n: 1
adder:
make_adder
Therefore, all
frames in this
environment must
be kept
adder(k):
return k + n
Associated with
an environment
make_adder(1)
def adder(k):
return k + n
return adder
7
Monday, October 17, 2011
Order of Growth
8
Monday, October 17, 2011
Order of Growth
A method for bounding the resources used by a function as the
"size" of a problem increases
8
Monday, October 17, 2011
Order of Growth
A method for bounding the resources used by a function as the
"size" of a problem increases
n: size of the problem
8
Monday, October 17, 2011
Order of Growth
A method for bounding the resources used by a function as the
"size" of a problem increases
n: size of the problem
R(n): Measurement of some resource used (time or space)
8
Monday, October 17, 2011
Order of Growth
A method for bounding the resources used by a function as the
"size" of a problem increases
n: size of the problem
R(n): Measurement of some resource used (time or space)
R(n) = Θ(f (n))
8
Monday, October 17, 2011
Order of Growth
A method for bounding the resources used by a function as the
"size" of a problem increases
n: size of the problem
R(n): Measurement of some resource used (time or space)
R(n) = Θ(f (n))
means that there are constants k1 and k2 such that
8
Monday, October 17, 2011
Order of Growth
A method for bounding the resources used by a function as the
"size" of a problem increases
n: size of the problem
R(n): Measurement of some resource used (time or space)
R(n) = Θ(f (n))
means that there are constants k1 and k2 such that
k1 · f (n) ≤ R(n) ≤ k2 · f (n)
8
Monday, October 17, 2011
Order of Growth
A method for bounding the resources used by a function as the
"size" of a problem increases
n: size of the problem
R(n): Measurement of some resource used (time or space)
R(n) = Θ(f (n))
means that there are constants k1 and k2 such that
k1 · f (n) ≤ R(n) ≤ k2 · f (n)
for sufficiently large values of n.
8
Monday, October 17, 2011
Iteration vs Memoized Tree Recursion
Iterative and memoized implementations are not the same.
Time
Space
def fib_iter(n):
prev, curr = 1, 0
for _ in range(n-1):
prev, curr = curr, prev + curr
return curr
@memo
def fib(n):
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
9
Monday, October 17, 2011
Iteration vs Memoized Tree Recursion
Iterative and memoized implementations are not the same.
Time
def fib_iter(n):
prev, curr = 1, 0
for _ in range(n-1):
prev, curr = curr, prev + curr
return curr
Space
Θ(n)
@memo
def fib(n):
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
9
Monday, October 17, 2011
Iteration vs Memoized Tree Recursion
Iterative and memoized implementations are not the same.
def fib_iter(n):
prev, curr = 1, 0
for _ in range(n-1):
prev, curr = curr, prev + curr
return curr
Time
Space
Θ(n)
Θ(1)
@memo
def fib(n):
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
9
Monday, October 17, 2011
Iteration vs Memoized Tree Recursion
Iterative and memoized implementations are not the same.
def fib_iter(n):
prev, curr = 1, 0
for _ in range(n-1):
prev, curr = curr, prev + curr
return curr
@memo
def fib(n):
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
Time
Space
Θ(n)
Θ(1)
Θ(n)
9
Monday, October 17, 2011
Iteration vs Memoized Tree Recursion
Iterative and memoized implementations are not the same.
def fib_iter(n):
prev, curr = 1, 0
for _ in range(n-1):
prev, curr = curr, prev + curr
return curr
@memo
def fib(n):
if n == 1:
return 0
if n == 2:
return 1
return fib(n-2) + fib(n-1)
Time
Space
Θ(n)
Θ(1)
Θ(n)
Θ(n)
9
Monday, October 17, 2011
Comparing orders of growth
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
Exponential growth!
Recursive fib takes
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
Exponential growth!
Recursive fib takes
√
1+ 5
≈ 1.61828
Θ(φ ) steps, where φ =
2
n
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
Exponential growth!
Recursive fib takes
√
1+ 5
≈ 1.61828
Θ(φ ) steps, where φ =
2
n
Incrementing the problem scales R(n) by a factor.
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
Exponential growth!
Recursive fib takes
√
1+ 5
≈ 1.61828
Θ(φ ) steps, where φ =
2
n
Incrementing the problem scales R(n) by a factor.
Θ(n)
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
Exponential growth!
Recursive fib takes
√
1+ 5
≈ 1.61828
Θ(φ ) steps, where φ =
2
n
Incrementing the problem scales R(n) by a factor.
Θ(n)
Linear growth.
Resources scale with the problem.
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
Exponential growth!
Recursive fib takes
√
1+ 5
≈ 1.61828
Θ(φ ) steps, where φ =
2
n
Incrementing the problem scales R(n) by a factor.
Θ(n)
Linear growth.
Resources scale with the problem.
Θ(log n)
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
Exponential growth!
Recursive fib takes
√
1+ 5
≈ 1.61828
Θ(φ ) steps, where φ =
2
n
Incrementing the problem scales R(n) by a factor.
Θ(n)
Θ(log n)
Linear growth.
Resources scale with the problem.
Logarithmic growth. These functions scale well.
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
Exponential growth!
Recursive fib takes
√
1+ 5
≈ 1.61828
Θ(φ ) steps, where φ =
2
n
Incrementing the problem scales R(n) by a factor.
Θ(n)
Θ(log n)
Linear growth.
Resources scale with the problem.
Logarithmic growth. These functions scale well.
Doubling the problem increments resources needed.
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
Exponential growth!
Recursive fib takes
√
1+ 5
≈ 1.61828
Θ(φ ) steps, where φ =
2
n
Incrementing the problem scales R(n) by a factor.
Θ(n)
Θ(log n)
Linear growth.
Resources scale with the problem.
Logarithmic growth. These functions scale well.
Doubling the problem increments resources needed.
Θ(1)
10
Monday, October 17, 2011
Comparing orders of growth
Θ(bn )
Exponential growth!
Recursive fib takes
√
1+ 5
≈ 1.61828
Θ(φ ) steps, where φ =
2
n
Incrementing the problem scales R(n) by a factor.
Θ(n)
Θ(log n)
Linear growth.
Resources scale with the problem.
Logarithmic growth. These functions scale well.
Doubling the problem increments resources needed.
Θ(1)
Constant. The problem size doesn't matter.
10
Monday, October 17, 2011
Exponentiation
11
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
11
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
11
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
�
1
n
b =
b · bn−1
if n = 0
otherwise
11
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
�
1
n
b =
b · bn−1


1
1
n
b = (b 2 n )2


b · bn−1
if n = 0
otherwise
if n = 0
if n is even
if n is odd
11
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
def square(x):
return x*x
�
1
n
b =
b · bn−1


1
1
n
b = (b 2 n )2


b · bn−1
if n = 0
otherwise
if n = 0
if n is even
if n is odd
11
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
def square(x):
return x*x
def fast_exp(b, n):
�
1
n
b =
b · bn−1


1
1
n
b = (b 2 n )2


b · bn−1
if n = 0
otherwise
if n = 0
if n is even
if n is odd
11
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
def square(x):
return x*x
def fast_exp(b, n):
if n == 0:
return 1
�
1
n
b =
b · bn−1


1
1
n
b = (b 2 n )2


b · bn−1
if n = 0
otherwise
if n = 0
if n is even
if n is odd
11
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
def square(x):
return x*x
�
1
n
b =
b · bn−1


1
1
n
b = (b 2 n )2


b · bn−1
def fast_exp(b, n):
if n == 0:
return 1
if n % 2 == 0:
return square(fast_exp(b, n//2))
if n = 0
otherwise
if n = 0
if n is even
if n is odd
11
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
def square(x):
return x*x
�
1
n
b =
b · bn−1


1
1
n
b = (b 2 n )2


b · bn−1
def fast_exp(b, n):
if n == 0:
return 1
if n % 2 == 0:
return square(fast_exp(b, n//2))
else:
return b * fast_exp(b, n-1)
if n = 0
otherwise
if n = 0
if n is even
if n is odd
11
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
Time
Space
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
def square(x):
return x*x
def fast_exp(b, n):
if n == 0:
return 1
if n % 2 == 0:
return square(fast_exp(b, n//2))
else:
return b * fast_exp(b, n-1)
12
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
Time
Space
Θ(n)
Θ(n)
def square(x):
return x*x
def fast_exp(b, n):
if n == 0:
return 1
if n % 2 == 0:
return square(fast_exp(b, n//2))
else:
return b * fast_exp(b, n-1)
12
Monday, October 17, 2011
Exponentiation
Goal: one more multiplication lets us double the problem size.
def exp(b, n):
if n == 0:
return 1
return b * exp(b, n-1)
Time
Space
Θ(n)
Θ(n)
Θ(log n)
Θ(log n)
def square(x):
return x*x
def fast_exp(b, n):
if n == 0:
return 1
if n % 2 == 0:
return square(fast_exp(b, n//2))
else:
return b * fast_exp(b, n-1)
12
Monday, October 17, 2011