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2

0

I came across the following homework problem:

My code for this problem was marked wrong and when I viewed the suggested solution, I couldn't understand where I went wrong. I ran the codes of both functions in Python IDLE compiler only to see that both functions return the same output as seen below:

>>> def dual_function(f,g,n): #Suggested solution
 def helper(x):
 f1,g1 = f,g
 if n%2==0:
 f1,g1=g1,f1
 for i in range(n):
 x=f1(x)
 f1,g1=g1,f1
 return x
 return helper

>>> def dual_function_two(f,g,n): #My solution
 def helper(x):
 if n%2==0:
 for i in range (n):
 if i%2==0:
 x = g(x)
 else:
 x = f(x)
 else:
 for i in range(n):
 if i%2==0:
 x = f(x)
 else:
 x = g(x)
 return x
 return helper

>>> add1 = lambda x: x+1
>>> add2 = lambda x: x+2
>>> dual_function(add1,add2,4)(3)
9
>>> dual_function_two(add1,add2,4)(3)
9
>>> 

I would appreciate it if someone could identify the mistake in my solution. Thank you.

python lambda functional-programming higher-order-functions

share|improve this question

asked Mar 27 at 4:02

PravPrav

3477 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Do you know the test cases used to grade your problem? I can't find any case where the two functions differ.
  
  – Sebastian
  Mar 27 at 4:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Those aren't good test functions, since addition is commutative -- if you called them in the wrong order you wouldn't notice the difference. But I tried mixing addition and multiplication, and I couldn't find a difference, either.
  
  – Barmar
  Mar 27 at 4:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  It's probably much slower than the suggested solution due to the constant modulo operations. Swapping is going to be much faster.
  
  – Rick Teachey
  Mar 27 at 5:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Sebastian Unfortunately, I do not have the test cases :(
  
  – Prav
  Mar 27 at 6:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Barmar Thank you for letting me know about using commutative operators in such functions, will remember them when working on similar problems in the future.
  
  – Prav
  Mar 27 at 6:20
  

  
  
  
  

 | 
show 1 more comment

2

0

I came across the following homework problem:

My code for this problem was marked wrong and when I viewed the suggested solution, I couldn't understand where I went wrong. I ran the codes of both functions in Python IDLE compiler only to see that both functions return the same output as seen below:

>>> def dual_function(f,g,n): #Suggested solution
 def helper(x):
 f1,g1 = f,g
 if n%2==0:
 f1,g1=g1,f1
 for i in range(n):
 x=f1(x)
 f1,g1=g1,f1
 return x
 return helper

>>> def dual_function_two(f,g,n): #My solution
 def helper(x):
 if n%2==0:
 for i in range (n):
 if i%2==0:
 x = g(x)
 else:
 x = f(x)
 else:
 for i in range(n):
 if i%2==0:
 x = f(x)
 else:
 x = g(x)
 return x
 return helper

>>> add1 = lambda x: x+1
>>> add2 = lambda x: x+2
>>> dual_function(add1,add2,4)(3)
9
>>> dual_function_two(add1,add2,4)(3)
9
>>> 

I would appreciate it if someone could identify the mistake in my solution. Thank you.

python lambda functional-programming higher-order-functions

share|improve this question

asked Mar 27 at 4:02

PravPrav

3477 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Do you know the test cases used to grade your problem? I can't find any case where the two functions differ.
  
  – Sebastian
  Mar 27 at 4:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Those aren't good test functions, since addition is commutative -- if you called them in the wrong order you wouldn't notice the difference. But I tried mixing addition and multiplication, and I couldn't find a difference, either.
  
  – Barmar
  Mar 27 at 4:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  It's probably much slower than the suggested solution due to the constant modulo operations. Swapping is going to be much faster.
  
  – Rick Teachey
  Mar 27 at 5:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Sebastian Unfortunately, I do not have the test cases :(
  
  – Prav
  Mar 27 at 6:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Barmar Thank you for letting me know about using commutative operators in such functions, will remember them when working on similar problems in the future.
  
  – Prav
  Mar 27 at 6:20
  

  
  
  
  

 | 
show 1 more comment

I came across the following homework problem:

My code for this problem was marked wrong and when I viewed the suggested solution, I couldn't understand where I went wrong. I ran the codes of both functions in Python IDLE compiler only to see that both functions return the same output as seen below:

>>> def dual_function(f,g,n): #Suggested solution
 def helper(x):
 f1,g1 = f,g
 if n%2==0:
 f1,g1=g1,f1
 for i in range(n):
 x=f1(x)
 f1,g1=g1,f1
 return x
 return helper

>>> def dual_function_two(f,g,n): #My solution
 def helper(x):
 if n%2==0:
 for i in range (n):
 if i%2==0:
 x = g(x)
 else:
 x = f(x)
 else:
 for i in range(n):
 if i%2==0:
 x = f(x)
 else:
 x = g(x)
 return x
 return helper

>>> add1 = lambda x: x+1
>>> add2 = lambda x: x+2
>>> dual_function(add1,add2,4)(3)
9
>>> dual_function_two(add1,add2,4)(3)
9
>>> 

I would appreciate it if someone could identify the mistake in my solution. Thank you.

python lambda functional-programming higher-order-functions

share|improve this question

asked Mar 27 at 4:02

PravPrav

3477 bronze badges

>>> def dual_function(f,g,n): #Suggested solution
 def helper(x):
 f1,g1 = f,g
 if n%2==0:
 f1,g1=g1,f1
 for i in range(n):
 x=f1(x)
 f1,g1=g1,f1
 return x
 return helper

>>> def dual_function_two(f,g,n): #My solution
 def helper(x):
 if n%2==0:
 for i in range (n):
 if i%2==0:
 x = g(x)
 else:
 x = f(x)
 else:
 for i in range(n):
 if i%2==0:
 x = f(x)
 else:
 x = g(x)
 return x
 return helper

>>> add1 = lambda x: x+1
>>> add2 = lambda x: x+2
>>> dual_function(add1,add2,4)(3)
9
>>> dual_function_two(add1,add2,4)(3)
9
>>> 

share|improve this question

asked Mar 27 at 4:02

PravPrav

3477 bronze badges

share|improve this question

asked Mar 27 at 4:02

PravPrav

3477 bronze badges

asked Mar 27 at 4:02

PravPrav

3477 bronze badges

asked Mar 27 at 4:02

PravPrav

3477 bronze badges

1 Answer
1

active

oldest

votes

3

The suggested solution is needlessly complex. Countless reassignments of variables and a loop are a recipe for a headache. Here's a simplified alternative -

def dual (f, g, n):
 if n == 0:
 return lambda x: x
 else:
 return lambda x: f(dual(g, f, n - 1)(x))

add1 = lambda x: 1 + x
add2 = lambda x: 2 + x

print(dual(add1,add2,4)(3))
# 9
# (1 + 2 + 1 + 2 + 3)

print(dual(add1,add2,9)(3))
# 16
# (1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 3)

print(dual(add1,add2,0)(3))
# 3

The reason this works is because in the recursive branch, we call dual with swapped arguments, dual(g,f,n-1). So f and g change places each time as n decrements down to 0, the base case, which returns the identity (no-op) function.

A slightly less readable version, but works identically -

def dual (f, g, n):
 return lambda x: 
 x if n == 0 else f(dual(g, f, n - 1)(x))

share|improve this answer

answered Mar 27 at 6:11

user633183user633183

76.5k2121 gold badges149149 silver badges190190 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Great, never knew I could be this simple. Thanks for the explanation! :)
  
  – Prav
  Mar 27 at 13:31
  

  
  
  
  

add a comment | 

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3

The suggested solution is needlessly complex. Countless reassignments of variables and a loop are a recipe for a headache. Here's a simplified alternative -

def dual (f, g, n):
 if n == 0:
 return lambda x: x
 else:
 return lambda x: f(dual(g, f, n - 1)(x))

add1 = lambda x: 1 + x
add2 = lambda x: 2 + x

print(dual(add1,add2,4)(3))
# 9
# (1 + 2 + 1 + 2 + 3)

print(dual(add1,add2,9)(3))
# 16
# (1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 3)

print(dual(add1,add2,0)(3))
# 3

The reason this works is because in the recursive branch, we call dual with swapped arguments, dual(g,f,n-1). So f and g change places each time as n decrements down to 0, the base case, which returns the identity (no-op) function.

A slightly less readable version, but works identically -

def dual (f, g, n):
 return lambda x: 
 x if n == 0 else f(dual(g, f, n - 1)(x))

share|improve this answer

answered Mar 27 at 6:11

user633183user633183

76.5k2121 gold badges149149 silver badges190190 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Great, never knew I could be this simple. Thanks for the explanation! :)
  
  – Prav
  Mar 27 at 13:31
  

  
  
  
  

add a comment | 

3

The suggested solution is needlessly complex. Countless reassignments of variables and a loop are a recipe for a headache. Here's a simplified alternative -

def dual (f, g, n):
 if n == 0:
 return lambda x: x
 else:
 return lambda x: f(dual(g, f, n - 1)(x))

add1 = lambda x: 1 + x
add2 = lambda x: 2 + x

print(dual(add1,add2,4)(3))
# 9
# (1 + 2 + 1 + 2 + 3)

print(dual(add1,add2,9)(3))
# 16
# (1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 3)

print(dual(add1,add2,0)(3))
# 3

The reason this works is because in the recursive branch, we call dual with swapped arguments, dual(g,f,n-1). So f and g change places each time as n decrements down to 0, the base case, which returns the identity (no-op) function.

A slightly less readable version, but works identically -

def dual (f, g, n):
 return lambda x: 
 x if n == 0 else f(dual(g, f, n - 1)(x))

share|improve this answer

answered Mar 27 at 6:11

user633183user633183

76.5k2121 gold badges149149 silver badges190190 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Great, never knew I could be this simple. Thanks for the explanation! :)
  
  – Prav
  Mar 27 at 13:31
  

  
  
  
  

add a comment | 

The suggested solution is needlessly complex. Countless reassignments of variables and a loop are a recipe for a headache. Here's a simplified alternative -

def dual (f, g, n):
 if n == 0:
 return lambda x: x
 else:
 return lambda x: f(dual(g, f, n - 1)(x))

add1 = lambda x: 1 + x
add2 = lambda x: 2 + x

print(dual(add1,add2,4)(3))
# 9
# (1 + 2 + 1 + 2 + 3)

print(dual(add1,add2,9)(3))
# 16
# (1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 3)

print(dual(add1,add2,0)(3))
# 3

The reason this works is because in the recursive branch, we call dual with swapped arguments, dual(g,f,n-1). So f and g change places each time as n decrements down to 0, the base case, which returns the identity (no-op) function.

A slightly less readable version, but works identically -

def dual (f, g, n):
 return lambda x: 
 x if n == 0 else f(dual(g, f, n - 1)(x))

share|improve this answer

answered Mar 27 at 6:11

user633183user633183

76.5k2121 gold badges149149 silver badges190190 bronze badges

def dual (f, g, n):
 if n == 0:
 return lambda x: x
 else:
 return lambda x: f(dual(g, f, n - 1)(x))

add1 = lambda x: 1 + x
add2 = lambda x: 2 + x

print(dual(add1,add2,4)(3))
# 9
# (1 + 2 + 1 + 2 + 3)

print(dual(add1,add2,9)(3))
# 16
# (1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 3)

print(dual(add1,add2,0)(3))
# 3

def dual (f, g, n):
 return lambda x: 
 x if n == 0 else f(dual(g, f, n - 1)(x))

share|improve this answer

answered Mar 27 at 6:11

user633183user633183

76.5k2121 gold badges149149 silver badges190190 bronze badges

answered Mar 27 at 6:11

user633183user633183

76.5k2121 gold badges149149 silver badges190190 bronze badges

answered Mar 27 at 6:11

user633183user633183

76.5k2121 gold badges149149 silver badges190190 bronze badges