On 11/12/13 12:36, uga...@talktalk.net wrote:
What I do not see is how?
Clearly, a = 3, it is constant throughout each iteration, and
if rest is equal to 3, then a + rest must be equal to 6.
Correct and the return value from the second invocation of mul() is 3.
You spoke of drilling down, and I see that mult(3, 2) drills down until
b == 0, I am less clear how it then begins to ascend,
It returns a value to the previous invocation of mul()
Remember we are calling mul() several times, each with a new set of values.
So mul(3,2) calls mul(3,1)
and mul(3,1) calls mul(3,0)
mul(3.0) returns 0 to mul(3,1)
mul(3,1) then returns 3+0 => 3 to mul(3,2)
mul(3,2) returns 3+3 => 6.
Let me try one more time but this time I will remove the recursion
and substitute the equivalent code with modified variable names.
how is rest assigned a single argument/value?
rest = a + mul(a,b-1)
rest takes on the sum of a and the return value from the
new call to mul()
I agree, rest equals 0 (zero) on the first iteration
No it doesn't. rest is never created in the b=0 case.
mul just returns 0 directly. See below...
It seems to me that rest has two arguments, not one.
I do not understand "+>", is it a typo perhaps?
Yes it was supposed to be => as in 3+2 => 5
Here goes breakdown number 2.
def mult1(a1, b1):
if b1 == 0:
return 0
rest1 = mult2(a1, b1 - 1)
value1 = a1 + rest1
return value1
def mult2(a2, b2):
if b2 == 0:
return 0
rest2 = mult3(a2, b2 - 1)
value2 = a2 + rest2
return value2
def mult3(a3, b3):
if b3 == 0:
return 0
mult1(3,2)
Try working through that set.
Then look at the code inside each function.
It is identical... So we only really need
one function mult().
Does that help?
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.flickr.com/photos/alangauldphotos
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