adamscott - 1 year ago 79

Ruby Question

`class Fibonacci`

def calc(n)

return n if n < 2

return calc(n - 1) + calc(n - 2)

end

end

puts Fibonacci.new.calc(40)

>> 102334155

I am trying to make sense of this program. I understand it is recursively calling the calc method twice if it is not 0 or 1, but I don't understand how it is working to get the correct number for the sequence.

I know this is my inexperience showing, but can someone explain to me how this is calculating the fib sequence correctly? It's just not clicking for me.

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Answer Source

When you are having trouble understanding what code is doing, you may have to resort to salting it with `puts`

statements. For recursions it's also helpful to vary indentation to show when a method calls itself and when a method returns. Here is one way you might do that for your code.

```
INDENT = 4
@tabs = 0
def calc(n)
pr_indent
puts "entered calc(#{n})"
if n < 2
pr_indent
puts "returning n=#{n}"
@tabs -= 1
return n
end
pr_indent
puts "calling calc(#{n-1})"
@tabs += 1
m1 = calc(n-1)
pr_indent
puts "calc(#{n-1}) returned m1=#{m1} to calc(#{n})"
pr_indent
puts "calling calc(#{n-2})"
@tabs += 1
m2 = calc(n-2)
pr_indent
puts "calc(#{n-2}) returned m2=#{m2} to calc(#{n-1})"
pr_indent
puts "returning m1+m2=#{m1+m2}"
@tabs -= 1
m1+m2
end
def pr_indent
print "#{' '*(@tabs*INDENT)}"
end
```

Now execute

```
calc(4)
```

causing the following to be printed. I will not provide running comments at it should be self-explanatory when you read what is printed as you step through the code.

```
entered calc(4)
calling calc(3)
entered calc(3)
calling calc(2)
entered calc(2)
calling calc(1)
entered calc(1)
returning n=1
calc(1) returned m1=1 to calc(2)
calling calc(0)
entered calc(0)
returning n=0
calc(0) returned m2=0 to calc(1)
returning m1+m2=1
calc(2) returned m1=1 to calc(3)
calling calc(1)
entered calc(1)
returning n=1
calc(1) returned m2=1 to calc(2)
returning m1+m2=2
calc(3) returned m1=2 to calc(4)
calling calc(2)
entered calc(2)
calling calc(1)
entered calc(1)
returning n=1
calc(1) returned m1=1 to calc(2)
calling calc(0)
entered calc(0)
returning n=0
calc(0) returned m2=0 to calc(1)
returning m1+m2=1
calc(2) returned m2=1 to calc(3)
returning m1+m2=3
```

and `calc(4)`

returns `3`

.

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