Considering the following code for Integer Partition:
int p (int n, int m)
{
if (n == m)
return 1 + p(n, m - 1);
if (m == 0 || n < 0)
return 0;
if (n == 0 || m == 1)
return 1;
return p(n, m - 1) + p(n - m, m);
}
If I take an example p(7,3), what happens after function becomes p(7,0) & p(4,3)?
If you have Python you can play around with this:
def p(n,m):
if n == m:
return 1 + p(n,m-1)
elif m == 0 or n < 0:
return 0
elif n == 0 or m == 1:
return 1
else:
return p(n,m-1) + p(n-m,m)
def tupleFromString(s):
#converts a string like `(3,7)` to the correspoding int tuple
s = s.strip()
arguments = s[1:len(s)-1].split(',')
return tuple(int(i) for i in arguments)
def toString(t):
#converts an int-tuple to a string, without the spaces
return str(t).replace(' ','')
def expandOnce(s):
s = s.strip()
if s.startswith('p'):
n,m = tupleFromString(s[1:])
if n == m:
return '1 + p' + toString((n,m-1))
elif m == 0 or n < 0:
return '0'
elif n == 0 or m == 1:
return '1'
else:
return 'p' + toString((n,m-1)) + ' + p' + toString((n-m,m))
else:
return s
def expandLine(line):
return ' + '.join(expandOnce(term) for term in line.split('+'))
def expand(s):
firstLine = True
k = len(s)
prefix = s + ' = '
while 'p' in s:
if firstLine:
firstLine = False
else:
prefix = ' '*k + ' = '
s = expandLine(s)
print(prefix + s)
print(prefix + str(sum(int(i) for i in s.split('+'))))
p(m,n) is a direct implementation of the function, and expand displays the steps as strings:
>>> p(4,3)
4
>>> expand('p(4,3)')
p(4,3) = p(4,2) + p(1,3)
= p(4,1) + p(2,2) + p(1,2) + p(-2,3)
= 1 + 1 + p(2,1) + p(1,1) + p(-1,2) + 0
= 1 + 1 + 1 + 1 + p(1,0) + 0 + 0
= 1 + 1 + 1 + 1 + 0 + 0 + 0
= 4
The logic of this is as follows. If you want to know what p(4,3) is, you consult the definition. p(4,3) has n = 4 and m = 3, so you need to use that last clause of the definition. This tells you that
p(4,3) = p(4,3-1) + p(4-3,3)
= p(4,2) + p(1,3)
That doesn't help unless you know what p(4,2) and p(1,3) are so you go back to the definition and find that p(4,2) = p(4,1) + p(2,2) and p(1,3) = p(1,2) + p(-1,2). Combining this with the above, you now know that
p(4,3) = p(4,3-1) + p(4-3,3)
= p(4,2) + p(1,3)
= p(4,1) + p(2,2) + p(1,3) + p(1,2)
at each stage, if there is a term which looks like p(m,n) -- you go back to the definition and see what it means. You eventually hit basis cases such as p(4,1) = 1. Once all of the p are evaluated -- just add what is left (just a bunch of ones and zeros).
Similarly,
p(7,3) = p(7,2) + p(4,3)
= p(7,1) + p(5,2) + p(4,2) + p(1,3)
= 1 + p(5,1) + p(3,2) + p(4,1) + p(2,2) + p(1,2) + p(-2,3)
= 1 + 1 + p(3,1) + p(1,2) + 1 + 1 + p(2,1) + p(1,1) + p(-1,2) + 0
= 1 + 1 + 1 + p(1,1) + p(-1,2) + 1 + 1 + 1 + 1 + p(1,0) + 0 + 0
= 1 + 1 + 1 + 1 + p(1,0) + 0 + 1 + 1 + 1 + 1 + 0 + 0 + 0
= 1 + 1 + 1 + 1 + 0 + 0 + 1 + 1 + 1 + 1 + 0 + 0 + 0
= 8