looping through possible combinations of McNuggets packs of 6, 9 and 20
Baba
raoulbia at gmail.com
Sat Aug 14 16:52:26 CEST 2010
On Aug 13, 8:25 pm, Ian Kelly <ian.g.ke... at gmail.com> wrote:
> It's not. You're not just trying to find the sixth value that can be
> bought in exact quantity, but a sequence of six values that can all be
> bought in exact quantity. The integers [6, 9, 12, 15, 18, 20] are not
> sequential.
Hi Ian,
Thanks for stating the obvious. I obviously hadn't understood a
fundamental part of the theorem which states that 6 SEQUENTIAL passes
must be found! That's a good lesson learned and will help me in future
exercises to make sure i understand the theory first. Thanks again!
Ok so with your and News123's help (no offence to all others but i
need to keep it simple at this stage)i was able to find the solution:
43
my code is probably not elegant but a huge step forward from where i
started:
def can_buy(n_nuggets):
for a in range (0,n_nuggets):
for b in range (0,n_nuggets):
for c in range (0,n_nuggets):
#print "trying for %d: %d %d %d" % (n_nuggets,a,b,c)
if 6*a+9*b+20*c==n_nuggets:
return [a,b,c]
return []
for n_nuggets in range(50):
result1 = can_buy(n_nuggets)
result2 = can_buy(n_nuggets+1)
result3 = can_buy(n_nuggets+2)
result4 = can_buy(n_nuggets+3)
result5 = can_buy(n_nuggets+4)
result6 = can_buy(n_nuggets+5)
if result1!=[] and result2!=[] and result3!=[] and result4!=[] and
result5!=[] and result6!=[]:
if (n_nuggets+5)-n_nuggets==5:
print n_nuggets-1
break
i suppose this can be tweaked to make it shorter? For instance i
wonder if i can do the same with less variable to be defined?
tnx
Baba
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