Problem Set 2

[elz]

Post and discuss your answers in the comments! (And yay for everybody who posted problem set 1 answers!)

Comments

[elz]

I've spent the last three days writing code for other things, so I'm just tackling the McNuggets now. :)

Problem 1:

6a,9b,20c = n
2,2,1 = 50
1,5,0 = 51
2,0,2 = 52
1,3,1 = 53
0,6,0 = 54
1,1,2 = 55

If you add 6 to those totals, you get 56,57,58,59,60,61.
If you add 6 to those totals, you get 62,63,64,65,66,67.

Problem 2:

So basically, once the condition is true for any run of six, it's true for that run of six + 6, which works out to every number greater than the start of the first run of six. If it were true for a run of nine, it would also be true for that run of nine + 9, etc.

[gchick]

Your other code allowed James T. Kirk to get laid. It us CRUCIALLY IMPORTANT!

As for this one, I had no problem with questions 1 & 2, stumbled into a nice solution for the original McNugget problem that solved it with math rather than exhaustive search, and then faceplanted in problem 4 because I'm pretty sure that assignment 2 isn't supposed to be NP-hard, even at MIT. I... need to step away from it and find some other practice problem with the same logical structure. Sigh.

[elz]

Heh. I thought so, anyway. *g*

I got a little intimidated looking at the problem set, because it's been a loooong time since I took math, but it really is mostly "can you loop through stuff until you come up with an answer?"

[gchick]

Yep. I tend to get stuck on "but what's the solution?" rather than "what am I practicing here?"

So with that in mind, I'm off to play with possible combinations of Hardison, Eliot, and Parker; they're better than McNuggets in so many ways. Thanks for posting your solution - it was a good clarifier when I was letting things get too complicated.

Problem #4

[elz]

This is my answer to #4, which is a lot like my answer to #3:

# Takes a tuple of three package sizes and a number of nuggets    
# Returns True or False depending on whether the exact
# given number of McNuggets can be purchased
def divisible_by_mcnuggets(packages, nuggets):
    # We shouldn't bother trying numbers that are greater than the
    # total number we're looking for. For example, if the total number
    # of nuggets is 50 and one of the packages contains 20 nuggets, we'll
    # never want more than 2 of them      
    maximum_a = nuggets/packages[0] + 1
    maximum_b = nuggets/packages[1] + 1
    maximum_c = nuggets/packages[2] + 1
    # Loop through all possible combinations of the three package sizes
    # from 0 to the maximum
    for a in range(0, maximum_a):
        for b in range(0, maximum_b):
            for c in range(0, maximum_c):
                if (packages[0]*a) + (packages[1]*b) + (packages[2]*c) == nuggets:
                    return True
    return False

# Prints the largest number of McNuggets that can't be bought up to 150
# for a given set of 3 package sizes 
def diophantine_mcnuggets():
    bestSoFar = 0           # variable that keeps track of largest number
                            # of McNuggets that cannot be bought in exact quantity
    packages = (6,9,20)     # variable that contains package sizes
    consecutive_divisibles = []
    while len(consecutive_divisibles) < 6:
        for n in range(1, 150): # only search for solutions up to size 150
            if divisible_by_mcnuggets(packages, n) == True:
                consecutive_divisibles.append(n)
            else:
                consecutive_divisibles = []
                bestSoFar = n         
    print "Given package sizes %i, %i, and %i, the largest number of McNuggets that cannot be bought in exact quantity is: %i" % (packages[0], packages[1], packages[2], bestSoFar)


diophantine_mcnuggets()

Re: Problem #4

[elz]

And to nitpick myself, I should have changed the line from:

while len(consecutive_divisibles) < 6:

to:

while len(consecutive_divisibles) < packages[0]:

Re: Problem #4

[(Anonymous)]

ELZ - My script wound up being very similar to yours. I couldn't figure out how to stop looking after 6 consecutive successful results, so I just put all successful results in a list. I then looked for numbers not in that list and used the highest number that was missing. I'm not sure if I'm happy to find this thread AFTER I finally figured it out (hours), or if I'm glad I didn't because I think I learned more.....

## This program looks for the maximum number that can't be
## created using combinations of package sizes.


# variables - change to manipulate how what numbers to try and max range
# to try within. If the result is close to your max range, you should
# probably increase the range.
packages = [6,9,20] # variable that contains package sizes
n = range(0, 201) # range to test within. Max number must be at least TWO more than expected result.

# tests all possible combinations. Saves numbers that were successfully computed to numworks.
numworks = []
for count in n:
for a in range(0, count/packages[0]+1): #instead of using all combinations, only combinations that
# wouldn't be higher than max range (so max range divided by package size). Adding one because
# of remainders. This cuts WAY down on the time to process the program vs just using "range (0, count)"
for b in range(0, count/packages[1]+1):
for c in range(0, count/packages[2]+1):
if ((a*packages[0] + b*packages[1] + c*packages[2]) == count):
numworks = numworks + [count]

### found that entire section is not needed to get correct results. It makes numworks look nicer, but isn't needed.
### Sort then Remove duplicates from numworks
##cycle = 0
##if numworks:
## numworks.sort()
## last = numworks[-1]
## for i in range(len(numworks)-2, -1, -1):
## if last == numworks[i]:
## del numworks[i]
## else:
## last = numworks[i]

# create numbad which is list of numbers that didn't work. This is based on numbers that are not in numworks.
numbad = [0]
for ii in range (0,numworks[-1]):
if ii not in numworks:
numbad = numbad + [ii]

# prints results. Also accounts for no results.
if numbad == [0]:
print "No numbers were generated. \nYou likely used a 1 (one) or a negitive number as a package size."
else:
print "Given package sizes %i, %i, and %i, the largest number of McNuggets that \ncannot be bought in exact quantity is: %i. \n \n*** Note: the highest number tested was %i ***" % (packages[0], packages[1], packages[2], numbad[-1], n[-1])

#4

[medrin]

Here is my solution! (but I could not get the indentations to work, anyone know how?)

If anyone have any questions, just ask!


#packets = (6,9,20)
no_right = 0
n=0
highest=0
corr_pack=False
print'Hi!'


#asks for the diffrent package sizes and checks that they are suitable (error handling)
while corr_pack==False:
print('What are the three packet sizes (in rising order)?')
packets=input()
if len(packets)!=3:
print ('There must be three and only three packet sizes')
elif type(packets[0])!=int or type(packets[1])!=int or type(packets[2])!=int:
print ('All packet sizes must be given in integrers')
elif packets[0]<0 or packets[1]<0 or packets[2]<0:
print ('The packet sizes must be positive numbers')
elif packets[0]>packets[1] or packets [0]>packets[2] or packets[1]>packets[2]:
print ('The packets must be given in a rising order')
else:
corr_pack=True

while n<200 and no_right< packets[0] : #n loops all possible combinations from 1 and up to a
works=False #consecutive number of possibles the same length as the minimum package
n+=1 #works turns true as soon as a solution is found,
c=0 #a,b,c is how many packages of each size are needed
while c*packets[2]<=n and works==False: #each loop goes until the combined number of nuggets is bigger
b=0 #than the wanted number.
while c*packets[2]+b*packets[1]<=n and works==False:
a=0
while c*packets[2]+b*packets[1]+a*packets[0]<=n and works==False:#Goes though all combinations
if c*packets[2]+b*packets[1]+a*packets[0]==n: #of a,b&c till a solution is
works=True #found or all combinations are tried.
else: #starts with a and works it way out.
a+=1 #if a correct solution are found works turns True
if c*packets[2]+b*packets[1]+a*packets[0]==n:
works=True
else:
b+=1
if c*packets[2]+b*packets[1]+a*packets[0]==n:
works=True
else:
c+=1
if works==True: #if there exists a correct solution the n_right counter incerases
no_right+=1
#print n,a,b,c
else: #if there isn't, it resets to zero and that number is stored in highest.
highest=n
no_right=0
#print n,no_right
print highest #the highest non-solvable number is printed.




Problem 3

[yhlee]

I'm pretty sure this is ugly and I know it's inefficient, but they asked for exhaustive search...

# Solving a Diophantine Equation.

# Helper function
def consecutiveValues(n1, n2, n3, n4, n5, n6):
	value = (n1 + 1 == n2) and (n2 + 1 == n3) and (n3 + 1 == n4) and (n4 + 1 == n5) and (n5 + 1 == n6)
	return value

# Helper function
def nuggetsTest(n):
	answer = False
	for a in range(0,11):
		for b in range(0,11):
			for c in range(0,11):
				if (a*6 + b*9 + c*20) == n:
					return True
	return answer

def mcnuggets():
	# Possible instance of number of McNuggets that cannot be purchased
	# exactly.
	n = 1

	# Stores successful values of n.
	successful_n = [0,0,0,0,0,0]

	# Stores failed values of n.
	failed_n = [0]

	# We already know from Problem 1 that 50, 51, ..., 55 (and all
	# instances thereafter) have exact solutions.  So let's set an
	# upper bound for a, b, and c as 10, which should catch all
	# feasible possibilities for each variable (6*10 = 60).

	while consecutiveValues(successful_n[-6], successful_n[-5], successful_n[-4], successful_n[-3], successful_n[-2], successful_n[-1]) == False:
		if nuggetsTest(n) == True:
			successful_n = successful_n + [n]
			print n
		else:
			failed_n = failed_n + [n]
		n = n + 1

	print "Largest number of McNuggets that cannot be bought in exact quantity: " + str(failed_n[-1])

Problem 4

[yhlee]

Very similar to problem 3. I didn't bother with the other cases they suggested trying, but it would be simple to make the changes.

packages = (6,9,20)	 # variable that contains package sizes

# helper function #1
def consecutiveValues(n1, n2, n3, n4, n5, n6):
	value = (n1 + 1 == n2) and (n2 + 1 == n3) and (n3 + 1 == n4) and (n4 + 1 == n5) and (n5 + 1 == n6)
	return value
	
# helper function #2
def nuggetsTest(n):
	answer = False
	for a in range(0,34):
		for b in range(0,34):
			for c in range(0,34):
				if (a*packages[0] + b*packages[1] + c*packages[2]) == n:
					return True
	return answer

def mcnuggets_search():
	n = 1
	
	# Stores successful values of n.
	successful_n = [0,0,0,0,0,0]
	
	bestSoFar = [0]	# variable that keeps track of largest
				# number of McNuggets that cannot be
				# bought in exact quantity

	while consecutiveValues(successful_n[-6], successful_n[-5], successful_n[-4], successful_n[-3], successful_n[-2], successful_n[-1]) == False and n < 200:
		if nuggetsTest(n) == True:
			successful_n = successful_n + [n]
			print n
		else:
			bestSoFar = bestSoFar + [n]
		n = n + 1
		
	print "Given package sizes " + str(packages[0]) + ", " + str(packages[1]) + ", and " + str(packages[2]) + ", the largest number of McNuggets that cannot be bought in exact quantity is: " + str(bestSoFar[-1]) + "."

nPtLQmypptMQSGzKShM

[(Anonymous)]

Sharp tihnikng! Thanks for the answer.

[bellerina]

Does anyone know if there is a limit to how many indentations/levels of nested loops is viable in Python? I'm having an issue with problem 3 in that my nested loops work fine until I add another while loop outside of them, and I can't figure out why.

[gchick]

There's no limit. Maybe if you posted a code snippet, someone might be able to spot something that you're not seeing?

(edit: on further research, I'm seeing a couple of references to limits between 20 and several hundred layers of indentation -- I'm not sure if the varying numbers are different versions or implementations or what, but even if it's 20, it's *way* more than anything you'd find in these problem sets, and possibly any sane software design.)

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