elz: (ada-tubes)
[personal profile] elz posting in [community profile] intro_to_cs
Post and discuss your answers in the comments! (And yay for everybody who posted problem set 1 answers!)

Date: 2009-11-19 02:25 am (UTC)
gchick: Small furry animal wearing a tin-foil hat (Default)
From: [personal profile] 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.

Date: 2009-11-19 02:55 am (UTC)
gchick: Small furry animal wearing a tin-foil hat (Default)
From: [personal profile] 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.

Re: Problem #4

Date: 2010-03-03 03:59 pm (UTC)
From: (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."
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])


Date: 2009-11-20 12:39 am (UTC)
medrin: matlab code with everything but 'hold on' blurred (Default)
From: [personal profile] 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

#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)?')
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')

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:
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:
if c*packets[2]+b*packets[1]+a*packets[0]==n:
if works==True: #if there exists a correct solution the n_right counter incerases
#print n,a,b,c
else: #if there isn't, it resets to zero and that number is stored in highest.
#print n,no_right
print highest #the highest non-solvable number is printed.

Problem 3

Date: 2009-11-20 02:04 am (UTC)
yhlee: wax seal (BPAL Peony Moon (julebug))
From: [personal profile] 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
			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

Date: 2009-11-20 02:05 am (UTC)
yhlee: wax seal (Default)
From: [personal profile] 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
			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]) + "."


Date: 2012-01-09 04:20 am (UTC)
From: (Anonymous)
Sharp tihnikng! Thanks for the answer.

Date: 2009-11-23 03:21 am (UTC)
bellerina: (Default)
From: [personal profile] 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.

Date: 2009-11-23 04:14 am (UTC)
gchick: Small furry animal wearing a tin-foil hat (Default)
From: [personal profile] 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.)
Edited Date: 2009-11-23 04:30 am (UTC)


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