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MIT OpenCourseWare 6.00 Introduction to Computer Science and Programming: Lesson 5, HW 1
it took me a while to finish this assignment, i had trouble understanding what they wanted me to do in the last two problems.....gladly i think i have done it properly :)
from string import *
# target strings
target1 = 'atgacatgcacaagtatgcat'
target2 = 'atgaatgcatggatgtaaatgcag'
# key strings
key10 = 'a'
key11 = 'atg'
key12 = 'atgc'
key13 = 'atgca'
#####################################
##### Problem Set 1 #####
##### Name: Wolfei #####
##### Collaborators: Internet #####
##### Time: pending #####
def countSubStringMatch(target, key):
instance = 0
w = 0
while find(target, key, w) != -1:
instance += 1
w = find(target, key, w) + 1
return instance
def countSubStringMatchRecursive(target, key):
instance = 0
for lmao in range(0,len(target)):
if target[lmao:len(key)+lmao] == key:
instance += 1
return instance
#####################################
##### Problem Set 2 #####
##### Name: Wolfei #####
##### Collaborators: Internet #####
##### Time: pending #####
def subStringMatchExact(target, key):
instance = 0
w = 0
group = ()
while find(target, key, w) != -1:
group = group + (find(target,key,w),)
instance += 1
w = find(target, key, w) + 1
return group
def subStringMatchExact2(target, key):
group = ()
instance = 0
for lmao in range(0,len(target)):
if target[lmao:len(key)+lmao] == key:
group = group + (lmao,)
instance += 1
return group
#####################################
##### Problem Set 3 #####
##### Name: Wolfei #####
##### Collaborators: Internet #####
##### Time: pending #####
### Function given to help complete problem 3
def subStringMatchOneSub(target,key):
"""search for all locations of key in target, with one substitution"""
allAnswers = ()
if len(key) != 1:
for miss in range(0,len(key)):
# miss picks location for missing element
# key1 and key2 are substrings to match
key1 = key[:miss]
key2 = key[miss+1:]
## print 'breaking key',key,'into',key1,key2
# match1 and match2 are tuples of locations of start of matches
# for each substring in target
match1 = subStringMatchExact(target,key1)
match2 = subStringMatchExact(target,key2)
## print match1, match2
# when we get here, we have two tuples of start points
# need to filter pairs to decide which are correct
filtered = constrainedMatchPair(match1,match2,len(key1))
allAnswers = allAnswers + filtered
## print 'match1',match1
## print 'match2',match2
## print 'possible matches for',key1,key2,'start at',filtered
return allAnswers
else: return " your key has only 1 character, therefore it could substitude all characters in the string"
def constrainedMatchPair(firstMatch,secondMatch,length):
group = ()
for lmao in secondMatch:
for lol in firstMatch:
if lol+length+1 == lmao:
group = group + (lol,)
return group
#####################################
##### Problem Set 4 #####
##### Name: Wolfei #####
##### Collaborators: Internet #####
##### Time: pending #####
def subStringMatchExactlyOneSub(target,key):
x = subStringMatchOneSub(target,key) #long one
y = subStringMatchExact(target,key) #short one
if len(key) != 1:
## print x,y
## print subStringMatchExactlyOneSub(x,y)
x, y = givenumbers(x,y)
if len(x) == 0:
print "there isn't an incorrect character in the string "
else:
print " This is your number"
return x
else:return " your key needs at least two characters in it"
def givenumbers(x,y):
## tuple(x)
## tuple(y)
for n in range(len(y)):
dummy03(x,y)
x, y = dummy03(x,y)
## print dummy03(x,y)
## print x, y
return x , y
def dummy03(x,y):
## a = subStringMatchOneSub(key,target) long one
## b = subStringMatchExact(target,key) short one
## group = ()
## x = 5,15,5,15,0,15,5
## y = 5,15
group = ()
## while len(y) != 0:
for XXX in x:
if y[0] != XXX:
group += (XXX,)
## print group, x
x = group
y = y[1:]
return x, y
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MIT OpenCourseWare 6.00 Introduction to Computer Science and Programming: Lesson 3, HW 1
##### Assignment 2 #####
#####################################
##### Problem Set 1 #####
##### Name: Wolfei #####
##### Collaborators: Internet #####
##### Time: pending #####
## 6 9 20
## 12 18 20 50
## 6 45 0 51
## 12 0 40 52
## 24 9 20 53
## 9 0 0 54
## 6 9 40 55
##
## We can easily derive the next amounts (56....65) like this
## (12 + 18 + 20) + 6 = 56
## (6 + 45 + 0 ) + 6 = 57
## (12 + 0 + 40) + 6 = 58
## (24 + 9 + 20) + 6 = 59
## (9 + 0 + 0) + 6 = 60
## (6 + 9 + 40) + 6 = 61
##
## Now we use the newly created and add 6 to them as we did with
## original ones
##
## [(12 + 18 + 20) + 6] + 6 = 62
## [(6 + 45 + 0 ) + 6] + 6 = 63
## [(12 + 0 + 40) + 6] + 6 = 64
## [(24 + 9 + 20) + 6] + 6 = 65
##### Problem Set 2 #####
##### Name: Wolfei #####
##### Collaborators: Internet #####
##### Time: pending #####
##The theorem is true because once we have found how to create x, x+1, ....x+5
##we can easily find the next combinations of macnuggets that will lead us to
##higher quantities by adding 6 to x, x+1, ...x+5
##### Problem Set 3 #####
##### Name: Wolfei #####
##### Collaborators: Internet #####
##### Time: 3 weeks #####
def Problem3():
x = []
n = 1
while n < 200+1:
if lolcakeA(n) or lolcakeB(n) or lolcakeC(n) == 1:
## print "saving n", n
n += 1
else:
## print "n isnt good", n
x += [n]
n += 1
print "Largest number of McNuggets that cannot be bought in exact quantity is :"
raw_input()
print x[-1]
def lolcakeA(x):
a = 0
b = 0
c = 0
for a in range (0,8+1):
if 6*a+9*b+20*c == x:
return 1
else:
for b in range (0,5+1):
if 6*a+9*b+20*c == x:
return 1
else:
for c in range (0,3+1):
if 6*a+9*b+20*c == x:
return 1
return 0
def lolcakeB(x):
a = 0
b = 0
c = 0
for b in range (0,40+1):
if 6*a+9*b+20*c == x:
return 1
else:
for c in range (0,40+1):
if 6*a+9*b+20*c == x:
return 1
else:
for a in range (0,40+1):
if 6*a+9*b+20*c == x:
return 1
return 0
def lolcakeC(x):
a = 0
b = 0
c = 0
for c in range (0,40+1):
if 6*a+9*b+20*c == x:
return 1
else:
for a in range (0,40+1):
if 6*a+9*b+20*c == x:
return 1
else:
for b in range (0,40+1):
if 6*a+9*b+20*c == x:
return 1
return 0
##### Problem Set 4 #####
##### Name: Wolfei #####
##### Collaborators: Internet #####
##### Time: pending #####
def Problem4():
lista = []
n = 1
x = input(" tell me the lowest number: ")
y = input(" tell me the number in the middle: ")
z = input(" tell me the highest number: ")
packages = (x, y, z)
while n < 200+1:
if lolcakeA2(n,x,y,z) or lolcakeB2(n,x,y,z) or lolcakeC2(n,x,y,z) == 1:
n += 1
else:
lista += [n]
n += 1
print ""
print "Given package sizes", x , y, "and", z,
print "the largest number of McNuggets that cannot be bought in exact quantity is:", lista[-1]
def lolcakeA2(n, x, y, z):
""" n = number you want to check, [xyz] are number of packages of McNuggets"""
x = 0
y = 0
z = 0
for x in range (0,40+1):
if 6*x+9*y+20*z == n:
return 1
else:
for y in range (0,40+1):
if 6*x+9*y+20*z == n:
return 1
else:
for z in range (0,40+1):
if 6*x+9*y+20*z == n:
return 1
return 0
def lolcakeB2(n, x, y, z):
""" n = number you want to check, [xyz] are number of packages of McNuggets"""
x = 0
y = 0
z = 0
for y in range (0,40+1):
if 6*x+9*y+20*z == n:
return 1
else:
for z in range (0,40+1):
if 6*x+9*y+20*z == n:
return 1
else:
for x in range (0,40+1):
if 6*x+9*y+20*z == n:
return 1
return 0
def lolcakeC2(n, x, y, z):
""" n = number you want to check, [xyz] are number of packages of McNuggets"""
x = 0
y = 0
z = 0
for z in range (0,40+1):
if 6*x+9*y+20*z == n:
return 1
else:
for x in range (0,40+1):
if 6*x+9*y+20*z == n:
return 1
else:
for y in range (0,40+1):
if 6*x+9*y+20*z == n:
return 1
return 0
wolfei
2 months ago
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MIT OpenCourseWare 6.00 Introduction to Computer Science and Programming: Lesson 2, HW 1
problem 1- done
problem 2- done too :D
##### Assignment 1 #####
#####################################
##### Problem Set 1 #####
##### Name: Wolfei #####
##### Collaborators: Internet #####
##### Time: 1 week #####
def PrimeSolver3():
a = 1
b = 1
c = a-1
d = 1
counter = 2
i = 1
divisors = 0
while counter <= 1000:#number of prime to stop
if ((a/2)*2) != a:#odd numbers go through
if divisorFinder(a) == 1:#numbers with 1 divisor go through
print a, counter
a += 1
counter += 1
else:a+=1
else: a += 1
print 'THE 1000TH PRIME NUMBER IS', (a-1)
def divisorFinder(x):
""" Finds the Divisors of any number"""
a = 1
numberofdivisors = 0
while(a<x):
if x%a == 0:#numbers that divide perfectly pass
a = a + 1
numberofdivisors += 1
else: a = a + 1#otherwise sent back to loop with an added 1
return numberofdivisors
#####################################
##### Problem Set 2 #####
##### Name: Wolfei #####
##### Collaborators: Internet #####
##### Time: 2days #####
from math import*
def SumsandRatio(n):
""" SumLog(primes) and ratio between <<< and n"""
a = 1
b = 1
c = a-1
d = 1
counter = 0
i = 1
divisors = 0
sumlogs = log(2)
while counter <= n:#number of prime to stop
if ((a/2)*2) != a:#odd numbers go through
if divisorFinder(a) == 1:#numbers with 1 divisor go through
logs = log(a)
sumlogs += logs
a += 1
counter += 1
else:a+=1
else: a += 1
counter += 1
print 'The sum of logs = ', sumlogs
print 'Number n = ', n
print 'Ratio between sumoflogs and n is :', sumlogs/n
wolfei
5 months ago
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