Computer Chocolate Code-off
Rules:
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Get into groups of 4
No computers or phones (or any other technology)
Complete as many problems as possible within the allotted time
You may ask me 4 yes/no questions (and they cannot be “Is this the answer?”)
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Problem 1: prints y y times (Answer: ‘333’)
def f1(x,y):
if (x == y):
return(str(y))
else:
return (str(y) + f1(x+1,y))
print(f1(1,3))
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Problem 2: Product of odd numbers (Answer: 15)
def f5 (a):
if (a <= 0):
return(1)
elif ((a%2) ==0):
return (f5(a - 1))
else:
return (a*f5(a-1))
print(f5(6))
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Problem 3: Counts numbers evenly divisible by 3 (Answer: 4)
def f7(a,b):
if (b <= 0):
return(a)
elif((b%3)== 0):
return(f7(a+1,b-1))
else:
return(f7(a,b-1))
print(f7(0,13))
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Problem 4: (Answer: 42 (had to be there somewhere))
def mk(x,y):
if (y == 0):
return("")
else:
return(str(x**y) + mk(x,y-1))
print(mk(2,2))
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1
Problem 5: prints numbers in order (a string) (Answer: ‘12345’)
def f6(x):
if (x <= 1):
return str(x)
else:
return(f6(x-1) + str(x) )
print(f6(5))
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Problem 6: Sums the numbers in a string (Answer: 13)
def f(x,y,z):
if y == 1:
return z + x
else:
return(f(x%y, y//10, z+x//y))
print(f(1354,1000,0))
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Problem 7: (Answer: “byte”)
def k(x):
if (x > 8):
return ('t')
elif (x > 6):
return('i')
elif (x > 5):
return('e')
elif (x > 3):
return('b')
elif (x > 2):
return('s')
elif (x > 0):
return('y')
else:
return('c')
def k2(x):
if (x == 0):
return ""
else:
return(k2(x//10) + k(x%10))
print(k2(4196))
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Problem 8: (Makes a number palindrome (Answer: 45654))
def fg(x,y,q,r):
if (x == 0):
return ""
else:
if (x == y):
return(str(r+x)+fg(x+q,y,-1,r))
else:
return(str(r+x)+fg(x+q,y,q,r))
print(fg(1,2,1,3))
2
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Problem 9: Determines whether a number is a perfect number or not (Answer: True)
def m(x,y,z):
if (x ==y):
return(z==y)
else:
if y%x == 0:
return(m(x+1,y,z+x))
else:
return(m(x+1,y,z))
print(m(1,28,0))
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Problem 10: (Replaces everything that’s not a multiple of 3 with dots (Answer: ..3..6.)
def h(x):
if (x == 0):
return("")
else:
if (x % 3) == 0:
return(h(x - 1) + str(x))
else:
return(h(x-1) + ".")
print(h(7))
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Problem 11: (Finds the GCD Answer: 3 )
def f4 (a, b):
if (b == 0):
return (a)
if (a < b):
return f4 (b, a)
else:
return f4 (b, a%b)
print(f4(27,12))
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Problem 12: Count the number of prime numbers between 2 and x (Answer: 5)
def f12(x,y):
if (y>(x//2)):
return(True)
elif (x%y==0):
return(False)
else:
return (f12(x,y+1))
def g12(x,y):
if (x >y):
return(0)
elif f12(x,2):
return (1 + g12(x+1,y))
else:
return(g12(x+1,y))
print(g12(2,11))
3
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Problem 13: Converts binary to decimal (Answer: 13)
def p2(y):
if y == 0:
return 1
else:
return (p2(y-1) * 2)
def p3(x,y):
if (x == 0):
return (0)
else:
if (x%2 == 1):
return(p2(y) + p3(x//10,y+1))
else:
return(p3(x//10,y+1))
print(p3(1101,0))
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Problem 14: (that takes an integer and returns the reverse of that integer (as an integer) Answer: 7423
def f14(x,y):
if (x//y)==0:
return(y//10)
else:
return(f14(x,y*10))
def g14(x,y):
if (f14(x,1) == 1):
return(int(str(x) + y))
else:
return g14(x % f14(x,1), str(x // f14(x,1) )+y)
print(g14(3247,""))
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Problem 15: Answer: 3332244441
def fn1(x,y):
if (x == y):
return("")
else:
return (str(y) + fn1(x+1,y))
def fn2(x):
if (x<9):
return(fn1(0,x))
else:
return(fn2(x//10) + fn1(0,x%10))
print(fn2(3241))
4