Difference between revisions of "CompSciWeek2"

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(Reading Assignment)
 
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print "The GCD of %d and %d is %d"%(a,b,b)
 
print "The GCD of %d and %d is %d"%(a,b,b)
  +
</source>
  +
  +
== Improved Version ==
  +
<source lang="python">
  +
doc = """
  +
Euclid's Algo.:
  +
reduce 7/30 - 8/18
  +
  +
30 = 6*5
  +
18 = 6*3
  +
(7*3 - 8*5)/(6*5*3)
  +
  +
gcd(a,b)
  +
  +
gcd(30,18) = 6
  +
= gcd(18,12) = 6
  +
= gcd(12,6)
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= gcd(6, 0) = 6 [termination]
  +
  +
show: gcd(a, b) = gcd(b, a mod b)
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... gcd(a, 0) = a
  +
  +
write:
  +
a = s*u, b = t*u
  +
u also divides r = a % b = a - b*q = (s - q*t)*u
  +
  +
  +
Properties:
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(a + b) mod c = a % c + b % c
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a - b mod c = a mod c - b mod c
  +
(a*b) mod c = (a mod c) * (b mod c)
  +
  +
a*(b/a)
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"""
  +
  +
# always return a <= b
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def order(a, b):
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if a < b:
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return a, b
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else: # a >= b
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return b, a
  +
  +
b = 1547
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a = 224
  +
  +
# Input integer a, b
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# Output: integer gcd(a,b)
  +
def gcd(a, b):
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b, a = order(b, a)
  +
  +
while b != 0:
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#print "%d, %d"%(a,b)
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a = a % b # a < b
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a, b = b, a
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return a
  +
  +
r = gcd(a, b)
  +
  +
str = "The GCD %d of %d and is %d "%(a, b, r)
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print str
  +
  +
# subtracting numbers in quotient form
  +
def qsub((a, b), (c, d)):
  +
return (a*c -b*d), (c*d)
 
</source>
 
</source>

Latest revision as of 10:56, 18 September 2014

(Wed. only)

Reading Assignment

  • Beginning Python, Chapter 5

Algorithms, Continued

  • Complexity notation, O(n), etc.
  • Loop Complexity
  • First algorithms (Horner, Euclid, Babylonian)
  • Code walk-through for a poorly designed Euclids algo.
  • The KISS, DRY, and incremental principles
  • Loading python modules

Poorly Designed Euclid's Algo.

<source lang="python"> a = 1547 b = 224

while(1):

 if a < b:
   c = a
   a = b
   b = c
 a = a % b
 if a < b:
   c = a
   a = b
   b = c
 if a == 0:
   break
 print "%d, %d"%(a,b)

print "The GCD of %d and %d is %d"%(a,b,b) </source>

Improved Version

<source lang="python"> doc = """ Euclid's Algo.: reduce 7/30 - 8/18

30 = 6*5 18 = 6*3 (7*3 - 8*5)/(6*5*3)

gcd(a,b)

gcd(30,18) = 6 = gcd(18,12) = 6 = gcd(12,6) = gcd(6, 0) = 6 [termination]

show: gcd(a, b) = gcd(b, a mod b)

 ... gcd(a, 0) = a

write:

 a = s*u, b = t*u
 u also divides r = a % b = a - b*q = (s - q*t)*u


Properties:

   (a + b) mod c = a % c + b % c
   a - b mod c = a mod c - b mod c
   (a*b) mod c = (a mod c) * (b mod c)

a*(b/a) """

  1. always return a <= b

def order(a, b):

 if a < b:
   return a, b
 else: # a >= b
   return b, a

b = 1547 a = 224

  1. Input integer a, b
  2. Output: integer gcd(a,b)

def gcd(a, b):

 b, a = order(b, a)
 while b != 0:
   #print "%d, %d"%(a,b)
   a = a % b # a < b
   a, b = b, a
 return a

r = gcd(a, b)

str = "The GCD %d of %d and is %d "%(a, b, r) print str

  1. subtracting numbers in quotient form

def qsub((a, b), (c, d)):

   return (a*c -b*d), (c*d)

</source>