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harpoo4n5
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A = B-1  
« on: Sep 25th, 2007, 12:05pm »		 Quote Quote Modify Modify
Using only the following 3 constructs:
 
X = 0  //A variable can be assigned an initial value of 0
X++    //A variable can be incremented
Loop(X) //A loop that will go through X times
 
How would you implement  
1. A = B.C  
2. A = B-1
 
for 1. I got:  
A= 0
Loop(B)
  Loop(C)
    A++
 
but can't figure out #2
 
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Re: A = B-1  
« Reply #1 on: Sep 25th, 2007, 12:42pm »		 Quote Quote Modify Modify
Hint: how would you implement A = B?
 
Hint 2: you need a third variable
 
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harpoo4n5
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Re: A = B-1  
« Reply #2 on: Sep 25th, 2007, 12:45pm »		 Quote Quote Modify Modify
Not sure how the hints help
 
A = 0
loop (B)  
  A++
 
If I have a 3rd variable C, then I could do:
C = 0
C++
 
which would make C = 1; but not sure how that helps?
« Last Edit: Sep 25th, 2007, 12:59pm by harpoo4n5 » IP Logged
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Re: A = B-1  
« Reply #3 on: Sep 25th, 2007, 1:05pm »		 Quote Quote Modify Modify
Hint 3: The main loop is C = B
 
Hint 4: Increment A one time fewer than C
 
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harpoo4n5
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Re: A = B-1  
« Reply #4 on: Sep 25th, 2007, 1:13pm »		 Quote Quote Modify Modify
Still stuck:
 
C = 0;  
loop B  
  C++
 
but to increment A one less time doesn't seem possible.
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Re: A = B-1  
« Reply #5 on: Sep 25th, 2007, 1:22pm »		 Quote Quote Modify Modify
Hint 5: At the end of each iteration you want A to be what C was at the end of the previous iteration
 
Give Away: Copy C to A before incrementing C
 
Solution:
Loop (B)
  A = 0
  Loop (C)
    A++
  C++
 
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Re: A = B-1  
« Reply #6 on: Sep 25th, 2007, 1:53pm »		 Quote Quote Modify Modify
Of course that implementation is pretty inefficient -- it takes O(B2) operations just to calculate B - 1.  It's possible to write an implementation which only takes O(B) operations. Wink
 
Hint: If X can only be 0 or 1, "Loop(X)" acts like "If X"
 
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Re: A = B-1  
« Reply #7 on: Sep 25th, 2007, 3:05pm »		 Quote Quote Modify Modify

A=0
C++
Loop(B)
   A++
   Loop(C)
      A=0
      C=0

Huh
« Last Edit: Sep 25th, 2007, 3:06pm by FiBsTeR » IP Logged
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Re: A = B-1  
« Reply #8 on: Sep 25th, 2007, 7:57pm »		 Quote Quote Modify Modify
That'll work, and is a few instructions shorter than what I had in mind:
 
A=0
C=0
D=0
D++
Loop(B)
  Loop(C)
    A++
  Loop(D)
    C++
    D=0
 
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