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Prove the algorithm for LCM

  1. Mar 5, 2005 #1
    algorithm to prove lcm:

    while a != b
    if a < b
    a:= a + m
    b:= b + n

    //postcondition: a is the lcm(m,n)

    what's the loop invariant?I thought it is(not sure):
    lcm(ak, bk) = lcm(ak/m, bk/n) *lcm(m,n)
    I am not sure and also impossible to prove my loop invariant....
    Another loop invariant I thought is "ak=qm and bk=rn",but I don't know how to
    show it's the lcm of m and n,
    Thanks in advance!!
  2. jcsd
  3. Mar 5, 2005 #2
    Forget it! problem soloved ...

    Forget it! problem soloved ...
  4. Mar 6, 2005 #3
    I'm curious, how do you solve that?
  5. Mar 7, 2005 #4
    loop invariant..

    the loop invariant should be
    1)ak >=m and bk> =m
    2)ak mod m=0 and bk mod n=0
    3)ak<=lcm(m,n) and bk<=lcm(m,n)

    U can prove a is the lcm(m,n) by these LI..
  6. Mar 8, 2005 #5
    Cool, I see.
  7. Mar 8, 2005 #6
    Wait, I think you also need ak > a(k-1) and bk > b(k-1)

    That's not really a loop invariant though because it's not true until the second iteration.

    I guess you could say something like "if k is > 1, then (that condition) is true."
    Last edited: Mar 8, 2005
  8. Mar 9, 2005 #7
    U right..

    I forget to mention that.. m,n>=1 is the precondition..
  9. Mar 10, 2005 #8
    No, that's not what I said. For example, what if the loop were:
    for(i = 0; i < 10; i++)
    a = m;
    b = n;

    That satisfies the loop invariants you gave, but it's not going to find the LCM.
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