in reply to Re^2: Algorithm for cancelling common factors between two lists of multiplicands

in thread Algorithm for cancelling common factors between two lists of multiplicands

I guess my approach is same as Limbic~Region as I was using subtraction of lower factorial terms with higher ones.

I copied the input list from one of your previous examples and I have ` 45700` instead of `4570` but aside from that the implementation should be easy to understand.

I guess if you sort numerator and denominator and subtract them you should be all set. I have not proved this formally but here is a stab at a simple proof that shows sorting and subtracting should work...

Let the fraction be

I don't think this is a rigourous proof this method but i sort of feel sorting and subtracting should give us what we need...X! Y! Z! ------- a! b! c! WLOG let's assume that X > Y > Z and a > b > c. Also let's assume that + b > Y , X > a (weaker assumption: Z > c)...<p> NOTE: if we divide X!/a! we will have (X-a) elements <p> To Prove: (X-a) + (Z-c) + (b-Y) is the shortest list one can find or in other words (X-a) + (Z-c) + (b-Y) <= (X-p) + (Z-q) + (r-Y) for any p,q,r in permut +ation(a,b,c) Proof: From the above equation since b > Y and Z > c, r should be equal to ei +ther a or b. If r = b then the solution is trivial<p> If r = a then we get (X-a) + (Z-c) + (b-Y) ?<= (X-b) + (Z-c) + (a-Y) canceling terms -a - c + b ?<= -b -c + a -a + b ?<= -b + a ====> YES since a > b we see that r = a is not the smallest list so r = b<p> Similarly we can also show that (X-a) + (Z-c) + (b-Y) <= (X-a) + (Y-c) + + (b-Z)

cheers

SK

PS: I think there will be 47448 elements and not 47444 as you suggested? as you need to count the first element too..

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