Talk:4-rings or 4-squares puzzle

Revision as of 17:38, 25 January 2017 by Nigel Galloway (talk | contribs) (Created page with "=Why 2860?= The equations are: a+b=X b+c+d=X d+e+f=X f+g=X which imply that d = a-c and d = g-e when d=9 the only values are a=9, c=0 when d=8 the values ar...")
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Why 2860?

The equations are:

a+b=X
b+c+d=X
d+e+f=X
f+g=X

which imply that d = a-c

            and d = g-e

when d=9 the only values are a=9, c=0 when d=8 the values are a=9, c=1; a=8, c=0. which for d=0..9 sums to 55.

So there are 55*55 cases to consider for a, c, d, e, and g. Fixing b fixes f so it should not be necessary to consider more than 55*55*10 (which is 25250) cases, which is rather less than the permutations that some solutions are testing!

The task is to determine the number of solutions that there are, for which I need to go a little further.

X must have a minimum value of d when b,c,e,f=0 and a maximum value of 18 when a,b=9. For d=0..9 I generate something Pascal Triangle like for the count Z of solutions:

d=9 Z9 = 10 -> 2*1 + 8*1*1 d=8 Z8 = 38 -> 2*1 + 2*1*1 + 7*2*2 d=7 Z7

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