# Talk:4-rings or 4-squares puzzle

# Why 2860?[edit]

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's Triangle like for the count Z of solutions: d=9 Z9 = 10 ->10*1 d=8 Z8 = 38 -> 2*1 + 9*2*2 d=7 Z7 = 82 -> 2*1 + 2*2*2 + 8*3*3 d=6 Z6 =140 -> 2*1 + 2*2*2 + 2*3*3 + 7*4*4 d=5 Z5 =210 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 6*5*5 d=4 Z4 =290 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 2*5*5 + 5*6*6 d=3 Z3 =378 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 2*5*5 + 2*6*6 + 4*7*7 d=2 Z2 =472 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 2*5*5 + 2*6*6 + 2*7*7 + 3*8*8 d=1 Z1 =570 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 2*5*5 + 2*6*6 + 2*7*7 + 2*8*8 + 2*9*9 d=0 Z0 =670 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 2*5*5 + 2*6*6 + 2*7*7 + 2*8*8 + 2*9*9 + 1*10*10 Sum of Z0 through Z9 is 2860

--Nigel Galloway (talk) 17:49, 25 January 2017 (UTC)

# Why nearly all write "non-unique solutions" ?[edit]

I think the unique solutions are part of the 2860 solutions.So I think calling them solutions is the right term --Horsth

There is a better algorithm for the solutions when they are unique: Take 1 from 7 as d and leave set of 6 remaining Take 1 from 6 as 'a' fixing c if part of the remaining set Take 1 from 4 as g fixing e if part of the remaining set which leaves 2 values for b. So maximum of 336 combinations (7*6*4*2) to test, rather than 5040 permutations of all 7.

--Nigel Galloway (talk) 18:29, 25 January 2017 (UTC)

~~Some missing solutions~~[edit]

~~It looks like some of the current implementations are missing some of the puzzle solutions.~~ --Rdm (talk) 12:55, 10 June 2017 (UTC)

4 5 3 1 6 2 7 7 2 6 1 3 5 4 3 7 2 1 5 4 6 6 4 1 5 2 3 7 6 4 5 1 2 7 3 7 3 2 5 1 4 6 3 8 1 2 4 5 6 <- 3 8 1 2 5 4 7 <- 3 8 2 1 4 6 5 <- 3 8 2 1 6 4 7 <- 4 7 1 3 2 6 5 5 6 2 3 1 7 4

- Re the J solution (reproduced above) Ummm. I think you have a bug in your list generation. Where does 8 fall in 1 through 7? --Thundergnat (talk) 13:34, 10 June 2017 (UTC)