Re: Re:Playing with numbers

*To*: mathgroup at smc.vnet.net*Subject*: [mg50228] Re: [mg50223] Re:[mg50135] Playing with numbers*From*: DrBob <drbob at bigfoot.com>*Date*: Sun, 22 Aug 2004 00:19:37 -0400 (EDT)*References*: <200408210704.DAA24819@smc.vnet.net> <opsc3o91yziz9bcq@monster.cox-internet.com> <opsc3siv10iz9bcq@monster.cox-internet.com>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

And now, let's look at only NINE cases. As in the 18-case solution, one of {a,d,i} must be 1 and {b,f,g} is a permutation of {2,3,4}, and we get these general solutions: conditions = Simplify@{h - f == i, f*g == c, e == f*b, a + b == h, c - d == a} chooseOne[x_] := {a, b, c, d, e, f, g, h, i} /. First@Solve[conditions, Complement[{a, c, d, e, h, i}, {x}]] /. x -> 1 chooseOne/@{a,d,i}//ColumnForm {{1,b,f g,-1+f g,b f,f,g,1+b,1+b-f}, {-1+f g,b,f g,1,b f,f,g,-1+b+f g,-1+b-f+f g}, {1-b+f,b,f g,-1+b-f+f g,b f,f,g,1+f,1}} Then, as in the 24-case solution, we argue that h>=6. In the first solution above, h==1+b, but b<=4, so that's not a solution. In the third solution above, h==1+f, but f<=4, so that's not a solution. That leaves only the middle solution, d==1. (This argument is actually evident from the original equations, WITHOUT using the solutions above.) Applying that to all six permutations of {2,3,4} makes 9 cases we've had to examine, and we get: Select[chooseOne[d] /. Thread[{b, f, g} -> #] & /@ Permutations@{2, 3, 4}, Union@# == Range@9 &] {{5,4,6,1,8,2,3,9,7}} Bobby On Sat, 21 Aug 2004 17:24:21 -0500, DrBob <drbob at bigfoot.com> wrote: > And here's a solution that checks only 18 cases that, still, is no more difficult to analyze than before. > > conditions = Simplify@{h - f == i, f*g == c, e == f*b, a + b == h, c - d == a} > {f + i == h, c == f g, e == b f, a + b == h, a + d == c} > > (1) Because all nine symbols must be distinct, those involved in product or quotient equations -- b, c, e, f, and g -- cannot be 1. Nor can the sums c and h. Hence one of {a, d, i} must be 1 (three possibilities). > > (2) Therefore, 2 <= g == c/f <=c/2 <= 4. The same argument applies to f and b, so {b,f,g} is a permutation of {2,3,4}. > > If one of {a,d,i} is chosen to be 1 we can solve for the other five variables (other than that variable and b, f, g). This reduces the number of cases to > > 3*3! > > 18 > > chooseOne[x_] := {a, b, c, d, e, f, g, h, i} /. First@Solve[conditions, Complement[{a, b, c, d, > e, f, g, h, i}, {b, f, g, x}]] /. x -> 1 > chooseOne /@ {a, d, i} > > {{1,b,f g,-1+f g,b f,f,g,1+b,1+b-f},{-1+f g,b,f g,1,b f,f,g,-1+b+f g,-1+ > b-f+f g},{1-b+f,b,f g,-1+b-f+f g,b f,f,g,1+f,1}} > > solution[s_List, {bx_, fx_, gx_}] := Module[{sol = s /. Thread[{b, f, g} -> {bx, fx, gx}]}, > If[Union@sol == Range@9, sol, {}] ] > Flatten[Outer[solution, chooseOne /@ {a, d, i} // > Evaluate, Permutations@{2, 3, 4}, 1], 1] /. {} -> Sequence[] > > {{5, 4, 6, 1, 8, 2, 3, 9, 7}} > > Bobby > > On Sat, 21 Aug 2004 16:14:15 -0500, DrBob <drbob at bigfoot.com> wrote: > >> That's pretty cool! It looks at 45 cases, though; I have a solution that looks at 24, with a pre-coding analysis that's no more onerous (I think). >> >> conditions = {h - f == i, f*g == c, e == f*b, a + b == h, c - d == a}; >> >> Note g == c/f <= c/2 <= 4. Also g >= 2 as otherwise c == f. The same argument applies to f and b. Hence {b,f,g} is a permutation of {2,3,4}. >> >> Given that, we try solving for the other six variables. But Solve can get only five (there are five equations). >> >> First[Solve[conditions, {a, c, d, e, i}]]; >> simpleSolve = {a, b, c, d, e, f, g, h, i} /. % >> {-b + h, b, f*g, b + f*g - h, b*f, f, g, h, -f + h} >> >> Note that four of these depend on h. The sums h==f+i==a+b imply that h is strictly greater than a, b, i, and f. Since {b, f, g} is a permutation of {2,3,4}, at least one of a and i must be bigger than four. Hence h>=6. >> >> Therefore, the number of possible solutions reduces to: >> >> 3!4 >> >> 24 >> >> In particular, the possible values of {b, f, g, h} are: >> >> bfgh = Flatten[Outer[Flatten@{#1, #2} &, Permutations@{2, 3, 4}, Range[6, 9], 1], 1]; >> >> Finally we check all 30 cases and list the solutions: >> >> check@{bx_, fx_, gx_, hx_} := Module[ >> {s = simpleSolve /. Thread[{b, f, g, h} -> {bx, fx, gx, hx}]}, >> If[Sort@s == Range@9, s, {}]] >> check /@ bfgh /. {} -> Sequence[] >> >> {{5, 4, 6, 1, 8, 2, 3, 9, 7}} >> >> That's a lot better than looking at 9!==362880 solutions!! >> >> Bobby >> >> On Sat, 21 Aug 2004 03:04:29 -0400 (EDT), Fred Simons <f.h.simons at tue.nl> wrote: >> >>> The question is to solve with Mathematica the set of equations >>> (1) H - F = I >>> (2) F x G = C >>> (3) E / B = F >>> (4) A + B = H >>> (5) C - D = A >>> where every variable from A to I stands for a number from 1 to 9 and each >>> number appears only once. >>> >>> Here is a solution where Mathematica does the computations that we otherwise >>> had to do by hand. >>> >>> Since the symbols represent different integers between 1 and 9, we conclude >>> from the equations i+f=h, a+b=h that h has to be at least 5. From the >>> equations c=f*g, e=f*b we conclude that f is between 2 and 4, that g is at >>> least 2, that e is at least 2, that c is at least 6 and that e is at least >>> 6. >>> >>> Given values for f, h, c and e, we can compute the other variables. The >>> result has to be a permutation of the numbers 1, ..., 9. >>> >>> In[21]:= solution = {a, b, c, d, e, f, g, h, i} /. >>> Solve[{h - f == i, f*g == c, e/b == f, >>> a + b == h, c - d == a}, {a, b, d, g, i}][[1]] >>> >>> Out[21]= >>> {-((e - f*h)/f), e/f, c, c + e/f - h, e, f, c/f, h, -f + h} >>> >>> In[23]:= >>> Timing[Do[If[Length[Union[solution]] == 9, >>> Print[solution]], {f, 2, 4}, >>> {c, Ceiling[5/f]*f, 9, f}, {e, Ceiling[5/f]*f, 9, f}, {h, 5, 9}]] >>> >>>> From In[23]:= >>> {5,4,6,1,8,2,3,9,7} >>> >>> Out[23]= >>> {0. Second,Null} >>> >>> So indeed there is only one solution. >>> >>> Fred Simons >>> Eindhoven University of Technology >>> >>> >>> >> >> >> > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Re:Playing with numbers***From:*"Fred Simons" <f.h.simons@tue.nl>