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arithmetic - Puzzle with Mathematica


puzzle


Hello everyone, This is a puzzle I got from someone via social media. Basically, we need to fill up the boxes with the numbers 1-9 (no repetitions) that fit the multiplication and addition operations.


I managed to solve this puzzle by using a brute force method in Excel+VBA. However, it would be very interesting if it can be solved in Mathematica with its specialty as computational software. Any idea will be appreciated.


Thanks.




Answer



A non brute-force approach is the following, similar to my answer for the Zebra Puzzle.


Both puzzles are examples of constrainst satisfaction problems, that can be solved with Reduce/Minimize/Maximize or, more efficiently, with LinearProgramming.


The good about this approach is that you can easily extend and apply to many similar problems.


The common part:



  • Assign an index i to each box from top left, i=1,2,…,9.

  • In each box you should put a digit k, k=1,…,9.

  • Assign an index l to the whole number/row, l=1,…,5.

  • the variable x[i,k] is 1 if there is the digit k in the cell i and 0 otherwise.


  • d[i] is the digit in cell i.

  • n[l] is the whole number in the row l (one or two cell).




The easier and slower approach is with Maximize. Build constraints and pass to Maximize with a constant objective function, so Maximize will try only to satisfy constraints. Constraints are:



  • n[1] * n[2] == n[3]

  • n[3] + n[4] == n[5]

  • each cell should be filled with exactly one digit

  • each digit should be placed in exactly one cell


  • 0 <= x[i,k] <= 1, x[i,k] \elem Integers


That's all.


d[i_] := Sum[x[i, k] k, {k, 9}]
n[l_] := FromDigits[d /@ {{1, 2}, {3}, {4, 5}, {6, 7}, {8, 9}}[[l]]]

solution = Last@Maximize[{0, {
n[1]*n[2] == n[3],
n[3] + n[4] == n[5],
Table[Sum[x[i, k], {k, 9}] == 1, {i, 9}],

Table[Sum[x[i, k], {i, 9}] == 1, {k, 9}],
Thread[0 <= Flatten@Array[x, {9, 9}] <= 1]}},
Flatten@Array[x, {9, 9}], Integers];

Array[n, 5] /. solution


{17, 4, 68, 25, 93}



Not fast (not linear).





A faster approach is to use LinearProgramming, but you need to:



  • change the first constraint so that it become linear

  • manually build matrix and vectors input for LinearProgramming (see docs)


The next piece of code do that. Please note that the single non-linear constraint n[1]*n[2] == n[3] has been replaced with 18 linear "conditional" constraints.


d[i_] := Sum[x[i, k] k, {k, 9}]
n[l_] := FromDigits[d /@ {{1, 2}, {3}, {4, 5}, {6, 7}, {8, 9}}[[l]]]


vars = Flatten@Array[x, {9, 9}];

constraints = Flatten@{
Table[{
k n[1] >= n[3] - 75 (1 - x[3, k]),
k n[1] <= n[3] + 859 (1 - x[3, k])
}, {k, 9}],
n[3] + n[4] == n[5],
Table[Sum[x[i, k], {k, 9}] == 1, {i, 9}],
Table[Sum[x[i, k], {i, 9}] == 1, {k, 9}]};


bm = CoefficientArrays[Equal @@@ constraints, vars];
solution = LinearProgramming[
Table[0, Length@vars],
bm[[2]],
Transpose@{-bm[[1]],
constraints[[All, 0]] /. {LessEqual -> -1, Equal -> 0,
GreaterEqual -> 1}},
Table[{0, 1}, Length@vars],
Integers

];

Array[n, 5] /. Thread[vars -> solution]


{17, 4, 68, 25, 93}



The execution is now about instantaneous.


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