A Lazier Way to Solve Knuth’s Skeleton Multiplication Puzzle

Last time, I described a solution to the following “skeleton multiplication puzzle” from Don Knuth’s The Art of Computer Programming:

        N N N  (1)
x N N N N (2)
N N N 7 (3)
N N N N (4)
N N N N (5)
N N N (6)
7 7 7 7 7 7 (N != 7)

The answer found by Knuth is the following:

        5 3 9  (1)
    x 1 4 4 3  (2)
      1 6 1 7  (3)
    2 1 5 6    (4)
  2 1 5 6      (5)
  5 3 9        (6)
  7 7 7 7 7 7 

I was able to find a second solution where there was a leading zero in row (4). To find it, I considered all the prime factors of 777777 and reasoned about which pairs could fit the pattern of the puzzle.

In this post I want to show you how I originally found this solution.

I wrote a bit of code in Python to express part of the problem as a constraint programming (CSP) model. Instead of trying to work out the solution myself, I write down the conditions that must be true, and let the constraint programming solver do the work.

The basic approach for solving a puzzle using a solver is:

  1. Write down the variables (unknowns) that I want the solver to find.
  2. Write down all of the constraints (logical conditions) that need to be true.
  3. Turn the above into the right API calls for the solver I am using.

Here’s how it works (and here’s a link to my code). I used the python_constraint package as my CSP solver. To specify the puzzle we need a Problem() object:

problem = Problem()

In a CSP I first define variables that I want the solver to find. The variables I care most about are n_1 and n_2. I know that n_1 < 1000 and n_2 < 10000. addVariable‘s first argument is the name of the variable and the second is the domain.

problem.addVariable("n_1", list(range(100, 999+1)))
problem.addVariable("n_2", list(range(1000, 9999+1))) 

I also am interested in the digits of each number. I will call the digits for the first (d_11, d_12, d_13), and the second (d_21, d_22, d_23, d_24). Remember that none of these digits can be a seven.

not_seven = [0, 1, 2, 3, 4, 5, 6, 8, 9] # no sevens allowed!                

# n_1 = d_13 d_12 d_11                                                      
problem.addVariable("d_13", not_seven)
problem.addVariable("d_12", not_seven)
problem.addVariable("d_11", not_seven)
# n_2 = d_24 d_23 d_22 d_21                                                 
problem.addVariable("d_24", not_seven)
problem.addVariable("d_23", not_seven)
problem.addVariable("d_22", not_seven)
problem.addVariable("d_21", not_seven) 

Now I start writing down conditions that I know to be true. The first is that the product is a bunch of sevens:

problem.addConstraint(lambda n_1, n_2: n_1 * n_2 == 777777, ['n_1', 'n_2'])

Even if you know Python this may be confusing because you may not be familiar with the python_constraint package. The point is that I am providing an anonymous (lambda) function that evaluates the product of n_1 and n_2 as the first argument, and the names of the variables I defined earlier as the second.

Next, if I put the digits d_1x together, I get n_1:

problem.addConstraint(lambda d_11, d_12, d_13, n_1:
                      100 * d_13 + 10 * d_12 + d_11 == n_0,
                      ['d_11', 'd_12', 'd_13', 'n_1'])

Same for n_2:

problem.addConstraint(lambda d_21, d_22, d_23, d_24, n_2:
                      1000 * d_24 + 100 * d_23 + 10 * d_22 + d_21 == n_2,
                      ['d_21', 'd_22', 'd_23', 'd_24', 'n_2']) 

Also, as considered earlier, when I multiply the last two digits of n_1 and n_2, the result must end in 7:

problem.addConstraint(lambda d_11, d_21: (d_11 * d_21) % 10 == 7, ['d_11', 'd_21'])

Now, if I wanted to be complete, I would write down a bunch more conditions for all of the Xs that appear in rows 3-6, using the rules of long multiplication. But why would I want to do that? The point is to be lazy.

Let’s just solve the problem, ask for all the solutions, and print them out.

s = problem.getSolutions()
# s is a list of solutions. Each solution is a dict of variable values.     
     for t in s:
         print(f"{t['n_0']} x {t['n_1']} = {t['n_0'] * t['n_1']}") 

Wow! The same four that I discovered above. This was “easier” for me in the sense that I already knew how to use the python_constraint package, so I could just do the coding above without having to “think”. Doing it “the old fashioned way” was in the end more fun, but requires more deliberation and focus for me.

Knuth skeleton multiplication puzzle.

Author: natebrix

Follow me on twitter at @natebrix.

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