This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

Title: Cliff Puzzle 7: 3x3 Recursion

From: cliff@watson.ibm.com

If you respond to this puzzle, if possible please send me your name,

address, affiliation, e-mail address, so I can properly credit you if

you provide unique information. PLEASE ALSO directly mail me a copy of

your response in addition to any responding you do in the newsgroup. I

will assume it is OK to describe your answer in any article or

publication I may write in the future, with attribution to you, unless

you state otherwise. Thanks, Cliff Pickover

* * *

Consider the 3x3 array below. All nine digits are used exactly once.

1 9 2 3 8 4 5 7 6

Notice that "384" is twice the number in the first row, and that

"576" is three times the number in the first row.

Questions:

1. Are there other ways of arranging the number to produce the same

result using each digit only once and the same rules?

Remember, the second row must be twice the first. The third row

must be 3 times the first row.

2. Start with the number in the last row (e.g "576" or any other

solution you may find) and continue to form another 3x3 matrix using the

same rules with the new starting number. In other words, the number in

the second row must be twice the first. The third row must be three

times the first. (For this problem you may truncate any digits in the

beginning. For example, 1384 would become 384.)

Keep going. How many matrices can you create before it is impossible

to continue. Again, each digit must be used only once

in each matrix.

pickover/pickover.07.s

-------------------------

> Title: Cliff Puzzle 7: 3x3 Recursion > Consider the 3x3 array below. All nine digits are used exactly once. > 1 9 2 > 3 8 4 > 5 7 6 > Questions: > 1. Are there other ways of arranging the numbers to produce the same > result using each digit only once and the same rules?

YES .

2 1 9 2 7 3 3 2 7 4 3 8 5 4 6 6 5 4 6 5 7 8 1 9 9 8 1

> same rules with the new starting number. In other words,

> the last number becomes the first, and the number in

> the new second row must be twice the first. The third row must be three

> times the first. (For this problem you may truncate any digits in the

> beginning. For example, 1384 would become 384.)

NONE. There is no solution to the continuation problem.

Bye.

Amit Agarwal

> to continue? Again, each digit must be used only once

> in each matrix.

>

>

-------------------------

By exhaustive search I found that there are only four such arrays.

Here they are:

1 9 2 2 1 9 2 7 3 3 2 7 3 8 4 4 3 8 5 4 6 6 5 4 5 7 6 6 5 7 8 1 9 9 8 1

Since these are the only four it is clear from inspection that

none of the last numbers ever begin another array with the desired

properties.

Bob Murphy (rmurphy@aludra.usc.edu)

-------------------------

Well, I think I have an answer to both parts. I did what should have been

a complete analysis of all possible column combinations, but it is

certainly possible that I missed a point somewhere. If you don't get any

answers contradicting this one, I'd be happy to send you my analysis.

Anyway - for part 1, I found the following three matrices (in additionn

to the one you gave):

2 1 9 2 7 3 3 2 7 4 3 8 5 4 6 6 5 4 6 5 7 8 1 9 9 8 1

Note that the first one of these can be generated from yours by moving

the third column to the first position, and the third one can be generated

from the second similarly. In both cases, one column does not receive

or produce any carryovers, so it can be placed at either end.

For part 2, there is obviously no solution, assuming that these are indeed

the only four matrices satisfying the requirements. In my analysis, I

included columns with carryovers in all positions, so if there were any

matrices that would satisfy the relaxed condition of part 2 I should

have found them.

Dan Blum

Institute for the

Learning Sciences

Northwestern U.

blum@ils.nwu.edu

-------------------------

Cliff,

In article <1blrk9INN10s@aludra.usc.edu> (Bob Murphy) writes:

>By exhaustive search I found that there are only 4 starting numbers

>which produce a 3x3 array with the desired property. Here they are:

>

> 1 9 2 2 1 9 2 7 3 3 2 7 > 3 8 4 4 3 8 5 4 6 6 5 4 > 5 7 6 6 5 7 8 1 9 9 8 1

For each of these solutions I happened to notice that the sum of each row

is a constant:

sum(row1) = 12 sum(row2) = 15 sum(row3) = 18

(necessary but not sufficient condition)

And the sums all differ by the same constant (3). I wonder if this

property may somehow be generalized to matrices of higher degree?

Regards,

-- Greg Schmidt schmidtg@iccgcc.decnet.ab.com

-------------------------

> If you respond to this puzzle, if possible please send me your name, address,

> affiliation, and e-mail address so I can credit you in the future if needed.

> If you like, tell me a little bit about yourself so I can cite you

> appropriately if you provide unique information. PLEASE ALSO directly mail

> me a copy of your response in addition to any responding you do in the

> newsgroup. I will assume it is OK to describe your answer in any article or

> publication I may write in the future, with attribution to you, unless you

> state otherwise.

> Thanks, Cliff Pickover

>

> Consider the 3x3 array below. All nine digits are used exactly once.

>

> 1 9 2 > 3 8 4 > 5 7 6

>

> Notice that "384" is twice the number in the first row, and that

> "576" is three times the number in the first row.

>

> Questions:

> 1. Are there other ways of arranging the numbers to produce the same

> result using each digit only once and the same rules?

> Remember, the second row must be twice the first. The third row

> must be 3 times the first row.

>

> 2. Start with the number in the last row (e.g "576" or any other

> solution you may find) and continue to form another 3x3 matrix using the

> same rules with the new starting number. In other words,

> the last number becomes the first, and the number in

> the new second row must be twice the first. The third row must be three

> times the first. (For this problem you may truncate any digits in the

> beginning. For example, 1384 would become 384.)

>

> Keep going. How many matrices can you create before it is impossible

> to continue? Again, each digit must be used only once

> in each matrix.

Well, this is probably not news to you by now, but I only get four solutions

to the original problem:

1 9 2 2 1 9 2 7 3 3 2 7 3 8 4 4 3 8 5 4 6 6 5 4 5 7 6 6 5 7 8 1 9 9 8 1

If we relax the rules slightly and allow zeroes, and just specify that the nine

numbers only have to be different, then we get two more solutions:

0 7 8 2 6 7 1 5 6 5 3 4 2 3 4 8 0 1

both of which use the digits 0-8, which may be of interest.

The second problem (in either form) has only the above solutions, with only one

matrix in each solution.

If we switch to base 9 (where we must use a zero), there is no solution to the

first, and only one solution to the second (with only one matrix):

4 8 1 0 7 2 5 6 3

In fact, I considered 3 versions of problem 2. In all cases zeroes were

allowed, but the 9 numbers had to be different. For each of them the first 3x3

matrix has to meet the original specifications; where they differ is in how the

succeeding matrices are constructed. In the ensuing discussion, the original

number is called 'n'. So in the example given with the problem, n is 192.

Version A: The second matrix consists of rows with n*2, n*3 and n*4 in them.

(The last three digits of those, anyway.) The next would have n*3,

n*4, and n*5, then n*4, n*5, n*6, etc.

Version B: The second matrix consists of n*3, n*6, n*9. (This is essentially

the second problem as given.)

Version C: The second matrix consists of n*4, n*5, n*6. The next would have

n*7, n*8, n*9 etc.

Results for various bases:

Base 9: A, B, C: 4 8 1 0 7 2 5 6 3 Base 10: A, B, C: 0 7 8 1 9 2 2 1 9 2 6 7 2 7 3 3 2 7 1 5 6 3 8 4 4 3 8 5 3 4 5 4 6 6 5 4 2 3 4 5 7 6 6 5 7 8 0 1 8 1 9 9 8 1 In addition, with version C, we get a second matrix for 219. 2 1 9 8 7 6 4 3 8 ==> 0 9 5 6 5 7 3 1 4 Base 11: (A, B, etc. represent 10, 11, etc..) A, B, C: 18 solutions. From now on, I'll only show the multiple matrix ones. A: 5 A 1 0 9 2 6 7 4 2 3 8 0 9 2 ==> 6 8 3 2 3 8 ==> 9 0 1 6 8 3 1 7 4 9 0 1 4 7 5 B: 9 3 4 5 A 1 7 6 8 ==> 0 9 2 5 A 1 6 8 3 C: 8 9 1 2 3 4 6 7 2 ==> 0 1 5 4 5 3 8 A 6 (Note that the B solution ends in an A solution matrix!) Base 12: 19 solutions A: 7 3 4 2 6 8 2 6 8 ==> 9 A 0 9 A 0 5 1 4 B: None C: 3 5 7 1 A 4 6 B 2 ==> 5 3 B A 4 9 8 9 6

Base 13: 71 solutions...and it rapidly increases from here....

The number of solutions rises rapidly, as we might expect, as the more possible

values for digits there are in the base, the more likely the set of 9 will be

distinct. If we look at solutions which only involve the digits 1-9, then the

following is a list of all solutions (for all bases):

Base 10: 1 9 2 2 1 9 2 7 3 3 2 7 3 8 4 4 3 8 5 4 6 6 5 4 5 7 6 6 5 7 8 1 9 9 8 1 Base 11: 7 8 3 8 4 6 8 9 1 4 5 6 5 9 1 6 7 2 1 2 9 3 2 7 4 5 3

(Tested all cases until base 17. After that, no solution can involve a carry.

But there are no solutions without carries. So, no more solutions.)

I hope this is of some interest.

Cheers,

Geoff.

------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. -------------------------------------------------------------------------------

-------------------------

> Ref: Your note of Mon, 19 Oct 92 22:24:47 EST

>

> Where are you from?

Whoops, knew I forgot to put something in. I'm currently a student at the

University of Sydney (Australia), doing Computer Science (Honours).

Cheers,

Geoff.

------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. -------------------------------------------------------------------------------

-------------------------

By the way, I tried searching for analogous solutions for other sizes and other

bases. So the new problems become:

Consider an n by n matrix containing the 'digits' from 1 to n^2 in a base b,

where b > n^2. The i'th row of the matrix consists of the last n 'digits' of

i*(first row). The versions differ in how succeeding matrices may be

constructed. Let f be the first row.

Version A: The next matrix has rows with 2*f, 3*f, ... , (n+1)*f

The j'th matrix has rows j*f, (j+1)*f, ... , (n+1-j)*f

Version B: The next matrix has rows with n*f, 2*n*f, ... , n*n*f

The j'th matrix has rows (n^(j-1))*f, 2*(n^(j-1))*f, .... , (n^j)*f

Version C: The next matrix has rows with (n+1)*f, (n+2)*f, ... , 2*n*f

The j'th matrix has rows (1+n*(j-1))*f, (2+n*(j-1))*f, ..., j*n*f

Basically these are analogies of the three versions I wrote to you about

before. The results I get are:

n: 1 base: any above 1 A, B, C: 1 n: 2 base: 5 A, B, C: 3 2 4 1 1 4 3 2 In case B, the second one extends: 4 1 ==> 3 2 3 2 1 4 In case C, the second one also extends: 4 1 ==> 2 3 3 2 1 4 base: 6 A, B, C: 1 4 3 4 3 2 1 2 Note that the only solution to the first problem (no overflow allowed) is 1 4 (in base 6) 3 2 n: 3 base: 10 A, B, C: 1 9 2 2 1 9 2 7 3 3 2 7 3 8 4 4 3 8 5 4 6 6 5 4 5 7 6 6 5 7 8 1 9 9 8 1 base: 11 A, B, C: 7 8 3 8 4 6 8 9 1 4 5 6 5 9 1 6 7 2 1 2 9 3 2 7 4 5 3 Note the base 10 solutions all solve the first problem, while none of the base 11 solutions do, and there is no second matrix for any of them. n: 4 base: 18 A, B, C: 1 15 14 4 1 15 16 2 2 1 15 16 2 3 13 16 3 13 10 8 3 13 14 4 4 3 13 14 4 7 9 14 5 11 6 12 5 11 12 6 6 5 11 12 6 11 5 12 7 9 2 16 7 9 10 8 8 7 9 10 8 15 1 10 3 13 14 4 3 13 16 2 4 1 15 14 4 3 13 14 7 9 10 8 7 9 14 4 8 3 13 10 8 7 9 10 11 5 6 12 11 5 12 6 12 5 11 6 12 11 5 6 15 1 2 16 15 1 10 8 16 7 9 2 16 15 1 2 In case C, two of them extend: 1 15 16 2 9 7 8 10 2 1 15 16 10 9 7 8 3 13 14 4 ==> 11 5 6 12 4 3 13 14 ==> 12 11 5 6 5 11 12 6 13 3 4 14 6 5 11 12 14 13 3 4 7 9 10 8 15 1 2 16 8 7 9 10 16 15 1 2

Note all of these solutions solve the first problem (no overflow).

Unfortunately, my algorithm is O((n!)^2), so any results for n = 5 are not

going to be forthcoming soon.

Cheers,

Geoff.

------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. -------------------------------------------------------------------------------

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