## Checking solved puzzles that have been printed

Everything about Sudoku that doesn't fit in one of the other sections
Animator wrote:So for example, replacing a number with 0 or 1.2 (one and a half) is impossible. (because it's not part of the range of numbers that should be used on in the grid)

Well, yes, of course. The assumption was that the integers 1 .. N are being used. My method also craps out if someone uses letters of the alphabet, shapes, or coloured tiddly-winks. ;)

Animator wrote:I know plenty of binary numbers and I immediatly reliased this when reading the original post

My explanation wasn't necessarily for your benefit. Based on the quality of your posts, I was quite sure you had a firm understanding of my method. I was hoping to clarify my method for others trying to understand why it works.

Animator wrote:Actually no, it should be a <= b and e <= f.

Sorry, I should have stated my constraints for the other cases. Here they are in full.

Find {a,b,e,f} such that
a³ + b³ = e³ + f³,
1 <= a < b <= 9,
1 <= e < f <= 9,
a <> e,
a <> f,
b <> e,
b <> f.

Find {a,b,c,e,f,g} such that
a³ + b³ + c³ = e³ + f³ + g³,
1 <= a < b < c <= 9,
1 <= e < f < g <= 9,
a <> e,
a <> f,
a <> g,
b <> e,
b <> f,
b <> g,
c <> e,
c <> f,
c <> g.

Find {a,b,c,d,e,f,g,h} such that
a³ + b³ + c³ + d³ = e³ + f³ + g³ + h³,
1 <= a < b < c < d <= 9,
1 <= e < f < g < h <= 9,
a <> e,
a <> f,
a <> g,
a <> h,
b <> e,
b <> f,
b <> g,
b <> h,
c <> e,
c <> f,
c <> g,
c <> h,
d <> e,
d <> f,
d <> g,
d <> h.

I believe it should be a < b (rather than a <= b) and it should be a < b < c (rather than a <= b <= c) and so on.

For a moment, consider if 1 <= a <= b <= c <= 9 and 1 <= e <= f <= g <= 9 while the other constraints remain the same. This means you could be examining something like 4³ + 4³ + 4³ = 7³ + 7³ + 7³ (which, of course, is not equal -- but that's not the point I'm trying to make). I thought the point of trying to find equivalencies was to be able to make substitutions, like how you pointed out that 4 5 6 can be replaced with 2 3 8 or 6² + 7² + 8² = 2² + 3² + 9². A valid row/column/block will not have three of any single digit, thus a < b < c.
scrose

Posts: 322
Joined: 31 May 2005

True...

It should be a < b, and e <= f. (or a < b < c and e <= f <= g)

For example replace 1 2 3 with 1 4 1. By using that replacement you will end up having the number 1 three times. (which was the point I was trying to make.)

And those are the combination my program isn't finding... and my guess now is that a combination like that is required for n³...
Animator

Posts: 469
Joined: 08 April 2005

Francisco wrote:I know that 1+...+9=45 can be made using the Gauss rule (It should be my rule since I made it up before learning it in school) =(1+9)*(9/2)=10*4.5=45

I'm not sure I understand what you mean. How do you use 1+...+9=(1+9)*(9/2)=10*4.5=45 to help you solve a puzzle? In particular, the (1+9)*(9/2) part?

Francisco wrote:The "my mind" rule is just PURE ATTENTION... I can check a row, column or box pretty fast and check which numbers are missing

Are you saying that you visually scan the row/column/block for missing/duplicate numbers (the same way I do) or do you mentally check if the row/column/block sums to 45? Or do you do both at the same time?
Last edited by scrose on Thu Jun 16, 2005 11:34 am, edited 1 time in total.
scrose

Posts: 322
Joined: 31 May 2005

Animator wrote:It should be a < b, and e <= f. (or a < b < c and e <= f <= g)

For example replace 1 2 3 with 1 4 1. By using that replacement you will end up having the number 1 three times. (which was the point I was trying to make.)

Ahhhhh! *ding* The lightbulb just appeared over my head! Now I understand what you meant, and I see that my constraints were too constraining. The set of replacement numbers can contain duplicates and the set of replacement numbers can contain some (but not all) of the original set of numbers.

Thank you for correcting me!
scrose

Posts: 322
Joined: 31 May 2005

Okay, based on the eased constraints, here are a few equivalencies. Animator, I will leave it up to you to find a grid that shows the method 1³ + 2³ + ... + 9³ does not work.

2³ + 3³ + 6³ = 1³ + 5³ + 5³
1³ + 2³ + 3³ + 6³ = 1³ + 1³ + 5³ + 5³
1³ + 2³ + 3³ + 9³ = 4³ + 4³ + 5³ + 8³
1³ + 2³ + 4³ + 9³ = 3³ + 6³ + 6³ + 7³
1³ + 2³ + 7³ + 8³ = 6³ + 6³ + 6³ + 6³
1³ + 4³ + 5³ + 8³ = 2³ + 2³ + 7³ + 7³
2³ + 3³ + 4³ + 6³ = 1³ + 4³ + 5³ + 5³
2³ + 3³ + 5³ + 6³ = 1³ + 5³ + 5³ + 5³
2³ + 3³ + 6³ + 7³ = 1³ + 5³ + 5³ + 7³
2³ + 3³ + 6³ + 8³ = 1³ + 5³ + 5³ + 8³
2³ + 3³ + 6³ + 9³ = 1³ + 5³ + 5³ + 9³
2³ + 5³ + 6³ + 9³ = 3³ + 3³ + 8³ + 8³
3³ + 4³ + 5³ + 9³ = 1³ + 6³ + 6³ + 8³
3³ + 4³ + 6³ + 8³ = 2³ + 5³ + 7³ + 7³
scrose

Posts: 322
Joined: 31 May 2005

Great job, now I don't have to look for the combinations myself :)

There are two special combinations in that list

First one:

1³ + 4³ + 5³ + 8³ = 2³ + 2³ + 7³ + 7³
1² + 4² + 5² + 8² = 2² + 2² + 7² + 7²
1 + 4 + 5 + 8 = 2 + 2 + 7 + 7

Second one:

2³ + 5³ + 6³ + 9³ = 3³ + 3³ + 8³ + 8³
2² + 5² + 6² + 9² = 3² + 3² + 8² + 8²
2 + 5 + 6 + 9 = 3 + 3 + 8 + 8

grid that is correct according to n³ (2 3 6 = 1 5 5) (it's also correct according to n, but that was not really intended):
1 5 5 | 1 4 5 | 7 8 9
1 4 5 | 7 8 9 | 2 3 6
7 8 9 | 2 3 6 | 1 4 5
-----------------------
5 5 1 | 4 5 1 | 8 9 7
4 5 1 | 8 9 7 | 3 6 2
8 9 7 | 3 6 2 | 4 5 1
-----------------------
5 1 5 | 5 1 4 | 9 7 8
5 1 4 | 9 7 8 | 6 2 3
9 7 8 | 6 2 3 | 5 1 4

grid that is correct according to n³, n², n (1 4 5 8 = 2 2 7 7):
2 2 7 | 7 2 3 | 6 7 9
7 2 3 | 6 7 9 | 2 2 7
6 7 9 | 2 2 7 | 7 2 3
-----------------------
2 7 2 | 2 3 7 | 7 9 6
2 3 7 | 7 9 6 | 2 7 2
7 9 6 | 2 7 2 | 2 3 7
-----------------------
7 2 2 | 3 7 7 | 9 6 2
3 7 7 | 9 6 2 | 7 2 2
9 6 2 | 7 2 2 | 3 7 7

(I was lazy when creating this grid... the problem is that you need to replace 4 numbers, which means it can't be limited to a single row/column, since you need to replace one in another row/box in order to make the box work. So I replaced all numbers, which would make it obvious that the grid is incorrect... as in, four numbers are missing)
Animator

Posts: 469
Joined: 08 April 2005

Well done! I suspected it wouldn't take you long!

Do we need to tackle 1^4 + 2^4 + ... + 9^4 or can we let this thread die a deserving death?
scrose

Posts: 322
Joined: 31 May 2005

I vote in favor of letting it die... it sure is intresting, and some intresting question remain to be answer, but is it useful for anything?

Some general observations I made (+/- a conclusion):

The higher the power the more numbers you need to replace (I guess!), ^2 can be fooled by using 2 numbers, ^3 needs 3 numbers...

If that pattern continues then ^ 10 should always be correct...

^ 9 might be incorrect when you use numbers multiple times, assuming ofcourse that the sum is still correct... ofcourse I have no idea if that's the case or not.

^ 9 might be correct if you combine it with the previous steps... fooling n, n² requires 3 numbers (to be changed), fooling n, n², n³ requires 4 numbers, so fooling n ^ 9 would requires 10 numbers, which ofcourse is impossible...
Animator

Posts: 469
Joined: 08 April 2005

I just can't leave well enough alone. Now I'm going correct my assumptions -- again!

scrose wrote:
Animator wrote:So for example, replacing a number with 0 or 1.2 (one and a half) is impossible. (because it's not part of the range of numbers that should be used on in the grid)

Well, yes, of course. The assumption was that the integers 1 .. N are being used. My method also craps out if someone uses letters of the alphabet, shapes, or coloured tiddly-winks. ;)

I believe my method works for all real numbers. Suppose we have a 3x3 grid where every row/column/block contains the set {-154/17, -1 * (square root of 19.9), -13/5, -1, 0, 5/9, pi, square root of 57, 18.26}. Using 2 as a base, the sum of every row/column/block should be approximately 314111.245940334.

I wonder if it works with imaginary numbers...
scrose

Posts: 322
Joined: 31 May 2005

Animator wrote:The higher the power the more numbers you need to replace (I guess!), ^2 can be fooled by using 2 numbers, ^3 needs 3 numbers...

If that pattern continues then ^ 10 should always be correct...

I agree. Let's leave the proof as an exercise for the reader. ;)

Of course, with a 2x5 grid, ^10 might not always be correct.
scrose

Posts: 322
Joined: 31 May 2005

Ha ha. Just for kicks and giggles, I looked back at the post that started this thread, just to see if we are still on topic. Surprisingly enough, we are, although I think it has gone far beyond what Trinkles expected. At the moment, Trinkles is still showing as having only one post. I hope we haven't scared him/her away with our babble.
scrose

Posts: 322
Joined: 31 May 2005

scrose wrote:I wonder if it works with imaginary numbers...

When the set of numbers is fixed, as it is with sudoku's, then your method works (or atleast that's what I believe).

The only way I can think about fooling it is to use something like: 1, 2, 4, 5, 6, 8, 9, 2, (log(132) / log(2)). (which is the same as log2 (132))

But obviously (log(132) / log(2)) isn't part of the number-set that is used, and therfor it cannot be fooled.
Animator

Posts: 469
Joined: 08 April 2005

Animator wrote:When the set of numbers is fixed, as it is with sudoku's, then your method works (or atleast that's what I believe).

Yes, I think you're right. As long as the replacement numbers come from the set of original numbers, my method cannot be fooled.

Good work, by the way, on finding that replacement set containing log(132) / log(2).
scrose

Posts: 322
Joined: 31 May 2005

Previous