When the aim is nothing
- What a nice square!
- Thank you. I made it myself!
- How did you do it?
- I put 1 or -1 in all the cells in the upper-left 2×2 square, the blue cells.
- And then?
- Then I multiplied the cells in each row and put the result at the end of the row, the yellow cells
- And you did the same with the columns?
- Correct.
- But what is the number in the brown cell at the bottom-right?
- It is the sum of the row and column products, i.e. the sum of the yellow cells.
- You mean you added up the numbers above it and to its left?
- Yeah, man!
- I would like to make a square like that!
- Try to make a blue 1997×1997 square (which becomes 1998×1998 when you calculate all the products and the sum in the lower-right).
- That sounds too easy.
- OK then. Make the lower-right cell equal 0 in your 1998×1998 square.
Problem source: Bamo Problems.
June 1st, 2010 at 10:17 am
Not easy, impossible. This is the same as getting a symmetric random walk with steps of (-1, +1) to return to its origin in an odd number of steps; you’re always off by at least one. Now a 2010×2010 I can do…
June 1st, 2010 at 2:47 pm
Sorry, but I don’t see the connection. If you have an odd number of -1s and 1s I agree that the sum can never be 0, but in this case we have an even number, 1997 x 2.
Mike, could you please explain?
June 1st, 2010 at 5:06 pm
Call the number of blue rows n.
The blue rows that individually contain even numbers of -1s contain a total of an even number of -1s.
If the number of blue rows that individually contain odd numbers of -1s is even (resp. odd), then those blue rows contain a total of an even (resp. odd) number of -1s.
What goes for the rows also applies to the columns. Thus the following 5 statements are logically equivalent:
1) The number of blue -1s is even.
2) The number of blue rows containing an odd number of -1s is even.
3) The number of -1s in the yellow column is even (by the construction of the yellow column).
4) The number of blue columns containing an odd number of -1s is even.
5) The number of -1s in the yellow row is even (by the construction of the yellow row).
Because statements 3 and 5 are logically equivalent, the total number of -1s in yellow cells is even. If the brown cell is to contain 0, there must be -1s in exactly half the yellow cells, i.e. n of the 2n yellow cells, which means that n must be even. Thus the task is impossible if n=1997.
June 2nd, 2010 at 8:50 am
Sorry Jan, I misunderstood the problem as having the same row and column totals (of zero). That’s what comes of looking at every rectangular table as though it were a cross tabulation!