Izzackly!
The problem as posed was for a 4x4 grid, IIRC (or maybe even just 3x3), but my approach worked for any size of grid. His became (I think) exponentially harder to code as N increased.
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The reason for wanting a mathematical approach is that this was given to a 10 year old as "maths" homework. So, we wondered if there was a mathematical approach.
As it happens, I used an almost identical approach to stickyfiddle. But none of that lets you work out the answer, it merely helps you narrow the options but ultimately you have to do a bit of brute force. It'd be nice if there were simultaneous equations that would help.
It's too tough as a purely mathematical challenge for a 10 year old, but it does provide a good problem which a mathematical and logical mind can explore.
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Sporky: It's been a bloody long time, but IIRC mapping a game space is big-O^n or possibly ^n^2, so enormously costly. I also seem to remember someone saying that mapping the game space of chess would require more bits of memory than there were atoms in the universe. Can't be fagged to work it out now, though.
"Brilliant things instead of weeping will be for the person who comes to the truthful one. But a long period of darkness, foul food, and the word 'woe' - to such an existence your religious view will lead you, O deceitful ones, of your own actions." - 1000 years BCE, Zoroaster predicts the coming of Keanu.
Are you perhaps... in sales?Originally Posted by frankus
仁慈的上帝，请带我走我将关闭自己的耳朵，我的心，我将一块石头
I got two out of those three suggestions using good old fashioned pen and paper....
Sometimes I really wish I had read further into the thread before I start working stuff out!!
Couple of things I noticed with this was that the top left "cell" is quite important because it's place value is as a "10" twice in the equation. Also, how many solutions are there to this? I can't imagine that there are many...
Well, I got a different solution to those offered so far and my mate looked up the solution and got a different one.
Another significant point is that the bottom right "cell" is used twice (doubled) which means that it will always cause an even number of units. As such, the top right and bottom left cells, which are used only once, must sum together to be an even number too, and the sum of all of that has to end with a zero.
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agreed. FWIW I liked sticky's approach
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What school inflicts this on 10 year olds!!!!!.
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Some time ago I taught a lunctime guitar class in a junior school. On the wall was a display about Fermat's last theorem. They'll be getting 8 years old science pupils to do Schrödinger next, but they still won't be able to write their name or tie their shoelaces.
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20 as far as I can figure. Using stickyfiddle's notation - if you select a value for a then there is only one possible value for d (since 100 - 20a - 2d must be divisible by 11 and d < 9). So possible combinations of a and d are
a = 0, d = 6
a = 1, d = 7
a = 2, d = 8
a = 3, d = 9
by rearranging the original formula you can show that b + c = (100 - 20a - 2d)/11 so for each combination of a and d you can work out how many combinations of b and c there are. For example, if a = 1 and d = 7 then b + c = 6 so there are 7 combinations but one of which has b = c so is discarded.
So we have
a = 0, d = 6 - 8 combinations
a = 1, d = 7 - 6 combinations
a = 2, d = 8 - 4 combinations
a = 3, d = 9 - 2 combinations
Giving 20 in total.
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