Can anyone please help me to solve this puzzle?

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- Aug 4th 2005, 03:29 PMDustinPuzzle--Help!
- Aug 16th 2005, 07:12 PMTheFarmer42Nasty
I plugged the (8) equations that you get by putting the letters a to p in the spaces into Mathematica and got some nasty answers.

Is there any other information - ie are the numbers all integers, and all consecutive?

I think i'll throw a genetic algoithm at it.

I could try some trial and error, but that isn't very exciting. - Aug 16th 2005, 08:20 PMDustin
Ops, I'm sorry for not giving that in the original..

Use the numbers 1-16 (all to be used, none to be used twice.) As always, the order of operations matters in each equation. - Aug 16th 2005, 08:46 PMTheFarmer42
Ok, so there is only 20 million million possible combinations of the numbers on the grid...

- Aug 17th 2005, 12:48 AMTheFarmer42
Genetic algorithm cranking away at the rate of 500,000 trials per hour. Now a genetic algorithm should find "good" solutions quite quickly, but unfortunately this is a problem where only the correct answer is any good.

We can only guarantee an answer through checking every possible combination which unfortunately will take about 4,500 years at the current rate. On the upside, there is a 50% chance that we would find it in 2,250 years using a random search, and i'd hope the GA would do a bit better than that.

Another source of hope is the doubling of computer speed, say every 18 months (moores "law"). In this case we can be guaranteed an answer in only 16.5 years.

By the way, are we reliably informed that it does have a solution?

An analytical solution would be preferred, but the integer quality of the problem makes it discrete in an unattractive way. - Aug 17th 2005, 02:40 AMCold
I've been wondering about this problem for ages - forgetting the fact that the numbers 1 - 16 are included only once makes a massive difference to the problem!

On the surface of it there are 16 unknowns with only 8 equations so it's not possible to crank out answers solving equations. But, using the order of operations (and the fact that the unknowns are all whole numbers), we can make progress. At least I think, I've not tried as yet.

Think about the divisions. Some of the numbers can't go in to some postions - or you'll end up with fractions. Now simply adding whole numbers to fractions won't equal a whole number answer.

I'll have another look at it now, I vaguely remember working out that two of the divisors were factors of another number - will have to look it up again.

I'm hopeful that the problem will reduce to trial and error of a few numbers in key positions.

If progress is to be made I'm sure that it'll be along these lines. It's a great puzzle. - Aug 17th 2005, 02:52 AMTheFarmer42
You make some good points. I thought some of them through, but didn't get the huge simplification i was hoping for.

Take row 3.

i.k/j -l=44

j can not be any larger than 5 with the limitations (each number can only be used once, largest 16, smallest 1)

In fact, if j is 5, there is only one solution for l

(i and k are interchangeable at this stage)

However there are still quite a few possible solutions when j is 1, 2 or 3.

Comparitively, sudoku is dead simple - linear programming solution solved in seconds.

This is much more interesting! - Aug 17th 2005, 03:43 AMCold
It's difficult. I don't know wether or not there is a unique solution. Just looking at C / G - K - O = -10 if

C / G > 1 if C>1 & G>1 and C not equal G. Largest possible c/g = 16/1 2<=c/g<=16

but if g not equal 1 then largest c/g is 8 so 2<=c/g<=8

but if g not equal 1 or 2 then largest c/g is 5 so 2<=c/g<=5

but if g not equal 1,2 or 3 then largest c/g is 4 so 2<=c/g<=4

but if g not equal 1,2,3 or 4 then largest c/g is 3 so 2<=c/g<=3

but if g not equal 1,2,3,4 or 5 then largest c/g is 2 so 2<=c/g<=2 or c/g = 2

If c/g = 16 then c=16, g=1 and -k - o = -26 or k + 0 = 26

K = 15 and O = 11 or vice versa

K = 14 and o = 12

Neither k nor o can = 13

I think I'm getting a headache! - Aug 17th 2005, 03:49 AMCold
Thinking about it, if g = 1 then j cannot = 1.

so 2<= I/J<=8 and 2<=F/J<=8 but j is a factor of I and F - Aug 17th 2005, 09:30 PMDustin
I'm normally good at math and so one of my friends asked me to do it....IT gave me a headache lol....I remember in HS Trig and HS Calculus I would do all sorts of critical thinking problems, but this one did get me stumped :(