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Need help with a discrete math problem.

Discrete Math
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I'm not familiar with these type of problems, seems like you need to use the bottom formul a for both, but I don't know how to fill it in.
So if I started filling in the bottom formula for A(3, 3) A(3-1, A(3,3-1)) A(2, A(3,2)) Plug the A(3,2) back into the formula: A(2, A(3-1, A(3, 2-1))) A(2, A(2, A(3, 1))) Now n = 1 so I can use the third formula to get A(3, 1) = 2 A(2, A(2, 2)) A(2, A(2-1, A(2, 2-1))) A(2, A(1, A(2, 1))) Use the third formula again: A(2, A(1, 2)) A(2, A(1-1,A(1, 2-1))) A(2, A(0, A(1, 1))) A(2, A(0, 2)) Use the first formula: A(2, 2n) A(2, 2(2) A(2, 4) A(1, A(2, 3)) A(1, A(1, A(2, 2))) A(1, A(1, A(1,A(2, 1)))) A(1, A(1, A(1, 2))) A(1, A(1, A(0, A(1, 1)))) A(1, A(1, A(0, 2))) A(1, A(1, 2(2))) A(1, A(1, 4)) A(1, A(0, A(1, 3))) A(1, A(0, A(0, A(1, 2)))) A(1, A(0, A(0, A(0, A(1, 1))))) A(1, A(0, A(0, A(0, 2)))) A(1, A(0, A(0, 4))) A(1, A(0, 8)) A(1, 16) A(0, A(1, 15)) A(0, A(0, A(1, 14)) lol, this is going to take a while.

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Anyone know if this is how you do it, or is there a faster way?
Sorry, this is WAY above my level of math :/ wish i could be more help
@amistre64 Do you know if I am doing this right?
it looks like a recurrsion alright
Yeah, that is what it is.
A( m-1, A(m,n-1) ) ; 3,3 A(m,n) = A( 3-1, A(3,3-1) ) A(3,3) =A( 2, A(3,2) ) A(3,2) = A( 2, A(3,1) ) A(3,1) = 2 A(3,2) = A( 2, 2) = A( 1, A(2,1) ) A(3,2) = A( 0, 2 ) = 4 A(3,3) =A( 2, 4 ) etc ....
multivariables seem to be a pain to do by hand, might help to program a simple loop for it
Well I have to show my work for the homework so I'll probably have to write it out. I guess it is basically just a lot of plugging in the numbers over and over again.
Makes sense though, the way this one is going, I bet the A(3,4) will be really long, haha.
if you can determine a pattern :) but ive never really tried for a multivariable
Oh that makes sense. It's like programming an if-then-else loop
Well I guess I could do a simple program that outputs each step and then just copy the results to the paper.
Having never seen anything like this I was like put what in what now? Lol
Ye, that would be faster I guess
Alrighty thanks for the help guys! I'm gonna hit up some python :D
Well I started on the program. I think it will take a little more work to get it to print all the steps. def A(m, n): print "(", m, ", " , n, ")" if m == 0: return (2*n) elif m >= 1 and n == 0: return 0 elif m >= 1 and n == 1: return 2 elif m >= 1 and n >= 2: return (m - 1, A(m, n - 1)) m = input('m: ') n = input('n: ') answer = A(m,n)
And, I don't think it is working quite right yet.
yeah, i was trying to do it in excel ... its tricky at best
oops, missed an A in the return
Here is my output: ( 3 , 3 ) ( 3 , 2 ) ( 3 , 1 ) ( 2 , 2 ) ( 2 , 1 ) ( 1 , 2 ) ( 1 , 1 ) ( 0 , 2 ) ( 2 , 4 ) ( 2 , 3 ) ( 2 , 2 ) ( 2 , 1 ) ( 1 , 2 ) ( 1 , 1 ) ( 0 , 2 ) ( 1 , 4 ) ( 1 , 3 ) ( 1 , 2 ) ( 1 , 1 ) ( 0 , 2 ) ( 0 , 4 ) ( 0 , 8 ) ( 1 , 16 ) ( 1 , 15 ) ( 1 , 14 ) ( 1 , 13 ) ( 1 , 12 ) ( 1 , 11 ) ( 1 , 10 ) ( 1 , 9 ) ( 1 , 8 ) ( 1 , 7 ) ( 1 , 6 ) ( 1 , 5 ) ( 1 , 4 ) ( 1 , 3 ) ( 1 , 2 ) ( 1 , 1 ) ( 0 , 2 ) ( 0 , 4 ) ( 0 , 8 ) ( 0 , 16 ) ( 0 , 32 ) ( 0 , 64 ) ( 0 , 128 ) ( 0 , 256 ) ( 0 , 512 ) ( 0 , 1024 ) ( 0 , 2048 ) ( 0 , 4096 ) ( 0 , 8192 ) ( 0 , 16384 ) ( 0 , 32768 )
I think one of the hardest parts will be getting it to output the results correctly
Any luck with excel? I'm gonna switch to C++ to try to get the output working right.
So here is what I came up with. Ended up using C. #include #include int m; int n; int answer; void PrintA(m,n) { printf("A(%d, %d) = ", m, n); printf("A(%d, A(%d, %d)\n", m-1, m , n-1); } int A(int m, int n) { if(m == 0) return (2*n); else if(m >= 1 && n == 0) return 0; else if(m >= 1 && n == 1) return 2; else if(m >= 1 && n >= 2) PrintA(m,n); return A(m - 1, A(m, n - 1)); } int main() { printf("m: "); scanf("%d", &m); printf("n: "); scanf("%d", &n); answer = A(m, n); printf("Answer: %d\n", answer); return 0; }
Here is what I get for the first one. A(3,3) Obviously the answer is wrong, but all the steps up to the answer should be correct. So I can calculate the answer myself from the last step, which is 4 I think.
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That works fine, now the problem is when I try A(3, 4) although I get the answer to be 0, it outputs for about 30 seconds, so I get like 1000s of lines of output which would be a bit too much to write. lol. Anyway, I'm thinking there must be some way to recognize a pattern in order to get the answer. :/
So yeah I found a pattern for A(3,3) problem when I got to A(1,16) and got the computer output you got.
which i will most likely leave as 2^16
So the stops working?
I seen 2^n*A(1,16-n)
what do you mean stops working?
So how would I use this pattern in finding a solution? mistyped, you said you found a problem earlier.
pattern not a problem i was referring to your problem/question
oh I get it now, lol.
wow yeah it looks like A(3,4) might take longer to find.
Ah right, so we would need a separate pattern for A(3,4) determined by some of the output?
You said you got 0. That is what I got too.
A(3,4) =A(2,A(3,3)) =A(2,2^16) =A(1,A(2,2^16-1)) ------------------ Let's look at A(2,2^16-1) A(2,2^16-1)=A(1,A(2,2^16-2)) There is like a pattern going on we have A(2,2^16-b) b keeps increasing eventually it will get to 2^16 Since this is happening in the n part of A then the output will be 0.
Is this what you were looking for?
Yes I think so, just trying to get my head around it. I think once I look at it for a bit I will understand.
Thanks so much for all the help though! Really appreciate it.
Like A(something bigger than 1,0)=0 eventually A(1,k) will be A(1,0)
And A(1,0)=0
so b will eventually get to 2^16?
and then it will be 0. I get it now, thank you so much :D
Like because we keep subtracting 1 each time.
I'm not doing all those steps. It is impossible to do all those steps. You have to look for a pattern.
n=2^16 n-1=2^16-1 (n-1)-1=n-2=2^16-2 (n-2)-1=n-3=2^16-3 ....
we keep subtracting 1 from the n before.
right right, exactly, until n becomes 0, then according to the second formula A(3,4) == 0
@myininaya Hi there, could you look over this one more time. According to my teacher, it is impossible to find the second number for A(3,4) by using a computer algorithm. And the students in my class are saying the answer is a really big number and not 0. Like a lot bigger then the number for A(3,3) Do you have any idea where that would be coming from?
Well it looked like the pattern was going toward it being 0. I can look at it again. Did you explain your thinking to the teacher person? Like why did they say the thinking was wrong?
The teacher hasn't looked at my work yet, he was just giving hints to help with the homework, and he said that this part cannot be done with a computer program because it will run out of memory and lose the answer.
I don't really understand why that would happen, but I just want to make sure to get it right since other students are getting really large numbers.
He didn't say a human couldn't do it though right?
No, he said it was possible to do by hand, but not with an algorithm.
Just because a computer doesn't have the space to use an algorithm to compute doesn't mean we don't have the space.
So the answer is still 0
n=2^16 n-1=2^16-1 (n-1)-1=n-2=2^16-2 (n-2)-1=n-3=2^16-3 remember when I wrote this?
2^16 steps is a lot. I don't know if it is too long for a computer or not. But it will actually take longer than 2^16 steps to compute.
Right, so there isn't some sort of pattern we could figure out to get the answer without the steps?
I said it would take longer than 2^16 because we did some other crap before to get there and we still have to do more stuff after that, but the pattern seems to be leading to the answer being zero.
Hmm, well I'll stick with that then. Perhaps my teacher is wrong. :D
What? You said your teacher said it was possible for a human I thought.
"No, he said it was possible to do by hand, but not with an algorithm."
Yeah, but I got 0 with the program I wrote.
So the answer you are saying is not zero?
Which is just a simple recursive function.
That is what I was led to believe by my fellow students and teacher.
So the teacher said the answer isn't 0 and does exist.
No, the teacher didn't say anything about the answer, just that it couldn't be done with a computer, and since I got the same number with I computer 0, as you got by hand, I figured it couldn't be right unless my teacher was wrong.
Can you copy it here? What you put in as the algorithm into your computer? I will try to follow it.
I just want to see if I agree with your algorithm.
Is that it above? The final draft of your program?
maybe you are accidentally getting 0. lol. i don't know.
like you did get the right answer with A(3,3)
maybe we can check another one.
Let's look at A(2,5) Use program and let's do it by hand.
darn that is long. lol.
haha, yeah. Runs for around 15 seconds.
that is probably like a 100th of the output there. lol
is the answer always 0 when n>m?
No, for m = 1 and n = 2 I get A(1, 2) = A(0, A(1, 1) Answer: 4
For 2 and 3
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A(1,2)=A(0,A(1,1)) =A(0,2) =2n =2n=2(2) since n=2. okay.did you computer also get 4 for A(1,2)?
hmmm... I get 16 by hand for A(2,3)
Yeah, computer seems to be doing it right.
Well I guess it is very possible that my teacher was wrong, idk.
I don't know. There is a possibility I'm wrong. There is always a chance for that.
Well there is a lot less possibility that you are wrong then I am wrong, so I guess I'll stick with that. :P
I don't see anything wrong with your program. Maybe for most computers the algorithm will run too long. Not sure.
Hard to tell, do you have time to look at one more thing really really quick?
ok i will look no guarantees I will be able to help but i certainly will try Alright, thanks, wish I could give you another medal :/

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