Diamond Subtraction

[BMAD]

Put four different whole numbers from 1 to 49, in the corners, then subtract the smaller from the larger and put the answers in between. Keep doing this until all four numbers become equal. How many steps can you make it last? The demo is "(14, 30, 18, 37) lasts 3 steps (don't count the step where they produce the same number)." Find the longest lasting case.

DEMO:

stage 0

14 30 ----> 30-14

37-14 30-18

37 18 37-18

stage 1

16 16-13 16-12

13 12

19 19-13 19-12

stage 2

3 4

6 7

stage 3

1

3 3

1

Stage 4 is stasis so it doesn't count

[BobbyGo]

I cheated; with less than 6 million different starting combinations, I wrote some code to brute force the answer.

1

8 45

21

Lasted for 12 iterations (possibly 11 or 13, counting isn't my strong suit )

[bonanova]

Is the example two steps?

14 30 16 7 4
=> 23 12 =>
37 18 19 4 7

Are they all just two steps?

a b |a-b| |b-c| |a-d|

=> |c-a| |b-d| =>

c d |d-c| |a-d| |b-c|

[bonanova]

I cheated; with less than 6 million different starting combinations, I wrote some code to brute force the answer.

1

8 45

21

Lasted for 12 iterations (possibly 11 or 13, counting isn't my strong suit )

13 DIAGONAL 1 8 21 45

1 8 21 45

7 13 24 44 - 1

6 11 20 37 - 2

5 9 17 31 - 3

4 8 14 26 - 4

4 6 12 22 - 5

2 6 10 18 - 6

4 4 8 16 - 7

0 4 8 12 - 8

4 4 4 12 - 9

0 0 8 8 - 10

0 8 0 8 - 11

8 8 8 8 - 12

0 0 0 0 - 13

[dark_magician_92]

I cheated; with less than 6 million different starting combinations, I wrote some code to brute force the answer.

1

8 45

21

Lasted for 12 iterations (possibly 11 or 13, counting isn't my strong suit )

Can you share the code?

[BobbyGo]

Can you share the code?

I'm sure there are more elegant ways of coding this, and it only gives back the first longest set it finds (it won't return 8,21,45,1 after it finds 1,8,21,45) so I'm sure there is more than one answer.

I think 5,29,42,49 might be another set, but haven't tested them yet.

int a,b,c,d,a1,b1,c1,d1,a2,b2,c2,d2,a3,b3,c3,d3;
int z3=0,z,x;

int main()
{
    for(a=1; a<50; a++)
    {
        for(b=1; b<50; b++)
        {
            for(c=1; c<50; c++)
            {
                for(d=1; d<50; d++)
                {
                    a1=a;b1=b;c1=c;d1=d;
                    x=1;z=0;

                    while(x==1)
                    {
                        if(a1==b1 and c1==d1 and a1==c1)
                        {
                            x=0;
                        }else
                        {
                            a2=sqrt ((a1-b1)*(a1-b1));
                            b2=sqrt ((c1-b1)*(c1-b1));
                            c2=sqrt ((c1-d1)*(c1-d1));
                            d2=sqrt ((a1-d1)*(a1-d1));
                            z+=1;
                        }

                        if(a2==b2 and c2==d2 and a2==c2)
                        {
                            x=0;
                        }else
                        {
                            a1=sqrt ((a2-b2)*(a2-b2));
                            b1=sqrt ((c2-b2)*(c2-b2));
                            c1=sqrt ((c2-d2)*(c2-d2));
                            d1=sqrt ((a2-d2)*(a2-d2));
                            z+=1;
                        }

                    }
                if(z>z3)
                {
                    z3=z;
                    a3=a;
                    b3=b;
                    c3=c;
                    d3=d;
                }

                }
            }
        }
    }

    cout<<z3<<endl<<endl<<"  "<<a3<<endl<<b3<<"  "<<d3<<endl<<"  "<<c3<<endl;
	return 0;
}

[BobbyGo]

Just realized the section:

a1=sqrt ((a2-b2)*(a2-b2));

b1=sqrt ((c2-b2)*(c2-b2));

c1=sqrt ((c2-d2)*(c2-d2));

d1=sqrt ((a2-d2)*(a2-d2));

should actually be:

a1=sqrt ((a2-d2)*(a2-d2));

b1=sqrt ((a2-b2)*(a2-b2));

c1=sqrt ((c2-b2)*(c2-b2));

d1=sqrt ((c2-d2)*(c2-d2));

I don't think it really it effects the outcome, since it stayed in the same order. But still.

[BMAD]

Is the example two steps?

14 30 16 7 4

=> 23 12 =>

37 18 19 4 7

Are they all just two steps?

a b |a-b| |b-c| |a-d|

=> |c-a| |b-d| =>

c d |d-c| |a-d| |b-c|

yes, your example is 2 steps.

[BMAD]

Is the example two steps?

14 30 16 7 4

=> 23 12 =>

37 18 19 4 7

Are they all just two steps?

a b |a-b| |b-c| |a-d|

=> |c-a| |b-d| =>

c d |d-c| |a-d| |b-c|

my example though shows 3 steps

[BMAD]

I cheated; with less than 6 million different starting combinations, I wrote some code to brute force the answer.

1

8 45

21

Lasted for 12 iterations (possibly 11 or 13, counting isn't my strong suit )

Anything special about these numbers?

[bonanova]

Is the example two steps?

14 30 16 7 4

=> 23 12 =>

37 18 19 4 7

Are they all just two steps?

a b |a-b| |b-c| |a-d|

=> |c-a| |b-d| =>

c d |d-c| |a-d| |b-c|

yes, your example is 2 steps.

It's not my example.

Check 37-14.

[BMAD]

Is the example two steps?

14 30 16 7 4

=> 23 12 =>

37 18 19 4 7

Are they all just two steps?

a b |a-b| |b-c| |a-d|

=> |c-a| |b-d| =>

c d |d-c| |a-d| |b-c|

yes, your example is 2 steps.

It's not my example.

Check 37-14.

silly me.