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]

  Reveal hidden contents

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

  Reveal hidden contents

1

8 45

21

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

[bonanova]

  Reveal hidden contents

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]

  On 5/10/2013 at 7:38 PM, BobbyGo said:

  Reveal hidden contents

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

  Reveal hidden contents

1

8 45

21

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

  Reveal hidden contents

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]

  On 5/10/2013 at 7:38 PM, BobbyGo said:

  Reveal hidden contents

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

  Reveal hidden contents

1

8 45

21

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

Can you share the code?

[BobbyGo]

  On 5/10/2013 at 8:38 PM, dark_magician_92 said:

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.

  Reveal hidden contents

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]

  On 5/10/2013 at 7:26 PM, bonanova said:

  Reveal hidden contents

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]

  On 5/10/2013 at 7:26 PM, bonanova said:

  Reveal hidden contents

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]

  On 5/10/2013 at 7:38 PM, BobbyGo said:

  Reveal hidden contents

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

  Reveal hidden contents

1

8 45

21

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

Anything special about these numbers?

[bonanova]

  On 5/10/2013 at 11:46 PM, BMAD said:

  On 5/10/2013 at 7:26 PM, bonanova said:

  Reveal hidden contents

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]

  On 5/12/2013 at 2:06 AM, bonanova said:

  On 5/10/2013 at 11:46 PM, BMAD said:

  On 5/10/2013 at 7:26 PM, bonanova said:

  Reveal hidden contents

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.