1 to 15 matrix (trivia)

slycrack

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Jul 25, 2019
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3
Hi guys need help.
I got this trivia and need to arrange it into the highest possible answer.
I need to arrange a number 1-15.

if the number you put in, is greater that the previous number then it will be multiplied by the difference between these two number.
Example: 6, and then 10 > the answer will be 24. (10-6=4, 6x4= 24).

but if the number is smaller than the previous one, it will be divided by the current number.
example: 14, and the 7 > the answer will be 2 (14/7 = 2).

now I need to arrange it the highest possible answer.
Example: 14, 7, and the 11 > the answer will be 8
14<7 then 14/7 = 2
7 > 11 then 11-7 = 4
4x2= 8.

please help. I'm really curious about the highest possible answer
 

Otis

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Apr 22, 2015
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1,872
… I need to arrange a number 1-15 … curious about the highest possible answer
Hello Sly. I had to read your post a few times, but I think I've figured out the puzzle steps. We arrange the numbers 1 through 15 to create an ordered list (sequence) and then follow a recipe (algorithm) that uses adjacent numbers (moving through the sequence from first to last) to cook up a final value. Regarding all possible sequences, you're curious to know the maximum final value.

What's the largest value that you've found, so far? (Please include the sequence, to help confirm the algorithm.)

Edit: Example deleted because I made a mistake.

By the way, do you have any algebra or programming experience? For example:

IF a<b THEN v=a*(b-a) ELSE v=a/(a-b)

😎
 
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slycrack

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Jul 25, 2019
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Hello Sly. I had to read your post a few times, but I think I've figured out the puzzle steps. We arrange the numbers 1 through 15 to create an ordered sequence (list) and then follow a recipe (algorithm) that uses adjacent numbers (moving through the sequence from first to last) to cook up a final value. Regarding all possible sequences, you're curious to know the maximum final value.

What's the largest value that you've found, so far? (Please include the sequence, to confirm the algorithm.)

Edit: Example deleted because I made a mistake.

By the way, do you have any algebra or programming experience? For example:

IF a<b THEN v=a*(b-a) ELSE v=a/(a-b)

😎
Sorry for my English Mr. Otis, it's not my native language. Yes we need to arrange the numbers 1 to 15,
the largest value I found is 15.680 with the sequence as following.

6, 11, 13, 2, 4, 5, 7, 8, 9, 3, 10, 12, 14, 1, and 15.

6 < 11 (11-6= 4, 6*5= 30)
11 < 13 (13-11=2, 30*2 = 60)
13 > 2 (60/2 = 30)
2 < 4 (30*2=60)
4 < 5 (60*1=60)
and so on and on resulting the number 15.680.

actually the first thing that came to mind is to programmed it, unfortunately I currently have no access to a decent computer.
thank you for your help good sir.

PS: if the sequence was 1,2,3...15, the result would be 1, and if it's backward (15,14,13...1) the result is 0.
 

Otis

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Apr 22, 2015
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1,872
Hello. Thank you for the additional information. I've deleted the example sequence from post #2 because I misinterpreted some wording and your first examples.

6, 11, 13, 2, 4, 5, 7, 8, 9, 3, 10, 12, 14, 1, 15

6 < 11 (11-6= 4, 6*5= 30)
11 < 13 (13-11=2, 30*2 = 60)
13 > 2 (60/2 = 30)
2 < 4 (30*2=60)
4 < 5 (60*1=60)
and so on and on resulting the number 15.680.
We compare two adjacent numbers in the sequence (a,b) and we create a new value (V) according to the algorithm. In the beginning, V equals the first number in the sequence. After that, each new V depends upon its previous value (i.e., V is recursive).

From the third line in your example above, I'm now thinking:

IF a<b THEN newV=oldV*(b-a) ELSE newV=oldV/b

I tried that algorithm on your sequence, and I got the same (15,680 -- comma delimiter, as used in the USA).

… if [the sequence is] backward (15,14,13, …,1) the result is 0.
I get a final result of 15/14! or 1/5811886080. That's about 0.000000000172. I'm guessing that you rounded to zero, yes?

Now that I have a clearer understanding of the puzzle, I'll play with it when I have time.

😎
 

slycrack

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Joined
Jul 25, 2019
Messages
3
Hello. Thank you for the additional information. I've deleted the example sequence from post #2 because I misinterpreted some wording and your first examples.


We compare two adjacent numbers in the sequence (a,b) and we create a new value (V) according to the algorithm. In the beginning, V equals the first number in the sequence. After that, each new V depends upon its previous value (i.e., V is recursive).

From the third line in your example above, I'm now thinking:

IF a<b THEN newV=oldV*(b-a) ELSE newV=oldV/b

I tried that algorithm on your sequence, and I got the same (15,680 -- comma delimiter, as used in the USA).


I get a final result of 15/14! or 1/5811886080. That's about 0.000000000172. I'm guessing that you rounded to zero, yes?

Now that I have a clearer understanding of the puzzle, I'll play with it when I have time.

😎
I finally got time to reply, yes we actually need to round up the numbers (make it a bit easier I guess), I've been trying difference sequences but the one that yield the highest value is currently that.

Thanks for helping man, if you have better sequence please post it here..
Thanks again
 

Otis

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Joined
Apr 22, 2015
Messages
1,872
5, 10, 14, 2, 6, 8, 15, 3, 13, 7, 9, 12, 1, 4, 11

6, 10, 4, 7, 12, 2, 15, 3, 5, 8, 13, 1, 9, 11, 14

Still playing …

:)
 

Otis

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Joined
Apr 22, 2015
Messages
1,872
Here are some patterns associated with the higher results I was getting. I'll use the two sequences above (in post #6) as examples.

5, 10, 14, 2, 6, 8, 15, 3, 13, 7, 9, 12, 1, 4, 11 → 168,000

6, 10, 4, 7, 12, 2, 15, 3, 5, 8, 13, 1, 9, 11, 14 → 280,800

The drops are shown in red, that is where a>b (when we divide previous V by b). The subsequent, increasing runs are shown in blue (when we're multiplying V by increases). Bold face denotes where the increase from b to a is four or more.

In the first sequence, the drops are 1, 2, 3, 7
The drops in the other sequence are 1, 2, 3, 4
Using the lowest numbers for b gave good results (that makes sense; less V-reduction, when dividing by smallest numbers).

The increasing runs in both sequences show groups of 1, 2, 2, 3, 3.
I'm thinking that no runs of 1 and as many runs of 3 and 4 are better (in general), but exceptions probably exist.

Both sequences have two big jumps (bold face).
Big jumps are good, of course, because we multiply by them.

So, I tried applying these thoughts: lowest bs, larger runs, and aim for three big jumps. After some more experimenting, I came up with:

4, 6, 10, 2, 7, 9, 14, 3, 8, 12, 1, 5, 11, 13, 15 → 512,000

I have a hunch that we can go much higher -- maybe exceed 750,000. Still pondering …


😎
 

Otis

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Joined
Apr 22, 2015
Messages
1,872
Taking it a couple steps further, I arranged the drops in decreasing order and I tried to spread out the runs as much as possible (while ending each with the highest numbers possible). This more than doubled the previous result (512000). I'm done playing with it, for now, but I might do some programming later. Thanks for posting the puzzle; it's an interesting (somewhat catch-22) situation.

4, 8, 10, 12, 3, 6, 9, 13, 2, 7, 11, 15, 1, 5, 14 → 1,105,920

😎
 
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