# 20 iterations

SwissGambit
Posers and Puzzles 20 Aug '07 21:58
1. SwissGambit
Caninus Interruptus
20 Aug '07 21:581 edit
a b c d

Fill in subsequent rows by taking the absolute value of the number above minus the one above and to the right:
|a-b| |b-c| |c-d| |d-a| (d has nothing to the right, so we 'wrap around' and use a)

The challenge is: Find numbers a, b, c, and d that yield 20 rows where at least one number on each row is not zero. (not counting the initial row)

Example:

1 1 1 3
0 0 2 2
0 2 0 2
2 2 2 2
0 0 0 0

This counts as 4 iterations.
2. agryson
AGW Hitman
20 Aug '07 22:12
Originally posted by SwissGambit
a b c d

Fill in subsequent rows by taking the absolute value of the number above minus the one above and to the right:
|a-b| |b-c| |c-d| |d-a| (d has nothing to the right, so we 'wrap around' and use a)

The challenge is: [b]Find numbers a, b, c, and d that yield 20 rows where at least one number on each row is not ...[text shortened]... 2, 3, 4, we get:
1 1 1 3
0 0 2 2
0 2 0 2
2 2 2 2
0 0 0 0

This counts as 4 iterations.
100, 3, 2, 1

?
did I understqnd the question right?
3. SwissGambit
Caninus Interruptus
20 Aug '07 22:271 edit
Originally posted by agryson
100, 3, 2, 1

?
did I understqnd the question right?
Using those numbers, i get

100 3 2 1
97 1 1 99
96 0 98 2
96 98 96 94
2 2 2 2
0 0 0 0

...which is only 4 iterations.

Order of operations is to subtract first, and then take the absolute value of that result.
4. agryson
AGW Hitman
21 Aug '07 00:05
Originally posted by SwissGambit
Using those numbers, i get

100 3 2 1
97 1 1 99
96 0 98 2
96 98 96 94
2 2 2 2
0 0 0 0

...which is only 4 iterations.

Order of operations is to subtract first, and then take the absolute value of that result.
hmm... ok, I'll work on it tomorrow,
5. invigorate
Only 1 F in Uckfield
21 Aug '07 13:59
Originally posted by SwissGambit
a b c d
Spot the mistake?
6. SwissGambit
Caninus Interruptus
21 Aug '07 19:13
Originally posted by invigorate
Spot the mistake?
The mistake is tackling a problem like this without a basic knowledge of algebra.
7. 21 Aug '07 20:55
Originally posted by invigorate
Spot the mistake?
I don't. What is the mistake?

I like this problem! Really!
Now I have something to think about during boring meetings.
8. 21 Aug '07 23:41
The mistake is a b c and d are letters not numbers!!!

HA HA HA!
9. 22 Aug '07 04:16
Originally posted by Gastel
The mistake is a b c and d are letters not numbers!!!

HA HA HA!
Who says they are letters?
He says from the beginning: "Start with four numbers: a b c d"
If he says the symbols are numbers they are, believe me. They perhaps look like letters but they are in fact numbers.

Can you solve the equation 2x=4 not realizing that the x stands for a number?
10. 22 Aug '07 09:14
I think he was trying to be funny.

Trying.
11. 22 Aug '07 09:37
Originally posted by mtthw
I think he was trying to be funny.

Trying.
Who? invigorate or Gastel or both? Exactly the same kind of humour? Strange...
12. 22 Aug '07 11:11
a=pi
b=e
c=sqrt(2)
d=sqrt(3)
13. 22 Aug '07 15:13
Originally posted by David113
a=pi
b=e
c=sqrt(2)
d=sqrt(3)
Doesn't work! Hits zero in four iterations ðŸ™‚. Try it out.
14. 22 Aug '07 16:37
I spent some time trying to come up with a pattern that worked. No luck so far.

So went for the brute force approach ðŸ™‚. Best so far is [4910, 4163, 2789, 262], which gives the 20 iterations you wanted.

I'm sure there's a better way.
15. SwissGambit
Caninus Interruptus
22 Aug '07 16:521 edit
Originally posted by David113
a=pi
b=e
c=sqrt(2)
d=sqrt(3)
Oops...the stipulation should allow only whole numbers.