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Aaron Meyerowitz
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Consider the convergents $3/2, 7/5, 17/12, 41/29, 99/70, \cdots$ to $\sqrt{2}.$ For the pair $(70,99)$ the fastest descent is

$$29\cdot 99-41\cdot 70=$$ $$17\cdot 29-12\cdot 41=$$ $$5\cdot 17- 7\cdot 12=$$ $$3 \cdot 5 - 2\cdot 7=$$

Starting from the convergents to a number $1 \lt r \lt 2$ this will pretty much be the case. At the worst you would need to skip every other convergent. This happens where there is a $1$ in the continued fraction.

So the worst case is the convergents $2/3,3/5,5/8,8/13,13/21,21/34,34/55,55/89,144/89\cdots$ to the golden ratio which lead to

$$34\cdot 144-55\cdot 89=$$ $$13\cdot 55-21\cdot 34=$$ $$5\cdot 21- 8\cdot 13=$$

Later Here I am making the assumption that the fastest descent is the one which minimizes $|a|+|b|$ at each stage. That is true but I did not prove it. I will not prove it here but I'll give an illustration. That isn't a proof, but it may be convincing and might suggest what the precise things to prove would be.

For fixed coprime $c,d \gt 1$ there are two solutions $a',b'$ and $a'',b''$ to $ac+bd=1$ with $|a|+|b| \lt c+d.$ One has $-d \lt a' \lt 0 \lt b' \lt c$ and the other, $a'',b''=a+d,c-b'$ has $-c \lt b'' \lt 0 \lt a'' \lt d.$ Some inequalities will be weak when $\min(c,d) \le 1.$

There are special aspects of the lovely example $70,99$ which might be distracting so let me use instead $70,129$

I'll write $$[70, 129] ==> [-293, 159], [-164, 89], [-35, 19], [94, -51], [223, -121]$$ to indicate the first few possible pairs $[a,b].$

The continuations are

$$ [35, 19] ==> [-32, 59], [-13, 24], [6, -11], [25, -46], [44, -81]$$

$$ [51, 94] ==> [-223, 121], [-129, 70], [-35, 19], [59, -32], [153, -83]$$

$$ [89, 164] ==> [-363, 197], [-199, 108], [-35, 19], [129, -70], [293, -159]$$

$$ [121, 223] ==> [-352, 191], [-129, 70], [94, -51], [317, -172], 540, -293]$$

$$ [159, 293] ==> [-457, 248], [-164, 89], [129, -70], [422, -229], [715, -388]$$

So I suppose a conjecture would be that, if one rejects the solution which minimizes $|a|+|b|$, then at the next stage you can't get a solution which does better than the one you rejected.

Consider the convergents $3/2, 7/5, 17/12, 41/29, 99/70, \cdots$ to $\sqrt{2}.$ For the pair $(70,99)$ the fastest descent is

$$29\cdot 99-41\cdot 70=$$ $$17\cdot 29-12\cdot 41=$$ $$5\cdot 17- 7\cdot 12=$$ $$3 \cdot 5 - 2\cdot 7=$$

Starting from the convergents to a number $1 \lt r \lt 2$ this will pretty much be the case. At the worst you would need to skip every other convergent. This happens where there is a $1$ in the continued fraction.

So the worst case is the convergents $2/3,3/5,5/8,8/13,13/21,21/34,34/55,55/89,144/89\cdots$ to the golden ratio which lead to

$$34\cdot 144-55\cdot 89=$$ $$13\cdot 55-21\cdot 34=$$ $$5\cdot 21- 8\cdot 13=$$

Consider the convergents $3/2, 7/5, 17/12, 41/29, 99/70, \cdots$ to $\sqrt{2}.$ For the pair $(70,99)$ the fastest descent is

$$29\cdot 99-41\cdot 70=$$ $$17\cdot 29-12\cdot 41=$$ $$5\cdot 17- 7\cdot 12=$$ $$3 \cdot 5 - 2\cdot 7=$$

Starting from the convergents to a number $1 \lt r \lt 2$ this will pretty much be the case. At the worst you would need to skip every other convergent. This happens where there is a $1$ in the continued fraction.

So the worst case is the convergents $2/3,3/5,5/8,8/13,13/21,21/34,34/55,55/89,144/89\cdots$ to the golden ratio which lead to

$$34\cdot 144-55\cdot 89=$$ $$13\cdot 55-21\cdot 34=$$ $$5\cdot 21- 8\cdot 13=$$

Later Here I am making the assumption that the fastest descent is the one which minimizes $|a|+|b|$ at each stage. That is true but I did not prove it. I will not prove it here but I'll give an illustration. That isn't a proof, but it may be convincing and might suggest what the precise things to prove would be.

For fixed coprime $c,d \gt 1$ there are two solutions $a',b'$ and $a'',b''$ to $ac+bd=1$ with $|a|+|b| \lt c+d.$ One has $-d \lt a' \lt 0 \lt b' \lt c$ and the other, $a'',b''=a+d,c-b'$ has $-c \lt b'' \lt 0 \lt a'' \lt d.$ Some inequalities will be weak when $\min(c,d) \le 1.$

There are special aspects of the lovely example $70,99$ which might be distracting so let me use instead $70,129$

I'll write $$[70, 129] ==> [-293, 159], [-164, 89], [-35, 19], [94, -51], [223, -121]$$ to indicate the first few possible pairs $[a,b].$

The continuations are

$$ [35, 19] ==> [-32, 59], [-13, 24], [6, -11], [25, -46], [44, -81]$$

$$ [51, 94] ==> [-223, 121], [-129, 70], [-35, 19], [59, -32], [153, -83]$$

$$ [89, 164] ==> [-363, 197], [-199, 108], [-35, 19], [129, -70], [293, -159]$$

$$ [121, 223] ==> [-352, 191], [-129, 70], [94, -51], [317, -172], 540, -293]$$

$$ [159, 293] ==> [-457, 248], [-164, 89], [129, -70], [422, -229], [715, -388]$$

So I suppose a conjecture would be that, if one rejects the solution which minimizes $|a|+|b|$, then at the next stage you can't get a solution which does better than the one you rejected.

Source Link
Aaron Meyerowitz
  • 30.1k
  • 1
  • 48
  • 104

Consider the convergents $3/2, 7/5, 17/12, 41/29, 99/70, \cdots$ to $\sqrt{2}.$ For the pair $(70,99)$ the fastest descent is

$$29\cdot 99-41\cdot 70=$$ $$17\cdot 29-12\cdot 41=$$ $$5\cdot 17- 7\cdot 12=$$ $$3 \cdot 5 - 2\cdot 7=$$

Starting from the convergents to a number $1 \lt r \lt 2$ this will pretty much be the case. At the worst you would need to skip every other convergent. This happens where there is a $1$ in the continued fraction.

So the worst case is the convergents $2/3,3/5,5/8,8/13,13/21,21/34,34/55,55/89,144/89\cdots$ to the golden ratio which lead to

$$34\cdot 144-55\cdot 89=$$ $$13\cdot 55-21\cdot 34=$$ $$5\cdot 21- 8\cdot 13=$$