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Softened the language in the opening of the question. Being blunt is rude!
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Sam Nead
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You are incorrectMosher's proof is, I think, fine. Re You should re-read the definition of $i(\delta, \{h\})$ given on line -14 of page 38 of Mosher's paper.

The sequence of flips produced by Mosher's algorithm is at most linear in the total intersection number. This worst-case bound is realized when the two triangulations differ by a power of a Dehn twist. (An average-case case bound is also possible; here the number of flips will be the log of the total intersection number.)

Another algorithm can be deduced from Hatcher's paper "Triangulations of surfaces".

In reply to the updated question:

Let $d_1$ be the first arc $h$ crosses, and let $d_2$ be the second. So $d_1$ is the diagonal of $Q$, and $d_1$ will be flipped to $d_1'$. After the flip, the first arc that $h$ meets is $d_2$. So the intersections of $h$ and $d_2$ are not counted. Thus $i(\delta',\{h\}) \leq i(\delta,\{h\}) - 1$, as desired.

You are incorrect. Re-read the definition of $i(\delta, \{h\})$ given on line -14 of page 38 of Mosher's paper.

The sequence of flips produced by Mosher's algorithm is at most linear in the total intersection number. This worst-case bound is realized when the two triangulations differ by a power of a Dehn twist. (An average-case case bound is also possible; here the number of flips will be the log of the total intersection number.)

Another algorithm can be deduced from Hatcher's paper "Triangulations of surfaces".

In reply to the updated question:

Let $d_1$ be the first arc $h$ crosses, and let $d_2$ be the second. So $d_1$ is the diagonal of $Q$, and $d_1$ will be flipped to $d_1'$. After the flip, the first arc that $h$ meets is $d_2$. So the intersections of $h$ and $d_2$ are not counted. Thus $i(\delta',\{h\}) \leq i(\delta,\{h\}) - 1$, as desired.

Mosher's proof is, I think, fine. You should re-read the definition of $i(\delta, \{h\})$ given on line -14 of page 38 of Mosher's paper.

The sequence of flips produced by Mosher's algorithm is at most linear in the total intersection number. This worst-case bound is realized when the two triangulations differ by a power of a Dehn twist. (An average-case case bound is also possible; here the number of flips will be the log of the total intersection number.)

Another algorithm can be deduced from Hatcher's paper "Triangulations of surfaces".

In reply to the updated question:

Let $d_1$ be the first arc $h$ crosses, and let $d_2$ be the second. So $d_1$ is the diagonal of $Q$, and $d_1$ will be flipped to $d_1'$. After the flip, the first arc that $h$ meets is $d_2$. So the intersections of $h$ and $d_2$ are not counted. Thus $i(\delta',\{h\}) \leq i(\delta,\{h\}) - 1$, as desired.

reply to the update
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Sam Nead
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You are incorrect. Re-read the definition of $i(\delta, \{h\})$ given on line -14 of page 38 of Mosher's paper.

The sequence of flips produced by Mosher's algorithm is at most linear in the total intersection number. This worst-case bound is realized when the two triangulations differ by a power of a Dehn twist. (An average-case case bound is also possible; here the number of flips will be the log of the total intersection number.)

Another algorithm can be deduced from Hatcher's paper "Triangulations of surfaces".

In reply to the updated question:

Let $d_1$ be the first arc $h$ crosses, and let $d_2$ be the second. So $d_1$ is the diagonal of $Q$, and $d_1$ will be flipped to $d_1'$. After the flip, the first arc that $h$ meets is $d_2$. So the intersections of $h$ and $d_2$ are not counted. Thus $i(\delta',\{h\}) \leq i(\delta,\{h\}) - 1$, as desired.

You are incorrect. Re-read the definition of $i(\delta, \{h\})$ given on line -14 of page 38 of Mosher's paper.

The sequence of flips produced by Mosher's algorithm is at most linear in the total intersection number. This worst-case bound is realized when the two triangulations differ by a power of a Dehn twist. (An average-case case bound is also possible; here the number of flips will be the log of the total intersection number.)

Another algorithm can be deduced from Hatcher's paper "Triangulations of surfaces".

You are incorrect. Re-read the definition of $i(\delta, \{h\})$ given on line -14 of page 38 of Mosher's paper.

The sequence of flips produced by Mosher's algorithm is at most linear in the total intersection number. This worst-case bound is realized when the two triangulations differ by a power of a Dehn twist. (An average-case case bound is also possible; here the number of flips will be the log of the total intersection number.)

Another algorithm can be deduced from Hatcher's paper "Triangulations of surfaces".

In reply to the updated question:

Let $d_1$ be the first arc $h$ crosses, and let $d_2$ be the second. So $d_1$ is the diagonal of $Q$, and $d_1$ will be flipped to $d_1'$. After the flip, the first arc that $h$ meets is $d_2$. So the intersections of $h$ and $d_2$ are not counted. Thus $i(\delta',\{h\}) \leq i(\delta,\{h\}) - 1$, as desired.

Hatcher
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Sam Nead
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You are incorrect. Re-read the definition of $i(\delta, \{h\})$ given on line -14 of page 38 of Mosher's paper.

The sequence of flips produced by Mosher's algorithm is at most linear in the total intersection number. This worst-case bound is realized when the two triangulations differ by a power of a Dehn twist. (An average-case case bound is also possible; here the number of flips will be the log of the total intersection number.)

Another algorithm can be deduced from Hatcher's paper "Triangulations of surfaces".

You are incorrect. Re-read the definition of $i(\delta, \{h\})$ given on line -14 of page 38 of Mosher's paper.

The sequence of flips produced by Mosher's algorithm is at most linear in the total intersection number. This worst-case bound is realized when the two triangulations differ by a power of a Dehn twist. (An average-case case bound is also possible; here the number of flips will be the log of the total intersection number.)

You are incorrect. Re-read the definition of $i(\delta, \{h\})$ given on line -14 of page 38 of Mosher's paper.

The sequence of flips produced by Mosher's algorithm is at most linear in the total intersection number. This worst-case bound is realized when the two triangulations differ by a power of a Dehn twist. (An average-case case bound is also possible; here the number of flips will be the log of the total intersection number.)

Another algorithm can be deduced from Hatcher's paper "Triangulations of surfaces".

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Sam Nead
  • 28.1k
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  • 131
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