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Brendan McKay
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  1. The mean number of fixed points is 1. This is very elementary.

  2. Consider the operation of rotating three values around: $p(i)\to p(j)\to p(k)\to p(i)$. Given a permutation with no fixed points, there are $n^2-O(n)$ rotations that create from it a permutation with exactly one fixed point. Given a permutation with exactly one fixed point, there are $n^2-O(n)$ rotations that create from it a permutation with no fixed points.

  3. From (2), the numbernumbers of permutations with 0 and 1 fixed points are the same to $O(1/n)$ relative error. Combining this with (1) shows that the fraction of permutations with no fixed points is at least $\frac13-O(1/n)$.

This is the switching method and it can be used in extremely many circumstances. By continuing to increase the number of fixed points by further switchings, step 1 could be avoided.

Incidentally, congratulations to Terry and Jean Bourgain.

  1. The mean number of fixed points is 1. This is very elementary.

  2. Consider the operation of rotating three values around: $p(i)\to p(j)\to p(k)\to p(i)$. Given a permutation with no fixed points, there are $n^2-O(n)$ rotations that create a permutation with exactly one fixed point. Given a permutation with exactly one fixed point, there are $n^2-O(n)$ rotations that create a permutation with no fixed points.

  3. From (2), the number of permutations with 0 and 1 fixed points are the same to $O(1/n)$ relative error. Combining this with (1) shows that the fraction of permutations with no fixed points is at least $\frac13-O(1/n)$.

This is the switching method and it can be used in extremely many circumstances. By continuing to increase the number of fixed points by further switchings, step 1 could be avoided.

Incidentally, congratulations to Terry and Jean Bourgain.

  1. The mean number of fixed points is 1. This is very elementary.

  2. Consider the operation of rotating three values around: $p(i)\to p(j)\to p(k)\to p(i)$. Given a permutation with no fixed points, there are $n^2-O(n)$ rotations that create from it a permutation with exactly one fixed point. Given a permutation with exactly one fixed point, there are $n^2-O(n)$ rotations that create from it a permutation with no fixed points.

  3. From (2), the numbers of permutations with 0 and 1 fixed points are the same to $O(1/n)$ relative error. Combining this with (1) shows that the fraction of permutations with no fixed points is at least $\frac13-O(1/n)$.

This is the switching method and it can be used in extremely many circumstances. By continuing to increase the number of fixed points by further switchings, step 1 could be avoided.

Incidentally, congratulations to Terry and Jean Bourgain.

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Brendan McKay
  • 37.7k
  • 3
  • 80
  • 147
  1. The mean number of fixed points is 1. This is very elementary.

  2. Consider the operation of rotating three values around: $p(i)\to p(j)\to p(k)\to p(i)$. Given a permutation with no fixed points, there are $n^2-O(n)$ rotations that create a permutation with exactly one fixed point. Given a permutation with exactly one fixed point, there are $n^2-O(n)$ rotations that create a permutation with no fixed points.

  3. From (2), the number of permutations with 0 and 1 fixed points are the same to $O(1/n)$ relative error. Combining this with (1) shows that the fraction of permutations with no fixed points is at least $\frac13-O(1/n)$.

This is the switching method and it can be used in extremely many circumstances. By continuing to increase the number of fixed points by further switchings, step 1 could be avoided.

Incidentally, congratulations to Terry and Jean Bourgain.

  1. The mean number of fixed points is 1. This is very elementary.

  2. Consider the operation of rotating three values around: $p(i)\to p(j)\to p(k)\to p(i)$. Given a permutation with no fixed points, there are $n^2-O(n)$ rotations that create a permutation with exactly one fixed point. Given a permutation with exactly one fixed point, there are $n^2-O(n)$ rotations that create a permutation with no fixed points.

  3. From (2), the number of permutations with 0 and 1 fixed points are the same to $O(1/n)$ relative error. Combining this with (1) shows that the fraction of permutations with no fixed points is at least $\frac13-O(1/n)$.

This is the switching method and it can be used in extremely many circumstances. By continuing to increase the number of fixed points by further switchings, step 1 could be avoided.

  1. The mean number of fixed points is 1. This is very elementary.

  2. Consider the operation of rotating three values around: $p(i)\to p(j)\to p(k)\to p(i)$. Given a permutation with no fixed points, there are $n^2-O(n)$ rotations that create a permutation with exactly one fixed point. Given a permutation with exactly one fixed point, there are $n^2-O(n)$ rotations that create a permutation with no fixed points.

  3. From (2), the number of permutations with 0 and 1 fixed points are the same to $O(1/n)$ relative error. Combining this with (1) shows that the fraction of permutations with no fixed points is at least $\frac13-O(1/n)$.

This is the switching method and it can be used in extremely many circumstances. By continuing to increase the number of fixed points by further switchings, step 1 could be avoided.

Incidentally, congratulations to Terry and Jean Bourgain.

Source Link
Brendan McKay
  • 37.7k
  • 3
  • 80
  • 147

  1. The mean number of fixed points is 1. This is very elementary.

  2. Consider the operation of rotating three values around: $p(i)\to p(j)\to p(k)\to p(i)$. Given a permutation with no fixed points, there are $n^2-O(n)$ rotations that create a permutation with exactly one fixed point. Given a permutation with exactly one fixed point, there are $n^2-O(n)$ rotations that create a permutation with no fixed points.

  3. From (2), the number of permutations with 0 and 1 fixed points are the same to $O(1/n)$ relative error. Combining this with (1) shows that the fraction of permutations with no fixed points is at least $\frac13-O(1/n)$.

This is the switching method and it can be used in extremely many circumstances. By continuing to increase the number of fixed points by further switchings, step 1 could be avoided.