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Goldbach hints

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rabbott Posts: 1649	Posted 18:14 Jan 30, 2019 \| A couple of suggestions to avoid some of Haskell's pickiness with numeric types. 1. Define `iSqrt` with `round` rather than `floor`: `iSqrt n = round (sqrt (fromIntegral n)). round` returns an `Integer; floor` returns a `Float.` Life is easier for us if `iSqrt` returns an `Integer.` `2.` Use (g-p) `div` 2 rather than `(g-p) / 2.` Using `/` returns a `Float`; the function `div` does integer division, which is what we want. With these changes `iSqrt` and `isASquare` deal strictly with integers, and your search will also deal strictly with integers. Last edited by rabbott at 19:24 Jan 30, 2019.
rabbott Posts: 1649	Posted 21:49 Jan 30, 2019 \| Other than 5777 and 5993, no other counterexamples have been found. See: http://oeis.org/A060003. (But your program should be written to look for them anyway!) Last edited by rabbott at 21:51 Jan 30, 2019.

rabbott

Posts: 1649

Posted 18:14 Jan 30, 2019 |

A couple of suggestions to avoid some of Haskell's pickiness with numeric types.

1. Define iSqrt with round rather than floor: iSqrt n = round (sqrt (fromIntegral n)). round returns an Integer; floor returns a Float. Life is easier for us if iSqrt returns an Integer.

2. Use (g-p) `div` 2 rather than (g-p) / 2. Using / returns a Float; the function div does integer division, which is what we want.

With these changes iSqrt and isASquare deal strictly with integers, and your search will also deal strictly with integers.

Last edited by rabbott at 19:24 Jan 30, 2019.

rabbott

Posts: 1649

Posted 21:49 Jan 30, 2019 |

Other than 5777 and 5993, no other counterexamples have been found. See: http://oeis.org/A060003. (But your program should be written to look for them anyway!)

Last edited by rabbott at 21:51 Jan 30, 2019.