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Given two unmarked jugs, one which holds 7 liters,

and another which holds 11 liters,

an unlimited supply of water, and no need to conserve,

how do you measure exactly 6 liters?

Problems of this type have been around for a long time,

but only recently did I realize that they are Diophantine equation

problems in disguise.

If we add 7's and subtract 11's, eventually we'll get to 6, or whatever other number is desired.

We just need to solve 7x - 11y = 6

I have another page which does that, but here I'll work this one out.

7 (x - y) - 4y = 6.

Let w = (x-y)

7w - 4y = 6

3w + 4 (w - y) = 6

By inspection, w = 2, y = 2 is a solution.

Then, x = w + y → x = 4.

If we fill the 7 liter jug 4 times,

and empty the 11 liter jug twice,

we'll have our 6 liters.

We proceed as follows:

Action | Count | 7L jug | 11L jug |
---|---|---|---|

Start | 0 | 0 | |

Fill 7 | (1) | 7 | 0 |

Pour into 11 | 0 | 7 | |

Fill 7 | (2) | 7 | 7 |

Pour into 11 | 3 | 11 | |

Empty 11 | [-1] | 3 | 0 |

Pour into 11 | 0 | 3 | |

Fill 7 | (3) | 7 | 3 |

Pour into 11 | 0 | 10 | |

Fill 7 | (4) | 7 | 10 |

Pour into 11 | 6 | 11 | |

At this point we have our 6, and don't need to carry out the last step, but it fits the equation better if we do. | |||

Empty 11 | [-2] | 6 | 0 |

We can also solve this the other way around, using

11x - 7y = 6.
In that case, our initial solution might be

x = -2, y = -4 (the negatives of the solution above)

but since you can't fill a jug "negative two times",

we can convert to positive values by adding 7 and 11, respectively,

to get

x = 5, y = 7,

which would mean fill the 11 five times, and empty the 7 seven times

Then, 5 * 11 - 7 * 7 = 55 - 49 = 6.

But that takes many more "fill, transfer, empty" steps than the above.

The general solution to 7x - 11y = 6 is

x = 4 + 11t,

y = 2 - 7t, for any integer t.

t | x | y |
---|---|---|

-3 | -29 | -19 |

-2 | -18 | -12 |

-1 | -7 | -5 |

0 | 4 | 2 |

1 | 15 | 9 |

2 | 26 | 16 |