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Rules of Powerdrain
Imagine you and your opponent are in a powerstation and your goal is
to control the powergrid. One player is positive energy (+) and the
other player is negative energy (-). Each player alternately places
powerplugs into the sections of the powergrid. After a total of 16
powerplugs are inserted, the powergrid adjusts the voltages of each
section of the powergrid and a winner is determined. The winner is
the player who has more of his own powerplugs still active. If a tie
occurs (both players have an equal number of active powerplugs at the
end of the game), then the tie is broken by adding up the power levels
of each players' active powerplugs. The player with the greater
absolute power level will win the game. There is no further tie breaker.
When two powerplugs of opposite polarities touch on their sides, they
drain the energy out of each other. A player's powerplug voltage may
drain to zero (0) but cannot change into the enemy's polarity. Two
powerplugs of the same polarity cannot affect each other (they do not
add nor subtract from each other's voltage). An active powerplug is
one that is not at zero (0) voltage at the end of the game when all
16 powerplugs have been inserted into the powergrid. A section of the
powergrid that does not have a powerplug inserted into it is not
affected.
The powergrid is given random voltage potential levels along the left
side and top of the grid. These random potentials are the 5 odd digits
1,3,5,7,9. The powerplugs are given two random even digits. Each
permutation of two even digits is represented by the 16 powerplugs.
When a powerplug is inserted into a section of the powergrid, the voltage
of the powerplug is determined by adding the absolute values of the
differences of the first digit with the left voltage potential and the
second digit with the top voltage potential. This is illustrated below:
5
--------- this is an example of two sections of a powergrid
7 | <2 4^ | the powerplug "24" was inserted into the section "75"
|+6 +2 | the voltage of this plug is the sums of the absolute
--------- differences (2 minus 7) plus (4 minus 5) which is
9 | <8 2^ | 5 + 1 = 6 and since the plug was owned by the positive
|-4 0 | player it was given a '+' value.
--------- The first number under the plug (+6) represents
the value of the plug itself, without any drain from
negative plugs that may be on any of its four sides
(above, below, on the left and right). The second number
below the plug is the voltage of the plug after
neighboring enemy plugs are taken into account. Notice
that a total of 4 volts was drained off of this plug by
negative plugs that were touching this plug.
Below is an example of a finished game between player1 and player2.
Player1 (positive player) won the game because he had 4 active plugs
of his voltage while player2 (negative player) only had 2 active plugs
left at the end of the game.
Player1 has four (4) active positive plugs left.
plug 46 at location 91 with a voltage of +10
plug 26 at location 93 with a voltage of +6 (4 volts were drained)
plug 28 at location 71 with a voltage of +8 (4 volts were drained)
plug 48 at location 15 with a voltage of +6
Player2 only has two (2) active negative plugs left in the grid.
plug 68 at location 31 with a voltage of -10
plug 84 at location 37 with a voltage of -8
1 3 7 5 9
----------------------------------------- Positive + player1
9 | <4 6^ | <2 6^ | <4 2^ | <8 2^ | <2 2^ |
|+10 +10|+10 +6 |+10 0 |-4 0 |+14 0 |
-----------------------------------------
7 | <2 8^ | <6 6^ | <4 4^ | | <2 4^ | Negative - player2
|+12 +8 |-4 0 |-6 0 | |-10 0 |
-----------------------------------------
1 | <8 8^ | <8 6^ | <6 2^ | <4 8^ | |
|+14 0 |-10 0 |+10 0 |+6 +6 | |
-----------------------------------------
5 | <6 4^ | | | | |
|-4 0 | | | | |
-----------------------------------------
3 | <6 8^ | | <8 4^ | | |
|-10 -10| |-8 -8 | | |
-----------------------------------------
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