Steady Average Timing
Definition
Steady Average Timing is the time system where
- one gets an initial period of time,
- time used for each move is deducted,
- and an extra period of time is added after exactly N moves were made in a period.
Extra periods are of same size, the initial one may differ.
A move is in a period if it ends before the period ends but not before the period starts.
In other words,
- one is required having made a minimum number of moves in each period.
Diagram
|---| time for N moves
|------| periods before
|--|------| periods after
Remarks
The difference to Canadian Timing is that resetting of time and stones happens when time runs out, not stones.
The drawback of Steady with a plain clock is a periodically falling flag. On the first look there seems to be no problem:
- some stones left -- you lost
- all stones gone -- next period
However, this is not correct. Why? Because a move is not complete when the stone hit the board, but when the clock was hit. Otherwise we had to decide if or not a stone hit the board earlier than the flag fell, which is virtually impossible. So, the criterion has to be something like
- lid closed -- you lost
- lid open -- next period
loading another duty on the player's neck: operating the lid.
With only a plain clock, Steady Average has to be emulated -- see below.
Emulation
The player
- gets the initial period,
- gets N stones,
- and closes his lid (shuts his supply).
If the player runs out of time with lid closed:
- the player loses on time.
If the player runs out of time with lid open:
- the opponent stops both clocks,
- the opponent puts one extra period on the player's clock,
- the player counts out N stones,
- the player closes his lid,
- they check,
- and the opponent starts the player's clock again.
If the player runs out of stones:
- (relaxed version)
- the player hits his clock and opens his lid.
- (strict version)
- the player stops both clocks,
- the player opens his lid,
- they check,
- and the player starts the opponent's clock.
Questions
Tderz: Fischer Timing would also fall under this definition, wouldn't it? The Fischer-seconds are added after exactly one move has been made in a period.
Robert Pauli: No, Total Average Timing is the one that covers Fischer! Under Steady Average any move beyond the N-th in the same period earns nothing. Under Total Average, however, you're already working towards your next extra period.
Tderz: Phew ... I might have my slow days -
or the definition is not yet foolproof?
I want to understand you - but do not succeed.
1
Steady Average Timing is the time system where
- one gets an initial period of time,
- time used for each move is deducted,
- and an extra period of time is added after exactly N moves were made in a period.
Total Average Timing is the time system where
- one gets an initial period of time,
- time used for each move is deducted,
- and an extra period of time is added after each N-th move.
Question/wondering: for N=1, how can this one move not be made in a period? Hence, my conclusion is until here: for N=1, the steady and total average timing have the same concept (thus both cover Fischer).
2
Could you explain your drawings? And, why, in the example has the player not lost after #step 8a ? (or got 3 minutes). Rather, s/he seems to have played 4 stones (3, 2, 1, 0 at 7 and 0 at 8a). Is the example flawed or am I simply not getting it?
Mutatis mutandis the player should have got his new 3 stones + 3 minutes at 4a, not at 4b .... sorry the penny fell just now ...
Robert Pauli: It's subtle.
- Total Average with N = 1 is Fischer, right, Steady Average with N = 1 not. Why? Because after you've done your move the unused time isn't added to the next period. However, it's neither spilled. You can use it for extra moves, "between" the periods.
-
- Sorry, the drawings don't reflect the fact that further moves beyond N are treated differently. For instance, if you make 2N moves in one period, Total Average grants you two further periods, whereas Steady Average still only one.
- Each step sees you at turn. Steps 4 - 7 therefore play four stones, right. You didn't lose in 8a because your charge of stones already was down to zero: condition met, but earned period still owed, and it stays owed under Steady Average until the period is done (4b, 8b).
Example
3 moves ("stones") in 3 minutes.
# Time Stones Used -------------------------- 1 3:00 3 0:40 2 2:20 2 1:50 3 0:30 1 0:29 4a 0:01 0 0:01 4b 3:00 3 1:10 5 1:50 2 1:45 6 0:05 1 0:03 7 0:02 0 0:01 8a 0:01 0 0:01 8b 3:00 3 2:10 9 0:50 2 0:50 10 0:00 1 lost