9-Darters
Bij darts levert iedere pijl die men gooit een score op. Bij sommige gebieden op het dartboard
horen echter speciale scores: men kan niet alleen de getallen 1 t/m 20 werpen, maar ook van
elk getal het dubbele en zelfs het drievoudige. Raakt men de roos dan levert dit 25 punten op;
midden in de roos telt dubbel en geeft 50 punten.
Een van de populairste spelvarianten bij darts is het spel 501. Daarbij start men op een totaal
van 501 punten en is het de bedoeling om zo snel mogelijk terug bij 0 te geraken. Elke score
die men gooit mag men van het totaal aftrekken. Maar om het spel uit te gooien dient men wel
precies op 0 te eindigen en de laatste pijl dient een dubbelteller te zijn: ofwel een dubbele van
1 t/m 20, ofwel de dubbele roos.
Het minimum aantal pijlen dat men nodig heeft om bij 0 te geraken is 9. Een dergelijke
combinatie heet een 9-darter, bijvoorbeeld: T20-T20-T20-T20-T19-T19-T19-D20-D25.
In deze notatie staat de letter T voor “triple” en de letter D voor “double”. Dus T20 levert
3x20=60 punten op, de maximale score voor een pijl. D25 geeft de dubbele roos aan en levert
2x25=50 punten op, de maximale score voor een dubbelteller.
OPGAVE: Bepaal alle essentieel verschillende 9-darters.
Hierbij zijn twee oplossingen NIET essentieel verschillend als ze alleen verschillen qua
volgorde van de eerste 8 pijlen: de combinatie T20-T19-T20-T19-T20-T19-T20-D20-D25
beschouwen we als hetzelfde als die in het voorbeeld hierboven. Maar twee mogelijkheden
zijn WEL essentieel verschillend als er met een andere dubbelteller wordt geeindigd, dus een
tweede voorbeeld van een 9-darter is: T20-T20-T20-T20-T19-T19-T19-D25-D20 ook al lijkt
het een permutatie van de eerdere oplossing.
9-Darters
When playing the game of darts, every dart yields a score. But some areas on the dartboard
yield special scores: it is not only possible to throw each of the values 1 to 20, but one can
also obtain double these values and even triple these values. If one hits the bull, 25 points are
scored; the middle area of the bull is called the double bull and yields 50 points.
One of the most popular variants of the game of darts is called 501. Here one starts from a
total of 501 points and it is the purpose of the game to bring this down to zero points as quick
as possible. Every score one throws is subtracted from the total. But to finish the game, one
should end exactly on zero points, and it is required that the last dart which achieves this must
yield a double score: either a double of 1 to 20, or the double bull.
The minimum number of darts required to reach 0 in this way, is 9. Such a combination is
called a 9-darter, for instance: T20-T20-T20-T20-T19-T19-T19-D20-D25.
In this notation, the letter T stands for “triple” and the letter D stands for “double”. So T20
yields 3x20=60 points, de maximum score for a single dart. D25 denotes the double bull and
yields 2x25=50 points, de maximum score for a double.
QUESTION: Find all essentially different 9-darters.
Here, two solutions ARE NOT considered to be essentially different if they only differ with
respect to the order of the first 8 darts: the combination T20-T19-T20-T19-T20-T19-T20D20-D25 is considered to be equivalent to the solution in the example given above. But two
solutions ARE considered to be essentially different if one finishes with a different double, so
a new second example of a 9-darter is: T20-T20-T20-T20-T19-T19-T19-D25-D20 even
though it may seem to be a permutation of the first solution.
9-Darters - SOLUTION
The maximum score for each dart is 60 (one needs to throw a T20). The maximum score for the last dart, which
is required to be a double, is 50 (one needs to throw a D25). Therefore, the maximum score for 9 darts together,
with the last dart a double, is 8x60+50=530. Which is 29 points too many. This raises the following question:
how can one adapt those maximum scores to yield 29 points less?
•
•
If the last dart, the D25, is replaced by another double, the alternatives are: D20, D19, D18, D17, …
which reduces the score by 10, 12, 14, 16, … points.
If one of the first 8 darts, a T20, is replaced, it can be replaced by:
(i)
another triple, such as: T19, T18, T17, … which reduces the score by 3, 6, 9, … points;
(ii)
a double, such as D25, D20, D19, … which reduces the score by 10, 20, 22, … points.
Clearly, replacing a T20 by a single reduces the score by at least 35 points, which is too much.
Thus, the required reduction of 29 points is to be composed of a multiple of 3 (obtained from replacing one or
more times T20 by other triples) and an even number which is at least as large as 10 (either obtained by
replacing the last D25 by another double, or by replacing some T20 by a double, or both). Note that, since 29 is
odd, the multiple of 3 also needs to be odd. Now, it is not too difficult to find all such compositions:
29 = 3+26 = 9+20 = 15+14.
Next, one can determine how these decompositions can be realized.
3+26:
There is only one way to make 3, namely by replacing once T20 by T19.
There are three ways to make 26, namely by: (1) replacing D25 by D12, (2) replacing T20 by D17, (3)
replacing D25 by D17 and T20 by D25.
9+20: There are three ways to make 9, namely by: (1) replacing 3 times T20 by T19, (2) replacing T20 by T19
and T20 by T18, (3) replacing T20 by T17.
There are four ways to make 20, namely by: (1) replacing D25 by D15, (2) replacing D25 by D20 and
T20 by D25, (3) replacing twice T20 by D25, (4) replacing T20 by D20.
15+14: There are seven ways to make 15, namely by: (1) replacing 5 times T20 by T19, (2) replacing 3 times
T20 by T19 and once T20 by T18, (3) replacing twice T20 by T19 and once T20 by T17, (4) replacing
once T20 by T19 and twice T20 by T18, (5) replacing once T20 by T19 and once T20 by T16, (6)
replacing once T20 by T18 and once T20 by T17, (7) replacing once T20 by T15.
There is only one way to make 14, namely by replacing D25 by D18.
Thus, the total number of possibilities is equal to 1x3 + 3x4 + 7x1 = 22.
Two of these possibilities have been given in the text, so there are 20 other solutions to be found...
An overview of all 22 possibilities is given in the following table.
T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20
T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20
T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20
T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T19 T19
T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T19 T19 T19 T19 T19 T19
T20 T20 T20 T20 T20 T20 T20 T20 T20 T20 T19 T19 T19 T19 T19 T17 T19 T19 T19 T18 T19 T19
T20 T20 T20 T19 T19 T19 T19 T18 T17 T17 T19 T19 T18 T18 T18 D25 T19 T19 T19 D25 T19 D25
T19 T17 T15 T18 D25 T16 D17 T17 D25 D20 T19 T17 T18 D25 D20 D25 D25 D20 T18 D25 T19 D25
D12 D15 D18 D15 D17 D18 D25 D18 D20 D25 D15 D18 D18 D20 D25 D25 D20 D25 D18 D25 D18 D25
PS: if one does regard different permutations of the first 8 darts to yield different solutions, then the total number
of possibilities for 9-darters can be computed as 3944.