Teaser 3020: Baz’s bizarre arrers
From The Sunday Times, 9th August 2020 [link]
“Bizarrers” dartboards have double and treble rings and twenty sectors ordered as on this normal dartboard [shown above]. However, a sector’s central angle (in degrees) is given by (100 divided by its basic score). The 20 sector incorporates the residual angle to complete 360º.
Each player starts on 501 and reduces this, eventually to 0 to win. After six three-dart throws, Baz’s seventh could win. His six totals were consecutive numbers. Each three-dart effort lodged a dart in each of three clockwise-adjacent sectors (hitting, in some order, a single zone, a double zone and a treble zone). [Each] three-sector angle sum (in degrees) exceeded that total.
The sectors scored in are calculable with certainty, but not how many times hit with certainty, except for one sector.
Which sector?
This puzzle uses a different spelling for “arraz” from the previous puzzle involving Baz (Teaser 2934).
[teaser3020]
GeoffR, 
Robert Brown, and 1 other are discussing. 
S2T2 
Jim Randell 5:21 pm on 7 August 2020 Permalink |
This Python program runs in 52ms.
Run: [ @replit ]
from fractions import Fraction as F from collections import defaultdict from itertools import product from enigma import tuples, subsets, multiset, seq_all_same, intersect, join, printf # sectors of the dartboard board = [ 20, 1, 18, 4, 13, 6, 10, 15, 2, 17, 3, 19, 7, 16, 8, 11, 14, 9, 12, 5 ] # map scores from 1 to 19 to sector angles (in degrees) angle = dict((n, F(100, n)) for n in board if n != 20) # the 20 sector completes the board angle[20] = 360 - sum(angle.values()) # consider 3 adjacent sectors d = defaultdict(list) # store the results by the score (single, double, treble) for ss in tuples(board + board[:2], 3): # calculate the sector angle sum a = sum(angle[s] for s in ss) # choose multipliers for the scores for ms in subsets(ss, size=len, select="P"): # calculate the total t = sum(s * m for (m, s) in enumerate(ms, start=1)) # which is less than the angle sum if not (t < a): continue d[t].append(ms) # look for 6 consecutive scores for ts in tuples(sorted(d.keys()), 6): if not (ts[0] + 5 == ts[-1]): continue # produce possible sets of scoring sectors mss = list(multiset.from_seq(*ss) for ss in product(*(d[t] for t in ts))) # each set of sectors should be the same if not seq_all_same(set(m.keys()) for m in mss): continue # look for common (sector, count) pairs for all possible scores ps = intersect(m.items() for m in mss) if len(ps) == 1: # output solution for (s, n) in ps: printf("sector = {s} [count = {n}]") printf(" scores = {ts}") for t in ts: printf(" {t}: {ss}", ss=join(d[t], sep=" "))Solution: We can say for certain that the 4 sector was hit twice.
Baz’s 6 scores were (in consecutive order): 58, 59, 60, 61, 62, 63.
So he has 138 left to score. (Which is achievable in many ways with 3 darts (assuming standard rules of darts apply), for example: treble 20, treble 20, double 9).
Possible ways to make the first 6 scores with consecutive sectors are:
We don’t know which of the alternatives makes up the 59, but whichever is chosen the 4 sector is used twice (in scoring 60 and 63), and the full list of sectors used is: 1, 2, 3, 4, 10, 15, 17, 18, 20.
There is only one other possible consecutive set of consecutive scores: (41, 42, 43, 44, 45, 46), but this leaves a 240 finish which is not possible with 3 darts (at least not using the standard rules of darts, but maybe there is a way to score at least 80 with one dart on this strange dartboard, or another rule that would allow Baz to win that we don’t know about). But it also turns out that with this set of scores we cannot be certain which sectors were definitely used to make the scores either, so this possibility is excluded without us needing to know the exact rules for the variation of darts being used in the puzzle.
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Frits 12:17 pm on 9 August 2020 Permalink |
It wasn’t easy for me to understand this puzzle. Only after seeing the code it became clear.
You don’t seem to check that after six three-dart throws Baz has scored at least 331 points.
With this extra check the 41-46 range can be eliminated earlier (ts[0] ≥ 52).
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Jim Randell 12:24 pm on 9 August 2020 Permalink |
@Frits: There are two possible sets of six consecutive scores, but one is eliminated by the requirement that we can be certain of the sectors that were definitely used to make the scores. It also turns out that it would not be possible to win with three darts from this position (at least on a normal dart board), and it is possible for the other set of six consecutive scores (in many ways, which can easily be seen), so I didn’t incorporate that test into my code, as it seemed to be superfluous (although probably useful in a manual solution).
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Frits 5:17 pm on 9 August 2020 Permalink |
Agree.
If the six totals had come out as f.i. 54-59 the score after six three-dart throws would have been 339 leaving 162 which is not a finish at darts.
If the program is intended to find a quick solution which has to be checked manually that’s OK with me.
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Jim Randell 8:30 am on 12 August 2020 Permalink |
Here is a modified version of my program that ensures that Baz can finish on his 7th throw [ @replit ].
Although as previously noted, this doesn’t change the solutions found.
The puzzle text only states that the remaining total should be reduced to 0, so that’s what I’ve implemented. No requirement for the final dart to be a double, and as no inner or outer bull is mentioned I have not included them either.
New Scientist Puzzle #06 explores scores achievable with n darts (on a standard dartboard).
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Robert Brown 1:07 am on 10 August 2020 Permalink |
That total doesn’t work. You need 58-63, which can be made 3 different ways. Total is 363 leaving 138 which can be finished with any 3 darts that sum to 46, all of which are trebles. It’s quite quick to manually check the 3 different sequences: there’s one sector that appears the same number of times in each sequence, which is the required answer.
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Jim Randell 8:55 am on 10 August 2020 Permalink |
@Robert: I’m not sure what total you are objecting to. Also I understood that in standard rules of darts you have to finish on a double. (Although in the version of the game used in the puzzle we are not told what the rules for finishing are. Fortunately though, of the two possible consecutive sequences of scores one of them is ruled out for reasons that are explained in the puzzle, so we don’t need to know the rules for finishing. We are just told that Baz could finish on his next throw, which is nice for him, but inconsequential for us).
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Robert Brown 8:22 pm on 10 August 2020 Permalink |
I don’t know the rules of darts! The total that I found (58,59,60,61,62,63) needs 138 to finish, so I assumed finishing on trebles was ok. But they tell me that a bull’s eye counts as 50, so you can finish with 2 bull’s eyes & a double 19.
I was objecting to Frits’s 6 totals (54-59) which I don’t see as a possibility. The important thing about my sequence is that there are 3 way to make 59, which changes the number of times that each sector is called, except for one which is the answer.
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Jim Randell 10:23 pm on 10 August 2020 Permalink |
@Robert: Right. I believe Frits only gave those numbers as a hypothetical example to illustrate his point.
You have found the right set of scores that leads to the answer. And fortunately (as I said above) you don’t need to know the rules of darts to eliminate the only other possible consecutive sequence of scores.
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GeoffR 8:40 am on 12 August 2020 Permalink |
I had the idea of defining a valid function for three adjacent sectors – i.e.valid if the six totals were less than the sum of the three angles. This enabled the code below to find the unique six consecutive totals, but not the single requested sector.
from itertools import permutations from collections import defaultdict D = defaultdict(set) L, L2 = [], [] # function to check if a three-sector is valid # a, b, c are adjacent sector numbers def is_valid(a, b, c): asum = sum(100 / x for x in (a, b, c)) ps = [] for p3 in permutations((a, b, c)): score = sum(m * x for m, x in zip((1, 2, 3), p3)) if score < asum: ps.append(score) return ps # the scores starting at north and running clockwise db = (20, 1, 18, 4, 13, 6, 10, 15, 2, 17, 3, 19, 7, 16, 8, 11, 14, 9, 12, 5) for p in range(20): p3 = tuple((p + i) % 20 for i in (0, 1, 2)) sc = tuple(db[i] for i in p3) L.append(sc) for x in range(20): a, b, c = L[x] s6 = is_valid(a, b, c) if s6: L2.append(s6) for p in permutations(L2,6): L1, L2, L3, L4, L5, L6 = p L22 = sorted(L1 + L2 + L3 + L4 + L5 + L6) L22 = sorted(list(set(L22))) for b in range(0, len(L22)-6): b1 = L22[b:b+6] # split slice into six totals a0, b0, c0, d0, e0, f0 = b1 # total of six scores s = a0 + b0 + c0 + d0 + e0 + f0 # total must be less than 180 to close with the 7th set of 3 darts # else there is another set of 6 consecutive numbers in the forties if 501 - s > 180: continue # check six numbers are consecutive if b0-a0 == c0-b0 == d0-c0 == e0-d0 == f0-e0 == 1: D[s].add ((a0,b0,c0,d0,e0,f0)) for k,v in D.items(): print(f"Sum of six 3 dart totals = {k}") print(f"Unique six totals are {v}") sc_left = 501 - k print(f"Seventh group of three darts score = {sc_left}") print(f"7th 3 darts group could be 3*18 + 3*20 + 2*12 = 138") print(f"Finishing on a double!") # Sum of six 3 dart totals = 363 # Unique six totals are {(58, 59, 60, 61, 62, 63)} # Seventh group of three darts score = 138 # 7th 3 darts group could be 3*18 + 3*20 + 2*12 = 138 # Finishing on a double!I looked at a print out of Jim’s dictionary which gives the same six consecutive totals and the triples used in each of the numbers 58,59,60,61, 62 and 63.
scores = (58, 59, 60, 61, 62, 63)
58: (3, 2, 17)
59: (20, 18, 1) (10, 2, 15) (2, 3, 17)
60: (4, 1, 18)
61: (18, 20, 1)
62: (2, 15, 10)
63: (1, 4, 18)
The number 59 has three triples and all these digits appear in the numbers 58,60,61,62 and 63. The total of 363 can therefore be made up three ways. All three ways require the numbers 60 and 63, which also contain the digit ‘4’, so we can say thet the digit ‘4’ definitely occurs twice.
A minor variation to my programme found the only eight valid sectors were:
(20, 1, 18), (1, 18, 4), (4, 13, 6), (10, 15, 2),
(15, 2, 17), (2, 17, 3), (3, 19, 7), (5, 20, 1)
Jim’s dictionary above show that only four of these eight triples were needed for the final solution;
i.e (2,3,17), (20,18,1), (10,2,15), (4,1,18)
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