Sunday Times Teaser 3238 – Any Colour, As Long As It’s Not Black

by BRG on 13 October 2024

by Stephen Hogg

Published Sunday October 13 2024 (link)

Henry learnt snooker scoring basics: pot a red (=1pt), not counting as a colour, then a colour (yellow=2pt, green=3pt, brown=4pt, blue=5pt, pink=6pt or black=7pt), alternately. Potted reds stay out of play, while a colour potted immediately after a red returns to play. Once the last red and accompanying colour are potted, then pot the colours in the aforementioned sequence. A miss or foul lets your opponent play. With fifteen reds available, Henry scored the first points (a prime number) as described, then missed a red with a foul, yielding four points to Ford. Only with these four points could Ford possibly win. Potting all the remaining reds, each followed by colours other than black, then the colours to pink, Ford needed the black to win by one point. He missed with a foul, which ended the game.

Give Ford’s score.

From → Uncategorized

2 Comments Leave one →

BRG permalink


from itertools import combinations_with_replacement as cwr

pr = {3, 5, 7, 11}
pr = {2} | {n for n in range(13, 149, 2) if all(n % p for p in pr)}

sols = set()
  
# the number of reds potted by Henry
for h_reds in range(1, 15):
  f_reds = 15 - h_reds 
  # consider the colours potted by Henry
  for h_clrs in cwr(range(2, 8), h_reds):
    # Henry's score is prime
    if (h_pts := h_reds + sum(h_clrs)) not in pr:
      continue
    
    # even if Ford scores with all blacks he cannot 
    # win but, with four more points, he can
    f_pmax = 8 * f_reds + sum(range(2, 8))
    if f_pmax <= h_pts < f_pmax + 4:
  
      # pick the colours potted by Ford (no blacks) ...
      for f_clrs in cwr(range(2, 7), f_reds):
        # ... who (with a four point foul) wins by one point
        # if he then pots all the colours except the black
        f_pts = f_reds + sum(f_clrs) + sum(range(2, 7)) + 4
        # he now needs a black to win by one point
        if h_pts + 1 == f_pts + 7:
          print(f"{f_pts = }, {f_clrs = }, {h_pts = }, {h_clrs = }")
          sols.add((f_pts, h_pts))

for f, h in sols:
  print(f"\nFord {f} points (Henry {h} points).")

from itertools import combinations_with_replacement as cwr

pr = {3, 5, 7, 11}

pr = {2} | {n for n in range(13, 149, 2) if all(n % p for p in pr)}

sols = set()

# the number of reds potted by Henry

for h_reds in range(1, 15):

f_reds = 15 - h_reds

# consider the colours potted by Henry

for h_clrs in cwr(range(2, 8), h_reds):

# Henry's score is prime

if (h_pts := h_reds + sum(h_clrs)) not in pr:

continue

# even if Ford scores with all blacks he cannot

# win but, with four more points, he can

f_pmax = 8 * f_reds + sum(range(2, 8))

if f_pmax <= h_pts < f_pmax + 4:

# pick the colours potted by Ford (no blacks) ...

for f_clrs in cwr(range(2, 7), f_reds):

# ... who (with a four point foul) wins by one point

# if he then pots all the colours except the black

f_pts = f_reds + sum(f_clrs) + sum(range(2, 7)) + 4

# he now needs a black to win by one point

if h_pts + 1 == f_pts + 7:

print(f"{f_pts = }, {f_clrs = }, {h_pts = }, {h_clrs = }")

sols.add((f_pts, h_pts))

for f, h in sols:

print(f"\nFord {f} points (Henry {h} points).")

Frits permalink

Maybe a bit complex but I am not a fan of simple brute force (so including some analysis).


from number_theory import is_prime

cols = [2, 3, 4, 5, 6, 7]
tcols = sum(cols)
(lo, hi) = (1 + cols[0], 1 + cols[-1])  # red + yellow, red + black
  
sols = set()

# Ford may drop 3 points at most with his first <r> colours otherwise 
# with the four points of the foul he can't win by one point anymore

# number of red balls for Ford
for rF in range(1, 4):
  # number of red balls for Henry
  rH = 15 - rF
  # Henry must have made at least rH * 3 points (red + yellow)
  if hi * rF + tcols < rH * lo: continue
  # possible scores for Henry
  Hs = [h for h in range(hi * rF + tcols, hi * rF + tcols + 4) 
        if is_prime(h)]
  # Ford must be able to drop at least 1 point per red/colour combination
  Fs = [str(f) for h in Hs 
        if (f := h + 1 - cols[-1]) - (tcols - cols[-1]) - 4 <= (hi - 1) * rF]
  sols.update(Fs)    

print("answer:", ' or '.join(sols))

from number_theory import is_prime

cols = [2, 3, 4, 5, 6, 7]

tcols = sum(cols)

(lo, hi) = (1 + cols[0], 1 + cols[-1]) # red + yellow, red + black

sols = set()

# Ford may drop 3 points at most with his first <r> colours otherwise

# with the four points of the foul he can't win by one point anymore

# number of red balls for Ford

for rF in range(1, 4):

# number of red balls for Henry

rH = 15 - rF

# Henry must have made at least rH * 3 points (red + yellow)

if hi * rF + tcols < rH * lo: continue

# possible scores for Henry

Hs = [h for h in range(hi * rF + tcols, hi * rF + tcols + 4)

if is_prime(h)]

# Ford must be able to drop at least 1 point per red/colour combination

Fs = [str(f) for h in Hs

if (f := h + 1 - cols[-1]) - (tcols - cols[-1]) - 4 <= (hi - 1) * rF]

sols.update(Fs)

print("answer:", ' or '.join(sols))

Sunday Times Teaser 3238 – Any Colour, As Long As It’s Not Black

by Stephen Hogg

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta