Sunday Times Teaser 3182 – Stand and be Counted

by BRG on September 15, 2023

by Colin Vout

Published Sunday September 17 2023 (link)

When bad light ended the first day’s play, the total score was 182 with three batters having been “out” during the day.

In the successive stands, the lower-numbered and the higher-numbered batters scored in the ratios 7:3, 10:7, 1:11 and 11:19, in that order. The total score was the sum of the scores of the five batters.

The third batter was the highest scorer.

[In cricket the batting team starts with batters 1 and 2 together, and each is able to score. When one of them is “out” the third batter comes “in” to replace them, when one of that pair is out the fourth batter comes in, and so on. The amount by which the team score has increased while a particular pair is together is called the “stand”.]

What were the scores of all the batters, in their batting order?

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BRG permalink

from number_theory import frobenius_solve

# stand ratios, ratio sums and total runs

ratios = ((7, 3), (10, 7), (1, 11), (11, 19))

ratio_sums = list(x + y for x, y in ratios)

runs = 182

# generate sequences of batter pairings in stands

def pairs(lwr, hgr, nxt, end, prs=()):

prs += ((lwr, hgr),)

if nxt == end:

yield prs

else:

yield from pairs(lwr, nxt, nxt + 1, end, prs)

yield from pairs(hgr, nxt, nxt + 1, end, prs)

# find multiples of ratio sums that sum to the

# total runs made

for s in frobenius_solve(ratio_sums, runs, min_m=1):

# consider possible pairings in stands

for prs in pairs(1, 2, 3, 6):

# compile batter scores

scrs = [0] * 6

for p, m, r in zip(prs, s, ratios):

scrs[p[0]] += m * r[0]

scrs[p[1]] += m * r[1]

# does batter three have the maximum score?

if max(scrs[1:]) == scrs[3]:

print(f"scores {scrs[1:]}, stands {prs}.")

John Z permalink

from itertools import product

# 'stands' from teaser statement

st = ((7, 3), (10, 7), (1, 11), (11, 19))

# find multiples of batter pair score ratios that sum to 182

for m in product(range(1, 10), repeat = 4):

if sum(m[i] * sum(st[i]) for i in range(4)) == 182:

# generate all possible batter pair sequences

for i in product((0, 1), repeat = 3):

# initial pair

bt = [(1, 2)]

for j in range(3):

bt += [(bt[j][0] if i[j] else bt[j][1], j + 3)]

# scores for each batter

sc = [0 for i in range(6)]

for k in range(4):

sc[bt[k][0]] += m[k] * st[k][0]

sc[bt[k][1]] += m[k] * st[k][1]

# third batter has maximum score

if sc[3] == max(sc):

print('batter scores:', *sc[1:])

Frits permalink

As requested a pure Python version.

from itertools import product

ratios = ((7, 3), (10, 7), (1, 11), (11, 19))
ratio_sums = [x + y for x, y in ratios]
N, T = len(ratios), 182

# decompose: choose numbers from <ns> so that sum(chosen numbers) equals <t>
def decompose(t, k, ns, s=[]):
  if t == 0:  
    yield s
  else:
    for n in ns:
      if n <= t and (k == N or n >= s[-1]):
        yield from decompose(t - n, k + 1, ns, s + [n])

scores = []
# for first three plays generate which of the 2 players wasn't "out"
for nos in product([0, 1], repeat = N - 1):
  scs = [[0] * N for _ in range(N + 2)]
  plays = []
  # for all five players determine the score (without multiplier)
  for p in range(1, N + 1):
    # let new player p + 1 play with "not out" player
    plays.append((p if p == 1 else plays[-1][nos[p - 2]], p + 1))
    scs[plays[-1][0]][p - 1] = ratios[p - 1][0]
    scs[plays[-1][1]][p - 1] = ratios[p - 1][1]
    
  scores.append(scs[1:])

# find a selection of ratio sum numbers that sum to the total runs made
for ms in decompose(T - sum(ratio_sums), N, ratio_sums, ratio_sums):
  multipliers = [ms.count(r) for r in ratio_sums]
  for scs in scores:
    # multiply scores by multiplier to get final scores
    final =  [sum(x * y for x, y in zip(s, multipliers)) for s in scs]
    if max(final) == final[2]:
      print("answer:", final)

from itertools import product

ratios = ((7, 3), (10, 7), (1, 11), (11, 19))

ratio_sums = [x + y for x, y in ratios]

N, T = len(ratios), 182

# decompose: choose numbers from <ns> so that sum(chosen numbers) equals <t>

def decompose(t, k, ns, s=[]):

if t == 0:

yield s

else:

for n in ns:

if n <= t and (k == N or n >= s[-1]):

yield from decompose(t - n, k + 1, ns, s + [n])

scores = []

# for first three plays generate which of the 2 players wasn't "out"

for nos in product([0, 1], repeat = N - 1):

scs = [[0] * N for _ in range(N + 2)]

plays = []

# for all five players determine the score (without multiplier)

for p in range(1, N + 1):

# let new player p + 1 play with "not out" player

plays.append((p if p == 1 else plays[-1][nos[p - 2]], p + 1))

scs[plays[-1][0]][p - 1] = ratios[p - 1][0]

scs[plays[-1][1]][p - 1] = ratios[p - 1][1]

scores.append(scs[1:])

# find a selection of ratio sum numbers that sum to the total runs made

for ms in decompose(T - sum(ratio_sums), N, ratio_sums, ratio_sums):

multipliers = [ms.count(r) for r in ratio_sums]

for scs in scores:

# multiply scores by multiplier to get final scores

final = [sum(x * y for x, y in zip(s, multipliers)) for s in scs]

if max(final) == final[2]:

print("answer:", final)

Sunday Times Teaser 3182 – Stand and be Counted

by Colin Vout

Published Sunday September 17 2023 (link)

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