Sunday Times Teaser 3009 – Head Count

by Victor Bryant

Published Sunday May 24 2020 (link)

My grandson and I play a simple coin game. In the first round we toss seven coins and I predict how many “heads” there will be whilst my grandson predicts the number of “tails”. After the tossing I score a point for each head plus a bonus of ten if my prediction was correct — my grandson scores likewise for the tails. We then repeat this with six coins, then five, and so on down to a single coin. After each round we keep cumulative totals of our scores. In one game, for over half of the pairs of cumulative scores, my grandson’s total was like mine but with the digits in reverse order. In fact he was ahead throughout and at one stage his cumulative total was double mine — and he had an even bigger numerical lead than that on just one earlier occasion and one later occasion.

List the number of heads in each of the seven rounds.

3 Comments Leave one →

Brian Gladman permalink


from itertools import product
from functools import lru_cache

@lru_cache
def are_reverse(x, y):
  return str(x) == str(y)[::-1]

# play the game: 
#   <n>   the number of rounds remaining 
#   <vs>  Victor's comulative score
#   <gs>  his grandson's cumulative score
#   <cs>  index in the returned sequence 
#         where Victor's cummulative score 
#         is half that of his grandson
#   <rc>  the 'digit reversed' scores count
def play(n, vs=0, gs=0, cs=0, rc=0, hvg=()):
  
  # in the last four rounds there must be at least 1, 2, 
  # 3 and 4 two digit 'digit reversed' cumulative scores
  if n < 4 and rc < 4 - n:
    return

  # return a solution if we have we seen a round in which
  # Victor's cumulative score half his grandson's?
  if not n:
    if cs:
      yield cs, hvg
  
  else:
    # try all combinations of heads and tails with Victor's 
    # and his grandson's winning bonuses 
    for h, v, g in product(range(n + 1), (0, 10), (0, 10)):
      # the cumulative scores
      vss = vs + h + v
      gss = gs + n - h + g
      
      # Victor is behind his grandson in all rounds
      if gss > vss:
        if gss == 2 * vss:
          # there is only one round where the grandson's score 
          # is twice Victor's score - skip if seen earlier
          if cs:
            continue
          # record the position of this round
          t = len(hvg)
        else:
          t = cs
        # update the 'digit reversed' scores count 
        rcc = rc + are_reverse(gss, vss)
        yield from play(n - 1, vss, gss, t, rcc, hvg + ((h, vss, gss),))

for eqi, hvg in play(7):
  hs, vs, gs = zip(*hvg)

  # find the numerical lead in the round in which Victor's
  # cumulative score is half that of his grandson
  t = hvg[eqi]
  nl = t[2] - t[1]
  
  # check that there was exacty one larger lead prior to this
  cnt_gnl = 0
  for i, (v, g) in enumerate(zip(vs, gs)):
    cnt_gnl += (g - v) > nl
    if i == eqi and cnt_gnl != 1:
      break
  else:
    # ... and exactly one after
    if cnt_gnl == 2:
      vgs = tuple(zip(vs, gs))
      print(f"Heads {tuple(hs)} Scores [{vgs}]")

from itertools import product

from functools import lru_cache

@lru_cache

def are_reverse(x, y):

return str(x) == str(y)[::-1]

# play the game:

# <n> the number of rounds remaining

# <vs> Victor's comulative score

# <gs> his grandson's cumulative score

# <cs> index in the returned sequence

# where Victor's cummulative score

# is half that of his grandson

# <rc> the 'digit reversed' scores count

def play(n, vs=0, gs=0, cs=0, rc=0, hvg=()):

# in the last four rounds there must be at least 1, 2,

# 3 and 4 two digit 'digit reversed' cumulative scores

if n < 4 and rc < 4 - n:

return

# return a solution if we have we seen a round in which

# Victor's cumulative score half his grandson's?

if not n:

if cs:

yield cs, hvg

else:

# try all combinations of heads and tails with Victor's

# and his grandson's winning bonuses

for h, v, g in product(range(n + 1), (0, 10), (0, 10)):

# the cumulative scores

vss = vs + h + v

gss = gs + n - h + g

# Victor is behind his grandson in all rounds

if gss > vss:

if gss == 2 * vss:

# there is only one round where the grandson's score

# is twice Victor's score - skip if seen earlier

if cs:

continue

# record the position of this round

t = len(hvg)

else:

t = cs

# update the 'digit reversed' scores count

rcc = rc + are_reverse(gss, vss)

yield from play(n - 1, vss, gss, t, rcc, hvg + ((h, vss, gss),))

for eqi, hvg in play(7):

hs, vs, gs = zip(*hvg)

# find the numerical lead in the round in which Victor's

# cumulative score is half that of his grandson

t = hvg[eqi]

nl = t[2] - t[1]

# check that there was exacty one larger lead prior to this

cnt_gnl = 0

for i, (v, g) in enumerate(zip(vs, gs)):

cnt_gnl += (g - v) > nl

if i == eqi and cnt_gnl != 1:

break

else:

# ... and exactly one after

if cnt_gnl == 2:

vgs = tuple(zip(vs, gs))

print(f"Heads {tuple(hs)} Scores [{vgs}]")

Robert Brown permalink

Simple manual solution:
For each section(7,6,5 etc) there’s an ongoing total of the number of heads & tails. The last digit is different for each section, unaffected by any bonuses.
We need at least 3 sections that end with a reversible pair of numbers. Only possible if the sum of that pair has a last digit that matches the above. This only works for 3 sections, the others are easily filled in following the teaser rules.- total number of bonuses can only stay the same or go up.

Reply
- Robert Brown permalink
  
  Error in above! We need at least 4 sections that end in a reversible pair. It only works for 4 sections. Mea Culpa
  
  Reply

Sunday Times Teaser 3009 – Head Count

by Victor Bryant

Published Sunday May 24 2020 (link)

Leave a comment to Brian Gladman Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta