Sunday Times Teaser 3234 – An Existential Threat Thwarted

by BRG on 15 September 2024

by Michael Fletcher

Published Sunday September 15 2024 (link)

“Merlin, the Saxons are on the rampage. A group of 90 soldiers is advancing towards Mount Badon. We only have 86 knights that we can muster. I think we are doomed”, said the King.

“Not necessarily, Sire. The strength of an army is proportional to the square of the number of its troops. An army of 100 would defeat an army of 80 yet still have 60 remaining because 100² – 80² = 60². Tell the knights to split into three groups of particular sizes and engage the Saxons in three simultaneous battles. If the Saxons split into three groups of 30, we can prevail.”

After the conflict there were 32 knights and 24 Saxons remaining.

What were the sizes of the three groups the knights split into?

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BRG permalink

Updated with suggestions from Jim Randell and Frits.

# if <r>^2 = <x>^2 - <y?^2 is a perfect square return <r> else None

d = {n * n: n for n in range(1, 50)}

def is_sqr(x, y):

return d.get(x * x - y * y, None)

# Assume that the Knights are split into groups of size a, b and c

# with a <= b <= c with an assumption that no battles end in draws.

# the Knights must lose at least one battle with

# a whole number of Saxon's remaining

for a in range(1, 30):

if not (s1 := is_sqr(30, a)):

continue

# the Saxons must also lose at least one battle

# with a whole number of Knights remaining

for c in range(31, 86 - a):

if not (k1 := is_sqr(c, 30)):

continue

# the third battle can go either way

b = 86 - (a + c)

k2 = s2 = 0

if b < 30 and (s2 := is_sqr(30, b)):

pass

elif b > 30 and (k2 := is_sqr(b, 30)):

pass

else:

continue

if k1 + k2 == 32 and s1 + s2 == 24:

f, s, l = f"{a}, {b}, {c}".rpartition(', ')

print(f"The Knight's group sizes are {f} and {l}.")

Jim permalink

Surely there is only one possible value for b = 86 − (a + c).

Reply
- BRG permalink
  
  Yes, a silly mistake.
  
  Reply

Frits permalink

sqs = {i * i for i in range(1, 85)}

# dictionary of remaining squares per group size

d = {i: r2 for i in range(1, 85) if (r2 := abs(900 - i * i)) in sqs}

# choose the three groups the knights split into

for g1 in d.keys():

for g2 in d.keys():

if g2 < g1: continue

if (g3 := 86 - (g1 + g2)) not in d or g3 < g2: continue

all3 = (g1, g2, g3)

# knights left

kl = sum(int(d[x]**.5) if x > 30 else 0 for x in all3)

if kl != 32: continue

# Saxons left

sl = sum(int(d[x]**.5) if x < 30 else 0 for x in all3)

if sl != 24: continue

print("answer:", all3)

John Z. permalink

Short and inefficient…

from itertools import product

for i3 in product(range(1,85), repeat = 3):

k3, s3 = 0, 0 # remaining soldiers in each tribe

if sum(i3) == 86 and i3[0] >= i3[1] >= i3[2]:

for k in i3:

x = abs((k ** 2 - 900)) ** 0.5

if x == int(x):

if k >= 30:

k3 += int(x)

else:

s3 += int(x)

if k3 == 32 and s3 == 24: #32 Knights and 24 Saxons remain

print(i3)

Sunday Times Teaser 3234 – An Existential Threat Thwarted

by Michael Fletcher

Published Sunday September 15 2024 (link)

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