Brainteaser 1675: The A-to-Z of sport
From The Sunday Times, 16th October 1994 [link]
The individual sport or sports of the 26 members of the Venerable Sports Club means that they are just enough to form:
a football team (11); or
a hockey team (11); or
a rugby team (15).The number of members playing only two of these sports is the same as the number of members playing rugby only, which is less than one third of the total membership.
How many members play all three sports?
This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.
[teaser1675]
John Crabtree and 
Robert Brown are discussing. 
Pete Good, 
GeoffR are discussing. 
S2T2 
Jim Randell 9:46 am on 5 September 2019 Permalink |
We can label the 7 non-empty parts of the Venn diagram, F, H, R, FH, FR, HR, FHR, according to which sports each member plays.
We can then express the constraints in terms of these 7 integer variables.
I expressed the constraints as a MiniZinc model, and then used the minizinc.py wrapper to collect the solutions and count them using Python.
This program runs in 263ms.
from minizinc import MiniZinc, var from enigma import multiset, printf # decision variables, the 7 non-empty areas of the venn diagram vs = "F H R FH FR HR FHR" # the MiniZinc model model = f""" % declare the variables {var('0..26', vs.split())}; % there are 26 members in total constraint F + H + R + FH + FR + HR + FHR = 26; % F's form a football team (11) constraint F + FH + FR + FHR = 11; % H's form a hockey team (11) constraint H + FH + HR + FHR = 11; % R's form a rugby team (15) constraint R + FR + HR + FHR = 15; % total number of two sports = number of only rugby constraint FH + FR + HR = R; % R is less than 1/3 of the total membership (= 26) constraint 3 * R < 26; solve satisfy; """ # create the model m = MiniZinc(model) # record FHR values r = multiset() # solve it for (F, H, R, FH, FR, HR, FHR) in m.solve(result=vs): printf("[F={F} H={H} R={R} FH={FH} FR={FR} HR={HR} FHR={FHR}]") r.add(FHR) # output solutions for (k, v) in r.most_common(): printf("FHR = {k} [{v} solutions]")Solution: 2 members play all three sports.
There are 7 possible solutions:
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Jim Randell 12:32 pm on 5 September 2019 Permalink |
And here’s a solution in Python. It runs in 93ms.
Run: [ @repl.it ]
from enigma import (irange, multiset, printf) # decompose t into k numbers (min value m) def decompose(t, k, m=0, s=[]): if k == 1: yield s + [t] else: for x in irange(m, t - k * m): yield from decompose(t - x, k - 1, m, s + [x]) # record FHR values r = multiset() # can form a football team of 11 for (F, FH, FR, FHR) in decompose(11, 4): # can form a hockey team of 11 for (H, HR) in decompose(11 - FH - FHR, 2): # number of rugby only = total number of two sports R = FH + FR + HR # can form a rugby team of 15 if not (R + FR + HR + FHR == 15): continue # R is less than 1/3 of the total membership (26) if not (3 * R < 26): continue # total membership is 26 if not (F + H + R + FH + FR + HR + FHR == 26): continue printf("[F={F} H={H} R={R} FH={FH} FR={FR} HR={HR} FHR={FHR}]") r.add(FHR) # output solutions for (k, v) in r.most_common(): printf("FHR = {k} [{v} solutions]")LikeLike
John Crabtree 12:41 am on 6 September 2019 Permalink |
Let R members play rugby only and let A members play all three sports.
Considering that there are 26 members and 37 players leads to 2A + R = 11,
R ≤ 7 and so A ≥ 2.
15 players play rugby, ie A + 2R ≥ 15, which leads to 3A ≤ 7.
And so 2 members play all three sports.
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