#! /usr/bin/env python3
#
def mcnuggets_test ( ):
#*****************************************************************************80
#
## mcnuggets_test() tests mcnuggets().
#
# Licensing:
#
# This code is distributed under the MIT license.
#
# Modified:
#
# 22 September 2022
#
# Author:
#
# John Burkardt
#
import platform
print ( '' )
print ( 'mcnuggets_test():' )
print ( ' Python version: ' + platform.python_version ( ) )
print ( ' Test mcnuggets()' )
mcnugget_solvable_test ( )
mcnugget_ways_test ( )
#
# Terminate.
#
print ( '' )
print ( 'mcnuggets_test():' )
print ( ' Normal end of execution.' )
return
def mcnugget_number_values ( n_data ):
#*****************************************************************************80
#
## mcnugget_number_values() returns McNugget numbers.
#
# Discussion:
#
# A restaurant offers Chicken McNuggets, but only in packages of 6, 9 or 20.
#
# A customer wishes to buy exactly N McNuggets.
#
# Presumably, there are M distinct ways to do this. M is known as the
# McNugget number of N.
#
# Licensing:
#
# This code is distributed under the MIT license.
#
# Modified:
#
# 22 September 2022
#
# Author:
#
# John Burkardt
#
# Reference:
#
# Scott Chapman, Chris O’Neill,
# Factoring in the Chicken McNugget Monoid,
# Mathematics Magazine,
# Volume 91, Number 5, 2018, pages 323-336.
#
# Input:
#
# integer N_DATA. The user sets N_DATA to 0 before the first call.
# Thereafter, it should simply be the value returned by the previous call.
#
# Output:
#
# integer N_DATA. On each call, the routine increments N_DATA by 1, and
# returns the corresponding data when there is no more data, the
# output value of N_DATA will be 0 again.
#
# integer N: the number of McNuggets desired.
#
# integer M: the number of different ways of getting N McNuggets.
#
import numpy as np
n_max = 101
n_vec = np.array ( [ \
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, \
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, \
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, \
30, 31, 32, 33, 34, 35, 36, 37, 38, 39, \
40, 41, 42, 43, 44, 45, 46, 47, 48, 49, \
50, 51, 52, 53, 54, 55, 56, 57, 58, 59, \
60, 61, 62, 63, 64, 65, 66, 67, 68, 69, \
70, 71, 72, 73, 74, 75, 76, 77, 78, 79, \
80, 81, 82, 83, 84, 85, 86, 87, 88, 89, \
90, 91, 92, 93, 94, 95, 96, 97, 98, 99, \
100 \
] )
m_vec = np.array ( [ \
1, 0, 0, 0, 0, 0, 1, 0, 0, 1, \
0, 0, 1, 0, 0, 1, 0, 0, 2, 0, \
1, 1, 0, 0, 2, 0, 1, 2, 0, 1, \
2, 0, 1, 2, 0, 1, 3, 0, 2, 2, \
1, 1, 3, 0, 2, 3, 1, 2, 3, 1, \
2, 3, 1, 2, 4, 1, 3, 3, 2, 2, \
5, 1, 3, 4, 2, 3, 5, 2, 3, 5, \
2, 3, 6, 2, 4, 5, 3, 3, 7, 2, \
5, 6, 3, 4, 7, 3, 5, 7, 3, 5, \
8, 3, 6, 7, 4, 5, 9, 3, 7, 8, \
5 \
] )
if ( n_data < 0 ):
n_data = 0
if ( n_max <= n_data ):
n_data = 0
n = 0
m = 0
else:
n = n_vec[n_data]
m = m_vec[n_data]
n_data = n_data + 1
return n_data, n, m
def mcnugget_solvable ( a, b ):
#*****************************************************************************80
#
## mcnugget_solvable() determines whether a McNuggets problem is solvable.
#
# Discussion:
#
# We are given a Diophantine equation
#
# a1 x1 + a2 x2 + ... + an * xn = b
#
# for which the coefficients a are positive integers, and
# the right hand side b is a nonnegative integer.
#
# We want to know whether there are nonnegative integers x which
# satisfy this equation. If so, we say the value b is "solvable".
#
# In particular, we might have a = [ 6, 9, 20 ] and b would be
# the desired number of Chicken McNuggets.
#
# Licensing:
#
# This code is distributed under the MIT license.
#
# Modified:
#
# 12 October 2022
#
# Author:
#
# John Burkardt
#
# Reference:
#
# Scott Chapman, Chris O’Neill,
# Factoring in the Chicken McNugget Monoid,
# Mathematics Magazine,
# Volume 91, Number 5, 2018, pages 323-336.
#
# Input:
#
# integer a(n): the number of McNuggets in each package size.
#
# integer b: the total number of McNuggets desired.
#
# Output:
#
# boolean solvable: true if there is a solution.
#
a_num = len ( a )
solvable = False
#
# Check whether b is immediately solvable.
#
if ( b == 0 ) :
solvable = True
return solvable
for i in range ( 0, a_num ):
if ( ( b % a[i] ) == 0 ):
solvable = True
return solvable
#
# Initially, the set "item" contains "b".
#
item = { b }
item_num = 1
#
# Loop on largest "descendant" of b.
#
while ( 0 < item_num ):
# n = item[-1]
n = max ( item )
item.remove ( n )
item_num = item_num - 1
for i in range ( 0, a_num ):
n_minus_a = n - a[i]
if ( n_minus_a == 0 ):
solvable = True
return solvable
if ( 0 <= n_minus_a ):
for j in range ( 0, a_num ):
if ( ( n_minus_a % a[j] ) == 0 ):
solvable = True
return solvable
#
# Add n_minus_a to set.
# Since it might already be a member, we use len() to update the set size.
#
item.add ( n_minus_a )
item_num = len ( item )
return solvable
def mcnugget_solvable_test ( ):
#*****************************************************************************80
#
## mcnugget_solvable_test() tests mcnugget_solvable().
#
# Licensing:
#
# This code is distributed under the MIT license.
#
# Modified:
#
# 12 October 2022
#
# Author:
#
# John Burkardt
#
print ( '' )
print ( 'mcnugget_solvable_test():' )
print ( ' mcnugget_solvable() reports whether you can order exactly' )
print ( ' N chicken McNuggets when they only come in packages' )
print ( ' of 6, 9, and 20.' )
print ( '' )
print ( ' N Solvable(N)' )
print ( '' )
a = [ 6, 9, 20 ]
for b in range ( 0, 101 ):
solvable = mcnugget_solvable ( a, b )
if ( solvable ):
print ( ' %3d True' % ( b ) )
else:
print ( ' %3d False' % ( b ) )
return
def mcnugget_ways ( a, nmax ):
#*****************************************************************************80
#
## mcnugget_ways() determines the number of solutions to a McNugget problem.
#
# Discussion:
#
# We are given a Diophantine equation
#
# a1 x1 + a2 x2 + ... + an * xn = n
#
# for which the coefficients a are positive integers, and
# the right hand side n is a nonnegative integer.
#
# In particular, we might have a = [ 6, 9, 20 ] and n would be
# the desired number of Chicken McNuggets.
#
# We want to know the number of distinct ways of ordering the
# desired number of McNuggets.
#
# Here, we assume that none of the values in a can be written
# as a linear combination of the other a values.
#
# We have to be careful so that we only count a given arrangement
# once, that is, the order in which we combine the values does not
# matter. Although 6+6+9 and 6+9+6 and 6+6+9 look "different",
# they should only count here as a single solution to the problem
# of ordering 21 McNuggets.
#
# Licensing:
#
# This code is distributed under the MIT license.
#
# Modified:
#
# 24 December 2022
#
# Author:
#
# John Burkardt
#
# Reference:
#
# Scott Chapman, Chris O’Neill,
# Factoring in the Chicken McNugget Monoid,
# Mathematics Magazine,
# Volume 91, Number 5, 2018, pages 323-336.
#
# Input:
#
# integer a(n): the number of McNuggets in each package size.
#
# integer nmax: the maximum number of McNuggets desired.
#
# Output:
#
# integer ways(nmax+1): the number of distinct ways of ordering 0, 1, 2, ...,
# nmax-1, nmax Chicken McNuggets.
#
import numpy as np
a_num = len ( a )
a = np.sort ( a )
#
# Count ways from 0 to nmax.
#
ways = np.zeros ( nmax + 1 )
#
# ways(i) = 0 + sum ( 1 <= j <= a_num ) ways(i-a(j)) if i-a(i) > 0
#
ways[0] = 1
for j in range ( 0, a_num ):
for i in range ( 1, nmax + 1 ):
pop = i - a[j]
if ( 0 <= pop ):
ways[i] = ways[i] + ways[pop]
return ways
def mcnugget_ways_test ( ):
#*****************************************************************************80
#
## mcnugget_ways_test() tests mcnugget_ways().
#
# Licensing:
#
# This code is distributed under the MIT license.
#
# Modified:
#
# 24 December 2022
#
# Author:
#
# John Burkardt
#
import numpy as np
print ( '' )
print ( 'mcnugget_ways_test():' )
print ( ' mcnugget_ways() reports the number of ways you can' )
print ( ' order exactly N chicken McNuggets when they only come in' )
print ( ' packages of 6, 9, and 20.' )
print ( '' )
print ( ' N Ways(N)' )
print ( '' )
a = np.array ( [ 6, 9, 20 ] )
nmax = 50
ways = mcnugget_ways ( a, nmax )
ways_exact = np.array ( [ \
1, 0, 0, 0, 0, 0, 1, 0, 0, 1, \
0, 0, 1, 0, 0, 1, 0, 0, 2, 0, \
1, 1, 0, 0, 2, 0, 1, 2, 0, 1, \
2, 0, 1, 2, 0, 1, 3, 0, 2, 2, \
1, 1, 3, 0, 2, 3, 1, 2, 3, 1, \
2, 3, 1, 2, 4, 1, 3, 3, 2, 2, \
5, 1, 3, 4, 2, 3, 5, 2, 3, 5, \
2, 3, 6, 2, 4, 5, 3, 3, 7, 2, \
5, 6, 3, 4, 7, 3, 5, 7, 3, 5, \
8, 3, 6, 7, 4, 5, 9, 3, 7, 8, \
5 \
] )
for n in range ( 0, nmax + 1 ):
print ( ' %2d %3d %3d' % ( n, ways[n], ways_exact[n] ) )
return
def timestamp ( ):
#*****************************************************************************80
#
## timestamp() prints the date as a timestamp.
#
# Licensing:
#
# This code is distributed under the MIT license.
#
# Modified:
#
# 21 August 2019
#
# Author:
#
# John Burkardt
#
import time
t = time.time ( )
print ( time.ctime ( t ) )
return
if ( __name__ == "__main__" ):
timestamp ( )
mcnuggets_test ( )
timestamp ( )