| k1masterm.mos |
(!******************************************************
Mosel Example Problems
======================
file k1mastm.mos
````````````````
Playing Mastermind
Given information:
Round Pos.1 Pos.2 Pos.3 Pos.4 Judgement
1 red yellow white green 1 color well positioned,
1 color on wrong position
2 blue green white blue 1 color well positioned
3 yellow black white black 1 color well positioned,
1 color on wrong position
4 blue yellow red black all colors on wrong position
(c) 2008 Fair Isaac Corporation
author: S. Heipcke, Mar. 2002
*******************************************************!)
model "K-1 Mastermind"
uses "mmxprs"
declarations
POS = 1..4 ! Positions in a row
COLORS = {"White","Blue","Yellow","Black","Red","Green"}
place: array(POS,COLORS) of mpvar ! 1 if color at the position, 0 otherwise
end-declarations
! Assign one color per position
forall(p in POS) sum(c in COLORS) place(p,c) = 1
! Round 1: 1 color well positioned, 1 color on wrong position
place(1,"Red") + place(2,"Yellow") + place(3,"White") + place(4,"Green") = 1
sum(p in POS | p<>1) place(p,"Red") +
sum(p in POS | p<>2) place(p,"Yellow") +
sum(p in POS | p<>3) place(p,"White") +
sum(p in POS | p<>4) place(p,"Green") = 1
! Round 2: 1 color well positioned
place(1,"Blue") + place(2,"Green") + place(3,"White") + place(4,"Blue") = 1
sum(p in POS | p<>1 and p<>4) place(p,"Blue") +
sum(p in POS | p<>2) place(p,"Green") +
sum(p in POS | p<>3) place(p,"White") = 0
! Round 3: 1 color well positioned, 1 color on wrong position
place(1,"Yellow") + place(2,"Black") + place(3,"White") + place(4,"Black") = 1
sum(p in POS | p<>1) place(p,"Yellow") +
sum(p in POS | p<>2 and p<>4) place(p,"Black") +
sum(p in POS | p<>3) place(p,"White") = 1
! Round 4: 4 colors on wrong position
place(1,"Blue") + place(2,"Yellow") + place(3,"Red") + place(4,"Black") = 0
sum(p in POS | p<>1) place(p,"Blue") +
sum(p in POS | p<>2) place(p,"Yellow") +
sum(p in POS | p<>3) place(p,"Red") +
sum(p in POS | p<>4) place(p,"Black") = 4
forall(p in POS, c in COLORS) place(p,c) is_binary
! Solve the problem: no objective
minimize(0)
! Solution printing
forall(c in COLORS) writeln(c,": ", getsol(sum(p in POS)p*place(p,c)) )
end-model
|
|
| k2marath.mos |
(!******************************************************
Mosel Example Problems
======================
file k2marath.mos
````````````````
Marathon puzzle
Dominique, Ignace, Naren, Olivier, Philippe, and Pascal
have arrived as the first six at the Paris marathon.
Reconstruct their arrival order from the following
information:
a) Olivier has not arrived last
b) Dominique, Pascal and Ignace have arrived before Naren
and Olivier
c) Dominique who was third last year has improved this year.
d) Philippe is among the first four.
e) Ignace has arrived neither in second nor third position.
f) Pascal has beaten Naren by three positions.
g) Neither Ignace nor Dominique are on the fourth position.
(c) 2008 Fair Isaac Corporation
author: S. Heipcke, Mar. 2002
*******************************************************!)
model "K-2 Marathon"
uses "mmxprs"
declarations
POS = 1..6 ! Arrival positions
RUNNERS = {"Dominique","Ignace","Naren","Olivier","Philippe","Pascal"}
arrive: array(RUNNERS,POS) of mpvar ! 1 if runner is p-th, 0 otherwise
end-declarations
! One runner per position
forall(p in POS) sum(r in RUNNERS) arrive(r,p) = 1
! One position per runner
forall(r in RUNNERS) sum(p in POS) arrive(r,p) = 1
! a: Olivier not last
arrive("Olivier",6) = 0
! b: Dominique, Pascal and Ignace before Naren and Olivier
sum(p in 5..6) (arrive("Dominique",p)+arrive("Pascal",p)+arrive("Ignace",p)) =
0
sum(p in 1..3) (arrive("Naren",p)+arrive("Olivier",p)) = 0
! c: Dominique better than third
arrive("Dominique",1)+arrive("Dominique",2) = 1
! d: Philippe is among the first four
sum(p in 1..4) arrive("Philippe",p) = 1
! e: Ignace neither second nor third
arrive("Ignace",2)+arrive("Ignace",3) = 0
! f: Pascal three places earlier than Naren
sum(p in 4..6) arrive("Pascal",p) = 0
sum(p in 1..3) arrive("Naren",p) = 0
! g: Neither Ignace nor Dominique on fourth position
arrive("Ignace",4)+arrive("Dominique",4) = 0
forall(p in POS, r in RUNNERS) arrive(r,p) is_binary
! Solve the problem: no objective
minimize(0)
! Solution printing
forall(r in RUNNERS) writeln(r,": ", getsol(sum(p in POS)p*arrive(r,p)) )
end-model
|
|
| k3congress.mos |
(!******************************************************
Mosel Example Problems
======================
file k3congress.mos
```````````````````
Congress puzzle
At an international congress taking place on 7-11 August
five researchers of different nationality are giving talks,
each on a different day. Find out the date and nationality
for every researcher from the following information:
a) Michael is not Japanese.
b) Eric is French and he talks before the 10th.
c) Arabinda talks on the 9th.
d) The Chinese who is not Hitoshi gives his talk on the 8th,
before Michael.
e) Hitoshi does his talk after the Indian and before the American.
(c) 2008 Fair Isaac Corporation
author: S. Heipcke, Mar. 2002
*******************************************************!)
model "K-3 Congress"
uses "mmxprs"
declarations
DAYS = 7..11
NAMES = {"Arabinda","Eric","Hitoshi","Michael","Zhicheng"}
NAT = {"Japanese","French","Chinese","American","Indian"}
talk: array(NAMES,NAT,DAYS) of mpvar ! (p,n,d) 1 if person p of nationality n
! gives a talk on day d, 0 otherwise
end-declarations
! One researcher per day and nationality
forall(p in NAMES) sum(n in NAT, d in DAYS) talk(p,n,d) = 1
! Every nationality exactly once
forall(n in NAT) sum(p in NAMES, d in DAYS) talk(p,n,d) = 1
! One talk per day
forall(d in DAYS) sum(p in NAMES, n in NAT) talk(p,n,d) = 1
! a: Michael is not Japanese
sum(d in DAYS) talk("Michael","Japanese",d) = 0
! b: Eric is French and he talks before the 10th
sum(d in 7..9) talk("Eric","French",d) = 1
! c: Arabinda talks on the 9th
sum(n in NAT) talk("Arabinda",n,9) = 1
! d: The Chinese who is not Hitoshi gives his talk on the 8th, before Michael
sum(d in DAYS) (talk("Hitoshi","Chinese",d)+talk("Michael","Chinese",d)) = 0
sum(p in NAMES) talk(p,"Chinese",8) = 1
sum(n in NAT, d in 9..11) talk("Michael",n,d) = 1
! e: Hitoshi does his talk after the Indian and before the American
sum(n in NAT) (talk("Hitoshi",n,7)+talk("Hitoshi",n,11)) = 0
sum(p in NAMES, d in 10..11) talk(p,"Indian",d) = 0
sum(p in NAMES, d in 7..8) talk(p,"American",d) = 0
sum(d in DAYS) (talk("Hitoshi","Indian",d)+talk("Hitoshi","American",d)) = 0
forall(p in NAMES, n in NAT, d in DAYS) talk(p,n,d) is_binary
! Solve the problem: no objective
minimize(0)
! Solution printing
forall(p in NAMES) do
write(p," (")
dp:=round(getsol(sum(n in NAT, d in DAYS)d*talk(p,n,d)))
forall(n in NAT) write( if(getsol(talk(p,n,dp))>0, n, "") )
writeln(") : ", dp)
end-do
end-model
|
|
| k4even.mos |
(!******************************************************
Mosel Example Problems
======================
file k4even.mos
```````````````
Placing a given number of chips on a 4x4 grid
such that every row and every column contains
an even number of chips.
- Enumerating multiple solutions -
(c) 2008 Fair Isaac Corporation
author: S. Heipcke, Mar. 2002, rev. June 2011
*******************************************************!)
model "K-4 Even"
uses "mmxprs"
declarations
POS = 1..4 ! Positions in a row/column
NCHIP = 10 ! Number of chips
place: array(POS,POS) of mpvar ! 1 if chip at the position, 0 otherwise
row, column: array(POS) of mpvar ! Chips per row/column divided by 2
end-declarations
! Total number of chips
sum(r,c in POS) place(r,c) = NCHIP
! Even number of chips in all rows and columns
forall(r in POS) sum(c in POS) place(r,c) = 2*row(r)
forall(c in POS) sum(r in POS) place(r,c) = 2*column(c)
forall(r,c in POS) place(r,c) is_binary
forall(r in POS) do
row(r) is_integer; row(r) <= 2
column(r) is_integer; column(r) <= 2
end-do
! Solve the problem: no objective
setparam("XPRS_enummaxsol", 100) ! Max. number of solutions to be saved
minimize(XPRS_ENUM,0)
! Solution printing
forall(i in 1..getparam("XPRS_enumsols")) do
selectsol(i) ! Select a solution from the pool
writeln("Solution ", i)
forall(r in POS) do
forall(c in POS) write( if(getsol(place(r,c))>0, "X ", ". ") )
writeln
end-do
end-do
end-model
|
|
| k5nephew.mos |
(!******************************************************
Mosel Example Problems
======================
file k5nephew.mos
`````````````````
Distributing wine barrels among nephews
A farmer wishes to distribute his 45 wine barrels among
his 5 nephews. 9 barrels are full, 9 filled to 3/4,
9 half-full, 9 filled to a quarter, and 9 are empty.
Each nephew is to receive the same number of barrels
and the same volume of wine. Furthermore, each nephew
must receive at least one barrel of every type and
no two of the nephews must be allocated exactly the
same number of barrels of every fill type.
- Enumerating multiple solutions -
(c) 2008 Fair Isaac Corporation
author: S. Heipcke, Mar. 2002, rev. June 2011
*******************************************************!)
model "K-5 Nephew"
uses "mmxprs"
declarations
NN = 5; NT = 5
NEPHEWS = 1..NN ! Nephews
TYPES = 1..NT ! Contents types of wine barrels
BARRELS = 45
COEF: array(TYPES) of integer ! Factors for "alldifferent" constraint
FILL: array(TYPES) of real ! Fill level of barrels
give: array(NEPHEWS,TYPES) of mpvar ! Barrels of type t for every nephew
p: array(NEPHEWS) of mpvar ! 'Pattern': value of the product
! c*give(n,t)
end-declarations
FILL:: [1, 0.75, 0.5, 0.25, 0]
COEF:: [10000, 1000, 100, 10, 1]
! Barrels per nephew
forall(n in NEPHEWS) sum(t in TYPES) give(n,t) = BARRELS/NN
forall(t in TYPES) sum(n in NEPHEWS) give(n,t) = BARRELS/NN
! Equilibrated contents
forall(n in NEPHEWS)
sum(t in TYPES) FILL(t)*give(n,t) = BARRELS/NN * (sum(t in TYPES) FILL(t))/NT
! Every nephew receives a different number of barrels of any given type
! (that is, a different pattern for every nephew)
forall(n in NEPHEWS) sum(t in TYPES) COEF(t)*give(n,t) = p(n)
forall(n in 2..NN) p(n) - p(n-1) >= 1
forall(n in NEPHEWS,t in TYPES) do
give(n,t) is_integer
1 <= give(n,t); give(n,t) <= NN
end-do
! Solve the problem: no objective
minimize(XPRS_ENUM,0)
! Solution printing
forall(i in 1..getparam("XPRS_enumsols")) do
selectsol(i) ! Select a solution from the pool
writeln("Solution ", i)
forall(n in NEPHEWS) do
write(n,": ")
forall(t in TYPES) write( getsol(give(n,t)), " " )
writeln
end-do
end-do
end-model
|
|
| k6queens.mos |
(!******************************************************
Mosel Example Problems
======================
file k6queens.mos
`````````````````
Placing N queens on an NxN chess board such that
they do not attack each other.
- Enumerating multiple solutions -
(c) 2008 Fair Isaac Corporation
author: S. Heipcke, Mar. 2002, rev. June 2011
*******************************************************!)
model "K-6 Queens"
uses "mmxprs"
declarations
NQ = 8 ! Number of rows and columns
POS = 1..NQ
queen: array(POS,POS) of mpvar ! 1 if queen at a position, 0 otherwise
end-declarations
! Objective: total number of queens
Total:= sum(r,c in POS) queen(r,c)
! Single queen per row and column
forall(r in POS) sum(c in POS) queen(r,c) = 1
forall(c in POS) sum(r in POS) queen(r,c) = 1
! Diagonals
forall(c in POS) sum(r in c..NQ) queen(r-c+1,r) <= 1
forall(r in 2..NQ) sum(c in r..NQ) queen(c,c-r+1) <= 1
forall(c in POS) sum(r in 1..c) queen(r,c-r+1) <= 1
forall(r in 2..NQ) sum(c in r..NQ) queen(c,NQ-c+r) <= 1
forall(r,c in POS) queen(r,c) is_binary
! Solve the problem
setparam("XPRS_enummaxsol", 100) ! Max. number of solutions to be saved
minimize(XPRS_ENUM,Total)
! Solution printing
forall(i in 1..getparam("XPRS_enumsols")) do
selectsol(i) ! Select a solution from the pool
writeln("Solution ", i, ". Total number of queens: ", getobjval)
forall(r in POS) do
forall(c in POS) write( if(getsol(queen(r,c))>0, "Q ", ". ") )
writeln
end-do
end-do
end-model
|
|
