Another update from ingydotnet^djgoku
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Ingy döt Net
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7604 changed files with 108452 additions and 22726 deletions
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@ -44,11 +44,11 @@ Adjust the supply and demand accordingly. If demand or supply becomes 0 for a gi
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Repeat until all supply and demand is met:
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<pre>
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2 2 2 0 1 2 3 1 0 - C-W(50)
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3 5 5 7 4 35 - 1 0 - E-X(50)
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2 2 2 0 3 2 3 1 0 - C-W(50)
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3 5 5 7 4 35 - 1 0 - E-X(10)
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4 5 5 7 4 - - 1 0 - C-X(20)
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5 5 5 - 4 - - 0 0 - A-X(30)
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6 - 19 - 23 - - - 19 - D-Y(30)
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6 - 19 - 23 - - - 4 - D-Y(30)
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- - - - - - - - - B-Y(20)
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</pre>
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Finally calculate the cost of your solution. In the example given it is £3100:
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@ -0,0 +1,28 @@
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vam=:1 :0
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:
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exceeding=. 0 <. -&(+/)
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D=. x,y exceeding x NB. x: demands
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S=. y,x exceeding y NB. y: sources
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C=. (m,.0),0 NB. m: costs
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B=. 1+>./,C NB. bigger than biggest cost
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mincost=. <./@-.&0 NB. smallest non-zero cost
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penalty=. |@(B * 2 -/@{. /:~ -. 0:)"1 - mincost"1
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R=. C*0
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while. 0 < +/D,S do.
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pS=. penalty C
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pD=. penalty |:C
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if. pS >&(>./) pD do.
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row=. (i. >./) pS
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col=. (i. mincost) row { C
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else.
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col=. (i. >./) pD
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row=. (i. mincost) col {"1 C
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end.
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n=. (row{S) <. col{D
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S=. (n-~row{S) row} S
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D=. (n-~col{D) col} D
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C=. C * S *&*/ D
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R=. n (<row,col)} R
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end.
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_1 _1 }. R
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)
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@ -0,0 +1,9 @@
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demand=: 30 20 70 30 60
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src=: 50 60 50 50
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cost=: 16 16 13 22 17,14 14 13 19 15,19 19 20 23 50,:50 12 50 15 11
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demand cost vam src
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0 0 50 0 0
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30 0 20 0 10
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0 20 0 30 0
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0 0 0 0 50
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@ -0,0 +1,48 @@
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immutable TProblem{T<:Integer,U<:String}
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sd::Array{Array{T,1},1}
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toc::Array{T,2}
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labels::Array{Array{U,1},1}
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tsort::Array{Array{T,2}, 1}
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end
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function TProblem{T<:Integer,U<:String}(s::Array{T,1},
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d::Array{T,1},
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toc::Array{T,2},
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slab::Array{U,1},
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dlab::Array{U,1})
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scnt = length(s)
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dcnt = length(d)
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size(toc) = (scnt,dcnt) || error("Supply, Demand, TOC Size Mismatch")
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length(slab) == scnt || error("Supply Label Size Labels")
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length(dlab) == dcnt || error("Demand Label Size Labels")
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0 <= minimum(s) || error("Negative Supply Value")
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0 <= minimum(d) || error("Negative Demand Value")
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sd = Array{T,1}[]
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push!(sd, s)
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push!(sd, d)
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labels = Array{U,1}[]
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push!(labels, slab)
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push!(labels, dlab)
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tsort = Array{T,2}[]
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push!(tsort, mapslices(sortperm, toc, 2))
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push!(tsort, mapslices(sortperm, toc, 1))
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TProblem(sd, toc, labels, tsort)
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end
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isbalanced(tp::TProblem) = sum(tp.sd[1]) == sum(tp.sd[2])
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type Resource{T<:Integer}
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dim::T
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i::T
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quant::T
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l::T
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m::T
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p::T
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q::T
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end
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function Resource{T<:Integer}(dim::T, i::T, quant::T)
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zed = zero(T)
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Resource(dim, i, quant, zed, zed, zed, zed)
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end
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isavailable(r::Resource) = 0 < r.quant
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Base.isless(a::Resource, b::Resource) = a.p < b.p || (a.p == b.p && b.q < a.q)
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@ -0,0 +1,47 @@
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function penalize!{T<:Integer,U<:String}(sd::Array{Array{Resource{T},1},1},
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tp::TProblem{T,U})
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avail = BitArray{1}[]
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for dim in 2:-1:1
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push!(avail, bitpack(map(isavailable, sd[dim])))
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end
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for dim in 1:2, r in sd[dim]
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if r.quant == 0
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r.l = r.m = r.p = r.q = 0
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continue
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end
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r.l == 0 || !avail[dim][r.l] || !avail[dim][r.m] || continue
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rsort = filter(x->avail[dim][x], vec(slicedim(tp.tsort[dim],dim,r.i)))
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rcost = vec(slicedim(tp.toc, dim, r.i))[rsort]
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if length(rsort) == 1
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r.l = r.m = rsort[1]
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r.p = r.q = rcost[1]
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else
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r.l, r.m = rsort[1:2]
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r.p = rcost[2] - rcost[1]
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r.q = rcost[1]
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end
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end
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nothing
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end
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function vogel{T<:Integer,U<:String}(tp::TProblem{T,U})
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sdcnt = collect(size(tp.toc))
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sol = spzeros(T, sdcnt[1], sdcnt[2])
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sd = Array{Resource{T},1}[]
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for dim in 1:2
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push!(sd, [Resource(dim, i, tp.sd[dim][i]) for i in 1:sdcnt[dim]])
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end
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while any(map(isavailable, sd[1])) && any(map(isavailable, sd[2]))
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penalize!(sd, tp)
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a = maximum([sd[1], sd[2]])
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b = sd[rem1(a.dim+1,2)][a.l]
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if a.dim == 2 # swap to make a supply and b demand
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a, b = b, a
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end
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expend = min(a.quant, b.quant)
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sol[a.i, b.i] = expend
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a.quant -= expend
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b.quant -= expend
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end
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return sol
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end
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@ -0,0 +1,27 @@
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sup = [50, 60, 50, 50]
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slab = ["W", "X", "Y", "Z"]
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dem = [30, 20, 70, 30, 60]
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dlab = ["A", "B", "C", "D", "E"]
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c = [16 16 13 22 17;
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14 14 13 19 15;
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19 19 20 23 50;
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50 12 50 15 11]
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tp = TProblem(sup, dem, c, slab, dlab)
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sol = vogel(tp)
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cost = sum(tp.toc .* sol)
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println("The solution is:")
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print(" ")
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for s in tp.labels[2]
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print(@sprintf "%4s" s)
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end
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println()
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for i in 1:size(tp.toc)[1]
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print(@sprintf " %4s" tp.labels[1][i])
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for j in 1:size(tp.toc)[2]
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print(@sprintf "%4d" sol[i,j])
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end
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println()
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end
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println("The total cost is: ", cost)
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