langs a-z
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Ingy döt Net
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11389 changed files with 98361 additions and 1020 deletions
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sub next_perm ( @a is copy ) {
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my $j = @a.end - 1;
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return Nil if --$j < 0 while [>] @a[ $j, $j+1 ];
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my $aj = @a[$j];
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my $k = @a.end;
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$k-- while [>] $aj, @a[$k];
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@a[ $j, $k ] .= reverse;
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my $r = @a.end;
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my $s = $j + 1;
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@a[ $r--, $s++ ] .= reverse while $r > $s;
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return @a;
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}
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# Contains only five floors. 5! = 120 permutations.
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for [1..5], &next_perm ...^ !* -> [ $b, $c, $f, $m, $s ] {
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say "Baker=$b Cooper=$c Fletcher=$f Miller=$m Smith=$s"
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if $b != 5 # Baker !live on top floor.
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and $c != 1 # Cooper !live on bottom floor.
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and $f != 1|5 # Fletcher !live on top or the bottom floor.
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and $m > $c # Miller lives on a higher floor than Cooper.
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and $s != $f-1|$f+1 # Smith !live adjacent to Fletcher
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and $f != $c-1|$c+1 # Fletcher !live adjacent to Cooper
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;
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}
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@ -0,0 +1,100 @@
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Prototype cond(Array t(1))
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Enumeration #Null
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#Baker
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#Cooper
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#Fletcher
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#Miller
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#Smith
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EndEnumeration
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Procedure checkTenands(Array tenants(1), Array Condions.cond(1))
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Protected i, j
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Protected.cond *f
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j=ArraySize(Condions())
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For i=0 To j
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*f=Condions(i) ; load the function pointer to the current condition
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If *f(tenants()) = #False
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ProcedureReturn #False
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EndIf
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Next
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ProcedureReturn #True
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EndProcedure
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Procedure C1(Array t(1))
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If Int(Abs(t(#Fletcher)-t(#Cooper)))<>1
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ProcedureReturn #True
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EndIf
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EndProcedure
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Procedure C2(Array t(1))
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If t(#Baker)<>5
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ProcedureReturn #True
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EndIf
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EndProcedure
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Procedure C3(Array t(1))
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If t(#Cooper)<>1
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ProcedureReturn #True
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EndIf
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EndProcedure
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Procedure C4(Array t(1))
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If t(#Miller) >= t(#Cooper)
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ProcedureReturn #True
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EndIf
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EndProcedure
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Procedure C5(Array t(1))
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If t(#Fletcher)<>1 And t(#Fletcher)<>5
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ProcedureReturn #True
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EndIf
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EndProcedure
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Procedure C6(Array t(1))
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If Int(Abs(t(#Smith)-t(#Fletcher)))<>1
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ProcedureReturn #True
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EndIf
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EndProcedure
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If OpenConsole()
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Dim People(4)
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Dim Conditions(5)
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Define a, b, c, d, e, i
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;
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;- Load all conditions
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Conditions(i)=@C1(): i+1
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Conditions(i)=@C2(): i+1
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Conditions(i)=@C3(): i+1
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Conditions(i)=@C4(): i+1
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Conditions(i)=@C5(): i+1
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Conditions(i)=@C6()
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;
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; generate and the all legal combinations
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For a=1 To 5
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For b=1 To 5
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If a=b: Continue: EndIf
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For c=1 To 5
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If a=c Or b=c: Continue: EndIf
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For d=1 To 5
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If d=a Or d=b Or d=c : Continue: EndIf
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For e=1 To 5
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If e=a Or e=b Or e=c Or e=d: Continue: EndIf
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People(#Baker)=a
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People(#Cooper)=b
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People(#Fletcher)=c
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People(#Miller)=d
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People(#Smith)=e
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If checkTenands(People(), Conditions())
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PrintN("Solution found;")
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PrintN("Baker="+Str(a)+#CRLF$+"Cooper="+Str(b)+#CRLF$+"Fletcher="+Str(c))
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PrintN("Miller="+Str(d)+#CRLF$+"Smith="+Str(e)+#CRLF$)
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EndIf
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Next
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Next
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Next
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Next
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Next
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Print("Press ENTER to exit"): Input()
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EndIf
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@ -0,0 +1,33 @@
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people$ = "Baler,Cooper,Fletcher,Miller,Smith"
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for baler = 1 to 4 ' can not be in room 5
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for cooper = 2 to 5 ' can not be in room 1
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for fletcher = 2 to 4 ' can not be in room 1 or 5
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for miller = 1 to 5 ' can be in any room
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for smith = 1 to 5 ' can be in any room
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if miller > cooper and abs(smith - fletcher) > 1 and abs(fletcher - cooper) > 1 then
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if baler + cooper + fletcher + miller + smith = 15 then ' that is 1 + 2 + 3 + 4 + 5
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rooms$ = baler;cooper;fletcher;miller;smith
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bad = 0
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for i = 1 to 5 ' make sure each room is unique
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rm$ = chr$(i + 48)
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r1 = instr(rooms$,rm$)
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r2 = instr(rooms$,rm$,r1+1)
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if r2 <> 0 then bad = 1
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next i
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if bad = 0 then goto [roomAssgn] ' if it is not bad it is a good assignment
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end if
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end if
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next smith
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next miller
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next fletcher
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next cooper
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next baler
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print "Cam't assign rooms" ' print this if it can not find a solution
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wait
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[roomAssgn]
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Print "Room Assignment"
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for i = 1 to 5
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print mid$(rooms$,i,1);" ";word$(people$,i,",");" "; ' print the room assignments
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next i
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@ -0,0 +1,71 @@
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#!/bin/bash
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NAMES=(Baker Cooper Fletcher Miller Smith)
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CRITERIA=(
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'Baker != TOP'
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'Cooper != BOTTOM'
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'Fletcher != TOP'
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'Fletcher != BOTTOM'
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'Miller > Cooper'
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'$(abs $(( Smith - Fletcher )) ) > 1'
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'$(abs $(( Fletcher - Cooper )) ) > 1'
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)
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# Code below here shouldn't need to change to vary parameters
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let BOTTOM=0
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let TOP=${#NAMES[@]}-1
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# Not available as a builtin
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function abs {
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let n=$1
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if (( n < 0 )); then let n=-n; fi
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echo "$n"
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}
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# Algorithm we use to iterate over the permutations
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# requires that we start with the array sorted lexically
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NAMES=($(printf "%s\n" "${NAMES[@]}" | sort))
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while true; do
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# set each name to its position in the array
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for (( i=BOTTOM; i<=TOP; ++i )); do
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eval "${NAMES[i]}=$i"
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done
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# check to see if we've solved the problem
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let solved=1
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for criterion in "${CRITERIA[@]}"; do
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if ! eval "(( $criterion ))"; then
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let solved=0
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break
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fi
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done
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if (( solved )); then
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echo "From bottom to top: ${NAMES[@]}"
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break
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fi
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# Bump the names list to the next permutation
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let j=TOP-1
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while (( j >= BOTTOM )) && ! [[ "${NAMES[j]}" < "${NAMES[j+1]}" ]]; do
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let j-=1
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done
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if (( j < BOTTOM )); then break; fi
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let k=TOP
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while (( k > j )) && [[ "${NAMES[k]}" < "${NAMES[j]}" ]]; do
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let k-=1
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done
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if (( k <= j )); then break; fi
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t="${NAMES[j]}"
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NAMES[j]="${NAMES[k]}"
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NAMES[k]="$t"
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for (( k=1; k<=(TOP-j); ++k )); do
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a=BOTTOM+j+k
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b=TOP-k+1
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if (( a < b )); then
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t="${NAMES[a]}"
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NAMES[a]="${NAMES[b]}"
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NAMES[b]="$t"
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fi
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done
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done
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@ -0,0 +1,17 @@
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include c:\cxpl\codes;
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int B, C, F, M, S;
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for B:= 1 to 4 do \Baker does not live on top (5th) floor
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for C:= 2 to 5 do \Cooper does not live on bottom floor
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if C#B then \Cooper & Baker live on different floors
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for F:= 2 to 4 do \Fletcher doesn't live on top or bottom
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if F#B & F#C & F#C-1 & F#C+1 then \ and she's not adjacent to Cooper
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for M:= 1 to 5 do
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if M#F & M#B & M>C then \Miller lives above Cooper
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for S:= 1 to 5 do \Smith is not adjacent to Fletcher
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if S#M & S#F & S#C & S#B & S#F-1 & S#F+1 then \show
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[Text(0, "Baker "); IntOut(0, B); CrLf(0); \all
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Text(0, "Cooper "); IntOut(0, C); CrLf(0); \possible
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Text(0, "Fletcher "); IntOut(0, F); CrLf(0); \solutions
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Text(0, "Miller "); IntOut(0, M); CrLf(0);
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Text(0, "Smith "); IntOut(0, S); CrLf(0);
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]
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