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globals[ all-colors colors-dissemination individualists ] turtles-own[ nei individualist? updates ] to setup ca set-patch-size (760 / world-size ) resize-world 0 (world-size - 1) 0 (world-size - 1) if set-random-seed? [random-seed random-seed-number] ;; setting the list with all possible colors set all-colors [5 15 25 35 45 55 65 75 85 95 105 115 125 135] let x all-colors let y  ;; preparing list Y with starting set of colors; LENGTH of Y equals to slider variable INITIAL-COLORS repeat initial-colors[ let z random (length x) set y lput (item z x) y set x remove-item z x ] let n (agents-density * world-size * world-size / 100) ask n-of n patches[ sprout 1[ set color one-of y ifelse (random-float 1) < individualist-chance [set individualist? true] [set individualist? false] ] ] ask turtles[ ifelse random-radius? [set nei turtles in-radius (1 + random-float (nei-radius - 0.9))] [set nei turtles in-radius nei-radius] ] ;; setting list COLORS-DISSEMINATION monitor-dissemintion set individualists count turtles with [individualist? = true] reset-ticks end to go ;; HOW WILL THE PROCESS OF BLAHA'S SOCIAL FUNCTION WORK? ;; 1] in one turn we randomly pick up one neigrborhood ;; 2] we check the color homogeneity of neighborhood, BUT also whether there are some individualists ;; 3] in case it is not homogenous, we find style with smallest overall effort needed for change ;; 4] style is picked out from list of neighbors styles, the individualists put new styles on the list, as well ;; 5] all neighborhood adopts the effortless style ;; (effortless = the least sum of distances from present style of turtle and the style on the list) ;; 6] tick, go to  ;; ad  ask one-of turtles[ ;ask turtles[ ;; preparing for ad  and ad  let x [color] of nei let i [individualist?] of nei ;; preparing ad  - actively used colors in the neighborhood AND ad  number of individualists set x remove-duplicates x set i remove false i ;print i ;; ad  if length x > 1 or length i > 0[ ;; ad  let unified-color effortless-style (x) (nei) ;; ad  ask nei[ set color unified-color set updates (updates + 1) ] ] ] ;; ad  tick ;; checking whether one color dominates all turtles monitor-dissemintion if ending? [stop] if minimal-updates? [if min [updates] of turtles > minimal-updates [stop]] end to monitor-dissemintion set colors-dissemination  foreach all-colors[ set colors-dissemination lput (count turtles with [color = ?]) colors-dissemination ] end ;; ad  and ad  to-report effortless-style [x y] ;; FINDING UNUSED COLORS: unused colors delimits the set in which the individualist could find new value ;let unused-colors all-colors ;foreach x[ ; set unused-colors remove ? unused-colors ;] ;; finding of new colors by individualists inside nei let using-colors x ask y with [individualist?][ ;; version where individualists prefere alternative solutions: ;ifelse length unused-colors > 0 [set color (one-of unused-colors)] [set color (one-of all-colors)] ;; version where individualists take one of possible solution regardless it is alternative or not: set color (one-of all-colors) set using-colors lput color using-colors ] set using-colors remove-duplicates using-colors set using-colors sort using-colors ;; counting effort of whole NEI in case of changing COLOR to every value from USING-COLORS ;; no other than USING-COLORS could be used, because Blaha said that all values should be lived and ;; only individuals are able to live value which is not present in the NEI (we take it that they ;; "invent" new value and immediately they live it and through this they present it to tohers in NEI) let effort  foreach using-colors[ ;; Z is effort needed for change to respective color from the USING-COLORS let z 0 ask y[ set z z + abs(color - ?) ] set effort lput z effort ] ;; now we find the position of minimal effort and doing so we find the effortless color let p position (min effort) effort ;; for control purposes we could print all used lists and found values ;print using-colors ;print effort ;print p ;print item p using-colors report item p using-colors end to-report ending? ;; value of ENDING? will be TRUE when all turtles will have same color, ;; it means that for one color will hold TRUE that number of tortles of this color equals to number of all turtles ;; for all the other colors it will hold FALSE ;; set number of all turtles as N let n count turtles ;; initializing list M where we will save logical value whether respective color is used by all turtles let m  ;; rolling over list M and saving logical values whether respective colors are used by all turtles foreach all-colors[ set m lput (n = count turtles with [color = ?]) m ] ;; reducing list M, we are only interested in whether all values are FALSE, or one value of them is TRUE set m remove-duplicates m ;print m ;; we know that only one color could be TRUE, it means used by all turtles, in that case LENGTH of lis M is 2 ;; otherwise list M consists of only FALSE values and after reduction its LENGTH equals to 1 report length m = 2 ;; so, in case one color is used by all turtles LENGTH of list M is 2 and procedure ENDING? reports value TRUE end