Companion Mathematica Notebook for "What is The 'Equal Weight View'?"

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Liliana Anderson
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1 Compnion Mthemtic Notebook for "Wht is The 'Equl Weight View'?" Dvid Jehle & Brnden Fitelson < July 9 The methods used in this notebook re specil cses of more generl decision procedure for the probbility clculus clled PrSAT which is described in this pper < nd which cn be downloded here < Tble Exmple: p q P P P P T T..55 d T F..5 b e F T..5 c f F F b - c - d - e - f Assume p & q nd p & ~q re only peer-propositions. Then pplying (SA) we ll hve: p q P P P P T T T F F T..5???? F F.4.5???? Tble Exmple: When we pply (SAMC) we end-up with: p q P P P P T T T F F T..5 c F F.4.5 b d Here is the minimiztion code tht yields the (unique) (SAMC)-solutions for b c d. Minimize@8EuclidenDistnce@8....4< b<d HPlus üü b<l == &&< < &&< b < < 8 b<d Ø.75 b Ø.75<< Minimize@8EuclidenDistnce@ < c d<d HPlus üü c d<l == &&< c < &&< d < < 8c d<d c Ø.75 d Ø.75<< Thus (SAMC) yields the following pir of updted distributions: p q P P P P T T T F F T F F

2 ew.nb Tble Exmple: Now if we dd the further ssumption tht p nd q re lso peer-propositions we get the following single distribution: p q P P P = P T T T F..5.5 F T..5.5 F F But this solution violtes (C) since: P i (q p) = P i Hq& pl.5 = =.5999 P P i HpL.5+.5 Hq pl+p Hq pl = P Hq& pl + P Hq& pl P HpL P HpL = =.547. Wht if we weken (SAMC) to only require pproximte stright verging (ASAMC)? There re three versions of this sort of rule tht we consider in the pper. The first is the wekest: it only requires tht ech peer get close to the stright verge. First we define º b iff b < e. Then on the wekest version of the rule we need to require tht ech peer ends-up stisfying (P) nd (C) nd lso ends-up close to the stright verge. More precisely in the exmple bove we re looking for pir of vectors <bcd> nd <efgh> such tht º.5 b º.5 c º.5 nd d º.5 nd (P) nd (C) re stisfied by <bcd> nd e º.5 f º.5 g º.5 nd h º.5 nd (P) nd (C) re stisfied by <efgh>. Moreover we wnt both vectors to be in between P nd P. p q P P P P T T..55 e T F..5 b f F T..5 c g F F.4.5 d h Here we see tht (even on this wekest rendition of ASAMC) ny solution requires tht e > 6. FindInstnceB.5 -e< <.5. < <.55 &&.5 -e<b <.5. < b <.5 &&.5 -e<c <.5.5 < c <. &&.5 -e<d <.5.5 < d <.4 && <e< 5 && && + b + c + d ã &&.5 -e<e <.5. < e <.55 &&.5 -e<f <.5. < f <.5 &&.5 -e<g <.5.5 < g <. && e<h <.5.5 < h <.4 && <e 5 && e e + f == e + f + g + h ã && e&&b f&&c g&&d h 8 b c d e f g h e<f 88 Ø b Ø c Ø.4686 d Ø.454 e Ø.66 f Ø g Ø.6878 h Ø.75 eø.6694<< &&

3 ew.nb FindInstnceB.5 -e< <.5. < <.55 &&.5 -e<b <.5. < b <.5 &&.5 -e<c <.5.5 < c <. &&.5 -e<d <.5.5 < d <.4 && <e< 5 && && + b + c + d ã &&.5 -e<e <.5. < e <.55 &&.5 -e<f <.5. < f <.5 &&.5 -e<g <.5.5 < g <. && e<h <.5.5 < h <.4 && <e 6 && e e + f == e + f + g + h ã && e&&b f&&c g&&d h 8 b c d e f g h e<f The strongest version of this pproximte strtegy forces the two peers to end-up in exct greement on ll peer-propositions nd to be close to splitting the difference s well. For instnce let e =.5. Then the question is this (for the strongest rendition). Is there vector <bcd> such tht º.5 b º.5 c º.5 nd d º.5 nd (P) nd (C) re stisfied by the resulting vector? Moreover we wnt these vlues to be in between P nd P. p q P P P = P T T..55 T F..5 b F T..5 c F F.4.5 d Once gin the only wy to stisfy this is if e > 6. && FindInstnceB.5 -e< <.5. < <.55 &&.5 -e<b <.5. < b <.5 &&.5 -e<c <.5.5 < c <. &&.5 -e<d < < d <.4 && <e< 5 && && + b + c + d ã 8 b c d e<f 88 Ø b Ø c Ø.4686 d Ø.454 eø.6694<< FindInstnceB.5 -e< <.5. < <.55 &&.5 -e<b <.5. < b <.5 &&.5 -e<c <.5.5 < c <. &&.5 -e<d <.5.5 < d <.4 && <e 6 && && + b + c + d ã 8 b c d e<f Wht if we lso require (PCI) to be stisfied s well? In the bove exmple both gents strt out greeing tht p nd q re negtively dependent. Thus (PCI) requires tht we ensure they continue to gree bout this. This dds n dditionl constrint. But so long s (C) is stisfied (i.e. if e > ) this constrint cn lso be met. In generl this will be the cse (t lest for the - 6 tomic-proposition cse).

4 4 ew.nb FindInstnceB.5 -e< <.5. < <.55 &&.5 -e && + b + c + d ã 8 b c d e<f + c 88 Ø.655 b Ø.4989 c Ø.7779 d Ø.884 eø << FindInstnceB.5 -e< <.5. < <.55 &&.5 -e Here s n exmple ner our Tble exmple which requires e >.5: p q P T T T F P 5 P = P F T..5 c F F.4.5 d b && + b + c + d ã 8 b c d e<f + c test@8e_ f_ g_ h_< 8i_ j_ k_ l_< x_d := FindInstnceB e + i -e< < e + i If@e > i i < < e e < < id && f + j -e f f g g < c < k k < c < gd && h + l -e<d < h + l If@l > h h < d < l l < d < hd && <e x&& e e+f + i i+j testb: && + b + c + d ã &&e+ f + g + h ã &&i+ j + k + l ã 8 b c d e<f; > : 5 >.5F 5 5 testb: > : >.6F 88 Ø b Ø.7494 c Ø.495 d Ø.8495 eø.6<< ü Code for lrger e cse serch ner our Tble exmple Reverse engineering lrger e cse ner Tble (for the strongest rendition of ASAMC): p q P P P = P T T x y T F z u b F T..5 c F F.4.5 d

5 ew.nb 5 FindExmple@x_ y_ z_ u b_ c_ d_ x_d := FindInstnceBx + z ã &&y+ u ã &&< x < &&< y < && x + y < z < && x x < < y y < < xd && z + u -e z z < b < u u < b < zd && + 5 -e<c < < c < && e<d < < d < 4 && + b ã x x + y + z x + u && + b + c + d ã 8x y z u< RelsF; FindExmpleBx y z 9 b F FindInstnceB 7 + x + z ã && + y ã &&< x < && 5 < y < &&< z < &&" 8e<<e< " 8b<H blœrels! x + y -e< < x + y If@y > x x < < y y < < xd && z + 5 -e z z Verifying the exmple: test@8e_ f_ g_ h_< 8i_ j_ k_ l_< x_d := FindInstnceB e + i -e< < e + i If@e > i i < < e e < < id && f + j -e f f g g < c < k k < c < gd && h + l -e<d < h + l If@l > h h < d < l l < d < hd && <e<x&& e e+f + i i+j && + b + c + d ã &&e+ f + g + h ã &&i+ j + k + l ã 8 b c d e<f;

6 6 ew.nb testb: testb: > : 5 > : 5 >.F 5 5 >.5F 5 5 testb: > : >.6F 88 Ø b Ø.7494 c Ø.49 d Ø.849 eø.5995<<

MATH 25 CLASS 5 NOTES, SEP

MATH 25 CLASS 5 NOTES, SEP 30 2011 Contents 1. A brief diversion: reltively prime numbers 1 2. Lest common multiples 3 3. Finding ll solutions to x + by = c 4 Quick links to definitions/theorems Euclid