A semi-formal approach Bridge and torch problem

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assume solution minimizes total number of crosses. gives total of 5 crosses - 3 pair crosses , 2 solo-crosses. also, assume choose fastest solo-cross. first, show if 2 slowest persons (c , d) cross separately, accumulate total crossing time of 15. done taking persons a, c, & d: c+a+b+a = 8+1+5+1=15. (here use because know using cross both c , d separately efficient.) but, time has elapsed , person , b still on starting side of bridge , must cross. not possible 2 slowest (c & d) cross separately. second, show in order c , d cross need cross on second pair-cross: i.e. not c or d, , b, must cross first. remember our assumption @ beginning states should minimize crosses , have 5 crosses - 3 pair-crossings , 2 single crossings. assume c , d cross first. c or d must cross bring torch other side, , whoever solo-crossed must cross again. hence, cross separately. also, impossible them cross last, since implies 1 of them must have crossed previously, otherwise there 3 persons total on start side. so, since there 3 choices pair-crossings , c , d cannot cross first or last, must cross on second, or middle, pair-crossing. putting together, , b must cross first, since know c , d cannot , minimizing crossings. then, must cross next, since assume should choose fastest make solo-cross. @ second, or middle, pair-crossing c , d must go. choose send fastest back, b. , b on start side , must cross last pair-crossing. gives us, b+a+d+b+b = 2+1+8+2+2 = 15.







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