95 lines
6.9 KiB
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95 lines
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<h2>Lowest-cost Search</h2><div class="info" style="cursor:help;width:200px;margin-bottom:10px;"><h3>Problem 328</h3><span style="width:300px;color:#666;">Published on Saturday, 12th March 2011, 10:00 pm; Solved by 259</span></div>
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<p>We are trying to find a hidden number selected from the set of integers {1, 2, ..., <var>n</var>} by asking questions.
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Each number (question) we ask, has a <u>cost equal to the number asked</u> and we get one of three possible answers:<br />
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<ul>
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<li> "Your guess is lower than the hidden number", or</li>
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<li> "Yes, that's it!", or</li>
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<li> "Your guess is higher than the hidden number".</li>
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</ul></p>
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<p>Given the value of <var>n</var>, an <i>optimal strategy</i> minimizes the total cost (i.e. the sum of all the questions asked) <u>for the worst possible case</u>. E.g.</p>
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<p>If <var>n</var>=3, the best we can do is obviously to ask the number "<b>2</b>". The answer will immediately lead us to find the hidden number (at a total cost = 2).</p>
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<p>If <var>n</var>=8, we might decide to use a "binary search" type of strategy: Our first question would be "<b>4</b>" and if the hidden number is higher than 4 we will need one or two additional questions.<br />
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Let our second question be "<b>6</b>". If the hidden number is still higher than 6, we will need a third question in order to discriminate between 7 and 8.<br />
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Thus, our third question will be "<b>7</b>" and the total cost for this worst-case scenario will be 4+6+7=<span style='color:red;'><b>17</b></span>.</p>
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<p>We can improve considerably the worst-case cost for <var>n</var>=8, by asking "<b>5</b>" as our first question.<br />
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If we are told that the hidden number is higher than 5, our second question will be "<b>7</b>", then we'll know for certain what the hidden number is (for a total cost of 5+7=<span style='color:blue;'><b>12</b></span>).<br />
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If we are told that the hidden number is lower than 5, our second question will be "<b>3</b>" and if the hidden number is lower than 3 our third question will be "<b>1</b>", giving a total cost of 5+3+1=<span style='color:blue;'><b>9</b></span>.<br />
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Since <span style='color:blue;'><b>12</b></span>><span style='color:blue;'><b>9</b></span>, the worst-case cost for this strategy is <span style='color:red;'><b>12</b></span>. That's better than what we achieved previously with the "binary search" strategy; it is also better than or equal to any other strategy.<br />
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So, in fact, we have just described an optimal strategy for <var>n</var>=8.</p>
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<p>Let C(<var>n</var>) be the worst-case cost achieved by an optimal strategy for <var>n</var>, as described above.<br />
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Thus C(1) = 0, C(2) = 1, C(3) = 2 and C(8) = 12.<br />
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Similarly, C(100) = 400 and <img style="vertical-align:middle" src="project/images/p_328_sum1.gif" />C(<var>n</var>) = 17575.</p>
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<p>Find <img style="vertical-align:middle" src="project/images/p_328_sum2.gif" />C(<var>n</var>).</p>
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