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copied!<p>It goes in two parts:</p> <p><strong>Which discount to pick?</strong> </p> <p>Suppose we choose to take any other discount (2 free for <code>m</code> items) than the smallest one (<code>n</code> items, <code>n<m</code>): Buying articles <code>a(1)</code> to <code>a(m)</code> we get articles <code>b</code> and <code>c</code> for free. We could just take the smallest discount, buying articles <code>a(1)</code> to <code>a(n)</code> to get <code>b</code> and <code>c</code> for free, and the buy the items <code>a(n+1)</code> to <code>a(m)</code> for the full price to end up in the same situation. Therefore picking the smallest discount is at worst on par with other choices.</p> <p><strong>Which free articles to pick?</strong></p> <p>Now suppose we apply the discount to items <code>a1</code> and <code>a2</code> instead of the most expensive ones <code>b1</code> an <code>b2</code> (<code>cost(a1)<cost(b1)</code> or <code>cost(a2)<cost(b2)</code>). Let's assume that <code>cost(a1)<cost(a2)</code> since the cases are symmetrical. Three cases arise:</p> <ol> <li>Somehow we still manage to get <code>a2</code> for free as part of a promotion: then we could also have gotten <code>a1</code> if we hadn't picked it already.</li> <li>We have to pay for it, as part of a discount. The discount allows us to buy items costing up to <code>c</code>. One of the following arises: <ul> <li><code>c<=cost(a1)</code>: we may swap <code>a1</code> and <code>a2</code>, which leads to a lower cost for this basket, without changing anything else.</li> <li><code>cost(a2)<=c<cost(a1)</code>: let's call <code>d</code> and <code>e</code> the items we get out of this second discount, <code>d</code> being the most expensive one. We could have picked <code>a2</code> as a free item in the first discount, replacing it by <code>d</code> in the second and picking <code>a1</code> as a free item in the second discount, resulting in a lower or equal cost.</li> </ul></li> <li>We buy it out of a discount: we could have picked <code>a2</code> and paid <code>a1</code> for a lower cost without changing anything else, hence picking <code>a1</code> was suboptimal.</li> </ol>

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