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Algoritmen en Datastructuren III Partim: Parallelle algoritmen - caagt

Algoritmen en Datastructuren III Partim: Parallelle algoritmen - caagt

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5.3. PRAM-variant<strong>en</strong> van klassieke sorteeralgoritm<strong>en</strong> 63<br />

De laatste twee rij<strong>en</strong> hebb<strong>en</strong> e<strong>en</strong> meerderheid aan <strong>en</strong><strong>en</strong> <strong>en</strong> zull<strong>en</strong> dus resulter<strong>en</strong> in e<strong>en</strong> rij <strong>en</strong><strong>en</strong>.<br />

Sam<strong>en</strong>g<strong>en</strong>om<strong>en</strong> hebb<strong>en</strong> we minst<strong>en</strong>s 4 rij<strong>en</strong> met ofwel alles null<strong>en</strong> ofwel alles <strong>en</strong><strong>en</strong> (in feite<br />

hebb<strong>en</strong> we er hier zelfs 5, door rij 5 <strong>en</strong> 6). Het resultaat van de kolomsorteerfase is:<br />

0 0 0 0 0 0 0 0 →<br />

0 0 0 0 0 0 0 0 ←<br />

0 0 0 0 0 1 0 0 →<br />

0 0 0 0 0 1 0 0 ←<br />

1 1 1 1 1 1 1 1 →<br />

1 1 1 1 1 1 1 1 ←<br />

1 1 1 1 1 1 1 1 →<br />

1 1 1 1 1 1 1 1 ←<br />

Na nog e<strong>en</strong> rij- <strong>en</strong> kolomsorteerfase bekom<strong>en</strong> we:<br />

0 0 0 0 0 0 0 0 →<br />

0 0 0 0 0 0 0 0 ←<br />

0 0 0 0 0 0 0 0 →<br />

1 0 0 0 0 0 0 1 ←<br />

1 1 1 1 1 1 1 1 →<br />

1 1 1 1 1 1 1 1 ←<br />

1 1 1 1 1 1 1 1 →<br />

1 1 1 1 1 1 1 1 ←<br />

Nog e<strong>en</strong> rijsorteerfase levert e<strong>en</strong> gesorteerde rij.<br />

5.3 PRAM-variant<strong>en</strong> van klassieke sorteeralgoritm<strong>en</strong><br />

<strong>Algoritm<strong>en</strong></strong> <strong>en</strong> Datastructur<strong>en</strong> <strong>III</strong> Veerle.Fack@UG<strong>en</strong>t.be

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