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Cryptology - Unofficial St. Mary's College of California Web Site

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220 CHAPTER 11. KNAPSACK CIPHERS<br />

Example: There are 8 piles <strong>of</strong> gold bars. Every bar in a given pile weighs<br />

the same. Each in pile 1 weighs 18oz. In pile 2 each weighs 29oz. Pile 3: 34oz,<br />

Pile 4: 41oz, Pile 5: 67oz, Pile 6: 88oz, Pile 7: 101oz and Pile 8: 119oz. Your<br />

backpack can carry a maximum <strong>of</strong> 300 oz (about 19 lbs). Which bars and how<br />

many should you pick 1<br />

⋄<br />

11.2 A Related Knapsack Problem<br />

An apparent simplification is to assume the cave contains exactly one bar <strong>of</strong><br />

any particular weight. So you only need to decide “yes” or “no” to each bar,<br />

rather than deciding how many bars from each pile to take. We also modify the<br />

problem to demand an exact amount <strong>of</strong> gold.<br />

Examples:<br />

(1) The weights <strong>of</strong> the bars are 4, 7, 12, 19, 22, and 25 lbs. If the total amount<br />

demanded is 55 lbs., this weight can be supplied: 4 + 7 + 19 + 25 = 55.<br />

However, there is no way to choose bars <strong>of</strong> total weight exactly 50 pounds.<br />

(2) The weights <strong>of</strong> the bars are 27, 31, 41, 48, 55, 59, 62, 65, 73, and 77. Total<br />

amount wanted is 257. Is this weight possible What about 364 2<br />

⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄<br />

This problem may seem easier than the first, but it is still quite hard. Why<br />

Think <strong>of</strong> the number <strong>of</strong> possibilities that must be considered. For each bar we<br />

must decide whether we want it or not:<br />

For one weight, say 27 lbs.: Do we want it yes or no (2 = 2 1 choices).<br />

For 27 and 31: yes/no and yes/no, or yy, yn, ny or nn (4 = 2 2 choices).<br />

For 27, 31, and 41: y/n and y/n and y/n gives yyy, yyn, yny, nyy, ynn,<br />

nyn, nny, and nnn (8 = 2 3 choices).<br />

By the time we have 5 weights we must make 2 5 = 32 choices, 8 weights demand<br />

2 8 = 256, and 10 weights, as in the last example, leads to 2 10 = 1024<br />

possibilities, too many for a quick answer if checking by hand. In fact, if there<br />

were 100 bars to choose from (a moderately rich dragon, in other words), there<br />

would be<br />

2 100 ≈ 1, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000<br />

1 I can make 295 as 2 × 18 + 2 × 34 + 3 × 41 + 67. Can you do better<br />

2 255 is possible (in seven different ways!) but 364 is not.

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