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

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Chapter 11<br />

Knapsack Ciphers<br />

Merkle had great confidence in even the single<br />

iteration knapsack system and posted a note on<br />

his <strong>of</strong>fice <strong>of</strong>fering a $100 reward to anyone who<br />

could break it. ...[After Shamir broke it,] Merkle,<br />

always one to put his money where his mouth<br />

was ... paid Shamir the $100 is prize money.<br />

... Merkle’s enthusiasm [wasn’t] dampened. He<br />

promptly raised his bet and <strong>of</strong>fered $1000 to anyone<br />

who could break a multiple iteration knapsack.<br />

It took two years, but in the end, Merkle<br />

had to pay.<br />

Whitfield Diffie<br />

The First Ten Years <strong>of</strong><br />

Public-Key Cryptography<br />

In cryptography one takes plaintext (that is easy to read) and attempts turn it<br />

into ciphertext (that is apparently hard to read). In other words, cryptography<br />

involves an easy and hard version <strong>of</strong> the same problem. There are several<br />

mathematical problems that also have both an easy version and a hard version.<br />

Can the parallel between an easy/hard math problem and the easy/hard to read<br />

text be made useful As it will turn out, yes.<br />

11.1 The Knapsack Problem<br />

One day, whilst out hiking, you discover a cave containing many gold bars.<br />

Being ethically challenged, you wish to carry out as much gold as possible.<br />

Unfortunately the owner <strong>of</strong> the gold, a dragon, will probably return soon and<br />

your backpack (knapsack) can only carry so much weight without breaking.<br />

How can you decide which gold bars to take This is the Knapsack problem.<br />

(To fit the knapsack problem into the cave-gold-dragon analogy, you can carry<br />

however much your knapsack can hold, you have time to make one trip, and,<br />

since the dragon will not the mistake <strong>of</strong> leaving his/her cave unguarded again,<br />

one trip only.)<br />

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