Example 1: Simple substitution

This is the simple letter-for-letter method found in Poe's ``The Gold Bug'' and many other stories. The key is a rearrangement of the 26 letters:

ABCDEFGHIJKLMNOPQRSTUVWXYZ

actqgwrzdevfbhinsymujxplok

Using this key, the plaintext:

  THE SECURITY OF THE RSA ENCODING SCHEME RELIES ON THE

FACT THAT NOBODY HAS BEEN ABLE TO DISCOVER HOW TO TAKE

CUBE ROOTS MOD N WITHOUT KNOWING NS FACTORS

becomes the ciphertext:

UZG MGTJYDUO IW UZG YMA GHTIQDHR MTZGBG YGFDGM IH UZG

WATU UZAU HICIQO ZAM CGGH ACFG UI QDMTIXGY ZIP UI UAVG

TJCG YIIUM BIQ H PDUZIJU VHIPDHR HM WATUIYM

The messages can be made harder to decode (but also harder to read!) by leaving out the spaces between words.

Most messages can be decoded by looking for frequently occuring pairs of letters (TH and HE are by far the most common), using these to identify a few letters to begin, and filling in the remaining letters one at a time (``The Gold Bug'' gives a good description, as do many books).

In a known-plaintext situation, the whole code is obtained almost immediately. However, in our example, the letters J, P, and others do not occur in the plaintext, so we could not tell how they are encoded. If we were allowed a chosen plaintext, we would use all the letters to get the entire key.